This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Notes for ECE 534
An Exploration of Random Processes for Engineers
Bruce Hajek
January 5, 2009 c 2008 by Bruce Hajek
All rights reserved. Permission is hereby given to freely print and circulate copies of these notes so long as the notes are left
intact and not reproduced for commercial purposes. Email to [email protected], pointing out errors or hard to understand
passages or providing comments, is welcome. Contents
1 Getting Started
1.1 The axioms of probability theory . . . . .
1.2 Independence and conditional probability
1.3 Random variables and their distribution .
1.4 Functions of a random variable . . . . . .
1.5 Expectation of a random variable . . . . .
1.6 Frequently used distributions . . . . . . .
1.7 Jointly distributed random variables . . .
1.8 Cross moments of random variables . . . .
1.9 Conditional densities . . . . . . . . . . . .
1.10 Transformation of random vectors . . . .
1.11 Problems . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. 2 Convergence of a Sequence of Random Variables
2.1 Four deﬁnitions of convergence of random variables .
2.2 Cauchy criteria for convergence of random variables
2.3 Limit theorems for sequences of independent random
2.4 Convex functions and Jensen’s inequality . . . . . .
2.5 Chernoﬀ bound and large deviations theory . . . . .
2.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. ......
......
variables
......
......
...... .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
. 3 Random Vectors and Minimum Mean Squared Error Estimation
3.1 Basic deﬁnitions and properties . . . . . . . . . . . . . . . . . . . . . . .
3.2 The orthogonality principle for minimum mean square error estimation .
3.3 Conditional expectation and linear estimators . . . . . . . . . . . . . . .
3.3.1 Conditional expectation as a projection . . . . . . . . . . . . . .
3.3.2 Linear estimators . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.3 Discussion of the estimators . . . . . . . . . . . . . . . . . . . . .
3.4 Joint Gaussian distribution and Gaussian random vectors . . . . . . . .
3.5 Linear Innovations Sequences . . . . . . . . . . . . . . . . . . . . . . . .
3.6 Discretetime Kalman ﬁltering . . . . . . . . . . . . . . . . . . . . . . .
3.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iii .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. 1
1
5
8
11
16
20
23
24
26
26
29 .
.
.
.
.
. 37
37
47
51
54
55
58 .
.
.
.
.
.
.
.
.
. 67
67
69
72
73
74
75
77
83
84
88 4 Random Processes
4.1 Deﬁnition of a random process . . . . . . . . . . . . . .
4.2 Random walks and gambler’s ruin . . . . . . . . . . . .
4.3 Processes with independent increments and martingales
4.4 Brownian motion . . . . . . . . . . . . . . . . . . . . . .
4.5 Counting processes and the Poisson process . . . . . . .
4.6 Stationarity . . . . . . . . . . . . . . . . . . . . . . . . .
4.7 Joint properties of random processes . . . . . . . . . . .
4.8 Conditional independence and Markov processes . . . .
4.9 Discretestate Markov processes . . . . . . . . . . . . . .
4.10 Spacetime structure of discretestate Markov processes
4.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. 5 Inference for Markov Models
5.1 A bit of estimation theory . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 The expectationmaximization (EM) algorithm . . . . . . . . . . . . . .
5.3 Hidden Markov models . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3.1 Posterior state probabilities and the forwardbackward algorithm
5.3.2 Most likely state sequence – Viterbi algorithm . . . . . . . . . .
5.3.3 The BaumWelch algorithm, or EM algorithm for HMM . . . . .
5.4 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
. 6 Dynamics of CountableState Markov Models
6.1 Examples with ﬁnite state space . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Classiﬁcation and convergence of discretetime Markov processes . . . . . . .
6.3 Classiﬁcation and convergence of continuoustime Markov processes . . . . .
6.4 Classiﬁcation of birthdeath processes . . . . . . . . . . . . . . . . . . . . . .
6.5 Time averages vs. statistical averages . . . . . . . . . . . . . . . . . . . . . .
6.6 Queueing systems, M/M/1 queue and Little’s law . . . . . . . . . . . . . . . .
6.7 Mean arrival rate, distributions seen by arrivals, and PASTA . . . . . . . . .
6.8 More examples of queueing systems modeled as Markov birthdeath processes
6.9 FosterLyapunov stability criterion and moment bounds . . . . . . . . . . . .
6.9.1 Stability criteria for discretetime processes . . . . . . . . . . . . . . .
6.9.2 Stability criteria for continuous time processes . . . . . . . . . . . . .
6.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7 Basic Calculus of Random Processes
7.1 Continuity of random processes . . . . . . . . . .
7.2 Mean square diﬀerentiation of random processes
7.3 Integration of random processes . . . . . . . . . .
7.4 Ergodicity . . . . . . . . . . . . . . . . . . . . . .
7.5 Complexiﬁcation, Part I . . . . . . . . . . . . . .
7.6 The KarhunenLo`ve expansion . . . . . . . . . .
e
7.7 Periodic WSS random processes . . . . . . . . . .
7.8 Problems . . . . . . . . . . . . . . . . . . . . . .
iv .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. 97
97
100
102
103
104
107
110
110
113
119
122 .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. 133
. 133
. 138
. 142
. 143
. 146
. 147
. 149
. 149 .
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
. 189
. 189
. 194
. 199
. 205
. 210
. 212
. 218
. 220 .
.
.
. .
.
.
.
.
.
.
. .
.
.
.
.
.
.
. 153
153
155
158
160
162
164
167
169
171
171
178
181 8 Random Processes in Linear Systems and Spectral Analysis
8.1 Basic deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.2 Fourier transforms, transfer functions and power spectral densities
8.3 Discretetime processes in linear systems . . . . . . . . . . . . . . .
8.4 Baseband random processes . . . . . . . . . . . . . . . . . . . . . .
8.5 Narrowband random processes . . . . . . . . . . . . . . . . . . . .
8.6 Complexiﬁcation, Part II . . . . . . . . . . . . . . . . . . . . . . .
8.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
. .
.
.
.
.
.
. .
.
.
.
.
.
. .
.
.
.
.
.
. .
.
.
.
.
.
. .
.
.
.
.
.
. .
.
.
.
.
.
. .
.
.
.
.
.
. .
.
.
.
.
.
. 227
. 227
. 231
. 237
. 239
. 241
. 248
. 249 9 Wiener ﬁltering
9.1 Return of the orthogonality principle . . . . . . . . . . . . . . . .
9.2 The causal Wiener ﬁltering problem . . . . . . . . . . . . . . . .
9.3 Causal functions and spectral factorization . . . . . . . . . . . .
9.4 Solution of the causal Wiener ﬁltering problem for rational power
9.5 Discrete time Wiener ﬁltering . . . . . . . . . . . . . . . . . . . .
9.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .....
.....
.....
spectral
.....
..... .....
.....
.....
densities
.....
..... .
.
.
.
.
. 257
257
260
260
265
268
273 10 Martingales
10.1 Conditional expectation revisited . . . . . . . . . .
10.2 Martingales with respect to ﬁltrations . . . . . . .
10.3 AzumaHoeﬀding inequaltity . . . . . . . . . . . .
10.4 Stopping times and the optional sampling theorem
10.5 Notes . . . . . . . . . . . . . . . . . . . . . . . . .
10.6 Problems . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. .
.
.
.
.
. 279
279
283
286
289
294
294 11 Appendix
11.1 Some notation . . . . . . . . . . . .
11.2 Convergence of sequences of numbers
11.3 Continuity of functions . . . . . . . .
11.4 Derivatives of functions . . . . . . .
11.5 Integration . . . . . . . . . . . . . .
11.5.1 Riemann integration . . . . .
11.5.2 Lebesgue integration . . . . .
11.5.3 RiemannStieltjes integration
11.5.4 LebesgueStieltjes integration
11.6 On the convergence of the mean . .
11.7 Matrices . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. 299
299
300
304
305
307
307
309
309
309
310
312 .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. 12 Solutions to Problems .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. .
.
.
.
.
.
.
.
.
.
. 317 v vi Preface
From an applications viewpoint, the main reason to study the subject of these notes is to help
deal with the complexity of describing random, timevarying functions. A random variable can
be interpreted as the result of a single measurement. The distribution of a single random variable is fairly simple to describe. It is completely speciﬁed by the cumulative distribution function
F (x), a function of one variable. It is relatively easy to approximately represent a cumulative
distribution function on a computer. The joint distribution of several random variables is much
more complex, for in general, it is described by a joint cumulative probability distribution function,
F (x1 , x2 , . . . , xn ), which is much more complicated than n functions of one variable. A random
process, for example a model of timevarying fading in a communication channel, involves many,
possibly inﬁnitely many (one for each time instant t within an observation interval) random variables. Woe the complexity!
These notes help prepare the reader to understand and use the following methods for dealing
with the complexity of random processes:
• Work with moments, such as means and covariances.
• Use extensively processes with special properties. Most notably, Gaussian processes are characterized entirely be means and covariances, Markov processes are characterized by onestep
transition probabilities or transition rates, and initial distributions. Independent increment
processes are characterized by the distributions of single increments.
• Appeal to models or approximations based on limit theorems for reduced complexity descriptions, especially in connection with averages of independent, identically distributed random
variables. The law of large numbers tells us that, in a certain context, a probability distribution can be characterized by its mean alone. The central limit theorem, similarly tells us
that a probability distribution can be characterized by its mean and variance. These limit
theorems are analogous to, and in fact examples of, perhaps the most powerful tool ever discovered for dealing with the complexity of functions: Taylor’s theorem, in which a function
in a small interval can be approximated using its value and a small number of derivatives at
a single point.
• Diagonalize. A change of coordinates reduces an arbitrary ndimensional Gaussian vector
into a Gaussian vector with n independent coordinates. In the new coordinates the joint
probability distribution is the product of n onedimensional distributions, representing a great
reduction of complexity. Similarly, a random process on an interval of time, is diagonalized by
the KarhunenLo`ve representation. A periodic random process is diagonalized by a Fourier
e
series representation. Stationary random processes are diagonalized by Fourier transforms.
vii • Sample. A narrowband continuous time random process can be exactly represented by its
samples taken with sampling rate twice the highest frequency of the random process. The
samples oﬀer a reduced complexity representation of the original process.
• Work with baseband equivalent. The range of frequencies in a typical radio transmission
is much smaller than the center frequency, or carrier frequency, of the transmission. The
signal could be represented directly by sampling at twice the largest frequency component.
However, the sampling frequency, and hence the complexity, can be dramatically reduced by
sampling a baseband equivalent random process.
These notes were written for the ﬁrst semester graduate course on random processes, oﬀered
by the Department of Electrical and Computer Engineering at the University of Illinois at UrbanaChampaign. Students in the class are assumed to have had a previous course in probability, which
is brieﬂy reviewed in the ﬁrst chapter of these notes. Students are also expected to have some
familiarity with real analysis and elementary linear algebra, such as the notions of limits, deﬁnitions
of derivatives, Riemann integration, and diagonalization of symmetric matrices. These topics are
reviewed in the appendix. Finally, students are expected to have some familiarity with transform
methods and complex analysis, though the concepts used are reviewed in the relevant chapters.
Each chapter represents roughly two weeks of lectures, and includes homework problems. Solutions to the even numbered problems without stars can be found at the end of the notes. Students
are encouraged to ﬁrst read a chapter, then try doing the even numbered problems before looking
at the solutions. Problems with stars, for the most part, investigate additional theoretical issues,
and solutions are not provided.
Hopefully some students reading these notes will ﬁnd them useful for understanding the diverse
technical literature on systems engineering, ranging from control systems, image processing, communication theory, and communication network performance analysis. Hopefully some students
will go on to design systems, and deﬁne and analyze stochastic models. Hopefully others will
be motivated to continue study in probability theory, going on to learn measure theory and its
applications to probability and analysis in general.
A brief comment is in order on the level of rigor and generality at which these notes are written.
Engineers and scientists have great intuition and ingenuity, and routinely use methods that are
not typically taught in undergraduate mathematics courses. For example, engineers generally have
good experience and intuition about transforms, such as Fourier transforms, Fourier series, and
z transforms, and some associated methods of complex analysis. In addition, they routinely use
generalized functions, in particular the delta function is frequently used. The use of these concepts
in these notes leverages on this knowledge, and it is consistent with mathematical deﬁnitions,
but full mathematical justiﬁcation is not given in every instance. The mathematical background
required for a full mathematically rigorous treatment of the material in these notes is roughly at
the level of a second year graduate course in measure theoretic probability, pursued after a course
on measure theory.
The author gratefully acknowledges the students and faculty (Todd Coleman, Christoforos Hadjicostis, Andrew Singer, R. Srikant, and Venu Veeravalli) in the past ﬁve years for their comments
and corrections. Bruce Hajek
January 2009
viii Organization
The ﬁrst four chapters of the notes are used heavily in the remaining chapters, so that most
readers should cover those chapters before moving on.
Chapter 1 is meant primarily as a review of concepts found in a typical ﬁrst course on probability
theory, with an emphasis on axioms and the deﬁnition of expectation.
Chapter 2 focuses on various ways in which a sequence of random variables can converge, and
the basic limit theorems of probability: law of large numbers, central limit theorem, and the
asymptotic behavior of large deviations.
Chapter 3 focuses on minimum mean square error estimation and the orthogonality principle.
Chapter 4 introduces the notion of random process, and brieﬂy covers several examples and classes
of random processes. Markov processes and martingales are introduced in this chapter, but
are covered in greater depth in later chapters.
The following four additional topics can be covered independently of each other.
Chapter 5 describes the use of Markov processes for modeling and statistical inference. Applications include natural language processing.
Chapter 6 describes the use of Markov processes for modeling and analysis of dynamical systems.
Applications include the modeling of queueing systems.
Chapter 79 These three chapters develop calculus for random processes based on mean square
convergence, moving to linear ﬁltering, orthogonal expansions, and ending with causal and
noncausal Wiener ﬁltering.
Chapter 10 This chapter explores martingales with respect to ﬁltrations.
In previous semesters, Chapters 14 and 79 were covered in ECE 534 at Illinois. In Fall 2008,
ECE 534 at Illinois covered Chapters 15, Sections 6.16.8, Chapter 7, Sections 8.18.4, and Section
9.1.
A number of background topics are covered in the appendix, including basic notation. ix Chapter 1 Getting Started
This chapter reviews many of the main concepts in a ﬁrst level course on probability theory with
more emphasis on axioms and the deﬁnition of expectation than is typical of a ﬁrst course. 1.1 The axioms of probability theory Random processes are widely used to model systems in engineering and scientiﬁc applications.
These notes adopt the most widely used framework of probability and random processes, namely
the one based on Kolmogorov’s axioms of probability. The idea is to assume a mathematically solid
deﬁnition of the model. This structure encourages a modeler to have a consistent, if not accurate,
model.
A probability space is a triplet (Ω, F , P ). The ﬁrst component, Ω, is a nonempty set. Each
element ω of Ω is called an outcome and Ω is called the sample space. The second component, F ,
is a set of subsets of Ω called events. The set of events F is assumed to be a σ algebra, meaning it
satisﬁes the following axioms: (See Appendix 11.1 for set notation).
A.1 Ω ∈ F
A.2 If A ∈ F then Ac ∈ F
A.3 If A, B ∈ F then A ∪ B ∈ F . Also, if A1 , A2 , . . . is a sequence of elements in F then
∞
i=1 Ai ∈ F
If F is a σ algebra and A, B ∈ F , then AB ∈ F by A.2, A.3 and the fact AB = (Ac ∪ B c )c . By
the same reasoning, if A1 , A2 , . . . is a sequence of elements in a σ algebra F , then ∞ Ai ∈ F .
i=1
Events Ai , i ∈ I , indexed by a set I are called mutually exclusive if the intersection Ai Aj = ∅
for all i, j ∈ I with i = j . The ﬁnal component, P , of the triplet (Ω, F , P ) is a probability measure
on F satisfying the following axioms:
P.1 P [A] ≥ 0 for all A ∈ F
P.2 If A, B ∈ F and if A and B are mutually exclusive, then P [A ∪ B ] = P [A] + P [B ]. Also,
if A1 , A2 , . . . is a sequence of mutually exclusive events in F then P ( ∞ Ai ) = ∞ P [Ai ].
i=1
i=1
P.3 P [Ω] = 1.
1 The axioms imply a host of properties including the following. For any subsets A, B , C of F :
• If A ⊂ B then P ]A] ≤ P [B ]
• P [A ∪ B ] = P [A] + P [B ] − P [AB ]
• P [A ∪ B ∪ C ] = P [A] + P [B ] + P [C ] − P [AB ] − P [AC ] − P [BC ] + P [ABC ]
• P [A] + P [Ac ] = 1
• P [∅] = 0.
Example 1.1.1 (Toss of a fair coin) Using “H ” for “heads” and “T ” for “tails,” the toss of a fair
coin is modelled by
Ω = {H, T }
F = {{H }, {T }, {H, T }, ∅} 1
P {H } = P {T } = ,
2 P {H, T } = 1, P [∅] = 0 Note that, for brevity, we omitted the square brackets and wrote P {H } instead of P [{H }]. Example 1.1.2 (Standard unitinterval probability space) Take Ω = {θ : 0 ≤ θ ≤ 1}. Imagine an
experiment in which the outcome ω is drawn from Ω with no preference towards any subset. In
particular, we want the set of events F to include intervals, and the probability of an interval [a, b]
with 0 ≤ a ≤ b ≤ 1 to be given by:
P [ [a, b] ] = b − a. (1.1) Taking a = b, we see that F contains singleton sets {a}, and these sets have probability zero. Since
F is to be a σ algrebra, it must also contain all the open intervals (a, b) in Ω, and for such an open
interval, P [ (a, b) ] = b − a. Any open subset of Ω is the union of a ﬁnite or countably inﬁnite set
of open intervals, so that F should contain all open and all closed subsets of Ω. Thus, F must
contain the intersection of any set that is the intersection of countably many open sets, and so on.
The speciﬁcation of the probability function P must be extended from intervals to all of F . It is
not a priori clear how large F can be. It is tempting to take F to be the set of all subsets of Ω.
However, that idea doesn’t work–see the starred homework showing that the length of all subsets
of R can’t be deﬁned in a consistent way. The problem is resolved by taking F to be the smallest
σ algebra containing all the subintervals of Ω, or equivalently, containing all the open subsets of Ω.
This σ algebra is called the Borel σ algebra for [0, 1], and the sets in it are called Borel sets. While
not every subset of Ω is a Borel subset, any set we are likely to encounter in applications is a Borel
set. The existence of the Borel σ algebra is discussed in an extra credit problem. Furthermore,
extension theorems of measure theory1 imply that P can be extended from its deﬁnition (1.1) for
interval sets to all Borel sets.
1 See, for example, H.L. Royden, Real Analysis, Third edition. Macmillan, New York, 1988, or S.R.S. Varadhan,
Probability Theory Lecture Notes, American Mathematical Society, 2001. The σ algebra F and P can be extended
somewhat further by requiring the following completeness property: if B ⊂ A ∈ F with P [A] = 0, then B ∈ F (and
also P [B ] = 0). 2 The smallest σ algebra, B , containing the open subsets of R is called the Borel σ algebra for R,
and the sets in it are called Borel sets. Similarly, the Borel σ algebra B n of subsets of Rn is the
smallest σ algebra containing all sets of the form [a1 , b1 ] × [a2 , b2 ] × · · · × [an , bn ]. Sets in B n are
called Borel subsets of Rn . The class of Borel sets includes not only rectangle sets and countable
unions of rectangle sets, but all open sets and all closed sets. Virtually any subset of Rn arising in
applications is a Borel set. Example 1.1.3 (Repeated binary trials) Suppose we’d like to represent an inﬁnite sequence of
binary observations, where each observation is a zero or one with equal probability. For example,
the experiment could consist of repeatedly ﬂipping a fair coin, and recording a one for heads and
a zero for tails, for each ﬂip. Then an outcome ω would be an inﬁnite sequence, ω = (ω1 , ω2 , · · · ),
such that for each i ≥ 1, ωi ∈ {0, 1}. Let Ω be the set of all such ω ’s. Certainly we’d want F to
contain any set that can be deﬁned in terms of only ﬁnitely many of the observations. In particular,
for any binary sequence (b1 , · · · , bn ) of some ﬁnite length n, the set {ω ∈ Ω : ωi = bi for 1 ≤ i ≤ n}
should be in F , and the probability of such a set should be 2−n . We take F to be the smallest
σ algebra of subsets of Ω containing all such sets. Again, the extension theorems of measure theory
can be used to show that P can be extended to a probability measure deﬁned on all of F .
For a speciﬁc example of a set in F , consider A = {ω ∈ Ω : ωi = 1 for all even i}. Let’s
check that A ∈ F . The key is to express A as A = ∪∞ An , where An = {ω ∈ Ω : ωi =
n=1
1 for all even i with i ≤ n} for each n. For any n, An is determined by only ﬁnitely many
observations, so An ∈ F . Therefore, A is the intersection of countably many sets in F , so A itself is
in F . Next, we identify P [A], by ﬁrst identifying P [A2n ] for n ≥ 1. For any sequence b1 , b2 , . . . , b2n
of length 2n, the probability of the set {ω ∈ Ω : ωi = bi for 1 ≤ i ≤ 2n} is 2−2n . The set A2n
is the union of 2n such sets, which are disjoint, due to the 2n possible choices of bi for odd i with
1 ≤ i ≤ 2n. So P [A2n ] = 2n 2−2n = 2−n . Furthermore, A ⊂ A2n so that P [A] ≤ P [A2n ] = 2−n .
Since n can be arbitrarily large, it follows that P [A] = 0. The following lemma gives a continuity property of probability measures which is analogous to
continuity of functions on Rn , reviewed in Appendix 11.3. If B1 , B2 , . . . is a sequence of events such
that B1 ⊂ B2 ⊂ B3 ⊂ · · · , then we can think that Bj converges to the set ∪∞ Bi as j → ∞. The
i=1
lemma states that in this case, P [Bj ] converges to the probability of the limit set as j → ∞. Lemma 1.1.4 (Continuity of Probability) Suppose B1 , B2 , . . . is a sequence of events.
(a) If B1 ⊂ B2 ⊂ · · · then limj →∞ P [Bj ] = P [ ∞ Bi ]
i=1
(b) If B1 ⊃ B2 ⊃ · · · then limj →∞ P [Bj ] = P [ ∞ Bi ]
i=1 Proof Suppose B1 ⊂ B2 ⊂ · · · . Let D1 = B1 , D2 = B2 − B1 , and, in general, let Di = Bi − Bi−1
for i ≥ 2, as shown in Figure 1.1. Then P [Bj ] = j=1 P [Di ] for each j ≥ 1, so
i
3 B1=D 1 D2 D3 ... Figure 1.1: A sequence of nested sets.
j lim P [Bj ] j →∞ = lim j →∞ P [Di ]
i=1 ∞ (a) = P [Di ]
i=1 (b) = ∞ ∞ P Di =P i=1 Bi
i=1 where (a) is true by the deﬁnition of the sum of an inﬁnite series, and (b) is true by axiom P.2. This
proves Lemma 1.1.4(a). Lemma 1.1.4(b) can be proved similarly, or can be derived by applying
c
Lemma 1.1.4(a) to the sets Bj .
Example 1.1.5 (Selection of a point in a square) Take Ω to be the square region in the plane,
Ω = {(x, y ) : 0 ≤ x, y ≤ 1}.
Let F be the Borel σ algebra for Ω, which is the smallest σ algebra containing all the rectangular
subsets of Ω that are aligned with the axes. Take P so that for any rectangle R,
P [R] = area of R.
(It can be shown that F and P exist.) Let T be the triangular region T = {(x, y ) : 0 ≤ y ≤ x ≤ 1}.
Since T is not rectangular, it is not immediately clear that T ∈ F , nor is it clear what P [T ] is.
That is where the axioms come in. For n ≥ 1, let Tn denote the region shown in Figure 1.2. Since Tn
0 1
n 2
n 1 Figure 1.2: Approximation of a triangular region.
4 Tn can be written as a union of ﬁnitely many mutually exclusive rectangles, it follows that Tn ∈ F
···
+1
and it is easily seen that P [Tn ] = 1+2+2 +n = n2n . Since T1 ⊃ T2 ⊃ T4 ⊃ T8 · · · and ∩j T2j = T , it
n
1
follows that T ∈ F and P [T ] = limn→∞ P [Tn ] = 2 .
The reader is encouraged to show that if C is the diameter one disk inscribed within Ω then
P [C ] = (area of C) = π .
4 1.2 Independence and conditional probability Events A1 and A2 are deﬁned to be independent if P [A1 A2 ] = P [A1 ]P [A2 ]. More generally, events
A1 , A2 , . . . , Ak are deﬁned to be independent if
P [Ai1 Ai2 · · · Aij ] = P [Ai1 ]P [Ai2 ] · · · P [Aij ]
whenever j and i1 , i2 , . . . , ij are integers with j ≥ 1 and 1 ≤ i1 < i2 < · · · < ij ≤ k .
For example, events A1 , A2 , A3 are independent if the following four conditions hold:
P [A1 A2 ] = P [A1 ]P [A2 ]
P [A1 A3 ] = P [A1 ]P [A3 ]
P [A2 A3 ] = P [A2 ]P [A3 ]
P [A1 A2 A3 ] = P [A1 ]P [A2 ]P [A3 ] A weaker condition is sometimes useful: Events A1 , . . . , Ak are deﬁned to be pairwise independent if Ai is independent of Aj whenever 1 ≤ i < j ≤ k . Independence of k events requires
that 2k − k − 1 equations hold: one for each subset of {1, 2, . . . , k } of size at least two. Pairwise
−
independence only requires that k = k(k2 1) equations hold.
2
If A and B are events and P [B ] = 0, then the conditional probability of A given B is deﬁned by
P [A  B ] = P [AB ]
.
P [B ] It is not deﬁned if P [B ] = 0, which has the following meaning. If you were to write a computer
routine to compute P [A  B ] and the inputs are P [AB ] = 0 and P [B ] = 0, your routine shouldn’t
simply return the value 0. Rather, your routine should generate an error message such as “input
error–conditioning on event of probability zero.” Such an error message would help you or others
ﬁnd errors in larger computer programs which use the routine.
As a function of A for B ﬁxed with P [B ] = 0, the conditional probability of A given B is itself
a probability measure for Ω and F . More explicitly, ﬁx B with P [B ] = 0. For each event A deﬁne
P [A] = P [A  B ]. Then (Ω, F , P ) is a probability space, because P satisﬁes the axioms P 1 − P 3.
(Try showing that).
If A and B are independent then Ac and B are independent. Indeed, if A and B are independent
then
P [Ac B ] = P [B ] − P [AB ] = (1 − P [A])P [B ] = P [Ac ]P [B ].
Similarly, if A, B , and C are independent events then AB is independent of C . More generally,
suppose E1 , E2 , . . . , En are independent events, suppose n = n1 + · · · + nk with ni > 1 for each i, and
5 suppose F1 is deﬁned by Boolean operations (intersections, complements, and unions) of the ﬁrst n1
events E1 , . . . , En1 , F2 is deﬁned by Boolean operations on the next n2 events, En1 +1 , . . . , En1 +n2 ,
and so on, then F1 , . . . , Fk are independent.
Events E1 , . . . , Ek are said to form a partition of Ω if the events are mutually exclusive and
Ω = E1 ∪ · · · ∪ Ek . Of course for a partition, P [E1 ] + · · · + P [Ek ] = 1. More generally, for any
event A, the law of total probability holds because A is the union of the mutually exclusive sets
AE1 , AE2 , . . . , AEk :
P [A] = P [AE1 ] + · · · + P [AEk ].
If P [Ei ] = 0 for each i, this can be written as
P [A] = P [A  E1 ]P [E1 ] + · · · + P [A  Ek ]P [Ek ].
Figure 1.3 illustrates the condition of the law of total probability.
E1 E2
E3 A E 4 Ω Figure 1.3: Partitioning a set A using a partition of Ω.
Judicious use of the deﬁnition of conditional probability and the law of total probability leads
to Bayes’ formula for P [Ei  A] (if P [A] = 0) in simple form
P [Ei  A] = P [AEi ]
P [A] = P [A  Ei ]P [Ei ]
,
P [A] or in expanded form:
P [Ei  A] = P [A  Ei ]P [Ei ]
.
P [A  E1 ]P [E1 ] + · · · + P [A  Ek ]P [Ek ] The remainder of this section gives the BorelCantelli lemma. It is a simple result based on
continuity of probability and independence of events, but it is not typically encountered in a ﬁrst
course on probability. Let (An : n ≥ 0) be a sequence of events for a probability space (Ω, F , P ).
Deﬁnition 1.2.1 The event {An inﬁnitely often} is the set of ω ∈ Ω such that ω ∈ An for inﬁnitely
many values of n.
Another way to describe {An inﬁnitely often} is that it is the set of ω such that for any k , there is
an n ≥ k such that ω ∈ An . Therefore,
{An inﬁnitely often} = ∩k≥1 (∪n≥k An ) .
For each k , the set ∪n≥k An is a countable union of events, so it is an event, and {An inﬁnitely often}
is an intersection of countably many such events, so that {An inﬁnitely often} is also an event.
6 Lemma 1.2.2 (BorelCantelli lemma) Let (An : n ≥ 1) be a sequence of events and let pn =
P [An ].
(a) If ∞
n=1 pn < ∞, then P {An inﬁnitely often} = 0. (b) If ∞
n=1 pn = ∞ and A1 , A2 , · · · are mutually independent, then P {An inﬁnitely often} = 1. Proof. (a) Since {An inﬁnitely often} is the intersection of the monotonically nonincreasing sequence of events ∪n≥k An , it follows from the continuity of probability for monotone sequences of
events (Lemma 1.1.4) that P {An inﬁnitely often} = limk→∞ P [∪n≥k An ]. Lemma 1.1.4, the fact
that the probability of a union of events is less than or equal to the sum of the probabilities of the
events, and the deﬁnition of the sum of a sequence of numbers, yield that for any k ≥ 1,
∞ m P [∪n≥k An ] = lim P [∪m k An ] ≤ lim
n=
m→∞ m→∞ pn =
n=k pn
n=k ∞
Combining the above yields P {An inﬁnitely often} ≤ limk→∞ ∞ k pn . If
n=1 pn < ∞, then
n=
limk→∞ ∞ k pn = 0, which implies part (a) of the lemma.
n=
(b) Suppose that ∞ pn = +∞ and that the events A1 , A2 , . . . are mutually independent. For
n=1
any k ≥ 1, using the fact 1 − u ≤ exp(−u) for all u,
m P [∪n≥k An ] = lim P [∪m k An ] = lim 1 −
n= m→∞ m→∞ (1 − pn )
n=k
∞ m ≥ lim 1 − exp(− m→∞ pn ) = 1 − exp(−
n=k pn ) = 1 − exp(−∞) = 1.
n=k Therefore, P {An inﬁnitely often} = limk→∞ P [∪n≥k An ] = 1. 1
Example 1.2.3 Consider independent coin tosses using biased coins, such that P [An ] = pn = n ,
∞1
th toss. Since
where An is the event of getting heads on the n
n=1 n = +∞, the part of the
BorelCantelli lemma for independent events implies that P {An inﬁnitely often} = 1. Example 1.2.4 Let (Ω, F , P ) be the standard unitinterval probability space deﬁned in Example
1
1
1.1.2, and let An = [0, n ]. Then pn = n and An+1 ⊂ An for n ≥ 1. The events are not independent,
1
because for m < n, P [Am An ] = P [An ] = n = P [Am ]P [An ]. Of course 0 ∈ An for all n. But for any
1
ω ∈ (0, 1], ω ∈ An for n > ω . Therefore, {An inﬁnitely often} = {0}. The single point set {0} has
probability zero, so P {An inﬁnitely often} = 0. This conclusion holds even though ∞ pn = +∞,
n=1
illustrating the need for the independence assumption in Lemma 1.2.2(b). 7 1.3 Random variables and their distribution Let a probability space (Ω, F , P ) be given. By deﬁnition, a random variable is a function X from
Ω to the real line R that is F measurable, meaning that for any number c,
{ω : X (ω ) ≤ c} ∈ F . (1.2) If Ω is ﬁnite or countably inﬁnite, then F can be the set of all subsets of Ω, in which case any
realvalued function on Ω is a random variable.
If (Ω, F , P ) is given as in the uniform phase example with F equal to the Borel subsets of [0, 2π ],
then the random variables on (Ω, F , P ) are called the Borel measurable functions on Ω. Since the
Borel σ algebra contains all subsets of [0, 2π ] that come up in applications, for practical purposes
we can think of any function on [0, 2π ] as being a random variable. For example, any piecewise
continuous or piecewise monotone function on [0, 2π ] is a random variable for the uniform phase
example.
The cumulative distribution function (CDF) of a random variable X is denoted by FX . It is
the function, with domain the real line R, deﬁned by
FX (c) = P {ω : X (ω ) ≤ c} (1.3) = P {X ≤ c} (for short) (1.4) If X denotes the outcome of the roll of a fair die (“die” is singular of “dice”) and if Y is uniformly
distributed on the interval [0, 1], then FX and FY are shown in Figure 1.4
FX 1 0 1 F 1 2 3 4 5 6 Y 0 1 Figure 1.4: Examples of CDFs.
The CDF of a random variable X determines P {X ≤ c} for any real number c. But what about
P {X < c} and P {X = c}? Let c1 , c2 , . . . be a monotone nondecreasing sequence that converges to
c from the left. This means ci ≤ cj < c for i < j and limj →∞ cj = c. Then the events {X ≤ cj }
are nested: {X ≤ ci } ⊂ {X ≤ cj } for i < j , and the union of all such events is the event {X < c}.
Thus, by Lemma 1.1.4
P {X < c} = lim P {X ≤ ci } = i→∞ lim FX (ci ) = FX (c−). i→∞ Therefore, P {X = c} = FX (c) − FX (c−) = FX (c), where FX (c) is deﬁned to be the size of
1
the jump of F at c. For example, if X has the CDF shown in Figure 1.5 then P {X = 0} = 2 .
The requirement that FX be right continuous implies that for any number c (such as c = 0 for this
example), if the value FX (c) is changed to any other value, the resulting function would no longer
be a valid CDF.
8 The collection of all events A such that P {X ∈ A} is determined by FX is a σ algebra containing
the intervals, and thus this collection contains all Borel sets. That is, P {X ∈ A} is determined by
FX for any Borel set A.
1 0.5 0 −1 Figure 1.5: An example of a CDF. Proposition 1.3.1 A function F is the CDF of some random variable if and only if it has the
following three properties:
F.1 F is nondecreasing
F.2 limx→+∞ F (x) = 1 and limx→−∞ F (x) = 0
F.3 F is right continuous
Proof The “only if” part is proved ﬁrst. Suppose that F is the CDF of some random variable X .
Then if x < y , F (y ) = P {X ≤ y } = P {X ≤ x} + P {x < X ≤ y } ≥ P {X ≤ x} = F (x) so that F.1
is true. Consider the events Bn = {X ≤ n}. Then Bn ⊂ Bm for n ≤ m. Thus, by Lemma 1.1.4,
∞ lim F (n) = lim P [Bn ] = P n→∞ n→∞ Bn = P [Ω] = 1. n=1 This and the fact F is nondecreasing imply the following. Given any > 0, there exists N so large
that F (x) ≥ 1 − for all x ≥ N . That is, F (x) → 1 as x → +∞. Similarly,
∞ lim F (n) = n→−∞ B−n = P [∅] = 0. lim P [B−n ] = P n→∞ n=1 so that F (x) → 0 as x → −∞. Property F.2 is proved.
The proof of F.3 is similar. Fix an arbitrary real number x. Deﬁne the sequence of events An
1
for n ≥ 1 by An = {X ≤ x + n }. Then An ⊂ Am for n ≥ m so
lim F (x + n→∞ 1
) = lim P [An ] = P
n→∞
n ∞ Ak = P {X ≤ x} = FX (x).
k=1 1
Convergence along the sequence x + n , together with the fact that F is nondecreasing, implies that
F (x+) = F (x). Property F.3 is thus proved. The proof of the “only if” portion of Proposition
1.3.1 is complete
To prove the “if” part of Proposition 1.3.1, let F be a function satisfying properties F.1F.3. It
must be shown that there exists a random variable with CDF F . Let Ω = R and let F be the set 9 ˜
˜
B of Borel subsets of R. Deﬁne P on intervals of the form (a, b] by P [(a, b]] = F (b) − F (a). It can
˜ can be extended to all of F so that
be shown by an extension theorem of measure theory that P
˜
the axioms of probability are satisﬁed. Finally, let X (ω ) = ω for all ω ∈ Ω.
Then
˜˜
˜
P [X ∈ (a, b]] = P [(a, b]] = F (b) − F (a).
˜
Therefore, X has CDF F .
The vast majority of random variables described in applications are one of two types, to be
described next. A random variable X is a discrete random variable if there is a ﬁnite or countably
inﬁnite set of values {xi : i ∈ I } such that P {X ∈ {xi : i ∈ I }} = 1. The probability mass function
(pmf) of a discrete random variable X , denoted pX (x), is deﬁned by pX (x) = P {X = x}. Typically
the pmf of a discrete random variable is much more useful than the CDF. However, the pmf and
CDF of a discrete random variable are related by pX (x) = FX (x) and conversely,
pX (y ), FX (x) = (1.5) y :y ≤x where the sum in (1.5) is taken only over y such that pX (y ) = 0. If X is a discrete random variable
with only ﬁnitely many mass points in any ﬁnite interval, then FX is a piecewise constant function.
A random variable X is a continuous random variable if the CDF is the integral of a function:
x FX (x) = fX (y )dy
−∞ The function fX is called the probability density function (pdf). If the pdf fX is continuous at a
point x, then the value fX (x) has the following nice interpretation:
1 x+ε
fX (y )dy
ε→0 ε x
1
= lim P {x ≤ X ≤ x + ε}.
ε→0 ε fX (x) = lim If A is any Borel subset of R, then
P { X ∈ A} = fX (x)dx. (1.6) A The integral in (1.6) can be understood as a Riemann integral if A is a ﬁnite union of intervals and
f is piecewise continuous or monotone. In general, fX is required to be Borel measurable and the
integral is deﬁned by Lebesgue integration.2
Any random variable X on an arbitrary probability space has a CDF FX . As noted in the proof
˜
of Proposition 1.3.1 there exists a probability measure PX (called P in the proof) on the Borel
subsets of R such that for any interval (a, b],
PX [(a, b]] = P {X ∈ (a, b]}.
We deﬁne the probability distribution of X to be the probability measure PX . The distribution PX
is determined uniquely by the CDF FX . The distribution is also determined by the pdf fX if X
is continuous type, or the pmf pX if X is discrete type. In common usage, the response to the
question “What is the distribution of X ?” is answered by giving one or more of FX , fX , or pX , or
possibly a transform of one of these, whichever is most convenient.
2 Lebesgue integration is deﬁned in Sections 1.5 and 11.5 10 1.4 Functions of a random variable Recall that a random variable X on a probability space (Ω, F , P ) is a function mapping Ω to the
real line R , satisfying the condition {ω : X (ω ) ≤ a} ∈ F for all a ∈ R. Suppose g is a function
mapping R to R that is not too bizarre. Speciﬁcally, suppose for any constant c that {x : g (x) ≤ c}
is a Borel subset of R. Let Y (ω ) = g (X (ω )). Then Y maps Ω to R and Y is a random variable.
See Figure 1.6. We write Y = g (X ).
X g Ω
X(ω) g(X(ω)) Figure 1.6: A function of a random variable as a composition of mappings.
Often we’d like to compute the distribution of Y from knowledge of g and the distribution of
X . In case X is a continuous random variable with known distribution, the following three step
procedure works well:
(1) Examine the ranges of possible values of X and Y . Sketch the function g .
(2) Find the CDF of Y , using FY (c) = P {Y ≤ c} = P {g (X ) ≤ c}. The idea is to express the
event {g (X ) ≤ c} as {X ∈ A} for some set A depending on c.
(3) If FY has a piecewise continuous derivative, and if the pmf fY is desired, diﬀerentiate FY .
If instead X is a discrete random variable then step 1 should be followed. After that the pmf of Y
can be found from the pmf of X using
pY (y ) = P {g (X ) = y } = pX (x)
x:g (x)=y Example 1.4.1 Suppose X is a N (µ = 2, σ 2 = 3) random variable (see Section 1.6 for the deﬁnition) and Y = X 2 . Let us describe the density of Y . Note that Y = g (X ) where g (x) = x2 . The
support of the distribution of X is the whole real line, and the range of g over this support is R+ .
Next we ﬁnd the CDF, FY . Since P {Y ≥ 0} = 1, FY (c) = 0 for c < 0. For c ≥ 0,
√
√
FY (c) = P {X 2 ≤ c} = P {− c ≤ X ≤ c}
√
√
− c−2
X −2
c−2
= P{ √
≤√
≤√}
3
3
3
√
√
− c−2
c−2
= Φ( √ ) − Φ( √
)
3
3
Diﬀerentiate with respect to c, using the chain rule and the fact, Φ (s) =
fY (c) = √
c
√ 1 {exp(−[ √−2 ]2 )
24πc
6 √1
2π 2 exp(− s2 ) to obtain √ −
+ exp(−[ − √c6 2 ]2 )} 0
11 if y ≥ 0
if y < 0 (1.7) Example 1.4.2 Suppose a vehicle is traveling in a straight line at speed a, and that a random
direction is selected, subtending an angle Θ from the direction of travel which is uniformly distributed over the interval [0, π ]. See Figure 1.7. Then the eﬀective speed of the vehicle in the B
Θ
a
Figure 1.7: Direction of travel and a random direction.
random direction is B = a cos(Θ). Let us ﬁnd the pdf of B .
The range of a cos(Θ) as θ ranges over [0, π ] is the interval [−a, a]. Therefore, FB (c) = 0 for
c ≤ −a and FB (c) = 1 for c ≥ a. Let now −a < c < a. Then, because cos is monotone nonincreasing
on the interval [0, π ],
c
}
a
c
= P {Θ ≥ cos−1 ( )}
a
c
cos−1 ( a )
= 1−
π FB (c) = P {a cos(Θ) ≤ c} = P {cos(Θ) ≤ 1 Therefore, because cos−1 (y ) has derivative, −(1 − y 2 )− 2 ,
fB (c) = √1
π a2 −c2 0  c < a
 c > a A sketch of the density is given in Figure 1.8. Example 1.4.3 Suppose Y = tan(Θ), as illustrated in Figure 1.9, where Θ is uniformly distributed
over the interval (− π , π ) . Let us ﬁnd the pdf of Y . The function tan(θ) increases from −∞ to ∞
22
as θ ranges over the interval (− π , π ). For any real c,
22
FY (c) = P {Y ≤ c}
= P {tan(Θ) ≤ c}
= P {Θ ≤ tan−1 (c)} =
12 tan−1 (c) +
π π
2 fB −a a 0 Figure 1.8: The pdf of the eﬀective speed in a uniformly distributed direction. Θ 0 Y Figure 1.9: A horizontal line, a ﬁxed point at unit distance, and a line through the point with
random direction.
Diﬀerentiating the CDF with respect to c yields that Y has the Cauchy pdf:
fY (c) = 1
π (1 + c2 ) −∞<c<∞ Example 1.4.4 Given an angle θ expressed in radians, let (θ mod 2π ) denote the equivalent angle
in the interval [0, 2π ]. Thus, (θ mod 2π ) is equal to θ + 2πn, where the integer n is such that
0 ≤ θ + 2πn < 2π .
Let Θ be uniformly distributed over [0, 2π ], let h be a constant, and let
˜
Θ = (Θ + h mod 2π )
˜
Let us ﬁnd the distribution of Θ.
˜ takes values in the interval [0, 2π ], so ﬁx c with 0 ≤ c < 2π and seek to ﬁnd
Clearly Θ
˜
P {Θ ≤ c}. Let A denote the interval [h, h + 2π ]. Thus, Θ + h is uniformly distributed over A. Let
˜
B = n [2πn, 2πn + c]. Thus Θ ≤ c if and only if Θ + h ∈ B . Therefore,
˜
P {Θ ≤ c} =
A T B 1
dθ
2π By sketching the set B , it is easy to see that A B is either a single interval of length c, or the
˜
˜
union of two intervals with lengths adding to c. Therefore, P {Θ ≤ c} = 2c , so that Θ is itself
π
uniformly distributed over [0, 2π ] 13 Example 1.4.5 Let X be an exponentially distributed random variable with parameter λ. Let
Y = X , which is the integer part of X , and let R = X − X , which is the remainder. We shall
describe the distributions of Y and R.
Clearly Y is a discrete random variable with possible values 0, 1, 2, . . . , so it is suﬃcient to ﬁnd
the pmf of Y . For integers k ≥ 0,
k+1 pY (k ) = P {k ≤ X < k + 1} = λe−λx dx = e−λk (1 − e−λ ) k and pY (k ) = 0 for other k .
Turn next to the distribution of R. Clearly R takes values in the interval [0, 1]. So let 0 < c < 1
and ﬁnd FR (c):
∞ FR (c) = P {X − X ≤ c} = P {X ∈ [k, k + c]}
k=0 ∞ ∞ e−λk (1 − e−λc ) = P {k ≤ X ≤ k + c} = =
k=0 k=0 where we used the fact 1 + α + α2 + · · · = 1
1−α fR (c) = 1 − e−λc
1 − e−λ for  α < 1. Diﬀerentiating FR yields the pmf: λe−λc
1−e−λ 0 0≤c≤1
otherwise What happens to the density of R as λ → 0 or as λ → ∞? By l’Hospital’s rule,
1 0≤c≤1
0 otherwise lim fR (c) = λ→ 0 That is, in the limit as λ → 0, the density of X becomes more and more “evenly spread out,” and
R becomes uniformly distributed over the interval [0, 1]. If λ is very large then the factor e−λ is
nearly zero, and the density of R is nearly the same as the exponential density with parameter λ.
An important step in many computer simulations of random systems is to generate a random
variable with a speciﬁed CDF, by applying a function to a random variable that is uniformly
distributed on the interval [0, 1]. Let F be a function satisfying the three properties required of a
CDF, and let U be uniformly distributed over the interval [0, 1]. The problem is to ﬁnd a function
g so that F is the CDF of g (U ). An appropriate function g is given by the inverse function of
F . Although F may not be strictly increasing, a suitable version of F −1 always exists, deﬁned for
0 < u < 1 by
F −1 (u) = min{x : F (x) ≥ u} (1.8) If the graphs of F and F −1 are closed up by adding vertical lines at jump points, then the graphs
are reﬂections of each other about the x = y line, as illustrated in Figure 1.10. It is not hard to
14 !1 F (u)
F(x)
1 u x
1 Figure 1.10: A CDF and its inverse.
check that for any real xo and uo with 0 < uo < 1,
F (xo ) ≥ uo if and only if xo ≥ F −1 (uo ) Thus, if X = F −1 (U ) then
P {F −1 (U ) ≤ x} = P {U ≤ F (x)} = F (x)
so that indeed F is the CDF of X
Example 1.4.6 Suppose F (x) = 1 − e−x for x ≥ 0 and F (x) = 0 for x < 0. Since F is continuously
increasing in this case, we can identify its inverse by solving for x as a function of u so that F (x) = u.
That is, for 0 < u < 1, we’d like 1 − e−x = u which is equivalent to e−x = 1 − u, or x = − ln(1 − u).
Thus, F −1 (u) = − ln(1 − u). So we can take g (u) = − ln(1 − u) for 0 < u < 1. That is, if U is
uniformly distributed on the interval [0, 1], then the CDF of − ln(1 − U ) is F . The choice of g is
not unique in general. For example, 1 − U has the same distribution as U , so the CDF of − ln(U )
is also F . To double check the answer, note that if x ≥ 0, then
P {− ln(1 − U ) ≤ x} = P {ln(1 − U ) ≥ −x} = P {1 − U ≥ e−x } = P {U ≤ 1 − e−x } = F (x). Example 1.4.7 Suppose F is the CDF for the experiment of rolling a fair die, shown on the left
half of Figure 1.4. One way to generate a random variable with CDF F is to actually roll a die.
To simulate that on a compute, we’d seek a function g so that g (U ) has the same CDF. Using
g = F −1 and using (1.8) or the graphical method illustrated in Figure 1.10 to ﬁnd F −1 , we get
i
that for 0 < u < 1, g (u) = i for i−1 < u ≤ 6 for 1 ≤ i ≤ 6. To double check the answer, note that
6
if 1 ≤ i ≤ 6, then
i−1
i
1
P {g (U ) = i} = P
<U ≤
=
6
6
6
so that g (U ) has the correct pmf, and hence the correct CDF. 15 1.5 Expectation of a random variable The expectation, alternatively called the mean, of a random variable X can be deﬁned in several
diﬀerent ways. Before giving a general deﬁnition, we shall consider a straight forward case. A
random variable X is called simple if there is a ﬁnite set {x1 , . . . , xm } such that X (ω ) ∈ {x1 , . . . , xm }
for all ω . The expectation of such a random variable is deﬁned by
m xi P {X = xi } E [X ] = (1.9) i=1 The deﬁnition (1.9) clearly shows that E [X ] for a simple random variable X depends only on the
pmf of X .
Like all random variables, X is a function on a probability space (Ω, F , P ). Figure 1.11 illustrates that the sum deﬁning E [X ] in (1.9) can be viewed as an integral over Ω. This suggests
writing
X (ω )P [dω ] E [X ] = (1.10) Ω X(ω )=x 1 X(ω )=x 2 Ω
X(ω )=x 3 Figure 1.11: A simple random variable with three possible values.
Let Y be another simple random variable on the same probability space as X , with Y (ω ) ∈
n
{y1 , . . . , yn } for all ω . Of course E [Y ] =
i=1 yi P {Y = yi }. One learns in any elementary
probability class that E [X + Y ] = E [X ] + E [Y ]. Note that X + Y is again a simple random
variable, so that E [X + Y ] can be deﬁned in the same way as E [X ] was deﬁned. How would you
prove E [X + Y ] = E [X ]+ E [Y ]? Is (1.9) helpful? We shall give a proof that E [X + Y ] = E [X ]+ E [Y ]
motivated by (1.10).
The sets {X = x1 }, . . . , {X = xm } form a partition of Ω. A reﬁnement of this partition consists
of another partition C1 , . . . , Cm such that X is constant over each Cj . If we let xj denote the value
of X on Cj , then clearly
E [X ] = xj P [Cj ]
j 16 Now, it is possible to select the partition C1 , . . . , Cm so that both X and Y are constant over each
Cj . For example, each Cj could have the form {X = xi } ∩ {Y = yk } for some i, k . Let yj denote
the value of Y on Cj . Then xj + yj is the value of X + Y on Cj . Therefore,
E [X + Y ] = (xj + yj )P [Cj ] =
j x j P [ Cj ] +
j yj P [Cj ] = E [X ] + E [Y ]
j While the expression (1.10) is rather suggestive, it would be overly restrictive to interpret it
as a Riemann integral over Ω. For example, if X is a random variable for the the standard unitinterval probability space deﬁned in Example 1.1.2, then it is tempting to deﬁne E [X ] by Riemann
integration (see the appendix):
1 X (ω )dω E [X ] = (1.11) 0 However, suppose X is the simple random variable such that X (w) = 1 for rational values of ω and
X (ω ) = 0 otherwise. Since the set of rational numbers in Ω is countably inﬁnite, such X satisﬁes
P {X = 0} = 1. Clearly we’d like E [X ] = 0, but the Riemann integral (1.11) is not convergent for
this choice of X .
The expression (1.10) can be used to deﬁne E [X ] in great generality if it is interpreted as a
Lebesgue integral, deﬁned as follows: Suppose X is an arbitrary nonnegative random variable.
Then there exists a sequence of simple random variables X1 , X2 , . . . such that for every ω ∈ Ω,
X1 (ω ) ≤ X2 (ω ) ≤ · · · and Xn (ω ) → X (ω ) as n → ∞. Then E [Xn ] is well deﬁned for each n and
is nondecreasing in n, so the limit of E [Xn ] as n → ∞ exists with values in [0, +∞]. Furthermore
it can be shown that the value of the limit depends only on (Ω, F , P ) and X , not on the particular
choice of the approximating simple sequence. We thus deﬁne E [X ] = limn→∞ E [Xn ]. Thus, E [X ]
is always well deﬁned in this way, with possible value +∞, if X is a nonnegative random variable.
Suppose X is an arbitrary random variable. Deﬁne the positive part of X to be the random
variable X+ deﬁned by X+ (ω ) = max{0, X (ω )} for each value of ω . Similarly deﬁne the negative
part of X to be the random variable X− (ω ) = max{0, −X (ω )}. Then X (ω ) = X+ (ω ) − X− (ω )
for all ω , and X+ and X− are both nonnegative random variables. As long as at least one of
E [X+ ] or E [X− ] is ﬁnite, deﬁne E [X ] = E [X+ ] − E [X− ]. The expectation E [X ] is undeﬁned
if E [X+ ] = E [X− ] = +∞. This completes the deﬁnition of E [X ] using (1.10) interpreted as a
Lebesgue integral.
We will prove that E [X ] deﬁned by the Lebesgue integral (1.10) depends only on the CDF of
X . It suﬃces to show this for a nonnegative random variable X . For such a random variable, and
n ≥ 1, deﬁne the simple random variable Xn by
Xn (ω ) = k 2−n
0 if k 2−n ≤ X (ω ) < (k + 1)2−n , k = 0, 1, . . . , 22n − 1
else Then
22n −1 k 2−n (FX ((k + 1)2−n ) − FX (k 2−n ) E [Xn ] =
k=0 so that E [Xn ] is determined by the CDF FX for each n. Furthermore, the Xn ’s are nondecreasing
in n and converge to X . Thus, E [X ] = limn→∞ E [Xn ], and therefore the limit E [X ] is determined
by FX .
17 In Section 1.3 we deﬁned the probability distribution PX of a random variable such that the
˜
canonical random variable X (ω ) = ω on (R, B , PX ) has the same CDF as X . Therefore E [X ] =
˜ ], or
E [X
∞ xPX (dx) E [X ] = (Lebesgue) (1.12) −∞ By deﬁnition, the integral (1.12) is the LebesgueStieltjes integral of x with respect to FX , so that
∞ xdFX (x) E [X ] = (LebesgueStieltjes) (1.13) −∞ Expectation has the following properties. Let X, Y be random variables and c a constant.
E.1 (Linearity) E [cX ] = cE [X ]. If E [X ], E [Y ] and E [X ] + E [Y ] are well deﬁned, then
E [X + Y ] is well deﬁned and E [X + Y ] = E [X ] + E [Y ].
E.2 (Preservation of order) If P {X ≥ Y } = 1 and E [Y ] is well deﬁned then E [X ] is well
deﬁned and E [X ] ≥ E [Y ].
E.3 If X has pdf fX then
∞ E [X ] = xfX (x)dx (Lebesgue) −∞ E.4 If X has pmf pX then
E [X ] = xpX (x) +
x>0 xpX (x).
x<0 E.5 (Law of the unconscious statistician) If g is Borel measurable,
E [g (X )] = g (X (ω ))P [dω ] (Lebesgue) Ω
∞ = g (x)dFX (x) (LebesgueStieltjes) −∞ and in case X is a continuous type random variable
∞ E [g (X )] = g (x)fX (x)dx (Lebesgue) −∞ E.6 (Integration by parts formula)
∞ 0 (1 − FX (x))dx − E [X ] = FX (x)dx, (1.14) −∞ 0 which is well deﬁned whenever at least one of the two integrals in (1.14) is ﬁnite. There is
a simple graphical interpretation of (1.14). Namely, E [X ] is equal to the area of the region
between the horizontal line {y = 1} and the graph of FX and contained in {x ≥ 0}, minus
the area of the region bounded by the x axis and the graph of FX and contained in {x ≤ 0},
as long as at least one of these regions has ﬁnite area. See Figure 1.12.
18 y
y=1 +
F (x)
X F (x)
X ! x 0 Figure 1.12: E [X ] is the diﬀerence of two areas.
Properties E.1 and E.2 are true for simple random variables and they carry over to general random
variables in the limit deﬁning the Lebesgue integral (1.10). Properties E.3 and E.4 follow from
the equivalent deﬁnition (1.12) and properties of LebesgueStieltjes integrals. Property E.5 can
be proved by approximating g by piecewise constant functions. Property E.6 can be proved by
−
integration by parts applied to (1.13). Alternatively, since FX 1 (U ) has the same distribution as
X, if U is uniformly distributed on the interval [0, 1], the law of the unconscious statistician yields
1−
that E [X ] = 0 FX 1 (u)du, and this integral can also be interpreted as the diﬀerence of the areas
of the same two regions.
Sometimes, for brevity, we write EX instead of E [X ]. The variance of a random variable X with
EX ﬁnite is deﬁned by Var(X ) = E [(X − EX )2 ]. By the linearity of expectation, if EX is ﬁnite, the
variance of X satisﬁes the useful relation: Var(X ) = E [X 2 − 2X (EX ) + (EX )2 ] = E [X 2 ] − (EX )2 .
The following two inequalities are simple and fundamental. The Markov inequality states that
if Y is a nonnegative random variable, then for c > 0,
P {Y ≥ c} ≤ E [Y ]
c To prove Markov’s inequality, note that I{Y ≥c} ≤ Y , and take expectations on each side. The
c
Chebychev inequality states that if X is a random variable with ﬁnite mean µ and variance σ 2 , then
for any d > 0,
σ2
P {X − µ ≥ d} ≤ 2
d
The Chebychev inequality follows by applying the Markov inequality with Y = X − µ2 and c = d2 .
The characteristic function ΦX of a random variable X is deﬁned by
ΦX (u) = E [ejuX ]
for real values of u, where j = √ −1. For example, if X has pdf f , then
∞ ΦX (u) = exp(jux)fX (x)dx,
−∞ which is 2π times the inverse Fourier transform of fX .
Two random variables have the same probability distribution if and only if they have the same
characteristic function. If E [X k ] exists and is ﬁnite for an integer k ≥ 1, then the derivatives of
ΦX up to order k exist and are continuous, and
(k) ΦX (0) = j k E [X k ]
19 For a nonnegative integervalued random variable X it is often more convenient to work with the
z transform of the pmf, deﬁned by
∞
X z k pX (k ) ΨX (z ) = E [z ] =
k=0 for real or complex z with  z ≤ 1. Two such random variables have the same probability distribution if and only if their z transforms are equal. If E [X k ] is ﬁnite it can be found from the
derivatives of ΨX up to the k th order at z = 1,
(k) ΨX (1) = E [X (X − 1) · · · (X − k + 1)] 1.6 Frequently used distributions The following is a list of the most basic and frequently used probability distributions. For each
distribution an abbreviation, if any, and valid parameter values are given, followed by either the
CDF, pdf or pmf, then the mean, variance, a typical example and signiﬁcance of the distribution.
The constants p, λ, µ, σ , a, b, and α are realvalued, and n and i are integervalued, except n
can be nonintegervalued in the case of the gamma distribution. Bernoulli: Be(p), 0 ≤ p ≤ 1 p
i=1
1−p i=0 0
else
z transform: 1 − p + pz
pmf: p(i) = mean: p variance: p(1 − p) Example: Number of heads appearing in one ﬂip of a coin. The coin is called fair if p =
biased otherwise. Binomial: Bi(n, p), n ≥ 1, 0 ≤ p ≤ 1
ni
p (1 − p)n−i
i
z transform: (1 − p + pz )n
pmf:p(i) = mean: np 0≤i≤n variance: np(1 − p) Example: Number of heads appearing in n independent ﬂips of a coin.
20 1
2 and Poisson: P oi(λ), λ ≥ 0
λi e−λ
i≥0
i!
z transform: exp(λ(z − 1))
pmf: p(i) = mean: λ variance: λ Example: Number of phone calls placed during a ten second interval in a large city.
Signiﬁcance: The Poisson pmf is the limit of the binomial pmf as n → +∞ and p → 0 in such a
way that np → λ.
Geometric: Geo(p), 0 < p ≤ 1
pmf: p(i) = (1 − p)i−1 p
i≥1
pz
z transform:
1 − z + pz
1
1−p
mean:
variance:
p
p2 Example: Number of independent ﬂips of a coin until heads ﬁrst appears.
Signiﬁcant property: If X has the geometric distribution, P {X > i} = (1 − p)i for integers i ≥ 1.
So X has the memoryless property:
P {X > i + j  X > i} = P {X > j } for i, j ≥ 1.
Any positive integervalued random variable with this property has a geometric distribution.
Gaussian (also called Normal): N (µ, σ 2 ), µ ∈ R, σ ≥ 0
pdf (if σ 2 > 0): f (x) = √
pmf (if σ 2 = 0): p(x) = 1 exp − 2πσ 2
1 x=µ
0
else characteristic function: exp(juµ −
mean: µ (x − µ)2
2σ 2 u2 σ 2
)
2 variance: σ 2 Example: Instantaneous voltage diﬀerence (due to thermal noise) measured across a resistor held
at a ﬁxed temperature.
Notation: The character Φ is often used to denote the CDF of a N (0, 1) random variable,3 and Q
is often used for the complementary CDF:
∞ Q(c) = 1 − Φ(c) =
c
3 x2
1
√ e− 2 dx
2π As noted earlier, Φ is also used to denote characteristic functions. The meaning should be clear from the context. 21 Signiﬁcant property (Central limit theorem): If X1 , X2 , . . . are independent and identically distributed with mean µ and nonzero variance σ 2 , then for any constant c,
X1 + · · · + Xn − nµ
√
≤c
nσ 2 lim P n→∞ Exponential: = Φ(c) Exp (λ), λ > 0
pdf: f (x) = λe−λx x≥0
λ
characteristic function:
λ − ju
1
1
mean:
variance: 2
λ
λ Example: Time elapsed between noon sharp and the ﬁrst telephone call placed in a large city, on
a given day.
Signiﬁcance: If X has the Exp(λ) distribution, P {X ≥ t} = e−λt for t ≥ 0. So X has the
memoryless property:
P {X ≥ s + t  X ≥ s} = P {X ≥ t} s, t ≥ 0 Any nonnegative random variable with this property is exponentially distributed.
Uniform: U (a, b) − ∞ < a < b < ∞
1
b−a pdf: f (x) = a≤x≤b
else 0 ejub − ejua
ju(b − a)
(b − a)2
variance:
12 characteristic function:
mean: a+b
2 Example: The phase diﬀerence between two independent oscillators operating at the same frequency may be modelled as uniformly distributed over [0, 2π ]
Signiﬁcance: Uniform is uniform.
Gamma(n, α): n, α > 0 (n real valued)
pdf: f (x) = αn xn−1 e−αx
Γ(n)
∞ where Γ(n) = x≥0 sn−1 e−s ds 0 α
α − ju characteristic function:
mean: n
α variance:
22 n
α2 n Signiﬁcance: If n is a positive integer then Γ(n) = (n − 1)! and a Gamma (n, α) random variable
has the same distribution as the sum of n independent, Exp(α) distributed random variables.
Rayleigh (σ 2 ): σ2 > 0
r
r2
r>0
exp − 2
σ2
2σ
r2
CDF : 1 − exp − 2
2σ
π
π
mean: σ
variance: σ 2 2 −
2
2
pdf: f (r) = Example: Instantaneous value of the envelope of a mean zero, narrow band noise signal.
1 Signiﬁcance: If X and Y are independent, N (0, σ 2 ) random variables, then (X 2 + Y 2 ) 2 has the
Rayleigh(σ 2 ) distribution. Also notable is the simple form of the CDF. 1.7 Jointly distributed random variables Let X1 , X2 , . . . , Xm be random variables on a single probability space (Ω, F , P ). The joint cumulative distribution function (CDF) is the function on Rm deﬁned by
FX1 X2 ···Xm (x1 , . . . , xm ) = P {X1 ≤ x1 , X2 ≤ x2 , . . . , Xm ≤ xm }
The CDF determines the probabilities of all events concerning X1 , . . . , Xm . For example, if R is
the rectangular region (a, b] × (a , b ] in the plane, then
P {(X1 , X2 ) ∈ R} = FX1 X2 (b, b ) − FX1 X2 (a, b ) − FX1 X2 (b, a ) + FX1 X2 (a, a )
We write +∞ as an argument of FX in place of xi to denote the limit as xi → +∞. By the
countable additivity axiom of probability,
FX1 X2 (x1 , +∞) = lim FX1 X2 (x1 , x2 ) = FX1 (x1 ) x2 →∞ The random variables are jointly continuous if there exists a function fX1 X2 ···Xm , called the
joint probability density function (pdf), such that
x1 FX1 X2 ···Xm (x1 , . . . , xm ) = xm ···
−∞ −∞ fX1 X2 ···Xm (u1 , . . . , um )dum · · · du1 . Note that if X1 and X2 are jointly continuous, then
FX1 (x1 ) = FX1 X2 (x1 , +∞)
x1 ∞ −∞ −∞ = fX1 X2 (u1 , u2 )du2 du1 . so that X1 has pdf given by
∞ fX1 (u1 ) = −∞ fX1 X2 (u1 , u2 )du2 . 23 If X1 , X2 , . . . , Xm are each discrete random variables, then they have a joint pmf pX1 X2 ···Xm
deﬁned by
pX1 X2 ···Xm (u1 , u2 , . . . , um ) = P [{X1 = u1 } ∩ {X2 = u2 } ∩ · · · ∩ {Xm = um }]
The sum of the probability masses is one, and for any subset A of Rm
P {(X1 , . . . , Xm ) ∈ A} = pX (u1 , u2 , . . . , um )
(u1 ,...,um )∈A The joint pmf of subsets of X1 , . . . Xm can be obtained by summing out the other coordinates of
the joint pmf. For example,
pX1 X2 (u1 , u2 ) pX1 (u1 ) =
u2 The joint characteristic function of X1 , . . . , Xm is the function on Rm deﬁned by
ΦX1 X2 ···Xm (u1 , u2 , . . . , um ) = E [ej (X1 u1 +X2 ux +···+Xm um ) ]
Random variables X1 , . . . , Xm are deﬁned to be independent if for any Borel subsets A1 , . . . , Am
of R, the events {X1 ∈ A1 }, . . . , {Xm ∈ Am } are independent. The random variables are independent if and only if the joint CDF factors.
FX1 X2 ···Xm (x1 , . . . , xm ) = FX1 (x1 ) · · · FXm (xm )
If the random variables are jointly continuous, independence is equivalent to the condition that the
joint pdf factors. If the random variables are discrete, independence is equivalent to the condition
that the joint pmf factors. Similarly, the random variables are independent if and only if the joint
characteristic function factors. 1.8 Cross moments of random variables Let X and Y be random variables on the same probability space with ﬁnite second moments. Three
important related quantities are:
the correlation: E [XY ]
the covariance: Cov(X, Y ) = E [(X − E [X ])(Y − E [Y ])]
Cov(X, Y )
the correlation coeﬃcient: ρXY =
Var(X )Var(Y )
A fundamental inequality is Schwarz’s inequality:
E [X 2 ]E [Y 2 ]  E [XY ]  ≤ (1.15) Furthermore, if E [Y 2 ] = 0, equality holds if and only if P [X = cY ] = 1 for some constant c.
Schwarz’s inequality (1.15) is equivalent to the L2 triangle inequality for random variables:
1 1 1 E [(X + Y )2 ] 2 ≤ E [X 2 ] 2 + E [Y 2 ] 2
24 (1.16) Schwarz’s inequality can be proved as follows. If P {Y = 0} = 1 the inequality is trivial, so suppose
E [Y 2 ] > 0. By the inequality (a + b)2 ≤ 2a2 + 2b2 it follows that E [(X − λY )2 ] < ∞ for any
constant λ. Take λ = E [XY ]/E [Y 2 ] and note that
0 ≤ E [(X − λY )2 ] = E [X 2 ] − 2λE [XY ] + λ2 E [Y 2 ]
E [XY ]2
= E [X 2 ] −
,
E [Y 2 ]
which is clearly equivalent to the Schwarz inequality. If P [X = cY ] = 1 for some c then equality
holds in (1.15), and conversely, if equality holds in (1.15) then P [X = cY ] = 1 for c = λ.
Application of Schwarz’s inequality to X − E [X ] and Y − E [Y ] in place of X and Y yields that
 Cov(X, Y )  ≤ Var(X )Var(Y ) Furthermore, if Var(Y ) = 0 then equality holds if and only if X = aY + b for some constants a and
b. Consequently, if Var(X ) and Var(Y ) are not zero, so that the correlation coeﬃcient ρXY is well
deﬁned, then  ρXY ≤ 1 with equality if and only if X = aY + b for some constants a, b.
The following alternative expressions for Cov(X, Y ) are often useful in calculations:
Cov(X, Y ) = E [X (Y − E [Y ])] = E [(X − E [X ])Y ] = E [XY ] − E [X ]E [Y ]
In particular, if either X or Y has mean zero then E [XY ] = Cov(X, Y ).
Random variables X and Y are called orthogonal if E [XY ] = 0 and are called uncorrelated
if Cov(X, Y ) = 0. If X and Y are independent then they are uncorrelated. The converse is far
from true. Independence requires a large number of equations to be true, namely FXY (x, y ) =
FX (x)FY (y ) for every real value of x and y . The condition of being uncorrelated involves only a
single equation to hold.
Covariance generalizes variance, in that Var(X ) = Cov(X, X ). Covariance is linear in each of
its two arguments:
Cov(X + Y, U + V ) = Cov(X, U ) + Cov(X, V ) + Cov(Y, U ) + Cov(Y, V )
Cov(aX + b, cY + d) = acCov(X, Y )
for constants a, b, c, d. For example, consider the sum Sm = X1 + · · · + Xm , such that X1 , · · · , Xm
are (pairwise) uncorrelated with E [Xi ] = µ and Var(Xi ) = σ 2 for 1 ≤ i ≤ n. Then E [Sm ] = mµ
and
Var(Sm ) = Cov(Sm , Sm )
= Var(Xi ) +
i = mσ 2 .
Therefore, S√ −mµ
m
mσ 2 Cov(Xi , Xj )
i,j :i=j has mean zero and variance one.
25 1.9 Conditional densities Suppose that X and Y have a joint pdf fXY . Recall that the pdf fY , the second marginal density
of fXY , is given by
∞ fY (y ) = fXY (x, y )dx
−∞ The conditional pdf of X given Y , denoted by fX Y (x  y ), is undeﬁned if fY (y ) = 0. It is deﬁned
for y such that fY (y ) > 0 by
fX Y (x  y ) = fXY (x, y )
fY (y ) − ∞ < x < +∞ If y is ﬁxed and fY (y ) > 0, then as a function of x, fX Y (x  y ) is itself a pdf.
The expectation of the conditional pdf is called the conditional expectation (or conditional
mean) of X given Y = y , written as
∞ E [X  Y = y ] = xfX Y (x  y )dx
−∞ If the deterministic function E [X  Y = y ] is applied to the random variable Y , the result is a
random variable denoted by E [X  Y ].
Note that conditional pdf and conditional expectation were so far deﬁned in case X and Y have
a joint pdf. If instead, X and Y are both discrete random variables, the conditional pmf pX Y and
the conditional expectation E [X  Y = y ] can be deﬁned in a similar way. More general notions of
conditional expectation are considered in a later chapter. 1.10 Transformation of random vectors A random vector X of dimension m has the form X= X1
X2
.
.
. Xm
where X1 , . . . , Xm are random variables. The joint distribution of X1 , . . . , Xm can be considered
to be the distribution of the vector X . For example, if X1 , . . . , Xm are jointly continuous, the joint
pdf fX1 X2 ···Xm (x1 , . . . , xn ) can as well be written as fX (x), and be thought of as the pdf of the
random vector X .
Let X be a continuous type random vector on Rn . Let g be a onetoone mapping from Rn
to Rn . Think of g as mapping xspace (here x is lower case, representing a coordinate value) into
y space. As x varies over Rn , y varies over the range of g . All the while, y = g (x) or, equivalently,
x = g −1 (y ).
∂y
Suppose that the Jacobian matrix of derivatives ∂x (x) is continuous in x and nonsingular for
all x. By the inverse function theorem of vector calculus, it follows that the Jacobian matrix of the
∂y
inverse mapping (from y to x) exists and satisﬁes ∂x (y ) = ( ∂x (x))−1 . Use  K  for a square matrix
∂y
K to denote det(K ) .
26 Proposition 1.10.1 Under the above assumptions, Y is a continuous type random vector and for
y in the range of g :
fY (y ) = fX (x)
∂y
∂x (x)  = fX (x)  ∂x
(y )
∂y Example 1.10.2 Let U , V have the joint pdf:
u + v 0 ≤ u, v ≤ 1
0
else fU V (u, v ) = and let X = U 2 and Y = U (1 + V ). Let’s ﬁnd the pdf fXY . The vector (U, V ) in the u − v plane is
transformed into the vector (X, Y ) in the x − y plane under a mapping g that maps u, v to x = u2
and y = u(1 + v ). The image in the x − y plane of the square [0, 1]2 in the u − v plane is the set A
given by
√
√
A = {(x, y ) : 0 ≤ x ≤ 1, and x ≤ y ≤ 2 x}
See Figure 1.13 The mapping from the square is one to one, for if (x, y ) ∈ A then (u, v ) can be
y
2 v
1 1 u x
1 Figure 1.13: Transformation from the u − v plane to the x − y plane.
recovered by u = √ x and v = y
√
x − 1. The Jacobian determinant is ∂x
∂u
∂y
∂u ∂x
∂v
∂y
∂v 2u
0
1+v u = = 2u2 Therefore, using the transformation formula and expressing u and V in terms of x and y yields
√ y
x+( √x −1) fXY (x, y ) = 2x 0 if (x, y ) ∈ A
else Example 1.10.3 Let U and V be independent continuous type random variables. Let X = U + V
and Y = V . Let us ﬁnd the joint density of X, Y and the marginal density of X . The mapping
g: uv → uv
27 = u+v v is invertible, with inverse given by u = x − y and v = y . The absolute value of the Jacobian
determinant is given by
∂x
∂u
∂y
∂u ∂x
∂v
∂y
∂v 11
01 = =1 Therefore
fXY (x, y ) = fU V (u, v ) = fU (x − y )fV (y )
The marginal density of X is given by
∞ ∞ fU (x − y )fV (y )dy fXY (x, y )dy = fX (x) = −∞ −∞ That is fX = fU ∗ fV . Example 1.10.4 Let X1 and X2 be independent N (0, σ 2 ) random variables, and let X = (X1 , X2 )T
denote the twodimensional random vector with coordinates X1 and X2 . Any point of x ∈ R2 can
1
be represented in polar coordinates by the vector (r, θ)T such that r = x = (x2 + x2 ) 2 and
1
2
x2
θ = tan−1 ( x1 ) with values r ≥ 0 and 0 ≤ θ < 2π . The inverse of this mapping is given by
x1 = r cos(θ)
x2 = r sin(θ)
We endeavor to ﬁnd the pdf of the random vector (R, Θ)T , the polar coordinates of X . The pdf of
X is given by
fX (x) = fX1 (x1 )fX2 (x2 ) = 1 − r22
e 2σ
2πσ 2 The range of the mapping is the set r > 0 and 0 < θ ≤ 2π . On the range,
∂x
r
∂
θ = ∂ x1
∂r
∂x2
∂r ∂x1
∂θ
∂x2
∂θ = cos(θ) −r sin(θ)
sin(θ) r cos(θ) =r Therefore for (r, θ)T in the range of the mapping, fR,Θ (r, θ) = fX (x) ∂x
r
∂
θ = r − r22
e 2σ
2πσ 2 Of course fR,Θ (r, θ) = 0 oﬀ the range of the mapping. The joint density factors into a function of
r and a function of θ, so R and Θ are independent. Moreover, R has the Rayleigh density with
parameter σ 2 , and Θ is uniformly distributed on [0, 2π ]. 28 1.11 Problems 1.1 Simple events
A register contains 8 random binary digits which are mutually independent. Each digit is a zero or
a one with equal probability. (a) Describe an appropriate probability space (Ω, F , P ) corresponding
to looking at the contents of the register.
(b) Express each of the following four events explicitly as subsets of Ω, and ﬁnd their probabilities:
E1=“No two neighboring digits are the same”
E2=“Some cyclic shift of the register contents is equal to 01100110”
E3=“The register contains exactly four zeros”
E4=“There is a run of at least six consecutive ones”
(c) Find P [E1 E3 ] and P [E2 E3 ].
1.2 Independent vs. mutually exclusive
(a) Suppose that an event E is independent of itself. Show that either P [E ] = 0 or P [E ] = 1.
(b) Events A and B have probabilities P [A] = 0.3 and P [B ] = 0.4. What is P [A ∪ B ] if A and B
are independent? What is P [A ∪ B ] if A and B are mutually exclusive?
(c) Now suppose that P [A] = 0.6 and P [B ] = 0.8. In this case, could the events A and B be
independent? Could they be mutually exclusive?
1.3 Congestion at output ports
Consider a packet switch with some number of input ports and eight output ports. Suppose four
packets simultaneously arrive on diﬀerent input ports, and each is routed toward an output port.
Assume the choices of output ports are mutually independent, and for each packet, each output
port has equal probability.
(a) Specify a probability space (Ω, F , P ) to describe this situation.
(b) Let Xi denote the number of packets routed to output port i for 1 ≤ i ≤ 8. Describe the joint
pmf of X1 , . . . , X8 .
(c) Find Cov(X1 , X2 ).
(d) Find P {Xi ≤ 1 for all i}.
(e) Find P {Xi ≤ 2 for all i}.
1.4 Frantic search
At the end of each day Professor Plum puts her glasses in her drawer with probability .90, leaves
them on the table with probability .06, leaves them in her briefcase with probability 0.03, and she
actually leaves them at the oﬃce with probability 0.01. The next morning she has no recollection
of where she left the glasses. She looks for them, but each time she looks in a place the glasses are
actually located, she misses ﬁnding them with probability 0.1, whether or not she already looked
in the same place. (After all, she doesn’t have her glasses on and she is in a hurry.)
(a) Given that Professor Plum didn’t ﬁnd the glasses in her drawer after looking one time, what is
the conditional probability the glasses are on the table?
(b) Given that she didn’t ﬁnd the glasses after looking for them in the drawer and on the table
once each, what is the conditional probability they are in the briefcase?
(c) Given that she failed to ﬁnd the glasses after looking in the drawer twice, on the table twice,
and in the briefcase once, what is the conditional probability she left the glasses at the oﬃce?
29 1.5 Conditional probability of failed device given failed attempts
A particular webserver may be working or not working. If the webserver is not working, any attempt
to access it fails. Even if the webserver is working, an attempt to access it can fail due to network
congestion beyond the control of the webserver. Suppose that the a priori probability that the server
is working is 0.8. Suppose that if the server is working, then each access attempt is successful with
probability 0.9, independently of other access attempts. Find the following quantities.
(a) P [ ﬁrst access attempt fails]
(b) P [server is working  ﬁrst access attempt fails ]
(c) P [second access attempt fails  ﬁrst access attempt fails ]
(d) P [server is working  ﬁrst and second access attempts fail ].
1.6 Conditional probabilities–basic computations of iterative decoding
Suppose B1 , . . . , Bn , Y1 , . . . , Yn are discrete random variables with joint pmf
p(b1 , . . . , bn , y1 , . . . , yn ) = n
i=1 qi (yi bi ) 2−n 0 if bi ∈ {0, 1} for 1 ≤ i ≤ n
else where qi (yi bi ) as a function of yi is a pmf for bi ∈ {0, 1}. Finally, let B = B1 ⊕· · ·⊕ Bn represent the
modulo two sum of B1 , · · · , Bn . Thus, the ordinary sum of the n +1 random variables B1 , . . . , Bn , B
is even. Express P [B = 1Y1 = y1 , · · · .Yn = yn ] in terms of the yi and the functions qi . Simplify
your answer.
(b) Suppose B and Z1 , . . . , Zk are discrete random variables with joint pmf
p(b, z1 , . . . , zk ) = 1
2 k
j =1 rj (zj b) 0 if b ∈ {0, 1}
else where rj (zj b) as a function of zj is a pmf for b ∈ {0, 1} ﬁxed. Express P [B = 1Z1 = z1 , . . . , Zk = zk ]
in terms of the zj and the functions rj .
1.7 Conditional lifetimes and the memoryless property of the geometric distribution
(a) Let X represent the lifetime, rounded up to an integer number of years, of a certain car battery.
Suppose that the pmf of X is given by pX (k ) = 0.2 if 3 ≤ k ≤ 7 and pX (k ) = 0 otherwise. (i)
Find the probability, P {X > 3}, that a three year old battery is still working. (ii) Given that the
battery is still working after ﬁve years, what is the conditional probability that the battery will
still be working three years later? (i.e. what is P [X > 8X > 5]?)
(b) A certain Illini basketball player shoots the ball repeatedly from half court during practice.
Each shot is a success with probability p and a miss with probability 1 − p, independently of the
outcomes of previous shots. Let Y denote the number of shots required for the ﬁrst success. (i)
Express the probability that she needs more than three shots for a success, P {Y > 3}, in terms of
p. (ii) Given that she already missed the ﬁrst ﬁve shots, what is the conditional probability that
she will need more than three additional shots for a success? (i.e. what is P [Y > 8Y > 5])?
(iii) What type of probability distribution does Y have?
1.8 Blue corners
Suppose each corner of a cube is colored blue, independently of the other corners, with some
probability p. Let B denote the event that at least one face of the cube has all four corners colored
blue. (a) Find the conditional probability of B given that exactly ﬁve corners of the cube are
colored blue. (b) Find P [B ], the unconditional probability of B.
30 1.9 Distribution of the ﬂow capacity of a network
A communication network is shown. The link capacities in megabits per second (Mbps) are given
by C1 = C3 = 5, C2 = C5 = 10 and C4 =8, and are the same in each direction. Information ﬂow
2 1
Source Destination
4 3 5 from the source to the destination can be split among multiple paths. For example, if all links are
working, then the maximum communication rate is 10 Mbps: 5 Mbps can be routed over links 1 and
2, and 5 Mbps can be routed over links 3 and 5. Let Fi be the event that link i fails. Suppose that
F1 , F2 , F3 , F4 and F5 are independent and P [Fi ] = 0.2 for each i. Let X be deﬁned as the maximum
rate (in Mbits per second) at which data can be sent from the source node to the destination node.
Find the pmf pX .
1.10 Recognizing cumulative distribution functions
Which of the following are valid CDF’s? For each that is not valid, state at least one reason why.
For each that is valid, ﬁnd P {X 2 > 5}.
2 (
F 1 (x ) =
1 e− x
4
−x2
−e4 x<0
x≥0 8
< 0
0.5 + e−x
F2 (x) =
:
1 x<0
0≤x<3
x≥3 F3 (x) = 8
< 0
0 .5 +
:
1 x
20 x≤0
0 < x ≤ 10
x ≥ 10 1.11 A CDF of mixed type
Let X have the CDF shown.
F
X
1.0 0.5 0 1 2 (a) Find P {X ≤ 0.8}.
(b) Find E [X ].
(c) Find Var(X ).
1.12 CDF and characteristic function of a mixed type random variable
Let X = (U − 0.5)+ , where U is uniformly distributed over the interval [0, 1]. That is, X = U − 0.5
if U − 0.5 ≥ 0, and X = 0 if U − 0.5 < 0.
(a) Find and carefully sketch the CDF FX . In particular, what is FX (0)?
(b) Find the characteristic function ΦX (u) for real values of u.
31 1.13 Poisson and geometric random variables with conditioning
Let Y be a Poisson random variable with mean µ > 0 and let Z be a geometrically distributed
random variable with parameter p with 0 < p < 1. Assume Y and Z are independent.
(a) Find P {Y < Z }. Express your answer as a simple function of µ and p.
(b) Find P [Y < Z Z = i] for i ≥ 1. (Hint: This is a conditional probability for events.)
(c) Find P [Y = iY < Z ] for i ≥ 0. Express your answer as a simple function of p, µ and i. (Hint:
This is a conditional probability for events.)
(d) Find E [Y Y < Z ], which is the expected value computed according to the conditional distribution found in part (c). Express your answer as a simple function of µ and p.
1.14 Conditional expectation for uniform density over a triangular region
Let (X, Y ) be uniformly distributed over the triangle with coordinates (0, 0), (1, 0), and (2, 1).
(a) What is the value of the joint pdf inside the triangle?
(b) Find the marginal density of X , fX (x). Be sure to specify your answer for all real values of x.
(c) Find the conditional density function fY X (y x). Be sure to specify which values of x the
conditional density is well deﬁned for, and for such x specify the conditional density for all y . Also,
for such x brieﬂy describe the conditional density of y in words.
(d) Find the conditional expectation E [Y X = x]. Be sure to specify which values of x this
conditional expectation is well deﬁned for.
1.15 Transformation of a random variable
Let X be exponentially distributed with mean λ−1 . Find and carefully sketch the distribution
functions for the random variables Y = exp(X ) and Z = min(X, 3).
1.16 Density of a function of a random variable
Suppose X is a random variable with probability density function
fX (x) = 2x 0 ≤ x ≤ 1
0 else (a) Find P [X ≥ 0.4X ≤ 0.8].
(b) Find the density function of Y deﬁned by Y = − log(X ).
1.17 Moments and densities of functions of a random variable
Suppose the length L and width W of a rectangle are independent and each uniformly distributed
over the interval [0, 1]. Let C = 2L + 2W (the length of the perimeter) and A = LW (the area).
Find the means, variances, and probability densities of C and A.
1.18 Functions of independent exponential random variables
Let X1 and X2 be independent random varibles, with Xi being exponentially distributed with
X1
parameter λi . (a) Find the pdf of Z = min{X1 , X2 }. (b) Find the pdf of R = X2 .
1.19 Using the Gaussian Q function
Express each of the given probabilities in terms of the standard Gaussian complementary CDF Q.
(a) P {X ≥ 16}, where X has the N (10, 9) distribution.
(b) P {X 2 ≥ 16}, where X has the N (10, 9) distribution.
(c) P {X − 2Y  > 1}, where X and Y are independent, N (0, 1) random variables. (Hint: Linear
combinations of independent Gaussian random variables are Gaussian.)
32 1.20 Gaussians and the Q function
Let X and Y be independent, N (0, 1) random variables.
(a) Find Cov(3X + 2Y, X + 5Y + 10).
(b) Express P {X + 4Y ≥ 2} in terms of the Q function.
(c) Express P {(X − Y )2 > 9} in terms of the Q function.
1.21 Correlation of histogram values
Suppose that n fair dice are independently rolled. Let
Xi = 1 if a 1 shows on the ith roll
0 else 1 if a 2 shows on the ith roll
0 else Yi = Let X denote the sum of the Xi ’s, which is simply the number of 1’s rolled. Let Y denote the sum
of the Yi ’s, which is simply the number of 2’s rolled. Note that if a histogram is made recording
the number of occurrences of each of the six numbers, then X and Y are the heights of the ﬁrst
two entries in the histogram.
(a) Find E [X1 ] and Var(X1 ).
(b) Find E [X ] and Var(X ).
(c) Find Cov(Xi , Yj ) if 1 ≤ i, j ≤ n (Hint: Does it make a diﬀerence if i = j ?)
(d) Find Cov(X, Y ) and the correlation coeﬃcient ρ(X, Y ) = Cov(X, Y )/ Var(X )Var(Y ).
(e) Find E [Y X = x] for any integer x with 0 ≤ x ≤ n. Note that your answer should depend on
x and n, but otherwise your answer is deterministic.
1.22 Working with a joint density
Suppose X and Y have joint density function fX,Y (x, y ) = c(1 + xy ) if 2 ≤ x ≤ 3 and 1 ≤ y ≤ 2,
and fX,Y (x, y ) = 0 otherwise. (a) Find c. (b) Find fX and fY . (c) Find fX Y .
1.23 A function of jointly distributed random variables
Suppose (U, V ) is uniformly distributed over the square with corners (0,0), (1,0), (1,1), and (0,1),
and let X = U V . Find the CDF and pdf of X .
1.24 Density of a diﬀerence
Let X and Y be independent, exponentially distributed random variables with parameter λ, such
that λ > 0. Find the pdf of Z = X − Y . 1.25 Working with a two dimensional density
L
et the random variables X and Y be jointly uniformly distributed over the region shown.
1 0
0 2 1 33 3 (a) Determine the value of fX,Y on the region shown.
(b) Find fX , the marginal pdf of X.
(c) Find the mean and variance of X.
(d) Find the conditional pdf of Y given that X = x, for 0 ≤ x ≤ 1.
(e) Find the conditional pdf of Y given that X = x, for 1 ≤ x ≤ 2.
(f) Find and sketch E [Y X = x] as a function of x. Be sure to specify which range of x this
conditional expectation is well deﬁned for.
1.26 Some characteristic functions
Find the mean and variance of random variables with the following characteristic functions: (a)
Φ(u) = exp(−5u2 + 2ju) (b) Φ(u) = (eju − 1)/ju, and (c) Φ(u) = exp(λ(eju − 1)).
1.27 Uniform density over a union of two square regions
Let the random variables X and Y be jointly uniformly distributed on the region {0 ≤ u ≤ 1, 0 ≤
v ≤ 1} ∪ {−1 ≤ u < 0, −1 ≤ v < 0}. (a) Determine the value of fXY on the region shown.
(b) Find fX , the marginal pdf of X .
(c) Find the conditional pdf of Y given that X = a, for 0 < a ≤ 1.
(d) Find the conditional pdf of Y given that X = a, for −1 ≤ a < 0.
(e) Find E [Y X = a] for a ≤ 1.
(f) What is the correlation coeﬃcient of X and Y ?
(g) Are X and Y independent?
(h) What is the pdf of Z = X + Y ?
1.28 A transformation of jointly continuous random variables
Suppose (U, V ) has joint pdf
fU,V (u, v ) = 9u2 v 2 if 0 ≤ u ≤ 1 & 0 ≤ v ≤ 1
0
else Let X = 3U and Y = U V . (a) Find the joint pdf of X and Y , being sure to specify where the joint
pdf is zero.
(b) Using the joint pdf of X and Y , ﬁnd the conditional pdf, fY X (y x), of Y given X . (Be sure to
indicate which values of x the conditional pdf is well deﬁned for, and for each such x specify the
conditional pdf for all real values of y .)
1.29 Transformation of densities
Let U and V have the joint pdf:
fU V (u, v ) = c(u − v )2 0 ≤ u, v ≤ 1
0
else for some constant c. (a) Find the constant c. (b) Suppose X = U 2 and Y = U 2 V 2 . Describe the
joint pdf fX,Y (x, y ) of X and Y . Be sure to indicate where the joint pdf is zero.
1.30 Jointly distributed variables
Let U and V be independent random variables, such that U is uniformly distributed over the
interval [0, 1], and V has the exponential probability density function
34 2 V
(a) Calculate E [ 1+U ].
(b) Calculate P {U ≤ V }.
(c) Find the joint probability density function of Y and Z, where Y = U 2 and Z = U V . 1.31 * Why not every set has a length
Suppose a length (actually, “onedimensional volume” would be a better name) of any subset A ⊂ R
could be deﬁned, so that the following three axioms are satisﬁed:
L0: 0 ≤ length(A) ≤ ∞ for any A ⊂ R
L1: length([a, b]) = b − a for a < b
L2: length(A) = length(A + y ), for any A ⊂ R and y ∈ R, where A + y represents the translation
of A by y , deﬁned by A + y = {x + y : x ∈ A}
L3: If A = ∪∞ Bi such that B1 , B2 , · · · are disjoint, then length(A) =
i=1 ∞
i=1 length(Bi ). The purpose of this problem is to show that the above supposition leads to a contradiction. Let
Q denote the set of rational numbers, Q = {p/q : p, q ∈ Z, q = 0}. (a) Show that the set of
rational numbers can be expressed as Q = {q1 , q2 , . . .}, which means that Q is countably inﬁnite.
Say that x, y ∈ R are equivalent, and write x ∼ y , if x − y ∈ Q. (b) Show that ∼ is an equivalence
relation, meaning it is reﬂexive (a ∼ a for all a ∈ R), symmetric (a ∼ b implies b ∼ a), and
transitive (a ∼ b and b ∼ c implies a ∼ c). For any x ∈ R, let Qx = Q + x. (c) Show that for
any x, y ∈ R, either Qx = Qy or Qx ∩ Qy = ∅. Sets of the form Qx are called equivalence classes
of the equivalence relation ∼. (d) Show that Qx ∩ [0, 1] = ∅ for all x ∈ R, or in other words, each
equivalence class contains at least one element from the interval [0, 1]. Let V be a set obtained
by choosing exactly one element in [0, 1] from each equivalence class (by accepting that V is well
deﬁned, you’ll be accepting what is called the Axiom of Choice). So V is a subset of [0, 1]. Suppose
q1 , q2 , . . . is an enumeration of all the rational numbers in the interval [−1, 1], with no number
appearing twice in the list. Let Vi = V + qi for i ≥ 1. (e) Verify that the sets Vi are disjoint, and
[0, 1] ⊂ ∪∞ Vi ⊂ [−1, 2]. Since the Vi ’s are translations of V , they should all have the same length
i=1
as V . If the length of V is deﬁned to be zero, then [0, 1] would be covered by a countable union
of disjoint sets of length zero, so [0, 1] would also have length zero. If the length of V were strictly
positive, then the countable union would have inﬁnite length, and hence the interval [−1, 2] would
have inﬁnite length. Either way there is a contradiction.
1.32 * On sigmaalgebras, random variables, and measurable functions
Prove the seven statements lettered (a)(g) in what follows.
Deﬁnition. Let Ω be an arbitrary set. A nonempty collection F of subsets of Ω is deﬁned to be
an algebra if: (i) Ac ∈ F whenever A ∈ F and (ii) A ∪ B ∈ F whenever A, B ∈ F .
(a) If F is an algebra then ∅ ∈ F , Ω ∈ F , and the union or intersection of any ﬁnite collection of
sets in F is in F .
Deﬁnition. F is called a σ algebra if F is an algebra such that whenever A1 , A2 , ... are each in F ,
so is the union, ∪Ai .
(b) If F is a σ algebra and B1 , B2 , . . . are in F , then so is the intersection, ∩Bi .
(c) Let U be an arbitrary nonempty set, and suppose that Fu is a σ algebra of subsets of Ω for
each u ∈ U . Then the intersection ∩u∈U Fu is also a σ algebra.
(d) The collection of all subsets of Ω is a σ algebra.
35 (e) If Fo is any collection of subsets of Ω then there is a smallest σ algebra containing Fo (Hint:
use (c) and (d).)
Deﬁnitions. B (R) is the smallest σ algebra of subsets of R which contains all sets of the form
(−∞, a]. Sets in B (R) are called Borel sets. A realvalued random variable on a probability space
(Ω, F , P ) is a realvalued function X on Ω such that {ω : X (ω ) ≤ a} ∈ F for any a ∈ R.
(f) If X is a random variable on (Ω, F , P ) and A ∈ B (R) then {ω : X (ω ) ∈ A} ∈ F . (Hint: Fix
a random variable X . Let D be the collection of all subsets A of B (R) for which the conclusion is
true. It is enough (why?) to show that D contains all sets of the form (−∞, a] and that D is a
σ algebra of subsets of R. You must use the fact that F is a σ algebra.)
Remark. By (f), P {ω : X (ω ) ∈ A}, or P {X ∈ A} for short, is well deﬁned for A ∈ B (R).
Deﬁnition. A function g mapping R to R is called Borel measurable if {x : g (x) ∈ A} ∈ B (R)
whenever A ∈ B (R).
(g) If X is a realvalued random variable on (Ω, F , P ) and g is a Borel measurable function, then
Y deﬁned by Y = g (X ) is also a random variable on (Ω, F , P ). 36 Chapter 2 Convergence of a Sequence of
Random Variables
Convergence to limits is a central concept in the theory of calculus. Limits are used to deﬁne
derivatives and integrals. We wish to consider derivatives and integrals of random functions, so it
is natural to begin by examining what it means for a sequence of random variables to converge.
See the Appendix for a review of the deﬁnition of convergence for a sequence of numbers. 2.1 Four deﬁnitions of convergence of random variables Recall that a random variable X is a function on Ω for some probability space (Ω, F , P ). A sequence
of random variables (Xn (ω ) : n ≥ 1) is hence a sequence of functions. There are many possible
deﬁnitions for convergence of a sequence of random variables.
One idea is to require Xn (ω ) to converge for each ﬁxed ω . However, at least intuitively, what
happens on an event of probability zero is not important. Thus, we use the following deﬁnition.
Deﬁnition 2.1.1 A sequence of random variables (Xn : n ≥ 1) converges almost surely to a
random variable X , if all the random variables are deﬁned on the same probability space, and
a.s.
P {limn→∞ Xn = X } = 1. Almost sure convergence is denoted by limn→∞ Xn = X a.s. or Xn → X.
Conceptually, to check almost sure convergence, one can ﬁrst ﬁnd the set {ω : limn→∞ Xn (ω ) =
X (ω )} and then see if it has probability one.
We shall construct some examples using the standard unitinterval probability space deﬁned
in Example 1.1.2. This particular choice of (Ω, F , P ) is useful for generating examples, because
random variables, being functions on Ω, can be simply speciﬁed by their graphs. For example,
consider the random variable X pictured in Figure 2.1. The probability mass function for such X
is given by P {X = 1} = P {X = 2} = 1 and P {X = 3} = 1 . Figure 2.1 is a bit sloppy, in that it
4
2
is not clear what the values of X are at the jump points, ω = 1/4 or ω = 1/2. However, each of
these points has probability zero, so the distribution of X is the same no matter how X is deﬁned
at those points.
Example 2.1.2 Let (Xn : n ≥ 1) be the sequence of random variables on the standard unitinterval
probability space deﬁned by Xn (ω ) = ω n , illustrated in Figure 2.2. This sequence converges for all
37 X(ω )
3
2
1 ω
1
4 0 3
4 1
2 1 Figure 2.1: A random variable on (Ω, F , P ). X1( ! ) X2( ! ) 1 X3( ! ) 1 ! 0
0 1 X4( ! ) 1 ! 0
0 1 ! 0 1 0 1 ! 0
0 1 Figure 2.2: Xn (ω ) = ω n on the standard unitinterval probability space.
ω ∈ Ω, with the limit
lim Xn (ω ) = n→∞ 0 if 0 ≤ ω < 1
1 if ω = 1. The single point set {1} has probability zero, so it is also true (and simpler to say) that (Xn : n ≥ 1)
converges a.s. to zero. In other words, if we let X be the zero random variable, deﬁned by X (ω ) = 0
a.s.
for all ω , then Xn → X . Example 2.1.3 (Moving, shrinking rectangles) Let (Xn : n ≥ 1) be the sequence of random
variables on the standard unitinterval probability space, as shown in Figure 2.3. The variable
1
X1 is identically one. The variables X2 and X3 are one on intervals of length 2 . The variables
1
X4 , X5 , X6 , and X7 are one on intervals of length 4 . In general, each n ≥ 1 can be written as
n = 2k + j where k = ln2 n and 0 ≤ j < 2k . The variable Xn is one on the length 2−k interval
(j 2−k , (j + 1)2−k ].
To investigate a.s. convergence, ﬁx an arbitrary value for ω . Then for each k ≥ 1, there
is one value of n with 2k ≤ n < 2k+1 such that Xn (ω ) = 1, and Xn (ω ) = 0 for all other n.
Therefore, limn→∞ Xn (ω ) does not exist. That is, {ω : limn→∞ Xn exists} = ∅, so of course,
P {limn→∞ Xn exists} = 0. Thus, Xn does not converge in the a.s. sense.
However, for large n, P {Xn = 0} is close to one. This suggests that Xn converges to the zero
random variable in some weaker sense.
Example 2.1.3 motivates us to consider the following weaker notion of convergence of a sequence
of random variables.
38 X1( ω )
1 ω 0
0 1 X2( ω ) X3( ω ) 1 1 ω 0
0 ω 0 1 0 X4( ω ) 1 X5( ω ) 1 X6( ω ) 1 ω 0
0 1 X7( ω ) 1 ω 0
0 1 ω 0 1 0 1 ω 0
0 1 Figure 2.3: A sequence of random variables on (Ω, F , P ).
Deﬁnition 2.1.4 A sequence of random variables (Xn ) converges to a random variable X in probability if all the random variables are deﬁned on the same probability space, and for any > 0,
limn→∞ P {X − Xn  ≥ } = 0. Convergence in probability is denoted by limn→∞ Xn = X p., or
p.
Xn → X.
Convergence in probability requires that X − Xn  be small with high probability (to be precise,
less than or equal to with probability that converges to one as n → ∞), but on the small
probability event that X − Xn  is not small, it can be arbitrarily large. For some applications that
is unacceptable. Roughly speaking, the next deﬁnition of convergence requires that X − Xn  be
small with high probability for large n, and even if it is not small, the average squared value has
to be small enough.
Deﬁnition 2.1.5 A sequence of random variables (Xn ) converges to a random variable X in the
2
mean square sense if all the random variables are deﬁned on the same probability space, E [Xn ] <
2 ] = 0. Mean square convergence is denoted by
+∞ for all n, and limn→∞ E [(Xn − X )
m.s.
limn→∞ Xn = X m.s. or Xn → X.
Although it isn’t explicitly stated in the deﬁnition of m.s. convergence, the limit random variable
must also have a ﬁnite second moment:
m.s. Proposition 2.1.6 If Xn → X , then E [X 2 ] < +∞.
m.s. 2
Proof. Suppose that Xn → X . By deﬁnition, E [Xn ] < ∞ for all n. Also by deﬁnition, there
2 ] < 1 for all n ≥ n . The L2 triangle inequality for random
exists some no so that E [(X − Xn )
o
1
1
2 ] 2 ≤ E [(X − X )2 ] 2 + E [X 2 ] 1 < +∞.
variables, (1.16), yields E [(X∞ )
∞
no
no 2 Example 2.1.7 (More moving, shrinking rectangles) This example is along the same lines as
Example 2.1.3, using the standard unitinterval probability space. Each random variable of the
sequence (Xn : n ≥ 1) is deﬁned as indicated in Figure 2.4. where the value an > 0 is some
39 Xn( ! ) an !
0 1/n 1 Figure 2.4: A sequence of random variables corresponding to moving, shrinking rectangles.
constant depending on n. The graph of Xn for n ≥ 1 has height an over some subinterval of Ω of
1
length n . We don’t explicitly identify the location of the interval, but we require that for any ﬁxed
ω , Xn (ω ) = an for inﬁnitely many values of n, and Xn (ω ) = 0 for inﬁnitely many values of n. Such
a choice of the locations of the intervals is possible because the sum of the lengths of the intervals,
∞1
n=1 n , is inﬁnite.
a.s.
Of course Xn → 0 if the deterministic sequence (an ) converges to zero. However, if there
is a constant > 0 such that an ≥ for all n (for example if an = 1 for all n), then {ω :
limn→∞ Xn (ω ) exists} = ∅, just as in Example 2.1.3. The sequence converges to zero in probability
for any choice of the constants (an ), because for any > 0,
P {Xn − 0 ≥ } ≤ P {Xn = 0} = 1
→ 0.
n Finally, to investigate mean square convergence, note that E [Xn − 02 ] =
a2
n a2
n
n. m.s. Hence, Xn → 0 if and only if the sequence of constants (an ) is such that limn→∞ n = 0. For example, if an = ln(n)
√
m.s.
for all n, then Xn → 0, but if an = n, then (Xn ) does not converge to zero in the m.s. sense.
(Proposition 2.1.13 below shows that a sequence can have only one limit in the a.s., p., or m.s.
p.
senses, so the fact Xn → 0, implies that zero is the only possible limit in the m.s. sense. So if
2
an
n → 0, then (Xn ) doesn’t converge to any random variable in the m.s. sense.) Example 2.1.8 (Anchored, shrinking rectangles) Let (Xn : n ≥ 1) be a sequence of random
variables deﬁned on the standard unitinterval probability space, as indicated in Figure 2.5, where
the value an > 0 is some constant depending on n. That is, Xn (ω ) is equal to an if 0 ≤ ω ≤ 1/n,
and to zero otherwise. For any nonzero ω in Ω, Xn (ω ) = 0 for all n such that n > 1/ω. Therefore,
a.s.
Xn → 0.
Whether the sequence (Xn ) converges in p. or m.s. sense for this example is exactly the same
as in Example 2.1.7. That is, for convergence in probability or mean square sense, the locations of
2
p.
m.s.
the shrinking intervals of support don’t matter. So Xn → 0. And Xn → 0 if and only if an → 0.
n
It is shown in Proposition 2.1.13 below that either a.s. or m.s. convergence imply convergence in
probability. Example 2.1.8 shows that a.s. convergence, like convergence in probability., can allow
40 Xn( ! ) an !
1 0 1/n Figure 2.5: A sequence of random variables corresponding to anchored, shrinking rectangles.
Xn (ω ) − X (ω ) to be extremely large for ω in a small probability set. So neither convergence in
probability, nor a.s. convergence, imply m.s. convergence, unless an additional assumption is made
to control the diﬀerence Xn (ω ) − X (ω ) everywhere on Ω.
Example 2.1.9 (Rearrangements of rectangles) Let (Xn : n ≥ 1) be a sequence of random variables deﬁned on the standard unitinterval probability space. The ﬁrst three random variables
in the sequence are indicated in Figure 2.6. Suppose that the sequence is periodic, with period
three, so that Xn+3 = Xn for all n ≥ 1. Intuitively speaking, the sequence of random variables
X1( ! ) X2( ! ) 1 X3( ! ) 1 ! 0
0 1 1 ! 0
0 1 ! 0
0 1 Figure 2.6: A sequence of random variables obtained by rearrangement of rectangles.
persistently jumps around. Obviously it does not converge in the a.s. sense. The sequence does
not settle down to converge, even in the sense of convergence in probability, to any one random
p.
variable. This can be proved as follows. Suppose for the sake of contradiction that Xn → X for
some random variable. Then for any > 0 and δ > 0, if n is suﬃciently large, P {Xn − X  ≥ } ≤ δ.
But because the sequence is periodic, it must be that P {Xn − X  ≥ } ≤ δ for 1 ≤ n ≤ 3. Since δ
is arbitrary it must be that P {Xn − X  ≥ } = 0 for 1 ≤ n ≤ 3. Since is arbitrary it must be that
P {X = Xn } = 1 for 1 ≤ n ≤ 3. Hence, P {X1 = X2 = X3 } = 1, which is a contradiction. Thus,
the sequence does not converge in probability. A similar argument shows it does not converge in
the m.s. sense, either.
Even though the sequence fails to converge in a.s., m.s., or p. senses, it can be observed that
all of the Xn ’s have the same probability distribution. The variables are only diﬀerent in that the
places they take their possible values are rearranged.
Example 2.1.9 suggests that it would be useful to have a notion of convergence that just depends
on the distributions of the random variables. One idea for a deﬁnition of convergence in distribution
is to require that the sequence of CDFs FXn (x) converge as n → ∞ for all n. The following example
shows such a deﬁnition could give unexpected results in some cases.
41 Example 2.1.10 Let U be uniformly distributed on the interval [0, 1], and for n ≥ 1, let Xn =
(−1)n U
. Let X denote the random variable such that X = 0 for all ω . It is easy to verify that
n
p.
a.s.
Xn → X and Xn → X. Does the CDF of Xn converge to the CDF of X ? The CDF of Xn is
graphed in Figure 2.7. The CDF FXn (x) converges to 0 for x < 0 and to one for x > 0. However,
FX n F
X n even n odd n −1 0 01
n n Figure 2.7: CDF of Xn = (−1)n
n. FXn (0) alternates between 0 and 1 and hence does not converge to anything. In particular, it
doesn’t converge to FX (0). Thus, FXn (x) converges to FX (x) for all x except x = 0.
Recall that the distribution of a random variable X has probability mass
at some value xo ,
i.e. P {X = xo } = > 0, if and only if the CDF has a jump of size at xo : F (xo ) − F (xo −) = .
Example 2.1.10 illustrates the fact that if the limit random variable X has such a point mass, then
even if Xn is very close to X , the value FXn (x) need not converge. To overcome this phenomenon,
we adopt a deﬁnition of convergence in distribution which requires convergence of the CDFs only
at the continuity points of the limit CDF. Continuity points are deﬁned for general functions in
Appendix 11.3. Since CDFs are rightcontinuous and nondecreasing, a point x is a continuity point
of a CDF F if and only if there is no jump of F at X : i.e. if FX (x) = FX (x−).
Deﬁnition 2.1.11 A sequence (Xn : n ≥ 1) of random variables converges in distribution to a
random variable X if
lim FXn (x) = FX (x) at all continuity points x of FX . n→∞ d. Convergence in distribution is denoted by limn→∞ Xn = X d. or Xn → X.
One way to investigate convergence in distribution is through the use of characteristic functions.
Proposition 2.1.12 Let (Xn ) be a sequence of random variables and let X be a random variable.
Then the following are equivalent:
d. (i) Xn → X
(ii) E [f (Xn )] → E [f (X )] for any bounded continuous function f .
(iii) ΦXn (u) → ΦX (u) for each u ∈ R (i.e. pointwise convergence of characteristic functions)
The relationships among the four types of convergence discussed in this section are given in
the following proposition, and are pictured in Figure 2.8. The deﬁnitions use diﬀering amounts of
42 information about the random variables (Xn : n ≥ 1) and X involved. Convergence in the a.s. sense
involves joint properties of all the random variables. Convergence in the p. or m.s. sense involves
only pairwise joint distributions–namely those of (Xn , X ) for all n. Convergence in distribution
involves only the individual distributions of the random variables to have a convergence property.
Convergence in the a.s., m.s., and p. senses require the variables to all be deﬁned on the same
probability space. For convergence in distribution, the random variables need not be deﬁned on
the same probability space.
a.s. p. m.s. d. h
by
ed wit .)
t
nat le
mi ariab men
o
s d m v d mo
e i do on
c
uen ran sec
te
seq gle
(If a sin a fini Figure 2.8: Relationships among four types of convergence of random variables.
a.s. p. Proposition 2.1.13 (a) If Xn → X then Xn → X.
p.
m.s.
(b) If Xn → X then Xn → X.
(c) If P {Xn  ≤ Y } = 1 for all n for some ﬁxed random variable Y with E [Y 2 ] < ∞, and if
p.
m.s.
Xn → X , then Xn → X.
p.
d.
(d) If Xn → X then Xn → X .
(e) Suppose Xn → X in the p., m.s., or a.s. sense and Xn → Y in the p., m.s., or a.s. sense.
Then P {X = Y } = 1. That is, if diﬀerences on sets of probability zero are ignored, a sequence of
random variables can have only one limit (if p., m.s., and/or a.s. senses are used).
d. d. (f ) Suppose Xn → X and Xn → Y. Then X and Y have the same distribution.
a.s. Proof. (a) Suppose Xn → X and let > 0. Deﬁne a sequence of events An by An = {ω : Xn (ω ) − X (ω ) < }
We only need to show that P [An ] → 1. Deﬁne Bn by
Bn = {ω : Xk (ω ) − X (ω ) < for all k ≥ n} Note that Bn ⊂ An and B1 ⊂ B2 ⊂ · · · so limn→∞ P [Bn ] = P [B ] where B = ∞
n=1 Bn . Clearly B ⊃ {ω : lim Xn (ω ) = X (ω )}
n→∞ so 1 = P [B ] = limn→∞ P [Bn ]. Since P [An ] is squeezed between P [Bn ] and 1, limn→∞ P [An ] = 1,
p.
so Xn → X .
m.s.
(b) Suppose Xn → X and let > 0. By the Markov inequality applied to X − Xn 2 ,
P { X − Xn ≥ } ≤
43 E [ X − Xn 2 ]
2 (2.1) The right side of (2.1), and hence the left side of (2.1), converges to zero as n goes to inﬁnity.
p.
Therefore Xn → X as n → ∞.
p.
(c) Suppose Xn → X . Then for any > 0,
P { X ≥ Y + } ≤ P { X − Xn ≥ } → 0
so that P { X ≥ Y + } = 0 for every > 0. Thus, P { X ≤ Y } = 1, so that P { X − Xn 2 ≤
4Y 2 } = 1. Therefore, with probability one, for any > 0,
 X − Xn 2 ≤ 4Y 2 I{X −Xn ≥ } + 2 so
E [ X − Xn 2 ] ≤ 4E [Y 2 I{X −Xn ≥ } ] + 2 In the special case that P {Y = L} = 1 for a constant L, the term E [Y 2 I{X −Xn ≥ } ] is equal to
L2 P {X − Xn  ≥ }, and by the hypotheses, P {X − Xn  ≥ } → 0. Even if Y is random, since
E [Y 2 ] < ∞ and P {X − Xn  ≥ } → 0, it still follows that E [Y 2 I{X −Xn ≥ } ] → 0 as n → ∞, by
m.s.
Corollary 11.6.5. So, for n large enough, E [X − Xn 2 ] ≤ 2 2 . Since was arbitrary, Xn → X.
p.
(d) Assume Xn → X. Select any continuity point x of FX . It must be proved that limn→∞ FXn (x) =
FX (x). Let > 0. Then there exists δ > 0 so that FX (x) ≤ FX (x − δ ) + 2 . (See Figure 2.9.) Now
F (x)
X
F (x!!)
X x! x
! Figure 2.9: A CDF at a continuity point.
{X ≤ x − δ } = {X ≤ x − δ, Xn ≤ x} ∪ {X ≤ x − δ, Xn > x}
⊂ {Xn ≤ x} ∪ {X − Xn  ≥ δ }
so
FX (x − δ ) ≤ FXn (x) + P { Xn − X ≥ δ }.
For all n suﬃciently large, P { Xn − X ≥ δ } ≤ 2 . This and the choice of δ yield, for all n suﬃciently
large, FX (x) ≤ FXn (x) + . Similarly, for all n suﬃciently large, FX (x) ≥ FXN (x) − . So for all n
suﬃciently large, FXn (x) − FX (x) ≤ . Since was arbitrary, limn→∞ FXn (x) = FX (x).
p.
p.
(e) By parts (a) and (b), already proved, we can assume that Xn → X and Xn → Y. Let > 0
and δ > 0, and select N so large that P {Xn − X  ≥ } ≤ δ and P {Xn − Y  ≥ } ≤ δ for all n ≥ N .
By the triangle inequality, X − Y  ≤ XN − X  + XN − Y . Thus,
{X − Y  ≥ 2 } ⊂ {XN − X  ≥ } ∪ {YN − X  ≥ } so that
P {X − Y  ≥ 2 } ≤ P {XN − X  ≥ } + P {XN − Y  ≥ } ≤ 2δ . We’ve proved that
P {X − Y  ≥ 2 } ≤ 2δ . Since δ was arbitrary, it must be that P {X − Y  ≥ 2 } = 0. Since was
arbitrary, it must be that P {X − Y  = 0} = 1.
44 d. d. (f) Suppose Xn → X and Xn → Y. Then FX (x) = FY (y ) whenever x is a continuity point of
both x and y . Since FX and FY are nondecreasing and bounded, they can have only ﬁnitely many
discontinuities of size greater than 1/n for any n, so that the total number of discontinuities is at
most countably inﬁnite. Hence, in any nonempty interval, there is a point of continuity of both
functions. So for any x ∈ R, there is a strictly decreasing sequence of numbers converging to x,
such that xn is a point of continuity of both FX and FY . So FX (xn ) = FY (xn ) for all n. Taking
the limit as n → ∞ and using the rightcontinuitiy of CDFs, we have FX (x) = FY (x).
√
Example 2.1.14 Suppose X0 is a random variable with P {X0 ≥ 0} = 1. Suppose Xn = 6+ Xn−1
for n ≥ 1. For example, if for some ω it happens that X0 (ω ) = 12, then
√
X1 (ω ) = 6 + 12 = 9.465 . . .
√
X2 (ω ) = 6 + 9.46 = 9.076 . . .
√
X3 (ω ) = 6 + 9.076 = 9.0127 . . .
Examining Figure 2.10, it is clear that for any ω with X0 (ω ) > 0, the sequence of numbers Xn (ω )
a.s.
converges to 9. Therefore, Xn → 9 The rate of convergence can be bounded as follows. Note that
6+ x
3 y
6+ x 9
6 x=y
0 x
0 9 Figure 2.10: Graph of the functions 6 +
for each x ≥ 0,  6 + √ x−9 ≤ 6+ x
3  Xn (ω ) − 9  ≤  6 + √ x and 6 + x .
3 − 9 . Therefore,
Xn−1 (ω )
−9 =
3 1
 Xn−1 (ω ) − 9 
3 so that by induction on n,
 Xn (ω ) − 9  ≤ 3−n  X0 (ω ) − 9 
a.s. (2.2) p. Since Xn → 9 it follows that Xn → 9.
2
Finally, we investigate m.s. convergence under the assumption that E [X0 ] < +∞. By the
inequality (a + b)2 ≤ 2a2 + 2b2 , it follows that
2
E [(X0 − 9)2 ] ≤ 2(E [X0 ] + 81) Squaring and taking expectations on each side of (2.10) and using (2.3) thus yields
2
E [ Xn − 9 2 ] ≤ 2 · 3−2n {E [X0 ] + 81}
m.s. Therefore, Xn → 9.
45 (2.3) Example 2.1.15 Let W0 , W1 , . . . be independent, normal random variables with mean 0 and variance 1. Let X−1 = 0 and
Xn = (.9)Xn−1 + Wn n≥0
In what sense does Xn converge as n goes to inﬁnity? For ﬁxed ω , the sequence of numbers
X0 (ω ), X1 (ω ), . . . might appear as in Figure 2.11. Xk
k Figure 2.11: A typical sample sequence of X .
Intuitively speaking, Xn persistently moves. We claim that Xn does not converge in probability
(so also not in the a.s. or m.s. senses). Here is a proof of the claim. Examination of a table for the
normal distribution yields that P {Wn ≥ 2} = P {Wn ≤ −2} ≥ 0.02. Then
P { Xn − Xn−1 ≥ 2} ≥ P {Xn−1 ≥ 0, Wn ≤ −2} + P {Xn−1 < 0, Wn ≥ 2}
= P {Xn−1 ≥ 0}P {Wn ≤ −2} + P {Xn−1 < 0}P {Wn ≥ 2}
= P {Wn ≥ 2} ≥ 0.02
Therefore, for any random variable X ,
P { Xn − X ≥ 1} + P { Xn−1 − X ≥ 1} ≥ P { Xn − X ≥ 1 or  Xn−1 − X ≥ 1}
≥ P { Xn − Xn−1 ≥ 2} ≥ 0.02
so P { Xn − X ≥ 1} does not converge to zero as n → ∞. So Xn does not converge in probability
to any random variable X . The claim is proved.
Although Xn does not converge in probability, or in the a.s. or m.s.) senses, it nevertheless seems
to asymptotically settle into an equilibrium. To probe this point further, let’s ﬁnd the distribution
of Xn for each n.
X0 = W0 is N (0, 1)
X1 = (.9)X0 + W1 is N (0, 1.81)
X2 = (.9)X1 + W2 is N (0, (.81)(1.81 + 1))
2
2
2
2
2
In general, Xn is N (0, σn ) where the variances satisfy the recursion σn = (0.81)σn−1 +1 so σn → σ∞
2 = 1 = 5.263. Therefore, the CDF of X converges everywhere to the CDF of any
where σ∞
n
0.19
d. 2
random variable X which has the N (0, σ∞ ) distribution. So Xn → X for any such X . 46 The previous example involved convergence in distribution of Gaussian random variables. The
limit random variable was also Gaussian. In fact, we close this section by showing that limits of
Gaussian random variables are always Gaussian. Recall that X is a Gaussian random variable with
mean µ and variance σ 2 if either σ 2 > 0 and FX (c) = Φ( c−µ ) for all c, where Φ is the CDF of the
σ
standard N (0, 1) distribution, or σ 2 = 0, in which case FX (c) = I{c≥µ} and P {X = µ} = 1.
Proposition 2.1.16 Suppose Xn is a Gaussian random variable for each n, and that Xn → X∞
as n → ∞, in any one of the four senses, a.s., m.s., p., or d. Then X∞ is also a Gaussian random
variable.
Proof. Since convergence in the other senses implies convergence in distribution, we can assume
2
that the sequence converges in distribution. Let µn and σn denote the mean and variance of Xn .
2 is bounded. Intuitively, if it weren’t bounded, the
The ﬁrst step is to show that the sequence σn
distribution of Xn would get too spread out to converge. Since FX∞ is a valid CDF, there exists
2
a value L so large that FX∞ (−L) < 1 and FX∞ (L) > 3 . By increasing L if necessary, we can also
3
assume that L and −L are continuity points of FX . So there exists no such that, whenever n ≥ no ,
FXn (−L) ≤ 1 and FXn (L) ≥ 2 . Therefore, for n ≥ no , P {Xn  ≤ L} ≥ FXn ( 2 ) − FXn ( 1 ) ≥ 1 . For
3
3
3
3
3
2
σn ﬁxed, the probability P {Xn  ≤ L} is maximized by µn = 0, so no matter what the value of µn
L
L
2
is, 2Φ( σn ) − 1 ≥ P {Xn  ≤ L}. Therefore, for n ≥ no , Φ( σn ) ≥ 3 , or equivalently, σn ≤ L/Φ−1 ( 2 ),
3
−1 is the inverse of Φ. The ﬁrst n − 1 terms of the sequence (σ 2 ) are ﬁnite. Therefore, the
where Φ
o
n
2
whole sequence (σn ) is bounded.
Constant random variables are considered to be Gaussian random variables–namely degenerate
ones with zero variance. So assume without loss of generality that X∞ is not a constant random
variable. Then there exists a value co so that FX∞ (co ) is strictly between zero and one. Since FX∞
is rightcontinuous, the function must lie strictly between zero and one over some interval of positive
length, with left endpoint co . The function can only have countably many points of discontinuity,
so it has inﬁnitely many points of continuity such that the function value is strictly between zero
and one. Let c1 and c2 be two distinct such points, and let p1 and p2 denote the values of FX∞ at
µ
those two points, and let bi = Φ−1 (pi ) for i = 1, 2. It follows that limn→∞ ci −n n = bi for i = 1, 2.
σ
−
The limit of the diﬀerence of the sequences is the diﬀerence of the limits, so limn→∞ c1σnc2 = b1 − b2 .
Since c1 − c2 = 0 and the sequence (σn ) is bounded, it follows that (σn ) has a ﬁnite limit, σ∞ , and
therefore also (µn ) has a ﬁnite limit, µ∞ . Therefore, the CDFs FXn converge pointwise to the CDF
2
2
for the N (µ∞ , σ∞ ) distribution. Thus, X∞ has the N (µ∞ , σ∞ ) distribution. 2.2 Cauchy criteria for convergence of random variables It is important to be able to show that a limit exists even if the limit value is not known. For
example, it is useful to determine if the sum of an inﬁnite series of numbers is convergent without
needing to know the value of the sum. One useful result for this purpose is that if (xn : n ≥ 1)
is monotone nondecreasing, i.e. x1 ≤ x2 ≤ · · · , and if it satisﬁes xn ≤ L for all n for some
ﬁnite constant L, then the sequence is convergent. This result carries over immediately to random
variables: if (Xn : n ≥ 1) is a sequence of random variables such P {Xn ≤ Xn+1 } = 1 for all n and
if there is a random variable Y such that P {Xn ≤ Y } = 1 for all n, then (Xn ) converges a.s.
47 For deterministic sequences that are not monotone, the Cauchy criteria gives a simple yet general
condition that implies convergence to a ﬁnite limit. A deterministic sequence (xn : n ≥ 1) is said
to be a Cauchy sequence if limm,n→∞ xm − xn  = 0. This means that, for any > 0, there exists N
suﬃciently large, such that xm − xn  < for all m, n ≥ N . If the sequence (xn ) has a ﬁnite limit
x∞ , then the triangle inequality for distances between numbers, xm − xn  ≤ xm − x∞  + xn − x∞ ,
implies that the sequence is a Cauchy sequence. More useful is the converse statement, called the
Cauchy criteria for convergence, or the completeness property of R: If (xn ) is a Cauchy sequence
then (xn ) converges to a ﬁnite limit as n → ∞. The following proposition gives similar criteria for
convergence of random variables.
Proposition 2.2.1 (Cauchy criteria for random variables) Let (Xn ) be a sequence of random
variables on a probability space (Ω, F , P ).
(a) Xn converges a.s. to some random variable if and only if
P {ω : lim Xm (ω ) − Xn (ω ) = 0} = 1. m,n→∞ (b) Xn converges m.s. to some random variable if and only if (Xn ) is a Cauchy sequence in the
2
m.s. sense, meaning E [Xn ] < +∞ for all n and
lim E [(Xm − Xn )2 ] = 0. (2.4) m,n→∞ (c) Xn converges p. to some random variable if and only if for every > 0, lim P {Xm − Xn  ≥ } = 0. (2.5) m,n→∞ Proof. (a) For any ω ﬁxed, (Xn (ω ) : n ≥ 1) is a sequence of numbers. So by the Cauchy criterion
for convergence of a sequence of numbers, the following equality of sets holds:
{ω : lim Xn (ω ) exists and is ﬁnite} = {ω :
n→∞ lim Xm (ω ) − Xn (ω ) = 0}. m,n→∞ Thus, the set on the left has probability one (i.e. X converges a.s. to a random variable) if and
only if the set on the right has probability one. Part (a) is proved.
m.s.
(b) First the “only if” part is proved. Suppose Xn → X∞ . By the L2 triangle inequality for
random variables,
1 1 1 E [(Xn − Xm )2 ] 2 ≤ E [(Xm − X∞ )2 ] 2 + E [(Xn − X∞ )2 ] 2
m.s. (2.6) Since Xn → X∞ . the right side of (2.6) converges to zero as m, n → ∞, so that (2.4) holds. The
“only if” part of (b) is proved.
Moving to the proof of the “if” part, suppose (2.4) holds. Choose the sequence k1 < k2 < . . .
recursively as follows. Let k1 be so large that E [(Xn − Xk1 )2 ] ≤ 1/2 for all n ≥ k1 . Once k1 , . . . , ki−1
are selected, let ki be so large that ki > ki−1 and E [(Xn − Xki )2 ] ≤ 2−i for all n ≥ ki . It follows from
this choice of the ki ’s that E [(Xki+1 − Xki )2 ] ≤ 2−i for all i ≥ 1. Let Sn = Xk1  + n−1 Xki+1 − Xki .
i=1
48 Note that Xki  ≤ Sn for 1 ≤ i ≤ k by the triangle inequality for diﬀerences of real numbers. By
the L2 triangle inequality for random variables (1.16),
n−1
1 1 1 2
2
E [Sn ] 2 ≤ E [Xk1 ] 2 + 1 2
E [(Xki+1 − Xki )2 ] 2 ≤ E [Xk1 ] 2 + 1.
i=1 Since Sn is monotonically increasing, it converges a.s. to a limit S∞ . Note that Xki  ≤ S∞ for
2
2
21
all i ≥ 1. By the monotone convergence theorem, E [S∞ ] = limn→∞ E [Sn ] ≤ (E [Xk1 ] 2 + 1)2 . So,
S∞ is in L2 (Ω, F , P ). In particular, S∞ is ﬁnite a.s., and for any ω such that S∞ (ω ) is ﬁnite, the
sequence of numbers (Xki (ω ) : i ≥ 1) is a Cauchy sequence. (See Example 11.2.3 in the appendix.)
By completeness of R, for ω in that set, the limit X∞ (ω ) exists. Let X∞ (ω ) = 0 on the zero
probability event that (Xki (ω ) : i ≥ 1) does not converge. Summarizing, we have limi→∞ Xki = X∞
a.s. and Xki  ≤ S∞ where S∞ ∈ L2 (Ω, F , P ). It therefore follows from Proposition 2.1.13(c) that
m.s.
Xki → X∞ .
The ﬁnal step is to prove that the entire sequence (Xn ) converges in the m.s. sense to X∞ .
For this purpose, let > 0. Select i so large that E [(Xn − Xki )2 ] < 2 for all n ≥ ki , and
E [(Xki − X∞ )2 ] ≤ 2 . Then, by the L2 triangle inequality, for any n ≥ ki ,
1 1 1 E [(Xn − X∞ )2 ] 2 ≤ E (Xn − Xki )2 ] 2 + E [(Xki − X∞ )2 ] 2 ≤ 2
m.s. Since was arbitrary, Xn → X∞ . The proof of (b) is complete.
p.
(c) First the “only if” part is proved. Suppose Xn → X∞ . Then for any > 0, P {Xm − Xn  ≥ 2 } ≤ P {Xm − X∞  ≥ } + P {Xm − X∞  ≥ } → 0
as m, n → ∞, so that (2.5) holds. The “only if” part is proved.
Moving to the proof of the “if” part, suppose (2.5) holds. Select an increasing sequence of
integers ki so that P {Xn − Xm  ≥ 2−i } ≤ 2−i for all m, n ≥ ki . It follows, in particular, that
P {Xki+1 − Xki  ≥ 2−i } ≤ 2−i . Since the sum of the probabilities of these events is ﬁnite, the probability that inﬁnitely many of the events is true is zero, by the BorelCantelli lemma (speciﬁcally,
Lemma 1.2.2(a)). Thus, P {Xki+1 − Xki  ≤ 2−i for all large enough i} = 1. Thus, for all ω is a
set with probability one, (Xki (ω ) : i ≥ 1) is a Cauchy sequence of numbers. By completeness of
R, for ω in that set, the limit X∞ (ω ) exists. Let X∞ (ω ) = 0 on the zero probability event that
p.
a.s.
(Xki (ω ) : i ≥ 1) does not converge. Then, Xki → X∞ . It follows that Xki → X∞ as well.
The ﬁnal step is to prove that the entire sequence (Xn ) converges in the p. sense to X∞ . For
this purpose, let > 0. Select i so large that P {Xn − Xki  ≥ } < for all n ≥ ki , and
P {Xki − X∞  ≥ } < . Then P {Xn − X∞  ≥ 2 } ≤ 2 for all n ≥ ki . Since was arbitrary,
p.
Xn → X∞ . The proof of (c) is complete.
The following is a corollary of Proposition 2.2.1(c) and its proof.
p. Corollary 2.2.2 If Xn → X∞ , then there is a subsequence (Xki : i ≥ 1) such that limi→∞ Xki =
X∞ a.s.
Proof. By Proposition 2.2.1(c), the sequence satisﬁes (2.2.1). By the proof of Proposition 2.2.1(c)
there is a subsequence (Xki ) that converges a.s. By uniqueness of limits in the p. or a.s. senses, the
49 limit of the subsequence is the same random variable, X∞ (up to diﬀerences on a set of measure
zero).
Proposition 2.2.1(b), the Cauchy criteria for mean square convergence, is used extensively in
these notes. The remainder of this section concerns a more convenient form of the Cauchy criteria
for m.s. convergence.
Proposition 2.2.3 (Correlation version of the Cauchy criterion for m.s. convergence) Let (Xn )
2
be a sequence of random variables with E [Xn ] < +∞ for each n. Then there exists a random
m.s.
variable X such that Xn → X if and only if the limit limm,n→∞ E [Xn Xm ] exists and is ﬁnite.
m.s.
Furthermore, if Xn → X , then limm,n→∞ E [Xn Xm ] = E [X 2 ].
proof The “if” part is proved ﬁrst. Suppose limm,n→∞ E [Xn Xm ] = c for a ﬁnite constant c. Then
E (Xn − Xm )2 = 2
2
E [Xn ] − 2E [Xn Xm ] + E [Xm ] → c − 2c + c = 0 as m, n → ∞
m.s. Thus, Xn is Cauchy in the m.s. sense, so Xn → X for some random variable X .
m.s.
To prove the “only if” part, suppose Xn → X . Observe next that
E [Xm Xn ] = E [(X + (Xm − X ))(X + (Xn − X ))]
= E [X 2 + (Xm − X )X + X (Xn − X ) + (Xm − X )(Xn − X )]
By the CauchySchwarz inequality,
1 1 E [ (Xm − X )X ] ≤ E [(Xm − X )2 ] 2 E [X 2 ] 2 → 0
1 1 E [ (Xm − X )(Xn − X ) ] ≤ E [(Xm − X )2 ] 2 E [(Xn − X )2 ] 2 → 0
and similarly E [ X (Xn − X ) ] → 0. Thus E [Xm Xn ] → E [X 2 ]. This establishes both the “only if”
part of the proposition and the last statement of the proposition. The proof of the proposition is
complete.
2
m.s. m.s. Corollary 2.2.4 Suppose Xn → X and Yn → Y . Then E [Xn Yn ] → E [XY ].
m.s. Proof. By the inequality (a + b)2 ≤ 2a2 + 2b2 , it follows that Xn + Yn → X + Y as n → ∞.
2
Proposition 2.2.3 therefore implies that E [(Xn + Yn )2 ] → E [(X + Y )2 ], E [Xn ] → E [X 2 ], and
2 ] → E [Y 2 ]. Since X Y = ((X + Y )2 − X 2 − Y 2 )/2, the corollary follows.
E [Yn
nn
n
n
n
n m.s. Corollary 2.2.5 Suppose Xn → X. Then E [Xn ] → E [X ].
Proof. Corollary 2.2.5 follows from Corollary 2.2.4 by taking Yn = 1 for all n. 50 Example 2.2.6 This example illustrates the use of Proposition 2.2.3. Let X1 , X2 , . . . be mean
zero random variables such that
1 if i = j
0 else E [Xi Xj ] = Does the series ∞ Xk converge in the mean square sense to a random variable with a ﬁnite second
k=1 k
moment? Let Yn = n=1 Xk . The question is whether Yn converges in the mean square sense to
k
k
a random variable with ﬁnite second moment. The answer is yes if and only if limm,n→∞ E [Ym Yn ]
exists and is ﬁnite. Observe that
min(m,n) E [Ym Yn ] =
k=1
∞ →
k=1 1
k2 1
as m, n → ∞
k2 ∞ 1
This sum is smaller than 1 + 1 x2 dx = 2 < ∞.1 Therefore, by Proposition 2.2.3, the series
∞ Xk
k=1 k indeed converges in the m.s. sense. 2.3 Limit theorems for sequences of independent random variables Sums of many independent random variables often have distributions that can be characterized
by a small number of parameters. For engineering applications, this represents a low complexity
method for describing the random variables. An analogous tool is the Taylor series approximation.
A continuously diﬀerentiable function f can be approximated near zero by the ﬁrst order Taylor’s
approximation
f (x) ≈ f (0) + xf (0)
A second order approximation, in case f is twice continuously diﬀerentiable, is
f (x) ≈ f (0) + xf (0) + x2
f (0)
2 Bounds on the approximation error are given by Taylor’s theorem, found in Appendix 11.4. In
essence, Taylor’s approximation lets us represent the function by the numbers f (0), f (0) and
f (0). We shall see that the law of large numbers and central limit theorem can be viewed not just
as analogies of the ﬁrst and second order Taylor’s approximations, but actually as consequences of
them.
Lemma 2.3.1 If xn → x as n → ∞ then (1 +
1 In fact, the sum is equal to
is the main point here. π2
,
6 xn n
n) → ex as n → ∞. but the technique of comparing the sum to an integral to show the sum is ﬁnite 51 Proof. The basic idea is to note that (1 + s)n = exp(n ln(1 + s)), and apply Taylor’s theorem to
ln(1+ s) about the point s = 0. The details are given next. Since ln(1+ s)s=0 = 0, ln(1+ s) s=0 = 1,
1
and ln(1 + s) = − (1+s)2 , the mean value form of Taylor’s Theorem (see the appendix) yields that
2 s
if s > −1, then ln(1 + s) = s − 2(1+y)2 , where y lies in the closed interval with endpoints 0 and s.
Thus, if s ≥ 0, then y ≥ 0, so that s− s2
≤ ln(1 + s) ≤ s
2 if s ≥ 0. More to the point, if it is only known that s ≥ − 1 , then y ≥ − 1 , so that
2
2
s − 2s2 ≤ ln(1 + s) ≤ s
xn
n, Letting s = if s ≥ − 1
2 multiplying through by n, and applying the exponential function, yields that
exp(xn − 2x2
xn n
n
) ≤ (1 +
) ≤ exp(xn )
n
n if xn ≥ − n
2 If xn → x as n → ∞ then the condition xn > − n holds for all large enough n, and xn −
2
yielding the desired result. 2x2
n
n → x, A sequence of random variables (Xn ) is said to be independent and identically distributed (iid)
if the Xi ’s are mutually independent and identically distributed.
Proposition 2.3.2 (Law of large numbers) Suppose that X1 , X2 , . . . is a sequence of random variables such that each Xi has ﬁnite mean m. Let Sn = X1 + · · · + Xn . Then
(a) p. Sn m.s.
n→ d. m. (hence also Sn → m and Sn → m.) if for some constant c, Var(Xi ) ≤ c for all i,
n
n
and Cov(Xi , Xj ) = 0 i = j (i.e. if the variances are bounded and the Xi ’s are uncorrelated). (b) Sn p.
n→ m if X1 , X2 , . . . are iid. (This version is the weak law of large numbers.) (c) Sn a.s.
n→ m if X1 , X2 , . . . are iid. (This version is the strong law of large numbers.) We give a proof of (a) and (b), but prove (c) only under an extra condition. Suppose the conditions
of (a) are true. Then
E Sn
−m
n 2 Sn
n = Var
= 1
n2 = 1
Var(Sn )
n2 Cov(Xi , Xj ) =
i j m.s. 1
n2 Var(Xi ) ≤
i c
n Therefore Sn → m.
n
Turn next to part (b). If in addition to the conditions of (b) it is assumed that Var(X1 ) < +∞,
then the conditions of part (a) are true. Since mean square convergence implies convergence in
probability, the conclusion of part (b) follows. An extra credit problem shows how to use the same
approach to verify (b) even if Var(X1 ) = +∞.
52 Here a second approach to proving (b) is given. The characteristic function of
E exp j uXi
n u
= E exp j ( )Xi
n = ΦX Xi
n is given by u
n where ΦX denotes the characteristic function of X1 . Since the characteristic function of the sum
of independent random variables is the product of the characteristic functions,
Φ Sn (u) = ΦX n u
n n . Since E (X1 ) = m it follows that ΦX is diﬀerentiable with ΦX (0) = 1, ΦX (0) = jm and Φ is
continuous. By Taylor’s theorem, for any u ﬁxed,
ΦX u
n = 1+ uΦX (un )
n u
for some un between 0 and n for all n. Since Φ (un ) → jm as n → ∞, Lemma 2.3.1 yields
u
ΦX ( n )n → exp(jum) as n → ∞. Note that exp(jum) is the characteristic function of a random
variable equal to m with probability one. Since pointwise convergence of characteristic functions to
d.
a valid characteristic function implies convergence in distribution, it follows that Sn → m. However,
n
convergence in distribution to a constant implies convergence in probability, so (b) is proved.
4
Part (c) is proved under the additional assumption that E [X1 ] < +∞. Without loss of generality
4 . There are n terms of the form X 4 and 3n(n − 1)
we assume that EX1 = 0. Consider expanding Sn
i
2
terms of the form Xi2 Xj with 1 ≤ i, j ≤ n and i = j . The other terms have the form Xi3 Xj , Xi2 Xj Xk
or Xi Xj Xk Xl for distinct i, j, k, l, and these terms have mean zero. Thus,
4
4
2
E [Sn ] = nE [X1 ] + 3n(n − 1)E [X1 ]2 Let Y = ∞ ( Sn )4 . The value of Y is well deﬁned but it is a priori possible that Y (ω ) = +∞ for
n=1 n
some ω . However, by the monotone convergence theorem, the expectation of the sum of nonnegative
random variables is the sum of the expectations, so that
∞ E [Y ] = E
n=1 Sn
n ∞ 4 =
n=1 4
2
nE [X1 ] + 3n(n − 1)E [X1 ]2
< +∞
n4 Therefore, P {Y < +∞} = 1. However, {Y < +∞} is a subset of the event of convergence
(
{w : Snnw) → 0 as n → ∞}, so the event of convergence also has probability one. Thus, part (c)
under the extra fourth moment condition is proved.
Proposition 2.3.3 (Central Limit Theorem) Suppose that X1 , X2 , . . . are i.i.d., each with mean
µ and variance σ 2 . Let Sn = X1 + · · · + Xn . Then the normalized sum
Sn − nµ
√
n
converges in distribution to the N (0, σ 2 ) distribution as n → ∞.
53 Proof. Without loss of generality, assume that µ = 0. Then the characteristic function of
S
u
the normalized sum √n is given by ΦX ( √n )n , where ΦX denotes the characteristic function of X1 .
n
Since X1 has mean 0 and ﬁnite second moment σ 2 , it follows that ΦX is twice diﬀerentiable with
ΦX (0) = 1, ΦX (0) = 0, ΦX (0) = −σ 2 , and ΦX is continuous. By Taylor’s theorem, for any u ﬁxed,
ΦX
for some un between 0 and u
√
n u
√
n = 1+ u2
Φ (un )
2n for all n. Since Φ (un ) → −σ 2 as n → ∞, Lemma 2.3.1 yields u2 σ 2 u
ΦX ( √n )n → exp(− 2 ) as n → ∞. Since pointwise convergence of characteristic functions to a
valid characteristic function implies convergence in distribution, the proposition is proved. 2.4 Convex functions and Jensen’s inequality Let ϕ be a function on R with values in R ∪ {+∞} such that φ(x) < ∞ for at least one value of x.
Then ϕ is said to be convex if for any a, b and λ with a < b and 0 ≤ λ ≤ 1
ϕ(aλ + b(1 − λ)) ≤ λϕ(a) + (1 − λ)ϕ(b).
This means that the graph of ϕ on any interval [a, b] lies below the line segment equal to ϕ at the
endpoints of the interval.
Proposition 2.4.1 Suppose f is twice continuously diﬀerentiable on R. Then f is convex if and
only if f (v ) ≥ 0 for all v .
Proof. Suppose that f is twice continuously diﬀerentiable. Given a < b, deﬁne Dab = λf (a) +
(1 − λ)f (b) − f (λa + (1 − λ)b). For any x ≥ a,
x x x f (u)du = f (a) + (x − a)f (a) + f (x) = f (a) +
a f (v )dvdu
a v v = f (a) + (x − a)f (a) + (x − v )f (v )dv
a Applying this to Dab yields
b min{λ(v − a), (1 − λ)(b − v )}f (v )dv Dab = (2.7) a On one hand, if f (v ) ≥ 0 for all v , then Dab ≥ 0 whenever a < b, so that f is convex. On the
other hand, if f (vo ) < 0 for some vo , then by the assumed continuity of f , there is a small open
interval (a, b) containing vo so that f (v ) < 0 for a < v < b. But then Da,b < 0, implying that f is
not convex.
Examples of convex functions include:
ax2 + bx + c
e λx for constants
for λ constant,
54 a, b, c with a ≥ 0, ϕ(x) = − ln x x > 0
+∞ x ≤ 0, x ln x x > 0
0
x=0
ϕ(x) = +∞ x < 0.
Theorem 2.4.2 (Jensen’s inequality) Let ϕ be a convex function and let X be a random variable
such that E [X ] is ﬁnite. Then E [ϕ(X )] ≥ ϕ(E [X ]).
For example, Jensen’s inequality implies that E [X 2 ] ≥ E [X ]2 , which also follows from the fact
Var(X ) = E [X 2 ] − E [X ]2 .
Proof. Since ϕ is convex, there is a tangent to the graph of ϕ at E [X ], meaning there is a
function L of the form L(x) = a + bx such that ϕ(x) ≥ L(x) for all x and ϕ(E [X ]) = L(E [X ]).
See the illustration in Figure 2.12. Therefore E [ϕ(X )] ≥ E [L(X )] = L(E [X ]) = ϕ(E [X ]), which
establishes the theorem. φ(x)
L(x)
x
E[X] Figure 2.12: A convex function and a tangent linear function. A function ϕ is called concave if −ϕ is convex. If ϕ is concave then E [ϕ(X )] ≤ ϕ(E [X ]). 2.5 Chernoﬀ bound and large deviations theory Let X1 , X2 , . . . be an iid sequence of random variables with ﬁnite mean µ, and let Sn = X1 +· · ·+Xn .
The weak law of large numbers implies that for ﬁxed a with a > µ, P { Sn ≥ a} → 0 as n → ∞. In
n
case the Xi ’s have ﬁnite variance, the central limit theorem oﬀers a reﬁnement of the law of large
numbers, by identifying the limit of P { Sn ≥ an }, where (an ) is a sequence that converges to µ in
n
c
the particular manner: an = µ + √n . For ﬁxed c, the limit is not zero. One can think of the central
limit theorem, therefore, to concern “normal” deviations of Sn from its mean. Large deviations
theory, by contrast, addresses P { Sn ≥ a} for a ﬁxed, and in particular it identiﬁes how quickly
n
P { Sn ≥ a} converges to zero as n → ∞. We shall ﬁrst describe the Chernoﬀ bound, which is a
n
simple upper bound on P { Sn ≥ a}. Then Cram´r’s theorem, to the eﬀect that the Chernoﬀ bound
e
n
is in a certain sense tight, is stated.
The moment generating function of X1 is deﬁned by M (θ) = E [eθX1 ], and ln M (θ) is called the
log moment generating function. Since eθX1 is a positive random variable, the expectation, and
55 hence M (θ) itself, is welldeﬁned for all real values of θ, with possible value +∞. The Chernoﬀ
bound is simply given as
P Sn
≥a
n ≤ exp(−n[θa − ln M (θ)]) for θ ≥ 0 (2.8) The bound (2.8), like the Chebychev inequality, is a consequence of Markov’s inequality applied to
an appropriate function. For θ ≥ 0:
P Sn
≥a
n = P {eθ(X1 +···+Xn −na) ≥ 1}
≤ E [eθ(X1 +···+Xn −na) ]
= E [eθX1 ]n e−nθa = exp(−n[θa − ln M (θ)]) To make the best use of the Chernoﬀ bound we can optimize the bound by selecting the best θ.
Thus, we wish to select θ ≥ 0 to maximize aθ − ln M (θ).
In general the log moment generating function ln M is convex. Note that ln M (0) = 0. Let us
suppose that M (θ) is ﬁnite for some θ > 0. Then
E [X1 eθX1 ]
E [eθX1 ] d ln M (θ)
θ=0 =
dθ = E [X1 ]
θ=0 The sketch of a typical case is shown in Figure 2.13. Figure 2.13 also shows the line of slope a.
ln M(! ) l(a)
! a! Figure 2.13: A log moment generating function and a line of slope a.
Because of the assumption that a > E [X1 ], the line lies strictly above ln M (θ) for small enough θ
and below ln M (θ) for all θ < 0. Therefore, the maximum value of θa − ln M (θ) over θ ≥ 0 is equal
to l(a), deﬁned by
l(a) = max −∞<θ<∞ θa − ln M (θ) (2.9) Thus, the Chernoﬀ bound in its optimized form, is
P Sn
≥a
n ≤ exp(−nl(a)) a > E [X1 ] There does not exist such a clean lower bound on the large deviation probability P { Sn ≥ a},
n
but by the celebrated theorem of Cram´r stated next without proof, the Chernoﬀ bound gives the
e
right exponent.
56 Theorem 2.5.1 (Cram´r’s theorem) Suppose E [X1 ] is ﬁnite, and that E [X1 ] < a. Then for
e
there exists a number n such that
Sn
≥a
n P >0 ≥ exp(−n(l(a) + )) for all n ≥ n . Combining this bound with the Chernoﬀ inequality yields
lim n→∞ 1
ln P
n Sn
≥a
n = −l(a) In particular, if l(a) is ﬁnite (equivalently if P {X1 ≥ a} > 0) then
Sn
≥a
n P
where ( n ) is a sequence with n = exp(−n(l(a) + ≥ 0 and limn→∞ n n )) = 0. Similarly, if a < E [X1 ] and l(a) is ﬁnite, then
1
P
n
where n is a sequence with n Sn
≤a
n = exp(−n(l(a) + ≥ 0 and limn→∞
Sn
∈ da
n P n n )) = 0. Informally, we can write for n large: ≈ e−nl(a) da (2.10) Example 2.5.2 Let X1 , X2 , . . . be independent and exponentially distributed with parameter λ =
1. Then
∞ ln M (θ) = ln − ln(1 − θ) θ < 1
+∞
θ≥1 eθx e−x dx = 0 See Figure 2.14
+ "" + "" ln M(! ) l(a) a !
0 0 1 1 Figure 2.14: ln M (θ) and l(a) for an Exp(1) random variable.
Therefore, for any a ∈ R,
l(a) = max{aθ − ln M (θ)}
θ = max{aθ + ln(1 − θ)}
θ<1 57 If a ≤ 0 then l(a) = +∞. On the other hand, if a > 0 then setting the derivative of aθ + ln(1 − θ)
1
to 0 yields the maximizing value θ = 1 − a , and therefore
l(a) = a − 1 − ln(a) a > 0
+∞
a≤0 The function l is shown in Figure 2.14. Example 2.5.3 Let X1 , X2 , . . . be independent Bernoulli random variables with parameter p satisfying 0 < p < 1. Thus Sn has the binomial distribution. Then ln M (θ) = ln(peθ + (1 − p)), which
has asymptotic slope 1 as θ → +∞ and converges to a constant as θ → −∞. Therefore, l(a) = +∞
(1−p
if a > 1 or if a < 0. For 0 ≤ a ≤ 1, we ﬁnd aθ − ln M (θ) is maximized by θ = ln( a(1−a) ), leading to
p
)
l(a) = a ln( a ) + (1 − a) ln( 1−a ) 0 ≤ a ≤ 1
p
1−p
+∞
else See Figure 2.15.
+ !! + !! ln M(" ) l(a) a "
0 0 p 1 Figure 2.15: ln M (θ) and l(a) for a Bernoulli distribution. 2.6 Problems 2.1 Limits and inﬁnite sums for deterministic sequences
(a) Using the deﬁnition of a limit, show that limθ→0 θ(1 + cos(θ)) = 0.
(b) Using the deﬁnition of a limit, show that limθ→0,θ>0 1+cos(θ) = +∞.
θ
(c) Determine whether the following sum is ﬁnite, and justify your answer: √
∞ 1+ n
n=1 1+n2 . 2.2 The limit of the product is the product of the limits
Consider two (deterministic) sequences with ﬁnite limits: limn→∞ xn = x and limn→∞ yn = y .
(a) Prove that the sequence (yn ) is bounded.
(b) Prove that limn→∞ xn yn = xy . (Hint: Note that xn yn − xy = (xn − x)yn + x(yn − y ) and use
part (a)).
2.3 The reciprocal of the limit is the limit of the reciprocal
Using the deﬁnition of converence for deterministic sequences, prove that if (xn ) is a sequence with
a nonzero ﬁnite limit x∞ , then the sequence (1/xn ) converges to 1/x∞ .
58 2.4 Limits of some deterministic series
Determine which of the following series are convergent (i.e. have partial sums converging to a ﬁnite
limit). Justify your answers.
∞ (a)
n=0 3n
n! ∞ (b)
n=1 (n + 2) ln n
(n + 5)3 ∞ (c)
n=1 1
.
(ln(n + 1))5 2.5 On convergence of deterministic sequences and functions
2
(a) Let xn = 8n n+n for n ≥ 1. Prove that limn→∞ xn = 8 .
3
32
(b) Suppose fn is a function on some set D for each n ≥ 1, and suppose f is also a function on
D. Then fn is deﬁned to converge to f uniformly if for any > 0, there exists an n such that
fn (x) − f (x) ≤ for all x ∈ D whenever n ≥ n . A key point is that n does not depend on x.
Show that the functions fn (x) = xn on the semiopen interval [0, 1) do not converge uniformly to
the zero function.
(c) The supremum of a function f on D, written supD f , is the least upper bound of f . Equivalently,
supD f satisﬁes supD f ≥ f (x) for all x ∈ D, and given any c < supD f , there is an x ∈ D such
that f (x) ≥ c. Show that  supD f − supD g  ≤ supD f − g . Conclude that if fn converges to f
uniformly on D, then supD fn converges to supD f .
2.6 Convergence of sequences of random variables
Let Θ be uniformly distributed on the interval [0, 2π ]. In which of the four senses (a.s., m.s., p.,
d.) do each of the following two sequences converge. Identify the limits, if they exist, and justify
your answers.
(a) (Xn : n ≥ 1) deﬁned by Xn = cos(nΘ).
(b) (Yn : n ≥ 1) deﬁned by Yn = 1 − Θ n .
π
2.7 Convergence of random variables on (0,1]
Let Ω = (0, 1], let F be the Borel σ algebra of subsets of (0, 1], and let P be the probability measure
on F such that P ([a, b]) = b − a for 0 < a ≤ b ≤ 1. For the following two sequences of random
variables on (Ω, F , P ), ﬁnd and sketch the distribution function of Xn for typical n, and decide in
which sense(s) (if any) each of the two sequences converges.
(a) Xn (ω ) = nω − nω , where x is the largest integer less than or equal to x.
(b) Xn (ω ) = n2 ω if 0 < ω < 1/n, and Xn (ω ) = 0 otherwise.
2.8 Convergence of random variables on (0,1], version 2
Let Ω = (0, 1], let F be the Borel σ algebra of subsets of (0, 1], and let P be the probability measure
on F such that P ([a, b]) = b − a for 0 < a ≤ b ≤ 1. Determine in which of the four senses (a.s., p.,
m.s, d.), if any, each of the following three sequences of random variables converges. Justify your
answers.
−1)n
(a) Xn (ω ) = (n√ω .
(b) Xn (ω ) = nω n .
(c) Xn (ω ) = ω sin(2πnω ). (Try at least for a heuristic justiﬁcation.)
2.9 On the maximum of a random walk with negative drift
Let X1 , X2 , . . . be independent, identically distributed random variables with mean E [Xi ] = −1.
Let S0 = 0, and for n ≥ 1, let Sn = X1 + · · · + Xn . Let Z = max{Sn : n ≥ 0}.
59 (a) Show that Z is well deﬁned with probability one, and P {Z < +∞} = 1.
(b) Does there exist a ﬁnite constant L, depending only on the above assumptions, such that
E [Z ] ≤ L? Justify your answer. (Hint: Z ≥ max{S0 , S1 } = max{0, X1 }.)
2.10 Convergence of a sequence of discrete random variables
Let Xn = X +(1/n) where P [X = i] = 1/6 for i = 1, 2, 3, 4, 5 or 6, and let Fn denote the distribution
function of Xn .
(a) For what values of x does Fn (x) converge to F (x) as n tends to inﬁnity?
(b) At what values of x is FX (x) continuous?
(c) Does the sequence (Xn ) converge in distribution to X ?
2.11 Convergence in distribution to a nonrandom limit
Let (Xn , n ≥ 1) be a sequence of random variables and let X be a random variable such that
P [X = c] = 1 for some constant c. Prove that if limn→∞ Xn = X d., then limn→∞ Xn = X p. That
is, prove that convergence in distribution to a constant implies convergence in probability to the
same constant.
2.12 Convergence of a minimum
Let U1 , U2 , . . . be a sequence of independent random variables, with each variable being uniformly
distributed over the interval [0, 1], and let Xn = min{U1 , . . . , Un } for n ≥ 1.
(a) Determine in which of the senses (a.s., m.s., p., d.) the sequence (Xn ) converges as n → ∞,
and identify the limit, if any. Justify your answers.
(b) Determine the value of the constant θ so that the sequence (Yn ) deﬁned by Yn = nθ Xn converges
in distribution as n → ∞ to a nonzero limit, and identify the limit distribution.
2.13 Convergence of a product
Let U1 , U2 , . . . be a sequence of independent random variables, with each variable being uniformly
distributed over the interval [0, 2], and let Xn = U1 U2 · · · Un for n ≥ 1.
(a) Determine in which of the senses (a.s., m.s., p., d.) the sequence (Xn ) converges as n → ∞,
and identify the limit, if any. Justify your answers.
(b) Determine the value of the constant θ so that the sequence (Yn ) deﬁned by Yn = nθ ln(Xn )
converges in distribution as n → ∞ to a nonzero limit.
2.14 Limits of functions of random variables
Let g and h be functions deﬁned as follows: −1 if x ≤ −1
x
if − 1 ≤ x ≤ 1
g (x) = 1
if x ≥ 1 h(x) = −1 if x ≤ 0
1
if x > 0. Thus, g represents a clipper and h represents a hard limiter. Suppose that (Xn : n ≥ 0) is a
sequence of random variables, and that X is also a random variable, all on the same underlying
probability space. Give a yes or no answer to each of the four questions below. For each yes answer,
identify the limit and give a justiﬁcation. For each no answer, give a counterexample.
(a) If limn→∞ Xn = X a.s., then does limn→∞ g (Xn ) a.s. necessarily exist?
(b) If limn→∞ Xn = X m.s., then does limn→∞ g (Xn ) m.s. necessarily exist?
(c) If limn→∞ Xn = X a.s., then does limn→∞ h(Xn ) a.s. necessarily exist?
(d) If limn→∞ Xn = X m.s., then does limn→∞ h(Xn ) m.s. necessarily exist?
60 2.15 Sums of i.i.d. random variables, I
A gambler repeatedly plays the following game: She bets one dollar and then there are three
possible outcomes: she wins two dollars back with probability 0.4, she gets just the one dollar back
with probability 0.1, and otherwise she gets nothing back. Roughly what is the probability that
she is ahead after playing the game one hundred times?
2.16 Sums of i.i.d. random variables, II
Let X1 , X2 , . . . be independent random variable with P [Xi = 1] = P [Xi = −1] = 0.5.
(a) Compute the characteristic function of the following random variables: X1 , Sn = X1 + · · · + Xn ,
√
and Vn = Sn / n.
(b) Find the pointwise limits of the characteristic functions of Sn and Vn as n → ∞.
(c) In what sense(s), if any, do the sequences (Sn ) and (Vn ) converge?
2.17 Sums of i.i.d. random variables, III
Fix λ > 0. For each integer n > λ, let X1,n , X2,n , . . . , Xn,n be independent random variables such
that P [Xi,n = 1] = λ/n and P [Xi,n = 0] = 1 − (λ/n). Let Yn = X1,n + X2,n + · · · + Xn,n .
(a) Compute the characteristic function of Yn for each n.
(b) Find the pointwise limit of the characteristic functions as n → ∞ tends. The limit is the
characteristic function of what probability distribution?
(c) In what sense(s), if any, does the sequence (Yn ) converge?
2.18 On the growth of the maximum of n independent exponentials
Suppose that X1 , X2 , . . . are independent random variables, each with the exponential distribution
,...,X
with parameter λ = 1. For n ≥ 2, let Zn = max{X1n) n } .
ln(
(a) Find a simple expression for the CDF of Zn .
(b) Show that (Zn ) converges in distribution to a constant, and ﬁnd the constant. (Note: It follows
immediately that Zn converges in p. to the same constant. It can also be shown that (Zn ) converges
in the a.s. and m.s. senses to the same constant.)
2.19 Normal approximation for quantization error
Suppose each of 100 real numbers are rounded to the nearest integer and then added. Assume the
individual roundoﬀ errors are independent and uniformly distributed over the interval [−0.5, 0.5].
Using the normal approximation suggested by the central limit theorem, ﬁnd the approximate
probability that the absolute value of the sum of the errors is greater than 5.
2.20 Limit behavior of a stochastic dynamical system
Let W1 , W2 , . . . be a sequence of independent, N (0, 0.5) random variables. Let X0 = 0, and deﬁne
2
X1 , X2 , . . . recursively by Xk+1 = Xk + Wk . Determine in which of the senses (a.s., m.s., p., d.)
the sequence (Xn ) converges as n → ∞, and identify the limit, if any. Justify your answer.
2.21 Applications of Jensen’s inequality
Explain how each of the inequalties below follows from Jensen’s inequality. Speciﬁcally, identify
the convex function and random variable used.
1
(a) E [ X ] ≥ E [1 ] , for a positive random variable X with ﬁnite mean.
X
(b) E [X 4 ] ≥ E [X 2 ]2 , for a random variable X with ﬁnite second moment.
(c) D(f g ) ≥ 0, where f and g are positive probability densities on a set A, and D is the divergence
(x)
distance deﬁned by D(f g ) = A f (x) ln f (x) dx. (The base used in the logarithm is not relevant.)
g
61 2.22 Convergence analysis of successive averaging
Let U1 , U2 , ... be independent random variables, each uniformly distributed on the interval [0,1].
Let X0 = 0 and X1 = 1, and for n ≥ 1 let Xn+1 = (1 − Un )Xn + Un Xn−1 . Note that given Xn−1
and Xn , the variable Xn+1 is uniformly distributed on the interval with endpoints Xn−1 and Xn .
(a) Sketch a typical sample realization of the ﬁrst few variables in the sequence.
(b) Find E [Xn ] for all n.
(c) Show that Xn converges in the a.s. sense as n goes to inﬁnity. Explain your reasoning. (Hint:
Let Dn = Xn − Xn−1 . Then Dn+1 = Un Dn , and if m > n then Xm − Xn  ≤ Dn .)
2.23 Understanding the Markov inequality
Suppose X is a random variable with E [X 4 ] = 30.
(a) Derive an upper bound on P [X  ≥ 10]. Show your work.
(b) (Your bound in (a) must be the best possible in order to get both parts (a) and (b) correct).
Find a distribution for X such that the bound you found in part (a) holds with equality.
2.24 Mean square convergence of a random series
The sum of inﬁnitely many random variables, X1 + X2 + · · · is deﬁned as the limit as n tends to
inﬁnity of the partial sums X1 + X2 + · · · + Xn . The limit can be taken in the usual senses (in
probability, in distribution, etc.). Suppose that the Xi are mutually independent with mean zero.
Show that X1 + X2 + · · · exists in the mean square sense if and only if the sum of the variances,
Var(X1 ) + Var(X2 ) + · · · , is ﬁnite. (Hint: Apply the Cauchy criteria for mean square convergence.)
2.25 Portfolio allocation
Suppose that you are given one unit of money (for example, a million dollars). Each day you bet
a fraction α of it on a coin toss. If you win, you get double your money back, whereas if you lose,
you get half of your money back. Let Wn denote the wealth you have accumulated (or have left)
after n days. Identify in what sense(s) the limit limn→∞ Wn exists, and when it does, identify the
value of the limit
(a) for α = 0 (pure banking),
(b) for α = 1 (pure betting),
(c) for general α.
(d) What value of α maximizes the expected wealth, E [Wn ]? Would you recommend using that
value of α?
(e) What value of α maximizes the long term growth rate of Wn (Hint: Consider ln(Wn ) and apply
the LLN.)
2.26 A large deviation
Let X1 , X2 , ... be independent, N(0,1) random variables. Find the constant b such that
2
2
2
P {X1 + X2 + . . . + Xn ≥ 2n} = exp(−n(b + where n n )) → 0 as n → ∞. What is the numerical value of the approximation exp(−nb) if n = 100. 2.27 Sums of independent Cauchy random variables
Let X1 , X2 , . . . be independent, each with the standard Cauchy density function. The standard
1
Cauchy density and its characteristic function are given by f (x) = π(1+x2 ) and Φ(u) = exp(−u).
Let Sn = X1 + X2 + · · · + Xn .
62 (a) Find the characteristic function of Sn for a constant θ.
nθ
(b) Does Sn converge in distribution as n → ∞? Justify your answer, and if the answer is yes,
n
identify the limiting distribution.
(c) Does Sn converge in distribution as n → ∞? Justify your answer, and if the answer is yes,
n2
identify the limiting distribution.
S
(d) Does √n converge in distribution as n → ∞? Justify your answer, and if the answer is yes,
n
identify the limiting distribution.
2.28 A rapprochement between the central limit theorem and large deviations
Let X1 , X2 , . . . be independent, identically distributed random variables with mean zero, variance
σ 2 , and probability density function f . Suppose the moment generating function M (θ) is ﬁnite for
θ in an open interval I containing zero.
(a) Show that for θ ∈ I , (ln M (θ)) is the variance for the “tilted” density function fθ deﬁned by
fθ (x) = f (x) exp(θx − ln M (θ)). In particular, since (ln M (θ)) is nonnegative, ln M is a convex
function. (The interchange of expectation and diﬀerentiation with respect to θ can be justiﬁed for
θ ∈ I . You needn’t give details.)
Let b > 0 and let Sn = X1 + · · · + Xn for n any positive integer. By the central limit the√
orem, P [Sn ≥ b n] → Q(b/σ ) as n → ∞. An upper bound on the Q function is given by
2
2
2
∞
∞
s
1
Q(u) = u √1 π e−s /2 ds ≤ u u√2π e−s /2 ds = u√2π e−u /2 . This bound is a good approximation
2
2 2 σ
if u is moderately large. Thus, Q(b/σ ) ≈ b√2π e−b /2σ if b/σ is moderately large.
√
√
(b) The large deviations upper bound yields P [Sn ≥ b n] ≤ exp(−n (b/ n)). Identify the limit
of the large deviations upper bound as n → ∞, and compare with the approximation given by the
central limit theorem. (Hint: Approximate ln M near zero by its second order Taylor’s approximation.) 2.29 Chernoﬀ bound for Gaussian and Poisson random variables
(a) Let X have the N (µ, σ 2 ) distribution. Find the optimized Chernoﬀ bound on P {X ≥ E [X ] + c}
for c ≥ 0.
(b) Let Y have the P oi(λ) distribution. Find the optimized Chernoﬀ bound on P {Y ≥ E [Y ] + c}
for c ≥ 0.
(c) (The purpose of this problem is to highlight the similarity of the answers to parts (a) and (b).)
c2
c
Show that your answer to part (b) can be expressed as P {Y ≥ E [Y ]+ c} ≤ exp(− 2λ ψ ( λ )) for c ≥ 0,
where ψ (u) = 2g (1 + u)/u2 , with g (s) = s(ln s − 1) + 1. (Note: Y has variance λ, so the essential
diﬀerence between the normal and Poisson bounds is the ψ term. The function ψ is strictly positive
and strictly decreasing on the interval [−1, +∞), with ψ (−1) = 2 and ψ (0) = 1. Also, uψ (u) is
strictly increasing in u over the interval [−1, +∞). )
2.30 Large deviations of a mixed sum
Let X1 , X2 , . . . have the Exp(1) distribution, and Y1 , Y2 , . . . have the P oi(1) distribution. Suppose
all these random variables are mutually independent. Let 0 ≤ f ≤ 1, and suppose Sn = X1 + · · · +
1
Xnf + Y1 + · · · + Y(1−f )n . Deﬁne l(f, a) = limn→∞ n ln P { Sn ≥ a} for a > 1. Cram´rs theorem can
e
n
be extended to show that l(f, a) can be computed by replacing the probability P { Sn ≥ a} by its
n
optimized Chernoﬀ bound. (For example, if f = 1/2, we simply view Sn as the sum of the n i.i.d.
2
1
random variables, X1 + Y1 , . . . , X n + Y n .) Compute l(f, a) for f ∈ {0, 3 , 2 , 1} and a = 4.
3
2
2
63 2.31 Large deviation exponent for a mixture distribution
Problem 2.30 concerns an example such that 0 < f < 1 and Sn is the sum of n independent random
variables, such that a fraction f of the random variables have a CDF FY and a fraction 1 − f have
a CDF FZ . It is shown in the solutions that the large deviations exponent for Sn is given by:
n
l(a) = max {θa − f MY (θ) − (1 − f )MZ (θ)}
θ where MY (θ) and MZ (θ) are the log moment generating functions for FY and FZ respectively.
Consider the following variation. Let X1 , X2 , . . . , Xn be independent, and identically distributed,
each with CDF given by FX (c) = f FY (c) + (1 − f )FZ (c). Equivalently, each Xi can be generated
by ﬂipping a biased coin with probability of heads equal to f , and generating Xi using CDF FY
if heads shows and generating Xi with CDF FZ if tails shows. Let Sn = X1 + · · · + Xn , and let l
e
denote the large deviations exponent for Sn .
n
(a) Express the function l in terms of f , MY , and MZ .
(b) Determine which is true and give a proof: l(a) ≤ l(a) for all a, or l(a) ≥ l(a) for all a. Can you
also oﬀer an intuitive explanation?
2.32 The limit of a sum of cumulative products of a sequence of uniform random
variables
Let A1 , A2 , . . . be a sequence of independent random variables, with
1
1
P [Ai = 1] = P [Ai = 2 ] = 2 for all i. Let Bk = A1 · · · Ak .
(a) Does limk→∞ Bk exist in the m.s. sense? Justify your anwswer.
(b) Does limk→∞ Bk exist in the a.s. sense? Justify your anwswer.
(c) Let Sn = B1 + . . . + Bn . Show that limm,n→∞ E [Sm Sn ] = 35 , which implies that limn→∞ Sn
3
exists in the m.s. sense.
(d) Find the mean and variance of the limit random variable.
(e) Does limn→∞ Sn exist in the a.s. sense? Justify your anwswer.
2.33 * Distance measures (metrics) for random variables
For random variables X and Y , deﬁne
d1 (X, Y ) = E [ X − Y  /(1+  X − Y )]
d2 (X, Y ) = min{ ≥ 0 : FX (x + ) + ≥ FY (x) and FY (x + ) + ≥ FX (x) for all x}
d3 (X, Y ) = (E [(X − Y )2 ])1/2 ,
where in deﬁning d3 (X, Y ) it is assumed that E [X 2 ] and E [Y 2 ] are ﬁnite.
(a) Show that di is a metric for i = 1, 2 or 3. Clearly di (X, X ) = 0 and di (X, Y ) = di (Y, X ).
Verify in addition the triangle inequality. (The only other requirement of a metric is that
di (X, Y ) = 0 only if X = Y . For this to be true we must think of the metric as being deﬁned
on equivalence classes of random variables.)
(b) Let X1 , X2 , . . . be a sequence of random variables and let Y be a random variable. Show that
Xn converges to Y
(i) in probability if and only if d1 (X, Y ) converges to zero,
(ii) in distribution if and only if d2 (X, Y ) converges to zero,
64 (iii) in the mean square sense if and only if d3 (X, Y ) converges to zero (assume E [Y 2 ] < ∞).
(Hint for (i): It helps to establish that
d1 (X, Y ) − /(1 + ) ≤ P { X − Y ≥ } ≤ d1 (X, Y )(1 + )/ .
The “only if” part of (ii) is a little tricky. The metric d2 is called the Levy metric.
2.34 * Weak Law of Large Numbers
Let X1 , X2 , . . . be a sequence of random variables which are independent and identically distributed.
Assume that E [Xi ] exists and is equal to zero for all i. If Var(Xi ) is ﬁnite, then Chebychev’s
inequality easily establishes that (X1 + · · · + Xn )/n converges in probability to zero. Taking that
result as a starting point, show that the convergence still holds even if Var(Xi ) is inﬁnite. (Hint:
Use “truncation” by deﬁning Uk = Xk I { Xk ≥ c} and Vk = Xk I { Xk < c} for some constant c.
E [ Uk ] and E [Vk ] don’t depend on k and converge to zero as c tends to inﬁnity. You might also
ﬁnd the previous problem helpful. 65 66 Chapter 3 Random Vectors and Minimum Mean
Squared Error Estimation
The reader is encouraged to review the section on matrices in the appendix before reading this
chapter. 3.1 Basic deﬁnitions and properties A random vector X of dimension m has the form X= X1
X2
.
.
. Xm
where the Xi ’s are random variables all on the same probability space. The expectation of X (also
called the mean of X ) is the vector EX (or E [X ]) deﬁned by EX = EX1
EX2
.
.
. EXm
Suppose Y is another random vector on the same probability space as X , with dimension n. The
cross correlation matrix of X and Y is the m × n matrix E [XY T ], which has ij th entry E [Xi Yj ].
The cross covariance matrix of X and Y , denoted by Cov(X, Y ), is the matrix with ij th entry
Cov(Xi , Yj ). Note that the correlation matrix is the matrix of correlations, and the covariance
matrix is the matrix of covariances.
In the particular case that n = m and Y = X , the cross correlation matrix of X with itself, is
simply called the correlation matrix of X , and is written as E [XX T ], and it has ij th entry E [Xi Xj ].
The cross covariance matrix of X with itself, Cov(X, X ), has ij th entry Cov(Xi , Xj ). This matrix
is called the covariance matrix of X , and it is also denoted by Cov(X ). So the notations Cov(X )
and Cov(X, X ) are interchangeable. While the notation Cov(X ) is more concise, the notation
67 Cov(X, X ) is more suggestive of the way the covariance matrix scales when X is multiplied by a
constant.
Elementary properties of expectation, correlation, and covariance for vectors follow immediately
from similar properties for ordinary scalar random variables. These properties include the following
(here A and C are nonrandom matrices and b and d are nonrandom vectors).
1. E [AX + b] = AE [X ] + b
2. Cov(X, Y ) = E [X (Y − EY )T ] = E [(X − EX )Y T ] = E [XY T ] − (EX )(EY )T
3. E [(AX )(CY )T ] = AE [XY T ]C T
4. Cov(AX + b, CY + d) = ACov(X, Y )C T
5. Cov(AX + b) = ACov(X )AT
6. Cov(W + X, Y + Z ) = Cov(W, Y ) + Cov(W, Z ) + Cov(X, Y ) + Cov(X, Z )
In particular, the second property above shows the close connection between correlation matrices
and covariance matrices. In particular, if the mean vector of either X or Y is zero, then the cross
correlation and cross covariance matrices are equal.
Not every square matrix is a correlation matrix. For example, the diagonal elements must be
nonnegative. Also, Schwarz’s inequality (see Section 1.8) must be respected, so that Cov(Xi , Xj ) ≤
Cov(Xi , Xi )Cov(Xj , Xj ). Additional inequalities arise for consideration of three or more random
variables at a time. Of course a square diagonal matrix is a correlation matrix if and only if its
diagonal entries are nonnegative, because only vectors with independent entries need be considered.
But if an m × m matrix is not diagonal, it is not a priori clear whether there are m random variables
with all m(m + 1)/2 correlations matching the entries of the matrix. The following proposition
neatly resolves these issues.
Proposition 3.1.1 Correlation matrices and covariance matrices are positive semideﬁnite. Conversely, if K is a positive semideﬁnite matrix, then K is the covariance matrix and correlation
matrix for some mean zero random vector X .
Proof. If K is a correlation matrix, then K = E [XX T ] for some random vector X . Given any
vector α, αT X is a scaler random variable, so
αT Kα = E [αT XX T α] = E [(αT X )(X T α)] = E [(αT X )2 ] ≥ 0.
Similarly, if K = Cov(X, X ) then for any vector α,
αT Kα = αT Cov(X, X )α = Cov(αT X, αT X ) = Var(αT X ) ≥ 0.
The ﬁrst part of the proposition is proved.
For the converse part, suppose that K is an arbitrary symmetric positive semideﬁnite matrix.
Let λ1 , . . . , λm and U be the corresponding set of eigenvalues and orthonormal matrix formed by
the eigenvectors. (See Section 11.7 in the appendix.) Let Y1 , . . . , Ym be independent, mean 0
random variables with Var(Yi ) = λi , and let Y be the random vector Y = (Y1 , . . . , Ym )T . Then
68 Cov(Y, Y ) = Λ, where Λ is the diagonal matrix with the λi ’s on the diagonal. Let X = U Y . Then
EX = 0 and
Cov(X, X ) = Cov(U Y, U Y ) = U ΛU T = K. Therefore, K is both the covariance matrix and the correlation matrix of X .
The characteristic function ΦX of X is the function on Rm deﬁned by
ΦX (u) = E [exp(juT X )]. 3.2 The orthogonality principle for minimum mean square error
estimation Let X be a random variable with some known distribution. Suppose X is not observed but that
we wish to estimate X . If we use a constant b to estimate X , the estimation error will be X − b.
The mean square error (MSE) is E [(X − b)2 ]. Since E [X − EX ] = 0 and EX − b is constant,
E [(X − b)2 ] = E [((X − EX ) + (EX − b))2 ]
= E [(X − EX )2 + 2(X − EX )(EX − b) + (EX − b)2 ]
= Var(X ) + (EX − b)2 .
From this expression it is easy to see that the mean square error is minimized with respect to b if
and only if b = EX . The minimum possible value is Var(X ).
Random variables X and Y are called orthogonal if E [XY ] = 0. Orthogonality is denoted by
“X ⊥ Y .”
The essential fact E [X − EX ] = 0 is equivalent to the following condition: X − EX is orthogonal
to constants: (X − EX ) ⊥ c for any constant c. Therefore, the choice of constant b yielding the
minimum mean square error is the one that makes the error X − b orthogonal to all constants. This
result is generalized by the orthogonality principle, stated next.
Fix some probability space and let L2 (Ω, F , P ) be the set of all random variables on the probability space with ﬁnite second moments. Let X be a random variable in L2 (Ω, F , P ), and let V be
a collection of random variables on the same probability space as X such that
V.1 V ⊂ L2 (Ω, F , P )
V.2 V is a linear class: If Z1 ∈ V and Z2 ∈ V and a1 , a2 are constants, then a1 Z1 + a2 Z2 ∈ V
V.3 V is closed in the mean square sense: If Z1 , Z2 , . . . is a sequence of elements of V and if
Zn → Z∞ m.s. for some random variable Z∞ , then Z∞ ∈ V .
That is, V is a closed linear subspace of L2 (Ω, F , P ). The problem of interest is to ﬁnd Z ∗ in V to
minimize the mean square error, E [(X − Z )2 ], over all Z ∈ V . That is, Z ∗ is the random variable in
V that is closest to X in the minimum mean square error (MMSE) sense. We call it the projection
of X onto V and denote it as ΠV (X ).
Estimating a random variable by a constant corresponds to the case that V is the set of constant
random variables: the projection of a random variable X onto the set of constant random variables
is EX . The orthogonality principle stated next is illustrated in Figure 3.1.
69 X
Z e
Z*
0 V Figure 3.1: Illustration of the orthogonality principle. Theorem 3.2.1 (The orthogonality principle) Let V be a closed, linear subspace of L2 (Ω, F , P ),
and let X ∈ L2 (Ω, F , P ), for some probability space (Ω, F , P ).
(a) (Existence and uniqueness) There exists a unique element Z ∗ (also denoted by ΠV (X )) in V
so that E [(X − Z ∗ )2 ] ≤ E [(X − Z )2 ] for all Z ∈ V . (Here, we consider two elements Z and
Z of V to be the same if P {Z = Z } = 1).
(b) (Characterization) Let W be a random variable. Then W = Z ∗ if and only if the following
two conditions hold:
(i) W ∈ V
(ii) (X − W ) ⊥ Z for all Z in V .
(c)(Error expression) The minimum mean square error (MMSE) is given by
E [(X − Z ∗ )2 ] = E [X 2 ] − E [(Z ∗ )2 ].
Proof. The proof of part (a) is given in an extra credit homework problem. The technical
condition V.3 on V is essential for the proof of existence. Here parts (b) and (c) are proved.
To establish the “if” half of part (b), suppose W satisﬁes (i) and (ii) and let Z be an arbitrary
element of V . Then W − Z ∈ V because V is a linear class. Therefore, (X − W ) ⊥ (W − Z ), which
implies that
E [(X − Z )2 ] = E [(X − W + W − Z )2 ]
= E [(X − W )2 + 2(X − W )(W − Z ) + (W − Z )2 ]
= E [(X − W )2 ] + E [(W − Z )2 ].
Thus E [(X − W )2 ] ≤ E [(X − Z )2 ]. Since Z is an arbitrary element of V , it follows that W = Z ∗ ,
and the “if” half of (b) is proved.
To establish the “only if” half of part (b), note that Z ∗ ∈ V by the deﬁnition of Z ∗ . Let Z ∈ V
and let c ∈ R. Then Z ∗ + cZ ∈ V , so that E [(X − (Z ∗ + cZ ))2 ] ≥ E [(X − Z ∗ )2 ]. But
E [(X − (Z ∗ + cZ ))2 ] = E [(X − Z ∗ ) − cZ )2 ] = E [(X − Z ∗ )2 ] − 2cE [(X − Z ∗ )Z ] + c2 E [Z 2 ],
70 so that
−2cE [(X − Z ∗ )Z ] + c2 E [Z 2 ] ≥ 0. (3.1) As a function of c the left side of (3.1) is a parabola with value zero at c = 0. Hence its derivative
with respect to c at 0 must be zero, which yields that (X − Z ∗ ) ⊥ Z . The “only if” half of (b) is
proved.
The expression of part (c) is proved as follows. Since X − Z ∗ is orthogonal to all elements of
V , including Z ∗ itself,
E [X 2 ] = E [((X − Z ∗ ) + Z ∗ )2 ] = E [(X − Z ∗ )2 ] + E [(Z ∗ )2 ].
This proves part (c).
The following propositions give some properties of the projection mapping ΠV , with proofs
based on the orthogonality principle.
Proposition 3.2.2 (Linearity of projection) Suppose V is a closed linear subspace of L2 (Ω, F , P ),
X1 and X2 are in L2 (Ω, F , P ), and a1 and a2 are constants. Then
ΠV (a1 X1 + a2 X2 ) = a1 ΠV (X1 ) + a2 ΠV (X2 ). (3.2) Proof. By the characterization part of the orthogonality principle (part (b) of Theorem 3.2.1),
the projection ΠV (a1 X1 + a2 X2 ) is characterized by two properties. So, to prove (3.2), it suﬃces
to show that a1 ΠV1 (X1 ) + a2 ΠV2 (X2 ) satisﬁes these two properties. First, we must check that
a1 ΠV1 (X1 ) + a2 ΠV2 (X2 ) ∈ V . This follows immediately from the fact that ΠV (Xi ) ∈ V , for i = 1, 2,
and V is a linear subspace, so the ﬁrst property is checked. Second, we must check that e ⊥ Z , where
e = a1 X1 + a2 X2 − (a1 ΠV (X1 )+ a2 ΠV (X2 )), and Z is an arbitrary element of V . Now e = a1 e1 + a2 e2 ,
where ei = Xi − ΠV (Xi ) for i = 1, 2, and ei ⊥ Z for i = 1, 2. So E [eZ ] = a1 E [e1 Z ] + a2 E [e2 Z ] = 0,
or equivalently, e ⊥ Z. Thus, the second property is also checked, and the proof is complete. Proposition 3.2.3 (Projections onto nested subspaces) Suppose V1 and V2 are closed linear subspaces of L2 (Ω, F , P ) such that V2 ⊂ V1 . Then for any X ∈ L2 (Ω, F , P ), ΠV2 (X ) = ΠV2 ΠV1 (X ). (In
words, the projection of X onto V2 can be found by ﬁrst projecting X onto V1 , and then projecting
the result onto V2 .) Furthermore,
E [(X − ΠV2 (X ))2 ] = E [(X − ΠV1 (X ))2 ] + E [(ΠV1 (X ) − ΠV2 (X ))2 ]. (3.3) In particular, E [(X − ΠV2 (X ))2 ] ≥ E [(X − ΠV1 (X ))2 ].
Proof. By the characterization part of the orthogonality principle (part (b) of Theorem 3.2.1), the
projection ΠV2 (X ) is characterized by two properties. So, to prove ΠV2 (X ) = ΠV2 ΠV1 (X ), it suﬃces
to show that ΠV2 ΠV1 (X ) satisﬁes the two properties. First, we must check that ΠV2 ΠV1 (X ) ∈ V2 .
This follows immediately from the fact that ΠV2 (X ) maps into V2 , so the ﬁrst property is checked.
Second, we must check that e ⊥ Z , where e = X − ΠV2 ΠV1 (X ), and Z is an arbitrary element of V2 .
Now e = e1 + e2 , where e1 = X − ΠV1 (X ) and e2 = ΠV1 (X ) − ΠV2 ΠV1 (X ). By the characterization
of ΠV1 (X ), e1 is perpendicular to any random variable in V1 . In particular, e1 ⊥ Z , because
Z ∈ V2 ⊂ V1 . The characterization of the projection of ΠV1 (X ) onto V2 implies that e2 ⊥ Z . Since
71 ei ⊥ Z for i = 1, 2, it follows that e ⊥ Z . Thus, the second property is also checked, so it is proved
that ΠV2 (X ) = ΠV2 ΠV1 (X ).
As mentioned above, e1 is perpendicular to any random variable in V1 , which implies that
e1 ⊥ e2 . Thus, E [e2 ] = E [e2 ] + E [e2 ], which is equivalent to (3.3). Therefore, (3.3) is proved. The
1
2
last inequality of the proposition follows, of course, from (3.3). The inequality is also equivalent to
the inequality minW ∈V2 E [(X − W )2 ] ≥ minW ∈V1 E [(X − W )2 ], and this inequality is true because
the minimum of a set of numbers cannot increase if more numbers are added to the set.
The following proposition is closely related to the use of linear innovations sequences, discussed
in Sections 3.5 and 3.6.
Proposition 3.2.4 (Projection onto the span of orthogonal subspaces) Suppose V1 and V2 are
closed linear subspaces of L2 (Ω, F , P ) such that V1 ⊥ V2 , which means that E [Z1 Z2 ] = 0 for any
Z1 ∈ V1 and Z2 ∈ V2 . Let V = V1 ⊕ V2 = {Z1 + Z2 : Zi ∈ Vi } denote the span of V1 and V2 . Then
for any X ∈ L2 (Ω, F , P ), ΠV (X ) = ΠV1 (X ) + ΠV2 (X ). The minimum mean square error satisﬁes
E [(X − ΠV (X ))2 ] = E [X 2 ] − E [(ΠV1 (X ))2 ] − E [(ΠV2 (X ))2 ].
Proof. The space V is also a closed linear subspace of L2 (Ω, F , P ) (see a starred homework
problem). By the characterization part of the orthogonality principle (part (b) of Theorem 3.2.1),
the projection ΠV (X ) is characterized by two properties. So to prove ΠV (X ) = ΠV1 (X ) + ΠV2 (X ),
it suﬃces to show that ΠV1 (X ) + ΠV2 (X ) satisﬁes these two properties. First, we must check that
ΠV1 (X )+ΠV2 (X ) ∈ V . This follows immediately from the fact that ΠVi (X ) ∈ Vi , for i = 1, 2, so the
ﬁrst property is checked. Second, we must check that e ⊥ Z , where e = X − (ΠV1 (X ) + ΠV2 (X )),
and Z is an arbitrary element of V . Now any such Z can be written as Z = Z1 + Z2 where Zi ∈ Vi
for i = 1, 2. Observe that ΠV2 (X ) ⊥ Z1 because ΠV2 (X ) ∈ V2 and Z1 ∈ V1 . Therefore,
E [eZ1 ] = E [(X − (ΠV1 (X ) + ΠV2 (X ))Z1 ]
= E [(X − ΠV1 (X ))Z1 ] = 0,
where the last equality follows from the characterization of ΠV1 (X ). Thus, e ⊥ Z1 , and similarly
e ⊥ Z2 , so e ⊥ Z. Thus, the second property is also checked, so ΠV (X ) = ΠV1 (X ) + ΠV2 (X ) is
proved.
Since ΠVi (X ) ∈ Vi for i = 1, 2, ΠV1 (X ) ⊥ ΠV2 (X ). Therefore, E [(ΠV (X ))2 ] = E [(ΠV1 (X ))2 ] +
E [(ΠV2 (X ))2 ], and the expression for the MMSE in the proposition follows from the error expression
in the orthogonality principle. 3.3 Conditional expectation and linear estimators In many applications, a random variable X is to be estimated based on observation of a random
variable Y . Thus, an estimator is a function of Y . In applications, the two most frequently
considered classes of functions of Y used in this context are essentially all functions, leading to the
best unconstrained estimator, or all linear functions, leading to the best linear estimator. These
two possibilities are discussed in this section.
72 3.3.1 Conditional expectation as a projection Suppose a random variable X is to be estimated using an observed random vector Y of dimension
m. Suppose E [X 2 ] < +∞. Consider the most general class of estimators based on Y , by setting
V = {g (Y ) : g : Rm → R, E [g (Y )2 ] < +∞}. (3.4) There is also the implicit condition that g is Borel measurable so that g (Y ) is a random variable.
The projection of X onto this class V is the unconstrained minimum mean square error (MMSE)
estimator of X given Y .
Let us ﬁrst proceed to identify the optimal estimator by conditioning on the value of Y , thereby
reducing this example to the estimation of a random variable by a constant, as discussed at the
beginning of Section 3.2. For technical reasons we assume for now that X and Y have a joint pdf.
Then, conditioning on Y ,
E [(X − g (Y ))2 ] = E [(X − g (Y ))2 Y = y ]fY (y )dy
Rm where
∞ E [(X − g (Y ))2 Y = y ] = (x − g (y ))2 fX Y (x  y )dx
−∞ Since the mean is the MMSE estimator of a random variable among all constants, for each ﬁxed y ,
the minimizing choice for g (y ) is
g ∗ (y ) = E [X Y = y ] = ∞ xfX Y (x  y )dx. (3.5) −∞ Therefore, the optimal estimator in V is g ∗ (Y ) which, by deﬁnition, is equal to the random variable
E [X Y ].
What does the orthogonality principle imply for this example? It implies that there exists an
optimal estimator g ∗ (Y ) which is the unique element of V such that
(X − g ∗ (Y )) ⊥ g (Y )
for all g (Y ) ∈ V . If X, Y have a joint pdf then we can check that E [X  Y ] satisﬁes the required
condition. Indeed,
E [(X − E [X  Y ])g (Y )] =
= (x − E [X  Y = y ])g (y )fX Y (x  y )fY (y )dxdy
(x − E [X  Y = y ])fX Y (x  y )dx g (y )fY (y )dy = 0,
because the expression within the braces is zero.
In summary, if X and Y have a joint pdf (and similarly if they have a joint pmf) then the
MMSE estimator of X given Y is E [X  Y ]. Even if X and Y don’t have a joint pdf or joint pmf,
we deﬁne the conditional expectation E [X  Y ] to be the MMSE estimator of X given Y. By the
73 orthogonality principle E [X  Y ] exists as long as E [X 2 ] < ∞, and it is the unique function of Y
such that
E [(X − E [X  Y ])g (Y )] = 0
for all g (Y ) in V .
Estimation of a random variable has been discussed, but often we wish to estimate a random
vector. A beauty of the MSE criteria is that it easily extends to estimation of random vectors,
because the MSE for estimation of a random vector is the sum of the MSEs of the coordinates:
m E [ X − g (Y ) 2 E [(Xi − gi (Y ))2 ]
=
i=1 Therefore, for most sets of estimators V typically encountered, ﬁnding the MMSE estimator of a
random vector X decomposes into ﬁnding the MMSE estimators of the coordinates of X separately.
Suppose a random vector X is to be estimated using estimators of the form g(Y), where here g
maps Rn into Rm . Assume E [ X 2 ] < +∞ and seek an estimator to minimize the MSE. As seen
above, the MMSE estimator for each coordinate Xi is E [Xi Y ], which is also the projection of Xi
onto the set of unconstrained estimators based on Y , deﬁned in (3.4). So the optimal estimator
g ∗ (Y ) of the entire vector X is given by E [X1  Y ] E [X2  Y ] g ∗ (Y ) = E [X  Y ] = .
. .
E [Xm  Y ]
Let the estimation error be denoted by e, e = X − E [X  Y ]. (Even though e is a random vector
we use lower case for it for an obvious reason.)
The mean of the error is given by Ee = 0. As for the covariance of the error, note that
E [Xj  Y ] is in V for each j , so ei ⊥ E [Xj  Y ] for each i, j . Since Eei = 0, it follows that
Cov(ei , E [Xj  Y ]) = 0 for all i, j . Equivalently, Cov(e, E [X  Y ]) = 0. Using this and the fact
X = E [X  Y ] + e yields
Cov(X ) = Cov(E [X  Y ] + e)
= Cov(E [X  Y ]) + Cov(e) + Cov(E [X Y ], e) + Cov(e, E [X Y ])
= Cov(E [X  Y ]) + Cov(e)
Thus, Cov(e) = Cov(X ) − Cov(E [X  Y ]).
In practice, computation of E [X  Y ] (for example, using (3.5) in case a joint pdf exists) may
be too complex or may require more information about the joint distribution of X and Y than
is available. For both of these reasons, it is worthwhile to consider classes of estimators that are
constrained to smaller sets of functions of the observations. A widely used set is the set of all linear
functions, leading to linear estimators, described next. 3.3.2 Linear estimators Let X and Y be random vectors with E [ X 2 ] < +∞ and E [ Y 2 ] < +∞. Seek estimators of
the form AY + b to minimize the MSE. Such estimators are called linear estimators because each
74 coordinate of AY + b is a linear combination of Y1 , Y2 , . . . , Ym and 1. Here “1” stands for the
random variable that is always equal to 1.
To identify the optimal linear estimator we shall apply the orthogonality principle for each
coordinate of X with
V = {c0 + c1 Y1 + c2 Y2 + . . . + cn Yn : c0 , c1 , . . . , cn ∈ R}
Let e denote the estimation error e = X − (AY + b). We must select A and b so that ei ⊥ Z for all
Z ∈ V . Equivalently, we must select A and b so that
ei ⊥ 1
e i ⊥ Yj all i
all i, j. The condition ei ⊥ 1, which means Eei = 0, implies that E [ei Yj ] = Cov(ei , Yj ). Thus, the
required orthogonality conditions on A and b become Ee = 0 and Cov(e, Y ) = 0. The condition
Ee = 0 requires that b = EX − AEY , so we can restrict our attention to estimators of the
form EX + A(Y − EY ), so that e = X − EX − A(Y − EY ). The condition Cov(e, Y ) = 0
becomes Cov(X, Y ) − ACov(Y, Y ) = 0. If Cov(Y, Y ) is not singular, then A must be given by
A = Cov(X, Y )Cov(Y, Y )−1 . In this case the optimal linear estimator, denoted by E [X  Y ], is
given by
E [X  Y ] = E [X ] + Cov(X, Y )Cov(Y, Y )−1 (Y − EY ) (3.6) Proceeding as in the case of unconstrained estimators of a random vector, we ﬁnd that the covariance
of the error vector satisﬁes
Cov(e) = Cov(X ) − Cov(E [X  Y ])
which by (3.6) yields
Cov(e) = Cov(X ) − Cov(X, Y )Cov(Y, Y )−1 Cov(Y, X ). 3.3.3 (3.7) Discussion of the estimators As seen above, the expectation E [X ], the MMSE linear estimator E [X Y , and the conditional
expectation E [X Y ], are all instances of projection mappings ΠV , for V consisting of constants,
linear estimators based on Y , or unconstrained estimators based on Y , respectively. Hence, the
orthogonality principle, and Propositions 3.2.23.2.4 all apply to these estimators.
Proposition 3.2.2 implies that these estimators are linear functions of X . In particular,
E [a1 X1 + a2 X2 Y ] = a1 E [X1 Y ] + a2 E [X2 Y ], and the same is true with “E ” replaced by “E .”
Proposition 3.2.3, regarding projections onto nested subspaces, implies an ordering of the mean
square errors:
E [(X − E [X  Y ])2 ] ≤ E [(X − E [X  Y ])2 ] ≤ Var(X ).
Furthermore, it implies that the best linear estimator of X based on Y is equal to the best linear
estimator of the estimator E [X Y ]: that is, E [X Y ] = E [E [X Y ]Y ]. It follows, in particular, that
E [X Y ] = E [X Y ] if and only if E [X Y ] has the linear form, AX + b. Similarly, E [X ], the best
constant estimator of X , is also the best constant estimator of E [X Y ] or of E [X Y ]. That is,
E [X ] = E [E [X Y ]] = E [E [X Y ]]. In fact, E [X ] = E [E [E [X Y ]Y ]].
75 Proposition 3.2.3 also implies relations among estimators based on diﬀerent sets of observations.
For example, suppose X is to be estimated and Y1 and Y2 are both possible observations. The space
of unrestricted estimators based on Y1 alone is a subspace of the space of unrestricted estimators
based on both Y1 and Y2 . Therefore, Proposition 3.2.3 implies that E [E [X Y1 , Y2 ]Y1 ] = E [X Y1 ], a
property that is sometimes called the tower property of conditional expectation. The same relation
holds true for the same reason for the best linear estimators: E [E [X Y1 , Y2 ]Y1 ] = E [X Y1 ].
Example 3.3.1 Let X, Y be jointly continuous random variables with the pdf
x + y 0 ≤ x, y ≤ 1
0
else fXY (x, y ) = Let us ﬁnd E [X  Y ] and E [X  Y ]. To ﬁnd E [X  Y ] we ﬁrst identify fY (y ) and fX Y (xy ).
∞ 1
2 fXY (x, y )dx = fY (y ) =
−∞ +y 0≤y ≤1
0
else Therefore, fX Y (x  y ) is deﬁned only for 0 ≤ y ≤ 1, and for such y it is given by
x+ y
1
+y
2 0≤x≤1 0 fX Y (x  y ) = else So for 0 ≤ y ≤ 1,
1 E [X  Y = y ] = xfX Y (x  y )dx =
0 2 + 3y
.
3 + 6y Therefore, E [X  Y ] = 2+3Y . To ﬁnd E [X  Y ] we compute EX = EY =
3+6Y
1
7
1
7
Cov(X, Y ) = − 144 so E [X  Y ] = 12 − 11 (Y − 12 ). 7
12 , Var(Y ) = 11
144 and Example 3.3.2 Suppose that Y = XU , where X and U are independent random variables, X has
the Rayleigh density
x −x2 /2σ 2
e
σ2 fX (x) = 0 x≥0
else and U is uniformly distributed on the interval [0, 1]. We ﬁnd E [X  Y ] and E [X  Y ]. To compute
E [X  Y ] we ﬁnd
∞ EX =
0 EY x2 −x2 /2σ2
1
e
dx =
σ2
σ = EXEU = σ
2 π
2 ∞ √
−∞ x2
2πσ 2 e−x 2 /2σ 2 dx = σ π
2 π
2 E [X 2 ] = 2σ 2
Var(Y ) = E [Y 2 ] − E [Y ]2 = E [X 2 ]E [U 2 ] − E [X ]2 E [U ]2 = σ 2
1
π
Cov(X, Y ) = E [U ]E [X 2 ] − E [U ]E [X ]2 = Var(X ) = σ 2 1 −
2
4
76 2π
−
38 Thus
(1 − π )
π
+2 4
2 (3 − π)
8 E [X  Y ] = σ Y− σ
2 π
2 To ﬁnd E [X  Y ] we ﬁrst ﬁnd the joint density and then the conditional density. Now
fXY (x, y ) = fX (x)fY X (y  x)
= 1 −x2 /2σ 2
e
σ2 0 0≤y≤x
else
∞ 1 −x2 /2σ 2
dx
y σ2 e ∞ fY (y ) = fXY (x, y )dx = √ = 2π
σQ 0 −∞ y
σ y≥0
y<0 where Q is the complementary CDF for the standard normal distribution. So for y ≥ 0
∞ E [X  Y = y ] =
= xfXY (x, y )dx/fY (y )
−∞
∞ x −x2 /2σ 2
dx
σ exp(−y 2 /2σ 2 )
y σ2 e
√
√
=
y
y
2π
2πQ( σ )
Q( σ )
σ Thus,
E [X  Y ] = σ exp(−Y 2 /2σ 2 )
√
2πQ( Y )
σ Example 3.3.3 Suppose that Y is a random variable and f is a Borel measurable function such
that E [f (Y )2 ] < ∞. Let us show that E [f (Y )Y ] = f (Y ). By deﬁnition, E [f (Y )Y ] is the random
variable of the form g (Y ) which is closest to f (Y ) in the mean square sense. If we take g (Y ) = f (Y ),
then the mean square error is zero. No other estimator can have a smaller mean square error. Thus,
E [f (Y )Y ] = f (Y ). Similarly, if Y is a random vector with E [Y 2 ] < ∞, and if A is a matrix and
b a vector, then E [AY + bY ] = AY + b. 3.4 Joint Gaussian distribution and Gaussian random vectors Recall that a random variable X is Gaussian (or normal) with mean µ and variance σ 2 > 0 if X
has pdf
fX (x) = √ 1
2πσ 2 e− (x−µ)2
2σ 2 . As a degenerate case, we say X is Gaussian with mean µ and variance 0 if P {X = µ} = 1.
Equivalently, X is Gaussian with mean µ and variance σ 2 ≥ 0 if its characteristic function is given
by
ΦX (u) = exp −
77 u2 σ 2
+ jµu .
2 Lemma 3.4.1 Suppose X1 , X2 , . . . , Xn are independent Gaussian random variables. Then any
linear combination a1 X1 + · · · + an Xn is a Gaussian random variable.
Proof. By an induction argument on n, it is suﬃcient to prove the lemma for n = 2. Also, if X
is a Gaussian random variable, then so is aX for any constant a, so we can assume without loss
of generality that a1 = a2 = 1. It remains to prove that if X1 and X2 are independent Gaussian
random variables, then the sum X = X1 + X2 is also a Gaussian random variable. Let µi = E [Xi ]
2
and σi = Var(Xi ). Then the characteristic function of X is given by
ΦX (u) = E [ejuX ] = E [ejuX1 ejuX2 ] = E [ejuX1 ]E [ejuX2 ]
2
2
u2 σ1
u2 σ2
u2 σ 2
= exp −
+ jµ1 u exp −
+ jµ2 u = exp −
+ jµu .
2
2
2
2
2
where µ = µ1 + µ2 and σ 2 = σ1 + σ2 . Thus, X is a N (µ, σ 2 ) random variable. Let (Xi : i ∈ I ) be a collection of random variables indexed by some set I , which possibly has
inﬁnite cardinality. A ﬁnite linear combination of (Xi : i ∈ I ) is a random variable of the form
a1 Xi1 + a2 Xi2 + · · · + an Xin where n is ﬁnite, ik ∈ I for each k, and ak ∈ R for each k .
Deﬁnition 3.4.2 A collection (Xi : i ∈ I ) of random variables has a joint Gaussian distribution
(and the random variables Xi : i ∈ I themselves are said to be jointly Gaussian) if every ﬁnite
linear combination of (Xi : i ∈ I ) is a Gaussian random variable. A random vector X is called
a Gaussian random vector if its coordinate random variables are jointly Gaussian. A collection of
random vectors is said to have a joint Gaussian distribution if all of the coordinate random variables
of all of the vectors are jointly Gaussian.
We write that X is a N (µ, K ) random vector if X is a Gaussian random vector with mean vector
µ and covariance matrix K .
Proposition 3.4.3 (a) If (Xi : i ∈ I ) has a joint Gaussian distribution, then each of the random
variables itself is Gaussian.
(b) If the random variables Xi : i ∈ I are each Gaussian and if they are independent, which
means that Xi1 , Xi2 , . . . , Xin are independent for any ﬁnite number of indices i1 , i2 , . . . , in ,
then (Xi : i ∈ I ) has a joint Gaussian distribution.
(c) (Preservation of joint Gaussian property under linear combinations and limits) Suppose
(Xi : i ∈ I ) has a joint Gaussian distribution. Let (Yj : j ∈ J ) denote a collection of random
variables such that each Yj is a ﬁnite linear combination of (Xi : i ∈ I ), and let (Zk : k ∈ K )
denote a set of random variables such that each Zk is a limit in probability (or in the m.s. or
a.s. senses) of a sequence from (Yj : j ∈ J ). Then (Yj : j ∈ J ) and (Zk : k ∈ K ) each have a
joint Gaussian distribution.
(c ) (Alternative version of (c)) Suppose (Xi : i ∈ I ) has a joint Gaussian distribution. Let Z
denote the smallest set of random variables that contains (Xi : i ∈ I ), is a linear class, and
is closed under taking limits in probability. Then Z has a joint Gaussian distribution.
78 TX (d) The characteristic function of a N (µ, K ) random vector is given by ΦX (u) = E [eju
1T
T
eju µ− 2 u Ku .
= (e) If X is a N (µ, K ) random vector and K is a diagonal matrix (i.e. cov(Xi , Xj ) = 0 for i = j ,
or equivalently, the coordinates of X are uncorrelated) then the coordinates X1 , . . . , Xm are
independent.
(f ) A N (µ, K ) random vector X such that K is nonsingular has a pdf given by
fX (x) = 1
m
2 (2π ) K  1
2 exp − (x − µ)T K −1 (x − µ)
2 . (3.8) Any random vector X such that Cov(X ) is singular does not have a pdf.
(g) If X and Y are jointly Gaussian vectors, then they are independent if and only if Cov(X, Y ) =
0.
Proof. (a) Supppose (Xi : i ∈ I ) has a joint Gaussian distribution, so that all ﬁnite linear
combinations of the Xi ’s are Gaussian random variables. Each Xi for i ∈ I is itself a ﬁnite linear combination of all the variables (with only one term). So each Xi is a Gaussian random variable.
(b) Suppose the variables Xi : i ∈ I are mutually independent, and each is Gaussian. Then any
ﬁnite linear combination of (Xi : i ∈ I ) is the sum of ﬁnitely many independent Gaussian random
variables (by Lemma 3.4.1), and is hence also a Gaussian random variable. So (Xi : i ∈ I ) has a
joint Gaussian distribution.
(c) Suppose the hypotheses of (c) are true. Let V be a ﬁnite linear combination of (Yj : j ∈ J ) :
V = b1 Yj1 + b2 Yj2 + · · · + bn Yjn . Each Yj is a ﬁnite linear combination of (Xi : i ∈ I ), so V can be
written as a ﬁnite linear combination of (Xi : i ∈ I ):
V = b1 (a11 Xi11 + a12 Xi12 + · · · + a1k1 Xi1k1 ) + · · · + bn (an1 Xin1 + · · · + ankn Xinkn ). Therefore V is thus a Gaussian random variable. Thus, any ﬁnite linear combination of (Yj : j ∈ J )
is Gaussian, so that (Yj : j ∈ J ) has a joint Gaussian distribution.
Let W be a ﬁnite linear combination of (Zk : k ∈ K ): W = a1 Zk1 + · · · + am Zkm . By assumption,
d. for 1 ≤ l ≤ m, there is a sequence (jl,n : n ≥ 1) of indices from J such that Yjl,n → Zkl as n → ∞.
Let Wn = a1 Yj1,n + · · · + am Yjm,n . Each Wn is a Gaussian random variable, because it is a ﬁnite
linear combination of (Yj : j ∈ J ). Also,
m W − Wn  ≤ al Zkl − Yjl,n . (3.9) l=1 Since each term on the righthand side of (3.9) converges to zero in probability, it follows that
p.
Wn → W as n → ∞. Since limits in probability of Gaussian random variables are also Gaussian
random variables (Proposition 2.1.16), it follows that W is a Gaussian random variable. Thus, an
arbitrary ﬁnite linear combination W of (Zk : k ∈ K ) is Gaussian, so, by deﬁnition, (Zk : k ∈ K )
has a joint Gaussian distribution.
79 (c ) Suppose (Xi : i ∈ I ) has a joint Gaussian distribution. Using the notation of part (c), let
(Yi : i ∈ I ) denote the set of all ﬁnite linear combinations of (Xi : i ∈ I ) and let (Zk : k ∈ K ) denote
the set of all random variables that are limits in probability of random variables in (Yi ; i ∈ I ). We
will show that Z = (Zk : k ∈ K ), which together with part (c) already proved, will establish (c ).
We begin by establishing that (Zk : k ∈ K ) satisﬁes the three properties required of Z :
(i) (Zk : k ∈ K ) contains (Xi : i ∈ I ).
(ii) (Zk : k ∈ K ) is a linear class
(iii) (Zk : k ∈ K ) is closed under taking limits in probability
Property (i) follows from the fact that for any io ∈ I , the random variable Xio is trivially a ﬁnite
linear combination of (Xi : i ∈ I ), and it is trivially the limit in probability of the sequence with all
entries equal to itself. Property (ii) is true because a linear combination of the form a1 Zk1 + a2 Zk2
is the limit in probability of a sequence of random variables of the form a1 Yjn,1 + a2 Yjn,2 , and,
since (Yj : j ∈ J ) is a linear class, a1 Yjn,1 + a2 Yjn2 is a random variable from (Yj : j ∈ J ) for
p. each n. To prove (iii), suppose Zkn → Z∞ as n → ∞ for some sequence k1 , k2 , . . . from K. By
passing to a subsequence if necessary, it can be assumed that P {Z∞ − Zkn  ≥ 2−(n+1) } ≤ 2−(n+1)
for all n ≥ 1. Since each Zkn is the limit in probability of a sequence of random variables from
(Yj : j ∈ J ), for each n there is a jn ∈ J so that P {Zkn − Yjn  ≥ 2−(n+1) } ≤ 2−(n+1) . Since
p
Z∞ − Yjn  ≤ Z∞ − Zkn  + Zkn − Yjn , it follows that P {Z∞ − Yjn  ≥ 2−n } ≤ 2−n . So Yjn → Z∞ .
Therefore, Z∞ is a random variable in (Zk : k ∈ K ), so (Zk : k ∈ K ) is closed under convergence
in probability. In summary, (Zk : k ∈ K ) has properties (i)(iii). Any set of random variables
with these three properties must contain (Yj : j ∈ J ), and hence must contain (Zk : k ∈ K ).
So (Zk : k ∈ K ) is indeed the smallest set of random variables with properties (i)(iii). That is,
(Zk : k ∈ K ) = Z , as claimed.
(d) Let X be a N (µ, K ) random vector. Then for any vector u with the same dimension as X ,
the random variable uT X is Gaussian with mean uT µ and variance given by
Var(uT X ) = Cov(uT X, uT X ) = uT Ku.
Thus, we already know the characteristic function of uT X . But the characteristic function of the
vector X evaluated at u is the characteristic function of uT X evaluated at 1:
TX ΦX (u) = E [eju T X) = E [ej (u T µ− 1 uT Ku
2 = ΦuT X (1) = eju , which establishes part (d) of the proposition.
(e) If X is a N (µ, K ) random vector and K is a diagonal matrix, then
m exp(jui µi − ΦX (u) =
i=1 kii u2
i
)=
2 Φi (ui )
i where kii denotes the ith diagonal element of K , and Φi is the characteristic function of a N (µi , kii )
random variable. By uniqueness of joint characteristic functions, it follows that X1 , . . . , Xm are
independent random variables. 80 (f) Let X be a N (µ, K ) random vector. Since K is positive semideﬁnite it can be written as
K = U ΛU T where U is orthonormal (so U U T = U T U = I ) and Λ is a diagonal matrix with
the eigenvalues λ1 , λ2 , . . . , λm of K along the diagonal. (See Section 11.7 of the appendix.) Let
Y = U T (X − µ). Then Y is a Gaussian vector with mean 0 and covariance matrix given by
Cov(Y, Y ) = Cov(U T X, U T X ) = U T KU = Λ. In summary, we have X = U Y + µ, and Y is a
vector of independent Gaussian random variables, the ith one being N (0, λi ). Suppose further that
K is nonsingular, meaning det(K ) = 0. Since det(K ) = λ1 λ2 · · · λm this implies that λi > 0 for
each i, so that Y has the joint pdf
m √ fY (y ) =
i=1 1
y2
exp − i
2λi
2πλi = 1
(2π ) m
2 det(K ) exp − y T Λ−1 y
2 . Since  det(U ) = 1 and U Λ−1 U T = K −1 , the joint pdf for the N (µ, K ) random vector X is given
by
fX (x) = fY (U T (x − µ)) = 1
m −
1 exp (2π ) 2 K  2 (x − µ)T K −1 (x − µ)
2 . Now suppose, instead, that X is any random vector with some mean µ and a singular covariance
matrix K . That means that det K = 0, or equivalently that λi = 0 for one of the eigenvalues of K , or
equivalently, that there is a vector α such that αT Kα = 0 (such an α is an eigenvector of K for eigenvalue zero). But then 0 = αT Kα = αT Cov(X, X )α = Cov(αT X, αT X ) = Var(αT X ). Therefore,
P {αT X = αT µ} = 1. That is, with probability one, X is in the subspace {x ∈ Rm : αT (x − µ) = 0}.
Therefore, X does not have a pdf.
(g) Suppose X and Y are jointly Gaussian vectors and uncorrelated (so Cov(X, Y ) = 0.) Let Z
denote the dimension m + n vector with coordinates X1 , . . . , Xm , Y1 , . . . , Yn . Since Cov(X, Y ) = 0,
the covariance matrix of Z is block diagonal:
Cov(Z ) = Cov(X )
0
0
Cov(Y ) . Therefore, for u ∈ Rm and v ∈ Rn ,
ΦZ u
v 1u
2v
= ΦX (u)ΦY (v ).
= exp − T Cov(Z ) u
u
+j
v
v T EZ Such factorization implies that X and Y are independent. The if part of part (f) is proved.
Conversely, if X and Y are jointly Gaussian and independent of each other, then the characteristic
function of the joint density must factor, which implies that Cov(Z ) is block diagonal as above.
That is, Cov(X, Y ) = 0.
Recall that in general, if X and Y are two random vectors on the same probability space, then
the mean square error for the MMSE linear estimator E [X Y  is greater than or equal to the mean
square error for the best unconstrained estimator, E [X Y . The tradeoﬀ, however, is that E [X Y 
can be much more diﬃcult to compute than E [X Y , which is determined entirely by ﬁrst and
81 second moments. As shown in the next proposition, if X and Y are jointly Gaussian, the two
estimators coincide. That is, the MMSE unconstrained estimator of Y is linear. We also know that
E [X Y = y ] is the mean of the conditional mean of X given Y = y . The proposition identiﬁes not
only the conditional mean, but the entire conditional distribution of X given Y = y , for the case
X and Y are jointly Gaussian.
Proposition 3.4.4 Let X and Y be jointly Gaussian vectors. Given Y = y , the conditional
distribution of X is N (E [X Y = y ], Cov(e)). In particular, the conditional mean E [X Y = y ] is
equal to E [X Y = y ]. That is, if X and Y are jointly Gaussian, then E [X Y ] = E [X Y ].
If Cov(Y ) is nonsingular,
E [X Y = y ] = E [X Y = y ] = EX + Cov(X, Y )Cov(Y )−1 (y − E [Y ])
Cov(e) = Cov(X ) − Cov(X, Y )Cov(Y ) −1 Cov(Y, X ), (3.10)
(3.11) and if Cov(e) is nonsingular,
fX Y (xy ) = 1
m
2 (2π ) Cov(e) 1
2 exp − 1
x − E [X Y = y ]
2 T Cov(e)−1 (x − E [X Y = y ]) . (3.12) Proof. Consider the MMSE linear estimator E [X Y ] of X given Y , and let e denote the
corresponding error vector: e = X − E [X Y ]. Recall that, by the orthogonality principle, Ee = 0
and Cov(e, Y ) = 0. Since Y and e are obtained from X and Y by linear transformations, they are
jointly Gaussian. Since Cov(e, Y ) = 0, the random vectors e and Y are also independent. For the
next part of the proof, the reader should keep in mind that if a is a deterministic vector of some
dimension m, and Z is a N (0, K ) random vector, for a matrix K that is not a function of a, then
Z + a has the N (a, K ) distribution.
Focus on the following rearrangement of the deﬁnition of e:
X = e + E [X Y ]. (3.13) (Basically, the whole proof of the proposition hinges on (3.13).) Since E [X Y ] is a function of Y
and since e is independent of Y with distribution N (0, Cov(e)), the following key observation can
be made. Given Y = y , the conditional distribution of e is the N (0, Cov(e)) distribution, which
does not depend on y , while E [X Y = y ] is completely determined by y . So, given Y = y , X can
be viewed as the sum of the N (0, Cov(e)) vector e and the determined vector E [X Y = y ]. So the
conditional distribution of X given Y = y is N (E [X Y = y ], Cov(e)). In particular, E [X Y = y ],
which in general is the mean of the conditional distribution of X given Y = y , is therefore the
mean of the N (E [X Y = y ], Cov(e)) distribution. Hence E [X Y = y ] = E [X Y = y ]. Since this is
true for all y , E [X Y ] = E [X Y ].
Equations (3.10) and (3.11), respectively, are just the equations (3.6) and (3.7) derived for the
MMSE linear estimator, E [X Y ], and its associated covariance of error. Equation (3.12) is just the
formula (3.8) for the pdf of a N (µ, K ) vector, with µ = E [X Y = y ] and K = Cov(e). Example 3.4.5 Suppose X and Y are jointly Gaussian mean zero random variables such that
X
43
the vector
has covariance matrix
. Let us ﬁnd simple expressions for the two
Y
39
82 random variables E [X 2 Y ] and P [X ≥ cY ]. Note that if W is a random variable with the N (µ, σ 2 )
distribution, then E [W 2 ] = µ2 + σ 2 and P {W ≥ c} = Q( c−µ ), where Q is the standard Gaussian
σ
complementary CDF. The idea is to apply these facts to the conditional distribution of X given Y .
2
Given Y = y , the conditional distribution of X is N ( Cov(X,Y ) y, Cov(X ) − Cov(X,Y)) ), or N ( y , 3).
3
Var(Y )
Var(Y
(y/
Therefore, E [X 2 Y = y ] = ( y )2 + 3 and P [X ≥ cY = y ] = Q( c−√3 3) ). Applying these two
3
(Y
functions to the random variable Y yields E [X 2 Y ] = ( Y )2 + 3 and P [X ≥ cY ] = Q( c−√3/3) ).
3 3.5 Linear Innovations Sequences Let X, Y1 , . . . , Yn be random vectors with ﬁnite second moments, all on the same probability space.
In general, computation of the joint projection E [X Y1 , . . . , Yn ] is considerably more complicated
than computation of the individual projections E [X Yi ], because it requires inversion of the covariance matrix of all the Y ’s. However, if E [Yi ] = 0 for all i and E [Yi YjT ] = 0 for i = j (i.e., all
coordinates of Yi are orthogonal to constants and to all coordinates of Yj for i = j ), then
n E [X − X Yi ], E [X Y1 , . . . , Yn ] = X + (3.14) i=1 where we write X for EX. The orthogonality principle can be used to prove (3.14) as follows. It
suﬃces to prove that the right side of (3.14) satisﬁes the two properties that together characterize
the left side of (3.14). First, the right side is a linear combination of 1, Y1 , . . . , Yn . Secondly, let e
denote the error when the right side of (3.14) is used to estimate X :
n e = X −X − E [X − X Yi ].
i=1 T
It must be shown that E [e(Y1T c1 + Y2T c2 + · · · + Yn cn + b)] = 0 for any constant vectors c1 , . . . , cn
and constant b. It is enough to show that E [e] = 0 and E [eYjT ] = 0 for all j . But E [X − X Yi ] has
the form Bi Yi , because X − X and Yi have mean zero. Thus, E [e] = 0. Furthermore, E [eYjT ] = E X − E [X Yj ] YjT − E [Bi Yi YjT ].
i :i = j Each term on the right side of this equation is zero, so E [eYjT ] = 0, and (3.14) is proved.
If 1, Y1 , Y2 , . . . , Yn have ﬁnite second moments but are not orthogonal, then (3.14) doesn’t directly apply. However, by orthogonalizing this sequence we can obtain a sequence 1, Y1 , Y2 , . . . , Yn
that can be used instead. Let Y1 = Y1 − E [Y1 ], and for k ≥ 2 let
Yk = Yk − E [Yk Y1 , . . . , Yk−1 ]. (3.15) Then E [Yi ] = 0 for all i and E [Yi YjT ] = 0 for i = j . In addition, by induction on k , we can prove
that the set of all random variables obtained by linear transformation of 1, Y1 , . . . , Yk is equal to
the set of all random variables obtained by linear transformation of 1, Y1 , . . . , Yk .
83 Thus, for any random variable X with ﬁnite second moments,
n E [X  Y1 , . . . , Yn ] = E [X Y1 , . . . , Yn ] = X + E [X − X Yi ]
i=1 n E [X Yi ]Yi = X+
i=1 E [(Yi )2 ] Moreover, this same result can be used to compute the innovations sequence recursively: Y1 =
Y1 − E [Y1 ], and
k−1 Yk = Yk − E [Yk ] − E [Yk Yi ]Yi i=1 E [(Yi )2 ] k ≥ 2. The sequence Y1 , Y2 , . . . , Yn is called the linear innovations sequence for Y1 , Y2 , . . . , Yn . 3.6 Discretetime Kalman ﬁltering Kalman ﬁltering is a statespace approach to the problem of estimating one random sequence
from another. Recursive equations are found that are useful in many realtime applications. For
notational convenience, because there are so many matrices in this section, lower case letters are
used for random vectors. All the random variables involved are assumed to have ﬁnite second
moments. The state sequence x0 , x1 , . . ., is to be estimated from an observed sequence y0 , y1 , . . ..
These sequences of random vectors are assumed to satisfy the following state and observation
equations.
State:
Observation: xk+1 = Fk xk + wk
yk = T
Hk xk + vk k≥0
k ≥ 0. It is assumed that
• x0 , v0 , v1 , . . . , w0 , w1 , . . . are pairwise uncorrelated.
• Ex0 = x0 , Cov(x0 ) = P0 , Ewk = 0, Cov(wk ) = Qk , Evk = 0, Cov(vk ) = Rk .
• Fk , Hk , Qk , Rk for k ≥ 0; P0 are known matrices.
• x0 is a known vector.
See Figure 3.2 for a block diagram of the state and observation equations. The evolution of the
state sequence x0 , x1 , . . . is driven by the random vectors w0 , w1 , . . ., while the random vectors v0 ,
v1 , . . . , represent observation noise.
Let xk = E [xk ] and Pk = Cov(xk ). These quantities are recursively determined for k ≥ 1 by
T
xk+1 = Fk xk and Pk+1 = Fk Pk Fk + Qk , (3.16) where the initial conditions x0 and P0 are given as part of the state model. The idea of the Kalman
ﬁlter equations is to recursively compute conditional expectations in a similar way.
84 vk
wk + xk+1 xk Delay T
Hk + yk F
k
Figure 3.2: Block diagram of the state and observations equations.
Let y k = (y0 , y1 , . . . , yk ) represent the observations up to time k . Deﬁne for nonnegative integers
i, j
x i j = E [xi y j ] and the associated covariance of error matrices
Σ i j = Cov(xi − xij ). The goal is to compute xk+1k for k ≥ 0. The Kalman ﬁlter equations will ﬁrst be stated, then
brieﬂy discussed, and then derived. The Kalman ﬁlter equations are given by
T
Fk − Kk Hk xkk−1 + Kk yk xk+1k = = Fk xkk−1 + Kk yk − (3.17) T
Hk xkk−1 with the initial condition x01 = x0 , where the gain matrix Kk is given by
T
Kk = Fk Σkk−1 Hk Hk Σkk−1 Hk + Rk −1 (3.18) and the covariance of error matrices are recursively computed by
T
Σk+1k = Fk Σkk−1 − Σkk−1 Hk Hk Σkk−1 Hk + Rk −1 T
T
Hk Σkk−1 Fk + Qk (3.19) with the initial condition Σ0−1 = P0 . See Figure 3.3 for the block diagram. yk Kk + xk+1 k Delay xk k−1 F −Kk HT
k
k
Figure 3.3: Block diagram of the Kalman ﬁlter.
We comment brieﬂy on the Kalman ﬁlter equations, before deriving them. First, observe what
happens if Hk is the zero matrix, Hk = 0, for all k . Then the Kalman ﬁlter equations reduce to
85 (3.16) with xkk−1 = xk , Σkk−1 = Pk and Kk = 0. Taking Hk = 0 for all k is equivalent to having
no observations available.
In many applications, the sequence of gain matrices can be computed ahead of time according
to (3.18) and (3.19). Then as the observations become available, the estimates can be computed
using only (3.17). In some applications the matrices involved in the state and observation models,
including the covariance matrices of the vk ’s and wk ’s, do not depend on k . The gain matrices
Kk could still depend on k due to the initial conditions, but if the model is stable in some sense,
then the gains converge to a constant matrix K , so that in steady state the ﬁlter equation (3.17)
becomes time invariant: xk+1k = (F − KH T )xkk−1 + Kyk .
In other applications, particularly those involving feedback control, the matrices in the state
and/or observation equations might not be known until just before they are needed.
The Kalman ﬁlter equations are now derived. Roughly speaking, there are two considerations
for computing xk+1k once xkk−1 is computed: (1) the time update, accounting for the change in
state from xk to xk+1 , and (2) the information update, accounting for the availability of the new
observation yk . Indeed, for k ≥ 0, we can write
xk+1k = xk+1k−1 + xk+1k − xk+1k−1 , (3.20) where the ﬁrst term on the right of (3.20), namely xk+1k−1 , represents the result of modifying xkk−1
to take into account the passage of time on the state, but with no new observation. The diﬀerence
in square brackets in (3.20) is the contribution of the new observation, yk , to the estimation of
xk+1 .
Time update: In view of the state update equation and the fact that wk is uncorrelated with
the random variables of y k−1 and has mean zero,
xk+1k−1 = E [Fk xk + wk y k−1 ]
= Fk E [xk y k−1 ] + E [wk y k−1 ]
= Fk xkk−1 (3.21) Thus, the time update consists of simply multiplying the previous estimate by Fk . If there were no
new observation, then this would be the entire Kalman ﬁlter. Furthermore, the covariance of error
matrix for predicting xk+1 by xk+1k−1 , is given by
Σk+1k−1 = Cov(xk+1 − xk+1k−1 )
= Cov(Fk (xk − xkk−1 ) + wk )
T
= Fk Σkk−1 Fk + Qk . (3.22) Information update: Consider next the new observation yk . The observation yk is not totally
new—for it can be predicted in part from the previous observations, or simply by its mean in
the case k = 0. Speciﬁcally, we can consider yk = yk − E [yk  y k−1 ] to be the new part of the
˜
observation yk . Here, y0 , y1 , . . . is the linear innovation sequence for the observation sequence
˜˜
y0 , y1 , . . ., as deﬁned in Section 3.5 (with the minor diﬀerence that here the vectors are indexed
from time k = 0 on, rather than from time k = 1). Since the linear span of the random variables in
(1, y k−1 , yk ) is the same as the linear span of the random variables in (1, y k−1 , yk ), for the purposes
˜
of incorporating the new observation we can pretend that yk is the new observation rather than yk .
˜
86 T
By the observation equation and the facts E [vk ] = 0 and E [y k−1 vk ] = 0, it follows that
T
E [yk  y k−1 ] = E Hk xk + wk y k−1
T
= Hk xkk−1 ,
T
so yk = yk − Hk xkk−1 . Since (1, y k−1 , yk ) and (1, y k−1 , yk ) have the same span and the random
˜
˜
˜
k−1 are orthogonal to the random variables in y , and all these random variables have
variables in y
˜
˜k
mean zero, xk+1k = E xk+1 y k−1 , yk
˜
˜
˜
˜
= xk+1 + E xk+1 − xk+1 y k−1 + E [xk+1 − xk+1 yk ]
= xk+1k−1 + E [xk+1 − xk+1 yk ] .
˜
Therefore, the term to be added for the information update (in square brackets in (3.20)) is
E [xk+1 − xk+1 yk ] . Since xk+1 − xk+1 and yk both have mean zero, the information update term
˜
˜
can be simpliﬁed to:
˜
˜
E [xk+1 − xk+1 yk ] = Kk yk , (3.23) where (use the fact Cov(xk+1 − xk+1 , yk ) = Cov(xk+1 , yk ) because E [˜k ] = 0)
˜
˜
y
Kk = Cov(xk+1 , yk )Cov(˜k )−1 .
˜
y (3.24) Putting it all together: Equation (3.20), with the time update equation (3.21) and the fact the
information update term is Kk yk , yields the main Kalman ﬁlter equation:
˜
xk+1k = Fk xkk−1 + Kk yk .
˜ (3.25) Taking into account the new observation yk , which is orthogonal to the previous observations, yields
˜
a reduction in the covariance of error:
Σk+1k = Σk+1k−1 − Cov(Kk yk ).
˜ (3.26) The Kalman ﬁlter equations (3.17), (3.18), and (3.19) follow easily from (3.25), (3.24), and (3.26),
as follows. To convert (3.24) into (3.18), use
T
Cov(xk+1 , yk ) = Cov(Fk xk + wk , Hk (xk − xkk−1 ) + vk )
˜ = T
Cov(Fk xk , Hk (xk − xkk−1 )) T
= Cov(Fk (xk − xkk−1 ), Hk (xk − xkk−1 )) = Fk Σkk−1 Hk
and
T
Cov(˜k ) = Cov(Hk (xk − xkk−1 ) + vk )
y
T
= Cov(Hk (xk − xkk−1 )) + Cov(vk )
T
= Hk Σkk−1 Hk + Rk 87 (3.27) To convert (3.26) into (3.19) use (3.22) and
T
Cov(Kk yk ) = Kk Cov(˜k )Kk
˜
y = Cov(xk+1 − Fk xkk−1 )Cov(˜k )−1 Cov(xk+1 − Fk xkk−1 )
y This completes the derivation of the Kalman ﬁltering equations. 3.7 Problems 3.1 Rotation of a joint normal distribution yielding independence
Let X be a Gaussian vector with
E [X ] = 10
5 Cov(X ) = 21
11 . (a) Write an expression for the pdf of X that does not use matrix notation.
(b) Find a vector b and orthonormal matrix U such that the vector Y deﬁned by Y = U T (X − b)
is a mean zero Gaussian vector such at Y1 and Y2 are independent.
3.2 Linear approximation of the cosine function over an interval
Let Θ be uniformly distributed on the interval [0, π ] (yes, [0, π ], not [0, 2π ]). Suppose Y = cos(Θ)
is to be estimated by an estimator of the form a + bΘ. What numerical values of a and b minimize
the mean square error?
3.3 Calculation of some minimum mean square error estimators
Let Y = X + N , where X has the exponential distribution with parameter λ, and N is Gaussian
with mean 0 and variance σ 2 . The variables X and N are independent, and the parameters λ and
1
1
σ 2 are strictly positive. (Recall that E [X ] = λ and Var(X ) = λ2 .)
(a) Find E [X Y ] and also ﬁnd the mean square error for estimating X by E [X Y ].
(b) Does E [X Y ] = E [X Y ]? Justify your answer. (Hint: Answer is yes if and only if there is no
estimator for X of the form g (Y ) with a smaller MSE than E [X Y ].)
3.4 Valid covariance matrix
For what real values of a and b is the following matrix the covariance matrix of some realvalued
random vector? 21b
K = a 1 0 .
b01
Hint: An symmetric n × n matrix is positive semideﬁnite if and only if the determinant of every
matrix obtained by deleting a set of rows and the corresponding set of columns, is nonnegative.
3.5 Conditional probabilities with joint Gaussians I
X
1ρ
Let
be a mean zero Gaussian vector with correlation matrix
Y
ρ1
(a) Express P [X ≤ 1Y ] in terms of ρ, Y , and the standard normal CDF, Φ.
(b) Find E [(X − Y )2 Y = y ] for real values of y .
88 , where ρ < 1. 3.6 Conditional probabilities with joint Gaussians II
Let X, Y be jointly Gaussian random variables with mean zero and covariance matrix
Cov X
Y = 46
6 18 .
u 2 You may express your answers in terms of the Φ function deﬁned by Φ(u) = −∞ √1 π e−s /2 ds.
2
(a) Find P [X − 1 ≥ 2].
(b) What is the conditional density of X given that Y = 3? You can either write out the density
in full, or describe it as a well known density with speciﬁed parameter values.
(c) Find P [X − E [X Y ] ≥ 1].
3.7 An estimation error bound
Suppose the random vector X
Y has mean vector 2
−2 and covariance matrix 83
32 . Let e = X − E [X  Y ].
(a) If possible, compute E [e2 ]. If not, give an upper bound.
(b) For what joint distribution of X and Y (consistent with the given information) is E [e2 ] maximized? Is your answer unique?
3.8 An MMSE estimation problem
(a) Let X and Y be jointly uniformly distributed over the triangular region in the x − y plane with
corners (0,0), (0,1), and (1,2). Find both the linear minimum mean square error (LMMSE) estimator estimator of X given Y and the (possibly nonlinear) MMSE estimator X given Y . Compute
the mean square error for each estimator. What percentage reduction in MSE does the MMSE
estimator provide over the LMMSE?
(b) Repeat part (a) assuming Y is a N (0, 1) random variable and X = Y .
3.9 Comparison of MMSE estimators for an example
1
Let X = 1+U , where U is uniformly distributed over the interval [0, 1].
(a) Find E [X U ] and calculate the MSE, E [(X − E [X U ])2 ].
(b) Find E [X U ] and calculate the MSE, E [(X − E [X U ])2 ].
3.10 Conditional Gaussian comparison
Suppose that X and Y are jointly Gaussian, mean zero, with Var(X ) = Var(Y ) = 10 and
Cov(X, Y ) = 8. Express the following probabilities in terms of the Q function.
(a) pa = P {X ≥ 2}.
(b) pb = P [X ≥ 2Y = 3].
(c) pc = P [X ≥ 2Y ≥ 3]. (Note: pc can be expressed as an integral. You need not carry out the
integration.)
(d) Indicate how pa , pb , and pc are ordered, from smallest to largest.
3.11 Diagonalizing a twodimensional Gaussian distribution
X1
1ρ
Let X =
be a mean zero Gaussian random vector with correlation matrix
,
X2
ρ1
where ρ < 1. Find an orthonormal 2 by 2 matrix U such that X = U Y for a Gaussian vector
Y1
Y=
such that Y1 is independent of Y2 . Also, ﬁnd the variances of Y1 and Y2 .
Y2
89 Note: The following identity might be useful for some of the problems that follow. If A, B, C,
and D are jointly Gaussian and mean zero, then E [ABCD] = E [AB ]E [CD] + E [AC ]E [BD] +
E [AD]E [BC ]. This implies that E [A4 ] = 3E [A2 ]2 , Var(A2 ) = 2E [A2 ], and Cov(A2 , B 2 ) =
2Cov(A, B )2 . Also, E [A2 B ] = 0. 3.12 An estimator of an estimator
Let X and Y be square integrable random variables and let Z = E [X  Y ], so Z is the MMSE
estimator of X given Y . Show that the LMMSE estimator of X given Y is also the LMMSE
estimator of Z given Y . (Can you generalize this result?).
3.13 Projections onto nested linear subspaces
(a) Use the Orthogonality Principle to prove the following statement: Suppose V0 and V1 are
two closed linear spaces of second order random variables, such that V0 ⊃ V1 , and suppose X
is a random variable with ﬁnite second moment. Let Zi∗ be the random variable in Vi with the
∗
minimum mean square distance from X . Then Z1 is the variable in V1 with the minimum mean
∗ . (b) Suppose that X, Y , and Y are random variables with ﬁnite second
square distance from Z0
1
2
moments. For each of the following three statements, identify the choice of subspace V0 and V1
such that the statement follows from part (a):
(i) E [X Y1 ] = E [ E [X Y1 , Y2 ] Y1 ].
(ii) E [X Y1 ] = E [ E [X Y1 , Y2 ] Y1 ]. (Sometimes called the “tower property.”)
(iii) E [X ] = E [E [X Y1 ]]. (Think of the expectation of a random variable as the constant closest to
the random variable, in the m.s. sense.
3.14 Some identities for estimators
Let X and Y be random variables with E [X 2 ] < ∞. For each of the following statements, determine
if the statement is true. If yes, give a justiﬁcation using the orthogonality principle. If no, give a
counter example.
(a) E [X cos(Y )Y ] = E [X Y ] cos(Y )
(b) E [X Y ] = E [X Y 3 ]
(c) E [X 3 Y ] = E [X Y ]3
(d) E [X Y ] = E [X Y 2 ]
(e) E [X Y ] = E [X Y 3 ]
3.15 Some identities for estimators, version 2
Let X, Y, and Z be random variables with ﬁnite second moments and suppose X is to be estimated.
For each of the following, if true, give a brief explanation. If false, give a counter example.
(a) E [(X − E [X Y ])2 ] ≤ E [(X − E [X Y, Y 2 ])2 ].
(b) E [(X − E [X Y ])2 ] = E [(X − E [X Y, Y 2 ]2 ] if X and Y are jointly Gaussian.
(c) E [ (X − E [E [X Z ] Y ])2 ] ≤ E [(X − E [X Y ])2 ].
(d) If E [(X − E [X Y ])2 ] = Var(X ), then X and Y are independent.
3.16 Some simple examples
Give an example of each of the following, and in each case, explain your reasoning.
(a) Two random variables X and Y such that E [X Y ] = E [X Y ], and such that E [X Y  is not
simply constant, and X and Y are not jointly Gaussian.
90 (b) A pair of random variables X and Y on some probability space such that X is Gaussian, Y is
Gaussian, but X and Y are not jointly Gaussian.
(c) Three random variables X, Y, and Z, which are pairwise independent, but all three together are
not independent. 3.17 The square root of a positivesemideﬁnite matrix
(a) True or false? If B is a square matrix over the reals, then BB T is positive semideﬁnite.
(b) True or false? If K is a symmetric positive semideﬁnite matrix over the reals, then there exists
a symmetric positive semideﬁnite matrix S over the reals such that K = S 2 . (Hint: What if K is
also diagonal?)
3.18 Estimating a quadratic
X
1ρ
Let
be a mean zero Gaussian vector with correlation matrix
, where ρ < 1.
Y
ρ1
(a) Find E [X 2 Y ], the best estimator of X 2 given Y.
(b) Compute the mean square error for the estimator E [X 2 Y ].
(c) Find E [X 2 Y ], the best linear (actually, aﬃne) estimator of X 2 given Y, and compute the mean
square error.
3.19 A quadratic estimator
Suppose Y has the N (0, 1) distribution and that X = Y . Find the estimator for X of the form
X = a + bY + cY 2 which minimizes the mean square error. (You can use the following numerical
values: E [Y ] = 0.8, E [Y 4 ] = 3, E [Y Y 2 ] = 1.6.)
(a) Use the orthogonality principle to derive equations for a, b, and c.
(b) Find the estimator X .
(c) Find the resulting minimum mean square error.
3.20 An innovations sequence and its application Y1
1
0.5 0.5
0
Y 0.5
1
0.5 0.25 .
Let 2 be a mean zero random vector with correlation matrix Y3 0.5 0.5
1
0.25 X
0 0.25 0.25
1 Y1
Y1 (a) Let Y1 , Y2 , Y3 denote the innovations sequence. Find the matrix A so that Y2 = A Y2 .
Y3
Y3 Y1
Y1 (b) Find the correlation matrix of Y2 and cross covariance matrix Cov(X, Y2 ).
Y3
Y3
(c) Find the constants a, b, and c to minimize E [(X − aY1 − bY2 − cY3 )2 ].
3.21 Estimation for an additive Gaussian noise model
Assume x and n are independent Gaussian vectors with means x, n and covariance matrices Σx
¯¯
and Σn . Let y = x + n. Then x and y are jointly Gaussian.
(a) Show that E [xy ] is given by either x + Σx (Σx + Σn )−1 (y − (¯ + n))
¯
x¯
91 or Σn (Σx + Σn )−1 x + Σx (Σx + Σn )−1 (y − n).
¯
¯
(b). Show that the conditional covariance matrix of x given y is given by any of the three expressions:
Σx − Σx (Σx + Σn )−1 Σx = Σx (Σx + Σn )−1 Σn = (Σ−1 + Σ−1 )−1 .
x
n
(Assume that the various inverses exist.)
3.22 A Kalman ﬁltering example
(a) Let σ 2 > 0 and let f be a real constant. Let x0 denote a N (0, σ 2 ) random variable and let f
be a realvalued constant. Consider the state and observation sequences deﬁned by:
(state) xk+1 = f xk + wk (observation) yk = xk + vk where w1 , w2 , . . . ; v1 , v2 , . . . are mutually independent N (0, 1) random variables. Write down the
Kalman ﬁlter equations for recursively computing the estimates xkk−1 , the (scaler) gains Kk , and
ˆ
2
the sequence of the variances of the errors (for brevity write σk for the covariance or error instead
of Σkk−1 ).
(b) For what values of f is the sequence of error variances bounded?
3.23 Steady state gains for onedimensional Kalman ﬁlter
This is a continuation of the previous problem.
2
(a) Show that limk→∞ σk exists.
2 , in terms of f .
(b) Express the limit, σ∞
2
(c) Explain why σ∞ = 1 if f = 0.
3.24 A variation of Kalman ﬁltering
(a) Let σ 2 > 0 and let f be a real constant. Let x0 denote a N (0, σ 2 ) random variable and let f
be a realvalued constant. Consider the state and observation sequences deﬁned by:
(state) xk+1 = f xk + wk (observation) y k = x k + wk where w1 , w2 , are mutually independent N (0, 1) random variables. Note that the state and observation equations are driven by the same sequence, so that some of the Kalman ﬁltering equations
derived in the notes do not apply. Derive recursive equations needed to compute xkk−1 , including
ˆ
recursive equations for any needed gains or variances of error. (Hints: What modiﬁcations need to
be made to the derivation for the standard model? Check that your answer is correct for f = 1.)
3.25 The Kalman ﬁlter for xkk
Suppose in a given application a Kalman ﬁlter has been implemented to recursively produce xk+1k
for k ≥ 0, as in class. Thus by time k , xk+1k , Σk+1k , xkk−1 , and Σkk−1 are already computed.
Suppose that it is desired to also compute xkk at time k . Give additional equations that can be
used to compute xkk . (You can assume as given the equations in the class notes, and don’t need
to write them all out. Only the additional equations are asked for here. Be as explicit as you can,
expressing any matrices you use in terms of the matrices already given in the class notes.)
92 3.26 An innovations problem
Let U1 , U2 , . . . be a sequence of independent random variables, each uniformly distributed on the
interval [0, 1]. Let Y0 = 1, and Yn = U1 U2 · · · Un for n ≥ 1.
(a) Find the variance of Yn for each n ≥ 1.
(b) Find E [Yn Y0 , . . . , Yn−1 ] for n ≥ 1.
(c) Find E [Yn Y0 , . . . , Yn−1 ] for n ≥ 1.
(d) Find the linear innovations sequence Y = (Y0 , Y1 , . . .).
(e) Fix a positive integer M and let XM = U1 + . . . + UM . Using the answer to part (d), ﬁnd
E [XM Y0 , . . . , YM ], the best linear estimator of XM given (Y0 , . . . , YM ).
3.27 Linear innovations and orthogonal polynomials for the normal distribution
(a) Let X be a N (0, 1) random variable. Show that for integers n ≥ 0,
E [X n ] = n!
(n/2)!2n/2 0 n even
n odd Hint: One approach is to apply the power series expansion for ex on each side of the identity
2
E [euX ] = eu /2 , and identify the coeﬃcients of un .
(b) Let X be a N (0, 1) random variable, and let Yn = X n for integers n ≥ 1. Express the ﬁrst four
terms, Y1 through Y4 , of the linear innovations sequence of Y in terms of U .
3.28 Linear innovations and orthogonal polynomials for the uniform distribution
(a) Let U be uniformly distributed on the interval [−1, 1]. Show that for integers n ≥ 0,
E [U n ] = 1
n+1 0 n even
n odd (b) Let Yn = U n for integers n ≥ 0. Note that Y0 ≡ 1. Express the ﬁrst four terms, Y1 through Y4 ,
of the linear innovations sequence of Y in terms of U .
3.29 Representation of three random variables with equal cross covariances
Let K be a matrix of the form 1aa
K = a 1 a ,
aa1
where a ∈ R.
(a) For what values of a is K the covariance matrix of some random vector?
(b) Let a have one of the values found in part (a). Fill in the missing entries of the matrix U, 1
∗ ∗ √3 1
U = ∗ ∗ √3 ,
1
∗ ∗ √3
to yield an orthonormal matrix, and ﬁnd a diagonal matrix Λ with nonnegative entries, so that
1
if Z is a three dimensional random vector with Cov(Z ) = I, then U Λ 2 Z has covariance matrix
K. (Hint: It happens that the matrix U can be selected independently of a. Also, 1 + 2a is an
eigenvalue of K.)
93 3.30 Kalman ﬁlter for a rotating state
Consider the Kalman state and observation equations for the following matrices, where θo = 2π/10
(the matrices don’t depend on time, so the subscript k is omitted):
F = (0.99) cos(θo ) − sin(θo )
sin(θo ) cos(θo ) H= 1
0 Q= 10
01 R=1 (a) Explain in words what successive iterates F n xo are like, for a nonzero initial state xo (this is
the same as the state equation, but with the random term wk left oﬀ).
(b) Write out the Kalman ﬁlter equations for this example, simplifying as much as possible (but
no more than possible! The equations don’t simplify all that much.)
3.31 * Proof of the orthogonality principle
Prove the seven statements lettered (a)(g) in what follows.
Let X be a random variable and let V be a collection of random variables on the same probability
space such that
(i) E [Z 2 ] < +∞ for each Z ∈ V
(ii) V is a linear class, i.e., if Z, Z ∈ V then so is aZ + bZ for any real numbers a and b.
(iii) V is closed in the sense that if Zn ∈ V for each n and Zn converges to a random variable Z in
the mean square sense, then Z ∈ V .
The Orthogonality Principle is that there exists a unique element Z ∗ ∈ V so that E [(X − Z ∗ )2 ] ≤
E [(X − Z )2 ] for all Z ∈ V . Furthermore, a random variable W ∈ V is equal to Z ∗ if and only if
(X − W ) ⊥ Z for all Z ∈ V . ((X − W ) ⊥ Z means E [(X − W )Z ] = 0.)
The remainder of this problem is aimed at a proof. Let d = inf {E [(X − Z )2 ] : Z ∈ V}. By deﬁnition
of inﬁmum there exists a sequence Zn ∈ V so that E [(X − Zn )2 ] → d as n → +∞.
(a) The sequence Zn is Cauchy in the mean square sense.
(Hint: Use the “parallelogram law”: E [(U − V )2 ] + E [(U + V )2 ] = 2(E [U 2 ] + E [V 2 ]). Thus, by the
Cauchy criteria, there is a random variable Z ∗ such that Zn converges to Z ∗ in the mean square
sense.
(b) Z ∗ satisﬁes the conditions advertised in the ﬁrst sentence of the principle.
(c) The element Z ∗ satisfying the condition in the ﬁrst sentence of the principle is unique. (Consider
two random variables that are equal to each other with probability one to be the same.) This
completes the proof of the ﬁrst sentence.
(d) (“if” part of second sentence). If W ∈ V and (X − W ) ⊥ Z for all Z ∈ V , then W = Z ∗ .
(The “only if” part of second sentence is divided into three parts:)
(e) E [(X − Z ∗ − cZ )2 ] ≥ E [(X − Z ∗ )2 ] for any real constant c.
(f) −2cE [(X − Z ∗ )Z ] + c2 E [Z 2 ] ≥ 0 for any real constant c.
(g) (X − Z ∗ ) ⊥ Z , and the principle is proved.
3.32 * The span of two closed subspaces is closed
Check that the span, V1 ⊕ V2 , of two closed linear spaces (deﬁned in Proposition 3.2.4) is also a
closed linear space. A hint for showing that V is closed is to use the fact that if (Zn ) is a m.s.
convergent sequence of random variables in V , then each variable in the sequence can be represented
as Zn = Zn,1 + Zn,2 , where Zn,i ∈ Vi , and E [(Zn − Zm )2 ] = E [(Zn,1 − Zm,1 )2 ] + E [(Zn,2 − Zm,2 )2 ].
3.33 * Von Neumann’s alternating projections algorithm
Let V1 and V2 be closed linear subspaces of L2 (Ω, F , P ), and let X ∈ L2 (Ω, F , P ). Deﬁne a sequence
94 (Zn : n ≥ 0) recursively, by alternating projections onto V1 and V2 , as follows. Let Z0 = X , and
for k ≥ 0, let Z2k+1 = ΠV1 (Z2k ) and Z2k+2 = ΠV2 (Z2k+1 ). The goal of this problem is to show that
m.s.
Zn → ΠV1 ∩V2 (X ). The approach will be to establish that (Zn ) converges in the m.s. sense, by
verifying the Cauchy criteria, and then use the orthogonality principle to identify the limit. Deﬁne
D(i, j ) = E [(Zi − Zj )]2 for i ≥ 0 and j ≥ 0, and let i = D(i + 1, i) for i ≥ 0.
(a) Show that i = E [(Zi )2 ] − E [(Zi+1 )2 ].
(b) Show that ∞ i ≤ E [Z 2 ] < ∞.
i=0
(c) Use the orthogonality principle to show that for n ≥ 1 and k ≥ 0:
D(n, n + 2k + 1) = n + D(n + 1, n + 2k + 1) D(n, n + 2k + 2) = D(n, n + 2k + 1) − n+2k+1 . (d) Use the above equations to show that for n ≥ 1 and k ≥ 0,
D(n, n + 2k + 1) = n + ··· + n+k −( n+k+1 + ··· + n+2k ) D(n, n + 2k + 2) = n + ··· + n+k −( n+k+1 + ··· + n+2k+1 ). Consequently, D(n, m) ≤ m−1 i for 1 ≤ n < m, and therefore (Zn : n ≥ 0) is a Cauchy sequence,
i= n
m.s.
so Zn → Z∞ for some random variable Z∞ .
(e) Verify that Z∞ ∈ V1 ∩ V2 .
(f) Verify that (X − Z∞ ) ⊥ Z for any Z ∈ V1 ∩ V2 . (Hint: Explain why (X − Zn ) ⊥ Z for all n,
and let n → ∞.)
By the orthogonality principle, (e) and (f) imply that Z∞ = ΠV1 ∩V2 (X ). 95 96 Chapter 4 Random Processes
4.1 Deﬁnition of a random process A random process X is an indexed collection X = (Xt : t ∈ T) of random variables, all on the
same probability space (Ω, F , P ). In many applications the index set T is a set of times. If T = Z,
or more generally, if T is a set of consecutive integers, then X is called a discretetime random
process. If T = R or if T is an interval of R, then X is called a continuoustime random process.
Three ways to view a random process X = (Xt : t ∈ T) are as follows:
• For each t ﬁxed, Xt is a function on Ω.
• X is a function on T × Ω with value Xt (ω ) for given t ∈ T and ω ∈ Ω.
• For each ω ﬁxed with ω ∈ Ω, Xt (ω ) is a function of t, called the sample path corresponding
to ω .
Example 4.1.1 Suppose W1 , W2 , . . . are independent random variables with
1
P {Wk = 1} = P {Wk = −1} = 2 for each k , and suppose X0 = 0 and Xn = W1 + · · · + Wn
for positive integers n. Let W = (Wk : k ≥ 1) and X = (Xn : n ≥ 0). Then W and X are
both discretetime random processes. The index set T for X is Z+ . A sample path of W and a
corresponding sample path of X are shown in Figure 4.1. X (ω) W (ω)
k k k Figure 4.1: Typical sample paths. 97 k The following notation is used:
µX (t) = E [Xt ]
RX (s, t) = E [Xs Xt ]
CX (s, t) = Cov(Xs , Xt )
FX,n (x1 , t1 ; . . . ; xn , tn ) = P {Xt1 ≤ x1 , . . . , Xtn ≤ xn }
and µX is called the mean function , RX is called the correlation function, CX is called the covariance
function, and FX,n is called the nth order CDF. Sometimes the preﬁx “auto,” meaning “self,” is
added to the words correlation and covariance, to emphasize that only one random process is
involved.
Deﬁnition 4.1.2 A second order random process is a random process (Xt : t ∈ T) such that
2
E [Xt ] < +∞ for all t ∈ T.
The mean, correlation, and covariance functions of a second order random process are all welldeﬁned and ﬁnite.
If Xt is a discrete random variable for each t, then the nth order pmf of X is deﬁned by
pX,n (x1 , t1 ; . . . ; xn , tn ) = P {Xt1 = x1 , . . . , Xtn = xn }.
Similarly, if Xt1 , . . . , Xtn are jointly continuous random variables for any distinct t1 , . . . , tn in T,
then X has an nth order pdf fX,n , such that for t1 , . . . , tn ﬁxed, fX,n (x1 , t1 ; . . . ; xn , tn ) is the joint
pdf of Xt1 , . . . , Xtn .
Example 4.1.3 Let A and B be independent, N (0, 1) random variables. Suppose Xt = A + Bt + t2
for all t ∈ R. Let us describe the sample functions, the mean, correlation, and covariance functions,
and the ﬁrst and second order pdf’s of X .
Each sample function corresponds to some ﬁxed ω in Ω. For ω ﬁxed, A(ω ) and B (ω ) are
numbers. The sample paths all have the same shape–they are parabolas with constant second
derivative equal to 2. The sample path for ω ﬁxed has t = 0 intercept A(ω ), and minimum value
ω2
A(ω ) − B (4 ) achieved at t = − B (w) . Three typical sample paths are shown in Figure 4.2. The
2 A(ω) −B(ω)
2 t
A(ω)− B(ω)
4 2 Figure 4.2: Typical sample paths.
98 various moment functions are given by
µX (t) = E [A + Bt + t2 ] = t2
RX (s, t) = E [(A + Bs + s2 )(A + Bt + t2 )] = 1 + st + s2 t2
CX (s, t) = RX (s, t) − µX (s)µX (t) = 1 + st.
As for the densities, for each t ﬁxed, Xt is a linear combination of two independent Gaussian random
variables, and Xt has mean µX (t) = t2 and variance Var(Xt ) = CX (t, t) = 1 + t2 . Thus, Xt is a
N (t2 , 1 + t2 ) random variable. That speciﬁes the ﬁrst order pdf fX,1 well enough, but if one insists
on writing it out in all detail it is given by
fX,1 (x, t) = 1
2π (1 + t2 ) exp − (x − t2 )2
2(1 + t2 ) . For s and t ﬁxed distinct numbers, Xs and Xt are jointly Gaussian and their covariance matrix
is given by
Cov Xs
Xt = 1 + s2 1 + st
1 + st 1 + t2 . The determinant of this matrix is (s − t)2 , which is nonzero. Thus X has a second order pdf fX,2 .
For most purposes, we have already written enough about fX,2 for this example, but in full detail
it is given by
fX,2 (x, s; y, t) = 1
1
exp −
2π s − t
2 x − s2
y − t2 T 1 + s2 1 + st
1 + st 1 + t2 −1 x − s2
y − t2 . The nth order distributions of X for this example are joint Gaussian distributions, but densities
don’t exist for n ≥ 3 because Xt1 , Xt2 , and Xt3 are linearly dependent for any t1 , t2 , t3 .
A random process (Xt : t ∈ T) is said to be Gaussian if the random variables Xt : t ∈ T
comprising the process are jointly Gaussian. The process X in the example just discussed is
Gaussian. All the ﬁnite order distributions of a Gaussian random process X are determined by the
mean function µX and autocorrelation function RX . Indeed, for any ﬁnite subset {t1 , t2 , . . . , tn }
of T, (Xt1 , . . . , Xtn )T is a Gaussian vector with mean (µX (t1 ), . . . , µX (tn ))T and covariance matrix
with ij th element CX (ti , tj ) = RX (ti , tj ) − µX (ti )µX (tj ). Two or more random processes are said
to be jointly Gaussian if all the random variables comprising the processes are jointly Gaussian.
Example 4.1.4 Let U = (Uk : k ∈ Z) be a random process such that the random variables
1
Uk : k ∈ Z are independent, and P [Uk = 1] = P [Uk = −1] = 2 for all k . Let X = (Xt : t ∈ R)
be the random process obtained by letting Xt = Un for n ≤ t < n + 1 for any n. Equivalently,
Xt = U t . A sample path of U and a corresponding sample path of X are shown in Figure 4.3.
Both random processes have zero mean, so their covariance functions are equal to their correlation
function and are given by
RU (k, l) = 1 if k = l
0 else RX (s, t) =
99 1 if s = t
0 else . Uk Xt
k t Figure 4.3: Typical sample paths.
The random variables of U are discrete, so the nth order pmf of U exists for all n. It is given by
pU,n (x1 , k1 ; . . . ; xn , kn ) = 2−n if (x1 , . . . , xn ) ∈ {−1, 1}n
0
else for distinct integers k1 , . . . , kn . The nth order pmf of X exists for the same reason, but it is a
bit more diﬃcult to write down. In particular, the joint pmf of Xs and Xt depends on whether
s = t . If s = t then Xs = Xt and if s = t then Xs and Xt are independent. Therefore,
the second order pmf of X is given as follows: fX,2 (x1 , t1 ; x2 , t2 ) = 4.2 1
2
1
4 if t1 = t2 and either x1 = x2 = 1 or x1 = x2 = −1 0 else. if t1 = t2 and x1 , x2 ∈ {−1, 1} Random walks and gambler’s ruin Suppose p is given with 0 < p < 1. Let W1 , W2 , . . . be independent random variables with
P {Wi = 1} = p and P {Wi = −1} = 1 − p for i ≥ 1. Suppose X0 is an integer valued random
variable independent of (W1 , W2 , . . .), and for n ≥ 1, deﬁne Xn by Xn = X0 + W1 + · · · + Wn .
A sample path of X = (Xn : n ≥ 0) is shown in Figure 4.4. The random process X is called a
random walk. Write Pk and Ek for conditional probabilities and conditional expectations given
that X0 = k . For example, Pk [A] = P [A  X0 = k ] for any event A. Let us summarize some of the
basic properties of X .
• Ek [Xn ] = k + n(2p − 1).
• Vark (Xn ) = Var(k + W1 + · · · + Wn ) = 4np(1 − p).
• limn→∞ Xn
n • limn→∞ Pk = 2p − 1 (a.s. and m.s. under Pk , k ﬁxed).
Xn −n(2p−1)
√
4np(1−p) ≤c • Pk {Xn = k + j − (n − j )} = = Φ(c).
n
j pj (1 − p)n−j for 0 ≤ j ≤ n. Almost all the properties listed are properties of the one dimensional distributions of X . In
fact, only the strong law of large numbers, giving the a.s. convergence in the third property listed,
depends on the joint distribution of the Xn ’s.
100 Xn (ω)
b k n Figure 4.4: A typical sample path.
The socalled Gambler’s Ruin problem is a nice example of the calculation of a probability
involving the joint distributions of the random walk X . Interpret Xn as the number of units of
money a gambler has at time n. Assume that the initial wealth k satisﬁes k ≥ 0, and suppose the
gambler has a goal of accumulating b units of money for some positive integer b ≥ k . While the
random walk (Xn : n ≥ 0) continues on forever, we are only interested in it until it hits either 0 or
b. Let Rb denote the event that the gambler is eventually ruined, meaning the random walk reaches
zero without ﬁrst reaching b. The gambler’s ruin probability is Pk [Rb ]. A simple idea allows us to
compute the ruin probability. The idea is to condition on the value of the ﬁrst step W1 , and then
to recognize that after the ﬁrst step is taken, the conditional probability of ruin is the same as the
unconditional probability of ruin for initial wealth k + W1 .
Let rk = Pk [Rb ] for 0 ≤ k ≤ b, so rk is the ruin probability for the gambler with initial wealth
k and target wealth b. Clearly r0 = 1 and rb = 0. For 1 ≤ k ≤ b − 1, condition on W1 to yield
rk = Pk {W1 = 1}Pk [Rb  W1 = 1] + Pk {W1 = −1}Pk [Rb  W1 = −1]
or rk = prk+1 + (1 − p)rk−1 . This yields b − 1 linear equations for the b − 1 unknowns r1 , . . . , rb−1 .
1
If p = 2 the equations become rk = 1 {rk−1 + rk+1 } so that rk = A + Bk for some constants A
2
and B . Using the boundary conditions r0 = 1 and rb = 0, we ﬁnd that rk = 1 − k in case p = 1 .
b
2
Note that, interestingly enough, after the gambler stops playing, he’ll have b units with probability
k
b and zero units otherwise. Thus, his expected wealth after completing the game is equal to his
initial capital, k .
k
k
If p = 1 , we seek a solution of the form rk = Aθ1 + Bθ2 , where θ1 and θ2 are the two roots of the
2
2 + (1 − p) and A, B are selected to meet the two boundary conditions.
quadratic equation θ = pθ
The roots are 1 and 1−p , and ﬁnding A and B yields, that if p = 1
p
2 rk = 1−p
p k 1− −
1−p
p 1−p
p
b 101 b 0 ≤ k ≤ b. Focus, now, on the case that p > 1 . By the law of large numbers, Xn → 2p − 1 a.s. as n → ∞.
2
n
This implies, in particular, that Xn → +∞ a.s. as n → ∞. Thus, unless the gambler is ruined
in ﬁnite time, his capital converges to inﬁnity. Let R be the event that the gambler is eventually
ruined. The events Rb increase with b because if b is larger the gambler has more possibilities to
be ruined before accumulating b units of money: Rb ⊂ Rb+1 ⊂ · · · and R = ∪∞ k Rb . Therefore by
b=
the countable additivity of probability,
Pk [R] = lim Pk [Rb ] = b→∞ lim rk b→∞ = 1−p
p k . Thus, the probability of eventual ruin decreases geometrically with the initial wealth k . 4.3 Processes with independent increments and martingales The increment of a random process X = (Xt : t ∈ T) over an interval [a, b] is the random variable
Xb − Xa . A random process is said to have independent increments if for any positive integer n and
any t0 < t1 < · · · < tn in T, the increments Xt1 − Xt0 , . . . , Xtn − Xtn−1 are mutually independent.
A random process (Xt : t ∈ T) is called a martingale if E [Xt ] is ﬁnite for all t and for any
positive integer n and t1 < t2 < · · · < tn < tn+1 ,
E [Xtn+1  Xt1 , . . . , Xtn ] = Xtn
or, equivalently,
E [Xtn+1 − Xtn  Xt1 , . . . , Xtn ] = 0.
If tn is interpreted as the present time, then tn+1 is a future time and the value of (Xt1 , . . . , Xtn )
represents information about the past and present values of X . With this interpretation, the
martingale property is that the future increments of X have conditional mean zero, given the past
and present values of the process.
An example of a martingale is the following. Suppose a gambler has initial wealth X0 . Suppose
the gambler makes bets with various odds, such that, as far as the past history of X can determine,
the bets made are all for fair games in which the expected net gains are zero. Then if Xt denotes
the wealth of the gambler at any time t ≥ 0, then (Xt : t ≥ 0) is a martingale.
Suppose (Xt ) is an independent increment process with index set T = R+ or T = Z+ , with X0
equal to a constant and with mean zero increments. Then X is a martingale, as we now show.
Let t1 < · · · < tn+1 be in T. Then (Xt1 , . . . , Xtn ) is a function of the increments Xt1 − X0 , Xt2 −
Xt1 , . . . , Xtn − Xtn−1 , and hence it is independent of the increment Xtn+1 − Xtn . Thus
E [Xtn+1 − Xtn  Xt1 , . . . , Xtn ] = E [Xtn+1 − Xtn ] = 0.
The random walk (Xn : n ≥ 0) arising in the gambler’s ruin problem is an independent increment
process, and if p = 1 it is also a martingale.
2
The following proposition is stated, without proof, to give an indication of some of the useful
deductions that follow from the martingale property.
102 Proposition 4.3.1 (a) Let X0 , X1 , X2 , . . . be nonnegative random variables such that
E [Xk+1  X0 , . . . , Xk ] ≤ Xk for k ≥ 0 (such X is a nonnegative supermartingale). Then
P max Xk 0≤k≤n ≥γ ≤ E [X0 ]
.
γ 2
(b) (Doob’s L2 Inequality) Let X0 , X1 , . . . be a martingale sequence with E [Xn ] < +∞ for some
n. Then
2 E 4.4 max Xk 0≤k≤n 2
≤ 4E [Xn ]. Brownian motion A Brownian motion, also called a Wiener process, is a random process W = (Wt : t ≥ 0) such that
B.0 P {W0 = 0} = 1.
B.1 W has independent increments.
B.2 Wt − Ws has the N (0, σ 2 (t − s)) distribution for t ≥ s.
B.3 P [Wt is a continuous function of t] = 1, or in other words, W is sample path continuous with
probability one.
A typical sample path of a Brownian motion is shown in Figure 4.5. A Brownian motion, being a
X (ω )
t t Figure 4.5: A typical sample path of Brownian motion.
mean zero independent increment process with P {W0 = 0} = 1, is a martingale.
The mean, correlation, and covariance functions of a Brownian motion W are given by
µW (t) = E [Wt ] = E [Wt − W0 ] = 0
and, for s ≤ t,
RW (s, t) = E [Ws Wt ]
= E [(Ws − W0 )(Ws − W0 + Wt − Ws )]
= E [(Ws − W0 )2 ] = σ 2 s
103 so that, in general,
CW (s, t) = RW (s, t) = σ 2 (s ∧ t).
A Brownian motion is Gaussian, because if 0 = t0 ≤ t1 ≤ · · · ≤ tn , then each coordinate of the
vector (Wt1 , . . . , Wtn ) is a linear combination of the n independent Gaussian random variables
(Wti − Wti−1 : 1 ≤ i ≤ n). Thus, properties B.0–B.2 imply that W is a Gaussian random process
with µW = 0 and RW (s, t) = σ 2 (s ∧ t). In fact, the converse is also true. If W = (Wt : t ≥ 0) is a
Gaussian random process with mean zero and RW (s, t) = σ 2 (s ∧ t), then B.0–B.2 are true.
Property B.3 does not come automatically. For example, if W is a Brownian motion and if U
is a Unif(0,1) distributed random variable independent of W , let W be deﬁned by
Wt = Wt + I{U =t} .
Then P {Wt = Wt } = 1 for each t ≥ 0 and W also satisﬁes B.0–B.2, but W fails to satisfy B.3.
Thus, W is not a Brownian motion. The diﬀerence between W and W is signiﬁcant if events
involving uncountably many values of t are investigated. For example,
P {Wt ≤ 1 for 0 ≤ t ≤ 1} = P {Wt ≤ 1 for 0 ≤ t ≤ 1}. 4.5 Counting processes and the Poisson process A function f on R+ is called a counting function if f (0) = 0, f is nondecreasing, f is right
continuous, and f is integer valued. The interpretation is that f (t) is the number of “counts”
observed during the interval (0, t]. An increment f (b) − f (a) is the number of counts in the interval
(a, b]. If ti denotes the time of the ith count for i ≥ 1, then f can be described by the sequence
(ti ). Or, if u1 = t1 and ui = ti − ti−1 for i ≥ 2, then f can be described by the sequence (ui ). See
Figure 4.6. The numbers t1 , t2 , . . . are called the count times and the numbers u1 , u2 , . . . are called f(t)
3
2
1
0 u1
0 u2 u3 t1 t2 t
t3 Figure 4.6: A counting function.
the intercount times. The following equations clearly hold:
∞ f (t) = I{t≥tn }
n=1 tn = min{t : f (t) ≥ n}
tn = u1 + · · · + un .
104 A random process is called a counting process if with probability one its sample path is a
counting function. A counting process has two corresponding random sequences, the sequence of
count times and the sequence of intercount times.
The most widely used example of a counting process is a Poisson process, deﬁned next.
Deﬁnition 4.5.1 Let λ ≥ 0. By deﬁnition, a Poisson process with rate λ is a random process
N = (Nt : t ≥ 0) such that
N.1 N is a counting process,
N.2 N has independent increments,
N.3 N (t) − N (s) has the P oi(λ(t − s)) distribution for t ≥ s.
Proposition 4.5.2 Let N be a counting process and let λ > 0. The following are equivalent:
(a) N is a Poisson process with rate λ.
(b) The intercount times U1 , U2 , . . . are mutually independent, Exp(λ) random variables.
(c) For each τ > 0, Nτ is a Poisson random variable with parameter λτ , and given {Nτ = n}, the
times of the n counts during [0, τ ] are the same as n independent, Unif[0, τ ] random variables,
reordered to be nondecreasing. That is, for any n ≥ 1, the conditional density of the ﬁrst n
count times, (T1 , . . . , Tn ), given the event {Nτ = n}, is:
n!
τn f (t1 , . . . , tn Nτ = n) = 0 0 < t1 < · · · < tn ≤ τ
else (4.1) Proof. It will be shown that (a) implies (b), (b) implies (c), and (c) implies (a).
(a) implies (b). Suppose N is a Poisson process. The joint pdf of the ﬁrst n count times
T1 , . . . , Tn can be found as follows. Let 0 < t1 < t2 < · · · < tn . Select > 0 so small that (t1 − , t1 ],
(t2 − , t2 ], . . . , (tn − , tn ] are disjoint intervals of R+ . Then the probability that (T1 , . . . , Tn ) is in
the ndimensional cube with upper corner t1 , . . . , tn and sides of length is given by
P {Ti ∈ (ti − , ti ] for 1 ≤ i ≤ n}
= P {Nt1 − = 0, Nt1 − Nt1 − = 1, Nt2 − − Nt1 = 0, . . . , Ntn − Ntn − = 1}
= (e−λ(t1 − ) )(λ e−λ )(e−λ(t2 − −t1 ) ) · · · (λ e−λ ) = (λ )n e−λtn .
The volume of the cube is n. Therefore (T1 , . . . , Tn ) has the pdf fT1 ···Tn (t1 , . . . , tn ) = λn e−λtn
0 if 0 < t1 < · · · < tn
else. (4.2) The vector (U1 , . . . , Un ) is the image of (T1 , . . . , Tn ) under the mapping (t1 , . . . , tn ) → (u1 , . . . , un )
deﬁned by u1 = t1 , uk = tk − tk−1 for k ≥ 2. The mapping is invertible, because tk = u1 + · · · + uk
105 for 1 ≤ k ≤ n, it has range Rn , and the Jacobian
+ 1 −1
1
∂u −1
1
= ∂t
.. . .. .
−1 1 has unit determinant. Therefore, by the formula for the transformation of random vectors (see
Section 1.10),
λn e−λ(u1 +···+un ) u ∈ Rn
+.
fU1 ...Un (u1 , . . . , un ) =
(4.3)
0
else
The joint pdf in (4.3) factors into the product of n pdfs, with each pdf being for an Exp(λ) random
variable. Thus the intercount times U1 , U2 , . . . are independent and each is exponentially distributed
with parameter λ. So (a) implies (b).
(b) implies (c). Suppose that N is a counting process such that the intercount times U1 , U2 , . . .
are independent, Exp(λ) random variables, for some λ > 0. Thus, for n ≥ 1, the ﬁrst n intercount
times have the joint pdf given in (4.3). Equivalently, appealing to the transformation of random
vectors in the reverse direction, the pdf of the ﬁrst n count times, (T1 , . . . , Tn ), is given by (4.2). Fix
τ > 0 and an integer n ≥ 1. The event {Nτ = n} is equivalent to the event (T1 , . . . , Tn+1 ) ∈ An,τ ,
where
n
An,τ = {t ∈ R++1 : 0 < t1 < · · · < tn ≤ τ < tn+1 }.
The conditional pdf of (T1 , . . . , Tn+1 ), given that {Nτ = n}, is obtained by starting with the joint
pdf of (T1 , . . . , Tn+1 ), namely λn+1 e−λ(tn+1 ) on {t ∈ Rn+1 : 0 < t1 < · · · < tn+1 }, setting it equal to
zero oﬀ of the set An,τ , and scaling it up by the factor 1/P {Nτ = n} on An,τ :
f (t1 , . . . , tn+1 Nτ = n) = λn+1 e−λtn+1
P {Nτ =n} 0 0 < t1 < · · · < tn ≤ τ < tn+1
else (4.4) The joint density of (T1 , . . . , Tn ), given that {Nτ = n}, is obtained for each (t1 , . . . , tn ) by integrating
the density in (4.4) with respect to tn+1 over R. If 0 < t1 < · · · < tn ≤ τ does not hold, the density
in (4.4) is zero for all values of tn+1 . If 0 < t1 < · · · < tn ≤ τ , then the density in (4.4) is nonzero
for tn+1 ∈ (τ, ∞). Integrating (4.4) with respect to tn+1 over (τ, ∞) yields:
f (t1 , . . . , tn Nτ = n) = λn e−λτ
P {Nτ =n} 0 0 < t1 < · · · < tn ≤ τ
else (4.5) The conditional density in (4.5) is constant over the set {t ∈ Rn : 0 < t1 < · · · < tn ≤ τ }. Since the
+
density must integrate to one, that constant must be the reciprocal of the ndimensional volume
of the set. The unit cube [0, τ ]n in Rn has volume τ n . It can be partitioned into n! equal volume
subsets corresponding to the n! possible orderings of the numbers t1 , . . . , tn . Therefore, the set
{t ∈ Rn : 0 ≤ t1 < · · · < tn ≤ τ }, corresponding to one of the orderings, has volume τ n /n!. Hence,
+
n −λτ (4.5) implies both that (4.1) holds and that P {Nτ = n} = (λτ )ne . These implications are for
!
n ≥ 1. Also, P {Nτ = 0} = P {U1 > τ } = e−λτ . Thus, Nτ is a Poi(λτ ) random variable.
106 (c) implies (a). Suppose t0 < t1 < . . . < tk and let n1 , . . . , nk be nonnegative integers. Set
n = n1 + . . . + nk and pi = (ti − ti−1 )/tk for 1 ≤ i ≤ k . Suppose (c) is true. Given there are n counts
in the interval [0, τ ], by (c), the distribution of the numbers of counts in each subinterval is as if
each of the n counts is thrown into a subinterval at random, falling into the ith subinterval with
probability pi . The probability that, for 1 ≤ i ≤ K , ni particular counts fall into the ith interval,
is pn1 · · · pnk . The number of ways to assign n counts to the intervals such that there are ki counts
1
k
n
in the ith interval is n1 ··· nk = n1 !n!nk ! . This thus gives rise to what is known as a multinomial
···
distribution for the numbers of counts per interval. We have
P {N (ti ) − N (ti−1 ) = ni for 1 ≤ i ≤ k }
= P {N (tk ) = n} P [N (ti ) − N (ti−1 ) = ni for 1 ≤ i ≤ k  N (tk ) = n]
n
(λtk )n e−λtk
pn1 · · · pnk
=
k
n!
n1 · · · nk 1
k =
i=1 (λ(ti − ti−1 ))ni e−λ(ti −ti−1 )
ni ! Therefore the increments N (ti ) − N (ti−1 ), 1 ≤ i ≤ k , are independent, with N (ti ) − N (ti−1 ) being
a Poisson random variable with mean λ(ti − ti−1 ), for 1 ≤ i ≤ k. So (a) is proved.
A Poisson process is not a martingale. However, if N is deﬁned by Nt = Nt − λt, then N is an
independent increment process with mean 0 and N0 = 0. Thus, N is a martingale. Note that N
has the same mean and covariance function as a Brownian motion with σ 2 = λ, which shows how
little one really knows about a process from its mean function and correlation function alone. 4.6 Stationarity Consider a random process X = (Xt : t ∈ T) such that either T = Z or T = R. Then X is said to be
stationary if for any t1 , . . . , tn and s in T, the random vectors (Xt1 , . . . , Xtn ) and (Xt1 +s , . . . , Xtn +s )
have the same distribution. In other words, the joint statistics of X of all orders are unaﬀected by
a shift in time. The condition of stationarity of X can also be expressed in terms of the CDF’s of
X : X is stationary if for any n ≥ 1, s, t1 , . . . , tn ∈ T, and x1 , . . . , xn ∈ R,
FX,n (x1 , t1 ; . . . ; xn , tn ) = FX,n (x1 , t1 + s; . . . ; xn ; tn + s).
Suppose X is a stationary second order random process. (Recall that second order means that
2
E [Xt ] < ∞ for all t.) Then by the n = 1 part of the deﬁnition of stationarity, Xt has the same
2
distribution for all t. In particular, µX (t) and E [Xt ] do not depend on t. Moreover, by the n = 2
2
part of the deﬁnition E [Xt1 Xt2 ] = E [Xt1 +s Xt2 +s ] for any s ∈ T. If E [Xt ] < +∞ for all t, then
E [Xt+s ] and RX (t1 + s, t2 + s) are ﬁnite and both do not depend on s.
A second order random process (Xt : t ∈ T) with T = Z or T = R is called wide sense stationary
(WSS) if
µX (t) = µX (s + t) and RX (t1 , t2 ) = RX (t1 + s, t2 + s)
for all t, s, t1 , t2 ∈ T. As shown above, a stationary second order random process is WSS. Wide
sense stationarity means that µX (t) is a ﬁnite number, not depending on t, and RX (t1 , t2 ) depends
107 on t1 , t2 only through the diﬀerence t1 − t2 . By a convenient and widely accepted abuse of notation,
if X is WSS, we use µX to be the constant and RX to be the function of one real variable such that
t∈T E [Xt ] = µX
E [Xt1 Xt2 ] = RX (t1 − t2 ) t1 , t2 ∈ T. The dual use of the notation RX if X is WSS leads to the identity RX (t1 , t2 ) = RX (t1 − t2 ). As a
practical matter, this means replacing a comma by a minus sign. Since one interpretation of RX
requires it to have two arguments, and the other interpretation requires only one argument, the
interpretation is clear from the number of arguments. Some brave authors even skip mentioning
that X is WSS when they write: “Suppose (Xt : t ∈ R) has mean µX and correlation function
RX (τ ),” because it is implicit in this statement that X is WSS.
Since the covariance function CX of a random process X satisﬁes
CX (t1 , t2 ) = RX (t1 , t2 ) − µX (t1 )µX (t2 ),
if X is WSS then CX (t1 , t2 ) is a function of t1 − t2 . The notation CX is also used to denote the
function of one variable such that CX (t1 − t2 ) = Cov(Xt1 , Xt2 ). Therefore, if X is WSS then
CX (t1 − t2 ) = CX (t1 , t2 ). Also, CX (τ ) = RX (τ ) − µ2 , where in this equation τ should be thought
X
of as the diﬀerence of two times, t1 − t2 .
In general, there is much more to know about a random vector or a random process than
the ﬁrst and second moments. Therefore, one can mathematically deﬁne WSS processes that
are spectacularly diﬀerent in appearance from any stationary random process. For example, any
random process (Xk : k ∈ Z) such that the Xk are independent with E [Xk ] = 0 and Var(Xk ) = 1
for all k is WSS. To be speciﬁc, we could take the Xk to be independent, with Xk being N (0, 1)
for k ≤ 0 and with Xk having pmf
pX,1 (x, k ) = P {Xk = x} = 1
2k2 1− 1
k2 x ∈ {k, −k }
if x = 0 0 else
for k ≥ 1. A typical sample path of this WSS random process is shown in Figure 4.7. Xk k Figure 4.7: A typical sample path.
The situation is much diﬀerent if X is a Gaussian process. Indeed, suppose X is Gaussian and
WSS. Then for any t1 , t2 , . . . , tn , s ∈ T, the random vector (Xt1 +s , Xt2 +s , . . . , Xtn +s )T is Gaussian
108 with mean (µ, µ, . . . , µ)T and covariance matrix with ij th entry CX ((ti + s) − (tj + s)) = CX (ti − tj ).
This mean and covariance matrix do not depend on s. Thus, the distribution of the vector does
not depend on s. Therefore, X is stationary.
In summary, if X is stationary then X is WSS, and if X is both Gaussian and WSS, then X is
stationary.
Example 4.6.1 Let Xt = A cos(ωc t +Θ), where ωc is a nonzero constant, A and Θ are independent
random variables with P {A > 0} = 1 and E [A2 ] < +∞. Each sample path of the random process
(Xt : t ∈ R) is a pure sinusoidal function at frequency ωc radians per unit time, with amplitude A
and phase Θ.
We address two questions. First, what additional assumptions, if any, are needed on the distributions of A and Θ to imply that X is WSS? Second, we consider two distributions for Θ which
each make X WSS, and see if they make X stationary.
To address whether X is WSS, the mean and correlation functions can be computed as follows.
Since A and Θ are independent and since cos(ωc t + Θ) = cos(ωc t) cos(Θ) − sin(ωc t) sin(Θ),
µX (t) = E [A] (E [cos(Θ)] cos(ωc t) − E [sin(Θ)] sin(ωc t)) .
Thus, the function µX (t) is a linear combination of cos(ωc t) and sin(ωc t). The only way such a
linear combination can be independent of t is if the coeﬃcients of both cos(ωc t) and sin(ωc t) are
zero (in fact, it is enough to equate the values of µX (t) at ωc t = 0, π , and π ). Therefore, µX (t)
2
does not depend on t if and only if E [cos(Θ)] = E [sin(Θ)] = 0.
Turning next to RX , using the trigonometric identity cos(a) cos(b) = (cos(a − b) + cos(a + b))/2
yields
RX (s, t) = E [A2 ]E [cos(ωc s + Θ) cos(ωc t + Θ)]
E [A2 ]
=
{cos(ωc (s − t)) + E [cos(ωc (s + t) + 2Θ)]} .
2
Since s + t can be arbitrary for s − t ﬁxed, in order that RX (s, t) be a function of s − t alone it is
necessary that E [cos(ωc (s + t) + 2Θ)] be a constant, independent of the value of s + t. Arguing just
as in the case of µX , with Θ replaced by 2Θ, yields that RX (s, t) is a function of s − t if and only
if E [cos(2Θ)] = E [sin(2Θ)] = 0.
Combining the ﬁndings for µX and RX , yields that X is WSS, if and only if,
E [cos(Θ)] = E [sin(Θ)] = E [cos(2Θ)] = E [sin(2Θ)] = 0.
There are many distributions for Θ in [0, 2π ] such that the four moments speciﬁed are zero. Two
possibilities are (a) Θ is uniformly distributed on the interval [0, 2π ], or, (b) Θ is a discrete random
π
variable, taking the four values 0, π , π , 32 with equal probability. Is X stationary for either
2
possibility?
We shall show that X is stationary if Θ is uniformly distributed over [0, 2π ]. Stationarity means
that for any ﬁxed constant s, the random processes (Xt : t ∈ R) and (Xt+s : t ∈ R) have the same
ﬁnite order distributions. For this example,
˜
Xt+s = A cos(ωc (t + s) + Θ) = A cos(ωc t + Θ)
˜
˜
where Θ = ((ωc s + Θ) mod 2π ). By Example 1.4.4, Θ is again uniformly distributed on the interval
˜ have the same joint distribution, so A cos(ωc t+Θ) and A cos(ωc t+Θ)
˜
[0, 2π ]. Thus (A, Θ) and (A, Θ)
109 have the same ﬁnite order distributions. Hence, X is indeed stationary if Θ is uniformly distributed
over [0, 2π ].
π
Assume now that Θ takes on each of the values of 0, π , π , and 32 with equal probability. Is X
2
stationary? If X were stationary then, in particular, Xt would have the same distribution for all t.
π
On one hand, P {X0 = 0} = P {Θ = π or Θ = 32 } = 1 . On the other hand, if ωc t is not an integer
2
2
π
multiple of 2 , then ωc t + Θ cannot be an integer multiple of π , so P {Xt = 0} = 0. Hence X is not
2
stationary.
(With more work it can be shown that X is stationary, if and only if, (Θ mod 2π ) is uniformly
distributed over the interval [0, 2π ].) 4.7 Joint properties of random processes Two random processes X and Y are said to be jointly stationary if their parameter set T is either
Z or R, and if for any t1 , . . . , tn , s ∈ T, the distribution of the random vector
(Xt1 +s , Xt2 +s , . . . , Xtn +s , Yt1 +s , Yt2 +s , . . . , Ytn +s )
does not depend on s.
The random processes X and Y are said to be jointly Gaussian if all the random variables
comprising X and Y are jointly Gaussian.
If X and Y are second order random processes on the same probability space, the cross correlation function, RXY , is deﬁned by RXY (s, t) = E [Xs Yt ], and the cross covariance function, CXY ,
is deﬁned by CXY (s, t) = Cov(Xs , Yt ).
The random processes X and Y are said to be jointly WSS, if X and Y are each WSS, and
if RXY (s, t) is a function of s − t. If X and Y are jointly WSS, we use RXY (τ ) for RXY (s, t)
where τ = s − t, and similarly CXY (s − t) = CXY (s, t). Note that CXY (s, t) = CY X (t, s), so
CXY (τ ) = CY X (−τ ). 4.8 Conditional independence and Markov processes Markov processes are naturally associated with the state space approach for modeling a system.
The idea of a state space model for a given system is to deﬁne the state of the system at any given
time t. The state of the system at time t should summarize everything about the system up to and
including time t that is relevant to the future of the system. For example, the state of an aircraft
at time t could consist of the position, velocity, and remaining fuel at time t. Think of t as the
present time. The state at time t determines the possible future part of the aircraft trajectory. For
example, it determines how much longer the aircraft can ﬂy and where it could possibly land. The
state at time t does not completely determine the entire past trajectory of the aircraft. Rather,
the state summarizes enough about the system up to the present so that if the state is known,
no more information about the past is relevant to the future possibilities. The concept of state is
inherent in the Kalman ﬁltering model discussed in Chapter 3. The notion of state is captured for
random processes using the notion of conditional independence and the Markov property, which
are discussed next.
Let X, Y, Z be random vectors. We shall deﬁne the condition that X and Z are conditionally
independent given Y . Such condition is denoted by X − Y − Z . If X, Y, Z are discrete, then
110 X − Y − Z is deﬁned to hold if
P [X = i, Z = k  Y = j ] = P [X = i  Y = j ]P [Z = k  Y = j ] (4.6) for all i, j, k with P {Y = j } > 0. Equivalently, X − Y − Z if
P {X = i, Y = j, Z = k }P {Y = j } = P {X = i, Y = j }P {Z = k, Y = j } (4.7) for all i, j, k . Equivalently again, X  Y  Z if
P [Z = k  X = i, Y = j ] = P [Z = k  Y = j ] (4.8) for all i, j, k with P {X = i, Y = j } > 0. The forms (4.6) and (4.7) make it clear that the condition
X − Y − Z is symmetric in X and Z : thus X − Y − Z is the same condition as Z − Y − X . The
form (4.7) does not involve conditional probabilities, so no requirement about conditioning events
having positive probability is needed. The form (4.8) shows that X − Y − Z means that knowing
Y alone is as informative as knowing both X and Y , for the purpose of determining conditional
probabilies of Z . Intuitively, the condition X − Y − Z means that the random variable Y serves
as a state.
If X, Y , and Z have a joint pdf, then the condition X − Y − Z can be deﬁned using the
pdfs and conditional pdfs in a similar way. For example, the conditional independence condition
X − Y − Z holds by deﬁnition if
fXZ Y (x, z y ) = fX Y (xy )fZ Y (z y ) whenever fY (y ) > 0 An equivalent condition is
fZ XY (z x, y ) = fZ Y (z y ) whenever fXY (x, y ) > 0. (4.9) Example 4.8.1 Suppose X, Y, Z are jointly Gaussian vectors. Let us see what the condition
X − Y − Z means in terms of the covariance matrices. Assume without loss of generality that the
vectors have mean zero. Because X, Y , and Z are jointly Gaussian, the condition (4.9) is equivalent
to the condition that E [Z X, Y ] = E [Z Y ] (because given X, Y , or just given Y , the conditional
distribution of Z is Gaussian, and in the two cases the mean and covariance of the conditional
distribution of Z is the same.) The idea of linear innovations applied to the length two sequence
˜
˜
(Y, X ) yields E [Z X, Y ] = E [Z Y ] + E [Z X ] where X = X − E [X Y ]. Thus X − Y − Z if and only if
˜ ] = 0, or equivalently, if and only if Cov(X, Z ) = 0. Since X = X − Cov(X, Y )Cov(Y )−1 Y ,
˜
˜
E [Z X
if follows that
˜
Cov(X, Z ) = Cov(X, Z ) − Cov(X, Y )Cov(Y )−1 Cov(Y, Z ).
Therefore, X − Y − Z if and only if
Cov(X, Z ) = Cov(X, Y )Cov(Y )−1 Cov(Y, Z ). (4.10) In particular, if X, Y , and Z are jointly Gaussian random variables with nonzero variances, the
condition X − Y − Z holds if and only if the correlation coeﬃcients satisfy ρXZ = ρXY ρY Z .
111 A general deﬁnition of conditional probabilities and conditional independence, based on the
general deﬁnition of conditional expectation given in Chapter 3, is given next. Recall that
P [F ] = E [IF ] for any event F , where IF denotes the indicator function of F . If Y is a random
vector, we deﬁne P [F Y ] to equal E [IF Y ]. This means that P [F Y ] is the unique (in the sense
that any two versions are equal with probability one) random variable such that
(1) P [F Y ] is a function of Y and it has ﬁnite second moments, and
(2) E [g (Y )P [F Y ]] = E [g (Y )IA ] for any g (Y ) with ﬁnite second moment.
Given arbitrary random vectors, we deﬁne X and Z to be conditionally independent given Y ,
(written X − Y − Z ) if for any Borel sets A and B ,
P [{X ∈ A}{Z ∈ B }Y ] = P [X ∈ AY ]P [Z ∈ B Y ].
Equivalently, X − Y − Z holds if for any Borel set B , P [Z ∈ B X, Y ] = P [Z ∈ B Y ].
Deﬁnition 4.8.2 A random process X = (Xt : t ∈ T) is said to be a Markov process if for any
t1 , . . . , tn+1 in T with t1 < · · · < tn , the following conditional independence condition holds:
(Xt1 , · · · , Xtn ) − Xtn − Xtn+1 (4.11) It turns out that the Markov property is equivalent to the following conditional independence
property: For any t1 , . . . , tn+m in T with t1 < · · · < tn+m ,
(Xt1 , · · · , Xtn ) − Xtn − (Xtn , · · · , Xtn+m ) (4.12) The deﬁnition (4.11) is easier to check than condition (4.12), but (4.12) is appealing because it is
symmetric in time. In words, thinking of tn as the present time, the Markov property means that
the past and future of X are conditionally independent given the present state Xtn . Example 4.8.3 (Markov property of independent increment processes) Let (Xt : t ≥ 0) be an
independent increment process such that X0 is a constant. Then for any t1 , . . . , tn+1 with 0 ≤ t1 ≤
· · · ≤ tn+1 , the vector (Xt1 , . . . , Xtn ) is a function of the n increments Xt1 − X0 , Xt2 − Xt1 , Xtn −
Xtn−1 , and is thus independent of the increment V = Xtn+1 − Xtn . But Xtn+1 is determined by
V and Xtn . Thus, X is a Markov process. In particular, random walks, Brownian motions, and
Poisson processes are Markov processes. Example 4.8.4 (Gaussian Markov processes) Suppose X = (Xt : t ∈ T) is a Gaussian random
process with Var(Xt ) > 0 for all t. By the characterization of conditional independence for jointly
Gaussian vectors (4.10), the Markov property (4.11) is equivalent to Xt1
Xt1 Xt Xt 2
2 Cov( . , Xtn+1 ) = Cov( . , Xtn )Var(Xtn )−1 Cov(Xtn , Xtn+1 )
.
.
.
.
Xtn Xtn
112 which, letting ρ(s, t) denote the correlation coeﬃcient between Xs and Xt , is equivalent to the
requirement ρ(t1 , tn )
ρ(t1 , tn+1 ) ρ(t2 , tn ) ρ(t2 , tn+1 )) ρ(tn , tn+1 )
= .
.
.
. .
.
ρ(tn , tn+1 ) ρ(tn , tn ) Therefore a Gaussian process X is Markovian if and only if
ρ(r, t) = ρ(r, s)ρ(s, t) whenever r, s, t ∈ T with r < s < t. (4.13) If X = (Xk : k ∈ Z ) is a discretetime stationary Gaussian process, then ρ(s, t) may be written
as ρ(k ), where k = s − t. Note that ρ(k ) = ρ(−k ). Such a process is Markovian if and only if
ρ(k1 + k2 ) = ρ(k1 )ρ(k2 ) for all positive integers k1 and k2 . Therefore, X is Markovian if and only if
ρ(k ) = bk for all k , for some constant b with b ≤ 1. Equivalently, a stationary Gaussian process
X = (Xk : k ∈ Z ) with V ar(Xk ) > 0 for all k is Markovian if and only if the covariance function
has the form CX (k ) = Abk for some constants A and b with A > 0 and b ≤ 1.
Similarly, if (Xt : t ∈ R) is a continuoustime stationary Gaussian process with V ar(Xt ) > 0
for all t, X is Markovian if and only if ρ(s + t) = ρ(s)ρ(t) for all s, t ≥ 0. The only bounded realvalued functions satisfying such a multiplicative condition are exponential functions. Therefore, a
stationary Gaussian process X with V ar(Xt ) > 0 for all t is Markovian if and only if ρ has the
form ρ(τ ) = exp(−ατ ), for some constant α ≥ 0, or equivalently, if and only if CX has the form
CX (τ ) = A exp(−ατ ) for some constants A > 0 and α ≥ 0. 4.9 Discretestate Markov processes This section delves further into the theory of Markov processes in the technically simplest case of
a discrete state space. Let S be a ﬁnite or countably inﬁnite set, called the state space. Given a
probability space (Ω, F , P ), an S valued random variable is deﬁned to be a function Y mapping Ω
to S such that {ω : Y (ω ) = s} ∈ F for each s ∈ S . Assume that the elements of S are ordered
so that S = {a1 , a2 , . . . , an } in case S has ﬁnite cardinality, or S = {a1 , a2 , a3 , . . .} in case S has
inﬁnite cardinality. Given the ordering, an S valued random variable is equivalent to a positive
integer valued random variable, so it is nothing exotic. Think of the probability distribution of
an S valued random variable Y as a row vector of possibly inﬁnite dimension, called a probability
vector: pY = (P {Y = a1 }, P {Y = a2 }, . . .). Similarly think of a deterministic function g on S as
a column vector, g = (g (a1 ), g (a2 ), . . .)T . Since the elements of S may not even be numbers, it
might not make sense to speak of the expected value of an S valued random variable. However,
if g is a function mapping S to the reals, then g (Y ) is a realvalued random variable and its
expectation is given by the inner product of the probability vector pY and the column vector g :
E [g (Y )] = i∈S pY (i)g (i) = pY g . A random process X = (Xt : t ∈ T) is said to have state space
S if Xt is an S valued random variable for each t ∈ T, and the Markov property of such a random
process is deﬁned just as it is for a real valued random process.
Let (Xt : t ∈ T) be a be a Markov process with state space S . For brevity we denote the ﬁrst
order pmf of X at time t as π (t) = (πi (t) : i ∈ S ). That is, πi (t) = pX (i, t) = P {X (t) = i}. The
113 following notation is used to denote conditional probabilities:
P [Xt1 = j1 , . . . , Xtn = jn Xs1 = i1 , . . . , Xsm = im ] = pX (j1 , t1 ; . . . ; jn , tn i1 , s1 ; . . . ; im , sm )
For brevity, conditional probabilities of the form P [Xt = j Xs = i] are written as pij (s, t), and are
called the transition probabilities of X .
The ﬁrst order pmfs π (t) and the transition probabilities pij (s, t) determine all the ﬁnite order
distributions of the Markov process as follows. Given
t1 < t2 < . . . < tn in T,
ii , i2 , ..., in ∈ S (4.14) one writes
pX (i1 , t1 ; · · · ; in , tn )
= pX (i1 , t1 ; · · · ; in−1 , tn−1 )pX (in , tn i1 , t1 ; · · · ; in−1 , tn−1 )
= pX (i1 , t1 ; · · · ; in−1 , tn−1 )pin−1 in (tn−1 , tn )
Application of this operation n − 2 more times yields that
pX (i1 , t1 ; · · · ; in , tn ) = πi1 (t1 )pi1 i2 (t1 , t2 ) · · · pin−1 in (tn−1 , tn ) (4.15) which shows that the ﬁnite order distributions of X are indeed determined by the ﬁrst order pmfs
and the transition probabilities. Equation (4.15) can be used to easily verify that the form (4.12)
of the Markov property holds.
Given s < t, the collection H (s, t) deﬁned by H (s, t) = (pij (s, t) : i, j ∈ S ) should be thought
of as a matrix, and it is called the transition probability matrix for the interval [s, t]. Let e denote
the column vector with all ones, indexed by S . Since π (t) and the rows of H (s, t) are probability
vectors, it follows that π (t)e = 1 and H (s, t)e = e. Computing the distribution of Xt by summing
over all possible values of Xs yields that πj (t) = i P [Xs = i, Xt = j ] = i πi (s)pij (s, t), which in
matrix form yields that π (t) = π (s)H (s, t) for s, t ∈ T, s ≤ t. Similarly, given s < τ < t, computing
the conditional distribution of Xt given Xs by summing over all possible values of Xτ yields
H (s, t) = H (s, τ )H (τ, t) s, τ, t ∈ T, s < τ < t. (4.16) The relations (4.16) are known as the ChapmanKolmogorov equations.
A Markov process is timehomogeneous if the transition probabilities pij (s, t) depend on s and
t only through t − s. In that case we write pij (t − s) instead of pij (s, t), and Hij (t − s) instead of
Hij (s, t). If the Markov process is timehomogeneous, then π (s + τ ) = π (s)H (τ ) for s, s + τ ∈ T and
τ ≥ 0. A probability distribution π is called an equilibrium (or invariant) distribution if πH (τ ) = π
for all τ ≥ 0.
Recall that a random process is stationary if its ﬁnite order distributions are invariant with
respect to translation in time. On one hand, referring to (4.15), we see that a timehomogeneous
Markov process is stationary if and only if π (t) = π for all t for some equilibrium distribution π .
On the other hand, a Markov random process that is stationary is time homogeneous.
Repeated application of the ChapmanKolmogorov equations yields that pij (s, t) can be expressed in terms of transition probabilities for s and t close together. For example, consider
Markov processes with index set the integers. Then H (n, k + 1) = H (n, k )P (k ) for n ≤ k , where
114 P (k ) = H (k, k + 1) is the onestep transition probability matrix. Fixing n and using forward recursion starting with H (n, n) = I , H (n, n + 1) = P (n), H (n, n + 2) = P (n)P (n + 1), and so forth
yields
H (n, l) = P (n)P (n + 1) · · · P (l − 1)
In particular, if the chain is timehomogeneous then H (k ) = P k for all k , where P is the time
independent onestep transition probability matrix, and π (l) = π (k )P l−k for l ≥ k . In this case a
probability distribution π is an equilibrium distribution if and only if πP = π . Example 4.9.1 Consider a twostage pipeline through which packets ﬂow, as pictured in Figure
4.8. Some assumptions about the pipeline will be made in order to model it as a simple discretetime Markov process. Each stage has a single buﬀer. Normalize time so that in one unit of time
a packet can make a single transition. Call the time interval between k and k + 1 the k th “time
slot,” and assume that the pipeline evolves in the following way during a given slot. a d1 d2 Figure 4.8: A twostage pipeline.
If at the beginning of the slot, there are no packets in stage one, then a new packet arrives to stage
one with probability a, independently of the past history of the pipeline and of the outcome
at stage two.
If at the beginning of the slot, there is a packet in stage one and no packet in stage two, then the
packet is transfered to stage two with probability d1 .
If at the beginning of the slot, there is a packet in stage two, then the packet departs from the
stage and leaves the system with probability d2 , independently of the state or outcome of
stage one.
These assumptions lead us to model the pipeline as a discretetime Markov process with the state
space S = {00, 01, 10, 11}, transition probability diagram shown in Figure 4.9 (using the notation
x = 1 − x) and onestep transition probability matrix P given by
¯ a
¯
0
a
0
¯ ad2 ad2 ad2 ad2 ¯
¯¯ P =
¯1
0
d1 d
0
¯
0
0
d2 d2
The rows of P are probability vectors. For example, the ﬁrst row is the probability distribution of
the state at the end of a slot, given that the state is 00 at the beginning of a slot. Now that the
model is speciﬁed, let us determine the throughput rate of the pipeline.
The equilibrium probability distribution π = (π00 , π01 , π10 , π11 ) is the probability vector satisfying the linear equation π = πP . Once π is found, the throughput rate η can be computed as
follows. It is deﬁned to be the rate (averaged over a long time) that packets transit the pipeline.
115 a ad 2 ad 2
01 00
d1 a ad 2 ad2
d1
10 11
d2 d 2 Figure 4.9: Onestep transition probability diagram.
Since at most two packets can be in the pipeline at a time, the following three quantities are all
clearly the same, and can be taken to be the throughput rate.
The rate of arrivals to stage one
The rate of departures from stage one (or rate of arrivals to stage two)
The rate of departures from stage two
Focus on the ﬁrst of these three quantities to obtain
η = P [an arrival at stage 1]
= P [an arrival at stage 1stage 1 empty at slot beginning]P [stage 1 empty at slot beginning]
= a(π00 + π01 ).
Similarly, by focusing on departures from stage 1, obtain η = d1 π10 . Finally, by focusing on
departures from stage 2, obtain η = d2 (π01 + π11 ). These three expressions for η must agree.
Consider the numerical example a = d1 = d2 = 0.5. The equation π = πP yields that π is
proportional to the vector (1, 2, 3, 1). Applying the fact that π is a probability distribution yields
that π = (1/7, 2/7, 3/7, 1/7). Therefore η = 3/14 = 0.214 . . .. In the remainder of this section we assume that X is a continuoustime, ﬁnitestate Markov
process. The transition probabilities for arbitrary time intervals can be described in terms of the
transition probabilites over arbitrarily short time intervals. By saving only a linearization of the
transition probabilities, the concept of generator matrix arises naturally, as we describe next.
Let S be a ﬁnite set. A purejump function for a ﬁnite state space S is a function x : R+ → S
such that there is a sequence of times, 0 = τ0 < τ1 < · · · with limi→∞ τi = ∞, and a sequence of
states with si = si+1 , i ≥ 0, such that that x(t) = si for τi ≤ t < τi+1 . A purejump Markov process
is an S valued Markov process such that, with probability one, the sample functions are purejump
functions.
Let Q = (qij : i, j ∈ S ) be such that
qij ≥ 0
qii = − j ∈S ,j =i qij 116 i, j ∈ S , i = j
i ∈ S. (4.17) An example for state space S = {1, 2, 3} is −1 0.5 0.5
Q = 1 −2 1 ,
0
1 −1 and this matrix Q can be represented by the transition rate diagram shown in Figure 4.10. A pure0.5
2 1 1
1 0.5
1
3 Figure 4.10: Transition rate diagram for a continuoustime Markov process.
jump, timehomogeneous Markov process X has generator matrix Q if the transition probabilities
(pij (τ )) satisfy
lim (pij (h) − I{i=j } )/h = qij
i, j ∈ S
(4.18)
h 0 or equivalently
pij (h) = I{i=j } + hqij + o(h) i, j ∈ S (4.19) where o(h) represents a quantity such that limh→0 o(h)/h = 0. For the example this means that
the transition probability matrix for a time interval of duration h is given by 1 − h 0.5h
0.5h
o(h) o(h) o(h)
h
1 − 2h
h + o(h) o(h) o(h) 0
h
1−h
o(h) o(h) o(h)
For small enough h, the rows of the ﬁrst matrix are probability distributions, owing to the assumptions on the generator matrix Q.
Proposition 4.9.2 Given a matrix Q satisfying (4.17), and a probability distribution
π (0) = (πi (0) : i ∈ S ), there is a purejump, timehomogeneous Markov process with generator
matrix Q and initial distribution π (0). The ﬁnite order distributions of the process are uniquely
determined by π (0) and Q.
The ﬁrst order distributions and the transition probabilities can be derived from Q and an
initial distribution π (0) by solving diﬀerential equations, derived as follows. Fix t > 0 and let h be
a small positive number. The ChapmanKolmogorov equations imply that
πj (t + h) − πj (t)
=
h πi (t)
i∈S pij (h) − I{i=j }
h . (4.20) Letting h converge to zero yields the diﬀerential equation:
∂πj (t)
=
∂t πi (t)qij
i∈S 117 (4.21) or, in matrix notation, ∂π(t) = π (t)Q. These equations, known as the Kolmogorov forward equa∂t
tions, can be rewritten as
∂πj (t)
=
∂t πi (t)qij −
i∈S ,i=j πj (t)qji , (4.22) i∈S ,i=j which shows that the rate change of the probability of being at state j is the rate of “probability
ﬂow” into state j minus the rate of probability ﬂow out of state j . Example 4.9.3 Consider the twostate, continuoustime Markov process with the transition rate
diagram shown in Figure 4.11 for some positive constants α and β . The generator matrix is given
α
1 2
β Figure 4.11: Transition rate diagram for a twostate continuoustime Markov process.
by
Q= −α α
β −β Let us solve the forward Kolmogorov equation for a given initial distribution π (0). The equation
for π1 (t) is
∂π1 (t)
= −απ1 (t) + βπ2 (t);
π1 (0) given
∂t
But π1 (t) = 1 − π2 (t), so
∂π1 (t)
= −(α + β )π1 (t) + β ;
∂t π1 (0) given By diﬀerentiation we check that this equation has the solution
t π1 (t) = π1 (0)e−(α+β )t + e−(α+β )(t−s) βds 0 = π1 (0)e−(α+β )t +
so that
π (t) = π (0)e−(α+β )t + β
(1 − e−(α+β )t ).
α+β β
α
,
α+β α+β (1 − e−(α+β )t ) For any initial distribution π (0),
lim π (t) = t→∞ β
α
,
α+β α+β . The rate of convergence is exponential, with rate parameter α + β , and the limiting distribution is
the unique probability distribution satisfying πQ = 0. 118 4.10 Spacetime structure of discretestate Markov processes The previous section showed that the distribution of a timehomogeneous, discretestate Markov
process can be speciﬁed by an initial probability distribution, and either a onestep transition
probability matrix P (for discretetime processes) or a generator matrix Q (for continuoustime
processes). Another way to describe these processes is to specify the spacetime structure, which is
simply the sequences of states visited and how long each state is visited. The spacetime structure
is discussed ﬁrst for discretetime processes, and then for continuoustime processes. One beneﬁt
is to show how little diﬀerence there is between discretetime and continuoustime processes.
Let (Xk : k ∈ Z+ ) be a timehomogeneous Markov process with onestep transition probability
matrix P . Let Tk denote the time that elapses between the k th and k + 1th jumps of X , and let
X J (k ) denote the state after k jumps. See Fig. 4.12 for illustration. More precisely, the holding X(k)
J
X (1) s
3
s2
s
1 J
X (2) J
X (3)
k !0 !2 !1 Figure 4.12: Illustration of jump process and holding times.
times are deﬁned by
T0 = min{t ≥ 0 : X (t) = X (0)} (4.23) Tk = min{t ≥ 0 : X (T0 + . . . + Tk−1 + t) = X (T0 + . . . + Tk−1 )} (4.24) and the jump process X J = (X J (k ) : k ≥ 0) is deﬁned by
X J (0) = X (0) and X J (k ) = X (T0 + . . . + Tk−1 ) (4.25) Clearly the holding times and jump process contain all the information needed to construct X , and
vice versa. Thus, the following description of the joint distribution of the holding times and the
jump process characterizes the distribution of X .
Proposition 4.10.1 Let X = (X (k ) : k ∈ Z+ ) be a timehomogeneous Markov process with onestep transition probability matrix P .
(a) The jump process X J is itself a timehomogeneous Markov process, and its onestep transition
probabilities are given by pJ = pij /(1 − pii ) for i = j , and pJ = 0, i, j ∈ S .
ij
ii
(b) Given X (0), X J (1) is conditionally independent of T0 .
(c) Given (X J (0), . . . , X J (n)) = (j0 , . . . , jn ), the variables T0 , . . . , Tn are conditionally independent, and the conditional distribution of Tl is geometric with parameter pjl jl :
P [Tl = k X J (0) = j0 , . . . , X J (n) = jn ] = pkl−l1 (1 − pjl jl )
jj
119 0 ≤ l ≤ n, k ≥ 1. Proof. Observe that if X (0) = i, then
{T0 = k, X J (1) = j } = {X (1) = i, X (2) = i, . . . , X (k − 1) = i, X (k ) = j },
so
P [T0 = k, X J (1) = j X (0) = i] = pk−1 pij = [(1 − pii )pk−1 ]pJ
ij
ii
ii (4.26) Because for i ﬁxed the last expression in (4.26) displays the product of two probability distributions,
conclude that given X (0) = i,
k
T0 has distribution ((1 − pii )pii−1 : k ≥ 1), the geometric distribution of mean 1/(1 − pii ) X J (1) has distribution (pJ : j ∈ S ) (i ﬁxed)
ij
T0 and X J (1) are independent
More generally, check that
P [X J (1) = j1 , . . . , X J (n) = jn , To = k0 , . . . , Tn = kn X J (0) = i] =
n pJ 1 pJ1 j2
ij j −
(pklljl 1 (1 − pjl jl ))
j . . . pJn−1 jn
j
l=0 This establishes the proposition.
Next we consider the spacetime structure of timehomogeneous continuoustime purejump
Markov processes. Essentially the only diﬀerence between the discrete and continuoustime Markov
processes is that the holding times for the continuoustime processes are exponentially distributed
rather than geometrically distributed. Indeed, deﬁne the holding times Tk , k ≥ 0 and the jump
process X J using (4.23)(4.25) as before.
Proposition 4.10.2 Let X = (X (t) : t ∈ R+ ) be a timehomogeneous, purejump Markov process
with generator matrix Q. Then
(a) The jump process X J is a discretetime, timehomogeneous Markov process, and its onestep
transition probabilities are given by
pJ =
ij −qij /qii for i = j
0
for i = j (4.27) (b) Given X (0), X J (1) is conditionally independent of T0 .
(c) Given X J (0) = j0 , . . . , X J (n) = jn , the variables T0 , . . . , Tn are conditionally independent,
and the conditional distribution of Tl is exponential with parameter −qjl jl :
P [Tl ≥ cX J (0) = j0 , . . . , X J (n) = jn ] = exp(cqjl jl ) 0 ≤ l ≤ n. Proof. Fix h > 0 and deﬁne the “sampled” process X (h) by X (h) (k ) = X (hk ) for k ≥ 0. See
Fig. 4.13. Then X (h) is a discretetime Markov process with onestep transition probabilities pij (h)
(h)
(the transition probabilities for the original process for an interval of length h). Let (Tk : k ≥ 0)
denote the sequence of holding times and (X J,h (k ) : k ≥ 0) the jump process for the process X (h) .
120 X(t) X(h)(1) (h)
X (2) (h)
X (3) s
3
s2
s
1 t Figure 4.13: Illustration of sampling of a purejump function.
The assumption that with probability one the sample paths of X are purejump functions,
implies that with probability one:
(h) (h) (
lim (X J,h (0), X J,h (1), . . . , X J,h (n), hT0 , hT1 , . . . , hTnh) ) = h→0 (X J (0), X J (1), . . . , X J (n), T0 , T1 , . . . , Tn ) (4.28) Since convergence with probability one implies convergence in distribution, the goal of identifying
the distribution of the random vector on the righthand side of (4.28) can be accomplished by
identifying the limit of the distribution of the vector on the left.
First, the limiting distribution of the process X J,h is identiﬁed. Since X (h) has onestep transition probabilities pij (h), the formula for the jump process probabilities for discretetime processes
(see Proposition 4.10.1, part a) yields that the one step transition probabilities pJ,h for X (J,h) are
ij
given by
pJ,h =
ij
= pij (h)
1 − pii (h)
pij (h)/h
qij
→
as h → 0
(1 − pii (h))/h
−qii (4.29) for i = j , where the limit indicated in (4.29) follows from the deﬁnition (4.18) of the generator matrix
Q. Thus, the limiting distribution of X J,h is that of a Markov process with onestep transition
probabilities given by (4.27), establishing part (a) of the proposition. The conditional independence
properties stated in (b) and (c) of the proposition follow in the limit from the corresponding
properties for the jump process X J,h guaranteed by Proposition 4.10.1. Finally, since log(1 + θ) =
θ + o(θ) by Taylor’s formula, we have for all c ≥ 0 that
(h) P [hTl > cX J,h (0) = j0 , . . . , X J,h = jn ] c/h = (pjl jl (h)) = exp( c/h log(pjl jl (h))) = exp( c/h (qjl jl h + o(h))) → exp(qjl jl c) as h → 0 which establishes the remaining part of (c), and the proposition is proved. 121 4.11 Problems 4.1 Event probabilities for a simple random process
Deﬁne the random process X by Xt = 2A + Bt where A and B are independent random variables
with P [A = 1] = P [A = −1] = P [B = 1] = P [B = −1] = 0.5. (a) Sketch the possible sample
functions. (b) Find P [Xt ≥ 0] for all t. (c) Find P [Xt ≥ 0 for all t].
4.2 Correlation function of a product
Let Y and Z be independent random processes with RY (s, t) = 2 exp(−s − t) cos(2πf (s − t)) and
RZ (s, t) = 9 + exp(−3s − t4 ). Find the autocorrelation function RX (s, t) where Xt = Yt Zt .
4.3 A sinusoidal random process
Let Xt = A cos(2πV t + Θ) where the amplitude A has mean 2 and variance 4, the frequency V in
Hertz is uniform on [0, 5], and the phase Θ is uniform on [0, 2π ]. Furthermore, suppose A, V and Θ
are independent. Find the mean function µX (t) and autocorrelation function RX (s, t). Is X WSS?
4.4 Another sinusoidal random process
Suppose that X1 and X2 are random variables such that EX1 = EX2 = EX1 X2 = 0 and Var(X1 ) =
Var(X2 ) = σ 2 . Deﬁne Yt = X1 cos(2πt) − X2 sin(2πt). (a) Is the random process Y necessarily
widesense stationary? (b) Give an example of random variables X1 and X2 satisfying the given
conditions such that Y is stationary. (c) Give an example of random variables X1 and X2 satisfying
the given conditions such that Y is not (strict sense) stationary.
4.5 A random line
Let X = (Xt : t ∈ R) be a random process such that Xt = R − St for all t, where R and S are
2
independent random variables, having the Rayleigh distribution with positive parameters σR and
2 , respectively.
σS
(a) Indicate three typical sample paths of X in a single sketch. Describe in words the set of possible
sample paths of X.
(b) Is X a Markov process? Why or why not?
(c) Does X have independent increments? Why or why not?
(d) Let A denote the area of the triangle bounded by portions of the coordinate axes and the graph
of X . Find E [A]. Simplify your answer as much as possible.
4.6 Brownian motion: Ascension and smoothing
Let W be a Brownian motion process and suppose 0 ≤ r < s < t.
(a) Find P {Wr ≤ Ws ≤ Wt }.
(b) Find E [Ws Wr , Wt ]. (This part is unrelated to part (a).)
4.7 Brownian bridge
Let W = (Wt : t ≥ 0) be a standard Brownian motion (i.e. a Brownian motion with paramter
σ 2 = 1.) Let Bt = Wt − tW1 for 0 ≤ t ≤ 1. The process B = (Bt : 0 ≤ t ≤ 1) is called a Brownian
bridge process. Like W , B is a mean zero Gaussian random process.
(a) Sketch a typical sample path of W , and the corresponding sample path of B.
(b) Find the autocorrelation function of B.
(c) Is B a Markov process?
122 (d) Show that B is independent of the random variable W1 . (This means that for any ﬁnite collection, t1 , . . . , tn ∈ [0, 1], the random vector (Bt1 , . . . , Btn )T is independent of W1 .)
(e) (Due to J.L. Doob.) Let Xt = (1 − t)W t , for 0 ≤ t < 1 and let X1 = 0. Let X denote the
1−t
random process X = (Xt : 0 ≤ t ≤ 1). Like W , X is a mean zero, Gaussian random process. Find
the autocorrelation function of X. Can you draw any conclusions?
4.8 Some Poisson process calculations
Let N = (Nt : t ≥ 0) be a Poisson process with rate λ > 0.
(a) Give a simple expression for P [N1 ≥ 1N2 = 2] in terms of λ.
(b) Give a simple expression for P [N2 = 2N1 ≥ 1] in terms of λ.
(c) Let Xt = Nt2 . Is X = (Xt : t ≥ 0) a timehomogeneous Markov process? If so, give the
transition probabilities pij (τ ). If not, explain.
4.9 A random process corresponding to a random parabola
Deﬁne a random process X by Xt = A + Bt + t2 , where A and B are independent, N (0, 1) random
ˆ
variables. (a) Find E [X5 X1 ], the linear minimum mean square error (LMMSE) estimator of X5
given X1 , and compute the mean square error. (b) Find the MMSE (possibly nonlinear) estimator
ˆ
of X5 given X1 , and compute the mean square error. (c) Find E [X5 X0 , X1 ] and compute the mean
square error. (Hint: Can do by inspection.)
4.10 MMSE prediction for a Gaussian process based on two observations
Let X be a stationary Gaussian process with mean zero and RX (τ ) = 5 cos( πτ )3−τ  . (a) Find the
2
covariance matrix of the random vector (X (2), X (3), X (4))T . (b) Find E [X (4)X (2)]. (c) Find
E [X (4)X (2), X (3)].
4.11 A simple discretetime random process
Let U = (Un : n ∈ Z) consist of independent random variables, each uniformly distributed on the
interval [0, 1]. Let X = (Xk : k ∈ Z} be deﬁned by Xk = max{Uk−1 , Uk }. (a) Sketch a typical
sample path of the process X . (b) Is X stationary? (c) Is X Markov? (d) Describe the ﬁrst order
distributions of X . (e) Describe the second order distributions of X .
4.12 Poisson process probabilities
Consider a Poisson process with rate λ > 0.
(a) Find the probability that there is (exactly) one count in each of the three intervals [0,1], [1,2],
and [2,3].
(b) Find the probability that there are two counts in the interval [0, 2] and two counts in the interval
[1, 3]. (Note: your answer to part (b) should be larger than your answer to part (a)).
(c) Find the probability that there are two counts in the interval [1,2], given that there are two
counts in the interval [0,2] and two counts in the the interval [1,3].
4.13 Sliding function of an i.i.d. Poisson sequence
Let X = (Xk : k ∈ Z ) be a random process such that the Xi are independent, Poisson random
variables with mean λ, for some λ > 0. Let Y = (Yk : k ∈ Z ) be the random process deﬁned by
Yk = Xk + Xk+1 .
(a) Show that Yk is a Poisson random variable with parameter 2λ for each k .
(b) Show that X is a stationary random process.
(c) Is Y a stationary random process? Justify your answer.
123 4.14 Adding jointly stationary Gaussian processes
Let X and Y be jointly stationary, jointly Gaussian random processes with mean zero, autocorrelation functions RX (t) = RY (t) = exp(−t), and crosscorrelation function
RXY (t) = (0.5) exp(−t − 3).
(a) Let Z (t) = (X (t) + Y (t))/2 for all t. Find the autocorrelation function of Z .
(b) Is Z a stationary random process? Explain.
(c) Find P [X (1) ≤ 5Y (2) + 1]. You may express your answer in terms of the standard normal
cumulative distribution function Φ.
4.15 Invariance of properties under transformations
Let X = (Xn : n ∈ Z), Y = (Yn : n ∈ Z), and Z = (Zn : n ∈ Z) be random processes such that
2
3
Yn = Xn for all n and Zn = Xn for all n. Determine whether each of the following statements is
always true. If true, give a justiﬁcation. If not, give a simple counter example.
(a) If X is Markov then Y is Markov.
(b) If X is Markov then Z is Markov.
(c) If Y is Markov then X is Markov.
(d) If X is stationary then Y is stationary.
(e) If Y is stationary then X is stationary.
(f) If X is wide sense stationary then Y is wide sense stationary.
(g) If X has independent increments then Y has independent increments.
(h) If X is a martingale then Z is a martingale.
4.16 A linear evolution equation with random coeﬃcients
2
Let the variables Ak , Bk , k ≥ 0 be mutually independent with mean zero. Let Ak have variance σA
2
and let Bk have variance σB for all k. Deﬁne a discretetime random process Y by
Y = (Yk : k ≥ 0), such that Y0 = 0 and Yk+1 = Ak Yk + Bk for k ≥ 0.
(a) Find a recursive method for computing Pk = E [(Yk )2 ] for k ≥ 0.
(b) Is Y a Markov process? Explain.
(c) Does Y have independent increments? Explain.
(d) Find the autocorrelation function of Y . ( You can use the second moments (Pk ) in expressing
your answer.)
(e) Find the corresponding linear innovations sequence (Yk : k ≥ 1).
4.17 On an M/D/inﬁnity system
Suppose customers enter a service system according to a Poisson point process on R of rate λ,
meaning that the number of arrivals, N (a, b], in an interval (a, b], has the Poisson distribution
with mean λ(b − a), and the numbers of arrivals in disjoint intervals are independent. Suppose
each customer stays in the system for one unit of time, independently of other customers. Because
the arrival process is memoryless, because the service times are deterministic, and because the
customers are served simultaneously, corresponding to inﬁnitely many servers, this queueing system
is called an M/D/∞ queueing system. The number of customers in the system at time t is given
by Xt = N (t − 1, t].
(a) Find the mean and autocovariance function of X .
(b) Is X stationary? Is X wide sense stationary?
(c) Is X a Markov process?
124 (d) Find a simple expression for P {Xt = 0 for t ∈ [0, 1]} in terms of λ.
(e) Find a simple expression for P {Xt > 0 for t ∈ [0, 1]} in terms of λ.
4.18 A ﬂy on a cube
Consider a cube with vertices 000, 001, 010, 100, 110, 101. 011, 111. Suppose a ﬂy walks along
edges of the cube from vertex to vertex, and for any integer t ≥ 0, let Xt denote which vertex the
ﬂy is at at time t. Assume X = (Xt : t ≥ 0) is a discretetime Markov process, such that given Xt ,
the next state Xt+1 is equally likely to be any one of the three vertices neighboring Xt .
(a) Sketch the one step transition probability diagram for X .
(b) Let Yt denote the distance of Xt , measured in number of hops, between vertex 000 and Xt . For
example, if Xt = 101, then Yt = 2. The process Y is a Markov process with states 0,1,2, and 3.
Sketch the onestep transition probability diagram for Y .
(c) Suppose the ﬂy begins at vertex 000 at time zero. Let τ be the ﬁrst time that X returns to
vertex 000 after time 0, or equivalently, the ﬁrst time that Y returns to 0 after time 0. Find E [τ ].
4.19 Time elapsed since Bernoulli renewals
Let U = (Uk : k ∈ Z) be such that for some p ∈ (0, 1), the random variables Uk are independent,
with each having the Bernoulli distribution with parameter p. Interpret Uk = 1 to mean that a
renewal, or replacement, of some part takes place at time k. For k ∈ Z, let
Xk = min{i ≥ 1 : Uk−i = 1}. In words, Xk is the time elapsed since the last renewal strictly before
time k.
(a) The process X is a timehomogeneous Markov process. Indicate a suitable state space, and
describe the onestep transition probabilities.
(b) Find the distribution of Xk for k ﬁxed.
(c) Is X a stationary random process? Explain.
(d) Find the k step transition probabilities, pi,j (k ) = P {Xn+k = j Xn = i}.
4.20 A random process created by interpolation
Let U = (Uk : k ∈ Z) such that the Uk are independent, and each is uniformly distributed on
the interval [0, 1]. Let X = (Xt : t ∈ R) denote the continuous time random process obtained by
linearly interpolating between the U ’s. Speciﬁcally, Xn = Un for any n ∈ Z, and Xt is aﬃne on
each interval of the form [n, n + 1] for n ∈ Z.
(a) Sketch a sample path of U and a corresponding sample path of X.
(b) Let t ∈ R. Find and sketch the ﬁrst order marginal density, fX,1 (x, t). (Hint: Let n = t and
a = t − n, so that t = n + a. Then Xt = (1 − a)Un + aUn+1 . It’s helpful to consider the cases
0 ≤ a ≤ 0.5 and 0.5 < a < 1 separately. For brevity, you need only consider the case 0 ≤ a ≤ 0.5.)
(c) Is the random process X WSS? Justify your answer.
(d) Find P {max0≤t≤10 Xt ≤ 0.5}.
4.21 Reinforcing samples
(Due to G. Polya) Suppose at time k = 2, there is a bag with two balls in it, one orange and one
blue. During each time step between k and k + 1, one of the balls is selected from the bag at
random, with all balls in the bag having equal probability. That ball, and a new ball of the same
color, are both put into the bag. Thus, at time k there are k balls in the bag, for all k ≥ 2. Let Xk
denote the number of blue balls in the bag at time k .
(a) Is X = (Xk : k ≥ 2) a Markov process?
125 (b) Let Mk = Xk . Thus, Mk is the fraction of balls in the bag at time k that are blue. Determine
k
whether M = (Mk : k ≥ 2) is a martingale.
(c) By the theory of martingales, since M is a bounded martingale, it converges a.s. to some
(
random variable M∞ . Let Vk = Mk (1 − Mk ). Show that E [Vk+1 Vk ] = kkk+2) Vk , and therefore that
( +1)2
+1)
1
E [Vk ] = (k6k . It follows that Var(limk→∞ Mk ) = 12 .
(d) More concretely, ﬁnd the distribution of Mk for each k , and then identify the distribution of
the limit random variable, M∞ . 4.22 Restoring samples
Suppose at time k = 2, there is a bag with two balls in it, one orange and one blue. During each
time step between k and k + 1, one of the balls is selected from the bag at random, with all balls
in the bag having equal probability. That ball, and a new ball of the other color, are both put into
the bag. Thus, at time k there are k balls in the bag, for all k ≥ 2. Let Xk denote the number of
blue balls in the bag at time k .
(a) Is X = (Xk : k ≥ 2) a Markov process? If so, describe the onestep transition probabilities.
(b) Compute E [Xk+1 Xk ] for k ≥ 2.
(c) Let Mk = Xk . Thus, Mk is the fraction of balls in the bag at time k that are blue. Determine
k
whether M = (Mk : k ≥ 2) is a martingale.
1
(d) Let Dk = Mk − 2 . Show that
2
E [Dk+1 Xk ] = 1
(k + 1)2 2
k (k − 2)Dk + 1
4 2
(e) Let vk = E [Dk ]. Prove by induction on k that vk ≤ 41 . What can you conclude about the limit
k
of Mk as k → ∞? (Be sure to specify what sense(s) of limit you mean.) 4.23 A spacetime transformation of Brownian motion
Suppose X = (Xt : t ≥ 0) is a realvalued, mean zero, independent increment process, and let
2
E [Xt ] = ρt for t ≥ 0. Assume ρt < ∞ for all t.
(a) Show that ρ must be nonnegative and nondecreasing over [0, ∞).
(b) Express the autocorrelation function RX (s, t) in terms of the function ρ for all s ≥ 0 and t ≥ 0.
(c) Conversely, suppose a nonnegative, nondecreasing function ρ on [0, ∞) is given. Let Yt = W (ρt )
for t ≥ 0, where W is a standard Brownian motion with RW (s, t) = min{s, t}. Explain why Y is
an independent increment process with E [Yt2 ] = ρt for all t ≥ 0.
(d) Deﬁne a process Z in terms of a standard Brownian motion W by Z0 = 0 and Zt = tW ( 1 ) for
t
t > 0. Does Z have independent increments? Justify your answer.
4.24 An M/M/1/B queueing system
Suppose X is a continuoustime Markov process with the transition rate diagram shown, for a
positive integer B and positive constant λ.
!
0 ! ! 1
1 2
1 ! ! ...
1 B !1
1 B
1 (a) Find the generator matrix, Q, of X for B = 4.
(b) Find the equilibrium probability distribution. (Note: The process X models the number of
customers in a queueing system with a Poisson arrival process, exponential service times, one
server, and a ﬁnite buﬀer.)
126 4.25 Identiﬁcation of special properties of two discretetime processes
Determine which of the properties:
(i) Markov property
(ii) martingale property
(iii) independent increment property
are possessed by the following two random processes. Justify your answers.
(a) X = (Xk : k ≥ 0) deﬁned recursively by X0 = 1 and Xk+1 = (1 + Xk )Uk for k ≥ 0, where
U0 , U1 , . . . are independent random variables, each uniformly distributed on the interval [0, 1].
(b) Y = (Yk : k ≥ 0) deﬁned by Y0 = V0 , Y1 = V0 + V1 , and Yk = Vk−2 + Vk−1 + Vk for k ≥ 2, where
Vk : k ∈ Z are independent Gaussian random variables with mean zero and variance one.
4.26 Identiﬁcation of special properties of two discretetime processes (version 2)
Determine which of the properties:
(i) Markov property
(ii) martingale property
(iii) independent increment property
are possessed by the following two random processes. Justify your answers.
(a) (Xk : k ≥ 0), where Xk is the number of cells alive at time k in a colony that evolves as
follows. Initially, there is one cell, so X0 = 1. During each discrete time step, each cell either dies
or splits into two new cells, each possibility having probability one half. Suppose cells die or split
independently. Let Xk denote the number of cells alive at time k .
(b) (Yk : k ≥ 0), such that Y0 = 1 and, for k ≥ 1, Yk = U1 U2 . . . Uk , where U1 , U2 , . . . are independent
random variables, each uniformly distributed over the interval [0, 2]
4.27 Identiﬁcation of special properties of two continuoustime processes
Answer as in the previous problem, for the following two random processes:
2
(a) Z = (Zt : t ≥ 0), deﬁned by Zt = exp(Wt − σ2 t ), where W is a Brownian motion with parameter
σ 2 . (Hint: Observe that E [Zt ] = 1 for all t.)
(b) R = (Rt : t ≥ 0) deﬁned by Rt = D1 + D2 + · · · + DNt , where N is a Poisson process with rate
λ > 0 and Di : i ≥ 1 is an iid sequence of random variables, each having mean 0 and variance σ 2 .
4.28 Identiﬁcation of special properties of two continuoustime processes (version 2)
Answer as in the previous problem, for the following two random processes:
(a) Z = (Zt : t ≥ 0), deﬁned by Zt = Wt3 , where W is a Brownian motion with parameter σ 2 .
(b) R = (Rt : t ≥ 0), deﬁned by Rt = cos(2πt + Θ), where Θ is uniformly distributed on the interval
[0, 2π ].
4.29 A branching process
Let p = (pi : i ≥ 0) be a probability distribution on the nonnegative integers with mean m.
Consider a population beginning with a single individual, comprising generation zero. The oﬀspring
of the initial individual comprise the ﬁrst generation, and, in general, the oﬀspring of the kth
generation comprise the k + 1st generation. Suppose the number of oﬀspring of any individual has
the probability distribution p, independently of how many oﬀspring other individuals have. Let
Y0 = 1, and for k ≥ 1 let Yk denote the number of individuals in the k th generation.
(a) Is Y = (Yk : k ≥ 0) a Markov process? Brieﬂy explain your answer.
k
(b) Find constants ck so that Yk is a martingale.
c
127 (c) Let am = P [Ym = 0], the probability of extinction by the mth generation. Express am+1 in terms
of the distribution p and am (Hint: condition on the value of Y1 , and note that the Y1 subpopulations
beginning with the Y1 individuals in generation one are independent and statistically identical to
the whole population.)
(d) Express the probability of eventual extinction, a∞ = limm→∞ am , in terms of the distribution
p. Under what condition is a∞ = 1?
(e) Find a∞ in terms of θ in case pk = θk (1 − θ) for k ≥ 0 and 0 ≤ θ < 1. (This distribution is
θ
similar to the geometric distribution, and it has mean m = 1−θ .)
4.30 Moving balls
Consider the motion of three indistinguishable balls on a linear array of positions, indexed by the
positive integers, such that one or more balls can occupy the same position. Suppose that at time
t = 0 there is one ball at position one, one ball at position two, and one ball at position three.
Given the positions of the balls at some integer time t, the positions at time t + 1 are determined
as follows. One of the balls in the left most occupied position is picked up, and one of the other
two balls is selected at random (but not moved), with each choice having probability one half. The
ball that was picked up is then placed one position to the right of the selected ball.
(a) Deﬁne a ﬁnitestate Markov process that tracks the relative positions of the balls. Try to
use a small number of states. (Hint: Take the balls to be indistinguishable, and don’t include
the position numbers.) Describe the signiﬁcance of each state, and give the onestep transition
probability matrix for your process.
(b) Find the equilibrium distribution of your process.
(c) As time progresses, the balls all move to the right, and the average speed has a limiting value,
with probability one. Find that limiting value. (You can use the fact that for a ﬁnitestate Markov
process in which any state can eventually be reached from any other, the fraction of time the process
is in a state i up to time t converges a.s. to the equilibrium probability for state i as t → ∞.
(d) Consider the following continuous time version of the problem. Given the current state at time
t, a move as described above happens in the interval [t, t + h] with probability h + o(h). Give the
generator matrix Q, ﬁnd its equilibrium distribution, and identify the long term average speed of
the balls.
4.31 Mean hitting time for a discretetime, discretestate Markov process
Let (Xk : k ≥ 0) be a timehomogeneous Markov process with the onestep transition probability
diagram shown.
0.4
0.6 1 0.2
2 0.8 0.4 3 0.6 (a) Write down the one step transition probability matrix P .
(b) Find the equilibrium probability distribution π .
(c) Let τ = min{k ≥ 0 : Xk = 3} and let ai = E [τ X0 = i] for 1 ≤ i ≤ 3. Clearly a3 = 0. Derive
equations for a1 and a2 by considering the possible values of X1 , in a way similar to the analysis
of the gambler’s ruin problem. Solve the equations to ﬁnd a1 and a2 .
4.32 Mean hitting time for a continuoustime, discretespace Markov process
Let (Xt : t ≥ 0) be a timehomogeneous Markov process with the transition rate diagram shown.
128 1
1 10 1
2 5 3 (a) Write down the rate matrix Q.
(b) Find the equilibrium probability distribution π .
(c) Let τ = min{t ≥ 0 : Xt = 3} and let ai = E [τ X0 = i] for 1 ≤ i ≤ 3. Clearly a3 = 0. Derive
equations for a1 and a2 by considering the possible values of Xt (h) for small values of h > 0 and
taking the limit as h → 0. Solve the equations to ﬁnd a1 and a2 .
4.33 Poisson merger
Summing counting processes corresponds to “merging” point processes. Show that the sum of K
independent Poisson processes, having rates λ1 , . . . , λK , respectively, is a Poisson process with rate
λ1 + . . . + λK . (Hint: First formulate and prove a similar result for sums of random variables,
and then think about what else is needed to get the result for Poisson processes. You can use the
deﬁnition of a Poisson process or one of the equivalent descriptions given by Proposition 4.5.2 in
the notes. Don’t forget to check required independence properties.)
4.34 Poisson splitting
Consider a stream of customers modeled by a Poisson process, and suppose each customer is one
of K types. Let (p1 , . . . , pK ) be a probability vector, and suppose that for each k , the k th customer
is type i with probability pi . The types of the customers are mutually independent and also
independent of the arrival times of the customers. Show that the stream of customers of a given
type i is again a Poisson stream, and that its rate is λpi . (Same hint as in the previous problem
applies.) Show furthermore that the K substreams are mutually independent.
4.35 Poisson method for coupon collector’s problem
(a) Suppose a stream of coupons arrives according to a Poisson process (A(t) : t ≥ 0) with rate
λ = 1, and suppose there are k types of coupons. (In network applications, the coupons could be
pieces of a ﬁle to be distributed by some sort of gossip algorithm.) The type of each coupon in the
1
stream is randomly drawn from the k types, each possibility having probability k , and the types of
diﬀerent coupons are mutually independent. Let p(k, t) be the probability that at least one coupon
of each type arrives by time t. (The letter “p” is used here because the number of coupons arriving
by time t has the Poisson distribution). Express p(k, t) in terms of k and t.
(b) Find limk→∞ p(k, k ln k + kc) for an arbitrary constant c. That is, ﬁnd the limit of the probability
that the collection is complete at time t = k ln k + kc. (Hint: If ak → a as k → ∞, then (1 + ak )k →
k
ea .)
(c) The rest of this problem shows that the limit found in part (b) also holds if the total number of
coupons is deterministic, rather than Poisson distributed. One idea is that if t is large, then A(t)
is not too far from its mean with high probability. Show, speciﬁcally, that
0 if c < c
limk→∞ P [A(k ln k + kc) ≥ k ln k + kc ] =
1 if c > c
(d) Let d(k, n) denote the probability that the collection is complete after n coupon arrivals. (The
letter “d” is used here because the number of coupons, n, is deterministic.) Show that for any k, t,
and n ﬁxed, d(k, n)P [A(t) ≥ n] ≤ p(k, t) ≤ P [A(t) ≥ n] + P [A(t) ≤ n]d(k, n).
(e) Combine parts (c) and (d) to identify limk→∞ d(k, k ln k + kc).
129 4.36 Some orthogonal martingales based on Brownian motion
(This problem is related to the problem on linear innovations and orthogonal polynomials in the
previous problem set.) Let W = (Wt : t ≥ 0) be a Brownian motion with σ 2 = 1 (called a standard
2
Brownian motion), and let Mt = exp(θWt − θ2 t ) for an arbitrary constant θ.
(a) Show that (Mt : t ≥ 0) is a martingale. (Hint for parts (a) and (b): For notational brevity, let
Ws represent (Wu : 0 ≤ u ≤ s) for the purposes of conditioning. If Zt is a function of Wt for each
t, then a suﬃcient condition for Z to be a martingale is that E [Zt Ws ] = Zs whenever 0 < s < t,
because then E [Zt Zu , 0 ≤ u ≤ s] = E [E [Zt Ws ]Zu , 0 ≤ u ≤ s] = E [Zs Zu , 0 ≤ u ≤ s] = Zs ).
(b) By the power series expansion of the exponential function,
exp(θWt − θ2 t
θ2
θ3
) = 1 + θWt + (Wt2 − t) + (Wt3 − 3tWt ) + · · ·
2
2
3!
∞
n
θ
=
Mn (t)
n!
n=0 √t
where Mn (t) = tn/2 Hn ( Wt ), and Hn is the nth Hermite polynomial. The fact that M is a martingale for any value of θ can be used to show that Mn is a martingale for each n (you don’t need to
supply details). Verify directly that Wt2 − t and Wt3 − 3tWt are martingales. (c) For ﬁxed t, (Mn (t) : n ≥ 0) is a sequence of orthogonal random variables, because it is the linear
innovations sequence for the variables 1, Wt , Wt2 , . . .. Use this fact and the martingale property of
the Mn processes to show that if n = m and s, t ≥ 0, then Mn (s) ⊥ Mm (t).
4.37 A state space reduction preserving the Markov property
Consider a timehomogeneous, discretetime Markov process X = (Xk : k ≥ 0) with state space
S = {1, 2, 3}, initial state X0 = 3, and onestep transition probability matrix 0.0 0.8 0.2
P = 0.1 0.6 0.3 .
0.2 0.8 0.0
(a) Sketch the transition probability diagram and ﬁnd the equilibrium probability distribution
π = (π1 , π2 , π3 ).
(b) Identify a function f on S so that f (s) = a for two choices of s and f (s) = b for the third
choice of s, where a = b, such that the process Y = (Yk : k ≥ 0) deﬁned by Yk = f (Xk ) is a Markov
process with only two states, and give the onestep transition probability matrix of Y . Brieﬂy
explain your answer.
4.38 * Autocorrelation function of a stationary Markov process
Let X = (Xk : k ∈ Z ) be a Markov process such that the state space, {ρ1 , ρ2 , ..., ρn }, is a ﬁnite
subset of the real numbers. Let P = (pij ) denote the matrix of onestep transition probabilities.
Let e be the column vector of all ones, and let π (k ) be the row vector
π (k ) = (P [X k = ρ1 ], ..., P [Xk = ρn ]).
(a) Show that P e = e and π (k + 1) = π (k )P .
(b) Show that if the Markov chain X is a stationary random process then π (k ) = π for all k , where
π is a vector such that π = πP .
(c) Prove the converse of part (b).
(m)
(m)
(d) Show that P [Xk+m = ρj Xk = ρi , Xk−1 = s1 , ..., Xk−m = sm ] = pij , where pij is the i, j th
130 element of the mth power of P , P m , and s1 , . . . , sm are arbitrary states.
(e) Assume that X is stationary. Express RX (k ) in terms of P , (ρi ), and the vector π of parts (b)
and (c). 131 132 Chapter 5 Inference for Markov Models
This chapter gives a glimpse of the theory of iterative algorithms for graphical models, as well as
an introduction to statistical estimation theory. It begins with a brief introduction to estimation
theory: maximum likelihood and Bayesian estimators are introduced, and an iterative algorithm,
known as the expectationmaximization algorithm, for computation of maximum likelihood estimators in certain contexts, is described. This general background is then focused on three inference
problems posed using Markov models. 5.1 A bit of estimation theory The two most commonly used methods for producing estimates of unknown quantities are the
maximum likelihood (ML) and Bayesian methods. These two methods are brieﬂy described in this
section, beginning with the ML method.
Suppose a parameter θ is to be estimated, based on observation of a random variable Y . An
estimator of θ based on Y is a function θ, which for each possible observed value y , gives the
estimate θ(y ). The ML method is based on the assumption that Y has a pmf pY (y θ) (if Y is
discrete type) or a pdf fY (y θ) (if Y is continuous type), where θ is the unknown parameter to be
estimated, and the family of functions pY (y θ) or fY (y θ), is known.
Deﬁnition 5.1.1 For a particular value y and parameter value θ, the likelihood of y for θ is
pY (y θ), if Y is discrete type, or fY (y θ), if Y is continuous type. The maximum likelihood estimate
of θ given Y = y for a particular y is the value of θ that maximizes the likelihood of y. That
is, the maximum likelihood estimator θM L is given by θM L (y ) = arg maxθ pY (y θ), or θM L (y ) =
arg maxθ fY (y θ).
Note that the maximum likelihood estimator is not deﬁned as one maximizing the likelihood
of the parameter θ to be estimated. In fact, θ need not even be a random variable. Rather, the
maximum likelihood estimator is deﬁned by selecting the value of θ that maximizes the likelihood
of the observation.
Example 5.1.2 Suppose Y is assumed to be a N (θ, σ 2 ) random variable, where σ 2 is known.
Equivalently, we can write Y = θ + W , where W is a N (0, σ 2 ) random variable. Given a value y is
−2
observed, the ML estimator is obtained by maximizing fY (y θ) = √ 1 2 exp(− (y2σθ) ) with respect
2
2πσ to θ. By inspection, θM L (y ) = y .
133 Example 5.1.3 Suppose Y is assumed to be a P oi(θ) random variable, for some θ > 0. Given the
observation Y = k for some ﬁxed k ≥ 0, the ML estimator is obtained by maximizing pY (k θ) =
e−θ θk
with respect to θ. Equivalently, dropping the constant k ! and taking the logarithm, θ is to
k!
be selected to maximize −θ + k ln θ. The derivative is −1 + k/θ, which is positive for θ < k and
negative for θ > k . Hence, θM L (k ) = k .
Note that in the ML method, the quantity to be estimated, θ, is not assumed to be random.
This has the advantage that the modeler does not have to come up with a probability distribution
for θ, and can still impose hard constraints on θ. But the ML method does not permit incorporation
of soft probabilistic knowledge the modeler may have about θ before any observation is used.
The Bayesian method is based on estimating a random quantity. Thus, in the end, the variable
to be estimated, say Z , and the observation, say Y , are jointly distributed random variables.
Deﬁnition 5.1.4 The Bayesian estimator of Z given Y, for jointly distributed random variables Z
and Y, and cost function C (z, y ), is the function Z = g (Y ) of Y which minimizes the average cost,
E [C (Z, Z )].
The assumed distribution of Z is called the prior or a priori distribution, whereas the conditional
distribution of Z given Y is called the posterior or a posteriori distribution. In particular, if Z is
discrete, there is a prior pmf, pZ , and a posterior pmf, pZ Y , or if Z and Y are jointly continuous,
there is a prior pdf, fZ , and a posterior pdf, fZ Y .
One of the most common choices of the cost function is the squared error, C (z, z ) = (z − z )2 , for
ˆ
ˆ
which the Bayesian estimators are the minimum mean squared error (MMSE) estimators, examined
in Chapter 3. Recall that the MMSE estimators are given by the conditional expectation, g (y ) =
E [Z Y = y ], which, given the observation Y = y , is the mean of the posterior distribution of Z
given Y = y.
A commonly used choice of C in case Z is a discrete random variable is C (z, z ) = I{z =z } . In
ˆ
b
ˆ to minimize P {Z = Z }, or equivalently, to maximize
ˆ
this case, the Bayesian objective is to select Z
ˆ
P {Z = Z }. For an estimator Z = g (Y ),
ˆ
P {Z = Z } = P [Z = g (y )Y = y ]pY (y ) =
y pZ Y (g (y )y )pY (y ).
y So a Bayesian estimator for C (z, z ) = I{z =z } is one such that g (y ) maximizes P [Z = g (y )Y = y ]
ˆ
b
for each y . That is, for each y , g (y ) is a maximizer of the posterior pmf of Z . The estimator, called
the maximum a posteriori probability (MAP) estimator, can be written concisely as
ZM AP (y ) = arg max pZ Y (z y ).
z Suppose there is a parameter θ to be estimated based on an observation Y, and suppose that
the pmf of Y, pY (y θ), is known for each θ. This is enough to determine the ML estimator, but
determination of a Bayesian estimator requires, in addition, a choice of cost function C and a prior
probability distribution (i.e. a distribution for θ). For example, if θ is a discrete variable, the
Bayesian method would require that a prior pmf for θ be selected. In that case, we can view the
134 parameter to be estimated as a random variable, which we might denote by the upper case symbol
Θ, and the prior pmf could be denoted by pΘ (θ). Then, as required by the Bayesian method, the
variable to be estimated, Θ, and the observation, Y , would be jointly distributed random variables.
The joint pmf would be given by pΘ,Y (θ, Y ) = pΘ (θ)pY (y θ). The posterior probability distribution
can be expressed as a conditional pmf, by Bayes’ formula:
pΘY (θy ) = pΘ (θ)pY (y θ)
pY (y ) (5.1) where pY (y ) =
θ pΘ,Y (θ , y ). Given y , the value of the MAP estimator is a value of θ that
maximizes pΘY (θy ) with respect to θ. For that purpose, the denominator in the righthand side
of (5.1) can be ignored, so that the MAP estimator is given by
ΘM AP (y ) = arg max pΘY (θy )
θ = arg max pΘ (θ)pY (y θ).
θ (5.2) The expression, (5.2), for ΘM AP (y ) is rather similar to the expression for the ML estimator,
θM L (y ) = arg maxθ pY (y θ). In fact, the two estimators agree if the prior pΘ (θ) is uniform, meaning
it is the same for all θ.
The MAP criterion for selecting estimators can be extended to the case that Y and θ are jointly
continuous variables, leading to the following:
ΘM AP (y ) = arg max fΘY (θy )
θ = arg max fΘ (θ)fY (y θ).
θ (5.3) In this case, the probability that any estimator is exactly equal to θ is zero, but taking ΘM AP (y )
to maximize the posterior pdf maximizes the probability that the estimator is within of the true
value of θ, in an asymptotic sense as → 0.
Example 5.1.5 Suppose Y is assumed to be a N (θ, σ 2 ) random variable, where the variance σ 2 is
known and θ is to be estimated. Using the Bayesian method, suppose the prior density of θ is the
N (0, b2 ) density for some known paramber b2 . Equivalently, we can write Y = Θ + W , where Θ is a
N (0, b2 ) random variable and W is a N (0, σ 2 ) random variable, independent of Θ. By the properties
of joint Gaussian densities given in Chapter 3, given Y = y , the posterior distribution (i.e. the
2
conditional distribution of Θ given y ) is the normal distribution with mean E [ΘY = y ] = b2b+y 2
σ
and variance b2 σ 2
.
b2 +σ 2 The mean and maximizing value of this conditional density are both equal to E [ΘY = y ]. Therefore, ΘM M SE (y ) = ΘM AP (y ) = E (ΘY = y ). It is interesting to compare
this example to Example 5.1.2. The Bayesian estimators (MMSE and MAP) are both smaller
b2
in magnitude than θM L (y ) = y, by the factor b2 +σ2 . If b2 is small compared to σ 2 , the prior
information indicates that θ is believed to be small, resulting in the Bayes estimators being
smaller in magnitude than the ML estimator. As b2 → ∞, the priori distribution gets increasingly
uniform, and the Bayes estimators coverge to the ML estimator. 135 Example 5.1.6 Suppose Y is assumed to be a P oi(θ) random variable. Using the Bayesian
method, suppose the prior distribution for θ is the uniformly distribution over the interval [0, θmax ],
for some known value θmax . Given the observation Y = k for some ﬁxed k ≥ 0, the MAP estimator
is obtained by maximizing
e−θ θk I{0≤θ≤θθmax }
pY (k θ)fΘ (θ) =
k!
θmax
with respect to θ. As seen in Example 5.1.3, the term
decreasing in θ for θ > k . Therefore, e−θ θk
k! is increasing in θ for θ < k and ΘM AP (k ) = min{k, θmax }.
It is interesting to compare this example to Example 5.1.3. Intuitively, the prior probability distribution indicates knowledge that θ ≤ θmax , but no more than that, because the prior restricted to
θ ≤ θmax is uniform. If θmax is less than k , the MAP estimator is strictly smaller than θM L (k ) = k .
As θmax → ∞, the MAP estimator converges to the ML estimator. Actually, deterministic prior
knowledge, such as θ ≤ θmax , can also be incorporated into ML estimation as a hard constraint.
The next example makes use of the following lemma.
Lemma 5.1.7 Suppose ci ≥ 0 for 1 ≤ i ≤ n and that c = n ci > 0. Then
i=1
maximized over all probability vectors p = (p1 . . . . , pn ) by pi = ci /c. n
i=1 ci log pi is Proof. If cj = 0 for some j , then clearly pj = 0 for the maximizing probability vector. By
eliminating such terms from the sum, we can assume without loss of generality that ci > 0 for
all i. The function to be maximized is a strictly concave function of p over a region with linear
constraints. The positivity constraints, namely pi ≥ 0, will be satisﬁed with strict inequality.
The remaining constraint is the equality constraint, n pi = 1. We thus introduce a Lagrange
i=1
multiplier λ for the equality constraint and seek the stationary point of the Lagrangian L(p, λ) =
n
n
i=1 ci log pi − λ(( i=1 pi ) − 1). By deﬁnition, the stationary point is the point at which the partial
c
∂L
derivatives with respect to the variables pi are all zero. Setting ∂pi = pi − λ = 0 yields that pi = ci
λ
i
for all i. To satisfy the linear constraint, λ must equal c. Example 5.1.8 Suppose b = (b1 , b2 , . . . , bn ) is a probability vector to be estimated by observing
Y = (Y1 , . . . , YT ). Assume Y1 , . . . , YT are independent, with each Yt having probability distribution
b: P {Yt = i} = bi for 1 ≤ t ≤ T and 1 ≤ i ≤ n. We shall determine the maximum likelihood
estimate, bM L (y ), given a particular observation y = (y1 , . . . , yT ). The likelihood to be maximized
with respect to b is p(y b) = by1 · · · byT = n bki where ki = {t : yt = i}. The log likelihood is
i=1 i
ln p(y b) = n ki ln(bi ). By Lemma 5.1.7, this is maximized by the empirical distribution of the
i=1
observations, namely bi = ki for 1 ≤ i ≤ n. That is, bM L = ( k1 , . . . , kn ).
T
T
T 136 Example 5.1.9 This is a Bayesian version of the previous example. Suppose b = (b1 , b2 , . . . , bn )
is a probability vector to be estimated by observing Y = (Y1 , . . . , YT ), and assume Y1 , . . . , YT are
independent, with each Yt having probability distribution b. For the Bayesian method, a distribution
of the unknown distribution b must be assumed. That is right, a distribution of the distribution
is needed. A convenient choice is the following. Suppose for some known numbers di ≥ 1 that
(b1 , . . . , bn−1 ) has the prior density:
di −1
i=1 bi Qn fB (b) = Z (d) 0 n−1
i=1 bi if bi ≥ 0 for 1 ≤ i ≤ n − 1, and
else ≤1 where bn = 1 − b1 − · · · − bn−1 , and Z (d) is a constant chosen so that fB integrates to one. A larger
value of di for a ﬁxed i expresses an a priori guess that the corresponding value bi may be larger. It
di
can be shown, in particular, that if B has this prior distribution, then E [Bi ] = d1 +···dn . The MAP
estimate, bM AP (y ), for a given observation vector y, is given by:
n bM AP (y ) = arg max ln (fB (b)p(y b)) = arg max − ln(Z (d)) +
b b (di − 1 + ki ) ln(bi )
i=1 By Lemma 5.1.7, bM AP (y ) = ( d1 −1+k1 , . . . , dn −1+kn ), where T = n (di − 1+ ki ) = T − n + n di .
i=1
i=1
e
e
T
T
Comparison with Example 5.1.8 shows that the MAP estimate is the same as the ML estimate,
except that di − 1 is added to ki for each i. If the di ’s are integers, the MAP estimate is the ML
estimate with some prior observations mixed in, namely, di − 1 prior observations of outcome i for
each i. A prior distribution such that the MAP estimate has the same algebraic form as the ML
estimate is called a conjugate prior, and the speciﬁc density fB for this example is a called the
Dirichlet density with parameter vector d. Example 5.1.10 Suppose that Y = (Y1 , . . . , YT ) is observed, and that it is assumed that the Yi
are independent, with the binomial distribution with parameters n and q. Suppose n is known, and
q is an unknown parameter to be estimated from Y . Let us ﬁnd the maximum likelihood estimate,
qM L (y ), for a particular observation y = (y1 , . . . , yT ). The likelihood is
T p(y q ) =
t=1 n yt
q (1 − q )3−yt = cq s (1 − q )nT −s ,
yt where s = y1 + · · · + yT , and c depends on y but not on q. The log likelihood is ln c + s ln(q ) +
s
(nT − s) ln(1 − q ). Maximizing over q yields qM L = nT . An alternative way to think about this is
ˆ
to realize that each Yt can be viewed as the sum of n independent Bernoulli(q ) random variables,
and s can be viewed as the observed sum of nT independent Bernoulli(q ) random variables. 137 5.2 The expectationmaximization (EM) algorithm The expectationmaximization algorithm is a computational method for computing maximum likelihood estimates in contexts where there are hidden random variables, in addition to observed data
and unknown parameters. The following notation will be used.
θ, a parameter to be estimated
X, the complete data
pcd (xθ), the pmf of the complete data, which is a known function for each value of θ
Y = h(X ), the observed random vector
Z, the unobserved data (This notation is used in the common case that X has the form X =
(Y, Z ).)
We write p(y θ) to denote the pmf of Y for a given value of θ. It can be expressed in terms of the
pmf of the complete data by:
p(y θ) =
pcd (xθ)
(5.4)
{x:h(x)=y } In some applications, there can be a very large number of terms in the sum in (5.4), making it
diﬃcult to numerically maximize p(y θ) with respect to θ (i.e. to compute θM L (y )).
Algorithm 5.2.1 (Expectationmaximization (EM) algorithm) An observation y is given, along
with an intitial estimate θ(0) . The algorithm is iterative. Given θ(k) , the next value θ(k+1) is computed in the following two steps:
(Expectation step) Compute Q(θθ(k) ) for all θ, where
Q(θθ(k) ) = E [ log pcd (X θ)  y, θ(k) ]. (5.5) (Maximization step) Compute θ(k+1) ∈ arg maxθ Q(θθ(k) ). In other words, ﬁnd a value θ(k+1) of
θ that maximizes Q(θθ(k) ) with respect to θ.
Some intuition behind the algorithm is the following. If a vector of complete data x could
be observed, it would be reasonable to estimate θ by maximizing the pmf of the complete data,
pcd (xθ), with respect to θ. This plan is not feasible if the complete data is not observed. The idea is
to estimate log pcd (X θ) by its conditional expectation, Q(θθ(k) ), and then ﬁnd θ to maximize this
conditional expectation. The conditional expectation is well deﬁned if some value of the parameter
θ is ﬁxed. For each iteration of the algorithm, the expectation step is completed using the latest
value of θ, θ(k) , in computing the expectation of log pcd (X θ).
In most applications there is some additional structure that helps in the computation of Q(θθ(k) ).
This typically happens when pcd factors into simple terms, such as in the case of hidden Markov
models discussed in this chapter, or when pcd has the form of an exponential raised to a low degree
polynomial, such as the Gaussian or exponential distribution. In some cases there are closed form
expressions for Q(θθ(k) ). In others, there may be an algorithm that generates samples of X with
the desired pmf pcd (xθ(k) ) using random number generators, and then log pcd (X θ) is used as an
approximation to Q(θθ(k) ).
138 Example 5.2.2 (Estimation of the variance of a signal) An observation Y is modeled as Y = S +N,
where the signal S is assumed to be a N (0, θ) random variable, where θ is an unknown parameter,
assumed to satisfy θ ≥ 0, and the noise N is a N (0, σ 2 ) random variable where σ 2 is known and
strictly positive. Suppose it is desired to estimate θ, the variance of the signal. Let y be a particular
observed value of Y. We consider two approaches to ﬁnding θM L : a direct approach, and the EM
algorithm.
For the direct approach, note that for θ ﬁxed, Y is a N (0, θ + σ 2 ) random variable. Therefore,
the pdf of Y evaluated at y , or likelihood of y , is given by
2 f (y θ) = exp(− 2(θy σ2 ) )
+
2π (θ + σ 2 ) . The natural log of the likelihood is given by
log f (y θ) = − y2
log(2π ) log(θ + σ 2 )
−
−
.
2
2
2(θ + σ 2 ) Maximizing over θ yields θM L = (y 2 − σ 2 )+ . While this onedimensional case is fairly simple, the
situation is diﬀerent in higher dimensions, as explored in Problem 5.5. Thus, we examine use of
the EM algorithm for this example.
To apply the EM algorithm for this example, take X = (S, N ) as the complete data. The
observation is only the sum, Y = S + N, so the complete data is not observed. For given θ, S and
N are independent, so the log of the joint pdf of the complete data is given as follows:
log pcd (s, nθ) = − log(2πθ) s2
log(2πσ 2 )
n2
−
−
− 2.
2
2θ
2
2σ For the estimation step, we ﬁnd
Q(θθ(k) ) = E [ log pcd (S, N θ) y, θ(k) ]
=− log(2πθ) E [S 2 y, θ(k) ] log(2πσ 2 ) E [N 2 y, θ(k) ]
−
−
−
.
2
2θ
2
2σ 2 For the maximization step, we ﬁnd
∂Q(θθ(k) )
1
E [S 2 y, θ(k) ]
=− +
∂θ
2θ
2θ2
from which we see that θ(k+1) = E [S 2 y, θ(k) ]. Computation of E [S 2 y, θ(k) ] is an exercise in
conditional Gaussian distributions, similar to Example 3.4.5. The conditional second moment is
the sum of the square of the conditional mean and the variance of the estimation error. Thus, the
EM algorithm becomes the following recursion:
θ (k+1) = θ(k)
θ(k) + σ 2 2 y2 + θ(k) σ 2
θ (k ) + σ 2 Problem 5.3 shows that if θ(0) > 0, then θ(k) → θM L as k → ∞.
139 (5.6) The following proposition shows that the likelihood p(y θ(k) ) is nondecreasing in k. In the ideal
case, the likelihood converges to the maximum possible value of the likelihood, and limk→∞ θ(k) =
θM L (y ). However, the sequence could converge to a local, but not global, maximizer of the likelihood,
or possibly even to an inﬂection point of the likelihood. This behavior is typical of gradient type
nonlinear optimization algorithms, which the EM algorithm is similar to. Note that even if the
parameter set is convex (as it is for the case of hidden Markov models), the corresponding sets
of probability distributions on Y are not convex. It is the geometry of the set of probability
distributions that really matters for the EM algorithm, rather than the geometry of the space of
the parameters.
Proposition 5.2.3 (Convergence of the EM algorithm) Suppose that the complete data pmf can
be factored as pcd (xθ) = p(y θ)k (xy, θ) such that
(i) log p(y θ) is diﬀerentiable in θ
(ii) E ¯
¯
¯
log k (X y, θ)  y, θ is ﬁnite for all θ (iii) D(k (·y, θ)k (·y, θ )) is diﬀerentiable with respect to θ for any θ ﬁxed.
and suppose that p(y θ(0) ) > 0. Then the likelihood p(y θ(k) ) is nondecreasing in k , and any limit
point θ∗ of the sequence (θ(k) ) is a stationary point of the objective function p(y θ), which by
deﬁnition means
∂p(y θ)
θ=θ∗ = 0.
(5.7)
∂θ
The proof of Proposition 5.2.3 is given after the notion of divergence between probability vectors
is introduced.
Deﬁnition 5.2.4 The divergence between probability vectors p = (p1 , . . . , pn ) and q = (q1 , . . . , qn ),
denoted by D(pq ), is deﬁned by D(pq ) = i pi log(pi /qi ), with the understanding that pi log(pi /qi ) =
0 if pi = 0 and pi log(pi /qi ) = +∞ if pi > qi = 0.
Lemma 5.2.5 (Basic properties of divergence)
(i) D(pq ) ≥ 0, with equality if and only if p = q
(ii) D is a convex function of the pair (p, q ).
Proof. Property (i) follows from Lemma 5.1.7. Here is another proof. In proving (i), we can
u log u u > 0
assume that qi > 0 for all i. The function φ(u) =
is convex. Thus, by Jensen’s
0
u=0
inequality,
pi
pi
D(pq ) =
φ( )qi ≥ φ(
· qi ) = φ(1) = 0,
qi
qi
i i so (i) is proved.
The proof of (ii) is based on the logsum inequality, which is the fact that for nonnegative
numbers a1 , . . . , an , b1 , . . . , bn :
ai
a
ai log
≥ a log ,
(5.8)
bi
b
i
140 where a = i ai and b = i bi . To verify (5.8), note that it is true if and only if it is true with each
ai replaced by cai , for any strictly positive constant c. So it can be assumed that a = 1. Similarly,
it can be assumed that b = 1. For a = b = 1, (5.8) is equivalent to the fact D(ab) ≥ 0, already
proved. So (5.8) is proved.
j
j
Let 0 < α < 1. Suppose pj = (pj , . . . , pj ) and q j = (q1 , . . . , qn ) are probability distributions for
n
1
1
2
j = 1, 2, and let pi = αp1 + (1 − α)p2 and qi = αqi + (1 − α)qi , for 1 ≤ i ≤ n. That is, (p1 , q 1 ) and
i
i
(p2 , q 2 ) are two pairs of probability distributions, and (p, q ) = α(p1 , q 1 ) + (1 − α)(p2 , q 2 ). For i ﬁxed
1
i
with 1 ≤ i ≤ n, the logsum inequality (5.8) with (a1 , a2 , b1 , b2 ) = (αp1 , (1 − α)p2 , αqi , (1 − α)q2 )
i
i
yields
αp1 log
i p2
p1
2
i
i
1 + (1 − α)pi log q 2
qi
i = αp1 log
i
≥ pi log αp1
(1 − α)p2
2
i
i
1 + (1 − α)pi log (1 − α)q 2
αqi
i pi
.
qi Summing each side of this inequality over i yields αD(p1 q 1 ) + (1 − α)D(p2 q 2 ) ≥ D(pq ), so that
D(pq ) is a convex function of the pair (p, q ).
Proof of Proposition 5.2.3 Using the factorization pcd (xθ) = p(y θ)k (xy, θ),
Q(θθ(k) ) = E [log pcd (X θ)y, θ(k) ]
= log p(y θ) + E [ log k (X y, θ) y, θ(k) ]
k (X y, θ)
= log p(y θ) + E [ log
y, θ(k) ] + R
k (X y, θ(k) )
= log p(y θ) − D(k (·y, θ(k) )k (·y, θ)) + R, (5.9) where
R = E [ log k (X y, θ(k) ) y, θ(k) ].
By assumption (ii), R is ﬁnite, and it depends on y and θ(k) , but not on θ. Therefore, the maximization step of the EM algorithm is equivalent to:
θ(k+1) = arg max log p(y θ) − D(k (·y, θ(k) )k (·y, θ))
θ (5.10) Thus, at each step, the EM algorithm attempts to maximize the log likelihood ratio log p(y θ) itself,
minus a term which penalizes large diﬀerences between θ and θ(k) .
The deﬁnition of θ(k+1) implies that Q(θ(k+1) θ(k) ) ≥ Q(θ(k) θ(k) ). Therefore, using (5.9) and
the fact D(k (·y, θ(k) )k (·y, θ(k) )) = 0, yields
log p(y θ(k+1) ) − D(k (·y, θ(k+1) )k (·y, θ(k) )) ≥ log p(y θ(k) ) (5.11) In particular, since the divergence is nonnegative, p(y θ(k) ) is nondecreasing in k. Therefore,
limk→∞ log p(y θ(k) ) exists.
Suppose now that the sequence (θk ) has a limit point, θ∗ . By continuity, implied by the diﬀerentiability assumption (i), limk→∞ p(y θk ) = p(y θ∗ ) < ∞. For each k ,
0 ≤ max log p(y θ) − D k (·y, θ(k) )  k (·y, θ)
θ − log p(y θ(k) ) ≤ log p(y θ(k+1) ) − log p(y θ(k) ) → 0 as k → ∞,
141 (5.12)
(5.13) where (5.12) follows from the fact that θ(k) is a possible value of θ in the maximization, and the
inequality in (5.13) follows from (5.10) and the fact that the divergence is always nonnegative.
Thus, the quantity on the righthand side of (5.12) converges to zero as k → ∞. So by continuity,
for any limit point θ∗ of the sequence (θk ),
max [log p(y θ) − D (k (·y, θ∗ )  k (·y, θ))] − log p(y θ∗ ) = 0
θ and therefore,
θ∗ ∈ arg max [log p(y θ) − D (k (·y, θ∗ )  k (·y, θ))]
θ So the derivative of log p(y θ) − D (k (·y, θ)  k (·y, θ∗ )) with respect to θ at θ = θ∗ is zero. The
same is true of the term D (k (·y, θ)  k (·y, θ∗ )) alone, because this term is nonnegative, it has
value 0 at θ = θ∗ , and it is assumed to be diﬀerentiable in θ. Therefore, the derivative of the ﬁrst
term, log p(y θ), must be zero at θ∗ .
Remark 5.2.6 In the above proposition and proof, we assume that θ∗ is unconstrained. If there
are inequality constraints on θ and if some of them are tight for θ∗ , then we still ﬁnd that if
θ∗ is a limit point of θ(k) , then it is a maximizer of f (θ) = log p(y θ) − D (k (·y, θ)  k (·y, θ∗ )) .
Thus, under regularity conditions implying the existence of Lagrange multipliers, the KuhnTucker
optimality conditions are satisﬁed for the problem of maximizing f (θ). Since the derivatives of
D (k (·y, θ)  k (·y, θ∗ )) with respect to θ at θ = θ∗ are zero, and since the KuhnTucker optimality
conditions only involve the ﬁrst derivatives of the objective function, those conditions for the
problem of maximizing the true log likelihood function, log p(y θ), also hold at θ∗ . 5.3 Hidden Markov models A popular model of onedimensional sequences with dependencies, explored especially in the context
of speech processing, are the hidden Markov models. Suppose that
X = (Y, Z ), where Z is unobserved data and Y is the observed data
Z = (Z1 , . . . , ZT ) is a timehomogeneous Markov process, with onestep transition probability
matrix A = (ai,j ), and with Z1 having the initial distribution π. Here, T , with T ≥ 1, denotes
the total number of observation times. The statespace of Z is denoted by S , and the number
of states of S is denoted by Ns .
Y = (Y1 , . . . , Yn ) is the observed data. It is such that given Z = z, for some z = (z1 , . . . , zn ), the
variables Y1 , · · · , Yn are conditionally independent with P [Yt = lZ = z ] = bzt ,l , for a given
observation generation matrix B = (bi,l ). The observations are assumed to take values in a
set of size No , so that B is an Ns × No matrix and each row of B is a probability vector.
The parameter for this model is θ = (π, A, B ). The model is illustrated in Figure 5.1. The pmf of
the complete data, for a given choice of θ, is
T −1 pcd (y, z θ) = πz1 T azt ,zt+1
t=1 142 bzt ,yt .
t=1 (5.14) Z
1 ! Z
A 2 A Z
3 A ... A ZT B B B B Y
1 Y
2 Y
3 Y
T Figure 5.1: Structure of hidden Markov model.
The correspondence between the pmf and the graph shown in Figure 5.1 is that each term on the
righthand side of (5.14) corresponds to an edge in the graph.
In what follows we consider the following three estimation tasks associated with this model:
1. Given the observed data and θ, compute the conditional distribution of the state (solved by
the forwardbackward algorithm)
2. Given the observed data and θ, compute the most likely sequence for hidden states (solved
by the Viterbi algorithm)
3. Given the observed data, compute the maximum likelihood (ML) estimate of θ (solved by the
BaumWelch/EM algorithm).
These problems are addressed in the next three subsections. As we will see, the ﬁrst of these
problems arises in solving the third problem. The second problem has some similarities to the ﬁrst
problem, but it can be addressed separately. 5.3.1 Posterior state probabilities and the forwardbackward algorithm In this subsection we assume that the parameter θ = (π, A, B ) of the hidden Markov model is
known and ﬁxed. We shall describe computationally eﬃcient methods for computing posterior
probabilites for the state at a given time t, or for a transition at a given pair of times t to t + 1,
of the hidden Markov process, based on past observations (case of causal ﬁltering) or based on
past and future observations (case of smoothing). These posterior probabilities would allow us to
compute, for example, MAP estimates of the state or transition of the Markov process at a given
time. For example, we have: ZttM AP = arg max P [Zt = iY1 = y1 , . . . , Yt = yt , θ] (5.15) ZtT M AP = arg max P [Zt = iY1 = y1 , . . . , YT = yT , θ] (5.16) (Zt , Zt+1 )TM AP i∈S i∈S = arg max (i,j )∈S×S P [Zt = i, Zt+1 = j Y1 = y1 , . . . , YT = yT , θ], (5.17) where the conventions for subscripts is similar to that used for Kalman ﬁltering: “tT ” denotes
that the state is to be estimated at time t based on the observations up to time T . The key to
eﬃcient computation is to recursively compute certain quantities through a recursion forward in
143 time, and others through a recursion backward in time. We begin by deriving a forward recursion
for the variables αi (t) deﬁned as follows:
αi (t) = P [Y1 = y1 , · · · , Yt = yt , Zt = iθ],
for i ∈ S and 1 ≤ t ≤ T. The intial value is αi (1) = πi biy1 . By the law of total probability, the
update rule is:
P [Y1 = y1 , · · · , Yt+1 = yt+1 , Zt = i, Zt+1 = j θ] αj (t + 1) =
i∈S P [Y1 = y1 , · · · , Yt = yt , Zt = iθ] =
i∈S · P [Zt+1 = j, Yt+1 = yt+1 Y1 = y1 , · · · , Yt = yt , Zt = i, θ]
αi (t)aij bjyt+1 . =
i∈S The righthand side of (5.15) can be expressed in terms of the α’s as follows.
P [Zt = iY1 = y1 , . . . , Yt = yt , θ] =
= P [Zt = i, Y1 = y1 , . . . , Yt = yt θ]
P [Y1 = y1 , . . . , Yt = yt θ]
αi (t)
j ∈S αj (t) (5.18) The computation of the α’s and the use of (5.18) is an alternative, and very similar to, the Kalman
ﬁltering equations. The diﬀerence is that for Kalman ﬁltering equations, the distributions involved
are all Gaussian, so it suﬃces to compute means and variances, and also the normalization in (5.18),
which is done once after the α’s are computed, is more or less done at each step in the Kalman
ﬁltering equations.
To express the posterior probabilities involving both past and future observations used in (5.16),
the following β variables are introduced:
βi (t) = P [Yt+1 = yt+1 , · · · , YT = yT Zt = i, θ],
for i ∈ S and 1 ≤ t ≤ T. The deﬁnition is not quite the time reversal of the deﬁnition of the α’s,
because the event Zt = i is being conditioned upon in the deﬁnition of βi (t). This asymmetry is
introduced because the presentation of the model itself is not symmetric in time. The backward
equation for the β ’s is as follows. The intial condition for the backward equations is βi (T ) = 1 for
all i. By the law of total probability, the update rule is
βi (t − 1) = P [Yt = yt , · · · , YT = yT , Zt = j Zt−1 = i, θ]
j ∈S P [Yt = yt , Zt = j Zt−1 = i, θ] =
j ∈S · P [Yt+1 = yt , · · · , YT = yT , Zt = j, Yt = yt , Zt−1 = i, θ]
= aij bjyt βj (t).
j ∈S 144 Note that
P [Zt = i, Y1 = y1 , . . . , YT = yT θ] = P [Zt = i, Y1 = y1 , . . . , Yt = yt θ]
· P [Yt+1 = yt+1 , . . . , YT = yT θ, Zt = i, Y1 = y1 , . . . , Yt = yt ]
= P [Zt = i, Y1 = y1 , . . . , Yt = yt θ]
· P [Yt+1 = yt+1 , . . . , YT = yT θ, Zt = i]
= αi (t)βi (t)
from which we derive the smoothing equation for the conditional distribution of the state at a time
t, given all the observations:
γi (t) = P [Zt = iY1 = y1 , . . . , YT = yT , θ]
P [Zt = i, Y1 = y1 , . . . , YT = yT θ]
=
P [Y1 = y1 , . . . , YT = yT θ]
αi (t)βi (t)
=
j ∈S αj (t)βj (t)
The variable γi (t) deﬁned here is the same as the probability in the righthand side of (5.16), so
that we have an eﬃcient way to ﬁnd the MAP smoothing estimator deﬁned in (5.16). For later
use, we note from the above that for any i such that γi (t) > 0,
P [Y1 = y1 , . . . , YT = yT θ] = αi (t)βi (t)
.
γi (t) (5.19) Similarly,
P [Zt = i, Zt+1 = j, Y1 = y1 , . . . , YT = yT θ]
= P [Zt = i, Y1 = y1 , . . . , Yt = yt θ]
· P [Zt+1 = j, Yt+1 = yt+1 θ, Zt = i, Y1 = y1 , . . . , Yt = yt ]
· P [Yt+2 = yt+2 , . . . , YT = yT θ, Zt = i, Zt+1 = j, Y1 = y1 , . . . , Yt+1 = yt+1 ]
= αi (t)aij bjyt+1 βj (t + 1),
from which we derive the smoothing equation for the conditional distribution of a statetransition
for some pair of consecutive times t and t + 1, given all the observations:
ξij (t) = P [Zt = i, Zt+1 = j Y1 = y1 , . . . , YT = yT , θ]
P [Zt = i, Zt+1 = j, Y1 = y1 , . . . , YT = yT θ]
=
P [Y1 = y1 , . . . , YT = yT θ]
αi (t)aij bjyt+1 βj (t + 1)
=
i ,j αi (t)ai j bj yt+1 βj (t + 1)
= γi (t)aij bjyt+1 βj (t + 1)
,
βi (t) where the ﬁnal expression is derived using (5.19). The variable ξi,j (t) deﬁned here is the same as
the probability in the righthand side of (5.17), so that we have an eﬃcient way to ﬁnd the MAP
smoothing estimator of a state transition, deﬁned in (5.17).
Summarizing, the forwardbackward or α − β algorithm for computing the posterior distribution
of the state or a transition is given by:
145 Algorithm 5.3.1 (The forwardbackward algorithm) The α’s can be recursively computed forward
in time, and the β ’s recursively computed backward in time, using:
αi (t)aij bjyt+1 , with initial condition αi (1) = πi biy1 αj (t + 1) =
i∈S βi (t − 1) = aij bjyt βj (t), with initial condition βi (T ) = 1.
j ∈S Then the posterior probabilities can be found:
P [Zt = iY1 = y1 , . . . , Yt = yt , θ] =
γi (t) = P [Zt = iY1 = y1 , . . . , YT = yT , θ] =
ξij (t) = P [Zt = i, Zt+1 = j Y1 = y1 , . . . , YT = yT , θ] =
= αi (t)
j ∈S αj (t) (5.20) αi (t)βi (t)
j ∈S αj (t)βj (t) (5.21) αi (t)aij bjyt+1 βj (t + 1)
(5.22)
i ,j αi (t)ai j bj yt+1 βj (t + 1)
γi (t)aij bjyt+1 βj (t + 1)
.
βi (t) (5.23) Remark 5.3.2 If the number of observations runs into the hundreds or thousands, the α’s and β ’s
can become so small that underﬂow problems can be encountered in numerical computation. However, the formulas (5.20), (5.21), and (5.22) for the posterior probabilities in the forwardbackward
algorithm are still valid if the α’s and β ’s are multiplied by time dependent (but state independent)
constants (for this purpose, (5.22) is more convenient than (5.23), because (5.23) invovles β ’s at
two diﬀerent times). Then, the α’s and β ’s can be renormalized after each time step of computation
to have sum equal to one. Moreover, the sum of the logarithms of the normalization factors for the
α’s can be stored in order to recover the log of the likelihood, log p(y θ) = log Ns −1 αi (T ).
i=0 5.3.2 Most likely state sequence – Viterbi algorithm Suppose the parameter θ = (π, A, B ) is known, and that Y = (Y1 , . . . , YT ) is observed. In some
applications one wishes to have an estimate of the entire sequence Z. Since θ is known, Y and Z
can be viewed as random vectors with a known joint pmf, namely pcd (y, z θ). For the remainder
of this section, let y denote a ﬁxed observed sequence, y = (y1 , . . . , yT ). We will seek the MAP
estimate, ZM AP (y, θ), of the entire state sequence Z = (Z1 , . . . , ZT ), given Y = y. By deﬁnition, it
is deﬁned to be the z that maximizes the posterior pmf p(z y, θ), and as shown in Section 5.1, it is
also equal to the maximizer of the joint pmf of Y and Z :
ZM AP (y, θ) = arg max pcd (y, z θ)
z The Viterbi algorithm (a special case of dynamic programming), described next, is a computationally eﬃcient algorithm for simultaneously ﬁnding the maximizing sequence z ∗ ∈ S T and computing
pcd (y, z ∗ θ). It uses the variables:
δi (t) = max (z1 ,...,zt−1 )∈S t−1 P [Z1 = z1 , . . . , Zt−1 = zt−1 , Zt = i, Y1 = y1 , · · · , Yt = yt θ].
146 The δ ’s can be computed by a recursion forward in time, using the initial values δi (1) = π (i)biy1
and the recursion derived as follows:
δj (t) = max
i = max
i max P [Z1 = z1 , . . . , Zt−2 = zt−2 , Zt−1 = i, Zt = j, Y1 = y1 , · · · , Yt = yt θ] max P [Z1 = z1 , . . . , Zt−2 = zt−2 , Zt−1 = i, Y1 = y1 , · · · , Yt−1 = yt−1 θ]ai,j bjyt {z1 ,...,zt−2 }
{z1 ,...,zt−2 } = max {δi (t − 1)ai,j bjyt }
i Note that δi (T ) = maxz :zT =i pcd (y, z θ). Thus, the following algorithm correctly ﬁnds ZM AP (y, θ).
Algorithm 5.3.3 (Viterbi algorithm) Compute the δ ’s and associated back pointers by a recursion
forward in time:
(initial condition)
(recursive step) δi (1) = π (i)biy1
δj (t) = max{δi (t − 1)aij bj,yt }
i (5.24) (storage of back pointers) φj (t) = arg max{δi (t − 1)ai,j bj,yt }
i Then z ∗ = ZM AP (y, θ) satisﬁes pcd (y, z ∗ θ) = maxi δi (T ), and z ∗ is given by tracing backward in
time:
∗
∗
∗
zT = arg max δi (T ) and zt−1 = φzt (t) for 2 ≤ t ≤ T.
(5.25)
i 5.3.3 The BaumWelch algorithm, or EM algorithm for HMM The EM algorithm, introduced in Section 5.2, can be usefully applied to many parameter estimation
problems with hidden data. This section shows how to apply it to the problem of estimating the
parameter of a hidden Markov model from an observed output sequence. This results in the BaumWelch algorithm, which was developed earlier than the EM algorithm, in the particular context of
HMMs.
The parameter to be estimated is θ = (π, A, B ). The complete data consists of (Y, Z ) whereas
the observed, incomplete data consists of Y alone. The initial parameter θ(0) = (π (0) , A(0) , B (0) )
should have all entries strictly positive, because any entry that is zero will remain zero at the end
of an iteration. Suppose θ(k) is given. The ﬁrst half of an iteration of the EM algorithm is to
compute, or determine in closed form, Q(θθ(k) ). Taking logarithms in the expression (5.14) for the
pmf of the complete data yields
T −1 log pcd (y, z θ) = log πz1 + T log azt ,zt+1 +
t=1 log bzt ,yt
t=1 Taking the expectation yields
Q(θθ(k) ) = E [log pcd (y, Z θ)y, θ(k) ]
T −1 = T γi (1) log πi +
i∈S ξij (t) log ai,j +
t=1 i,j 147 γi (t) log bi,yt ,
t=1 i∈S where the variables γi (t) and ξi,j (t) are deﬁned using the model with parameter θ(k) . In view of this
closed form expression for Q(θθ(k) ), the expectation step of the EM algorithm essentially comes
down to computing the γ ’s and the ξ ’s. This computation can be done using the forwardbackward
algorithm, Algorithm 5.3.1, with θ = θ(k) .
The second half of an iteration of the EM algorithm is to ﬁnd the value of θ that maximizes
Q(θθ(k) ), and set θ(k+1) equal to that value. The parameter θ = (π, A, B ) for this problem can be
viewed as a set of probability vectors. Namely, π is a probability vector, and, for each i ﬁxed, aij
as j varies, and bil as l varies, are probability vectors. Therefore, Example 5.1.8 and Lemma 5.1.7
will be of use. Motivated by these, we rewrite the expression found for Q(θθ(k) ) to get
T −1 Q(θθ (k) )= γi (1) log πi +
i∈S T ξij (t) log ai,j + γi (t) log bi,yt
i∈S t=1 i,j t=1
T −1 = γi (1) log πi +
i∈S ξij (t) log ai,j
i,j t=1
T + γi (t)I{yt =l}
i∈S l log bi,l (5.26) t=1 The ﬁrst summation in (5.26) has the same form as the sum in Lemma 5.1.7. Similarly, for each i
ﬁxed, the sum over j involving ai,j , and the sum over l involving bi,l , also have the same form as
the sum in Lemma 5.1.7. Therefore, the maximization step of the EM algorithm can be written in
the following form:
(k+1) πi = γi (1) (k+1) = (k+1) = ai,j
bi,l T −1
t=1 ξi,j (t)
T −1
t=1 γi (t)
T
t=1 γi (t)I{yt =l}
T
t=1 γi (t) (5.27)
(5.28)
(5.29) The update equations (5.27)(5.29) have a natural interpretation. Equation (5.27) means that the
new value of the distribution of the initial state, (k+1) , is simply the posterior distribution of the
initial state, computed assuming θ(k) is the true parameter value. The other two update equations
are similar, but are more complicated because the transition matrix A and observation generation
matrix B do not change with time. The denominator of (5.28) is the posterior expected number of
times the state is equal to i up to time T − 1, and the numerator is the posterior expected number
of times two consecutive states are i, j. Thus, if we think of the time of a jump as being random,
the righthand side of (5.28) is the timeaveraged posterior conditional probability that, given the
state at the beginning of a transition is i at a typical time, the next state will be j. Similarly, the
righthand side of (5.29) is the timeaveraged posterior conditional probability that, given the state
is i at a typical time, the observation will be l.
Algorithm 5.3.4 (BaumWelch algorithm, or EM algorithm for HMM) Select the state space S ,
and in particular, the cardinality, Ns , of the state space, and let θ(0) denote a given initial choice
of parameter. Given θ(k) , compute θ(k+1) by using the forwardbackward algorithm (Algorithm
5.3.1) with θ = θ(k) to compute the γ ’s and ξ ’s. Then use (5.27)(5.29) to compute θ(k+1) =
(π (k+1) , A(k+1) , B (k+1) ).
148 5.4 Notes The EM algorithm is due to A.P. Dempster, N.M. Laird, and B.D. Rubin [3]. The paper includes
examples and a proof that the likelihood is increased with each iteration of the algorithm. An
article on the convergence of the EM algorithm is given in [16]. Earlier related work includes that
of Baum et al. [2], giving the BaumWelch algorithm. A tutorial on inference for HMMs and
applications to speech recognition is given in [11]. 5.5 Problems 5.1 Estimation of a Poisson parameter
Suppose Y is assumed to be a P oi(θ) random variable. Using the Bayesian method, suppose the
prior distribution of θ is the exponential distribution with some known parameter λ > 0. (a) Find
ΘM AP (k ), the MAP estimate of θ given that Y = k is observed, for some k ≥ 0.
(b) For what values of λ is ΘM AP (k ) ≈ θM L (k )? (The ML estimator was found in Example 5.1.3.)
Why should that be expected?
5.2 A variance estimation problem with Poisson observation
The input voltage to an optical device is X and the number of photons observed at a detector is
N . Suppose X is a Gaussian random variable with mean zero and variance σ 2 , and that given
X , the random variable N has the Poisson distribution with mean X 2 . (Recall that the Poisson
distribution with mean λ has probability mass function λn e−λ /n! for n ≥ 0.)
(a) Express P {N = n]} in terms of σ 2 . You can express this is the last candidate. You do not have
to perform the integration.
(b) Find the maximum likelihood estimator of σ 2 given N . (Caution: Estimate σ 2 , not X . Be as
explicit as possible–the ﬁnal answer has a simple form. Hint: You can ﬁrst simplify your answer to
e2n (2n
part (a) by using the fact that if X is a N (0, σ 2 ) random variable, then E [X 2n ] = σ n!2n )! . )
5.3 Convergence of the EM algorithm for an example
The purpose of this exercise is to verify for Example 5.2.2 that if θ(0) > 0, then θ(k) → θM L
as k → ∞. As shown in the example, θM L = (y 2 − σ 2 )+ . Let F (θ) = θ
θ+σ 2 2 y2 + θσ 2
θ+σ 2 so that the recursion (5.6) has the form θ(k+1) = F (θ(k) ). Clearly, over R+ , F is increasing and
bounded. (a) Show that 0 is the only nonnegative solution of F (θ) = θ if y ≤ σ 2 and that 0 and
y − σ 2 are the only nonnegative solutions of F (θ) = θ if y > σ 2 . (b) Show that for small θ > 0,
2 2− 2
θ
θ
1
θ
θ
θ
F (θ) = θ + θ (yσ4 σ ) + o(θ3 ). (Hint: For 0 < θ < σ 2 , θ+σ2 = σ2 1+θ/σ2 = σ2 (1 − σ2 + ( σ2 )2 − . . .).
(c) Sketch F and argue, using the above properties of F, that if θ(0) > 0, then θ(k) → θM L .
5.4 Transformation of estimators and estimators of transformations
Consider estimating a parameter θ ∈ [0, 1] from an observation Y . A prior density of θ is available
for the Bayesian estimators, MAP and MMSE, and the conditional density of Y given θ is known.
Answer the following questions and brieﬂy explain your answers.
(a) Does 3 + 5θM L = (3 + 5θ)M L ?
(b) Does (θM L )3 = (θ3 )M L ?
(c) Does 3 + 5θM AP = (3 + 5θ)M AP ?
149 (d) Does (θM AP )3 = (θ3 )M AP ?
(e) Does 3 + 5θM M SE = (3 + 5θ)M M SE ?
(f) Does (θM M SE )3 = (θ3 )M M SE ?
5.5 Using the EM algorithm for estimation of a signal variance
This problem generalizes Example 5.2.2 to vector observations. Suppose the observation is Y =
S + N , such that the signal S and noise N are independent random vectors in Rd . Assume that
S is N (0, θI ), and N is N (0, ΣN ), where θ, with θ > 0, is the parameter to be estimated, I is the
identity matrix, and ΣN is known.
(a) Suppose θ is known. Find the MMSE estimate of S , SM M SE , and ﬁnd an espression for the
covariance matrix of the error vector, S − SM M SE .
(b) Suppose now that θ is unknown. Describe a direct approach to computing θM L (Y ).
(c) Describe how θM L (Y ) can be computed using the EM algorithm.
(d) Consider how your answers to parts (b) and (c) simplify in case d = 2 and the covariance matrix
of the noise, ΣN , is the identity matrix.
5.6 Finding a most likely path
Consider an HMM with state space S = {0, 1}, observation space {0, 1, 2}, and parameter
θ = (π, A, B ) given by:
π = (a, a3 ) A= a a3
a3 a B= ca ca2 ca3
ca2 ca3 ca Here a and c are positive constants. Their actual numerical values aren’t important, other than
the fact that a < 1. Find the MAP state sequence for the observation sequence 021201, using the
Viterbi algorithm. Show your work.
5.7 An underconstrained estimation problem
Suppose the parameter θ = (π, A, B ) for an HMM is unknown, but that it is assumed that the
number of states Ns in the statespace S for (Zt ) is equal to the number of observations, T . Describe
a trivial choice of the ML estimator θM L (y ) for a given observation sequence y = (y1 , . . . , yT ). What
is the likelihood of y for this choice of θ?
5.8 Specialization of BaumWelch algorithm for no hidden data
(a) Determine how the BaumWelch algorithm simpliﬁes in the special case that B is the identity
matrix, so that Xt = Yt for all t. (b) Still assuming that B is the identity matrix, suppose that
S = {0, 1} and the observation sequence is 0001110001110001110001. Find the ML estimator for π
and A.
5.9 Free energy and the Boltzmann distribution
Let S denote a ﬁnite set of possible states of a physical system, and suppose the (internal) energy
of any state s ∈ S is given by V (s) for some function V on S . Let T > 0. The Helmholtz free
energy of a probability distribution Q on S is deﬁned to be the average (internal) energy minus the
temperature times entropy: F (Q) = i Q(i)V (i) + T i Q(i) log Q(i). Note that F is a convex
function of Q. (We’re assuming Boltzmann’s constant is normalized to one, so that T should
actually be in units of energy, but by abuse of notation we will call T the temperature.)
(a) Use the method of Lagrange multipliers to show that the Boltzmann distribution deﬁned by
150 1
BT (i) = Z (T ) exp(−V (i)/T ) minimizes F (Q). Here Z (T ) is the normalizing constant required to
make BT a probability distribution.
(b) Describe the limit of the Boltzmann distribution as T → ∞.
(c) Describe the limit of the Boltzmann distribution as T → 0. If it is possible to simulate a random
variable with the Boltzmann distribution, does this suggest an application?
(d) Show that F (Q) = T D(QBT ) + (term not depending on Q). Therefore, given an energy
function V on S and temperature T > 0, minimizing free energy over Q in some set is equivalent
to minimizing the divergence D(QBT ) over Q in the same set. 5.10 BaumWelch saddlepoint
Suppose that the BaumWelch algorithm is run on a given data set with initial parameter θ(0) =
(π (0) , A(0) , B (0) ) such that π (0) = π (0) A(0) (i.e., the initial distribution of the state is an equilibrium
distribution of the state) and every row of B (0) is identical. Explain what happens, assuming an
ideal computer with inﬁnite precision arithmetic is used.
5.11 Inference for a mixture model
(a) An observed random vector Y is distributed as a mixture of Gaussian distributions in d dimensions. The parameter of the mixture distribution is θ = (θ1 , . . . , θJ ), where θj is a ddimensional
vector for 1 ≤ j ≤ J . Speciﬁcally, to generate Y a random variable Z, called the class label for
the observation, is generated. The variable Z is uniformly distributed on {1, . . . , J }, and the conditional distribution of Y given (θ, Z ) is Gaussian with mean vector θZ and covariance the d × d
identity matrix. The class label Z is not observed. Assuming that θ is known, ﬁnd the posterior
pmf p(z y, θ). Give a geometrical interpretation of the MAP estimate Z for a given observation
Y = y.
(b) Suppose now that the parameter θ is random with the uniform prior over a very large region
and suppose that given θ, n random variables are each generated as in part (a), independently, to
produce
(Z (1) , Y (1) , Z (2) , Y (2) , . . . , Z (n) , Y (n) ). Give an explicit expression for the joint distribution
P (θ, z (1) , y (1) , z (2) , y (2) , . . . , z (n) , y (n) ).
(c) The iterative conditional modes (ICM) algorithm for this example corresponds to taking turns
maximizing P (θ, z (1) , y (1) , z (2) , y (2) , . . . , z (n) , y (n) ) with respect to θ for z ﬁxed and with respect to z
for θ ﬁxed. Give a simple geometric description of how the algorithm works and suggest a method
to initialize the algorithm (there is no unique answer for the later).
(d) Derive the EM algorithm for this example, in an attempt to compute the maximum likelihood
estimate of θ given y (1) , y (2) , . . . , y (n) .
5.12 Constraining the BaumWelch algorithm
The BaumWelch algorithm as presented placed no prior assumptions on the parameters π , A, B ,
other than the number of states Ns in the state space of (Zt ). Suppose matrices A and B are given
with the same dimensions as the matrices A and B to be esitmated, with all elements of A and
B having values 0 and 1. Suppose that A and B are constrained to satisfy A ≤ A and B ≤ B , in
the elementbyelement ordering (for example, aij ≤ aij for all i, j.) Explain how the BaumWelch
algorithm can be adapted to this situation.
5.13 * Implementation of algorithms
Write a computer program to (a) simulate a HMM on a computer for a speciﬁed value of the
151 paramter θ = (π, A, B ), (b) To run the forwardbackward algorithm and compute the α’s, β ’s, γ ’s,
and ξ ’s , (c) To run the BaumWelch algorithm. Experiment a bit and describe your results. For
example, if T observations are generated, and then if the BaumWelch algorithm is used to estimate
the paramter, how large does T need to be to insure that the estimates of θ are pretty accurate. 152 Chapter 6 Dynamics of CountableState Markov
Models
Markov processes are useful for modeling a variety of dynamical systems. Often questions involving
the longtime behavior of such systems are of interest, such as whether the process has a limiting
distribution, or whether timeaverages constructed using the process are asymptotically the same
as statistical averages. 6.1 Examples with ﬁnite state space Recall that a probability distribution π on S is an equilibrium probability distribution for a timehomogeneous Markov process X if π = πH (t) for all t. In the discretetime case, this condition
reduces to π = πP . We shall see in this section that under certain natural conditions, the existence
of an equilibrium probability distribution is related to whether the distribution of X (t) converges
as t → ∞. Existence of an equilibrium distribution is also connected to the mean time needed for
X to return to its starting state. To motivate the conditions that will be imposed, we begin by
considering four examples of ﬁnite state processes. Then the relevant deﬁnitions are given for ﬁnite
or countablyinﬁnite state space, and propositions regarding convergence are presented.
Example 6.1.1 Consider the discretetime Markov process with the onestep probability diagram
shown in Figure 6.1. Note that the process can’t escape from the set of states S1 = {a, b, c, d, e},
so that if the initial state X (0) is in S1 with probability one, then the limiting distribution is
supported by S1 . Similarly if the initial state X (0) is in S2 = {f, g, h} with probability one, then
the limiting distribution is supported by S2 . Thus, the limiting distribution is not unique for this
process. The natural way to deal with this problem is to decompose the original problem into two
problems. That is, consider a Markov process on S1 , and then consider a Markov process on S2 .
Does the distribution of X (0) necessarily converge if X (0) ∈ S1 with probability one? The
answer is no. For example, note that if X (0) = a, then X (k ) ∈ {a, c, e} for all even values of k ,
whereas X (k ) ∈ {b, d} for all odd values of k . That is, πa (k ) + πc (k ) + πe (k ) is one if k is even and
is zero if k is odd. Therefore, if πa (0) = 1, then π (k ) does not converge as k → ∞.
Basically speaking, the Markov process of Example 6.1.1 fails to have a unique limiting distribution independent of the initial state for two reasons: (i) the process is not irreducible, and (ii)
153 1
a 0.5 b c 1
0.5 0.5 0.5 0.5
e d f 0.5
1 0.5 1
g h 0.5 Figure 6.1: A onestep transition probability diagram with eight states.
the process is not aperiodic.
Example 6.1.2 Consider the twostate, continuous time Markov process with the transition rate
diagram shown in Figure 6.2 for some positive constants α and β . This was already considered in
Example 4.9.3, where we found that for any initial distribution π (0),
!
1 2
" Figure 6.2: A transition rate diagram with two states. lim π (t) = lim π (0)H (t) = ( t→∞ t→∞ β
α
,
).
α+β α+β The rate of convergence is exponential, with rate parameter α + β , which happens to be the nonzero
eigenvalue of Q. Note that the limiting distribution is the unique probability distribution satisfying
πQ = 0. The periodicity problem of Example 6.1.1 does not arise for continuoustime processes. Example 6.1.3 Consider the continuoustime Markov process with the transition rate diagram in
Figure 6.3. The Q matrix is the blockdiagonal matrix given by
!
1 !
2 3 " 4
" Figure 6.3: A transition rate diagram with four states. −α α
0
0 β −β 0
0 Q=
0
0 −α α 0
0
β −β
154 This process is not irreducible, but rather the transition rate diagram can be decomposed into two
parts, each equivalent to the diagram for Example 6.1.2. The equilibrium probability distributions
β
β
α
α
are the probability distributions of the form π = (λ α+β , λ α+β , (1 − λ) α+β , (1 − λ) α+β ), where λ is
the probability placed on the subset {1, 2}. Example 6.1.4 Consider the discretetime Markov process with the transition probability diagram
in Figure 6.4. The onestep transition probability matrix P is given by
1 2 1 1 1
3 Figure 6.4: A onestep transition probability diagram with three states. 010
P = 0 0 1 100
Solving the equation π = πP we ﬁnd there is a unique equilibrium probability vector, namely
π = ( 1 , 1 , 1 ). On the other hand, if π (0) = (1, 0, 0), then
333 (1, 0, 0) if k ≡ 0 mod 3
k
(0, 1, 0) if k ≡ 1 mod 3
π (k ) = π (0)P = (0, 0, 1) if k ≡ 2 mod 3
Therefore, π (k ) does not converge as k → ∞. 6.2 Classiﬁcation and convergence of discretetime Markov processes The following deﬁnition applies for either discrete time or continuous time.
Deﬁnition 6.2.1 Let X be a timehomogeneous Markov process on the countable state space S .
The process is said to be irreducible if for all i, j ∈ S , there exists s > 0 so that pij (s) > 0.
The next deﬁnition is relevant only for discretetime processes.
Deﬁnition 6.2.2 The period of a state i is deﬁned to be GCD{k ≥ 0 : pii (k ) > 0}, where “GCD”
stands for greatest common divisor. The set {k ≥ 0 : pii (k ) > 0} is closed under addition, which by
a result in elementary algebra1 implies that the set contains all suﬃciently large integer multiples
of the period. The Markov process is called aperiodic if the period of all the states is one.
1 Such as the Euclidean algorithm, Chinese remainder theorem, or Bezout theorem 155 Proposition 6.2.3 If X is irreducible, all states have the same period.
Proof. Let i and j be two states. By irreducibility, there are integers k1 and k2 so that pij (k1 ) > 0
and pji (k2 ) > 0. For any integer n, pii (n + k1 + k2 ) ≥ pij (k1 )pjj (n)pji (k2 ), so the set {k ≥ 0 :
pii (k ) > 0} contains the set {k ≥ 0 : pjj (k ) > 0} translated up by k1 + k2 . Thus the period of i is
less than or equal to the period of j . Since i and j were arbitrary states, the proposition follows.
For a ﬁxed state i, deﬁne τi = min{k ≥ 1 : X (k ) = i}, where we adopt the convention
that the minimum of an empty set of numbers is +∞. Let Mi = E [τi X (0) = i]. If P [τi <
+∞X (0) = i] < 1, then state i is called transient (and by convention, Mi = +∞). Otherwise
P[τi < +∞X (0) = i] = 1, and i is then said to be positive recurrent if Mi < +∞ and to be null
recurrent if Mi = +∞.
Proposition 6.2.4 Suppose X is irreducible and aperiodic.
(a) All states are transient, or all are positive recurrent, or all are null recurrent.
(b) For any initial distribution π (0), limt→∞ πi (t) = 1/Mi , with the understanding that the limit
is zero if Mi = +∞.
(c) An equilibrium probability distribution π exists if and only if all states are positive recurrent.
(d) If it exists, the equilibrium probability distribution π is given by πi = 1/Mi . (In particular, if
it exists, the equilibrium probability distribution is unique).
Proof. (a) Suppose state i is recurrent. Given X (0) = i, after leaving i the process returns to
state i at time τi . The process during the time interval {0, . . . , τi } is the ﬁrst excursion of X from
state 0. From time τi onward, the process behaves just as it did initially. Thus there is a second
excursion from i, third excursion from i, and so on. Let Tk for k ≥ 1 denote the length of the k th
excursion. Then the Tk ’s are independent, and each has the same distribution as T1 = τi . Let j be
another state and let denote the probability that X visits state j during one excursion from i.
Since X is irreducible, > 0. The excursions are independent, so state j is visited during the k th
excursion with probability , independently of whether j was visited in earlier excursions. Thus, the
number of excursions needed until state j is reached has the geometric distribution with parameter
, which has mean 1/ . In particular, state j is eventually visited with probability one. After j
is visited the process eventually returns to state i, and then within an average of 1/ additional
excursions, it will return to state j again. Thus, state j is also recurrent. Hence, if one state is
recurrent, all states are recurrent.
The same argument shows that if i is positive recurrent, then j is positive recurrent. Given
X (0) = i, the mean time needed for the process to visit j and then return to i is Mi / , since on
average 1/ excursions of mean length Mi are needed. Thus, the mean time to hit j starting from
i, and the mean time to hit i starting from j , are both ﬁnite. Thus, j is positive recurrent. Hence,
if one state is positive recurrent, all states are positive recurrent.
(b) Part (b) of the proposition follows by an application of the renewal theorem, which can be
found in [1].
(c) Suppose all states are positive recurrent. By the law of large numbers, for any state j , the
long run fraction of time the process is in state j is 1/Mj with probability one. Similarly, for any
states i and j , the long run fraction of time the process is in state j is γij /Mi , where γij is the
156 mean number of visits to j in an excursion from i. Therefore 1/Mj = γij /Mi . This implies that
i 1/Mi = 1. That is, π deﬁned by πi = 1/Mi is a probability distribution. The convergence for
each i separately given in part (b), together with the fact that π is a probability distribution, imply
that i πi (t) − πi  → 0. Thus, taking s to inﬁnity in the equation π (s)H (t) = π (s + t) yields
πH (t) = π , so that π is an equilibrium probability distribution.
Conversely, if there is an equilibrium probability distribution π , consider running the process
with initial state π . Then π (t) = π for all t. So by part (b), for any state i, πi = 1/Mi . Taking a
state i such that πi > 0, it follows that Mi < ∞. So state i is positive recurrent. By part (a), all
states are positive recurrent.
(d) Part (d) was proved in the course of proving part (c).
We conclude this section by describing a technique to establish a rate of convergence to the
equilibrium distribution for ﬁnitestate Markov processes. Deﬁne δ (P ) for a onestep transition
probability matrix P by
δ (P ) = min
pij ∧ pkj ,
i,k j where a ∧ b = min{a, b}. The number δ (P ) is known as Dobrushin’s coeﬃcient of ergodicity. Since
a + b − 2(a ∧ b) = a − b for a, b ≥ 0, we also have
1 − 2δ (P ) = min pij − pkj . i,k j
i µi . Let µ for a vector µ denote the L1 norm: µ = Proposition 6.2.5 For any probability vectors π and σ , π P − σP ≤ (1 − δ (P )) π − σ . Furthermore, if δ (P ) > 0 then there is a unique equilibrium distribution π ∞ , and for any other probability
distribution π on S , π P l − π ∞ ≤ 2(1 − δ (P ))l .
Proof. Let πi = πi − πi ∧ σi and σi = σi − πi ∧ σi . Note that if πi ≥ σi then πi = πi − σi
˜
˜
˜
and σi = 0, and if πi ≤ σi then σi = σi − πi and πi = 0. Also, π and σ are both equal to
˜
˜
˜
˜
˜
1 − i πi ∧ σi . Therefore, π − σ = π − σ = 2 π = 2 σ . Furthermore,
˜˜
˜
˜
π P − σP = πP − σP
˜
˜
πi Pij −
˜ =
j i  = (1/ π )
˜
j ≤ (1/ π )
˜ πi σk (Pij − Pkj )
˜˜
i,k Pij − Pkj  π i σk
˜˜
i,k ≤ σk Pkj
˜
k j π (2 − 2δ (P )) = π − σ (1 − δ (P )),
˜ which proves the ﬁrst part of the proposition. Iterating the inequality just proved yields that
π P l − σP l ≤ (1 − δ (P ))l π − σ ≤ 2(1 − δ (P ))l . (6.1) This inequality for σ = πP n yields that π P l − πP l+n ≤ 2(1 − δ (P ))l . Thus the sequence πP l is
a Cauchy sequence and has a limit π ∞ , and π ∞ P = π ∞ . Finally, taking σ in (6.1) equal to π ∞
yields the last part of the proposition.
157 Proposition 6.2.5 typically does not yield the exact asymptotic rate that π l − π ∞ tends to
zero. The asymptotic behavior can be investigated by computing (I − zP )−1 , and then matching
powers of z in the identity (I − zP )−1 = ∞ z n P n .
n=0 6.3 Classiﬁcation and convergence of continuoustime Markov processes Chapter 4 discusses Markov processes in continuous time with a ﬁnite number of states. Here
we extend the coverage of continuoustime Markov processes to include countably inﬁnitely many
states. For example, the state of a simple queue could be the number of customers in the queue,
and if there is no upper bound on the number of customers that can be waiting in the queue, the
state space is Z+ . One possible complication, that rarely arises in practice, is that a continuous
time process can make inﬁnitely many jumps in a ﬁnite amount of time.
Let S be a ﬁnite or countably inﬁnite set, and let
∈ S . A purejump function is a function
x : R+ → S ∪ { } such that there is a sequence of times, 0 = τ0 < τ1 < . . . , and a sequence of
states, s0 , s1 , . . . with si ∈ S , and si = si+1 , i ≥ 0, so that
si x(t) = if τi ≤ t < τi+1 i ≥ 0
if t ≥ τ ∗ (6.2) where τ ∗ = limi→∞ τi . If τ ∗ is ﬁnite it is said to be the explosion time of the function x, and if
τ ∗ = +∞ the function is said to be nonexplosive. The example corresponding to S = {0, 1, . . .},
τi = i/(i + 1) and si = i is pictured in Fig. 6.5. Note that τ ∗ = 1 for this example.
x(t) !0 !1 !2. . . !" t Figure 6.5: A purejump function with an explosion time.
Deﬁnition 6.3.1 A purejump Markov process (Xt : t ≥ 0) is a Markov process such that, with
probability one, its sample paths are purejump functions. Such a process is said to be nonexplosive
if its sample paths are nonexplosive, with probability one.
Generator matrices are deﬁned for countablestate Markov processes just as they are for ﬁnitestate Markov processes. A purejump, timehomogeneous Markov process X has generator matrix
Q = (qij : i, j ∈ S ) if
lim (pij (h) − I{i=j } )/h = qij
i, j ∈ S
(6.3)
h 0 158 or equivalently
pij (h) = I{i=j } + hqij + o(h) i, j ∈ S (6.4) where o(h) represents a quantity such that limh→0 o(h)/h = 0.
The spacetime properties for continuoustime Markov processes with a countably inﬁnite number of states are the same as for a ﬁnite number of states. There is a discretetime jump process,
and the holding times, given the jump process, are exponentially distributed. Also, the following
holds.
Proposition 6.3.2 Given a matrix Q = (qij : i, j ∈ S ) satisfying qij ≥ 0 for distinct states i and j ,
and qii = − j ∈S ,j =i qij for each state i, and a probability distribution π (0) = (πi (0) : i ∈ S ), there
is a purejump, timehomogeneous Markov process with generator matrix Q and initial distribution
π (0). The ﬁnitedimensional distributions of the process are uniquely determined by π (0) and
Q. The ChapmanKolmogorov equations, H (s, t) = H (s, τ )H (τ, t), and the Kolmogorov forward
∂πj (t)
equations, ∂t = i∈S πi (t)qij , hold.
Example 6.3.3 (Birthdeath processes) A useful class of countablestate Markov processes is the
set of birthdeath processes. A (continuoustime) birthdeath process with parameters (λ0 , λ2 , . . .)
and (µ1 , µ2 , . . .) (also set λ−1 = µ0 = 0) is a purejump Markov process with state space S = Z+
and generator matrix Q deﬁned by qkk+1 = λk , qkk = −(µk + λk ), and qkk−1 = µk for k ≥ 0, and
qij = 0 if i − j  ≥ 2. The transition rate diagram is shown in Fig. 6.6. The spacetime structure, as !0
0 µ1 !1
1 !3 !2 µ2 2 µ3 3
µ4 Figure 6.6: Transition rate diagram of a birthdeath process.
deﬁned in Section 4.10, of such a process is as follows. Given the process is in state k at time t, the
next state visited is k +1 with probability λk /(λk + µk ) and k − 1 with probability µk /(λk + µk ). The
holding time of state k is exponential with parameter λk + µk . The Kolmogorov forward equations
for birthdeath processes are
∂πk (t)
= λk−1 πk−1 (t) − (λ + µ)πk (t) + µk+1 πk+1 (t)
∂t (6.5) Example 6.3.4 (Description of a Poisson process as a Markov process) Let λ > 0 and consider a
birthdeath process N with λk = λ and µk = 0 for all k , with initial state zero with probability
one. The spacetime structure of this Markov process is then rather simple. Each transition is an
upward jump of size one, so the jump process is deterministic: N J (k ) = k for all k . Ordinarily,
the holding times are only conditionally independent given the jump process, but since the jump
process is deterministic, the holding times are independent. Also, since qk,k = −λ for all k , each
holding time is exponentially distributed with parameter λ. Therefore, N satisﬁes condition (b) of
Proposition 4.5.2, so that N is a Poisson process with rate λ.
159 Deﬁne, for i ∈ S , τio = min{t > 0 : X (t) = i}, and τi = min{t > τio : X (t) = i}. Thus, if
X (0) = i, τi is the ﬁrst time the process returns to state i, with the exception that τi = +∞ if the
process never returns to state i. The following deﬁnitions are the same as when X is a discretetime
process. Let Mi = E [τi X (0) = i]. If P [τi < +∞] < 1, then state i is called transient. Otherwise
P [τi < +∞] = 1, and i is then said to be positive recurrent if Mi < +∞ and to be null recurrent if
Mi = +∞. The following propositions are analogous to those for discretetime Markov processes.
Proofs can be found in [1, 10].
Proposition 6.3.5 Suppose X is irreducible.
(a) All states are transient, or all are positive recurrent, or all are null recurrent.
(b) For any initial distribution π (0), limt→+∞ πi (t) = 1/(−qii Mi ), with the understanding that the
limit is zero if Mi = +∞.
Proposition 6.3.6 Suppose X is irreducible and nonexplosive.
(a) A probability distribution π is an equilibrium distribution if and only if πQ = 0.
(b) An equilibrium probability distribution exists if and only if all states are positive recurrent.
(c) If all states are positive recurrent, the equilibrium probability distribution is given by πi =
1/(−qii Mi ). (In particular, if it exists, the equilibrium probability distribution is unique).
The assumption that X be nonexplosive is needed for Proposition 6.3.6(a) (per Problem 6.9),
but the following proposition shows that the Markov processes encountered in most applications
are nonexplosive.
Proposition 6.3.7 Suppose X is irreducible. Fix a state io and for k ≥ 1 let Sk denote the set of
states reachable from io in k jumps. Suppose for each k ≥ 1 there is a constant γk so that the jump
intensities on Sk are bounded by γk , that is, suppose −qii ≤ γk for i ∈ Sk . If ∞ γ1 = +∞, then
k=1 k
the process X is nonexplosive. 6.4 Classiﬁcation of birthdeath processes The classiﬁcation of birthdeath processes, introduced in Example 6.3.3, is relatively simple. To
avoid trivialities, consider a birthdeath process such that the birth rates, (λi : i ≥ 0) and death
rates (µi : i ≥ 1) are all strictly positive. Then the process is irreducible.
First, investigate whether the process is nonexplosive, because this is a necessary condition for
both recurrence and positive recurrence. This is usually a simple matter, because if the rates are
bounded or grow at most linearly, the process is nonexplosive by Proposition 6.3.7. In some cases,
even if Proposition 6.3.7 doesn’t apply, it can be shown by some other means that the process
is nonexplosive. For example, a test is given below for the process to be recurrent, and if it is
recurrent, it is not explosive.
Next, investigate whether X is positive recurrent. Suppose we already know that the process
is nonexplosive. Then the process is positive recurrent if and only if πQ = 0 for some probability
160 distribution π, and if it is positive recurrent, π is the equilibrium distribution. Now πQ = 0 if and
only if ﬂow balance holds for any state k :
(λk + µk )πk = λk−1 πk−1 + µk+1 πk+1 . (6.6) Equivalently, ﬂow balance must hold for all sets of the form {0, . . . , n − 1} (just sum each side
of (6.6) over k ∈ {1, . . . , n − 1}). Therefore, πQ = 0 if and only if πn−1 λn−1 = πn µn for n ≥ 1,
which holds if and only if there is a probability distribution π with πn = π0 λ0 . . . λn−1 /(µ1 . . . µn )
for n ≥ 1. Thus, a probability distribution π with πQ = 0 exists if and only if S1 < +∞, where
∞ S1 =
i=0 λ0 . . . λi−1
,
µ 1 . . . µi (6.7) with the understanding that the i = 0 term in the sum deﬁning S1 is one. Thus, under the
assumption that X is nonexplosive, X is positive recurrent if and only if S1 < ∞, and if X is
positive recurrent, the equilibrium distribution is given by πn = (λ0 . . . λn−1 )/(S1 µ1 . . . µn ).
Finally, investigate whether X is recurrent. This step is not necessary if we already know that
X is positive recurrent, because a positive recurrent process is recurrent. The following test for
recurrence is valid whether or not X is explosive. Since all states have the same classiﬁcation,
the process is recurrent if and only if state 0 is recurrent. Thus, the process is recurrent if the
probability the process never hits 0, for initial state 1, is zero. We shall ﬁrst ﬁnd the probability of
never hitting state zero for a modiﬁed process, which stops upon reaching a large state n, and then
let n → ∞ to ﬁnd the probability the original process never reaches state 0. Let bin denote the
probability, for initial state i, the process does not reach zero before reaching n. Set the boundary
conditions, b0n = 0 and bnn = 1. Fix i with 1 ≤ i ≤ n − 1, and derive an expression for bin by ﬁrst
conditioning on the state reached by the ﬁrst jump of the process, starting from state i. By the
spacetime structure, the probability the ﬁrst jump is up is λi /(λi + µi ) and the probability the
ﬁrst jump is down is µi /(λi + µi ). Thus,
bin = λi
µi
bi+1,n +
bi−1,n ,
λ i + µi
λi + µ i which can be rewritten as µi (bin − bi−1,n ) = λi (bi+1,n − bi,n ). In particular, b2n − b1n = b1n µ1 /λ1
and b3n − b2n = b1n µ1 µ2 /(λ1 λ2 ), and so on, which upon summing yields the expression
k−1 bkn = b1n
i=0 µ 1 µ 2 . . . µi
.
λ1 λ2 . . . λi with the convention that the i = 0 term in the sum is one. Finally, the condition bnn = 1 yields
the solution
1
b1n = n−1 µ1 µ2 ...µi .
(6.8)
i=0 λ1 λ2 ...λi Note that b1n is the probability, for initial state 1, of the event Bn that state n is reached without
an earlier visit to state 0. Since Bn+1 ⊂ Bn for all n ≥ 1,
P [∩n≥1 Bn X (0) = 1] = lim b1n = 1/S2
n→∞ 161 (6.9) where ∞ S2 =
i=0 µ 1 µ 2 . . . µi
,
λ1 λ2 . . . λi with the understanding that the i = 0 term in the sum deﬁning S2 is one. Due to the deﬁnition
of pure jump processes used, whenever X visits a state in S the number of jumps up until that
time is ﬁnite. Thus, on the event ∩n≥1 Bn , state zero is never reached. Conversely, if state zero is
never reached, then either the process remains bounded (which has probability zero) or ∩n≥1 Bn is
true. Thus, P [zero is never reachedX (0) = 1] = 1/S2 . Consequently, X is recurrent if and only if
S2 = ∞.
In summary, the following proposition is proved.
Proposition 6.4.1 Suppose X is a continuoustime birthdeath process with strictly positive birth
rates and death rates. If X is nonexplosive (for example, if the rates are bounded or grow at most
linearly with n, or if S2 = ∞) then X is positive recurrent if and only if S1 < +∞. If X is positive
recurrent the equilibrium probability distribution is given by πn = (λ0 . . . λn−1 )/(S1 µ1 . . . µn ).
The process X is recurrent if and only if S2 = ∞.
Discretetime birthdeath processes have a similar characterization. They are discretetime,
timehomogeneous Markov processes with state space equal to the set of nonnegative integers. Let
nonnegative birth probabilities (λk : k ≥ 0) and death probabilities (µk : k ≥ 1) satisfy λ0 ≤ 1, and
λk + µk ≤ 1 for k ≥ 1. The onestep transition probability matrix P = (pij : i, j ≥ 0) is given by λi
if j = i + 1 µi
if j = i − 1 1 − λi − µi if j = i ≥ 1
pij =
(6.10) 1 − λ0
if j = i = 0 0
else.
Implicit in the speciﬁcation of P is that births and deaths can’t happen simultaneously. If the birth
and death probabilities are strictly positive, Proposition 6.4.1 holds as before, with the exception
that the discretetime process cannot be explosive.2 6.5 Time averages vs. statistical averages Let X be a positive recurrent, irreducible, timehomogeneous Markov process with equilibrium
probability distribution π . To be deﬁnite, suppose X is a continuoustime process, with purejump
sample paths and generator matrix Q. The results of this section apply with minor modiﬁcations
to the discretetime setting as well. Above it is noted that limt→∞ πi (t) = πi = 1/(−qii Mi ), where
Mi is the mean “cycle time” of state i. A related consideration is convergence of the empirical
distribution of the Markov process, where the empirical distribution is the distribution observed
over a (usually large) time interval.
For a ﬁxed state i, the fraction of time the process spends in state i during [0, t] is
1
t
2 t I{X (s)=i} ds
0 If in addition λi + µi = 1 for all i, then the discretetime process has period 2. 162 Let T0 denote the time that the process is ﬁrst in state i, and let Tk for k ≥ 1 denote the time
that the process jumps to state i for the k th time after T0 . The cycle times Tk+1 − Tk , k ≥ 0 are
independent and identically distributed, with mean Mi . Therefore, by the law of large numbers,
with probability one,
1
lim Tk /(kMi ) = lim
k→∞
k→∞ kMi k −1 (Tl+1 − Tl )
l=0 =1
Furthermore, during the k th cycle interval [Tk , Tk+1 ), the amount of time spent by the process
in state i is exponentially distributed with mean −1/qii , and the time spent in the state during
disjoint cycles is independent. Thus, with probability one,
Tk 1
lim
k→∞ kMi 0 1
I{X (s)=i} ds = lim
k→∞ kMi k −1 Tl+1 I{X (s)=i} ds
Tl l=0 T1
1
E
I{X (s)=i} ds
Mi
T0
= 1/(−qii Mi ) = Combining these two observations yields that
1
t→∞ t t I{X (s)=i} ds = 1/(−qii Mi ) = πi lim (6.11) 0 with probability one. In short, the limit (6.11) is expected, because the process spends on average
−1/qii time units in state i per cycle from state i, and the cycle rate is 1/Mi . Of course, since state
i is arbitrary, if j is any other state then
1
t→∞ t t I{X (s)=j } ds = 1/(−qjj Mj ) = πj lim (6.12) 0 By considering how the time in state j is distributed among the cycles from state i, it follows that
the mean time spent in state j per cycle from state i is Mi πj .
So for any nonnegative function φ on S ,
1
t→∞ t t lim Tk
1
φ(X (s))ds
k→∞ kMi 0
T1
1
=
E
φ(X (s))ds
Mi
T0 T1
1
=
E
φ(j )
I{X (s)=j } ds
Mi
T0 φ(X (s))ds =
0 lim j ∈S = 1
Mi = T1 φ(j )E
j ∈S I{X (s)=j } ds
T0 φ(j )πj (6.13) j ∈S Finally, if φ is a function of S such that either j ∈S φ+ (j )πj < ∞ or
since (6.13) holds for both φ+ and φ− , it must hold for φ itself.
163 j ∈S φ− (j )πj < ∞, then queue server system
Figure 6.7: A single server queueing system. 6.6 Queueing systems, M/M/1 queue and Little’s law Some basic terminology of queueing theory will now be explained. A simple type of queueing system
is pictured in Figure 6.7. Notice that the system is comprised of a queue and a server. Ordinarily
whenever the system is not empty, there is a customer in the server, and any other customers in
the system are waiting in the queue. When the service of a customer is complete it departs from
the server and then another customer from the queue, if any, immediately enters the server. The
choice of which customer to be served next depends on the service discipline. Common service
disciplines are ﬁrstcome ﬁrstserved (FCFS) in which customers are served in the order of their
arrival, or lastcome ﬁrstserved (LCFS) in which the customer that arrived most recently is served
next. Some of the more complicated service disciplines involve priority classes, or the notion of
“processor sharing” in which all customers present in the system receive equal attention from the
server.
Often models of queueing systems involve a stochastic description. For example, given positive
parameters λ and µ, we may declare that the arrival process is a Poisson process with rate λ,
and that the service times of the customers are independent and exponentially distributed with
parameter µ. Many queueing systems are given labels of the form A/B/s, where “A” is chosen to
denote the type of arrival process, “B” is used to denote the type of departure process, and s is
the number of servers in the system. In particular, the system just described is called an M/M/1
queueing system, sonamed because the arrival process is memoryless (i.e. a Poisson arrival process),
the service times are memoryless (i.e. are exponentially distributed), and there is a single server.
Other labels for queueing systems have a fourth descriptor and thus have the form A/B/s/b, where
b denotes the maximum number of customers that can be in the system. Thus, an M/M/1 system
is also an M/M/1/∞ system, because there is no ﬁnite bound on the number of customers in the
system.
A second way to specify an M/M/1 queueing system with parameters λ and µ is to let A(t) and
D(t) be independent Poisson processes with rates λ and µ respectively. Process A marks customer
arrival times and process D marks potential customer departure times. The number of customers in
the system, starting from some initial value N (0), evolves as follows. Each time there is a jump of
A, a customer arrives to the system. Each time there is a jump of D, there is a potential departure,
meaning that if there is a customer in the server at the time of the jump then the customer departs.
If a potential departure occurs when the system is empty then the potential departure has no eﬀect
164 on the system. The number of customers in the system N can thus be expressed as
t N (t) = N (0) + A(t) + I{N (s−)≥1} dD(s)
0 It is easy to verify that the resulting process N is Markov, which leads to the third speciﬁcation of
an M/M/1 queueing system.
A third way to specify an M/M/1 queuing system is that the number of customers in the system
N (t) is a birthdeath process with λk = λ and µk = µ for all k , for some parameters λ and µ. Let
ρ = λ/µ. Using the classiﬁcation criteria derived for birthdeath processes, it is easy to see that
the system is recurrent if and only if ρ ≤ 1, and that it is positive recurrent if and only if ρ < 1.
Moreover, if ρ < 1 then the equilibrium distribution for the number of customers in the system is
given by πk = (1 − ρ)ρk for k ≥ 0. This is the geometric distribution with zero as a possible value,
and with mean
∞ ∞ ρk−1 k = (1 − ρ)ρ( kπk = (1 − ρ)ρ N=
k=0 k=1 ρ
1
)=
1−ρ
1−ρ The probability the server is busy, which is also the mean number of customers in the server, is
1 − π0 = ρ. The mean number of customers in the queue is thus given by ρ/(1 − ρ) − ρ = ρ2 /(1 − ρ).
This third speciﬁcation is the most commonly used way to deﬁne an M/M/1 queueing process.
Since the M/M/1 process N (t) is positive recurrent, the Markov ergodic convergence theorem
implies that the statistical averages just computed, such as N , are also equal to the limit of the
timeaveraged number of customers in the system as the averaging interval tends to inﬁnity.
An important performance measure for a queueing system is the mean time spent in the system
or the mean time spent in the queue. Littles’ law, described next, is a quite general and useful
relationship that aids in computing mean transit time.
Little’s law can be applied in a great variety of circumstances involving ﬂow through a system
with delay. In the context of queueing systems we speak of a ﬂow of customers, but the same
principle applies to a ﬂow of water through a pipe. Little’s law is that λT = N where λ is the
mean ﬂow rate, T is the mean delay in the system, and N is the mean content of the system. For
example, if water ﬂows through a pipe with volume one cubic meter at the rate of two cubic meters
per minute, then the mean time (averaged over all drops of water) that water spends in the pipe is
T = N /λ = 1/2 minute. This is clear if water ﬂows through the pipe without mixing, for then the
transit time of each drop of water is 1/2 minute. However, mixing within the pipe does not eﬀect
the average transit time.
Little’s law is actually a set of results, each with somewhat diﬀerent mathematical assumptions.
The following version is quite general. Figure 6.8 pictures the cumulative number of arrivals (α(t))
and the cumulative number of departures (δ (t)) versus time, for a queueing system assumed to be
initially empty. Note that the number of customers in the system at any time s is given by the
diﬀerence N (s) = α(s) − δ (s), which is the vertical distance between the arrival and departure
graphs in the ﬁgure. On the other hand, assuming that customers are served in ﬁrstcome ﬁrstserved order, the horizontal distance between the graphs gives the times in system for the customers.
Given a (usually large) t > 0, let γt denote the area of the region between the two graphs over the
interval [0, t]. This is the shaded region indicated in the ﬁgure. It is then natural to deﬁne the
timeaveraged values of arrival rate and system content as
λt = α(t)/t and Nt =
165 1
t t N (s)ds = γt /t
0 ! s
N(s) "s s
t Figure 6.8: A single server queueing system.
Finally, the average, over the α(t) customers that arrive during the interval [0, t], of the time spent
in the system up to time t, is given by
T t = γt /α(t).
Once these deﬁnitions are accepted, we have the following obvious proposition.
Proposition 6.6.1 (Little’s law, expressed using averages over time) For any t > 0,
N t = λt T t (6.14) Furthermore, if any two of the three variables in (6.14) converge to a positive ﬁnite limit as t → ∞,
then so does the third variable, and the limits satisfy N ∞ = λ∞ T ∞ .
For example, the number of customers in an M/M/1 queue is a positive recurrent Markov
process so that
lim N t = N = ρ/(1 − ρ) t→∞ where calculation of the statistical mean N was previously discussed. Also, by the law of large
numbers applied to interarrival times, we have that the Poisson arrival process for an M/M/1
queue satisﬁes limt→∞ λt = λ with probability one. Thus, with probability one,
lim T t = N /λ = t→∞ 1
.
µ−λ In this sense, the average waiting time in an M/M/1 system is 1/(µ − λ). The average time in service
is 1/µ (this follows from the third description of an M/M/1 queue, or also from Little’s law applied
to the server alone) so that the average waiting time in queue is given by W = 1/(µ − λ) − 1/µ =
ρ/(µ − λ). This ﬁnal result also follows from Little’s law applied to the queue alone.
166 6.7 Mean arrival rate, distributions seen by arrivals, and PASTA The mean arrival rate for the M/M/1 system is λ, the parameter of the Poisson arrival process.
However for some queueing systems the arrival rate depends on the number of customers in the
system. In such cases the mean arrival rate is still typically meaningful, and it can be used in
Little’s law.
Suppose the number of customers in a queuing system is modeled by a birth death process
with arrival rates (λk ) and departure rates (µk ). Suppose in addition that the process is positive
recurrent. Intuitively, the process spends a fraction of time πk in state k and while in state k the
arrival rate is λk . Therefore, the average arrival rate is
∞ λ= πk λk
k=0 Similarly the average departure rate is
∞ µ= πk µk
k=1 and of course λ = µ because both are equal to the throughput of the system.
Often the distribution of a system at particular systemrelated sampling times are more important than the distribution in equilibrium. For example, the distribution seen by arriving customers
may be the most relevant distribution, as far as the customers are concerned. If the arrival rate
depends on the number of customers in the system then the distribution seen by arrivals need not
be the same as the equilibrium distribution. Intuitively, πk λk is the longterm frequency of arrivals
which occur when there are k customers in the system, so that the fraction of customers that see
k customers in the system upon arrival is given by
rk = πk λk
.
λ The following is an example of a system with variable arrival rate.
Example 6.7.1 (Singleserver, discouraged arrivals) Suppose λk = α/(k + 1) and µk = µ for all
k , where µ and α are positive constants. Then
∞ S2 =
k=0 (k + 1)!µk
= ∞ and S1 =
αk ∞
k=0 αk
α
= exp( ) < ∞
k
µ
k !µ so that the number of customers in the system is a positive recurrent Markov process, with no
additional restrictions on α and µ. Moreover, the equilibrium probability distribution is given by
πk = (α/µ)k exp(−α/µ)/k !, which is the Poisson distribution with mean N = α/µ. The mean
arrival rate is
∞ λ=
k=0 α
πk α
= µ exp(− )
k+1
µ ∞
k=0 (α/µ)k+1
α
α
α
= µ exp(− )(exp( ) − 1) = µ(1 − exp(− )). (6.15)
(k + 1)!
µ
µ
µ
167 This expression derived for λ is clearly equal to µ, because the departure rate is µ with probability
1 − π0 and zero otherwise. The distribution of the number of customers in the system seen by
arrivals, (rk ) is given by
rk = πk α
(α/µ)k+1 exp(−α/µ)
=
for k ≥ 0
(k + 1)!(1 − exp(−α/µ))
(k + 1)λ which in words can be described as the result of removing the probability mass at zero in the
Poisson distribution, shifting the distribution down by one, and then renormalizing. The mean
number of customers in the queue seen by a typical arrival is therefore (α/µ − 1)/(1 − exp(−α/µ)).
This mean is somewhat less than N because, roughly speaking, the customer arrival rate is higher
when the system is more lightly loaded.
The equivalence of timeaverages and statistical averages for computing the mean arrival rate
and the distribution seen by arrivals can be shown by application of ergodic properties of the
processes involved. The associated formal approach is described next, in slightly more generality.
Let X denote an irreducible, positiverecurrent purejump Markov process. If the process makes a
jump from state i to state j at time t, say that a transition of type (i, j ) occurs. The sequence of
transitions of X forms a new Markov process, Y . The process Y is a discretetime Markov process
with state space {(i, j ) ∈ S × S : qij > 0}, and it can be described in terms of the jump process for
X , by Y (k ) = (X J (k − 1), X J (k )) for k ≥ 0. (Let X J (−1) be deﬁned arbitrarily.)
J
The onestep transition probability matrix of the jump process X J is given by πij = qij /(−qii ),
and X J is recurrent because X is recurrent. Its equilibrium distribution π J (if it exists) is proportional to −πi qii (see Problem 6.3), and X J is positive recurrent if and only if this distribution can
be normalized to make a probability distribution, i.e. if and only if R = − i πi qii < ∞. Assume
J
for simplicity that X J is positive recurrent. Then πi = −πi qii /R is the equilibrium probability
distribution of X J . Furthermore, Y is positive recurrent and its equilibrium distribution is given
by
Y
πij J
= πi pJ
ij
−πi qii qij
=
R −qii
πi qij
=
R
Since limiting time averages equal statistical averages for Y , lim (number of ﬁrst n transitions of X that are type (i, j ))/n = πi qij /R n→∞ with probability one. Therefore, if A ⊂ S × S , and if (i, j ) ∈ A, then
number of ﬁrst n transitions of X that are type (i, j )
=
n→∞
number of ﬁrst n transitions of X with type in A
lim πi qij
(i ,j )∈A πi qi j (6.16) To apply this setup to the special case of a queueing system in which the number of customers
in the system is a Markov birthdeath processes, let the set A be the set of transitions of the form
(i, i + 1). Then deduce that the fraction of the ﬁrst n arrivals that see i customers in the system
upon arrival converges to πi λi / j πj λj with probability one.
Note that if λi = λ for all i, then λ = λ and π = r. The condition λi = λ also implies that the
arrival process is Poisson. This situation is called “Poisson Arrivals See Time Averages” (PASTA). 168 6.8 More examples of queueing systems modeled as Markov birthdeath processes For each of the four examples of this section it is assumed that new customers are oﬀered to the
system according to a Poisson process with rate λ, so that the PASTA property holds. Also, when
there are k customers in the system then the service rate is µk for some given numbers µk . The
number of customers in the system is a Markov birthdeath process with λk = λ for all k . Since
the number of transitions of the process up to any given time t is at most twice the number of
customers that arrived by time t, the Markov process is not explosive. Therefore the process is
positive recurrent if and only if S1 is ﬁnite, where
∞ S1 =
k=0 λk
µ 1 µ 2 . . . µk Special cases of this example are presented in the next four examples.
Example 6.8.1 (M/M/m systems) An M/M/m queueing system consists of a single queue and
m servers. The arrival process is Poisson with some rate λ and the customer service times are
independent and exponentially distributed with mean µ for some µ > 0. The total number of
customers in the system is a birthdeath process with µk = µ min(k, m). Let ρ = λ/(mµ). Since
µk = mµ for all k large enough it is easy to check that the process is positive recurrent if and only
if ρ < 1. Assume now that ρ < 1. Then the equilibrium distribution is given by
(λ/µ)k
for 0 ≤ k ≤ m
S1 k !
= πm ρj for j ≥ 1 πk =
πm+j where S1 is chosen to make the probabilities sum to one (use the fact 1 + ρ + ρ2 . . . = 1/(1 − ρ)):
m−1 S1 =
k=0 (λ/µ)k
k! + (λ/µ)m
.
m!(1 − ρ) An arriving customer must join the queue (rather that go directly to a server) if and only if the
system has m or more customers in it. By the PASTA property, this is the same as the equilibrium
probability of having m or more customers in the system:
∞ πm+j = πm /(1 − ρ) PQ =
j =0 This formula is called the Erlang C formula for probability of queueing. Example 6.8.2 (M/M/m/m systems) An M/M/m/m queueing system consists of m servers. The
arrival process is Poisson with some rate λ and the customer service times are independent and
exponentially distributed with mean µ for some µ > 0. Since there is no queue, if a customer
arrives when there are already m customers in the system, then the arrival is blocked and cleared
169 from the system. The total number of customers in the system is a birth death process, but with
the state space reduced to {0, 1, . . . , m}, and with µk = kµ for 1 ≤ k ≤ m. The unique equilibrium
distribution is given by
πk = (λ/µ)k
S1 k ! for 0 ≤ k ≤ m where S1 is chosen to make the probabilities sum to one.
An arriving customer is blocked and cleared from the system if and only if the system already
has m customers in it. By the PASTA property, this is the same as the equilibrium probability of
having m customers in the system: PB = πm = (λ/µ)m
m!
m (λ/µ)j
j =0
j! This formula is called the Erlang B formula for probability of blocking. Example 6.8.3 (A system with a discouraged server) The number of customers in this system is a
birthdeath process with constant birth rate λ and death rates µk = 1/k . It is is easy to check that
all states are transient for any positive value of λ (to verify this it suﬃces to check that S2 < ∞).
It is not diﬃcult to show that N (t) converges to +∞ with probability one as t → ∞. Example 6.8.4 (A barely stable system) The number of customers in this system is a birthdeath
λ(1+k2
process with constant birth rate λ and death rates µk = 1+(k−1))2 for all k ≥ 1. Since the departure
rates are barely larger than the arrival rates, this system is near the borderline between recurrence
and transience. However, we see that
∞ S1 =
k=0 1
<∞
1 + k2 so that N (t) is positive recurrent with equilibrium distribution πk = 1/(S1 (1 + k 2 )). Note that the
mean number of customers in the system is
∞ k/(S1 (1 + k 2 )) = ∞ N=
k=0 By Little’s law the mean time customers spend in the system is also inﬁnity. It is debatable whether
this system should be thought of as “stable” even though all states are positive recurrent and all
waiting times are ﬁnite with probability one. 170 6.9 FosterLyapunov stability criterion and moment bounds Communication network models can become quite complex, especially when dynamic scheduling,
congestion, and physical layer eﬀects such as fading wireless channel models are included. It is thus
useful to have methods to give approximations or bounds on key performance parameters. The
criteria for stability and related moment bounds discussed in this chapter are useful for providing
such bounds.
Aleksandr Mikhailovich Lyapunov (18571918) contributed signiﬁcantly to the theory of stability of dynamical systems. Although a dynamical system may evolve on a complicated, multiple
dimensional state space, a recurring theme of dynamical systems theory is that stability questions
can often be settled by studying the potential of a system for some nonnegative potential function
V . Potential functions used for stability analysis are widely called Lyapunov functions. Similar
stability conditions have been developed by many authors for stochastic systems. Below we present
the well known criteria due to Foster [4] for recurrence and positive recurrence. In addition we
present associated bounds on the moments, which are expectations of some functions on the state
space, computed with respect to the equilibrium probability distribution.3
Subsection 6.9.1 discusses the discretetime tools, and presents examples involving load balancing routing, and input queued crossbar switches. Subsection 6.9.2 presents the continuous time
tools, and an example. 6.9.1 Stability criteria for discretetime processes Consider an irreducible discretetime Markov process X on a countable state space S , with onestep
transition probability matrix P . If f is a function on S , then P f represents the function obtained
by multiplication of the vector f by the matrix P : P f (i) = j ∈S pij f (j ). If f is nonnegative,
then P f is well deﬁned, with the understanding that P f (i) = +∞ is possible for some, or all,
values of i. An important property of P f is that P f (i) = E [f (X (t + 1)X (t) = i]. Let V be
a nonnegative function on S , to serve as the Lyapunov function. The drift vector of V (X (t)) is
deﬁned by d(i) = E [V (X (t + 1))X (t) = i] − V (i). That is, d = P V − V . Note that d(i) is always
welldeﬁned, if the value +∞ is permitted. The drift vector is also given by
pij (V (j ) − V (i)). d(i) = (6.17) j :j = i Proposition 6.9.1 (FosterLyapunov stability criterion) Suppose V : S → R+ and C is a ﬁnite
subset of S .
(a) If {i : V (i) ≤ K } is ﬁnite for all K , and if P V − V ≤ 0 on S − C , then X is recurrent.
(b) If > 0 and b is a constant such that P V − V ≤ − + bIC , then X is positive recurrent.
Proposition 6.9.2 (Moment bound) Suppose V , f , and g are nonnegative functions on S and
suppose
P V (i) − V (i) ≤ −f (i) + g (i)
for all i ∈ S
(6.18)
3 A version of these moment bounds was given by Tweedie [15], and a version of the moment bound method was
used by Kingman [5] in a queueing context. As noted in [9], the moment bound method is closely related to Dynkin’s
formula. The works [13, 14, 6, 12], and many others, have demonstrated the wide applicability of the stability methods
in various queueing network contexts, using quadratic Lyapunov functions. 171 d
queue 1
a 1 u
u d
queue 2 2 Figure 6.9: Two queues fed by a single arrival stream.
In addition, suppose X is positive recurrent, so that the means, f = πf and g = πg are welldeﬁned.
Then f ≤ g . (In particular, if g is bounded, then g is ﬁnite, and therefore f is ﬁnite.)
Corollary 6.9.3 (Combined FosterLyapunov stability criterion and moment bound) Suppose V, f,
and g are nonnegative functions on S such that
P V (i) − V (i) ≤ −f (i) + g (i) for all i ∈ S (6.19) In addition, suppose for some > 0 that the set C deﬁned by C = {i : f (i) < g (i)+ } is ﬁnite. Then
X is positive recurrent and f ≤ g . (In particular, if g is bounded, then g is ﬁnite, and therefore f
is ﬁnite.)
Proof. Let b = max{g (i) + − f (i) : i ∈ C }. Then V, C, b, and satisfy the hypotheses of
Proposition 6.9.1(b), so that X is positive recurrent. Therefore the hypotheses of Proposition 6.9.2
are satisﬁed, so that f ≤ g .
The assumptions in Propositions 6.9.1 and 6.9.2 and Corollary 6.9.3 do not imply that V is
ﬁnite. Even so, since V is nonnegative, for a given initial state X (0), the long term average drift of
V (X (t)) is nonnegative. This gives an intuitive reason why the mean downward part of the drift,
f , must be less than or equal to the mean upward part of the drift, g . Example 6.9.4 (Probabilistic routing to two queues) Consider the routing scenario with two
queues, queue 1 and queue 2, fed by a single stream of packets, as pictured in Figure 6.9. Here,
0 ≤ a, u, d1 , d2 ≤ 1, and u = 1 − u. The state space for the process is S = Z2 , where the state
+
x = (x1 , x2 ) denotes x1 packets in queue 1 and x2 packets in queue 2. In each time slot, a new
arrival is generated with probability a, and then is routed to queue 1 with probability u and to
queue 2 with probability u. Then each queue i, if not empty, has a departure with probability di .
Note that we allow a packet to arrive and depart in the same slot. Thus, if Xi (t) is the number of
packets in queue i at the beginning of slot t, then the system dynamics can be described as follows:
Xi (t + 1) = Xi (t) + Ai (t) − Di (t) + Li (t) for i ∈ {0, 1} (6.20) where
• A(t) = (A1 (t), A2 (t)) is equal to (1, 0) with probability au, (0, 1) with probability au, and
A(t) = (0, 0) otherwise.
172 • Di (t) : t ≥ 0, are Bernoulli(di ) random variables, for i ∈ {0, 1}
• All the A(t)’s, D1 (t)’s, and D2 (t)’s are mutually independent
• Li (t) = (−(Xi (t) + Ai (t) − Di (t)))+ (see explanation next)
If Xi (t) + Ai (t) = 0, there can be no actual departure from queue i. However, we still allow Di (t)
to equal one. To keep the queue length process from going negative, we add the random variable
Li (t) in (6.20). Thus, Di (t) is the potential number of departures from queue i during the slot, and
Di (t) − Li (t) is the actual number of departures. This completes the speciﬁcation of the onestep
transition probabilities of the Markov process.
A necessary condition for positive recurrence is, for any routing policy, a < d1 + d2 , because the
total arrival rate must be less than the total depature rate. We seek to show that this necessary
condition is also suﬃcient, under the random routing policy.
Let us calculate the drift of V (X (t)) for the choice V (x) = (x2 + x2 )/2. Note that (Xi (t +1))2 =
1
2
(Xi (t) + Ai (t) − Di (t) + Li (t))2 ≤ (Xi (t) + Ai (t) − Di (t))2 , because addition of the variable Li (t)
can only push Xi (t) + Ai (t) − Di (t) closer to zero. Thus,
P V (x) − V (x) = E [V (X (t + 1))X (t) = x] − V (x)
≤ 2 1
2 E [(xi + Ai (t) − Di (t))2 − x2 X (t) = x]
i
i=1 2 = 1
xi E [Ai (t) − Di (t)X (t) = x] + E [(Ai (t) − Di (t))2 X (t) = x] (6.21)
2 i=1
2 ≤ xi E [Ai (t) − Di (t)X (t) = x] +1 i=1 = − (x1 (d1 − au) + x2 (d2 − au)) + 1 (6.22) Under the necessary condition a < d1 + d2 , there are choices of u so that au < d1 and au < d2 , and
for such u the conditions of Corollary 6.9.3 are satisﬁed, with f (x) = x1 (d1 − au) + x2 (d2 − au),
g (x) = 1, and any > 0, implying that the Markov process is positive recurrent. In addition, the
ﬁrst moments under the equlibrium distribution satisfy:
(d1 − au)X1 + (d2 − au)X 2 ≤ 1. (6.23) In order to deduce an upper bound on X 1 + X 2 , we select u∗ to maximize the minimum of the
two coeﬃcients in (6.23). Intuitively, this entails selecting u to minimize the absolute value of the
diﬀerence between the two coeﬃcients. We ﬁnd:
= max min{d1 − au, d2 − au} 0≤u≤1 = min{d1 , d2 ,
and the corresponding value u∗ of u is given by 0 ∗
−
1
u=
+ d12ad2
2
1 d1 + d2 − a
}
2 if d1 − d2 < −a
if d1 − d2  ≤ a
if d1 − d2 > a 173 For the system with u = u∗ , (6.23) yields
1
X1 + X2 ≤ . (6.24) We remark that, in fact,
X1 + X2 ≤ 2
d1 + d2 − a (6.25) If d1 − d2  ≤ a then the bounds (6.24) and (6.25) coincide, and otherwise, the bound (6.25) is
strictly tighter. If d1 − d2 < −a then u∗ = 0, so that X1 = 0, and (6.23) becomes (d2 − a)X2 ≤ 1
, which implies (6.25). Similarly, if d1 − d2 > a, then u∗ = 1, so that X2 = 0, and (6.23) becomes
(d1 − a)X1 ≤ 1, which implies (6.25). Thus, (6.25) is proved. Example 6.9.5 (Routetoshorter policy) Consider a variation of the previous example such that
when a packet arrives, it is routed to the shorter queue. To be deﬁnite, in case of a tie, the packet
is routed to queue 1. Then the evolution equation (6.20) still holds, but with with the description
of the arrival variables changed to the following:
• Given X (t) = (x1 , x2 ), A(t) = (I{x1 ≤x2 } , I{x1 >x2 } ) with probability a, and A(t) = (0, 0)
otherwise.
Let P RS denote the onestep transition probability matrix when the routetoshorter policy is used.
Proceeding as in (6.21) yields:
2 P RS V (x) − V (x) ≤ xi E [Ai (t) − Di (t))X (t) = x] + 1
i=1 = a x1 I{x1 ≤x2 } + x2 I{x1 >x2 } − d1 x1 − d2 x2 + 1 Note that x1 I{x1 ≤x2 } + x2 I{x1 >x2 } ≤ ux1 + ux2 for any u ∈ [0, 1], with equality for u = I{x1 ≤x2 } .
Therefore, the drift bound for V under the routetoshorter policy is less than or equal to the drift
bound (6.22), for V for any choice of probabilistic splitting. In fact, routetoshorter routing can
be viewed as a controlled version of the independent splitting model, for which the control policy is
selected to minimize the bound on the drift of V in each state. It follows that the routetoshorter
process is positive recurrent as long as a < d1 + d2 , and (6.23) holds for any value of u such that
au < d1 and au ≤ d2 . In particular, (6.24) holds for the routetoshorter process.
We remark that the stronger bound (6.25) is not always true for the routetoshorter policy.
The problem is that even if d1 − d2 < −a, the routetoshorter policy can still route to queue 1,
and so X 1 = 0. In fact, if a and d2 are ﬁxed with 0 < a < d2 < 1, then X1 → ∞ as d1 → 0 for
the routetoshorter policy. Intuitively, that is because occasionally there will be a large number of
customers in the system due to statistical ﬂuctuations, and then there will be many customers in
queue 1. But if d2 << 1, those customers will remain in queue 2 for a very long time. 174 input 1 1,1 output 1 1,2
1,3
1,4 input 2 2,1 output 2 2,2
2,3
2,4 input 3 3,1 output 3 3,2
3,3
3,4 input 4 4,1 output 4 4,2
4,3
4,4 Figure 6.10: A 4 × 4 input queued switch. Example 6.9.6 (An input queued switch with probabilistic switching) 4 Consider a packet switch
with N inputs and N outputs, as pictured in Figure 6.10. Suppose there are N 2 queues – N at
each input – with queue i, j containing packets that arrived at input i and are destined for output
j , for i, j ∈ E , where E = {1, · · · , N }. Suppose the packets are all the same length, and adopt
a discretetime model, so that during one time slot, a transfer of packets can occur, such that at
most one packet can be transferred from each input, and at most one packet can be transferred to
each output. A permutation σ of E has the form σ = (σ1 , . . . , σN ), where σ1 , . . . , σN are distinct
elements of E . Let Π denote the set of all N ! such permutations. Given σ ∈ Π, let R(σ ) be the
N × N switching matrix deﬁned by Rij = I{σi =j } . Thus, Rij (σ ) = 1 means that under permutation
σ , input i is connected to output j , or, equivalently, a packet in queue i, j is to depart, if there is
any such packet. A state x of the system has the form x = (xij : i, j ∈ E ), where xij denotes the
number of packets in queue i, j .
The evolution of the system over a time slot [t, t + 1) is described as follows:
Xij (t + 1) = Xij (t) + Aij (t) − Rij (σ (t)) + Lij (t)
where
• Aij (t) is the number of packets arriving at input i, destined for output j , in the slot. Assume
that the variables (Aij (t) : i, j ∈ E, t ≥ 0) are mutually independent, and for each i, j , the
random variables (Aij (t) : t ≥ 0) are independent, identically distributed, with mean λij and
E [A2 ] ≤ Kij , for some constants λij and Kij . Let Λ = (λij : i, j ∈ E ).
ij
• σ (t) is the switch state used during the slot
• Lij = (−(Xij (t) + Aij (t) − Rij (σ (t)))+ , which takes value one if there was an unused potential
departure at queue ij during the slot, and is zero otherwise.
The number of packets at input i at the beginning of the slot is given by the row sum
j ∈E Xij (t), its mean is given by the row sum
j ∈E λij , and at most one packet at input i
4 Tassiulas [12] originally developed the results of Examples 6.9.6 and 6.9.7, in the context of wireless networks.
The paper [8] presents similiar results in the context of a packet switch. 175 can be served in a time slot. Similarly, the set of packets waiting for output j , called the virtual
queue for output j , has size given by the column sum i∈E Xij (t). The mean number of arrivals
to the virtual queue for output j is i∈E λij (t), and at most one packet in the virtual queue can
be served in a time slot. These considerations lead us to impose the following restrictions on Λ:
λij < 1 for all i and
j ∈E λij < 1 for all j (6.26) i∈ E Except for trivial cases involving deterministic arrival sequences, the conditions (6.26) are necessary
for stable operation, for any choice of the switch schedule (σ (t) : t ≥ 0).
Let’s ﬁrst explore random, independent and identically distributed (i.i.d.) switching. That is,
given a probability distribution u on Π, let (σ (t) : t ≥ 0) be independent with common probability
distribution u. Once the distributions of the Aij ’s and u are ﬁxed, we have a discretetime Markov
process model. Given Λ satisfying (6.26), we wish to determine a choice of u so that the process
with i.i.d. switch selection is positive recurrent.
Some standard background from switching theory is given in this paragraph. A line sum of a
matrix M is either a row sum, j Mij , or a column sum, i Mij . A square matrix M is called
doubly stochastic if it has nonnegative entries and if all of its line sums are one. Birkhoﬀ’s theorem,
celebrated in the theory of switching, states that any doubly stochastic matrix M is a convex
combination of switching matrices. That is, such an M can be represented as M = σ∈Π R(σ )u(σ ),
where u = (u(σ ) : σ ∈ Π) is a probability distribution on Π. If M is a nonnegative matrix with all
line sums less than or equal to one, then if some of the entries of M are increased appropriately,
a doubly stochastic matrix can be obtained. That is, there exists a doubly stochastic matrix M
so that Mij ≤ Mij for all i, j . Applying Birkhoﬀ’s theorem to M yields that there is a probability
distribution u so that Mij ≤ σ∈Π R(σ )u(σ ) for all i, j .
Suppose Λ satisﬁes the necessary conditions (6.26). That is, suppose that all the line sums of
Λ are less than one. Then with deﬁned by
= 1 − (maximum line sum of Λ)
,
N each line sum of (λij + : i, j ∈ E ) is less than or equal to one. Thus, by the observation at the
end of the previous paragraph, there is a probability distribution u∗ on Π so that λij + ≤ µij (u∗ ),
where
µij (u) =
Rij (σ )u(σ ).
σ ∈Π We consider the system using probability distribution u∗ for the switch states. That is, let (σ (t) :
t ≥ 0) be independent, each with distribution u∗ . Then for each ij , the random variables Rij (σ (t))
are independent, Bernoulli(µij (u∗ )) random variables.
1
Consider the quadratic Lyapunov function V given by V (x) = 2 i,j x2 . As in (6.21),
ij
P V (x) − V (x) ≤ xij E [Aij (t) − Rij (σ (t))Xij (t) = x] +
i,j 1
2 E [(Aij (t) − Rij (σ (t)))2 X (t) = x].
i,j Now
E [Aij (t) − Rij (σ (t))Xij (t) = x] = E [Aij (t) − Rij (σ (t))] = λij − µij (u∗ ) ≤ −
176 and 1
2 E [(Aij (t) − Rij (σ (t)))2 X (t) = x] ≤
i,j 1
where K = 2 (N + i,j 1
2 E [(Aij (t))2 + (Rij (σ (t)))2 ] ≤ K
i,j Kij ). Thus, P V (x) − V (x) ≤ − xij + K (6.27) ij Therefore, by Corollary 6.9.3, the process is positive recurrent, and
K X ij ≤ (6.28) ij That is, the necessary condition (6.26) is also suﬃcient for positive recurrence and ﬁnite mean queue
length in equilibrium, under i.i.d. random switching, for an appropriate probability distribution u∗
on the set of permutations. Example 6.9.7 (An input queued switch with maximum weight switching) The random switching
policy used in Example 2a depends on the arrival rate matrix Λ, which may be unknown a priori.
Also, the policy allocates potential departures to a given queue ij , whether or not the queue is
empty, even if other queues could be served instead. This suggests using a dynamic switching
policy, such as the maximum weight switching policy, deﬁned by σ (t) = σ M W (X (t)), where for a
state x,
σ M W (x) = arg max
xij Rij (σ ).
(6.29)
σ ∈Π ij The use of “arg max” here means that σ M W (x) is selected to be a value of σ that maximizes the
sum on the right hand side of (6.29), which is the weight of permutation σ with edge weights xij . In
order to obtain a particular Markov model, we assume that the set of permutations Π is numbered
from 1 to N ! in some fashion, and in case there is a tie between two or more permutations for having
the maximum weight, the lowest numbered permutation is used. Let P M W denote the onestep
transition probability matrix when the routetoshorter policy is used.
Letting V and K be as in Example 2a, we ﬁnd under the maximum weight policy that
P M W V (x) − V (x) ≤ xij (λij − Rij (σ M W (x))) + K
ij The maximum of a function is greater than or equal to the average of the function, so that for any
probability distribution u on Π
xij Rij (σ M W (t)) ≥ u(σ )
σ ij = ij xij µij (u)
ij 177 xij Rij (σ ) (6.30) with equality in (6.30) if and only if u is concentrated on the set of maximum weight permutations.
In particular, the choice u = u∗ shows that
xij Rij (σ M W (t)) ≥
ij xij µij (u∗) ≥
ij xij (λij + )
ij Therefore, if P is replaced by P M W , (6.27) still holds. Therefore, by Corollary 6.9.3, the process
is positive recurrent, and the same moment bound, (6.28), holds, as for the randomized switching
strategy of Example 2a. On one hand, implementing the maximum weight algorithm does not
require knowledge of the arrival rates, but on the other hand, it requires that queue length information be shared, and that a maximization problem be solved for each time slot. Much recent
work has gone towards reduced complexity dynamic switching algorithms. 6.9.2 Stability criteria for continuous time processes Here is a continuous time version of the FosterLyapunov stability criteria and the moment bounds.
Suppose X is a timehomegeneous, irreducible, continuoustime Markov process with generator
matrix Q. The drift vector of V (X (t)) is the vector QV . This deﬁnition is motivated by the fact
that the mean drift of X for an interval of duration h is given by
dh (i) =
= E [V (X (t + h))X (t) = i] − V (i)
h
pij (h) − δij
V (j )
h j ∈S = qij +
j ∈S o(h)
h V (j ), (6.31) so that if the limit as h → 0 can be taken inside the summation in (6.31), then dh (i) → QV (i) as
h → 0. The following useful expression for QV follows from the fact that the row sums of Q are
zero:
QV (i) =
qij (V (j ) − V (i)).
(6.32)
j :j = i Formula (6.32) is quite similar to the formula (6.17) for the drift vector for a discretetime process.
Proposition 6.9.8 (FosterLyapunov stability criterion–continuous time) Suppose V : S → R+
and C is a ﬁnite subset of S .
(a) If QV ≤ 0 on S − C , and {i : V (i) ≤ K } is ﬁnite for all K then X is recurrent.
(b) Suppose for some b > 0 and > 0 that
QV (i) ≤ − + bIC (i) for all i ∈ S . (6.33) Suppose further that {i : V (i) ≤ K } is ﬁnite for all K , or that X is nonexplosive. Then X is
positive recurrent.
178 ! 1 queue 1 u
1
m ! 2 u
1 queue 2 u
2 ! 3 1 m 2 u
2 queue 3 Figure 6.11: A system of three queues with two servers. Example 6.9.9 Suppose X has state space S = Z+ , with qi0 = µ for all i ≥ 1, qii+1 = λi for
all i ≥ 0, and all other oﬀdiagonal entries of the rate matrix Q equal to zero, where µ > 0 and
1
λi > 0 such that
i≥0 λi < +∞. Let C = {0}, V (0) = 0, and V (i) = 1 for i ≥ 0. Then
QV = −µ + (λ0 + µ)IC , so that (6.33) is satisﬁed with = µ and b = λ0 + µ. However, X is not
µ
λ
positive recurrent. In fact, X is explosive. To see this, note that pJ +1 = µ+iλi ≥ exp(− λi ). Let
ii
δ be the probability that, starting from state 0, the jump process does not return to zero. Then
1
δ = ∞ pJ +1 ≥ exp(−µ ∞ λi ) > 0. Thus, X J is transient. After the last visit to state zero, all
i=0 ii
i=0
the jumps of X J are up one. The corresponding mean holding times of X are λi1 µ which have a
+
ﬁnite sum, so that the process X is explosive. This example illustrates the need for the assumption
just after (6.33) in Proposition 6.9.8.
As for the case of discrete time, the drift conditions imply moment bounds.
Proposition 6.9.10 (Moment bound–continuous time) Suppose V , f , and g are nonnegative functions on S , and suppose QV (i) ≤ −f (i) + g (i) for all i ∈ S . In addition, suppose X is positive
recurrent, so that the means, f = πf and g = πg are welldeﬁned. Then f ≤ g .
Corollary 6.9.11 (Combined FosterLyapunov stability criterion and moment bound–continuous
time) Suppose V , f , and g are nonnegative functions on S such that QV (i) ≤ −f (i) + g (i) for all
i ∈ S , and, for some > 0, the set C deﬁned by C = {i : f (i) < g (i) + } is ﬁnite. Suppose also
that {i : V (i) ≤ K } is ﬁnite for all K . Then X is positive recurrent and f ≤ g .
Example 6.9.12 (Random server allocation with two servers) Consider the system shown in Figure 6.11. Suppose that each queue i is fed by a Poisson arrival process with rate λi , and suppose
there are two potential departure processes, D1 and D2 , which are Poisson processes with rates
m1 and m2 , respectively. The ﬁve Poisson processes are assumed to be independent. No matter
how the potential departures are allocated to the permitted queues, the following conditions are
necessary for stability:
λ1 < m1 , λ3 < m2 , and λ1 + λ2 + λ3 < m1 + m2
179 (6.34) That is because server 1 is the only one that can serve queue 1, server 2 is the only one that can
serve queue 3, and the sum of the potential service rates must exceed the sum of the potential
arrival rates for stability. A vector x = (x1 , x2 , x2 ) ∈ Z3 corresponds to xi packets in queue i for
+
each i. Let us consider random selection, so that when Di has a jump, the queue served is chosen at
random, with the probabilities determined by u = (u1 , u2 ). As indicated in Figure 6.11, a potential
service by server 1 is given to queue 1 with probability u1 , and to queue 2 with probability u1 .
Similarly, a potential service by server 2 is given to queue 2 with probability u2 , and to queue 3
with probability u2 . The rates of potential service at the three stations are given by
µ1 (u) = u1 m1
µ2 (u) = u1 m1 + u2 m2
µ3 (u) = u2 m2 .
1
Let V (x) = 2 (x2 + x2 + x2 ). Using (6.32), we ﬁnd that the drift vector QV is given by
1
2
3 1
QV (x) =
2 3
2 ((xi + 1) − x2 )λi
i i=1 1
+
2 3 ((xi − 1)2 − x2 )µi (u)
+
i
i=1 Now (xi − 1)2 ≤ (xi − 1)2 , so that
+
3 QV (x) ≤ xi (λi − µi (u)) + i=1 γ
2 (6.35) where γ is the total rate of events, given by γ = λ1 + λ2 + λ3 + µ1 (u)+ µ2 (u)+ µ3 (u), or equivalently,
γ = λ1 + λ2 + λ3 + m1 + m2 . Suppose that the necessary condition (6.34) holds. Then there exists
some > 0 and choice of u so that
λi + ≤ µi (u) for 1≤i≤3 λ
and the largest such choice of is = min{m1 − λ1 , m2 − λ3 , m1 +m2 −3 1 −λ2 −λ3 }. (See excercise.)
So QV (x) ≤ − (x1 + x2 + x3 ) + γ for all x, so Corollary 6.9.11 implies that X is positive recurrent
γ
and X 1 + X 2 + X 3 ≤ 2 . Example 6.9.13 (Longer ﬁrst server allocation with two servers) This is a continuation of Example
6.9.12, concerned with the system shown in Figure 6.11. Examine the right hand side of (6.35).
Rather than taking a ﬁxed value of u, suppose that the choice of u could be speciﬁed as a function
of the state x. The maximum of a function is greater than or equal to the average of the function,
so that for any probability distribution u,
3 xi µi (u) ≤ max
i=1 u xi µi (u )
i = max m1 (x1 u1 + x2 u1 ) + m2 (x2 u2 + x3 u2 )
u = m1 (x1 ∨ x2 ) + m2 (x2 ∨ x3 )
180 (6.36) with equality in (6.36) for a given state x if and only if a longer ﬁrst policy is used: each service
opportunity is allocated to the longer queue connected to the server. Let QLF denote the onestep
transition probability matrix when the longest ﬁrst policy is used. Then (6.35) continues to hold
for any ﬁxed u, when Q is replaced by QLF . Therefore if the necessary condition (6.34) holds,
can be taken as in Example 6.9.12, and QLF V (x) ≤ − (x1 + x2 + x3 ) + γ for all x. So Corollary
γ
6.9.11 implies that X is positive recurrent under the longer ﬁrst policy, and X 1 + X 2 + X 3 ≤ 2 .
(Note: We see that
3 QLF V (x) ≤ xi λi − m1 (x1 ∨ x2 ) − m2 (x2 ∨ x3 ) + i=1 γ
,
2 but for obtaining a bound on X 1 + X 2 + X 3 it was simpler to compare to the case of random service
allocation.) 6.10 Problems 6.1 Mean hitting time for a simple Markov process
Let (X (n) : n ≥ 0) denote a discretetime, timehomogeneous Markov chain with state space
{0, 1, 2, 3} and onestep transition probability matrix 0
100 1−a 0 a 0 P =
0
0.5 0 0.5 0
010
for some constant a with 0 ≤ a ≤ 1. (a) Sketch the transition probability diagram for X and
give the equilibrium probability vector. If the equilibrium vector is not unique, describe all the
equilibrium probability vectors.
(b) Compute E [min{n ≥ 1 : X (n) = 3}X (0) = 0].
6.2 A two station pipeline in continuous time
This is a continuoustime version of Example 4.9.1. Consider a pipeline consisting of two singlebuﬀer stages in series. Model the system as a continuoustime Markov process. Suppose new packets
are oﬀered to the ﬁrst stage according to a rate λ Poisson process. A new packet is accepted at
stage one if the buﬀer in stage one is empty at the time of arrival. Otherwise the new packet is
lost. If at a ﬁxed time t there is a packet in stage one and no packet in stage two, then the packet is
transfered during [t, t + h) to stage two with probability hµ1 + o(h). Similarly, if at time t the second
stage has a packet, then the packet leaves the system during [t, t + h) with probability hµ2 + o(h),
independently of the state of stage one. Finally, the probability of two or more arrival, transfer, or
departure events during [t, t + h) is o(h). (a) What is an appropriate statespace for this model?
(b) Sketch a transition rate diagram. (c) Write down the Q matrix. (d) Derive the throughput,
assuming that λ = µ1 = µ2 = 1. (e) Still assuming λ = µ1 = µ2 = 1. Suppose the system starts
with one packet in each stage. What is the expected time until both buﬀers are empty?
6.3 Equilibrium distribution of the jump chain
Suppose that π is the equilibrium distribution for a timehomogeneous Markov process with transition rate matrix Q. Suppose that B −1 = i −qii πi , where the sum is over all i in the state space,
181 is ﬁnite. Show that the equilibrium distribution for the jump chain (X J (k ) : k ≥ 0) (deﬁned in
J
Section 4.10) is given by πi = −Bqii πi . (So π and π J are identical if and only if qii is the same for
all i.)
6.4 A simple Poisson process calculation
Let (N (t) : t ≥ 0) be a Poisson random process with rate λ > 0. Compute P [N (s) = iN (t) = k ]
where 0 < s < t and i and k are nonnegative integers. (Caution: note order of s and t carefully).
6.5 A simple question of periods
Consider a discretetime Markov process with the nonzero onestep transition probabilities indicated by the following graph. 1 2 3 4 5 6 7 8 (a) What is the period of state 4?
(b) What is the period of state 6?
6.6 A mean hitting time problem
Let (X (t) : t ≥ 0) be a timehomogeneous, purejump Markov process with state space {0, 1, 2}
and Q matrix −4 2
2
Q = 1 −2 1 2
0 −2.
(a) Write down the state transition diagram and compute the equilibrium distribution.
(b) Compute ai = E [min{t ≥ 0 : X (t) = 1}X (0) = i] for i = 0, 1, 2. If possible, use an approach
that can be applied to larger state spaces.
(c) Derive a variation of the Kolmogorov forward diﬀerential equations for the quantities: αi (t) =
P [X (s) = 2 for 0 ≤ s ≤ t and X (t) = iX (0) = 0] for 0 ≤ i ≤ 2. (You need not solve the equations.)
(d) The forward Kolmogorov equations describe the evolution of an initial probability distribution
going forward in time, given an initial. In other problems, a boundary condition is given at a
ﬁnal time, and a diﬀerential equation working backwards in time from a ﬁnal condition is called
for (called Kolmogorov backward equations). Derive a backward diﬀerential equation for: βj (t) =
P [X (s) = 2 for t ≤ s ≤ tf X (t) = j ], for 0 ≤ j ≤ 2 and t ≤ tf for some ﬁxed time tf . (Hint:
Express βi (t − h) in terms of the βj (t) s for t ≤ tf , and let h → 0. You need not solve the equations.)
6.7 A birthdeath process with periodic rates
Consider a single server queueing system in which the number in the system is modeled as a
continuous time birthdeath process with the transition rate diagram shown, where λa , λb , µa , and
µb are strictly positive constants.
182 !a
0 !b
2 1
µ a !a µ !
3 µ b a !a b ... 4
µ b µ a (a) Under what additional assumptions on these four parameters is the process positive recurrent?
(b) Assuming the system is positive recurrent, under what conditions on λa , λb , µa , and µb is it true
that the distribution of the number in the system at the time of a typical arrival is the same as the
equilibrium distribution of the number in the system?
6.8 Markov model for a link with resets
Suppose that a regulated communication link resets at a sequence of times forming a Poisson
process with rate µ. Packets are oﬀered to the link according to a Poisson process with rate λ.
Suppose the link shuts down after three packets pass in the absence of resets. Once the link is
shut down, additional oﬀered packets are dropped, until the link is reset again, at which time the
process begins anew.
! µ (a) Sketch a transition rate diagram for a ﬁnite state Markov process describing the system state.
(b) Express the dropping probability (same as the long term fraction of packets dropped) in terms
of λ and µ.
6.9 An unusual birthdeath process
Consider the birthdeath process X with arrival rates λk = (p/(1 − p))k /ak and death rates µk =
(p/(1 − p))k−1 /ak , where .5 < p < 1, and a = (a0 , a1 , . . .) is a probability distribution on the
nonnegative integers with ak > 0 for all k . (a) Classify the states for the process X as transient,
null recurrent or positive recurrent. (b) Check that aQ = 0. Is a an equilibrium distribution for
X ? Explain. (c) Find the onestep transition probabilities for the jumpchain, X J (d) Classify the
states for the process X J as transient, null recurrent or positive recurrent.
6.10 A queue with decreasing service rate
Consider a queueing system in which the arrival process is a Poisson process with rate λ. Suppose
the instantaneous completion rate is µ when there are K or fewer customers in the system, and µ/2
when there are K + 1 or more customers in the system. The number in the system is modeled as a
birthdeath Markov process. (a) Sketch the transition rate diagram. (b) Under what condition on
λ and µ are all states positive recurrent? Under this condition, give the equilibrium distribution.
(c) Suppose that λ = (2/3)µ. Describe in words the typical behavior of the system, given that it is
initially empty.
6.11 Limit of a distrete time queueing system
Model a queue by a discretetime Markov chain by recording the queue state after intervals of q
seconds each. Assume the queue evolves during one of the atomic intervals as follows: There is an
arrival during the interval with probability αq , and no arrival otherwise. If there is a customer in
the queue at the beginning of the interval then a single departure will occur during the interval
183 with probability βq . Otherwise no departure occurs. Suppose that it is impossible to have an
arrival and a departure in a single atomic interval. (a) Find ak =P[an interarrival time is kq ] and
bk =P[a service time is kq ]. (b) Find the equilibrium distribution, p = (pk : k ≥ 0), of the number
of customers in the system at the end of an atomic interval. What happens as q → 0?
6.12 An M/M/1 queue with impatient customers
Consider an M/M/1 queue with parameters λ and µ with the following modiﬁcation. Each customer
in the queue will defect (i.e. depart without service) with probability αh + o(h) in an interval of
length h, independently of the other customers in the queue. Once a customer makes it to the
server it no longer has a chance to defect and simply waits until its service is completed and then
departs from the system. Let N (t) denote the number of customers in the system (queue plus
server) at time t. (a) Give the transition rate diagram and generator matrix Q for the Markov
chain N = (N (t) : t ≥ 0). (b) Under what conditions are all states positive recurrent? Under this
condition, ﬁnd the equilibrium distribution for N . (You need not explicitly sum the series.) (c)
Suppose that α = µ. Find an explicit expression for pD , the probability that a typical arriving
customer defects instead of being served. Does your answer make sense as λ/µ converges to zero
or to inﬁnity?
6.13 Statistical multiplexing
Consider the following scenario regarding a oneway link in a storeandforward packet communication network. Suppose that the link supports eight connections, each generating traﬃc at 5
kilobits per second (kbps). The data for each connection is assumed to be in packets exponentially
distributed in length with mean packet size 1 kilobit. The packet lengths are assumed mutually
independent and the packets for each stream arrive according to a Poisson process. Packets are
queued at the beginning of the link if necessary, and queue space is unlimited. Compute the mean
delay (queueing plus transmission time–neglect propagation delay) for each of the following three
scenarios. Compare your answers. (a) (Full multiplexing) The link transmit speed is 50 kbps. (b)
The link is replaced by two 25 kbps links, and each of the two links carries four sessions. (Of course
the delay would be larger if the sessions were not evenly divided.) (c) (Multiplexing over two links)
The link is replaced by two 25 kbps links. Each packet is transmitted on one link or the other, and
neither link is idle whenever a packet from any session is waiting.
6.14 A queue with blocking
(M/M/1/5 system) Consider an M/M/1 queue with service rate µ, arrival rate λ, and the modiﬁcation that at any time, at most ﬁve customers can be in the system (including the one in service,
if any). If a customer arrives and the system is full (i.e. already has ﬁve customers in it) then the
customer is dropped, and is said to be blocked. Let N (t) denote the number of customers in the
system at time t. Then (N (t) : t ≥ 0) is a Markov chain. (a) Indicate the transition rate diagram of
the chain and ﬁnd the equilibrium probability distribution. (b) What is the probability, pB , that a
typical customer is blocked? (c) What is the mean waiting time in queue, W , of a typical customer
that is not blocked? (d) Give a simple method to numerically calculate, or give a simple expression
for, the mean length of a busy period of the system. (A busy period begins with the arrival of a
customer to an empty system and ends when the system is again empty.)
6.15 Three queues and an autonomously traveling server
Consider three stations that are served by a single rotating server, as pictured. 184 " 1 station 1
! " 2 station 2
$ " 3 # station 3 Customers arrive to station i according to a Poisson process of rate λi for 1 ≤ i ≤ 3, and the total
service requirement of each customer is exponentially distributed, with mean one. The rotation
of the server is modelled by a three state Markov process with the transition rates α, β, and γ as
indicated by the dashed lines. When at a station, the server works at unit rate, or is idle if the
station is empty. If the service to a customer is interrupted because the server moves to the next
station, the service is resumed when the server returns.
(a) Under what condition is the system stable? Brieﬂy justify your answer.
(b) Identify a method for computing the mean customer waiting time at station one.
6.16 On two distibutions seen by customers
Consider a queueing system in which the number in the system only changes in steps of plus one
or minus one. Let D(k, t) denote the number of customers that depart in the interval [0,t] that
leave behind exactly k customers, and let R(k,t) denote the number of customers that arrive in the
interval [0,t] to ﬁnd exactly k customers already in the system. (a) Show that D(k, t) − R(k, t) ≤ 1
for all k and t. (b) Let αt (respectively δt ) denote the number of arrivals (departures) up to time
t. Suppose that αt → ∞ and αt /δt → 1 as t → ∞. Show that if the following two limits exist for
a given value k , then they are equal: rk = limt→∞ R(k, t)/αt and dk = limt→∞ D(k, t)/δt .
6.17 Recurrence of mean zero random walks
(a) Suppose B1 , B2 , . . . is a sequence of independent, mean zero, integer valued random variables,
which are bounded, i.e. P [Bi  ≤ M ] = 1 for some M .
(a) Let X0 = 0 and Xn = B1 + · · · + Bn for n ≥ 0. Show that X is recurrent.
(b) Suppose Y0 = 0 and Yn+1 = Yn + Bn + Ln , where Ln = (−(Yn + Bn ))+ . The process Y is a
reﬂected version of X . Show that Y is recurrent.
6.18 Positive recurrence of reﬂected random walk with negative drift
Suppose B1 , B2 , . . . is a sequence of independent, integer valued random variables, each with mean
B < 0 and second moment B 2 < +∞. Suppose X0 = 0 and Xn+1 = Xn + Bn + Ln , where
Ln = (−(Xn + Bn ))+ . Show that X is positive recurrent, and give an upper bound on the mean
under the equilibrium distribution, X . (Note, it is not assumed that the B ’s are bounded.)
6.19 Routing with two arrival streams
(a) Generalize Example 6.9.4 to the scenario shown.
185 d
queue 1
a 1 u 1 u 1 d
queue 2 a 2 u
u 1 2 2
2 d
queue 3 3 where ai , dj ∈ (0, 1) for 1 ≤ i ≤ 2 and 1 ≤ j ≤ 3. In particular, determine conditions on a1 and
a2 that insure there is a choice of u = (u1 , u2 ) which makes the system positive recurrent. Under
those conditions, ﬁnd an upper bound on X1 + X 2 + X3 , and select u to mnimize the bound.
(b) Generalize Example 1.b to the scenario shown. In particular, can you ﬁnd a version of routetoshorter routing so that the bound found in part (a) still holds? 6.20 An inadequacy of a linear potential function
Consider the system of Example 6.9.5 (a discretetime model, using the route to shorter policy,
with ties broken in favor of queue 1, so u = I{x1 ≤x2 } ):
d
queue 1
a 1 u
u d
queue 2 2 Assume a = 0.7 and d1 = d2 = 0.4. The system is positive recurrent. Explain why the function
V (x) = x1 + x2 does not satisfy the FosterLyapunov stability criteria for positive recurrence, for
any choice of the constant b and the ﬁnite set C . 6.21 Allocation of service
Prove the claim in Example 6.9.12 about the largest value of . 6.22 Opportunistic scheduling
(Based on [14]) Suppose N queues are in parallel, and suppose the arrivals to a queue i form an
independent, identically distributed sequence, with the number of arrivals in a given slot having
mean ai > 0 and ﬁnite second moment Ki . Let S (t) for each t be a subset of E = {1, . . . , N } and
t ≥ 0. The random sets S (t) : t ≥ 0 are assumed to be independent with common distribution w.
The interpretation is that there is a single server, and in slot i, it can serve one packet from one
of the queues in S (t). For example, the queues might be in the base station of a wireless network
with packets queued for N mobile users, and S (t) denotes the set of mobile users that have working
channels for time slot [t, t + 1). See the illustration:
186 a a 1 2 queue 1 Fading queue 2 channel .
.
.
a N state s queue N (a) Explain why the following condition is necessary for stability: For all s ⊂ E with s = ∅,
ai <
i∈s w (B ) (6.37) B :B ∩s=∅ (b) Consider u of the form u = (u(i, s) : i ∈ E, s ⊂ E ), with u(i, s) ≥ 0, u(i, s) = 0 if i ∈ s, and
i∈E u(i, s) = I{s=∅} . Suppose that given S (t) = s, the queue that is given a potential service
opportunity has probability distribution (u(i, s) : i ∈ E ). Then the probability of a potential service
at queue i is given by µi (u) = s u(i, s)w(s) for i ∈ E . Show that under the condition (6.37), for
some > 0, u can be selected to that ai + ≤ µi (u) for i ∈ E . (Hint: Apply the mincut, maxﬂow
theorem to an appropriate graph.)
(c) Show that using the u found in part (b) that the process is positive recurrent.
(d) Suggest a dynamic scheduling method which does not require knowledge of the arrival rates or
the distribution w, which yields the same bound on the mean sum of queue lengths found in part
(b).
6.23 Routing to two queues – continuous time model
Give a continuous time analog of Examples 6.9.4 and 6.9.5. In particular, suppose that the arrival
process is Poisson with rate λ and the potential departure processes are Poisson with rates µ1 and
µ2 .
6.24 Stability of two queues with transfers
Let (λ1 , λ2 , ν, µ1 , µ2 ) be a vector of strictly positve parameters, and consider a system of two service
stations with transfers as pictured.
" " 1 station 1
u! station 2
2 µ µ 1 2 Station i has Possion arrivals at rate λi and an exponential type server, with rate µi . In addition,
customers are transferred from station 1 to station 2 at rate uν , where u is a constant with u ∈
U = [0, 1]. (Rather than applying dynamic programming here, we will apply the method of FosterLyapunov stability theory in continuous time.) The system is described by a continuoustime
Markov process on Z2 with some transition rate matrix Q. (You don’t need to write out Q.)
+
(a) Under what condition on (λ1 , λ2 , ν, µ1 , µ2 ) is there a choice of the constant u such that the
Markov process describing the system is positive recurrent?
x2
x2
(b) Let V be the quadratic Lyapunov function, V (x1 , x2 ) = 21 + 22 . Compute the drift vector QV .
187 (c) Under the condition of part (a), and using the moment bound associated with the FosterLyapunov criteria, ﬁnd an upper bound on the mean number in the system in equilibrium, X1 + X2 .
(The smaller the bound the better.)
6.25 Stability of a system with two queues and modulated server
Consider two queues, queue 1 and queue 2, such that in each time slot, queue i receives a new
packet with probability ai , where 0 < a1 < 1 and 0 < a2 < 1. Suppose the server is described by a
three state Markov process, as shown.
a 1 1 queue 1 0 ! server longer
a 2
queue 2 2 If the server process is in state i for i ∈ {1, 2} at the beginning of a slot, then a potential service
is given to station i. If the server process is in state 0 at the beginning of a slot, then a potential
service is given to the longer queue (with ties broken in favor of queue 1). Then during the slot,
the server state jumps with the probabilities indicated. (Note that a packet can arrive and depart
in one time slot.) For what values of a1 and a2 is the process stable? Brieﬂy explain your answer
(but rigorous proof is not required). 188 Chapter 7 Basic Calculus of Random Processes
The calculus of deterministic functions revolves around continuous functions, derivatives, and integrals. These concepts all involve the notion of limits. See the appendix for a review of continuity,
diﬀerentiation and integration. In this chapter the same concepts are treated for random processes.
We’ve seen four diﬀerent senses in which a sequence of random variables can converge: almost
surely (a.s.), in probability (p.), in mean square (m.s.), and in distribution (d.). Of these senses,
we will use the mean square sense of convergence the most, and make use of the correlation version
of the Cauchy criterion for m.s. convergence, and the associated facts that for m.s. convergence,
the means of the limits are the limits of the means, and correlations of the limits are the limits of
correlations (Proposition 2.2.3 and Corollaries 2.2.4 and 2.2.5). As an application of integration
of random processes, ergodicity and the KarhunenLo´ve expansion are discussed. In addition,
e
notation for complexvalued random processes is introduced. 7.1 Continuity of random processes The topic of this section is the deﬁnition of continuity of a continuoustime random process, with
a focus on continuity deﬁned using m.s. convergence. Chapter 2 covers convergence of sequences.
Limits for deterministic functions of a continuous variable can be deﬁned in either of two equivalent
ways. Speciﬁcally, a function f on R has a limit y at to , written as lims→to f (s) = y , if either of
the two equivalent conditions is true:
(1) (Deﬁnition based on
s − to  ≤ δ . and δ ) Given > 0, there exists δ > 0 so that  f (s) − y ≤ whenever (2) (Deﬁnition based on sequences) f (sn ) → y for any sequence (sn ) such that sn → to .
Let’s check that (1) and (2) are equivalent. Suppose (1) is true, and let (sn ) be such that sn → to .
Let > 0 and then let δ be as in condition (1). Since sn → to , it follows that there exists no so
that sn − to  ≤ δ for all n ≥ no . But then f (sn ) − y  ≤ by the choice of δ. Thus, f (sn ) → y.
That is, (1) implies (2).
For the converse direction, it suﬃces to prove the contrapositive: if (1) is not true then (2) is
not true. Suppose (1) is not true. Then there exists an > 0 so that, for any n ≥ 1, there exists a
1
value sn with sn − to  ≤ n such that f (sn ) − y  > . But then sn → to , and yet f (sn ) → y, so (2)
is false. That is, not (1) implies not (2). This completes the proof that (1) and (2) are equivalent.
189 Similarly, and by essentially the same reasons, convergence for a continuoustime random process
can be deﬁned using either and δ , or using sequences, at least for limits in the p., m.s., or d.
senses. As we will see, the situation is slightly diﬀerent for a.s. limits. Let X = (Xt : t ∈ T) be a
random process such that the index set T is equal to either all of R, or an interval in R, and ﬁx
to ∈ T.
Deﬁnition 7.1.1 (Limits for continuoustime random processes.) The process (Xt : t ∈ T) has
limit Y at to :
(i) in the m.s. sense, written lims→to Xs = Y m.s., if for any > 0, there exists δ > 0 so that
m.s.
E [(Xs − Y )2 ] < whenever s ∈ T and s − to  < δ. An equivalent condition is Xsn → Y as
n → ∞, whenever sn → to .
(ii) in probability, written lims→to Xs = Y p., if given any > 0, there exists δ > 0 so that
p.
P {Xs − Y  ≥ ] < whenever s ∈ T and s − to  < δ. An equivalent condition is Xsn → Y as
n → ∞, whenever sn → to .
(iii) in distribution, written lims→to Xs = Y d., if given any continuity point c of FY and any > 0,
there exists δ > 0 so that FX,1 (c, s) − FY (c) < whenever s ∈ T and s − to  < δ. An equivalent
d. condition is Xsn → Y as n → ∞, whenever sn → to . (Recall that FX,1 (c, s) = P {Xs ≤ c}.)
(iv) almost surely, written lims→to Xs = Y a.s., if there is an event Fto having probability one such
that Fto ⊂ {ω : lims→to Xs (ω ) = Y (ω )}.1
The relationship among the above four types of convergence in continuous time is the same as
the relationship among the four types of convergence of sequences, illustrated in Figure 2.8. That
is, the following is true:
a.s. Proposition 7.1.2 The following statements hold as s → to for a ﬁxed to in T : If either Xs → Y
p.
p.
m.s.
d.
or Xs → Y then Xs → Y. If Xs → Y. then Xs → Y. Also, if there is a random variable Z with
p.
m.s.
E [Z ] < ∞ and Xt  ≤ Z for all t, and if Xs → Y then Xs → Y
Proof. As indicated in Deﬁnition 7.1.1, the ﬁrst three types of convergence are equivalent to
convergence along sequences, in the corresponding senses. The fourth type of convergence, namely
a.s. convergence as s → to , implies convergence along sequences (Example 7.1.3 shows that the
converse is not true). That is true because if (sn ) is a sequence converging to to ,
{ω : lim Xt (ω ) = Y (ω )} ⊂ {ω : lim Xsn (ω ) = Y (ω )}.
n→∞ s→to Therefore, if the ﬁrst of these sets contains an event which has probability one, the second of these
sets is an event which has probability one. The proposition then follows from the same relations
for convergence of sequences. In particular, a.s. convergence for continuous time implies a.s.
convergence along sequences (as just shown), which implies convergence in p. along sequences, which
1
This deﬁnition is complicated by the fact that the set {ω : lims→to Xs (ω ) = Y (ω )} involves uncountably many
random variables, and it is not necessarily an event. There is a way to simplify the deﬁnition as follows, but it
requires an extra assumption. A probability space (Ω, F , P ) is complete, if whenever N is an event having probability
zero, all subsets of N are events. If (Ω, F , P ) is complete, the deﬁnition of lims→to Xs = Y a.s., is equivalent to the
requirement that {ω : lims→to Xs (ω ) = Y (ω )} be an event and have probability one. 190 is the same as convergence in probability. The other implications of the proposition follow directly
from the same implications for sequences, and the fact the ﬁrst three deﬁnitions of convergence for
continuous time have a form based on sequences.
The following example shows that a.s. convergence as s → to is strictly stronger than a.s.
convergence along sequences.
Example 7.1.3 Let U be uniformly distributed on the interval [0, 1]. Let Xt = 1 if t − U is a
rational number, and Xt = 0 otherwise. Each sample path of X takes values zero and one in any
ﬁnite interval, so that X is not a.s. convergent at any to . However, for any ﬁxed t, P {Xt = 0} = 1.
Therefore, for any sequence sn , since there are only countably many terms, P {Xsn = 0 for all n} = 1
so that Xsn → 0 a.s.
Deﬁnition 7.1.4 (Four types of continuity at a point for a random process) For each to ∈ T ﬁxed,
the random process X = (Xt : t ∈ T) is continuous at to in any one of the four senses: m.s., p.,
a.s., or d., if lims→to Xs = Xto in the corresponding sense.
The following is immediately implied by Proposition 7.1.2. It shows that for convergence of a
random process at a single point, the relations illustrated in Figure 2.8 again hold.
Corollary 7.1.5 If X is continuous at to in either the a.s. or m.s. sense, then X is continuous at
to in probability. If X is continuous at to in probability, then X is continuous at to in distribution.
Also, if there is a random variable Z with E [Z ] < ∞ and Xt  ≤ Z for all t, and if X is continuous
at to in probability, then it is continuous at to in the m.s. sense.
A deterministic function f on R is simply called continuous if it is continuous at all points. Since
we have four senses of continuity at a point for a random process, this gives four types of continuity
for random processes. Before stating them formally, we describe a ﬁfth type of continuity of random
processes, which is often used in applications. Recall that for a ﬁxed ω ∈ Ω, the random process
X gives a sample path, which is a function on T. Continuity of a sample path is thus deﬁned as
it is for any deterministic function. The subset of Ω, {ω : Xt (ω ) is a continuous function of t}, or
more concisely, {Xt is a continuous function of t}, is the set of ω such that the sample path for ω
is continuous. The ﬁfth type of continuity requires that the sample paths be continuous, if a set
probability zero is ignored. 191 Deﬁnition 7.1.6 (Five types of continuity for a whole random process) A random process
X = (Xt : t ∈ T) is said to be
m.s. continuous if it is m.s. continuous at each t
continuous in p. if it is continuous in p. at each t
continuous in d. if it is continuous in d. at each t
a.s. continuous at each t, if, as the phrase indicates, it is a.s. continuous at each t.2
a.s. samplepath continuous, if F ⊂ {Xt is continuous in t} for some event F with P (F ) = 1.
The relationship among the ﬁve types of continuity for a whole random process is pictured in
Figure 7.1, and is summarized in the following proposition.
a.s. continuous at each t
p. a.s. sample!path continuous m.s. d. ith
by
ted le w t.)
ina ariab men
dom v
mo
s is dom nd
ces e ran seco
te
pro gl
(If a sin a fini Figure 7.1: Relationships among ﬁve types of continuity of random processes. Proposition 7.1.7 If a process is a.s. samplepath continuous it is a.s. continuous at each t. If
a process is a.s. continuous at each t or m.s. continuous, it is continuous in p. If a process is
continuous in p. it is continuous in d. Also, if there is a random variable Y with E [Y ] < ∞ and
Xt  ≤ Y for all t, and if X is continuous in p., then X is m.s. continuous.
Proof. Suppose X is a.s. samplepath continuous. Then for any to ∈ T,
{ω : Xt (ω ) is continuous at all t ∈ T} ⊂ {ω : Xt (ω ) is continuous at to } (7.1) Since X is a.s. samplepath continuous, the set on the lefthand side of (7.1) contains an event F
with P (F ) = 1 and F is also a subset of the set on the the righthand side of (7.1). Thus, X is
a.s. continuous at to . Since to was an arbitrary element of T, if follows that X is a.s. continuous at
each t. The remaining implications of the proposition follow from Corollary 7.1.5. Example 7.1.8 (Shows a.s. samplepath continuity is strictly stronger than a.s. continuity at each
t.) Let X = (Xt : 0 ≤ t ≤ t) be given by Xt = I{t≥U } for 0 ≤ t ≤ 1, where U is uniformly distributed
2 We avoid using the terminology “a.s. continuous” for the whole random process, because such terminology could
too easily be confused with a.s. samplepath continuous 192 over [0, 1]. Thus, each sample path of X has a single upward jump of size one, at a random time U
uniformly distributed over [0, 1]. So every sample path is discontinuous, and therefore X is not a.s.
samplepath continuous. For any ﬁxed t and ω , if U (ω ) = t (i.e. if the jump of X is not exactly at
time t) then Xs (ω ) → Xt (ω ) as s → t. Since P {U = t} = 1, it follows that X is a.s. continuous at
each t. Therefore X is also continuous in p. and d. senses. Finally, since Xt  ≤ 1 for all t and X is
continuous in p., it is also m.s. continuous.
The remainder of this section focuses on m.s. continuity. Recall that the deﬁnition of m.s.
convergence of a sequence of random variables requires that the random variables have ﬁnite second
moments, and consequently the limit also has a ﬁnite second moment. Thus, in order for a random
process X = (Xt : t ∈ T) to be continuous in the m.s. sense, it must be a second order process:
2
E [Xt ] < ∞ for all t ∈ T. Whether X is m.s. continuous depends only on the correlation function
RX , as shown in the following proposition.
Proposition 7.1.9 Suppose (Xt : t ∈ T) is a second order process. The following are equivalent:
(i) RX is continuous at all points of the form (t, t) (This condition involves RX for points in and
near the set of points of the form (t, t). It is stronger than requiring RX (t, t) to be continuous
in t–see example 7.1.10.)
(ii) X is m.s. continuous
(iii) RX is continuous over T × T.
If X is m.s. continuous, then the mean function, µX (t), is continuous. If X is wide sense stationary,
the following are equivalent:
(i ) RX (τ ) is continuous at τ = 0
(ii ) X is m.s. continuous
(iii ) RX (τ ) is continuous over all of R.
Proof. ((i) implies (ii)) Fix t ∈ T and suppose that RX is continuous at the point (t, t). Then
RX (s, s), RX (s, t), and RX (t, s) all converge to RX (t, t) as s → t. Therefore, lims→t E [(Xs − Xt )2 ] =
lims→t (RX (s, s) − RX (s, t) − RX (t, s) + RX (t, t)) = 0. So X is m.s. continuous at t. Therefore if
RX is continuous at all points of the form (t, t) ∈ T × T, then X is m.s. continuous at all t ∈ T.
Therefore (i) implies (ii).
((ii) implies (iii)) Suppose condition (ii) is true. Let (s, t) ∈ T × T, and suppose (sn , tn ) ∈ T × T
for all n ≥ 1 such that limn→∞ (sn , tn ) = (s, t). Therefore, sn → s and tn → t as n → ∞. By
m.s.
m.s.
condition (b), it follows that Xsn → Xs and Xtn → Xt as n → ∞. Since the limit of the
correlations is the correlation of the limit for a pair of m.s. convergent sequences (Corollary 2.2.4)
it follows that RX (sn , tn ) → RX (s, t) as n → ∞. Thus, RX is continuous at (s, t), where (s, t) was
an arbitrary point of T × T. Therefore RX is continuous over T × T, proving that (ii) implies (iii).
Obviously (iii) implies (i), so the proof of the equivalence of (i)(iii) is complete.
m.s.
If X is m.s. continuous, then, by deﬁnition, for any t ∈ T, Xs → Xt as s → t. It thus follows
that µX (s) → µX (t), because the limit of the means is the mean of the limit, for a m.s. convergent
sequence (Corollary 2.2.5). Thus, m.s. continuity of X implies that the deterministic mean function,
µX , is continuous.
193 Finally, if X is WSS, then RX (s, t) = RX (τ ) where τ = s − t, and the three conditions (i)(iii)
become (i )(iii ), so the equivalence of (i)(iii) implies the equivalence of (i )(iii ). Example 7.1.10 Let X = (Xt : t ∈ R) be deﬁned by Xt = U for t < 0 and Xt = V for t ≥ 0,
where U and V are independent random variables with mean zero and variance one. Let tn be a
sequence of strictly negative numbers converging to 0. Then Xtn = U for all n and X0 = V . Since
P {U − V  ≥ } = 0 for small enough, Xtn does not converge to X0 in p. sense. So X is not
continuous in probability at zero. It is thus not continuous in the m.s or a.s. sense at zero either. The
only one of the ﬁve senses that the whole process could be continuous is continuous in distribution.
The process X is continuous in distribution if and only if U and V have the same distribution.
Finally, let us check the continuity properties of the autocorrelation function. The autocorrelation
function is given by RX (s, t) = 1 if either s, t < 0 or if s, t ≥ 0, and RX (s, t) = 0 otherwise. So RX
1
1
1
1
is not continuous at (0, 0), because R( n , − n ) = 0 for all n ≥ 1, so R( n , − n ) → RX (0, 0) = 1. as
n → ∞. However, it is true that RX (t, t) = 1 for all t, so that RX (t, t) is a continuous function of
t. This illustrates the fact that continuity of the function of two variables, RX (s, t), at a particular
ﬁxed point (to , to ), is a stronger requirement than continuity of the function of one variable, RX (t, t),
at t = to . Example 7.1.11 Let W = (Wt : t ≥ 0) be a Brownian motion with parameter σ 2 . Then E [(Wt −
Ws )2 ] = σ 2 t − s → 0 as s → t. Therefore W is m.s. continuous. Another way to show W is
m.s. continuous is to observe that the autocorrelation function, RW (s, t) = σ 2 (s ∧ t), is continuous.
Since W is m.s. continuous, it is also continuous in the p. and d. senses. As we stated in deﬁning
W , it is a.s. samplepath continuous, and therefore a.s. continuous at each t ≥ 0, as well. Example 7.1.12 Let N = (Nt : t ≥ 0) be a Poisson process with rate λ > 0. Then for ﬁxed t,
E [(Nt − Ns )2 ] = λ(t − s) + (λ(t − s))2 → 0 as s → t. Therefore N is m.s. continuous. As required,
RN , given by RN (s, t) = λ(s ∧ t) + λ2 st, is continuous. Since N is m.s. continuous, it is also
continuous in the p. and d. senses. N is also a.s. continuous at any ﬁxed t, because the probability
of a jump at exactly time t is zero for any ﬁxed t. However, N is not a.s. sample continuous. In
fact, P [N is continuous on [0, a]] = e−λa and so P [N is continuous on R+ ] = 0.
Deﬁnition 7.1.13 A random process (Xt : t ∈ T), such that T is a bounded interval (open, closed,
or mixed) in R with endpoints a < b, is piecewise m.s. continuous, if there exist n ≥ 1 and
a = t0 < t1 < · · · < tn = b, such that, for 1 ≤ k ≤ n: X is m.s. continuous over (tk−1 , tk ) and has
m.s. limits at the endpoints of (tk−1 , tk ).
More generally, if T is all of R or an interval in R, X is piecewise m.s. continuous over T if it is
piecewise m.s. continuous over every bounded subinterval of T. 7.2 Mean square diﬀerentiation of random processes Before considering the m.s. derivative of a random process, we review the deﬁnition of the derivative
of a function (also, see Appendix 11.4). Let the index set T be either all of R or an interval in
194 R. Suppose f is a deterministic function on T. Recall that for a ﬁxed t in T, f is diﬀerentiable
)
at t if lims→t f (ss−f (t) exists and is ﬁnite, and if f is diﬀerentiable at t, the value of the limit is
−t
the derivative, f (t). The whole function f is called diﬀerentiable if it is diﬀerentiable at all t. The
function f is called continuously diﬀerentiable if f is diﬀerentiable, and the derivative function f
is continuous.
In many applications of calculus, it is important that a function f be not only diﬀerentiable,
but continuously diﬀerentiable. In much of the applied literature, when there is an assumption
that a function is diﬀerentiable, it is understood that the function is continuously diﬀerentiable.
For example, by the fundamental theorem of calculus,
b f (b) − f (a) = f (s)ds (7.2) a holds if f is a continuously diﬀerentiable function with derivative f . Example 11.4.2 shows that
(7.2) might not hold if f is simply assumed to be diﬀerentiable.
Let X = (Xt : t ∈ T) be a second order random process such that the index set T is equal to
either all of R or an interval in R. The following deﬁnition for m.s. derivatives is analogous to the
deﬁnition of derivatives for deterministic functions.
Deﬁnition 7.2.1 For each t ﬁxed, the random process X = (Xt : t ∈ T) is mean square (m.s.)
diﬀerentiable at t if the following limit exists
lim s→t Xs −Xt
s−t m.s. The limit, if it exists, is the m.s. derivative of X at t, denoted by Xt . The whole random process X is
said to be m.s. diﬀerentiable if it is m.s. diﬀerentiable at each t, and it is said to be m.s. continuously
diﬀerentiable if it is m.s. diﬀerentiable and the derivative process X is m.s. continuous.
Let ∂i denote the operation of taking the partial derivative with respect to the ith argument.
For example, if f (x, y ) = x2 y 3 then ∂2 f (x, y ) = 3x2 y 2 and ∂1 ∂2 f (x, y ) = 6xy 2 . The partial
derivative of a function is the same as the ordinary derivative with respect to one variable, with
the other variables held ﬁxed. We shall be applying ∂1 and ∂2 to an autocorrelation function
RX = (RX (s, t) : (s, t) ∈ T × T}, which is a function of two variables.
Proposition 7.2.2 (a) (The derivative of the mean is the mean of the derivative) If X is m.s.
diﬀerentiable, then the mean function µX is diﬀerentiable, and µX (t) = µX (t). (i.e. the
operations of (i) taking expectation, which basically involves integrating over ω , and (ii) differentiation with respect to t, can be done in either order.)
(b) If X is m.s. diﬀerentiable, the cross correlation functions are given by RX X = ∂1 RX and
RXX = ∂2 RX , and the autocorrelation function of X is given by RX = ∂1 ∂2 RX = ∂2 ∂1 RX .
(In particular, the indicated partial derivatives exist.)
(c) X is m.s. diﬀerentiable at t if and only if the following limit exists and is ﬁnite:
lim s,s →t RX (s, s ) − RX (s, t) − RX (t, s ) + RX (t, t)
.
(s − t)(s − t) (7.3) (Therefore, the whole process X is m.s. diﬀerentiable if and only if the limit in (7.3) exists
and is ﬁnite for all t ∈ T.)
195 (d) X is m.s. continuously diﬀerentiable if and only if RX , ∂2 RX , and ∂1 ∂2 RX exist and are
continuous. (By symmetry, if X is m.s. continuously diﬀerentiable, then also ∂1 RX is continuous.)
(e) (Specialization of (d) for WSS case) Suppose X is WSS. Then X is m.s. continuously diﬀerentiable if and only if RX (τ ), RX (τ ), and RX (τ ) exist and are continuous functions of τ . If
X is m.s. continuously diﬀerentiable then X and X are jointly WSS, X has mean zero (i.e.
µX = 0) and autocorrelation function given by RX (τ ) = −RX (τ ), and the cross correlation
functions are given by RX X (τ ) = RX (τ ) and RXX (τ ) = −RX (τ ).
(f ) (A necessary condition for m.s. diﬀerentiability) If X is WSS and m.s. diﬀerentiable, then
RX (0) exists and RX (0) = 0.
(g) If X is a m.s. diﬀerentiable Gaussian process, then X and its derivative process X are jointly
Gaussian.
Proof. (a) Suppose X is m.s. diﬀerentiable. Then for any t ﬁxed,
Xs − Xt m.s.
→ Xt as s → t.
s−t
It thus follows that
µX (s) − µX (t)
→ µX (t) as s → t,
s−t (7.4) because the limit of the means is the mean of the limit, for a m.s. convergent sequence (Corollary 2.2.5). But (7.4) is just the deﬁnition of the statement that the derivative of µX at t is equal
to µX (t). That is, dµX (t) = µX (t) for all t, or more concisely, µX = µX .
dt
(b) Suppose X is m.s. diﬀerentiable. Since the limit of the correlations is the correlation of the
limits for m.s. convergent sequences (Corollary 2.2.4), for t, t ∈ T,
RX X (t, t ) = lim E
s→t X (s) − X (t)
s−t X (t ) = lim s→t RX (s, t ) − RX (t, t )
= ∂1 RX (t, t )
s−t Thus, RX X = ∂1 RX , and in particular, the partial derivative ∂1 RX exists. Similarly, RXX =
∂2 RX . Also, by the same reasoning,
X (s ) − X (t )
s −t
RX X (t, s ) − RX X (t, t )
= lim
s →t
s −t
= ∂2 RX X (t, t ) = ∂2 ∂1 RX (t, t ), RX (t, t ) = lim E X (t) s →t so that RX = ∂2 ∂1 RX . Similarly, RX = ∂1 ∂1 RX .
(c) By the correlation form of the Cauchy criterion, (Proposition 2.2.3), X is m.s. diﬀerentiable
at t if and only if the following limit exists and is ﬁnite:
lim E s,s →t X (s) − X (t)
s−t
196 X (s ) − X (t)
s −t . (7.5) Multiplying out the terms in the numerator in the right side of (7.5) and using E [X (s)X (s )] =
RX (s, s ), E [X (s)X (t)] = RX (s, t), and so on, shows that (7.5) is equivalent to (7.3). So part (c)
is proved.
(d) The numerator in (7.3) involves RX evaluated at the four courners of the rectangle [t, s] ×
[t, s ], shown in Figure 7.2. Suppose RX , ∂2 RX and ∂1 ∂2 RX exist and are continuous functions.
! + + ! s!
t t s Figure 7.2: Sampling points of RX .
Then by the fundamental theorem of calculus,
s (RX (s, s ) − RX (s, t)) − (RX (t, s ) − RX (t, t)) = s ∂2 RX (s, v )dv −
t ∂2 RX (t, v )dv
t s [∂2 RX (s, v ) − ∂2 RX (t, v )] dv =
t
s s = ∂1 ∂2 RX (u, v )dudv.
t (7.6) t Therefore, the ratio in (7.3) is the average value of ∂1 ∂2 RX over the rectangle [t, s] × [t, s ]. Since
∂1 ∂2 RX is assumed to be continuous, the limit in (7.3) exists and it is equal to ∂1 ∂2 RX (t, t). Therefore, by part (c) already proved, X is m.s. diﬀerentiable. By part (b), the autocorrelation function
of X is ∂1 ∂2 RX . Since this is assumed to be continuous, it follows that X is m.s. continuous. Thus,
X is m.s. continuously diﬀerentiable.
(e) If X is WSS, then RX (s − t) = RX (τ ) where τ = s − t. Suppose RX (τ ), RX (τ ) and RX (τ )
exist and are continuous functions of τ . Then
∂1 RX (s, t) = RX (τ ) ∂2 ∂1 RX (s, t) = −RX (τ ). and (7.7) The minus sign in (7.7) appears because RX (s, t) = RX (τ ) where τ = s − t, and the derivative of
with respect to t is −1. So, the hypotheses of part (d) hold, so that X is m.s. diﬀerentiable. Since
X is WSS, its mean function µX is constant, which has derivative zero, so X has mean zero. Also
by part (c) and (7.7), RX X (τ ) = RX (τ ) and RX X = −RX . Similarly, RXX (τ ) = −RX (τ ). Note
that X and X are each WSS and the cross correlation functions depend on τ alone, so X and X
are jointly WSS.
(f ) If X is WSS then
E X (t) − X (0)
t 2 =−
197 2(RX (t) − RX (0))
t2 (7.8) Therefore, if X is m.s. diﬀerentiable then the right side of (7.8) must converge to a ﬁnite limit as
t → 0, so in particular it is necessary that (RX (t) − RX (0))/t → 0 as t → 0. Therefore RX (0) = 0.
(g) The derivative process X is obtained by taking linear combinations and m.s. limits of
random variables in X = (Xt ; t ∈ T). Therefore, (g) follows from the fact that the joint Gaussian
property is preserved under linear combinations and limits (Proposition 3.4.3(c)). Example 7.2.3 Let f (t) = t2 sin(1/t2 ) for t = 0 and f (0) = 0 as in Example 11.4.2, and let
X = (Xt : t ∈ R) be the deterministic random process such that X (t) = f (t) for all t ∈ R. Since X
is diﬀerentiable as an ordinary function, it is also m.s. diﬀerentiable, and its m.s. derivative X is
equal to f . Since X , as a deterministic function, is not continuous at zero, it is also not continuous
at zero in the m.s. sense. We have RX (s, t) = f (s)f (t) and ∂2 RX (s, t) = f (s)f (t), which is not
continuous. So indeed the conditions of Proposition 7.2.2(d) do not hold, as required. Example 7.2.4 A Brownian motion W = (Wt : t ≥ 0) is not m.s. diﬀerentiable. If it were, then
)−W
for any ﬁxed t ≥ 0, W (ss−t (t) would converge in the m.s. sense as s → t to a random variable
with a ﬁnite second moment. For a m.s. convergent seqence, the second moments of the variables
in the sequence converge to the second moment of the limit random variable, which is ﬁnite. But
W (s) − W (t) has mean zero and variance σ 2 s − t, so that
lim E s→t W (s) − W (t)
s−t 2 σ2
s→t s − t = lim = +∞. (7.9) Thus, W is not m.s. diﬀerentiable at any t. For another approach, we could appeal to Proposition
7.2.2 to deduce this result. The limit in (7.9) is the same as the limit in (7.5), but with s and s
restricted to be equal. Hence (7.5), or equivalently (7.3), is not a ﬁnite limit, implying that W is
not diﬀerentiable at t.
Similarly, a Poisson process is not m.s. diﬀerentiable at any t. A WSS process X with RX (τ ) =
1
e−ατ  is not m.s. diﬀerentiable because RX (0) does not exist. A WSS process X with RX (τ ) = 1+τ 2
is m.s. diﬀerentiable, and its derivative process X is WSS with mean 0 and covariance function
RX (τ ) = − 1
1 + τ2 = 2 − 6τ 2
.
(1 + τ 2 )3 Proposition 7.2.5 Suppose X is a m.s. diﬀerentiable random process and f is a diﬀerentiable
function. Then the product Xf = (X (t)f (t) : t ∈ R) is mean square diﬀerentiable and (Xf ) =
X f + Xf .
Proof: Fix t. Then for each s = t,
X (s)f (s) − X (t)f (t)
s−t =
m.s. → (X (s) − X (t))f (s) X (t)(f (s) − f (t))
+
s−t
s−t
X (t)f (t) + X (t)f (t) as s → t.
198 Deﬁnition 7.2.6 A random process X on a bounded interval (open, closed, or mixed) with endpoints a < b is continuous and piecewise continuously diﬀerentiable in the m.s. sense, if X is m.s.
continuous over the interval, and if there exists n ≥ 1 and a = t0 < t1 < · · · < tn = b, such that,
for 1 ≤ k ≤ n: X is m.s. continuously diﬀerentiable over (tk−1 , tk ) and X has ﬁnite limits at the
endpoints of (tk−1 , tk ).
More generally, if T is all of R or a subinterval of R, then a random process X = (Xt : t ∈ T)
is continuous and piecewise continuously diﬀerentiable in the m.s. sense if its restriction to any
bounded interval is continuous and piecewise continuously diﬀerentiable in the m.s. sense. 7.3 Integration of random processes Let X = (Xt : a ≤ t ≤ b) be a random process and let h be a function on a ﬁnite interval [a, b].
How shall we deﬁne the following integral?
b
a Xt h(t)dt. (7.10) One approach is to note that for each ﬁxed ω , Xt (ω ) is a deterministic function of time, and so
the integral can be deﬁned as the integral of a deterministic function for each ω . We shall focus
on another approach, namely mean square (m.s.) integration. An advantage of m.s. integration is
that it relies much less on properties of sample paths of random processes.
As for integration of deterministic functions, the m.s. Riemann integrals are based on Riemann
sums, deﬁned as follows. Given:
• A partition of (a, b] of the form (t0 , t1 ], (t1 , t2 ], · · · , (tn−1 , tn ], where n ≥ 0 and
a = t0 ≤ t1 · · · < tn = b
• A sampling point from each subinterval, vk ∈ (tk−1 , tk ], for 1 ≤ k ≤ n,
the corresponding Riemann sum for Xh is deﬁned by
n Xvk h(vk )(tk − tk−1 ).
k=1 The norm of the partition is deﬁned to be maxk tk − tk−1 .
b Deﬁnition 7.3.1 The Riemann integral a Xt h(t)dt is said to exist in the m.s. sense and its
value is the random variable I if the following is true. Given any > 0, there is a δ > 0 so that
E [( n=1 Xvk h(vk )(tk − tk−1 ) − I )2 ] ≤ whenever the norm of the partition is less than or equal to
k
δ . This deﬁnition is equivalent to the following condition, expressed using convergence of sequences.
The m.s. Riemann integral exists and is equal to I , if for any sequence of partitions, speciﬁed by
m
m
((tm , tm , . . . , tm ) : m ≥ 1), with corresponding sampling points ((v1 , . . . , vnm ) : m ≥ 1), such that
nm
1
2
th partition converges to zero as m → ∞, the corresponding sequence of Riemann
norm of the m
sums converges in the m.s. sense to I as m → ∞. The process Xt h(t) is said to be m.s. Riemann
b
integrable over (a, b] if the integral a Xt h(t)dt exists and is ﬁnite.
Next, suppose Xt h(t) is deﬁned over the whole real line. If Xt h(t) is m.s. Riemann integrable over
every bounded interval [a, b], then the Riemann integral of Xt h(t) over R is deﬁned by
∞ b Xt h(t)dt =
−∞ lim a,b→∞ −a 199 Xt h(t)dt m.s. provided that the indicated limit exist as a, b jointly converge to +∞.
Whether an integral exists in the m.s. sense is determined by the autocorrelation function of
the random process involved, as shown next. The condition involves Riemann integration of a
deterministic function of two variables. As reviewed in Appendix 11.5, a twodimensional Riemann
integral over a bounded rectangle is deﬁned as the limit of Riemann sums corresponding to a
partition of the rectangle into subrectangles and choices of sampling points within the subrectangles.
If the sampling points for the Riemann sums are required to be horizontally and vertically alligned,
then we say the twodimensional Riemann integral exists with aligned sampling.
Proposition 7.3.2 The integral b
a Xt h(t)dt exists in the m.s. Riemann sense if and only if bb
a a RX (s, t)h(s)h(t)dsdt (7.11) exists as a two dimensional Riemann integral with aligned sampling. The m.s. integral exists, in
particular, if X is m.s. piecewise continuous over [a, b] and h is piecewise continuous over [a, b].
Proof. By deﬁnition, the m.s. integral of Xt h(t) exists if and only if the Riemann sums converge
in the m.s. sense for an arbitary sequence of partitions and sampling points, such that the norms
of the partitions converge to zero. So consider an arbitrary sequence of partitions of (a, b] into
intervals speciﬁed by the collection of endpoints, ((tm , tm , . . . , tm ) : m ≥ 1), with corresponding
nm
0
1
m
sampling point vk ∈ (tm 1 , tm ] for each m and 1 ≤ k ≤ nm , such that the norm of the mth partition
k−
k
converges to zero as m → ∞. For each m ≥ 1, let Sm denote the corresponding Riemann sum:
nm
m
m
Xvk h(vk )(tm − tm 1 ).
k−
k Sm =
k=1 By the correlation form of the Cauchy criterion for m.s. convergence (Proposition 2.2.3), (Sm :
m ≥ 1) converges in the m.s. sense if and only if limm,m →∞ E [Sm Sm ] exists and is ﬁnite. Now
nm nm
mm
m
m
RX (vj , vk )h(vj )h(vk )(tm − tm 1 )(tm − tm 1 ),
j
j−
k
k− E [Sm Sm ] = (7.12) j =1 k=1 and the righthand side of (7.12) is the Riemann sum for the integral (7.11), for the partition of
mm
(a, b] × (a, b] into rectangles of the form (tm 1 , tm ] × (tm 1 , tm ] and the sampling points (vj , vk ).
j−
j
k−
k
Note that the mm sampling points are aligned, in that they are determined by the m + m numm
m
m
m
bers v1 , . . . , vnm , v1 , . . . , vnm . Moreover, any Riemann sum for the integral (7.11) with aligned
sampling can arise in this way. Further, as m, m → ∞, the norm of this partition, which is the
maximum length or width of any rectangle of the partition, converges to zero. Thus, the limit
limm,m →∞ E [Sm Sm ] exists for any sequence of partitions and sampling points if and only if the
integral (7.11) exists as a twodimensional Riemann integral with aligned sampling.
Finally, if X is piecewise m.s. continuous over [a, b] and h is piecewise continuous over [a, b], then
there is a partition of [a, b] into intervals of the form (sk−1 , sk ] such that X is m.s. continuous over
(sk−1 , sk ) with m.s. limits at the endpoints, and h is continuous over (sk−1 , sk ) with ﬁnite limits
at the endpoints. Therefore, RX (s, t)h(s)h(t) restricted to each rectangle of the form (sj −1 , sj ) ×
(sk−1 , sk ), is the restriction of a continuous function on [sj −1 , sj ] × [sk−1 , sk ]. Thus RX (s, t)h(s)h(t)
is Riemann integrable over [a, b] × [a, b]. 200 Proposition 7.3.3 Suppose Xt h(t) and Yt k (t) are both m.s. integrable over [a, b]. Then
b b E Xt h(t)dt
2 b b Xt h(t)dt RX (s, t)h(s)h(t)dsdt b Xs h(s)ds RXY (s, t)h(s)k (t)dsdt b
a Yt k (t)dt Xs h(s)ds, a b (7.17) a
b = a a b b b b Cov a = a (7.16) CXY (s, t)h(s)k (t)dsdt a Yt k (t)dt a b = a
b (7.15) a
b Xt h(t)dt (7.14) CX (s, t)h(s)h(t)dsdt. a
b E b = a Var (7.13) a a E µX (t)h(t)dt = a
b Xt h(t) + Yt k (t)dt =
a b Xt h(t)dt +
a Yt k (t))dt (7.18) a Proof. Let (Sm ) denote the sequence of Riemann sums appearing in the proof of Proposition
7.3.2. Since the mean of a m.s. convergent sequence of random variables is the limit of the means
(Corollary 2.2.5),
b E Xt h(t)dt = a lim E [Sm ] m→∞ nm = m
m
µX (vk )h(vk )(tm − tm 1 ).
k−
k lim m→∞ The righthand side of (7.19) is a limit of Riemann sums for the integral
limit exists and is equal to E b
a Xt h(t)dt (7.19) k=1
b
a µX (t)h(t)dt. Since this for any sequence of partitions and sample points, it b b follows that a µX (t)h(t)dt exists as a Riemann integral, and is equal to E a Xt h(t)dt , so (7.13)
is proved.
The second moment of the m.s. limit of (Sm : m ≥ 0) is equal to limm,m →∞ E [Sm Sm ], by the
correlation form of the Cauchy criterion for m.s. convergence (Proposition 2.2.3), which implies
(7.14). It follows from (7.13) that
2 b Xt h(t)dt E
a b b = µX (s)µX (t)h(s)h(t)dsdt
a a Subtracting each side of this from the corresponding side of (7.14) yields (7.15). The proofs of
(7.16) and (7.17) are similar to the proofs of (7.14) and (7.15), and are left to the reader.
For any partition of [a, b] and choice of sampling points, the Riemann sums for the three integrals
appearing (7.17) satisfy the corresponding additivity condition, implying (7.17).
The fundamental theorem of calculus, stated in Appendix 11.5, states the increments of a
continuous, piecewise continuously diﬀerentiable function are equal to integrals of the derivative of
the function. The following is the generalization of the fundamental theorem of calculus to the m.s.
calculus.
201 Theorem 7.3.4 (Fundamental Theorem of m.s. Calculus) Let X be a m.s. continuously diﬀerentiable random process. Then for a < b,
b Xb − Xa = Xt dt (m.s. Riemann integral) (7.20) a More generally, if X is continuous and piecewise continuously diﬀerentiable, (11.4) holds with Xt
replaced by the righthand derivative, D+ Xt . (Note that D+ Xt = Xt whenever Xt is deﬁned.)
Proof. The m.s. Riemann integral in (7.20) exists because X is assumed to be m.s. continuous.
b
Let B = Xb − Xa − a Xt dt, and let Y be an arbitrary random variable with a ﬁnite second moment.
It suﬃces to show that E [Y B ] = 0, because a possible choice of Y is B itself. Let φ(t) = E [Y Xt ].
Then for s = t,
φ(s) − φ(t)
Xs − Xt
=E Y
s−t
s−t
Taking a limit as s → t and using the fact the correlation of a limit is the limit of the correlations
for m.s. convergent sequences, it follows that φ is diﬀerentiable and φ (t) = E [Y Xt ]. Since X is
m.s. continuous, it similarly follows that φ is continuous.
Next, we use the fact that the integral in (7.20) is the m.s. limit of Riemann sums, with each
Riemann sum corresponding to a partition of (a, b] speciﬁed by some n ≥ 1 and a = t0 < · · · < tn = b
and sampling points vk ∈ (tk−1 , tk ] for a ≤ k ≤ n. Since the limit of the correlation is the correlation
of the limt for m.s. convergence,
n b EY Xt dt = a lim tk −tk−1 →0 Xvk (tk − tk−1 ) EY
k=1
n = tk −tk−1 →0 b φ (vk )(tk − tk−1 ) = lim k=1 φ (t)dt
a b Therefore, E [Y B ] = φ(b) − φ(a) − a φ (t)dt, which is equal to zero by the fundamental theorem
of calculus for deterministic continuously diﬀerentiable functions. This establishes (7.20) in case
X is m.s. continuously diﬀerentiable. If X is m.s. continuous and only piecewise continuously
diﬀerentiable, we can use essentially the same proof, observing that φ is continuous and piecewise
b
continuously diﬀerentiable, so that E [Y B ] = φ(b) − φ(a) − a φ (t)dt = 0 by the fundamental
theorem of calculus for deterministic continuous, piecewise continuously diﬀerential functions. Proposition 7.3.5 Suppose X is a Gaussian random process. Then X , together with all mean
square derivatives of X that exist, and all m.s. Riemann integrals of X of the form I (a, b) =
b
a Xt h(t)dt that exist, are jointly Gaussian.
Proof. The m.s. derivatives and integrals of X are obtained by taking m.s. limits of linear
combinations of X = (Xt ; t ∈ T). Therefore, the proposition follows from the fact that the joint
Gaussian property is preserved under linear combinations and limits (Proposition 3.4.3(c)). 202 Theoretical Exercise Suppose X = (Xt : t ≥ 0) is a random process such that RX is continuous.
t
Let Yt = 0 Xs ds. Show that Y is m.s. diﬀerentiable, and P [Yt = Xt ] = 1 for t ≥ 0.
Example 7.3.6 Let (Wt : t ≥ 0) be a Brownian motion with σ 2 = 1, and let Xt =
t ≥ 0. Let us ﬁnd RX and P [Xt  ≥ t] for t > 0. Since RW (u, v ) = u ∧ v ,
s for t RX (s, t) = E Wu du
0
s t
0 Ws ds Wv dv
0 t (u ∧ v )dvdu. =
0 0 To proceed, consider ﬁrst the case s ≥ t and partition the region of integration into three parts as
shown in Figure 7.3. The contributions from the two triangular subregions is the same, so v
t
u<v
u>v u t s Figure 7.3: Partition of region of integration.
t u RX (s, t) = 2
0 = s t vdvdu + t3
+
3 0
t2 (s vdvdu
t − t)
2 = 0 t2 s t3
−.
2
6 Still assuming that s ≥ t, this expression can be rewritten as
RX (s, t) = st(s ∧ t) (s ∧ t)3
−
.
2
6 (7.21) Although we have found (7.21) only for s ≥ t, both sides are symmetric in s and t. Thus (7.21)
holds for all s, t.
Since W is a Gaussian process, X is a Gaussian process. Also, E [Xt ] = 0 (because W is mean
3
2
zero) and E [Xt ] = RX (t, t) = t3 . Thus, Xt
P [Xt  ≥ t] = 2P ≥
t3
3 Note that P [Xt  ≥ t] → 1 as t → +∞. 203 t
= 2Q
t3
3 3
t . Example 7.3.7 Let N = (Nt : t ≥ 0) be a second order process with a continuous autocorrelation
function RN and let x0 be a constant. Consider the problem of ﬁnding a m.s. diﬀerentiable random
process X = (Xt : t ≥ 0) satisfying the linear diﬀerential equation
Xt = −Xt + Nt , X0 = x0 . (7.22) Guided by the case that Nt is a smooth nonrandom function, we write
t Xt = x0 e−t + e−(t−v) Nv dv (7.23) 0 or
t Xt = x0 e−t + e−t ev Nv dv. (7.24) 0 Using Proposition 7.2.5, it is not diﬃcult to check that (7.24) indeed gives the solution to (7.22).
Next, let us ﬁnd the mean and autocovariance functions of X in terms of those of N . Taking
the expectation on each side of (7.23) yields
t µX (t) = x0 e−t + e−(t−v) µN (v )dv. (7.25) 0 A diﬀerent way to derive (7.25) is to take expectations in (7.22) to yield the deterministic linear
diﬀerential equation:
µX (t) = −µX (t) + µN (t); µX (0) = x0 which can be solved to yield (7.25). To summarize, we found two methods to start with the
stochastic diﬀerential equation (7.23) to derive (7.25), thereby expressing the mean function of the
solution X in terms of the mean function of the driving process N . The ﬁrst is to solve (7.22) to
obtain (7.23) and then take expectations, the second is to take expectations ﬁrst and then solve
the deterministic diﬀerential equation for µX .
The same two methods can be used to express the covariance function of X in terms of the
covariance function of N . For the ﬁrst method, we use (7.23) to obtain
s x0 e−s + CX (s, t) = Cov 0
s t =
0 t e−(s−u) Nu du, x0 e−t + e−(t−v) Nv dv 0 e−(s−u) e−(t−v) CN (u, v )dvdu. (7.26) 0 The second method is to derive deterministic diﬀerential equations. To begin, note that
∂1 CX (s, t) = Cov (Xs , Xt ) = Cov (−Xs + Ns , Xt ) so
∂1 CX (s, t) = −CX (s, t) + CN X (s, t).
204 (7.27) For t ﬁxed, this is a diﬀerential equation in s. Also, CX (0, t) = 0. If somehow the cross covariance
function CN X is found, (7.27) and the boundary condition CX (0, t) = 0 can be used to ﬁnd CX .
So we turn next to ﬁnding a diﬀerential equation for CN X .
∂2 CN X (s, t) = Cov(Ns , Xt ) = Cov(Ns , −Xt + Nt ) so
∂2 CN X (s, t) = −CN X (s, t) + CN (s, t). (7.28) For s ﬁxed, this is a diﬀerential equation in t with initial condition CN X (s, 0) = 0. Solving (7.28)
yields
t CN X (s, t) = e−(t−v) CN (s, v )dv. (7.29) 0 Using (7.29) to replace CN X in (7.27) and solving (7.27) yields (7.26). 7.4 Ergodicity Let X be a stationary or WSS random process. Ergodicity generally means that certain time
averages are asymptotically equal to certain statistical averages. For example, suppose
X = (Xt : t ∈ R) is WSS and m.s. continuous. The mean µX is deﬁned as a statistical average:
µX = E [Xt ] for any t ∈ R.
The time average of X over the interval [0, t] is given by
1
t t
0 Xu du. Of course, for t ﬁxed, the time average is a random variable, and is typically not equal to the
statistical average µX . The random process X is called mean ergodic (in the m.s. sense) if
t 1
t→∞ t
lim Xu du = µX m.s. 0 A discrete time WSS random process X is similarly called mean ergodic (in the m.s. sense) if
1
n→∞ n n lim Xi = µX m.s. (7.30) i=1 For example, by the m.s. version of the law of large numbers, if X = (Xn : n ∈ Z) is WSS with
CX (n) = I{n=0} (so that the Xi ’s are uncorrelated) then (7.30) is true. For another example, if
CX (n) = 1 for all n, it means that X0 has variance one and P {Xk = X0 } = 1 for all k (because
equality holds in the Schwarz inequality: CX (n) ≤ CX (0)). Then for all n ≥ 1,
1
n n Xk = X0 .
k=1 Since X0 has variance one, the process X is not ergodic if CX (n) = 1 for all n. In general, whether
X is m.s. ergodic in the m.s. sense is determined by the autocovariance function, CX . The result
is stated and proved next for continuous time, and the discretetime version is true as well.
205 Proposition 7.4.1 Let X be a realvalued, WSS, m.s. continuous random process. Then X is
mean ergodic (in the m.s. sense) if and only if
t 2
t→∞ t t−τ
t lim 0 CX (τ )dτ = 0. (7.31) Suﬃcient conditions are
(a) limτ →∞ CX (τ ) = 0. (This condition is also necessary if limτ →∞ CX (τ ) exists.)
(b) ∞
−∞ CX (τ )dτ < +∞. (c) limτ →∞ RX (τ ) = 0.
(d) ∞
−∞ RX (τ )dτ < +∞. Proof. By the deﬁnition of m.s. convergence, X is mean ergodic if and only if
lim E t→∞
t 1
t 2 t Xu du − µX = 0. t Since E 1 0 Xu du = 1 0 µX du = µX , (7.32) is the same as Var
t
t
the properties of m.s. integrals,
Var 1
t t Xu du
0 =
=
=
= 1
t2
1
t2
1
t2
1
t
2
t t 1
t = Cov
= (7.32) 0 t Xu du,
0 1
t 1
t t
0 Xu du → 0 as t → ∞. By t Xv dv
0 t CX (u − v )dudv
0 (7.33) 0
t t−v t −v
t−τ CX (τ )dτ dv
0 (7.34)
0 t −t −τ CX (τ )dvdτ +
0
t
−t
t
0 0 t − τ 
t
t−τ
t CX (τ )dvdτ (7.35) CX (τ )dτ
CX (τ )dτ, where for v ﬁxed the variable τ = u − v was introduced, and we use the fact that in both (7.34)
and (7.35), the pair (v, τ ) ranges over the region pictured in Figure 7.4. This establishes the ﬁrst
statement of the proposition.
For the remainder of the proof, it is important to keep in mind that the integral in (7.33) is
simply the average of CX (u − v ) over the square [0, t] × [0, t]. The function CX (u − v ) is equal to
CX (0) along the diagonal of the square, and the magnitude of the function is bounded by CX (0)
everywhere in the square. Thus, if CX (u, v ) is small for u − v larger than some constant, if t is
large, the average of CX (u − v ) over the square will be small. The integral in (7.31) is equivalent
to the integral in (7.33), and both can be viewed as a weighted average of CX (τ ), with a triangular
weighting function.
206 ! t t v !t Figure 7.4: Region of integration for (7.34) and (7.35).
It remains to prove the assertions regarding (a)(d). Suppose CX (τ ) → c as τ → ∞. We claim
the left side of (7.31) is equal to c. Indeed, given ε > 0 there exists L > 0 so that CX (τ ) − c ≤ ε
whenever τ ≥ L. For 0 τ ≤ L we can use the Schwarz inequality to bound CX (τ ), namely
CX (τ ) ≤ CX (0). Therefore for t ≥ L,
2
t t
0 t−τ
t CX (τ )dτ − c =
≤ 2
t
2
t t
0
t
0 t−τ
(CX (τ ) − c) dτ
t
t−τ
CX (τ ) − c dτ
t 2L
2ε t t − τ
≤
dτ
(CX (0) + c) dτ +
t0
tLt
2L (CX (0) + c) 2ε t t − τ
2L (CX (0) + c)
≤
+
dτ =
+ε
t
L0
t
t
≤ 2ε for t large enough
Thus the left side of (7.31) is equal to c, as claimed. Hence if limτ →∞ CX (τ ) = c, (7.31) holds if
and only if c = 0. It remains to prove that (b), (c) and (d) each imply (7.31).
Suppose condition (b) holds. Then
2
t t
0 t−τ
t CX (τ )dτ ≤
≤ 2
t
1
t t CX (τ )dτ
0
∞ CX (τ )dτ → 0 as t → ∞
−∞ so that (7.31) holds.
Suppose either condition (c) or condition (d) holds. By the same arguments applied to CX for
parts (a) and (b), it follows that
2
t t
0 t−τ
t RX (τ )dτ → 0 as t → ∞. Since the integral in (7.31) is the variance of a random variable, it is nonnegative. Also, the integral
is a weighted average of CX (τ ), and CX (τ ) = RX (τ ) − µ2 . Therefore,
X
0≤ 2
t t
0 t−τ
t CX (τ )dt = −µ2 +
X 2
t 207 t
0 t−τ
t RX (τ )dτ → −µ2 as t → ∞.
X Thus, (7.31) holds, so that X is mean ergodic in the m.s. sense. In addition, we see that conditions
(c) and (d) also each imply that µX = 0. Example 7.4.2 Let fc be a nonzero constant, let Θ be a random variable such that cos(Θ), sin(Θ),
cos(2Θ), and sin(2Θ) have mean zero, and let A be a random variable independent of Θ such that
E [A2 ] < +∞. Let X = (Xt : t ∈ R) be deﬁned by Xt = A cos(2πfc t + Θ). Then X is WSS with
2
µX = 0 and RX (τ ) = CX (τ ) = E [A ] cos(2πfc τ ) . Condition (7.31) is satisﬁed, so X is mean ergodic.
2
Mean ergodicity can also be directly veriﬁed:
1
t t Xu du = 0 =
≤ At
cos(2πfc u + Θ)du
t0
A(sin(2πfc t + Θ) − sin(Θ))
2πfc t
A
→ 0 m.s. as t → ∞.
πfc t Example 7.4.3 (Composite binary source) A student has two biased coins, each with a zero on one
side and a one on the other. Whenever the ﬁrst coin is ﬂipped the outcome is a one with probability
3
1
4 . Whenever the second coin is ﬂipped the outcome is a one with probability 4 . Consider a random
process (Wk : k ∈ Z) formed as follows. First, the student selects one of the coins, each coin being
selected with equal probability. Then the selected coin is used to generate the Wk ’s — the other
coin is not used at all.
This scenario can be modelled as in Figure 7.5, using the following random variables: Uk
S=0
S=1 Vk Wk Figure 7.5: A composite binary source. • (Uk : k ∈ Z) are independent Be 3
4 random variables • (Vk : k ∈ Z) are independent Be 1
4 random variables • S is a Be 1
2 random variable • The above random variables are all independent
• Wk = (1 − S )Uk + SVk .
208 The variable S can be thought of as a switch state. Value S = 0 corresponds to using the coin with
3
probability of heads equal to 4 for each ﬂip.
Clearly W is stationary, and hence also WSS. Is W mean ergodic? One approach to answering
this is the direct one. Clearly
µW = E [Wk ] = E [Wk S = 0]P [S = 0] + E [Wk  S = 1]P [S = 1] = 31 11
·+·
42 42 = 1
.
2 So the question is whether
1
n n 1
m.s.
2 ? Wk →
k=1 But by the strong law of large numbers
1
n n Wk = k=1 =
m.s. → 1
n n ((1 − S )Uk + SVk )
k=1 (1 − S ) 1
n n Uk +S k=1 3
1
(1 − S ) + S
4
4 = 1
n n Vk
k=1 3S
−.
4
2 1
Thus, the limit is a random variable, rather than the constant 2 . Intuitively, the process W has
such strong memory due to the switch mechanism that even averaging over long time intervals does
not diminish the randomness due to the switch.
Another way to show that W is not mean ergodic is to ﬁnd the covariance function CW and use
2
the necessary and suﬃcient condition (7.31) for mean ergodicity. Note that for k ﬁxed, Wk = Wk
2 ] = 1 . If k = l, then
with probability one, so E [Wk
2 E [Wk Wl ] = E [Wk Wl  S = 0]P [S = 0] + E [Wk Wl  S = 1]P [S = 1]
1
1
= E [Uk Ul ] + E [Vk Vl ]
2
2
1
1
= E [Uk ]E [Ul ] + E [Vk ]E [Vl ]
2
2
2
2
3
1
1
1
5
=
+
=
.
4
2
4
2
16
Therefore,
CW (n) = 1
4
1
16 if n = 0
if n = 0 Since limn→∞ CW (n) exists and is not zero, W is not mean ergodic.
In many applications, we are interested in averages of functions that depend on multiple random
variables. We discuss this topic for a discrete time stationary random process, (Xn : n ∈ Z). Let h
209 be a bounded, Borel measurable function on Rk for some k . What time average would we expect
to be a good approximation to the statistical average E [h(X1 , . . . , Xk )]? A natural choice is
1
n n
j =1 h(Xj , Xj +1 , . . . , Xj +k−1 ). We deﬁne a stationary random process (Xn : n ∈ Z) to be ergodic if
1
n→∞ n n lim h(Xj , . . . , Xj +k−1 ) = E [h(X1 , . . . , Xk )]
j =1 for every k ≥ 1 and for every bounded Borel measurable function h on Rk , where the limit is taken
in any of the three senses a.s., p. or m.s.3 An interpretation of the deﬁnition is that if X is ergodic
then all of its ﬁnite dimensional distributions are determined as time averages.
As an example, suppose
h(x1 , x2 ) = 1 if x1 > 0 ≥ x2
.
0 else Then h(X1 , X2 ) is one if the process (Xk ) makes a “down crossing” of level 0 between times one
and two. If X is ergodic then with probability 1,
lim n→∞ 1
(number of down crossings between times 1 and n + 1) = P [X1 > 0 ≥ X2 ]. (7.36)
n Equation (7.36) relates quantities that are quite diﬀerent in nature. The left hand side of (7.36)
is the long timeaverage downcrossing rate, whereas the right hand side of (7.36) involves only the
joint statistics of two consecutive values of the process.
Ergodicity is a strong property. Two types of ergodic random processes are the following:
• a process X = (Xk ) such that the Xk ’s are iid.
• a stationary Gaussian random process X such that limn→∞ RX (n) = 0 or, limn→∞ CX (n) = 0. 7.5 Complexiﬁcation, Part I In some application areas, primarily in connection with spectral analysis as we shall see, complex
valued random variables naturally arise. Vectors and matrices over C are reviewed in the appendix.
A complex random variable X = U + jV can be thought of as essentially a two dimensional random
variable with real coordinates U and V . Similarly, a random complex ndimensional vector X can be
written as X = U + jV , where U and V are each ndimensional real vectors. As far as distributions
are concerned, a random vector in ndimensional complex space Cn is equivalent to a random vector
with 2n real dimensions. For example, if the 2n real variables in U and V are jointly continuous,
then X is a continuous type complex random vector and its density is given by a function fX (x)
for x ∈ Cn . The density fX is related to the joint density of U and V by fX (u + jv ) = fU V (u, v )
for all u, v ∈ Rn .
3 The mathematics literature uses a diﬀerent deﬁnition of ergodicity for stationary processes, which is equivalent.
There are also deﬁnitions of ergodicity that do not require stationarity. 210 As far as moments are concerned, all the second order analysis covered in the notes up to this
point can be easily modiﬁed to hold for complex random variables, simply by inserting complex
conjugates in appropriate places. To begin, if X and Y are complex random variables, we deﬁne
their correlation by E [XY ∗ ] and similarly their covariance as E [(X − E [X ])(Y − E [Y ])∗ ]. The
Schwarz inequality becomes E [XY ∗ ] ≤ E [X 2 ]E [Y 2 ] and its proof is essentially the same as
for real valued random variables. The cross correlation matrix for two complex random vectors
X and Y is given by E [XY ∗ ], and similarly the cross covariance matrix is given by Cov(X, Y ) =
E [(X − E [X ])(Y − E [Y ])∗ ]. As before, Cov(X ) = Cov(X, X ). The various formulas for covariance
still apply. For example, if A and C are complex matrices and b and d are complex vectors, then
Cov(AX + b, CY + d) = ACov(X, Y )C ∗ . Just as in the case of real valued random variables, a
matrix K is a valid covariance matrix (in other words, there exits some random vector X such that
K = Cov(X )) if and only if K is Hermitian symmetric and positive semideﬁnite.
Complex valued random variables X and Y with ﬁnite second moments are said to be orthogonal
if E [XY ∗ ] = 0, and with this deﬁnition the orthogonality principle holds for complex valued random
variables. If X and Y are complex random vectors, then again E [X Y ] is the MMSE estimator of
X given Y , and the covariance matrix of the error vector is given by Cov(X ) − Cov(E [X Y ]). The
MMSE estimator for X of the form AY + b (i.e. the best linear estimator of X based on Y ) and
the covariance of the corresponding error vector are given just as for vectors made of real random
variables:
ˆ
E [X Y ] = E [X ] + Cov(X, Y )Cov(Y )−1 (Y − E [Y ])
ˆ
Cov(X − E [X Y ]) = Cov(X ) − Cov(X, Y )Cov(Y )−1 Cov(Y, X )
By deﬁnition, a sequence X1 , X2 , . . . of complex valued random variables converges in the m.s.
sense to a random variable X if E [Xn 2 ] < ∞ for all n and if limn→∞ E [Xn − X 2 ] = 0. The
various Cauchy criteria still hold with minor modiﬁcation. A sequence with E [Xn 2 ] < ∞ for all
n is a Cauchy sequence in the m.s. sense if limm,n→∞ E [Xn − Xm 2 ] = 0. As before, a sequence
converges in the m.s. sense if and only if it is a Cauchy sequence. In addition, a sequence X1 , X2 , . . .
of complex valued random variables with E [Xn 2 ] < ∞ for all n converges in the m.s. sense if
∗
and only if limm,n→∞ E [Xm Xn ] exits and is a ﬁnite constant c. If the m.s. limit exists, then the
limiting random variable X satisﬁes E [X 2 ] = c.
Let X = (Xt : t ∈ T) be a complex random process. We can write Xt = Ut + jVt where U and
V are each real valued random processes. The process X is deﬁned to be a second order process
if E [Xt 2 ] < ∞ for all t. Since Xt 2 = Ut2 + Vt2 for each t, X being a second order process is
equivalent to both U and V being second order processes. The correlation function of a second
∗
order complex random process X is deﬁned by RX (s, t) = E [Xs Xt ]. The covariance function is
given by CX (s, t) = Cov(Xs , Xt ) where the deﬁnition of Cov for complex random variables is used.
The deﬁnitions and results given for m.s. continuity, m.s. diﬀerentiation, and m.s. integration all
carry over to the case of complex processes, because they are based on the use of the Cauchy criteria
for m.s. convergence which also carries over. For example, a complex valued random process is m.s.
continuous if and only if its correlation function RX is continuous. Similarly the cross correlation
function for two second order random processes X and Y is deﬁned by RXY (s, t) = E [Xs Yt∗ ]. Note
∗
that RXY (s, t) = RY X (t, s).
Let X = (Xt : t ∈ T) be a complex random process such that T is either the real line or
the set of integers, and write Xt = Ut + jVt where U and V are each real valued random processes. By deﬁnition, X is stationary if and only if for any t1 , . . . , tn ∈ T, the joint distribution of
211 (Xt1 +s , . . . , Xtn +s ) is the same for all s ∈ T. Equivalently, X is stationary if and only if U and V
are jointly stationary. The process X is deﬁned to be WSS if X is a second order process such that
E [Xt ] does not depend on t, and RX (s, t) is a function of s − t alone. If X is WSS we use RX (τ ) to
denote RX (s, t), where τ = s − t. A pair of complexvalued random processes X and Y are deﬁned
to be jointly WSS if both X and Y are WSS and if the cross correlation function RXY (s, t) is a
∗
function of s − t. If X and Y are jointly WSS then RXY (−τ ) = RY X (τ ).
In summary, everything we’ve discussed in this section regarding complex random variables,
vectors, and processes can be considered a simple matter of notation. One simply needs to add
lines to indicate complex conjugates, to use X 2 instead of X 2 , and to use a star “∗ ” for Hermitian
transpose in place of “T ” for transpose. We shall begin using the notation at this point, and return
to a discussion of the topic of complex valued random processes in a later section. In particular,
we will examine complex normal random vectors and their densities, and we shall see that there is
somewhat more to complexiﬁcation than just notation. 7.6 The KarhunenLo`ve expansion
e We’ve seen that under a change of coordinates, an ndimensional random vector X is transformed
into a vector Y = U ∗ X such that the coordinates of Y are orthogonal random variables. Here
U is the unitary matrix such that E [XX ∗ ] = U ΛU ∗ . The columns of U are eigenvectors of the
Hermitian symmetric matrix E [XX ∗ ] and the corresponding nonnegative eigenvalues of E [XX ∗ ]
comprise the diagonal of the diagonal matrix Λ. The columns of U form an orthonormal basis for
Cn . The KarhunenLo`ve expansion gives a similar change of coordinates for a random process on
e
a ﬁnite interval, using an orthonormal basis of functions instead of an othonormal basis of vectors.
Fix an interval [a, b]. The L2 norm of a real or complex valued function f on the interval [a, b]
is deﬁned by
b f (t)2 dt f  =
a We write L2 [a, b] for the set of all functions on [a, b] which have ﬁnite L2 norm. The inner product
of two functions f and g in L2 [a, b] is deﬁned by
b (f, g ) = f (t)g ∗ (t)dt a The functions f and g are said to be orthogonal if (f, g ) = 0. Note that f  = (f, f ) and the
Schwarz inequality holds: (f, g ) ≤ f  · g . An orthonormal basis for L2 [a, b] is a sequence of
functions φ1 , φ2 , . . . which are mutually orthogonal and have norm one, or in other words, (φi , φj ) =
I{i=j } for all i and j .
In many applications it is useful to use representations of the form
N f (t) = cn φn (t). (7.37) n=1 In such a case, we think of (c1 , . . . , cN ) as the coordinates of f relative to the basis (φn ), and we
might write f ↔ (c1 , , . . . , cN ). For example, transmitted signals in many digital communication
systems have this form, where the coordinate vector (c1 , , . . . , cN ) represents a data symbol. The
212 geometry of the space of all functions f of the form (7.37) for the ﬁxed orthonormal basis (φn ) is
equivalent to the geometry of the coordinates vectors. For example, if g has a similar representation,
N g (t) = dn φn (t),
n=1 or equivalently g ↔ (d1 , . . . , dN ), then f + g ↔ (c1 , . . . , cN ) + (d1 , . . . , dN ) and
N b N (f, g ) = d∗ φ∗ (t) dt
nn cm φm (t)
a m=1
N N n=1
b cm d∗
n = φm (t)φ∗ (t)dt
n a m=1 n=1
N
N cm d∗ (φm , φn )
n =
m=1 n=1
N cm d∗
m =
m=1 That is, the inner product of the functions, (f, g ), is equal to the inner product of their coordinate
vectors. Note that for 1 ≤ n ≤ N , φn ↔ (0, . . . , 0, 1, 0, . . . , 0), such that the one is in the nth
position. If f ↔ (c1 , . . . , cN ), the coordinates of f can be obtained by computing inner products
between f and the basis functions:
N N b cm φm (t) φ∗ (t)dt =
n (f, φn ) =
a cm (φm , φn ) = cn .
m=1 m=1 Another way to derive that (f, φn ) = cn is to note that f ↔ (c1 , . . . , cN ) and φn ↔ (0, . . . , 0, 1, 0, . . . , 0),
so (f, φn ) is the inner product of (c1 , . . . , cN ) and (0, . . . , 0, 1, 0, . . . , 0), or cn . Thus, the coordinate
vector for f is given by f ↔ ((f, φ1 ), . . . , (f, φN )).
The dimension of the space L2 [a, b] is inﬁnite, so there are orthonormal bases (φn : n ≥ 1) with
inﬁnitely many functions (i.e. N = ∞). For such a basis, a function f can have the representation
∞ f (t) = cn φn (t) (7.38) n=1 In many instances encountered in practice, the sum (7.38) converges for each t, but in general what
is meant is that the convergence is in the sense of the L2 [a, b] norm:
N b lim N →∞ a cn φn (t)2 dt = 0, f (t) −
n=1 or equivalently,
N lim f − N →∞ cn φn  = 0
n=1 213 An orthonormal basis (φn ) is deﬁned to be complete if any function f in L2 [a, b] can be written as
in (7.38). A commonly used example of a complete orthonormal basis for an interval [0, T ] is given
by
1
√,
T
2
=T φ1 (t) =
φ2k (t) φ2 (t) = 2
T πt
cos( 2T ),
2
T cos( 2πkt ), φ2k+1 (t) =
T φ3 (t) = 2
T πt
sin( 2T ), sin( 2πkt ) for k ≥ 1.
T (7.39) What happens if we replace f by a random process X = (Xt : a ≤ t ≤ b)? Suppose (φn : 1 ≤
n ≤ N ) is an orthonormal basis consisting of continuous functions, with N ≤ ∞. (The basis does
not have to be complete, and we allow N to be ﬁnite or to equal inﬁnity. Of course, if the basis is
complete, then N = ∞.) Suppose that X is m.s. continuous, or equivalently, that RX is continuous
b
b
as a function on [a, b] × [a, b]. In particular, RX is bounded. Then E [ a Xt 2 dt] = a RX (t, t)dt < ∞,
b
so that a Xt 2 dt is ﬁnite with probability one. Suppose that X can be represented as
N Xt = Cn φn (t). (7.40) n=1 Such a representation exists if the basis (φn ) is complete, but some random processes have the form
(7.40) even if N is ﬁnite or if N is inﬁnite but the basis is not complete. The representation (7.40)
reduces the description of the continuoustime random process to the description of the coeﬃcients,
(Cn ). This representation of X is much easier to work with if the coordinate random variables are
orthogonal.
Deﬁnition 7.6.1 A KarhunenLo`ve (KL) expansion for a random process X = (Xt : a ≤ t ≤ b)
e
is a representation of the form (7.40) such that:
(1) the functions (φn ) are orthonormal: (φm , φn ) = I{m=n} , and
∗
(2) the coordinate random variables Cn are mutually orthogonal: E [Cm Cn ] = 0.
Example 7.6.2 Let Xt = A for 0 ≤ 1 ≤ T, where A is a random variable with 0 < E [A2 ] < ∞.
√
I ≤≤
Then X has the form in (7.40) for N = 1, C1 = A T , and φ1 (t) = {0√tT T } . This is trivially a KL
expansion, with only one term. Example 7.6.3 Let Xt = A cos(2πt/T +Θ) for 0 ≤ t ≤ T, where A is a realvalued random variable
with 0 < E [A2 ] < ∞, and Θ is a random variable uniformly distributed on [0, 2π ] and independent
of A. By the cosine angle addition formula, Xt = A cos(Θ) cos(2πt/T ) − A sin(Θ) sin(2πt/T ). Then
X has the form in (7.40) for N = 2,
√
√
cos(2πt/T )
sin(2πt/T )
√
√
C1 = A 2T cos(Θ), C2 = −A 2T sin(Θ), φ1 (t) =
φ2 (t) =
.
2T
2T
In particular, φ1 and φ2 form an orthonormal basis with N = 2 elements. To check whether
∗
∗
this is a KL expansion, we see if E [C1 C2 ] = 0. Since E [C1 C2 ] = −2T E [A2 ]E [cos(Θ) sin(Θ)] =
2 ]E [sin(2Θ)] = 0, this is indeed a KL expansion, with two terms.
−T E [A 214 Proposition 7.6.4 Suppose X = (Xt : a ≤ t ≤ b) is m.s. continuous and (φn ) is an orthonormal
basis of continuous functions. If (7.40) holds for some random variables (Cn ), it is a KL expansion (i.e., the coordinate random variables are orthogonal) if and only if the basis functions are
eigenfunctions of RX :
RX φn = λn φn ,
(7.41)
where for a function φ ∈ L2 [a, b], RX φ denotes the function (RX φ)(s) =
(7.40) is a KL expansion, the eigenvalues are given by λn = E [Cn 2 ].
Proof. Suppose (7.40) holds. Then Cn = (X, φn ) = b
∗
a Xt φn (t)dt, b
a RX (s, t)φ(t)dt. In case so that ∗
E [Cm Cn ] = E [(X, φm )(X, φn )∗ ]
b =E
a
b b =
a b Xs φ∗ (s)ds
m ∗ Xt φ∗ (t)dt
n a RX (s, t)φ∗ (s)φn (t)dsdt
m a = (RX φn , φm ) (7.42) ∗
Now, if the basis functions (φn ) are eigenfunctions of RX , then E [Cm Cn ] = (RX φn , φm ) =
∗ ] = 0 if n = m, so that (7.40) is
(λn φn , φm ) = λn (φn , φm ) = λn I{m=n} . In particular, E [Cm Cn
a KL expansion. Also, taking m = n yields E [Cn 2 ] = λn .
Conversely, suppose (7.40) is a KL expansion. Without loss of generality, suppose that the basis
(φn ) is complete. (If it weren’t complete, it could be extended to a complete basis by augmenting it
with functions from a complete basis and applying the GrammSchmidt method of orthogonalizing.)
Then for n ﬁxed, (RX φn , φm ) = 0 for all m = n. By completeness of the (φn ), the function RX φn has
an expansion of the form (7.38), but all terms except possibly the nth are zero. Hence, Rn φn = λn φn
for some constant λn , so the eigenrelations (7.41) hold. Again, E [Cn 2 ] = λn by the computation
above. The following theorem is stated without proof.
Theorem 7.6.5 (Mercer’s theorem) If X = (Xt : a ≤ t ≤ b) is m.s. continuous, or, equivalently,
that RX is continuous. Then there exists a complete orthonormal basis (φn : n ≥ 1) of continuous
eigenfunctions and corresponding nonnegative eigenvalues (λn : n ≥ 1) for RX , and RX is given
by the following series expansion:
∞ λn φn (s)φ∗ (t).
n RX (s, t) = (7.43) n=1 The series converges uniformly in s, t, meaning that
N lim λn φn (s)φ∗ (t) = 0
n max RX (s, t) − N →∞ a≤s,t≤b n=1 Theorem 7.6.6 ( KarhunenLo`ve expansion) Any process X satisfying the conditions of Mercer’s
e
theorem has a KL expansion,
∞ Xt = φn (t)(X, φn ),
n=1 215 and the series converges in the m.s. sense, uniformly in t.
Proof. Use the complete orthonormal basis (φn ) guaranteed by Mercer’s theorem. By (7.42),
E [(X, φm )∗ (X, φn )] = (RX φn , φm ) = λn I{n=m} . Also,
b E [Xt (X, φn )∗ ] = E [Xt ∗
Xs φn (s)ds] a
b RX (t, s)φn (s)ds = λn φn (t). =
a These facts imply that for ﬁnite N, 2 N E Xt − N λn φn (t)2 , φn (t)(X, φn ) = RX (t, t) −
n=1 (7.44) n=1 which, since the series on the right side of (7.44) converges uniformly in t as n → ∞, implies the
stated convergence property for the representation of X .
Remarks (1) The means of the coordinates of X in a KL expansion can be expressed using the
mean function µX (t) = E [Xt ] as follows:
b E [(X, φn )] = µX (t)φ∗ (t)dt = (µX , φn )
n a Thus, the mean of the nth coordinate of X is the nth coordinate of the mean function of X.
(2) Symbolically, mimicking matrix notation, we can write the representation (7.43) of RX as ∗ λ1
φ1 (t) φ∗ (t) λ2 2 . λ3
RX (s, t) = [φ1 (s)φ2 (s) · · · ] . . .. .
(3) If f ∈ L2 [a, b] and f (t) represents a voltage or current across a resistor, then the energy
dissipated during the interval [a, b] is, up to a multiplicative constant, given by
∞ b
2 2 (Energy of f ) = f  = (f, φn )2 . f (t) dt =
a n=1 The mean total energy of (Xt : a < t < b) is thus given by
b b Xt 2 dt] = E[
a RX (t, t)dt
a
b∞ λn φn (t)2 dt =
a n=1
∞ = λn
n=1 216 (4) If (Xt : a ≤ t ≤ b) is a real valued mean zero Gaussian process and if the orthonormal basis
functions are real valued, then the coordinates (X, φn ) are uncorrelated, real valued, jointly Gaussian random variables, and therefore are independent. Example 7.6.7 Let W = (Wt : t ≥ 0) be a Brownian motion with parameter σ 2 . Let us ﬁnd the
KL expansion of W over the interval [0, T ]. Substituting RX (s, t) = σ 2 (s ∧ t) into the eigenrelation
(7.41) yields
t T σ 2 sφn (s)ds +
0 σ 2 tφn (s)ds = λn φn (t) (7.45) t Diﬀerentiating (7.45) with respect to t yields
T σ 2 tφn (t) − σ 2 tφn (t) + σ 2 φn (s)ds = λn φn (t), (7.46) t and diﬀerentiating a second time yields that the eigenfunctions satisfy the diﬀerential equation
λφ = −σ 2 φ. Also, setting t = 0 in (7.45) yields the boundary condition φn (0) = 0, and setting
t = T in (7.46) yields the boundary condition φn (T ) = 0. Solving yields that the eigenvalue and
eigenfunction pairs for W are
λn = 4σ 2 T 2
(2n + 1)2 π 2 φn (t) = 2
sin
T (2n + 1)πt
2T n≥0 These functions form a complete orthonormal basis for L2 [0, T ]. Example 7.6.8 Let X be a white noise process. Such a process is not a random process as deﬁned
in these notes, but can be deﬁned as a generalized process in the same way that a delta function can
be deﬁned as a generalized function. Generalized random processes, just like generalized functions,
only make sense when multiplied by a suitable function and then integrated. For example, the
delta function δ is deﬁned by the requirement that for any function f that is continuous at t = 0,
∞ f (t)δ (t)dt = f (0)
−∞
∞ A white noise process X is such that integrals of the form −∞ f (t)X (t)dt exist for functions f with
ﬁnite L2 norm f . The integrals are random variables with ﬁnite second moments, mean zero and
correlations given by
∞ f (s)Xs ds E
−∞ ∗ ∞ g (t)Xt dt
−∞ ∞ = σ2 f (t)g ∗ (t)dt −∞ In a formal or symbolic sense, this means that X is a WSS process with mean µX = 0 and
∗
autocorrelation function RX (s, t) = E [Xs Xt ] given by RX (τ ) = σ 2 δ (τ ).
What would the KL expansion be for a white noise process over some ﬁxed interval [a,b]?
The eigenrelation (7.41) becomes simply σ 2 φ(t) = λn φ(t) for all t in the interval. Thus, all the
eigenvalues of a white noise process are equal to σ 2 , and any function f with ﬁnite norm is an
217 eigenfunction. Thus, if (φn : n ≥ 1) is an arbitrary complete orthonormal basis for the square
integrable functions on [a, b], then the coordinates of the white noise process X , formally given by
Xn = (X, φn ), satisfy
∗
E [Xn Xm ] = σ 2 I{n=m} .
(7.47)
This oﬀers a reasonable interpretation of white noise. It is a generalized random process such that
its coordinates (Xn : n ≥ 1) relative to an arbitrary orthonormal basis for a ﬁnite interval have
mean zero and satisfy (7.47). 7.7 Periodic WSS random processes Let X = (Xt : t ∈ R) be a WSS random process and let T be a positive constant.
Proposition 7.7.1 The following three conditions are equivalent:
(a) RX (T ) = RX (0)
(b) P [XT +τ = Xτ ] = 1 for all τ ∈ R
(c) RX (T + τ ) = RX (τ ) for all τ ∈ R (i.e. RX (τ ) is periodic with period T.)
Proof. Suppose (a) is true. Since RX (0) is real valued, so is RX (T ), yielding
∗
∗
∗
∗
E [XT +τ − Xτ 2 ] = E [XT +τ XT +τ − XT +τ Xτ − Xτ XT +τ + Xτ Xτ ]
∗
= RX (0) − RX (T ) − RX (T ) + RX (0) = 0 Therefore, (a) implies (b). Next, suppose (b) is true and let τ ∈ R. Since two random variables
that are equal with probability one have the same expectation, (b) implies that
∗
∗
RX (T + τ ) = E [XT +τ X0 ] = E [Xτ X0 ] = RX (τ ). Therefore (b) imples (c). Trivially (c) implies (a), so the equivalence of (a) through (c) is proved. Deﬁnition 7.7.2 We call X a periodic, WSS process of period T if X is WSS and any of the three
equivalent properties (a), (b), or (c) of Proposition 7.7.1 hold.
Property (b) almost implies that the sample paths of X are periodic. However, for each τ it can
be that Xτ = Xτ +T on an event of probability zero, and since there are uncountably many real
numbers τ , the sample paths need not be periodic. However, suppose (b) is true and deﬁne a
process Y by Yt = X(t mod T ) . (Recall that by deﬁnition, (t mod T ) is equal to t + nT , where n
is selected so that 0 ≤ t + nT < T .) Then Y has periodic sample paths, and Y is a version of X ,
which by deﬁnition means that P [Xt = Yt ] = 1 for any t ∈ R. Thus, the properties (a) through (c)
are equivalent to the condition that X is WSS and there is a version of X with periodic sample
paths of period T .
Suppose X is a m.s. continuous, periodic, WSS random process. Due to the periodicity of X ,
it is natural to consider the restriction of X to the interval [0, T ]. The KarhunenLo`ve expansion
e
of X restricted to [0, T ] is described next. Let φn be the function on [0, T ] deﬁned by
φn (t) = e2πjnt/T
√
T 218 The functions (φn : n ∈ Z) form a complete orthonormal basis for L2 [0, T ].4 In addition, for any n
ﬁxed, both RX (τ ) and φn are periodic with period dividing T , so
T T RX (s − t)φn (t)dt RX (s, t)φn (t)dt =
0 0 s RX (t)φn (s − t)dt =
s− T
T RX (t)φn (s − t)dt =
0 T
1
√
RX (t)e2πjns/T e−2πjnt/T dt
T0
= λn φn (s). = where λn is given by T λn = RX (t)e−2πjnt/T dt = √
T (RX , φn ). (7.48) 0 Therefore φn is an eigenfunction of RX with eigenvalue λn . The KarhunenLo`ve expansion (5.20)
e
of X over the interval [0, T ] can be written as
∞
ˆ
Xn e2πjnt/T Xt = (7.49) n=−∞ ˆ
where Xn is deﬁned by
1
1
ˆ
Xn = √ (X, φn ) =
T
T T Xt e−2πjnt/T dt 0 Note that 1
λn
ˆ ˆ∗
E [Xm Xn ] = E [(X, φm )(X, φn )∗ ] =
I
T
T {m=n}
Although the representation (7.49) has been derived only for 0 ≤ t ≤ T , both sides of (7.49) are
periodic with period T . Therefore, the representation (7.49) holds for all t. It is called the spectral
representation of the periodic, WSS process X .
By (7.48), the series expansion (7.38) applied to the function RX over the interval [0, T ] can be
written as
∞ RX (t) = λn 2πjnt/T
e
T
n=−∞
pX (ω )ejωt , = (7.50) ω where pX is the function on the real line R = (ω : −∞ < ω < ∞),5 deﬁned by
pX (ω ) = λn /T
0 ω=
else 2πn
T for some integer n 4
Here it is more convenient to index the functions by the integers, rather than by the nonnegative integers. Sums
P
P
of the form ∞ −∞ should be interpreted as limits of N=−N as N → ∞.
n=
n
5
The Greek letter ω is used here as it is traditionally used for frequency measured in radians per second. It is
related to the frequency f measured in cycles per second by ω = 2πf . Here ω is not the same as a typical element
of the underlying space of all outcomes, Ω. The meaning of ω should be clear from the context. 219 and the sum in (7.50) is only over ω such that pX (ω ) = 0. The function pX is called the power
spectral mass function of X . It is similar to a probability mass function, in that it is positive for
πn
at most a countable inﬁnity of values. The value pX ( 2T ) is equal to the power of the nth term in
the representation (7.49):
ˆ
ˆ
E [Xn e2πjnt/T 2 ] = E [Xn 2 ] = pX ( 2πn
)
T and the total mass of pX is the total power of X , RX (0) = E [Xt ]2 .
Periodicity is a rather restrictive assumption to place on a WSS process. In the next chapter we
shall further investigate spectral properties of WSS processes. We shall see that many WSS random
processes have a power spectral density. A given random variable might have a pmf or a pdf, and
it deﬁnitely has a CDF. In the same way, a given WSS process might have a power spectral mass
function or a power spectral density function, and it deﬁnitely has a cumulative power spectral
distribution function. The periodic WSS processes of period T are precisely those WSS processes
that have a power spectral mass function that is concentrated on the integer multiples of 2π .
T 7.8 Problems 7.1 Calculus for a simple Gaussian random process
Deﬁne X = (Xt : t ∈ R) by Xt = A + Bt + Ct2 , where A, B, and C are independent, N (0, 1)
1
random variables. (a) Verify directly that X is m.s. diﬀerentiable. (b) Express P { 0 Xs ds ≥ 1} in
terms of Q, the standard normal complementary CDF.
7.2 Lack of sample path continuity of a Poisson process
Let N = (Nt : t ≥ 0) be a Poisson process with rate λ > 0. (a) Find the following two probabilities,
explaining your reasoning: P {N is continuous over the interval [0,T] } for a ﬁxed T > 0, and
P {N is continuous over the interval [0, ∞)}. (b) Is N sample path continuous a.s.? Is N m.s.
continuous?
7.3 Properties of a binary valued process
Let Y = (Yt : t ≥ 0) be given by Yt = (−1)Nt , where N is a Poisson process with rate λ > 0.
(a) Is Y a Markov process? If so, ﬁnd the transition probability function pi,j (s, t) and the transition
rate matrix Q. (b) Is Y mean square continuous? (c) Is Y mean square diﬀerentiable? (d) Does
1T
limT →∞ T 0 yt dt exist in the m.s. sense? If so, identify the limit.
7.4 Some statements related to the basic calculus of random processes
Classify each of the following statements as either true (meaning always holds) or false, and justify
your answers.
(a) Let Xt = Z , where Z is a Gaussian random variable. Then X = (Xt : t ∈ R) is mean ergodic
in the m.s. sense.
σ 2 τ  ≤ 1
(b) The function RX deﬁned by RX (τ ) =
is a valid autocorrelation function.
0 τ >1
(c) Suppose X = (Xt : t ∈ R) is a mean zero stationary Gaussian random process, and suppose X
is m.s. diﬀerentiable. Then for any ﬁxed time t, Xt and Xt are independent.
220 7.5 Diﬀerentiation of the square of a Gaussian random process
(a) Show that if random variables (An : n ≥ 0) are mean zero and jointly Gaussian and if
limn→∞ An = A m.s., then limn→∞ A2 = A2 m.s. (Hint: If A, B, C , and D are mean zero and
n
jointly Gaussian, then E [ABCD] = E [AB ]E [CD] + E [AC ]E [BD] + E [AD]E [BC ].)
(b) Show that if random variables (An , Bn : n ≥ 0) are jointly Gaussian and limn→∞ An = A m.s.
and limn→∞ Bn = B m.s. then limn→∞ An Bn = AB m.s. (Hint: Use part (a) and the identity
2− 2
2
ab = (a+b) 2 a −b .)
2
(c) Let X be a mean zero, m.s. diﬀerentiable Gaussian random process, and let Yt = Xt for all t.
Is Y m.s. diﬀerentiable? If so, justify your answer and express the derivative in terms of Xt and
Xt .
7.6 Cross correlation between a process and its m.s. derivative
Suppose X is a m.s. diﬀerentiable random process. Show that RX X = ∂1 RX . (It follows, in
particular, that ∂1 RX exists.)
7.7 Fundamental theorem of calculus for m.s. calculus
Suppose X = (Xt : t ≥ 0) is a m.s. continuous random process. Let Y be the process deﬁned by
t
Yt = 0 Xu du for t ≥ 0. Show that X is the m.s. derivative of Y . (It follows, in particular, that Y
is m.s. diﬀerentiable.)
7.8 A windowed Poisson process
Let N = (Nt : t ≥ 0) be a Poisson process with rate λ > 0, and let X = (Xt : t ≥ 0) be deﬁned by
Xt = Nt+1 − Nt . Thus, Xt is the number of counts of N during the time window (t, t + 1].
(a) Sketch a typical sample path of N , and the corresponding sample path of X .
(b) Find the mean function µX (t) and covariance function CX (s, t) for s, t ≥ 0. Express your
answer in a simple form.
(c) Is X Markov? Why or why not?
(d) Is X meansquare continuous? Why or why not?
t
(e) Determine whether 1 0 Xs ds converges in the mean square sense as t → ∞.
t
7.9 An integral of white noise times an exponential
t
Let Xt = 0 Zu e−u du, for t ≥ 0, where Z is white Gaussian noise with autocorrelation function
δ (τ )σ 2 , for some σ 2 > 0. (a) Find the autocorrelation function, RX (s, t) for s, t ≥ 0. (b) Is X
mean square diﬀerentiable? Justify your answer. (c) Does Xt converge in the mean square sense
as t → ∞? Justify your answer.
7.10 A singular integral with a Brownian motion
1
Consider the integral 0 wt dt, where w = (wt : t ≥ 0) is a standard Brownian motion. Since
t
1 wt
Var( wt ) = 1 diverges as t → 0, we deﬁne the integral as lim →0
t
t
t dt m.s. if the limit exists.
(a) Does the limit exist? If so, what is the probability distribution of the limit?
∞
T
(b) Similarly, we deﬁne 1 wt dt to be limT →∞ 1 wt dt m.s. if the limit exists. Does the limit
t
t
exist? If so, what is the probability distribution of the limit?
7.11 An integrated Poisson process
t
Let N = (Nt : t ≥ 0) denote a Poisson process with rate λ > 0, and let Yt = 0 Ns ds for s ≥ 0. (a)
Sketch a typical sample path of Y . (b) Compute the mean function, µY (t), for t ≥ 0. (c) Compute
Var(Yt ) for t ≥ 0. (d) Determine the value of the limit, limt→∞ P [Yt < t].
221 7.12 Recognizing m.s. properties
Suppose X is a mean zero random process. For each choice of autocorrelation function shown, indicate which of the following properties X has: m.s. continuous, m.s. diﬀerentiable, m.s. integrable
over ﬁnite length intervals, and mean ergodic in the the m.s. sense.
(a) X is WSS with RX (τ ) = (1 − τ )+ ,
(b) X is WSS with RX (τ ) = 1 + (1 − τ )+ ,
(c) X is WSS with RX (τ ) = cos(20πτ ) exp(−10τ ),
1 if s = t
(d) RX (s, t) =
, (not WSS, you don’t need to check for mean ergodic property)
0 else
√
(e) RX (s, t) = s ∧ t for s, t ≥ 0. (not WSS, you don’t need to check for mean ergodic property)
7.13 A random Taylor’s approximation
Suppose X is a mean zero WSS random process such that RX is twice continuously diﬀerentiable.
Guided by Taylor’s approximation for deterministic functions, we might propose the following
estimator of Xt given X0 and X0 : Xt = X0 + tX0 .
(a) Express the covariance matrix for the vector (X0 , X0 , Xt )T in terms of the function RX and its
derivatives.
(b) Express the mean square error E [(Xt − Xt )2 ] in terms of the function RX and its derivatives.
(c) Express the optimal linear estimator E [Xt X0 , X0 ] in terms of X0 , X0 , and the function RX and
its derivatives.
(d) (This part is optional  not required.) Compute and compare limt→0 (mean square error)/t4
for the two estimators, under the assumption that RX is four times continuously diﬀerentiable.
7.14 A stationary Gaussian process
1
Let X = (Xt : t ∈ Z) be a real stationary Gaussian process with mean zero and RX (t) = 1+t2 .
Answer the following unrelated questions.
(a) Is X a Markov process? Justify your anwer.
(b) Find E [X3 X0 ] and express P {X3 − E [X3 X0 ] ≥ 10} in terms of Q, the standard Gaussian
complementary cumulative distribution function.
(c) Find the autocorrelation function of X , the m.s. derivative of X .
(d) Describe the joint probability density of (X0 , X0 , X1 )T . You need not write it down in detail.
7.15 Integral of a Brownian bridge
A standard Brownian bridge B can be deﬁned by Bt = Wt − tW1 for 0 ≤ t ≤ 1, where W is a
Brownian motion with parameter σ 2 = 1. A Brownian bridge is a mean zero, Gaussian random
process which is a.s. sample path continuous, and has autocorrelation function RB (s, t) = s(1 − t)
for 0 ≤ s ≤ t ≤ 1.
1
(a) Why is the integral X = 0 Bt dt well deﬁned in the m.s. sense?
(b) Describe the joint distribution of the random variables X and W1 .
7.16 Correlation ergodicity of Gaussian processes
A WSS random process X is called correlation ergodic (in the m.s. sense) if for any constant h,
lim m.s. t→∞ 1
t t Xs+h Xs ds = E [Xs+h Xs ]
0 Suppose X is a mean zero, realvalued Gaussian process such that RX (τ ) → 0 as τ  → ∞. Show
that X is correlation ergodic. (Hints: Let Yt = Xt+h Xt . Then correlation ergodicity of X is
222 equivalent to mean ergodicity of Y . If A, B, C, and D are mean zero, jointly Gaussian random
variables, then E [ABCD] = E [AB ]E [CD] + E [AC ]E [BD] + E [AD]E [BC ].
7.17 A random process which changes at a random time
Let Y = (Yt : t ∈ R) and Z = (Zt : t ∈ R) be stationary Gaussian Markov processes with mean zero
and autocorrelation functions RY (τ ) = RZ (τ ) = e−τ  . Let U be a realvalued random variable and
suppose Y , Z , and U , are mutually independent. Finally, let X = (Xt : t ∈ R) be deﬁned by
Xt = Yt t < U
Zt t ≥ U (a) Sketch a typical sample path of X .
(b) Find the ﬁrst order distributions of X .
(c) Express the mean and autocorrelation function of X in terms of the CDF, FU , of U .
(d) Under what condition on FU is X m.s. continuous?
(e) Under what condition on FU is X a Gaussian random process?
7.18 Gaussian review question
Let X = (Xt : t ∈ R) be a realvalued stationary GaussMarkov process with mean zero and
autocorrelation function CX (τ ) = 9 exp(−τ ).
(a) A fourth degree polynomial of two variables is given by p(x, y ) = a+bx+cy +dxy +ex2 y +f xy 2 +...
such that all terms have the form cxi y j with i + j ≤ 4. Suppose X2 is to be estimated by an
estimator of the form p(X0 , X1 ). Find the fourth degree polynomial p to minimize the MSE:
E [(X2 − p(X0 , X1 ))2 ] and ﬁnd the resulting MMSE. (Hint: Think! Very little computation is
needed.)
1
(b) Find P [X2 ≥ 4X0 = π , X1 = 3]. You can express your answer using the Gaussian Q function
2 /2
∞1
Q(c) = c √2π e−u du. (Hint: Think! Very little computation is needed.)
7.19 First order diﬀerential equation driven by Gaussian white noise
Let X be the solution of the ordinary diﬀerential equation X = −X + N , with initial condition x0 ,
where N = (Nt : t ≥ 0) is a real valued Gaussian white noise with RN (τ ) = σ 2 δ (τ ) for some constant σ 2 > 0. Although N is not an ordinary random process, we can interpret this as the condition
that N is a Gaussian random process with mean µN = 0 and correlation function RN (τ ) = σ 2 δ (τ ).
(a) Find the mean function µX (t) and covariance function CX (s, t).
(b) Verify that X is a Markov process by checking the necessary and suﬃcient condition: CX (r, s)CX (s, t) =
CX (r, t)CX (s, s) whenever r < s < t. (Note: The very deﬁnition of X also suggests that X is a
Markov process, because if t is the “present time,” the future of X depends only on Xt and the
future of the white noise. The future of the white noise is independent of the past (Xs : s ≤ t).
Thus, the present value Xt contains all the information from the past of X that is relevant to the
future of X . This is the continuoustime analog of the discretetime Kalman state equation.)
(c) Find the limits of µX (t) and RX (t + τ, t) as t → ∞. (Because these limits exist, X is said to be
asymptotically WSS.)
7.20 KL expansion of a simple random process
Let X be a WSS random process with mean zero and autocorrelation function
RX (τ ) = 100(cos(10πτ ))2 = 50 + 50 cos(20πτ ).
(a) Is X mean square diﬀerentiable? (Justify your answer.)
223 (b) Is X mean ergodic in the m.s. sense? (Justify your answer.)
(c) Describe a set of eigenfunctions and corresponding eigenvalues for the KarhunenLo`ve expane
sion of (Xt : 0 ≤ t ≤ 1).
7.21 KL expansion of a ﬁnite rank process
Suppose Z = (Zt : 0 ≤ t ≤ T ) has the form Zt = N=1 Xn ξn (t) such that the functions ξ1 , . . . , ξN
n
are orthonormal over the interval [0, T ], and the vector X = (X1 , ..., XN )T has a correlation matrix
K with det(K ) = 0. The process Z is said to have rank N . Suppose K is not diagonal. Describe
the KarhunenLo`ve expansion of Z . That is, describe an orthornormal basis (φn : n ≥ 1), and
e
eigenvalues for the KL expansion of X , in terms of the given functions (ξn ) and correlation matrix
K . Also, describe how the coordinates (Z, φn ) are related to X .
7.22 KL expansion for derivative process
Suppose that X = (Xt : 0 ≤ t ≤ 1) is a m.s. continuously diﬀerentiable random process on the
interval [0, 1]. Diﬀerentiating the KL expansion of X yields X (t) = n (X, φn )φn (t), which looks
similar to a KL expansion for X , but it may be that the functions φn are not orthonormal. For some
cases it is not diﬃcult to identify the KL expansion for X . To explore this, let (φn (t)), ((X, φn )),
and (λn ) denote the eigenfunctions, coordinate random variables, and eigenvalues, for the KL
expansion of X over the interval [0, 1]. Let (ψk (t)), ((X , ψk )), and (µk ), denote the corresponding
quantities for X . For each of the following choices of (φn (t)), express the eigenfunctions, coordinate
random variables, and eigenvalues, for X in terms of those for X :
(a) φn (t) = e2πjnt , n ∈√
Z
√
(b) φ1 (t) = 1, φ2k (t) = 2 cos(2πkt), and φ2k+1 (t) = 2 sin(2πkt) for k ≥ 1.
√
(c) φn (t) = 2 sin( (2n+1)πt ), n ≥ 0. (Hint: Sketch φn and φn for n = 1, 2, 3.)
√2
√
(d) φ1 (t) = c1 (1 + 3t) and φ2 (t) = c2 (1 − 3t). (Suppose λn = 0 for n ∈ {1, 2}. The constants cn
should be selected so that φn  = 1 for n = 1, 2, but there is no need to calculate the constants for
this problem.)
7.23 An inﬁnitely diﬀerentiable process
2
Let X = (Xt : t ∈ R) be WSS with autocorrelation function RX (τ ) = e−τ /2 . (a) Show that X is
k times diﬀerentiable in the m.s. sense, for all k ≥ 1. (b) Let X (k) denote the k th derivative process
of X, for k ≥ 1. Is X (k) mean ergodic in the m.s. sense for each k ? Justify your answer.
7.24 Mean ergodicity of a periodic WSS random process
Let X be a mean zero periodic WSS random process with period T > 0. Recall that X has a power
spectral representation
Xt =
Xn e2πjnt/T .
n∈Z where the coeﬃcients Xn are orthogonal random variables. The power spectral mass function of X
πn
is the discrete mass function pX supported on frequencies of the form 2T , such that E [Xn 2 ] =
2πn
pX ( T ). Under what conditions on pX is the process X mean ergodic in the m.s. sense? Justify
your answer.
7.25 Application of the KL expansion to estimation
Let X = (Xt : 0 ≤ T ) be a random process given by Xt = AB sin( πt ), where A and T are positive
T
constants and B is a N (0, 1) random variable. Think of X as an amplitude modulated random
224 signal.
(a) What is the expected total energy of X ?
(b) What are the mean and covariance functions of X ?
(c) Describe the KarhunenLo´ve expansion of X . (Hint: Only one eigenvalue is nonzero, call it
e
λ1 . What are λ1 , the corresponding eigenfunction φ1 , and the ﬁrst coordinate X1 = (X, φ1 )? You
don’t need to explicitly identify the other eigenfunctions φ2 , φ3 , . . .. They can simply be taken to
ﬁll out a complete orthonormal basis.)
(d) Let N = (Xt : 0 ≤ T ) be a realvalued Gaussian white noise process independent of X with
RN (τ ) = σ 2 δ (τ ), and let Y = X + N . Think of Y as a noisy observation of X . The same basis
functions used for X can be used for the KarhunenLo`ve expansions of N and Y . Let N1 = (N, φ1 )
e
and Y1 = (Y, φ1 ). Note that Y1 = X1 + N1 . Find E [B Y1 ] and the resulting mean square error.
(Remark: The other coordinates Y2 , Y3 , . . . are independent of both X and Y1 , and are thus useless
for the purpose of estimating B . Thus, E [B Y1 ] is equal to E [B Y ], the MMSE estimate of B given
the entire observation process Y .)
7.26 * An autocorrelation function or not?
Let RX (s, t) = cosh(a(s − t − 0.5)) for −0.5 ≤ s, t ≤ 0.5 where a is a positive constant. Is RX
the autocorrelation function of a random process of the form X = (Xt : −0.5 ≤ t ≤ 0.5)? If not,
explain why not. If so, give the KarhunenLo`ve expansion for X .
e
7.27 * On the conditions for m.s. diﬀerentiability
t2 sin(1/t2 ) t = 0
. Sketch f and show that f is diﬀerentiable over all of R, and
(a) Let f (t) =
0
t=0
1
ﬁnd the derivative function f . Note that f is not continuous, and −1 f (t)dt is not well deﬁned,
whereas this integral would equal f (1) − f (−1) if f were continuous.
(b) Let Xt = Af (t), where A is a random variable with mean zero and variance one. Show that X
is m.s. diﬀerentiable.
(c) Find RX . Show that ∂1 RX and ∂2 ∂1 RX exist but are not continuous. 225 226 Chapter 8 Random Processes in Linear Systems
and Spectral Analysis
Random processes can be passed through linear systems in much the same way as deterministic
signals can. A timeinvariant linear system is described in the time domain by an impulse response
function, and in the frequency domain by the Fourier transform of the impulse response function.
In a sense we shall see that Fourier transforms provide a diagonalization of WSS random processes,
just as the KarhunenLo`ve expansion allows for the diagonalization of a random process deﬁned
e
on a ﬁnite interval. While a m.s. continuous random process on a ﬁnite interval has a ﬁnite average
energy, a WSS random process has a ﬁnite mean average energy per unit time, called the power.
Nearly all the deﬁnitions and results of this chapter can be carried through in either discrete
time or continuous time. The set of frequencies relevant for continuoustime random processes is all
of R, while the set of frequencies relevant for discretetime random processes is the interval [−π, π ].
For ease of notation we shall primarily concentrate on continuoustime processes and systems in
the ﬁrst two sections, and give the corresponding deﬁnition for discrete time in the third section.
Representations of baseband random processes and narrowband random processes are discussed
in Sections 8.4 and 8.5. Roughly speaking, baseband random processes are those which have power
only in low frequencies. A baseband random process can be recovered from samples taken at a
sampling frequency that is at least twice as large as the largest frequency component of the process.
Thus, operations and statistical calculations for a continuoustime baseband process can be reduced
to considerations for the discrete time sampled process. Roughly speaking, narrowband random
processes are those processes which have power only in a band (i.e. interval) of frequencies. A
narrowband random process can be represented as baseband random processes that is modulated
by a deterministic sinusoid. Complex random processes naturally arise as baseband equivalent
processes for realvalued narrowband random processes. A related discussion of complex random
processes is given in the last section of the chapter. 8.1 Basic deﬁnitions The output (Yt : t ∈ R) of a linear system with impulse response function h(s, t) and a random
process input (Xt : t ∈ R) is deﬁned by
∞ Ys = h(s, t)Xt dt
−∞ 227 (8.1) See Figure 8.1. For example, the linear system could be a simple integrator from time zero, deﬁned X Y h Figure 8.1: A linear system with input X , impulse response function h, and output Y.
by s
0 Xt dt Ys = 0 s≥0
s < 0, in which case the impulse response function is
1 s≥t≥0
0 otherwise. h(s, t) = The integral (8.1) deﬁning the output Y will be interpreted in the m.s. sense. Thus, the integral
deﬁning Ys for s ﬁxed exists if and only if the following Riemann integral exists and is ﬁnite:
∞ ∞ −∞ −∞ h∗ (s, τ )h(s, t)RX (t, τ )dtdτ (8.2) A suﬃcient condition for Ys to be well deﬁned is that RX is a bounded continuous function, and
∞
h(s, t) is continuous in t with −∞ h(s, t)dt < ∞. The mean function of the output is given by
∞ ∞ µY (s) = E [ h(s, t)Xt dt] =
−∞ h(s, t)µX (t)dt (8.3) −∞ As illustrated in Figure 8.2, the mean function of the output is the result of passing the mean
function of the input through the linear system. The cross correlation function between the output µX µY h Figure 8.2: A linear system with input µX and impulse response function h.
and input processes is given by
∞ RY X (s, τ ) = E [
−∞
∞ = ∗
h(s, t)Xt dtXτ ] h(s, t)RX (t, τ )dt (8.4) −∞ and the correlation function of the output is given by
∗ ∞ RY (s, u) = E Ys h(u, τ )Xτ dτ
−∞ ∞ = h∗ (u, τ )RY X (s, τ )dτ −∞
∞ ∞ −∞ −∞ h∗ (u, τ )h(s, t)RX (t, τ )dtdτ = 228 (8.5)
(8.6) Recall that Ys is well deﬁned as a m.s. integral if and only if the integral (8.2) is well deﬁned and
ﬁnite. Comparing with (8.6), it means that Ys is well deﬁned if and only if the right side of (8.6)
with u = s is well deﬁned and gives a ﬁnite value for E [Ys 2 ].
The linear system is time invariant if h(s, t) depends on s, t only through s − t. If the system is
time invariant we write h(s − t) instead of h(s, t), and with this substitution the deﬁning relation
(8.1) becomes a convolution: Y = h ∗ X .
A linear system is called bounded input bounded output (bibo) stable if the output is bounded
whenever the input is bounded. In case the system is time invariant, bibo stability is equivalent to
the condition
∞
h(τ )dτ < ∞. (8.7) −∞ In particular, if (8.7) holds and if an input signal x satisﬁes xs  < L for all s, then the output
signal y = x ∗ h satisﬁes
∞ ∞ h(t − s)Lds = L y (t) ≤
−∞ h(τ )dτ
−∞ for all t. If X is a WSS random process then by the Schwarz inequality, RX is bounded by RX (0).
Thus, if X is WSS and m.s. continuous, and if the linear system is timeinvariant and bibo stable,
the integral in (8.2) exists and is bounded by
∞ ∞ ∞ h(τ )dτ )2 < ∞ h(s − τ )h(s − t)dtdτ = RX (0)( RX (0)
−∞ −∞ −∞ Thus, the output of a linear, timeinvariant bibo stable system is well deﬁned in the m.s. sense if
the input is a stationary, m.s. continuous process.
A paragraph about convolutions is in order. It is useful to be able to recognize convolution
integrals in disguise. If f and g are functions on R, the convolution is the function f ∗ g deﬁned by
∞ f ∗ g (t) = f (s)g (t − s)ds
−∞ or equivalently ∞ f ∗ g (t) = f (t − s)g (s)ds
−∞ or equivalently, for any real a and b
∞ f ∗ g (a + b) = f (a + s)g (b − s)ds.
−∞ A simple change of variable shows that the above three expressions are equivalent. However, in
order to immediately recognize a convolution, the salient feature is that the convolution is the
integral of the product of f and g , with the arguments of both f and g ranging over R in such a
way that the sum of the two arguments is held constant. The value of the constant is the value at
which the convolution is being evaluated. Convolution is commutative: f ∗ g = g ∗ f and associative:
(f ∗ g ) ∗ k = f ∗ (g ∗ k ) for three functions f, g, k . We simply write f ∗ g ∗ k for (f ∗ g ) ∗ k . The
convolution f ∗ g ∗ k is equal to a double integral of the product of f ,g , and k , with the arguments
229 of the three functions ranging over all triples in R3 with a constant sum. The value of the constant
is the value at which the convolution is being evaluated. For example,
∞ ∞ −∞ −∞ f (a + s + t)g (b − s)k (c − t)dsdt. f ∗ g ∗ k (a + b + c) = Suppose that X is WSS and that the linear system is time invariant. Then (8.3) becomes
∞ ∞ h(s − t)µX dt = µX µY (s) = h(t)dt
−∞ −∞ Observe that µY (s) does not depend on s. Equation (8.4) becomes
∞ h(s − t)RX (t − τ )dt RY X (s, τ ) =
−∞ = h ∗ RX (s − τ ), (8.8) which in particular means that RY X (s, τ ) is a function of s − τ alone. Equation (8.5) becomes
∞ RY (s, u) = h∗ (u − τ )RY X (s − τ )dτ. (8.9) −∞ The right side of (8.9) looks nearly like a convolution, but as τ varies the sum of the two arguments
is u − τ + s − τ , which is not constant as τ varies. To arrive at a true convolution, deﬁne the new
function h by h(v ) = h∗ (−v ). Using the deﬁnition of h and (8.8) in (8.9) yields
∞ h(τ − u)(h ∗ RX )(s − τ )dτ RY (s, u) =
−∞ = h ∗ (h ∗ RX )(s − u) = h ∗ h ∗ RX (s − u)
which in particular means that RY (s, u) is a function of s − u alone.
To summarize, if X is WSS and if the linear system is time invariant, then X and Y are jointly
WSS with
∞
µY = µ X
h(t)dt
RY X = h ∗ RX
RY = h ∗ h ∗ RX .
(8.10)
−∞ The convolution h ∗ h, equal to h ∗ h, can also be written as
∞ h ∗ h(t) = h(s)h(t − s)ds
−∞
∞ = h(s)h∗ (s − t)ds (8.11) −∞ The expression shows that h ∗ h(t) is the correlation between h and h∗ translated by t from the
origin.
The equations derived in this section for the correlation functions RX , RY X and RY also hold for
the covariance functions CX , CY X , and CY . The derivations are the same except that covariances
rather than correlations are computed. In particular, if X is WSS and the system is linear and
time invariant, then CY X = h ∗ CX and CY = h ∗ h ∗ CX .
230 8.2 Fourier transforms, transfer functions and power spectral densities Fourier transforms convert convolutions into products, so this is a good point to begin using Fourier
transforms. The Fourier transform of a function g mapping R to the complex numbers C is formally
deﬁned by
∞ g (ω ) = e−jωt g (t)dt (8.12) −∞ Some important properties of Fourier transforms are stated next.
Linearity: ag + bh = ag + bh
Inversion: g (t) = ∞ jωt
g (ω ) dω
2π
−∞ e Convolution to multiplication: g ∗ h = g h and g ∗ h = 2π g h
Parseval’s identity: ∞
∗
−∞ g (t)h (t)dt = ∞
dω
∗
−∞ g (ω )h (ω ) 2π Transform of time reversal: h = h∗ , where h(t) = h∗ (−t)
Diﬀerentiation to multiplication by jω : dg
dt (ω ) = (jω )g (ω ) Pure sinusoid to delta function: For ωo ﬁxed: ejωo t (ω ) = 2πδ (ω − ωo )
Delta function to pure sinusoid: For to ﬁxed: δ (t − to )(ω ) = e−jωto
The inversion formula above shows that a function g can be represented as an integral (basically
a limiting form of linear combination) of sinusoidal functions of time ejωt , and g (ω ) is the coeﬃcient
in the representation for each ω . Paresval’s identity applied with g = h yields that the total
energy of g (the square of the L2 norm) can be computed in either the time or frequency domain:
∞
∞
g 2 = −∞ g (t)2 dt = −∞ g (ω )2 dω . The factor 2π in the formulas can be attributed to the use
2π
of frequency ω in radians. If ω = 2πf , then f is the frequency in Hertz (Hz) and dω is simply df .
2π
The Fourier transform can be deﬁned for a very large class of functions, including generalized
functions such as delta functions. In these notes we won’t attempt a systematic treatment, but will
use Fourier transforms with impunity. In applications, one is often forced to determine in what
senses the transform is well deﬁned on a casebycase basis. Two suﬃcient conditions for the Fourier
transform of g to be well deﬁned are mentioned in the remainder of this paragraph. The relation
(8.12) deﬁning a Fourier transform of g is well deﬁned if, for example, g is a continuous function
∞
which is integrable: −∞ g (t)dt < ∞, and in this case the dominated convergence theorem implies
that g is a continuous function. The Fourier transform can also be naturally deﬁned whenever g
has a ﬁnite L2 norm, through the use of Parseval’s identity. The idea is that if g has ﬁnite L2 norm,
then it is the limit in the L2 norm of a sequence of functions gn which are integrable. Owing to
Parseval’s identity, the Fourier transforms gn form a Cauchy sequence in the L2 norm, and hence
have a limit, which is deﬁned to be g .
Return now to consideration of a linear timeinvariant system with an impulse response function
h = (h(τ ) : τ ∈ R). The Fourier transform of h is used so often that a special name and notation
is used: it is called the transfer function and is denoted by H (ω ).
231 The output signal y = (yt : t ∈ R) for an input signal x = (xt : t ∈ R) is given in the time
domain by the convolution y = x ∗ h. In the frequency domain this becomes y (ω ) = H (ω )x(ω ).
For example, given a < b let H[a,b] (ω ) be the ideal bandpass transfer function for frequency band
[a, b], deﬁned by
1 a≤ω≤b
H[a,b] (ω ) =
(8.13)
0 otherwise.
If x is the input and y is the output of a linear system with transfer function H[a,b] , then the
relation y (ω ) = H[a,b] (ω )x(ω ) shows that the frequency components of x in the frequency band
[a, b] pass through the ﬁlter unchanged, and the frequency components of x outside of the band are
completely nulled. The total energy of the output function y can therefore be interpreted as the
energy of x in the frequency band [a, b]. Therefore,
∞ H[a,b] (ω )2 x(ω )2 Energy of x in frequency interval [a, b] = y 2 =
−∞ b dω
=
2π x(ω )2
a dω
.
2π Consequently, it is appropriate to call x(ω )2 the energy spectral density of the deterministic signal
x.
Given a WSS random process X = (Xt : t ∈ R), the Fourier transform of its correlation function
RX is denoted by SX . For reasons that we will soon see, the function SX is called the power spectral
density of X . Similarly, if Y and X are jointly WSS, then the Fourier transform of RY X is denoted
by SY X , called the cross power spectral density function of Y and X . The Fourier transform of
the time reverse complex conjugate function h is equal to H ∗ , so H (ω )2 is the Fourier transform
of h ∗ h. With the above notation, the second moment relationships in (8.10) become:
SY X (ω ) = H (ω )SX (ω ) SY (ω ) = H (ω )2 SX (ω )
∞ Let us examine some of the properties of the power spectral density, SX . If −∞ RX (t)dt < ∞
then SX is well deﬁned and is a continuous function. Because RY X = RXY , it follows that
∗
∗
SY X = SXY . In particular, taking Y = X yields RX = RX and SX = SX , meaning that SX is
realvalued.
∞
The Fourier inversion formula applied to SX yields that RX (τ ) = −∞ ejωτ SX (ω ) dω . In partic2π
ular,
∞
dω
E [Xt 2 ] = RX (0) =
SX (ω ) .
(8.14)
2π
−∞
The expectation E [Xt 2 ] is called the power (or total power) of X , because if Xt is a voltage or
current accross a resistor, Xt 2 is the instantaneous rate of dissipation of heat energy. Therefore,
(8.14) means that the total power of X is the integral of SX over R. This is the ﬁrst hint that the
name power spectral density for SX is justiﬁed.
Let a < b and let Y denote the output when the WSS process X is passed through the linear
timeinvariant system with transfer function H[a,b] deﬁned by (8.13). The process Y represents the
part of X in the frequency band [a, b]. By the relation SY = H[a,b] 2 SX and the power relationship
(8.14) applied to Y , we have
∞ Power of X in frequency interval [a, b] = E [Yt 2 ] = SY (ω )
−∞ 232 dω
=
2π b SX (ω )
a dω
2π (8.15) Two observations can be made concerning (8.15). First, the integral of SX over any interval [a, b]
is nonnegative. If SX is continuous, this implies that SX is nonnegative. Even if SX is not continuous, we can conclude that SX is nonnegative except possibly on a set of zero measure. The
second observation is that (8.15) fully justiﬁes the name “power spectral density of X ” given to SX . Example 8.2.1 Suppose X is a WSS process and that Y is a moving average of X with averaging
window duration T for some T > 0:
Yt = t 1
T Xs ds
t−T Equivalently, Y is the output of the linear timeinvariant system with input X and impulse response
function h given by
1
0≤τ ≤T
T
h(τ ) =
0 else
The output correlation function is given by RY = h ∗ h ∗ RX . Using (8.11) and referring to Figure
8.3 we ﬁnd that h ∗ h is a triangular shaped waveform:
h ∗ h(τ ) = τ 
1
(1 −
)+
T
T Similarly, CY = h ∗ h ∗ CX . Let’s ﬁnd in particular an expression for the variance of Yt in terms
h(s) h(s−t) 1
T ~
h*h s
t 0 T −T T Figure 8.3: Convolution of two rectangle functions.
of the function CX .
∞ (h ∗ h)(0 − τ )CX (τ )dτ Var(Yt ) = CY (0) =
−∞ = T 1
T (1 −
−T τ 
)CX (τ )dτ
T (8.16) The expression in (8.16) arose earlier in these notes, in the section on mean ergodicity.
Let’s see the eﬀect of the linear system on the power spectral density of the input. Observe
that
∞ H (ω ) = e−jωt h(t)dt = −∞ = 2e−jωT /2
Tω = e−jωT /2 1
T e−jωT − 1
−jω ejωT /2 − e−jωT /2
2j
sin( ωT )
2
ωT
2 233 Equivalently, using the substitution ω = 2πf ,
H (2πf ) = e−jπf T sinc(f T )
where in these notes the sinc function is deﬁned by
sinc(u) = sin(πu)
πu u=0
u = 0. 1 (8.17) (Some authors use somewhat diﬀerent deﬁnitions for the sinc function.) Therefore H (2πf )2 =
sinc(f T )2 , so that the output power spectral density is given by SY (2πf ) = SX (2πf )sinc(f T )2 .
See Figure 8.4.
H(2π f) sinc( u) f u
0 1 2 0 1
T 2
T Figure 8.4: The sinc function and the impulse response function. Example 8.2.2 Consider two linear timeinvariant systems in parallel as shown in Figure 8.5. The X
Y h
k U
V Figure 8.5: Parallel linear systems.
ﬁrst has input X , impulse response function h, and output U . The second has input Y , impulse
response function k , and output V . Suppose that X and Y are jointly WSS. We can ﬁnd RU V as
follows. The main trick is notational: to use enough diﬀerent variables of integration so that none
are used twice.
∞ ∞ −∞
∞ =
−∞
∞ = ∗ ∞ h(t − s)Xs ds RU V (t, τ ) = E k (τ − v )Yv dv
−∞ h(t − s)RXY (s − v )k ∗ (τ − v )dsdv −∞ {h ∗ RXY (t − v )} k ∗ (τ − v )dv −∞ = h ∗ k ∗ RXY (t − τ ).
Note that RU V (t, τ ) is a function of t − τ alone. Together with the fact that U and V are individually
WSS, this implies that U and V are jointly WSS, and RU V = h ∗ k ∗ RXY . The relationship is
234 expressed in the frequency domain as SU V = HK ∗ SXY , where K is the Fourier transform of k .
Special cases of this example include the case that X = Y or h = k . Example 8.2.3 Consider the circuit with a resistor and a capacitor shown in Figure 8.6. Take as R
+ C x(t) q (t) +
− y(t) − Figure 8.6: An RC circuit modeled as a linear system.
the input signal the voltage diﬀerence on the left side, and as the output signal the voltage across
the capacitor. Also, let qt denote the charge on the upper side of the capacitor. Let us ﬁrst identify
the impulse response function by assuming a deterministic input x and a corresponding output y .
The elementary equations for resistors and capacitors yield
dq
1
= (xt − yt )
dt
R
Therefore and yt = qt
C dy
1
=
(xt − yt )
dt
RC which in the frequency domain is
jω y (ω ) = 1
(x(ω ) − y (ω ))
RC so that y = H x for the system transfer function H given by
H (ω ) = 1
1 + RCjω Suppose, for example, that the input X is a realvalued, stationary Gaussian Markov process, so
that its autocorrelation function has the form RX (τ ) = A2 e−ατ  for some constants A2 and α > 0.
Then
2 A2 α
SX (ω ) = 2
ω + α2
and
2 A2 α
SY (ω ) = SX (ω )H (ω )2 = 2
(ω + α2 )(1 + (RCω )2 ) Example 8.2.4 A random signal, modeled by the input random process X , is passed into a linear
timeinvariant system with feedback and with noise modeled by the random process N , as shown
235 Nt
Xt + H (!)
1 + Yt H2(!) H3(!) Figure 8.7: A feedback system.
in Figure 8.7. The output is denoted by Y . Assume that X and N are jointly WSS and that the
random variables comprising X are orthogonal to the random variables comprising N : RXN = 0.
Assume also, for the sake of system stability, that the magnitude of the gain around the loop
satisﬁes H3 (ω )H1 (ω )H2 (ω ) < 1 for all ω such that SX (ω ) > 0 or SN (ω ) > 0. We shall express
the output power spectral density SY in terms the power spectral densities of X and N , and the
three transfer functions H1 , H2 , and H3 . An expression for the signaltonoise power ratio at the
output will also be computed.
Under the assumed stability condition, the linear system can be written in the equivalent form
˜
˜
shown in Figure 8.8. The process X is the output due to the input signal X , and N is the output Xt H2(!)H1(!) ~ Xt 1!H (!)H ( )H2(!)
3
1! ~~~ +
Nt H2(!)
!
1!H (!)H ( )H2(!)
3
1 Yt =Xt+Nt ~ Nt Figure 8.8: An equivalent representation.
due to the input noise N . The structure in Figure 8.8 is the same as considered in Example 8.2.2.
Since RXN = 0 it follows that RX N = 0, so that SY = SX + SN . Consequently,
˜˜
˜
˜
SY (ω ) = SX (ω ) + SN (ω ) =
˜
˜ H2 (ω )2  H1 (ω )2 SX (ω ) + SN (ω )
1 − H3 (ω )H1 (ω )H2 (ω )2 The output signaltonoise ratio is the ratio of the power of the signal at the output to the power
of the noise at the output. For this example it is given by
˜
E [Xt 2 ]
=
˜
E [Nt 2 ] ∞
H2 (ω )H1 (ω )2 SX (ω )
−∞ 1−H3 (ω )H1 (ω )H2 (ω )2
∞
H2 (ω )2 SN (ω )
−∞ 1−H3 (ω )H1 (ω )H2 (ω )2 236 dω
2π
dω
2π Example 8.2.5 Consider the linear timeinvariant system deﬁned as follows. For input signal x
the output signal y is deﬁned by y + y + y = x + x . We seek to ﬁnd the power spectral density of
the output process if the input is a white noise process X with RX (τ ) = σ 2 δ (τ ) and SX (ω ) = σ 2
for all ω . To begin, we identify the transfer function of the system. In the frequency domain, the
system is described by ((jω )3 + jω + 1)y (ω ) = (1 + jω )x(ω ), so that
H (ω ) = 1 + jω
1 + jω
=
3
1 + jω + (jω )
1 + j (ω − ω 3 ) Hence,
SY (ω ) = SX (ω )H (ω )2 = σ 2 (1 + ω 2 )
σ 2 (1 + ω 2 )
=
.
1 + ( ω − ω 3 )2
1 + ω 2 − 2ω 4 + ω 6 Observe that ∞ output power = SY (ω )
−∞ 8.3 dω
< ∞.
2π Discretetime processes in linear systems The basic deﬁnitions and use of Fourier transforms described above carry over naturally to discrete
time. In particular, if the random process X = (Xk : k ∈ Z) is the input of a linear, discretetime
system with impule response function h, then the output Y is the random process given by
∞ h(k, n)Xn . Yk =
n=−∞ The equations in Section 8.1 can be modiﬁed to hold for discrete time simply by replacing integration
over R by summation over Z. In particular, if X is WSS and if the linear system is timeinvariant
then (8.10) becomes
∞ µ Y = µX h(n) RY X = h ∗ RX RY = h ∗ h ∗ RX , (8.18) n=−∞ where the convolution in (8.18) is deﬁned for functions g and h on Z by
∞ g ∗ h(n) = g (n − k )h(k )
k=−∞ Again, Fourier transforms can be used to convert convolution to multiplication. The Fourier transform of a function g = (g (n) : n ∈ Z) is the function g on [−π, π ] deﬁned by
∞ e−jωn g (n). g (ω ) =
−∞ Some of the most basic properties are:
237 Linearity: ag + bh = ag + bh
Inversion: g (n) = π
jωn g (ω ) dω
2π
−π e Convolution to multiplication: g ∗ h = g h and g ∗ h =
Parseval’s identity: ∞
∗
n=−∞ g (n)h (n) = π
−π 1
2π g h g (ω )h∗ (ω ) dω
2π Transform of time reversal: h = h∗ , where h(t) = h(−t)∗
Pure sinusoid to delta function: For ωo ∈ [−π, π ] ﬁxed: ejωo n (ω ) = 2πδ (ω − ωo )
Delta function to pure sinusoid: For no ﬁxed: I{n=no } (ω ) = e−jωno
The inversion formula above shows that a function g on Z can be represented as an integral
(basically a limiting form of linear combination) of sinusoidal functions of time ejωn , and g (ω ) is
the coeﬃcient in the representation for each ω . Paresval’s identity applied with g = h yields that
the total energy of g (the square of the L2 norm) can be computed in either the time or frequency
π
domain: g 2 = ∞ −∞ g (n)2 = −π g (ω )2 dω .
n=
2π
The Fourier transform and its inversion formula for discretetime functions are equivalent to
the Fourier series representation of functions in L2 [−π, π ] using the complete orthogonal basis
(ejωn : n ∈ Z) for L2 [−π, π ], as discussed in connection with the KarhunenLo`ve expansion. The
e
functions in this basis all have norm 2π . Recall that when we considered the KarhunenLo`ve
e
expansion for a periodic WSS random process of period T , functions on a time interval were
1
important and the power was distributed on the integers Z scaled by T . In this section, Z is
considered to be the time domain and the power is distributed over an interval. That is, the role
of Z and a ﬁnite interval are interchanged. The transforms used are essentially the same, but with
j replaced by −j .
Given a linear timeinvariant system in discrete time with an impulse response function h =
(h(τ ) : τ ∈ Z), the Fourier transform of h is denoted by H (ω ). The deﬁning relation for the
system in the time domain, y = h ∗ x, becomes y (ω ) = H (ω )x(ω ) in the frequency domain. For
−π ≤ a < b ≤ π ,
b
dω
Energy of x in frequency interval [a, b] =
x(ω )2 .
2π
a
so it is appropriate to call x(ω )2 the energy spectral density of the deterministic, discretetime
signal x.
Given a WSS random process X = (Xn : n ∈ Z), the Fourier transform of its correlation
function RX is denoted by SX , and is called the power spectral density of X . Similarly, if Y and
X are jointly WSS, then the Fourier transform of RY X is denoted by SY X , called the cross power
spectral density function of Y and X . With the above notation, the second moment relationships
in (8.18) become:
SY X (ω ) = H (ω )SX (ω )
SY (ω ) = H (ω )2 SX (ω )
The Fourier inversion formula applied to SX yields that RX (n) =
ular,
π
dω
E [Xn 2 ] = RX (0) =
SX (ω ) .
2π
−π
238 π
jωn S (ω ) dω .
X
2π
−π e In partic The expectation E [Xn 2 ] is called the power (or total power) of X , and for −π < a < b ≤ π we
have
b
dω
SX (ω )
Power of X in frequency interval [a, b] =
2π
a 8.4 Baseband random processes Deterministic baseband signals are considered ﬁrst. Let x be a continuoustime signal (i.e. a
∞
function on R) such that its energy, −∞ x(t)2 dt, is ﬁnite. By the Fourier inversion formula, the
signal x is an integral, which is essentially a sum, of sinusoidal functions of time, ejωt . The weights
are given by the Fourier transform x(w). Let fo > 0 and let ωo = 2πfo . The signal x is called a
baseband signal, with onesided band limit fo Hz , or equivalently ωo radians/second, if x(ω ) = 0
for ω  ≥ ωo . For such a signal, the Fourier inversion formula becomes
ωo x(t) = ejωt x(ω ) −ωo dω
2π (8.19) Equation (8.19) displays the baseband signal x as a linear combination of the functions ejωt indexed
by ω ∈ [−ωo , ωo ].
A celebrated theorem of Nyquist states that the baseband signal x is completely determined by
1
its samples taken at sampling frequency 2fo . Speciﬁcally, deﬁne T by T = 2fo . Then
∞ x(t) = x(nT ) sinc
n=−∞ t − nT
T . (8.20) where the sinc function is deﬁned by (8.17). Nyquist’s equation (8.20) is indeed elegant. It obviously
holds by inspection if t = mT for some integer m, because for t = mT the only nonzero term in
the sum is the one indexed by n = m. The equation shows that the sinc function gives the correct
interpolation of the narrowband signal x for times in between the integer multiples of T . We shall
give a proof of (8.20) for deterministic signals, before considering its extension to random processes.
A proof of (8.20) goes as follows. Henceforth we will use ωo more often than fo , so it is worth
remembering that ωo T = π . Taking t = nT in (8.19) yields
ωo x(nT ) =
−ωo
ωo = ejωnT x(ω ) dω
2π x(ω )(e−jωnT )∗ −ωo dω
2π (8.21) Equation (8.21) shows that x(nT ) is given by an inner product of x and e−jωnT . The functions
ˆ
e−jωnT , considered on the interval −ωo < ω < ωo and indexed by n ∈ Z, form a complete orthogonal
ωo
basis for L2 [−ωo , ωo ], and −ωo T e−jωnT 2 dω = 1. Therefore, x over the interval [−ωo , ωo ] has the
2π
following Fourier series representation:
∞ e−jωnT x(nT ) x(ω ) = T
n=−∞ 239 ω ∈ [−ωo , ωo ] (8.22) Plugging (8.22) into (8.19) yields
∞ ωo x(t) = x(nT )T ejωt e−jωnT −ωo n=−∞ dω
.
2π (8.23) The integral in (8.23) can be simpliﬁed using
ωo T ejωτ −ωo τ
dω
= sinc
2π
T . (8.24) with τ = t − nT to yield (8.20) as desired.
The sampling theorem extends naturally to WSS random processes. A WSS random process
X with spectral density SX is said to be a baseband random process with onesided band limit ωo
if SX (ω ) = 0 for  ω ≥ ωo .
Proposition 8.4.1 Suppose X is a WSS baseband random process with onesided band limit ωo
and let T = π/ωo . Then for each t ∈ R
∞ Xt = XnT sinc
n=−∞ t − nT
T m.s. (8.25) If B is the process of samples deﬁned by Bn = XnT , then the power spectral densities of B and X
are related by
ω
1
SX
T
T SB (ω ) = Proof. Fix t ∈ R. It must be shown that
zero as N → ∞: εN N for  ω ≤ π (8.26) deﬁned by the following expectation converges to N = E Xt − XnT sinc
n=−N 2 t − nT
t ∗
When the square is expanded, terms of the form E [Xa Xb ] arise, where a and b take on the values
t or nT for some n. But
∗
E [Xa Xb ] = RX (a − b) = ∞ ejωa (ejωb )∗ SX (ω ) −∞ dω
.
2π Therefore, εN can be expressed as an integration over ω rather than as an expectation:
N ∞ εN jωt e = − −∞ e jωnT sinc n=−N t − nT
T 2 SX (ω ) dω
.
2π (8.27) For t ﬁxed, the function (ejωt : −ωo < ω < ωo ) has a Fourier series representation (use (8.24))
∞ ejωt = T
= ωo ejωnT −∞
∞
jωnT e ejωt e−jωnT −ωo sinc −∞ 240 t − nT
T . dω
2π so that the quantity inside the absolute value signs in (8.27) is the approximation error for the
N th partial Fourier series sum for ejωt . Since ejωt is continuous in ω , a basic result in the theory
of Fourier series yields that the Fourier approximation error is bounded by a single constant for
all N and ω , and as N → ∞ the Fourier approximation error converges to 0 uniformly on sets of
the form  ω ≤ ωo − ε. Thus εN → 0 as N → ∞ by the dominated convergence theorem. The
representation (8.25) is proved.
Clearly B is a WSS discrete time random process with µB = µX and
∞ dω
2π
−∞
ωo
dω
ejnT ω SX (ω ) ,
2π
−ωo
ejnT ω SX (ω ) RB (n) = RX (nT ) =
= so, using a change of variable ν = T ω and the fact T =
π ejnν RB (n) =
−π π
ωo yields ν dν
1
SX ( ) .
T
T 2π (8.28) But SB (ω ) is the unique function on [−π, π ] such that
π ejnω SB (ω ) RB (n) =
−π dω
2π so (8.26) holds. The proof of Proposition 8.4.1 is complete.
As a check on (8.26), we note that B (0) = X (0), so the processes have the same total power.
Thus, it must be that
π SB (ω )
−π dω
2π ∞ = SX (ω )
−∞ dω
,
2π (8.29) which is indeed consistent with (8.26).
Example 8.4.2 If µX = 0 and the spectral density SX of X is constant over the interval [−ωo , ωo ],
then µB = 0 and SB (ω ) is constant over the interval [−π, π ]. Therefore RB (n) = CB (n) = 0 for
n = 0, and the samples (B (n)) are mean zero, uncorrelated random variables. Theoretical Exercise What does (8.26) become if X is W SS and has a power spectral density,
but X is not a baseband signal? 8.5 Narrowband random processes As noted in the previous section, a signal – modeled as either a deterministic ﬁnite energy signal
or a WSS random process – can be reconstructed from samples taken at a sampling rate twice the
highest frequency of the signal. For example, a typical voice signal may have highest frequency 5
KHz. If such a signal is multiplied by a signal with frequency 109 Hz, the highest frequency of the
resulting product is about 200,000 times larger than that of the original signal. Na¨ application
ıve
241 of the sampling theorem would mean that the sampling rate would have to increase by the same
factor. Fortunately, because the energy or power of such a modulated signal is concentrated in a
narrow band, the signal is nearly as simple as the original baseband signal. The motivation of this
section is to see how signals and random processes with narrow spectral ranges can be analyzed in
terms of equivalent baseband signals. For example, the eﬀects of ﬁltering can be analyzed using
baseband equivalent ﬁlters. As an application, an example at the end of the section is given which
describes how a narrowband random process (to be deﬁned) can be simulated using a sampling rate
equal to twice the onesided width of a frequency band of a signal, rather than twice the highest
frequency of the signal.
Deterministic narrowband signals are considered ﬁrst, and the development for random processes follows a similar approach. Let ωc > ωo > 0. A narrowband signal (relative to ωo and
ωc ) is a signal x such that x(ω ) = 0 unless ω is in the union of two intervals: the upper band,
(ωc − ωo , ωc + ωo ), and the lower band, (−ωc − ωo , −ωc + ωo ). More compactly, x(ω ) = 0 if
 ω  −ωc  ≥ ωo .
A narrowband signal arises when a sinusoidal signal is modulated by a narrowband signal, as
shown next. Let u and v be realvalued baseband signals, each with onesided bandwidth less than
ωo , as deﬁned at the beginning of the previous section. Deﬁne a signal x by
x(t) = u(t) cos(ωc t) − v (t) sin(ωc t). (8.30) Since cos(ωc t) = (ejωc t + e−jωc t )/2 and − sin(ωc t) = (jejωc t − je−jωc t )/2, (8.30) becomes
x(ω ) = 1
{u(ω − ωc ) + u(ω + ωc ) + j v (ω − ωc ) − j v (ω + ωc )}
2 (8.31) j
1
Graphically, x is obtained by sliding 2 u to the right by ωc , 1 u to the left by ωc , 2 v to the right by
2
−j
ωc , and 2 v to the left by ωc , and then adding. Of course x is realvalued by its deﬁnition. The
reader is encouraged to verify from (8.31) that x(ω ) = x∗ (−ω ). Equation (8.31) shows that indeed
x is a narrowband signal.
A convenient alternative expression for x is obtained by deﬁning a complex valued baseband
signal z by z (t) = u(t) + jv (t). Then x(t) = Re(z (t)ejωc t ). It is a good idea to keep in mind the
case that ωc is much larger than ωo (written ωc
ωo ). Then z varies slowly compared to the
complex sinusoid ejωc t . In a small neighborhood of a ﬁxed time t, x is approximately a sinusoid
with frequency ωc , peak amplitude z (t), and phase given by the argument of z (t). The signal z is
called the complex envelope of x and z (t) is called the real envelope of x.
So far we have shown that a realvalued narrowband signal x results from modulating sinusoidal
functions by a pair of realvalued baseband signals, or equivalently, modulating a complex sinusoidal
function by a complexvalued baseband signal. Does every realvalued narrowband signal have such
a representation? The answer is yes, as we now show. Let x be a realvalued narrowband signal
with ﬁnite energy. One attempt to obtain a baseband signal from x is to consider e−jωc t x(t). This
has Fourier transform x(ω + ωc ), and the graph of this transform is obtained by sliding the graph
of x(ω ) to the left by ωc . As desired, that shifts the portion of x in the upper band to the baseband
interval (−ωo , ωo ). However, the portion of x in the lower band gets shifted to an interval centered
about −2ωc , so that e−jωc t x(t) is not a baseband signal.
An elegant solution to this problem is to use the Hilbert transform of x, denoted by x. By
ˇ 242 deﬁnition, x(ω ) is the signal with Fourier transform −j sgn(ω )x(ω ), where
ˇ 1 ω>0
0 ω=0
sgn(ω ) = −1 ω < 0
Therefore x can be viewed as the result of passing x through a linear, timeinvariant system with
ˇ
transfer function −j sgn(ω ) as pictured in Figure 8.9. Since this transfer function satisﬁes H ∗ (ω ) =
H (−ω ), the output signal x is again realvalued. In addition, H (ω ) = 1 for all ω , except ω = 0, so
ˇ x −j sgn(ω) x Figure 8.9: The Hilbert transform as a linear, timeinvariant system.
that the Fourier transforms of x and x have the same magnitude for all nonzero ω . In particular,
ˇ
x and x have equal energies.
ˇ
Consider the Fourier transform of x + j x. It is equal to 2x(ω ) in the upper band and it is zero
ˇ
elsewhere. Thus, z deﬁned by z (t) = (x(t) + j x(t))e−jωc t is a baseband complex valued signal. Note
ˇ
that x(t) = Re(x(t)) = Re(x(t) + j x(t)), or equivalently
ˇ
x(t) = Re z (t)ejωc t (8.32) If we let u(t) = Re(z (t)) and v (t) = Im(z (t)), then u and v are realvalued baseband signals such
that z (t) = u(t) + jv (t), and (8.32) becomes (8.30).
In summary, any ﬁnite energy realvalued narrowband signal x can be represented as (8.30) or
(8.32), where z (t) = u(t) + jv (t). The Fourier transform z can be expressed in terms of x by
z (ω ) = 2x(ω + ωc ) ω  ≤ ωo
0
else, (8.33) and u is the Hermetian symmetric part of z and v is −j times the Hermetian antisymmetric part
of z :
u(ω ) = 1
(z (ω ) + z ∗ (−ω ))
2 v (ω ) = −j
(z (ω ) − z ∗ (−ω ))
2 In the other direction, x can be expressed in terms of u and v by (8.31).
If x1 and x2 are each narrowband signals with corresponding complex envelope processes z1 and
z2 , then the convolution x = x1 ∗ x2 is again a narrowband signal, and the corresponding complex
1
envelope is 2 z1 ∗ z2 . To see this, note that the Fourier transform, z , of the complex envelope z for
x is given by (8.33). Similar equations hold for zi in terms of xi for i = 1, 2. Using these equations
1
and the fact x(ω ) = x1 (ω )x2 (ω ), it is readily seen that z (ω ) = 2 z1 (ω )z2 (ω ) for all ω , establishing
the claim. Thus, the analysis of linear, time invariant ﬁltering of narrowband signals can be carried
out in the baseband equivalent setting.
A similar development is considered next for WSS random processes. Let U and V be jointly
WSS realvalued baseband random processes, and let X be deﬁned by
Xt = Ut cos(ωc t) − Vt sin(ωc t)
243 (8.34) or equivalently, deﬁning Zt by Zt = Ut + jVt ,
Xt = Re Zt ejωc t (8.35) In some sort of generalized sense, we expect that X is a narrowband process. However, such
an X need not even be WSS. Let us ﬁnd the conditions on U and V that make X WSS. First, in
order that µX (t) not depend on t, it must be that µU = µV = 0.
Using the notation ct = cos(ωc t), st = sin(ωc t), and τ = a − b,
RX (a, b) = RU (τ )ca cb − RU V (τ )ca sb − RV U (τ )sa cb + RV (τ )sa sb .
Using the trigonometric identities such as ca cb = (ca−b + ca+b )/2, this can be rewritten as
RX (a, b) =
+ RU (τ ) + RV (τ )
2
RU (τ ) − RV (τ )
2 ca−b +
ca+b − RU V (τ ) − RV U (τ )
2
RU V (τ ) + RV U (τ )
2 sa−b
sa+b . Therefore, in order that RX (a, b) is a function of a − b, it must be that RU = RV and RU V = −RV U .
Since in general RU V (τ ) = RV U (−τ ), the condition RU V = −RV U means that RU V is an odd
function: RU V (τ ) = −RU V (−τ ).
We summarize the results as a proposition.
Proposition 8.5.1 Suppose X is given by (8.34) or (8.35), where U and V are jointly WSS. Then
X is WSS if and only if U and V are mean zero with RU = RV and RU V = −RV U . Equivalently,
X is WSS if and only if Z = U + jV is mean zero and E [Za Zb ] = 0 for all a, b. If X is WSS then
RX (τ ) = RU (τ ) cos(ωc τ ) + RU V (τ ) sin(ωc τ )
1
SX (ω ) =
[SU (ω − ωc ) + SU (ω + ωc ) − jSU V (ω − ωc ) + jSU V (ω + ωc )]
2
∗
and, with RZ (τ ) deﬁned by RZ (a − b) = E [Za Zb ], RX (τ ) = 1
Re(RZ (τ )ejωc τ )
2 The functions SX , SU , and SV are nonnegative, even functions, and SU V is a purely imaginary
odd function (i.e. SU V (ω ) = Im(SU V (ω )) = −SU V (−ω ).)
Let X by any WSS realvalued random process with a spectral density SX , and continue to
let ωc > ωo > 0. Then X is deﬁned to be a narrowband random process if SX (ω ) = 0 whenever
 ω  − ωc ≥ ωo . Equivalently, X is a narrowband random process if RX (t) is a narrowband
function. We’ve seen how such a process can be obtained by modulating a pair of jointly WSS
baseband random processes U and V . We show next that all narrowband random processes have
such a representation.
To proceed as in the case of deterministic signals, we ﬁrst wish to deﬁne the Hilbert transform
ˇ
ˇ
of X , denoted by X . A slight concern about deﬁning X is that the function −j sgn(ω ) does not
have ﬁnite energy. However, we can replace this function by the function given by
H (ω ) = −j sgn(ω )Iω≤ωo +ωc ,
244 ˇ
which has ﬁnite energy and it has a realvalued inverse transform h. Deﬁne X as the output when
X is passed through the linear system with impulse response h. Since X and h are real valued, the
ˇ
random process X is also real valued. As in the deterministic case, deﬁne random processes Z , U ,
ˇ
and V by Zt = (Xt + j Xt )e−jωc t , Ut = Re(Zt ), and Vt = Im(Zt ).
Proposition 8.5.2 Let X be a narrowband WSS random process, with spectral density SX satisfying SX (ω ) = 0 unless ωc − ωo ≤ ω  ≤ ωc + ωo , where ωo < ωc . Then µX = 0 and the following
representations hold
Xt = Re(Zt ejωc t ) = Ut cos(ωc t) − Vt sin(ωc t)
where Zt = Ut + jVt , and U and V are jointly WSS realvalued random processes with mean zero
and
SU (ω ) = SV (ω ) = [SX (ω − ωc ) + SX (ω + ωc )] Iω≤ωo
(8.36)
and
SU V (ω ) = j [SX (ω + ωc ) − SX (ω − ωc )] Iω≤ωo (8.37) ˇ
RU (τ ) = RV (τ ) = RX (τ ) cos(ωc τ ) + RX (τ ) sin(ωc τ ) (8.38) ˇ
RU V (τ ) = RX (τ ) sin(ωc τ ) − RX (τ ) cos(ωc τ ) (8.39) Equivalently, and .
Proof To show that µX = 0, consider passing X through a linear, timeinvariant system with
transfer function K (ω ) = 1 if ω is in either the upper band or lower band, and K (ω ) = 0 otherwise.
∞
Then µY = µX −∞ h(τ )dτ = µX K (0) = 0. Since K (ω ) = 1 for all ω such that SX (ω ) > 0, it
follows that RX = RY = RXY = RY X . Therefore E [Xt − Yt 2 ] = 0 so that Xt has the same mean
as Yt , namely zero, as claimed.
By the deﬁnitions of the processes Z , U , and V , using the notation ct = cos(ωc t) and st =
sin(ωc t), we have
ˇ
Ut = Xt ct + Xt st ˇ
Vt = −Xt st + Xt ct The remainder of the proof consists of computing RU , RV , and RU V as functions of two variables,
because it is not yet clear that U and V are jointly WSS.
ˇ
ˇ
By the fact X is WSS and the deﬁnition of X , the processes X and X are jointly WSS, and
the various spectral densities are given by
SXX = HSX
ˇ SX X = H ∗ SX = −HSX
ˇ SX = H 2 SX = SX
ˇ Therefore,
ˇ
RXX = RX
ˇ ˇ
RX X = −RX
ˇ
245 RX = RX
ˇ Thus, for real numbers a and b,
ˇ
X (a)ca + X (a)sa ˇ
X (b)cb + X (b)sb
ˇ
= RX (a − b)(ca cb + sa sb ) + RX (a − b)(sa cb − ca sb )
ˇ
= RX (a − b)ca−b + RX (a − b)sa−b RU (a, b) = E Thus, RU (a, b) is a function of a − b, and RU (τ ) is given by the right side of (8.38). The proof that
RV also satisﬁes (8.38), and the proof of (8.39) are similar. Finally, it is a simple matter to derive
(8.36) and (8.37) from (8.38) and (8.39), respectively. 2
Equations (8.36) and (8.37) have simple graphical interpretations, as illustrated in Figure 8.10.
Equation (8.36) means that SU and SV are each equal to the sum of the upper lobe of SX shifted SX SU=SV
+ = SUV
+ j = j j Figure 8.10: A narrowband power spectral density and associated baseband spectral densities.
to the left by ωc and the lower lobe of SX shifted to the right by ωc . Similarly, equation (8.36)
means that SU V is equal to the sum of j times the upper lobe of SX shifted to the left by ωc and
−j times the lower lobe of SX shifted to the right by ωc . Equivalently, SU and SV are each twice
the symmetric part of the upper lobe of SX , and SU V is j times the antisymmetric part of the
upper lobe of SX . Since RU V is an odd function of τ , if follows that RU V (0) = 0. Thus, for any
ﬁxed time t, Ut and Vt are uncorrelated. That does not imply that Us and Vt are uncorrelated for
all s and t, for the cross correlation function RXY is identically zero if and only if the upper lobe
of SX is symmetric about ωc .
Example 8.5.3 ( Baseband equivalent ﬁltering of a random process) As noted above, ﬁltering of
narrowband deterministic signals can be described using equivalent baseband signals, namely the
complex envelopes. The same is true for ﬁltering of narrowband random processes. Suppose X is
a narrowband WSS random process, suppose g is a ﬁnite energy narrowband signal, and suppose
Y is the output process when X is ﬁltered using impulse response function g . Then Y is also a
WSS narrowband random process. Let Z denote the complex envelope of X , given in Proposition
8.5.2, and let zg denote the complex envelope signal of g , meaning that zg is the complex baseband
246 signal such that g (t) = Re(zg (t)ejωc t . It can be shown that the complex envelope process of Y is
1
1
1
2 zg ∗ Z . Thus, the ﬁltering of X by g is equivalent to the ﬁltering of Z by 2 zg . Example 8.5.4 (Simulation of a narrowband random process) Let ωo and ωc be postive numbers
with 0 < ωo < ωc . Suppose SX is a nonnegative function which is even (i.e. SX (ω ) = SX (−ω ) for
all ω ) with SX (ω ) = 0 if ω  − ωc  ≥ ωo . We discuss brieﬂy the problem of writing a computer
simulation to generate a realvalued WSS random process X with power spectral density SX .
By Proposition 8.5.1, it suﬃces to simulate baseband random processes U and V with the
power spectral densities speciﬁed by (8.36) and cross power spectral density speciﬁed by (8.37).
For increased tractability, we impose an additional assumption on SX , namely that the upper lobe
of SX is symmetric about ωc . This assumption is equivalent to the assumption that SU V vanishes,
and therefore that the processes U and V are uncorrelated with each other. Thus, the processes U
and V can be generated independently.
In turn, the processes U and V can be simulated by ﬁrst generating sequences of random
1
variables UnT and VnT for sampling frequency T = 2fo = ωo . A discrete time random process
π
with power spectral density SU can be generated by passing a discretetime white noise sequence
with unit variance through a discretetime linear timeinvariant system with realvalued impulse
response function such that the transfer function H satisﬁes SU = H 2 . For example, taking
H (ω ) =
SU (ω ) works, though it might not be the most well behaved linear system. (The
problem of ﬁnding a transfer function H with additional properties such that SU = H 2 is called
the problem of spectral factorization, which we shall return to in the next chapter.) The samples
VkT can be generated similarly.
For a speciﬁc example, suppose that (using kHz for kilohertz, or thousands of Hertz)
SX (2πf ) = 1 9, 000 kHz < f  < 9, 020 kHz
.
0 else (8.40) Notice that the parameters ωo and ωc are not uniquely determined by SX . They must simply be
positive numbers with ωo < ωc such that
(9, 000 kHz, 9, 020 kHz ) ⊂ (fc − fo , fc + fo )
However, only the choice fc = 9, 010 kHz makes the upper lobe of SX symmetric around fc .
Therefore we take fc = 9, 010 kHz . We take the minmum allowable value for fo , namely fo =
10 kHz . For this choice, (8.36) yields
SU (2πf ) = SV (2πf ) = 2 f  < 10 kHz
0 else (8.41) and (8.37) yields SU V (2πf ) = 0 for all f . The processes U and V are continuoustime baseband
random processes with onesided bandwidth limit 10 kHz . To simulate these processes it is therefore
enough to generate samples of them with sampling period T = 0.5 × 10−4 , and then use the Nyquist
1 An elegant proof of this fact is based on spectral representation theory for WSS random processes, covered for
example in Doob, Stochastic Processes, Wiley, 1953. The basic idea is to deﬁne the Fourier transform of a WSS
random process, which, like white noise, is a generalized random process. Then essentially the same method we
described for ﬁltering of deterministic narrowband signals works. 247 sampling representation described in Section 8.4. The processes of samples will, according to (8.26),
have power spectral density equal to 4 × 104 over the interval [−π, π ]. Consequently, the samples
can be taken to be uncorrelated with E [Ak 2 ] = E [Bk 2 ] = 4 × 104 . For example, these variables
can be taken to be independent real Gaussian random variables. Putting the steps together, we
ﬁnd the following representation for X :
∞ Xt = cos(ωc t) An sinc
n=−∞ 8.6 ∞ t − nT
T − sin(ωc t) Bn sinc
n=−∞ t − nT
T Complexiﬁcation, Part II A complex random variable Z is said to be circularly symmetric if Z has the same distribution
as ejθ Z for every real value of θ. If Z has a pdf fZ , circular symmetry of Z means that fZ (z )
is invariant under rotations about zero, or, equivalently, fZ (z ) depends on z only through z . A
collection of random variables (Zi : i ∈ I ) is said to be jointly circularly symmetric if for every real
value of θ, the collection (Zi : i ∈ I ) has the same ﬁnite dimensional distributions as the collection
(Zi ejθ : i ∈ I ). Note that if (Zi : i ∈ I ) is jointly circularly symmetric, and if (Yj : j ∈ J ) is another
collection of random variables such that each Yj is a linear combination of Zi ’s (with no constants
added in) then the collection (Yj : j ∈ J ) is also jointly circularly symmetric.
Recall that a complex random vector Z , expressed in terms of real random vectors U and V as
Z = U + jV , has mean EZ = EU + jEV and covariance matrix Cov(Z ) = E [(Z − EZ )(Z − EZ )∗ ].
The pseudocovariance matrix of Z is deﬁned by Covp (Z ) = E [(Z − EZ )(Z − EZ )T ], and it diﬀers
from the covariance of Z in that a transpose, rather than a Hermitian transpose, is involved. Note
that Cov(Z ) and Covp (Z ) are readily expressed in terms of Cov(U ), Cov(V ), and Cov(U, V ) as:
Cov(Z ) = Cov(U ) + Cov(V ) + j (Cov(V, U ) − Cov(U, V ))
Covp (Z ) = Cov(U ) − Cov(V ) + j (Cov(V, U ) + Cov(U, V ))
where Cov(V, U ) = Cov(U, V )T . Conversely,
Cov(U ) = Re (Cov(Z ) + Covp (Z )) /2, Cov(V ) = Re (Cov(Z ) − Covp (Z )) /2, and
Cov(U, V ) = Im (−Cov(Z ) + Covp (Z )) /2.
The vector Z is deﬁned to be Gaussian if the random vectors U and V are jointly Gaussian.
Suppose that Z is a complex Gaussian random vector. Then its distribution is fully determined
by its mean and the matrices Cov(U ), Cov(V ), and Cov(U, V ), or equivalently by its mean and
the matrices Cov(Z ) and Covp (Z ). Therefore, for a real value of θ, Z and ejθ Z have the same
distribution if and only if they have the same mean, covariance matrix, and pseudocovariance
matrix. Since E [ejθ Z ] = ejθ EZ , Cov(ejθ Z ) = Cov(Z ), and Covp (ejθ Z ) = ej 2θ Covp (Z ), Z and ejθ Z
have the same distribution if and only if (ejθ − 1)EZ = 0 and (ej 2θ − 1)Covp (Z ) = 0. Hence, if θ is
not a multiple of π , Z and ejθ Z have the same distribution if and only if EZ = 0 and Covp (Z ) = 0.
Consequently, a Gaussian random vector Z is circularly symmetric if and only if its mean vector
and pseudocovariance matrix are zero.
248 The joint density function of a circularly symmetric complex random vector Z with n complex
dimensions and covariance matrix K , with det K = 0, has the particularly elegant form:
fZ (z ) = exp(−z ∗ K −1 z )
.
π n det(K ) (8.42) Equation (8.42) can be derived in the same way the density for Gaussian vectors with real components is derived. Namely, (8.42) is easy to verify if K is diagonal. If K is not diagonal, the
Hermetian symmetric positive deﬁnite matrix K can be expressed as K = U ΛU ∗ , where U is a
unitary matrix and Λ is a diagonal matrix with strictly positive diagonal entries. The random
vector Y deﬁned by Y = U ∗ Z is Gaussian and circularly symmetric with covariance matrix Λ, and
∗ −1
since det(Λ) = det(K ), it has pdf fY (y ) = exp(−y ΛK ) y) . Since  det(U ) = 1, fZ (z ) = fY (U ∗ x),
π n det(
which yields (8.42).
Let us switch now to random processes. Let Z be a complexvalued random process and let U
and V be the realvalued random processes such that Zt = Ut + jVt . Recall that Z is Gaussian if U
and V are jointly Gaussian, and the covariance function of Z is deﬁned by CZ (s, t) = Cov(Zs , Zt ).
p
The pseudocovariance function of Z is deﬁned by CZ (s, t) = Covp (Zs , Zt ). As for covariance
p
matrices of vectors, both CZ and CZ are needed to determine CU , CV , and CU V .
Following the vast majority of the literature, we deﬁne Z to be wide sense stationary (WSS)
if µZ (t) is constant and if CZ (s, t) (or RZ (s, t)) is a function of s − t alone. Some authors use
a stronger deﬁnition of WSS, by deﬁning Z to be WSS if either of the following two equivalent
conditions is satisﬁed:
p
• µZ (t) is constant, and both CZ (s, t) and CZ (s, t) are functions of s − t • U and V are jointly WSS
If Z is Gaussian then it is stationary if and only if it satisﬁes the stronger deﬁnition of WSS.
A complex random process Z = (Zt : t ∈ T) is called circularly symmetric if the random
variables of the process, (Zt : t ∈ T), are jointly circularly symmetric. If Z is a complex Gaussian
random process, it is circularly symmetric if and only if it has mean zero and Covp (s, t) = 0 for
Z
all s, t. Proposition 8.5.2 shows that the baseband equivalent process Z for a Gaussian realvalued
narrowband WSS random process X is circularly symmetric. Nearly all complex valued random
processes in applications arise in this fashion. For circularly symmetric complex random processes,
the deﬁnition of WSS we adopted, and the stronger deﬁnition mentioned in the previous paragraph,
are equivalent. A circularly symmetric complex Gaussian random process is stationary if and only
if it is WSS.
The interested reader can ﬁnd more related to the material in this section in Neeser and Massey,
“Proper Complex Random Processes with Applications to Information Theory,” IEEE Transactions
on Information Theory, vol. 39, no. 4, July 1993. 8.7 Problems 8.1 On ﬁltering a WSS random process
Suppose Y is the output of a linear timeinvariant system with WSS input X , impulse response
function h, and transfer function H . Indicate whether the following statements are true or false.
Justify your answers. (a) If H (ω ) ≤ 1 for all ω then the power of Y is less than or equal to the
249 power of X . (b) If X is periodic (in addition to being WSS) then Y is WSS and periodic. (c) If X
has mean zero and strictly positive total power, and if h2 > 0, then the output power is strictly
positive.
8.2 On the cross spectral density
Suppose X and Y are jointly WSS such that the power spectral densities SX , SY , and SXY are
continuous. Show that for each ω , SXY (ω )2 ≤ SX (ω )SY (ω ). Hint: Fix ωo , let > 0, and let J
denote the interval of length centered at ωo . Consider passing both X and Y through a linear
timeinvariant system with transfer function H (ω ) = IJ (ω ). Apply the Schwarz inequality to the
output processes sampled at a ﬁxed time, and let → 0.
8.3 Modulating and ﬁltering a stationary process
2
Let X = (Xt : t ∈ Z ) be a discretetime meanzero stationary random process with power E [X0 ] =
1. Let Y be the stationary discrete time random process obtained from X by modulation as follows:
Yt = Xt cos(80πt + Θ),
where Θ is independent of X and is uniformly distributed over [0, 2π ]. Let Z be the stationary
discrete time random process obtained from Y by the linear equations:
Zt+1 = (1 − a)Zt + aYt+1
for all t, where a is a constant with 0 < a < 1. (a) Why is the random process Y stationary?
(b) Express the autocorrelation function of Y , RY (τ ) = E [Yτ Y0 ], in terms of the autocorrelation
function of X . Similarly, express the power spectral density of Y , SY (ω ), in terms of the power
spectral density of X , SX (ω ). (c) Find and sketch the transfer function H (ω ) for the linear system
describing the mapping from Y to Z . (d) Can the power of Z be arbitrarily large (depending on
a)? Explain your answer. (e) Describe an input X satisfying the assumptions above so that the
power of Z is at least 0.5, for any value of a with 0 < a < 1.
8.4 Filtering a Gauss Markov process
Let X = (Xt : −∞ < t < +∞) be a stationary Gauss Markov process with mean zero and
autocorrelation function RX (τ ) = exp(−τ ). Deﬁne a random process Y = (Yt : t ∈ R) by the
˙
diﬀerential equation Yt = Xt − Yt .
(a) Find the cross correlation function RXY . Are X and Y jointly stationary?
(b) Find E [Y5 X5 = 3]. What is the approximate numerical value?
(c) Is Y a Gaussian random process? Justify your answer.
(d) Is Y a Markov process? Justify your answer.
8.5 Slight smoothing
Suppose Y is the output of the linear timeinvariant system with input X and impulse response
1
function h, such that X is WSS with RX (τ ) = exp(−τ ), and h(τ ) = a I{τ ≤ a } for a > 0. If a
2
is small, then h approximates the delta function δ (τ ), and consequently Yt ≈ Xt . This problem
explores the accuracy of the approximation.
(a) Find RY X (0), and use the power series expansion of eu to show that RY X (0) = 1 − a + o(a) as
4
a → 0. Here, o(a) denotes any term such that o(a)/a → 0 as a → 0.
(b) Find RY (0), and use the power series expansion of eu to show that RY (0) = 1 − a + o(a) as
3
a → 0.
(c) Show that E [Xt − Yt 2 ] = a + o(a) as a → 0.
6
250 8.6 A stationary twostate Markov process
Let X = (Xk : k ∈ Z) be a stationary Markov process with state space S = {1, −1} and onestep
transition probability matrix
1−p
p
P=
,
p
1−p
where 0 < p < 1. Find the mean, correlation function and power spectral density function of X .
Hint: For nonnegative integers k :
Pk = 1
2
1
2 1
2
1
2 1
2 + (1 − 2p)k −1
2 −1
2
1
2 . 8.7 A stationary twostate Markov process in continuous time
Let X = (Xt : t ∈ R) be a stationary Markov process with state space S = {1, −1} and Q matrix
Q= −α α
α −α , where α > 0. Find the mean, correlation function and power spectral density function of X . (Hint:
Recall from the example in the chapter on Markov processes that for s < t, the matrix of transition
probabilities pij (s, t) is given by H (τ ), where τ = t − s and
H (τ ) = 1+e−2ατ
2
1−e−2ατ
2 1−e−2ατ
2
1+e−2ατ
2 . 8.8 A linear estimation problem
Suppose X and Y are possibly complex valued jointly WSS processes with known autocorrelation
functions, crosscorrelation function, and associated spectral densities. Suppose Y is passed through
a linear timeinvariant system with impulse response function h and transfer function H , and let
Z be the output. The mean square error of estimating Xt by Zt is E [Xt − Zt 2 ].
(a) Express the mean square error in terms of RX , RY , RXY and h.
(b) Express the mean square error in terms of SX , SY , SXY and H .
(c) Using your answer to part (b), ﬁnd the choice of H that minimizes the mean square error. (Hint:
Try working out the problem ﬁrst assuming the processes are real valued. For the complex case,
2
note that for σ 2 > 0 and complex numbers z and zo , σ 2 z 2 − 2Re(z ∗ zo ) is equal to σz − zo 2 − zo2 ,
σ
σ
z
which is minimized with respect to z by z = σo .)
2
8.9 Linear time invariant, uncorrelated scattering channel
A signal transmitted through a scattering environment can propagate over many diﬀerent paths
on its way to a receiver. The channel gains along distinct paths are often modeled as uncorrelated.
The paths may diﬀer in length, causing a delay spread. Let h = (hu : u ∈ Z) consist of uncorrelated,
possibly complex valued random variables with mean zero and E [hu 2 ] = gu . Assume that G =
u gu < ∞. The variable hu is the random complex gain for delay u, and g = (gu : u ∈ Z) is
the energy gain delay mass function with total gain G. Given a deterministic signal x, the channel
output is the random signal Y deﬁned by Yi = ∞ −∞ hu xi−u .
u=
(a) Determine the mean and autocorrelation function for Y in terms of x and g .
(b) Express the average total energy of Y : E [ i Yi2 ], in terms of x and g .
(c) Suppose instead that the input is a WSS random process X with autocorrelation function RX .
251 The input X is assumed to be independent of the channel h. Express the mean and autocorrelation
function of the output Y in terms of RX and g . Is Y WSS?
(d) Since the impulse response function h is random, so is its Fourier transform, H = (H (ω ) : −π ≤
ω ≤ π ). Express the autocorrelation function of the random process H in terms of g .
8.10 The accuracy of approximate diﬀerentiation
Let X be a WSS baseband random process with power spectral density SX , and let ωo be the
onesided band limit of X . The process X is m.s. diﬀerentiable and X can be viewed as the
output of a timeinvariant linear system with transfer function H (ω ) = jω .
(a) What is the power spectral density of X ?
−
(b) Let Yt = Xt+a2aXt−a , for some a > 0. We can also view Y = (Yt : t ∈ R) as the output of
a timeinvariant linear system, with input X . Find the impulse response function k and transfer
function K of the linear system. Show that K (ω ) → jω as a → 0.
(c) Let Dt = Xt − Yt . Find the power spectral density of D.
(d) Find a value of a, depending only on ωo , so that E [Dt 2 ] ≤ (0.01)E [Xt ]2 . In other words, for
such a, the m.s. error of approximating Xt by Yt is less than one percent of E [Xt 2 ]. You can use
2
the fact that 0 ≤ 1 − sin(u) ≤ u for all real u. (Hint: Find a so that SD (ω ) ≤ (0.01)SX (ω ) for
u
6
ω  ≤ ωo .)
8.11 Some linear transformations of some random processes
Let U = (Un : n ∈ Z) be a random process such that the variables Un are independent, identically
distributed, with E [Un ] = µ and Var(Un ) = σ 2 , where µ = 0 and σ 2 > 0. Please keep in mind that
µ = 0. Let X = (Xn : n ∈ Z) be deﬁned by Xn = ∞ Un−k ak , for a constant a with 0 < a < 1.
k=0
(a) Is X stationary? Find the mean function µX and autocovariance function CX for X .
(b) Is X a Markov process ? (Hint: X is not necessarily Gaussian. Does X have a state representation driven by U ?)
(c) Is X mean ergodic in the m.s. sense?
Let U be as before, and let Y = (Yn : n ∈ Z) be deﬁned by Yn = ∞ Un−k Ak , where A is a
k=0
random variable distributed on the interval (0, 0.5) (the exact distribution is not speciﬁed), and A
is independent of the random process U .
(d) Is Y stationary? Find the mean function µY and autocovariance function CY for Y . (Your
answer may include expectations involving A.)
(e) Is Y a Markov process? (Give a brief explanation.)
(f) Is Y mean ergodic in the m.s. sense?
8.12 Filtering Poisson white noise
A Poisson random process N = (Nt : t ≥ 0) has indpendent increments. The derivative of N ,
written N , does not exist as an ordinary random process, but it does exist as a generalized random
process. Graphically, picture N as a superposition of delta functions, one at each arrival time of the
Poisson process. As a generalized random process, N is stationary with mean and autocovariance
functions given by E [Nt ] = λ, and CN (s, t) = λδ (s − t), repectively, because, when integrated,
t
these functions give the correct values for the mean and covariance of N : E [Nt ] = 0 λds and
st
CN (s, t) = 0 0 λδ (u − v )dvdu. The random process N can be extended to be deﬁned for negative
times by augmenting the original random process N by another rate λ Poisson process for negative
times. Then N can be viewed as a stationary random process, and its integral over intervals gives
rise to a process N (a, b] as described in Problem 4.17. (The process N − λ is a white noise process,
252 in that it is a generalized random process which is stationary, mean zero, and has autocorrelation
function λδ (τ ). Both N and N − λ are called Poisson shot noise processes. One application for
such processes is modeling noise in small electronic devices, in which eﬀects of single electrons can
be registered. For the remainder of this problem, N is used instead of the mean zero version.) Let
X be the output when N is passed through a linear timeinvariant ﬁlter with an impulse response
∞
function h, such that −∞ h(t)dt is ﬁnite. (Remark: In the special case that h(t) = I{0≤t<1} , X is
the M/D/∞ process of Problem 4.17.)
(a) Find the mean function and covariance functions of X .
(b) Consider the special case that h(t) = e−t I{t≥0} . Explain why X is a Markov process in this
case. (Hint: What is the behavior of X between the arrival times of the Poisson process? What
does X do at the arrival times?)
8.13 A linear system with a feedback loop
The system with input X and output Y involves feedback with the loop transfer function shown.
X Y +
!
1+j " (a) Find the transfer function K of the system describing the mapping from X to Y.
(b) Find the corresponding impulse response function.
(c) The power of Y divided by the power of X , depends on the power spectral density, SX . Find
the supremum of this ratio, over all choices of SX , and describe what choice of SX achieves this
supremum.
8.14 Linear and nonlinear reconstruction from samples
Suppose Xt = ∞ −∞ g (t − n − U )Bn , where the Bn ’s are independent with mean zero and variance
n=
σ 2 > 0, g is a function with ﬁnite energy g (t)2 dt and Fourier transform G(ω ), U is a random
variable which is independent of B and uniformly distributed on the interval [0, 1]. The process X
is a typical model for a digital baseband signal, where the Bn ’s are random data symbols.
(a) Show that X is WSS, with mean zero and RX (t) = σ 2 g ∗ g (t).
(b) Under what conditions on G and T can the sampling theorem be used to recover X from its
samples of the form (X (nT ) : n ∈ Z)?
(c) Consider the particular case g (t) = (1 − t)+ and T = 0.5. Although this falls outside the
conditions found in part (b), show that by using nonlinear operations, the process X can be
recovered from its samples of the form (X (nT ) : n ∈ Z). (Hint: Consider a sample path of X )
8.15 Sampling a cubed Gaussian process
Let X = (Xt : t ∈ R) be a baseband mean zero stationary real Gaussian random process with
1
3
onesided band limit fo Hz. Thus, Xt = ∞ −∞ XnT sinc t−nT where T = 2fo . Let Yt = Xt for
n=
T
each t.
(a) Is Y stationary? Express RY in terms of RX , and SY in terms of SX and/or RX . (Hint: If
A, B are jointly Gaussian and mean zero, Cov(A3 , B 3 ) = 6Cov(A, B )3 + 9E [A2 ]E [B 2 ]Cov(A, B ).)
1
(b) At what rate T should Y be sampled in order that Yt = ∞ −∞ YnT sinc t−nT
?
n=
T
(c) Can Y be recovered with fewer samples than in part (b)? Explain.
253 8.16 An approximation of white noise
White noise in continuous time can be approximated by a piecewise constant process as follows.
Let T be a small positive constant, let AT be a positive scaling constant depending on T , and let
(Bk : k ∈ Z) be a discretetime white noise process with RB (k ) = σ 2 I{k=0} . Deﬁne (Nt : t ∈ R) by
Nt = Bk for t ∈ [kT, (k + 1)T ).
1
(a) Sketch a typical sample path of N and express E [ 0 Ns ds2 ] in terms of AT , T and σ 2 . For
1
simplicity assume that T = K for some large integer K .
(b) What choice of AT makes the expectation found in part (a) equal to σ 2 ? This choice makes
N a good approximation to a continuoustime white noise process with autocorrelation function
σ 2 δ (τ ).
(c) What happens to the expectation found in part (a) as T → 0 if AT = 1 for all T ?
8.17 Simulating a baseband random process
Suppose a realvalued Gaussian baseband process X = (Xt : t ∈ R) with mean zero and power
spectral density
1 if f  ≤ 0.5
SX (2πf ) =
0 else
is to be simulated over the time interval [−500, 500] through use of the sampling theorem with
sampling time T = 1. (a) What is the joint distribution of the samples, Xn : n ∈ Z ? (b) Of course
a computer cannot generate inﬁnitely many random variables in a ﬁnite amount of time. Therefore,
consider approximating X by X (N ) deﬁned by
N
(N )
Xt Xn sinc(t − n) =
n=−N (N ) Find a condition on N to guarantee that E [(Xt − Xt )2 ] ≤ 0.01 for t ∈ [−500, 500]. (Hint: Use
1
sinc(τ ) ≤ πτ  and bound the series by an integral. Your choice of N should not depend on t
because the same N should work for all t in the interval [−500, 500] ).
8.18 Filtering to maximize signal to noise ratio
Let X and N be continuous time, mean zero WSS random processes. Suppose that X has power
spectral density SX (ω ) = ω I{ω≤ωo } , and that N has power spectral density SN (ω ) = σ 2 for all
ω . Suppose also that X and N are uncorrelated with each other. Think of X as a signal, and N
as noise. Suppose the signal plus noise X + N is passed through a linear timeinvariant ﬁlter with
transfer function H , which you are to specify. Let X denote the output signal and N denote the
output noise. What choice of H , subject the constraints (i) H (ω ) ≤ 1 for all ω , and (ii) (power of
X ) ≥ (power of X )/2, minimizes the power of N ?
8.19 Finding the envelope of a deterministic signal
(a) Find the complex envelope z (t) and real envelope z (t) of x(t) = cos(2π (1000)t)+cos(2π (1001)t),
using the carrier frequency fc = 1000.5Hz . Simplify your answer as much as possible.
(b) Repeat part (a), using fc = 995Hz . (Hint: The real envelope should be the same as found in
part (a).)
(c) Explain why, in general, the real envelope of a narrowband signal does not depend on which
frequency fc is used to represent the signal (as long as fc is chosen so that the upper band of the
signal is contained in an interval [fc − a, fc + a] with a << fc .)
254 8.20 Sampling a signal or process that is not band limited
(a) Fix T > 0 and let ωo = π/T . Given a ﬁnite energy signal x, let xo be the bandlimited signal
with Fourier transform xo (ω ) = I{ω≤ωo } ∞ −∞ x(ω + 2nωo ). Show that x(nT ) = xo (nT ) for all
n=
integers n. (b) Explain why xo (t) = ∞ −∞ x(nT )sinc t−nT .
n=
T
o
(c) Let X be a mean zero WSS random process, and let RX be the autocorrelation function for
o
o
power spectral density SX (ω ) deﬁned by SX (ω ) = I{ω≤ωo } ∞ −∞ SX (ω + 2nωo ). Show that
n=
o (nT ) for all integers n. (d) Explain why the random process Y deﬁned by Y =
RX (nT ) = RX
t
∞
o
o
XnT sinc t−nT is WSS with autocorrelation function RX . (e) Find SX in case SX (ω ) =
n=−∞
T
exp(−αω ) for ω ∈ R.
8.21 A narrowband Gaussian process
Let X be a realvalued stationary Gaussian process with mean zero and RX (τ ) = cos(2π (30τ ))(sinc(6τ ))2 .
(a) Find and carefully sketch the power spectral density of X . (b) Sketch a sample path of X . (c)
The process X can be represented by Xt = Re(Zt e2πj 30t ), where Zt = Ut + jVt for jointly stationary
narrowband realvalued random processes U and V . Find the spectral densities SU , SV , and SU V .
(d) Find P [Z33  > 5]. Note that Zt  is the real envelope process of X .
8.22 Another narrowband Gaussian process
Suppose a realvalued Gaussian random process R = (Rt : t ∈ R) with mean 2 and power spectral
4
density SR (2πf ) = e−f /10 is fed through a linear timeinvariant system with transfer function
H (2πf ) = 0.1 5000 ≤ f  ≤ 6000
0 else (a) Find the mean and power spectral density of the output process X = (Xt : t ∈ R). (b) Find
P [X25 > 6]. (c) The random process X is a narrowband random process. Find the power spectral
densities SU , SV and the cross spectral density SU V of jointly WSS baseband random processes U
and V so that
Xt = Ut cos(2πfc t) − Vt sin(2πfc t),
using fc = 5500. (d) Repeat part (c) with fc = 5000.
8.23 Another narrowband Gaussian process (version 2)
Suppose a realvalued Gaussian white noise process N (we assume white noise has mean zero) with
power spectral density SN (2πf ) ≡ No for f ∈ R is fed through a linear timeinvariant system with
2
transfer function H speciﬁed as follows, where f represents the frequency in gigahertz (GHz) and
a gigahertz is 109 cycles per second. 1
19.10 ≤ f  ≤ 19.11 19.12−f 
H (2πf ) =
19.11 ≤ f  ≤ 19.12
0.01 0
else
(a) Find the mean and power spectral density of the output process X = (Xt : t ∈ R).
(b) Express P [X25 > 2] in terms of No and the standard normal complementary CDF function Q.
(c) The random process X is a narrowband random process. Find and sketch the power spectral
densities SU , SV and the cross spectral density SU V of jointly WSS baseband random processes U
and V so that
Xt = Ut cos(2πfc t) − Vt sin(2πfc t),
255 using fc = 19.11 GHz.
(d) The complex envelope process is given by Z = U + jV and the real envelope process is given
by Z . Specify the distributions of Zt and Zt  for t ﬁxed.
8.24 Declaring the center frequency for a given random process
Let a > 0 and let g be a nonnegative function on R which is zero outside of the interval [a, 2a].
Suppose X is a narrowband WSS random process with power spectral density function SX (ω ) =
g (ω ), or equivalently, SX (ω ) = g (ω ) + g (−ω ). The process X can thus be viewed as a narrowband
signal for carrier frequency ωc , for any choice of ωc in the interval [a, 2a]. Let U and V be the
baseband random processes in the usual complex envelope representation: Xt = Re((Ut + jVt )ejωc t ).
(a) Express SU and SU V in terms of g and ωc .
∞
(b) Describe which choice of ωc minimizes −∞ SU V (ω )2 dω . (Note: If g is symmetric arround
dπ
some frequency ν , then ωc = ν . But what is the answer otherwise?)
8.25 * Cyclostationary random processes
A random process X = (Xt : t ∈ R) is said to be cyclostationary with period T , if whenever s is
an integer multiple of T , X has the same ﬁnite dimensional distributions as (Xt+s : t ∈ R). This
property is weaker than stationarity, because stationarity requires equality of ﬁnite dimensional
distributions for all real values of s.
(a) What properties of the mean function µX and autocorrelation function RX does any second
order cyclostationary process possess? A process with these properties is called a wide sense
cyclostationary process.
(b) Suppose X is cyclostationary and that U is a random variable independent of X that is uniformly
distributed on the interval [0, T ]. Let Y = (Yt : t ∈ R) be the random process deﬁned by Yt = Xt+U .
Argue that Y is stationary, and express the mean and autocorrelation function of Y in terms of
the mean function and autocorrelation function of X . Although X is not necessarily WSS, it is
reasonable to deﬁne the power spectral density of X to equal the power spectral density of Y .
(c) Suppose B is a stationary discretetime random process and that g is a deterministic function.
Let X be deﬁned by
∞ g (t − nT )Bn . Xt =
−∞ Show that X is a cyclostationary random process. Find the mean function and autocorrelation
function of X in terms g , T , and the mean and autocorrelation function of B . If your answer is
complicated, identify special cases which make the answer nice.
(d) Suppose Y is deﬁned as in part (b) for the speciﬁc X deﬁned in part (c). Express the mean
µY , autocorrelation function RY , and power spectral density SY in terms of g , T , µB , and SB .
8.26 * Zero crossing rate of a stationary Gaussian process
Consider a zeromean stationary Gaussian random process X with SX (2πf ) = f  − 50 for 50 ≤
f  ≤ 60, and SX (2πf ) = 0 otherwise. Assume the process has continuous sample paths (it can be
shown that such a version exists.) A zero crossing from above is said to occur at time t if X (t) = 0
and X (s) > 0 for all s in an interval of the form [t − , t) for some > 0. Determine the mean rate
of zero crossings from above for X . If you can ﬁnd an analytical solution, great. Alternatively, you
can estimate the rate (aim for three signiﬁcant digits) by Monte Carlo simulation of the random
process. 256 Chapter 9 Wiener ﬁltering
9.1 Return of the orthogonality principle Consider the problem of estimating a random process X at some ﬁxed time t given observation of
a random process Y over an interval [a, b]. Suppose both X and Y are mean zero second order
random processes and that the minimum mean square error is to be minimized. Let Xt denote the
best linear estimator of Xt based on the observations (Ys : a ≤ s ≤ b). In other words, deﬁne
Vo = {c1 Ys1 + · · · + cn Ysn : for some constants n, a ≤ s1 , . . . , sn ≤ b, and c1 , . . . , cn }
and let V be the m.s. closure of V , which includes Vo and any random variable that is the m.s.
limit of a sequence of random variables in Vo . Then Xt is the random variable in V that minimizes
the mean square error, E [Xt − Xt 2 ]. By the orthogonality principle, the estimator Xt exists and it
is unique in the sense that any two solutions to the estimation problem are equal with probability
one.
Perhaps the most useful part of the orthogonality principle is that a random variable W is equal
to Xt if and only if (i) W ∈ V and (ii) (Xt − W ) ⊥ Z for all Z ∈ V . Equivalently, W is equal to
Xt if and only if (i) W ∈ V and (ii) (Xt − W ) ⊥ Yu for all u ∈ [a, b]. Furthermore, the minimum
mean square error (i.e. the error for the optimal estimator Xt ) is given by E [Xt 2 ] − E [Xt 2 ].
b
Note that m.s. integrals of the form a h(t, s)Ys ds are in V , because m.s. integrals are m.s.
limits of ﬁnite linear combinations of the random variables of Y . Typically the set V is larger than
the set of all m.s. integrals of Y . For example, if u is a ﬁxed time in [a, b] then Yu ∈ V . In addition,
if Y is m.s. diﬀerentiable, then Yu is also in V . Typically neither Yu nor Yu can be expressed as a
m.s. integral of (Ys : s ∈ R). However, Yu can be obtained as an integral of the process Y multiplied
by a delta function, though the integration has to be taken in a generalized sense.
b
The integral a h(t, s)Ys ds is the linear MMSE estimator if and only if
b Xt − h(t, s)Ys ds ⊥ Yu for u ∈ [a, b] a or equivalently
b E [(Xt − ∗
h(t, s)Ys ds)Yu ] = 0 for u ∈ [a, b] a or equivalently
b RXY (t, u) = h(t, s)RY (s, u)ds
a 257 for u ∈ [a, b]. Suppose now that the observation interval is the whole real line R and suppose that X and Y
are jointly WSS. Then for t and v ﬁxed, the problem of estimating Xt from (Ys : s ∈ R) is the
same as the problem of estimating Xt+v from (Ys+v : s ∈ R). Therefore, if h(t, s) for t ﬁxed is the
optimal function to use for estimating Xt from (Ys : s ∈ R), then it is also the optimal function to
use for estimating Xt+v from (Ys+v : s ∈ R). Therefore, h(t, s) = h(t + v, s + v ), so that h(t, s) is a
function of t − s alone, meaning that the optimal impulse response function h corresponds to a time∞
ˆ
invariant system. Thus, we seek to ﬁnd an optimal estimator of the form Xt = −∞ h(t − s)Ys ds.
The optimality condition becomes
∞ h(t − s)Ys ds ⊥ Yu Xt − for u ∈ R −∞ which is equivalent to the condition
∞ RXY (t − u) = h(t − s)RY (s − u)ds for u ∈ R
−∞ or RXY = h ∗ RY . In the frequency domain the optimality condition becomes SXY (ω ) = H (ω )SY (ω )
for all ω . Consequently, the optimal ﬁlter H is given by
H (ω ) = SXY (ω )
SY (ω ) and the corresponding minimum mean square error is given by
∞ E [Xt − Xt 2 ] = E [Xt 2 ] − E [Xt 2 ] = SX (ω ) −
−∞ SXY (ω )2
SY (ω ) dω
2π Example 9.1.1 Consider estimating a random process from observation of the random process
plus noise, as shown in Figure 9.1. Assume that X and N are jointly WSS with mean zero. Suppose
X Y + h X N Figure 9.1: An estimator of a signal from signal plus noise, as the output of a linear ﬁlter.
X and N have known autocorrelation functions and suppose that RXN ≡ 0, so the variables of the
process X are uncorrelated with the variables of the process N . The observation process is given
by Y = X + N . Then SXY = SX and SY = SX + SN , so the optimal ﬁlter is given by
H (ω ) = SX (ω )
SXY (ω )
=
SY (ω )
SX (ω ) + SN (ω ) The associated minimum mean square error is given by
∞ E [Xt − Xt 2 ] = SX (ω )2
SX (ω ) + SN (ω )
SX (ω )SN (ω ) dω
SX (ω ) + SN (ω ) 2π
SX (ω ) − −∞
∞ =
−∞ 258 dω
2π Example 9.1.2 This example is a continuation of the previous example, for a particular choice
of power spectral densities. Suppose that the signal process X is WSS with mean zero and power
1
spectral density SX (ω ) = 1+ω2 , suppose the noise process N is WSS with mean zero and power
−τ  4
spectral density 4+ω2 , and suppose SXN ≡ 0. Equivalently, RX (τ ) = e 2 , RN (τ ) = e−2τ  and
RXN ≡ 0. We seek the optimal linear estimator of Xt given (Ys : s ∈ R), where Y = X + N .
Seeking an estimator of the form
∞ h(t − s)Ys ds Xt =
−∞ we ﬁnd from the previous example that the transform H of h is given by
H (ω ) = 1
1+ω 2 SX (ω )
=
SX (ω ) + SN (ω ) 1
1+ω 2 + 4
4+ω 2 = 4 + ω2
8 + 5ω 2 We will ﬁnd h by ﬁnding the inverse transform of H . First, note that
8
12
12
+ ω2
4 + ω2
1
5
5
=5
+
=+
8 + 5ω 2
8 + 5ω 2 8 + 5ω 2
5 8 + 5ω 2
1
We know that 5 δ (t) ↔ 1 . Also, for any α > 0,
5 e−αt ↔ ω2 2α
,
+ α2 (9.1) so
1
=
8 + 5ω 2 1
5
8
5 + ω2 = 1
5·2 5
8 8
5 2 ( 8 + ω2)
5 ↔ 1
√
4 10 e q
− 8 t
5 Therefore the optimal ﬁlter is given in the time domain by
1
h(t) = δ (t) +
5 3
√
5 10 e q
− 8 t
5 The associated minimum mean square error is given by (one way to do the integration is to use the
∞
fact that if k ↔ K then −∞ K (ω ) dω = k (0)):
2π
∞ E [Xt − Xt 2 ] =
−∞ SX (ω )SN (ω ) dω
=
SX (ω ) + SN (ω ) 2π ∞
−∞ 4
dω
=4
2 2π
8 + 5ω 1
√
4 10 1
=√
10 In an example later in this chapter we will return to the same random processes, but seek the best
linear estimator of Xt given (Ys : s ≤ t).
259 9.2 The causal Wiener ﬁltering problem A linear system is causal if the value of the output at any given time does not depend on the
future of the input. That is to say that the impulse response function satisﬁes h(t, s) = 0 for s > t.
In the case of a linear, timeinvariant system, causality means that the impulse response function
satisﬁes h(τ ) = 0 for τ < 0. Suppose X and Y are mean zero and jointly WSS. In this section we
will consider estimates of X given Y obtained by passing Y through a causal linear timeinvariant
system. For convenience in applications, a ﬁxed parameter T is introduced. Let Xt+T t be the
minimum mean square error linear estimate of Xt+T given (Ys : s ≤ t). Note that if Y is the same
process as X and T > 0, then we are addressing the problem of predicting Xt+T from (Xs : s ≤ t).
∞
An estimator of the form −∞ h(t − s)Ys ds is sought such that h corresponds to a causal system.
Once again, the orthogonality principle implies that the estimator is optimal if and only if it satisﬁes
∞ h(t − s)Ys ds ⊥ Yu Xt+T − for u ≤ t −∞ which is equivalent to the condition
∞ RXY (t + T − u) = h(t − s)RY (s − u)ds for u ≤ t −∞ or RXY (t + T − u) = h ∗ RY (t − u). Setting τ = t − u and combining this optimality condition
with the constraint that h is a causal function, the problem is to ﬁnd an impulse response function
h satisfying:
RXY (τ + T ) = h ∗ RY (τ ) for τ ≥ 0 (9.2) h(v ) = 0 for v < 0 (9.3) Equations (9.2) and (9.3) are called the WienerHopf equations. We shall show how to solve them
in the case the power spectral densities are rational functions by using the method of spectral
factorization. The next section describes some of the tools needed for the solution. 9.3 Causal functions and spectral factorization A function h on R is said to be causal if h(τ ) = 0 for τ < 0, and it is said to be anticausal if
h(τ ) = 0 for τ > 0. Any function h on R can be expressed as the sum of a causal function and
an anticausal function as follows. Simply let u(t) = I{t≥0} and notice that h(t) is the sum of the
causal function u(t)h(t) and the anticausal function (1 − u(t))h(t). More compactly, we have the
representation h = uh + (1 − u)h.
A transfer function H is said to be of positive type if the corresponding impulse response function
h is causal, and H is said to be of negative type if the corresponding impulse response function is
anticausal. Any transfer function can be written as the sum of a positive type transfer function
and a negative type transfer function. Indeed, suppose H is the Fourier transform of an impulse
response function h. Deﬁne [H ]+ to be the Fourier transform of uh and [H ]− to be the Fourier
transform of (1 − u)h. Then [H ]+ is called the positive part of H and [H ]− is called the negative
part of H . The following properties hold:
• H = [H ]+ + [H ]− (because h = uh + (1 − u)h)
260 • [H ]+ = H if and only if H is positive type
• [H ]− = 0 if and only if H is positive type
• [[H ]+ ]− = 0 for any H
• [[H ]+ ]+ = [H ]+ and [[H ]− ]− = [H ]−
• [H + G]+ = [H ]+ + [G]+ and [H + G]− = [H ]− + [G]−
Note that uh is the casual function that is closest to h in the L2 norm. That is, uh is the
projection of h onto the space of causal functions. Indeed, if k is any causal function, then
∞ ∞ 0 h(t) − k (t)2 dt = h(t)2 dt + −∞ h(t) − k (t)2 dt −∞
0 0 h(t)2 dt ≥ (9.4) −∞ and equality holds in (9.4) if and only if k = uh (except possibly on a set of measure zero). By
Parseval’s relation, it follows that [H ]+ is the positive type function that is closest to H in the L2
norm. Equivalently, [H ]+ is the projection of H onto the space of positive type functions. Similarly,
[H ]− is the projection of H onto the space of negative type functions. Up to this point in these
notes, Fourier transforms have been deﬁned for real values of ω only. However, for the purposes
of factorization to be covered later, it is useful to consider the analytic continuation of the Fourier
transforms to larger sets in C. We use the same notation H (ω ) for the function H deﬁned for real
values of ω only, and its continuation deﬁned for complex ω . The following examples illustrate the
use of the projections [ ]+ and [ ]− , and consideration of transforms for complex ω . Example 9.3.1 Let g (t) = e−αt for a constant α > 0. The functions g , ug and (1 − u)g are
g(t)
t
u(t)g(t)
t
(1−u(t))g(t)
t Figure 9.2: Decomposition of a twosided exponential function.
pictured in Figure 9.2. The corresponding transforms are given by:
∞ 1
jω + α
0
0
1
[G]− (ω ) =
eαt e−jωt dt =
−jω + α
−∞
2α
G(ω ) = [G]+ (ω ) + [G]− (ω ) = 2
ω + α2
[G]+ (ω ) = e−αt e−jωt dt = 261 Note that [G]+ has a pole at ω = jα, so that the imaginary part of the pole of [G]+ is positive.
Equivalently, the pole of [G]+ is in the upper half plane.
More generally, suppose that G(ω ) has the representation
N1 G( ω ) =
n=1 γn
+
jω + αn N
n=N1 +1 γn
−jω + αn where Re(αn ) > 0 for all n. Then
N1 [G]+ (ω ) =
n=1 N γn
jω + αn [G]− (ω ) =
n=N1 +1 γn
−jω + αn Example 9.3.2 Let G be given by
G(ω ) = 1 − ω2
(jω + 1)(jω + 3)(jω − 2) Note that G has only three simple poles. The numerator of G has no factors in common with the
denominator, and the degree of the numerator is smaller than the degree of the denominator. By
the theory of partial fraction expansions in complex analysis, it therefore follows that G can be
written as
γ1
γ2
γ3
G(ω ) =
+
+
jω + 1 jω + 3 jω − 2
In order to identify γ1 , for example, multiply both expressions for G by (jω + 1) and then let
jω = −1. The other constants are found similarly. Thus
γ1 = = 1−
(jω + 1)(jω − 2) γ3 = 1 + 32
=1
(−3 + 1)(−3 − 2) j ω =−1 ω2 γ2 = 1 + (−1)2
1
=−
(−1 + 3)(−1 − 2)
3 = 1 − ω2
(jω + 3)(jω − 2) 1 − ω2
(jω + 1)(jω + 3) j ω =−3 =
j ω =2 1 + 22
1
=
(2 + 1)(2 + 3)
3 Consequently,
[G]+ (ω ) = − 1
1
+
3(jω + 1) jω + 3 and [G]− (ω ) = −jωT 1
3(jω − 2) Example 9.3.3 Suppose that G(ω ) = (e +α) . Multiplication by e−jωT in the frequency domain
jω
represents a shift by T in the time domain, so that
g (t) = e−α(t−T ) t ≥ T
,
0
t<T
262 g(t) T>0: t
T
T<0: g(t)
t
T Figure 9.3: Exponential function shifted by T.
as pictured in Figure 9.3. Consider two cases. First, if T ≥ 0, then g is causal, G is positive type,
and therefore [G]+ = G and [G]− = 0. Second, if T ≤ 0 then
eαT e−αt t ≥ 0
0
t<0 g (t)u(t) = −jωT αT αT e
−e
so that [G]+ (ω ) = jω+α and [G]− (ω ) = G(ω ) − [G]+ (ω ) = e (jω+α) . We can also ﬁnd [G]− by
computing the transform of (1 − u(t))g (t) (still assuming that T ≤ 0):
0 [G]− (ω ) = eα(T −t) e−jωt dt = T eαT −(α+jω)t
−(α + jω ) 0 =
t=T e−jωT − eαT
(jω + α) Example 9.3.4 Suppose H is the transfer function for impulse response function h. Let us unravel
the notation and express
∞
2 dω
ejωT H (ω ) +
2π
−∞
in terms of h and T . (Note that the factor ejωT is used, rather than e−jωT as in the previous
example.) Multiplication by ejωT in the frequency domain corresponds to shifting by −T in the
time domain, so that
ejωT H (ω ) ↔ h(t + T )
and thus
ejωT H (ω ) + ↔ u(t)h(t + T ) Applying Parseval’s identity, the deﬁnition of u, and a change of variables yields
∞ ejωT H (ω )
−∞ 2
+ dω
2π ∞ u(t)h(t + T )2 dt =
−∞
∞ h(t + T )2 dt =
0 ∞ h(t)2 dt =
T 263 The integral decreases from the energy of h to zero as T ranges from −∞ to ∞. Example 9.3.5 Suppose [H ]− = [K ]− = 0. Let us ﬁnd [HK ]− . As usual, let h denote the inverse
transform of H , and k denote the inverse transform of K . The supposition implies that h and k
are both causal functions. Therefore the convolution h ∗ k is also a causal function. Since HK is
the transform of h ∗ k , it follows that HK is a positive type function. Equivalently, [HK ]− = 0.
The decomposition H = [H ]+ + [H ]− is an additive one. Next we turn to multiplicative
decomposition, concentrating on rational functions. A function H is said to be rational if it can
be written as the ratio of two polynomials. Since polynomials can be factored over the complex
numbers, a rational function H can be expressed in the form
H (ω ) = γ (jω + β1 )(jω + β2 ) · · · (jω + βK )
(jω + α1 )(jω + α2 ) · · · (jω + αN ) for complex constants γ, α1 , . . . , αN , β1 , . . . , βK . Without loss of generality, we assume that {αi } ∩
{βj } = ∅. We also assume that the real parts of the constants α1 , . . . , αN , β1 , . . . , βK are nonzero.
The function H is positive type if and only if Re(αi ) > 0 for all i, or equivalently, if and only if all
the poles of H (ω ) are in the upper half plane Im(ω ) > 0.
A positive type function H is said to have minimum phase if Re(βi ) > 0 for all i. Thus, a
positive type function H is minimum phase if and only if 1/H is also positive type.
Suppose that SY is the power spectral density of a WSS random process and that SY is a
∗
rational function. The function SY , being nonnegative, is also realvalued, so SY = SY . Thus, if
the denominator of SY has a factor of the form jω + α then the denominator must also have a
factor of the form −jω + α∗ . Similarly, if the numerator of SY has a factor of the form jω + β then
the numerator must also have a factor of the form −jω + β ∗ . Example 9.3.6 The function SY given by
SY (ω ) = 8 + 5ω 2
(1 + ω 2 )(4 + ω 2 ) can be factored as SY (ω ) = √ (jω +
5 8
5) (jω + 2)(jω + 1) √ (−jω +
5 8
5) (−jω + 2)(−jω + 1) (9.5) −
SY (ω ) +
SY (ω ) +
−
where SY is a positive type, minimum phase function and SY is a negative type function with
−
+∗
SY = (SY ) . Note that the operators [ ]+ and [ ]− give us an additive decomposition of a function H into
the sum of a positive type and a negative type function, whereas spectral factorization has to do
264 with products. At least formally, the factorization can be accomplished by taking a logarithm,
doing an additive decomposition, and then exponentiating:
SX (ω ) = exp([ln SX (ω )]+ ) exp([ln SX (ω )]− ) . (9.6) −
SX (ω ) +
SX (ω ) Notice that if h ↔ H then, formally,
1+h+ h∗h h∗h∗h
H2 H2
+
· · · ↔ exp(H ) = 1 + H +
+
···
2!
3!
2!
3! +
so that if H is positive type, then exp(H ) is also positive type. Thus, the factor SX in (9.6) is
−
indeed a positive type function, and the factor SX is a negative type function. Use of (9.6) is called
the cepstrum method. Unfortunately, there is a host of problems, both numerical and analytical,
in using the method, so that it will not be used further in these notes. 9.4 Solution of the causal Wiener ﬁltering problem for rational
power spectral densities The WienerHopf equations (9.2) and ( 9.3) can be formulated in the frequency domain as follows:
Find a positive type transfer function H such that
ejωT SXY − HSY + =0 (9.7) +
+−
Suppose SY is factored as SY = SY SY such that SY is a minimum phase, positive type transfer
−
+
−
function and SY = (SY )∗ . Then SY and S1 are negative type functions. Since the product of
−
Y two negative type functions is again negative type, (9.7) is equivalent to the equation obtained by
multiplying the quantity within square brackets in (9.7) by S1 , yielding the equivalent problem:
−
Y Find a positive type transfer function H such that
ejωT SXY
+
− HSY
−
SY =0 (9.8) + +
The function HSY , being the product of two positive type functions, is itself positive type. Thus
(9.8) becomes
ejωT SXY
+
− HSY = 0
−
SY
+ Solving for H yields that the optimal transfer function is given by
H= 1 ejωT SXY
+
−
SY
SY (9.9)
+ The orthogonality principle yields that the mean square error satisﬁes
E [Xt+T − Xt+T t 2 ] = E [Xt+T 2 ] − E [Xt+T t 2 ]
∞ H (ω )2 SY (ω ) = RX (0) −
−∞
∞ = RX (0) −
−∞ 265 ejωT SXY
−
SY dω
2π
2 + dω
2π (9.10) +
where we used the fact that SY 2 = SY . Another expression for the MMSE, which involves the optimal ﬁlter h, is the following:
MMSE = E [(Xt+T − Xt+T t )(Xt+T − Xt+T t )∗ ]
∗
= E [(Xt+T − Xt+T t )Xt+T ] = RX (0) − RXX (t, t + T )
b
∞ = RX (0) −
−∞ ∗
h(s)RXY (s + T )ds. Exercise Evaluate the limit as T → −∞ and the limit as T → ∞ in (9.10). Example 9.4.1 This example involves the same model as in an example in Section 9.1, but here
a causal estimator is sought. The observed random process is Y = X + N , were X is WSS with
1
mean zero and power spectral density SX (ω ) = 1+ω2 , N is WSS with mean zero and power spectral
4
density SN (ω ) = 4+ω2 , and SXN = 0. We seek the optimal casual linear estimator of Xt given
(Ys : s ≤ t). The power spectral density of Y is given by
SY (ω ) = SX (ω ) + SN (ω ) = 8 + 5ω 2
(1 + ω 2 )(4 + ω 2 ) −
+
and its spectral factorization is given by (9.5), yielding SY and SY . Since RXN = 0 it follows that SXY (ω ) = SX (ω ) = 1
.
(jω + 1)(−jω + 1) Therefore
SXY (ω )
−
SY (ω ) =
= (−jω + 2)
√ 5(jω + 1)(−jω +
γ1
γ2
+
jω + 1 −jω + 8 8
5) 5 where
γ1 = γ2 = −jω + 2
√ 5(−jω + −jω + 2
√
5(jω + 1) =√
8
5 ) j ω =−1 −
=√
q
j ω= 8
5 3
√
5+ 8 8
5 +2
√
5+ 8 Therefore
SXY (ω )
−
SY (ω ) =
+ 266 γ1
jω + 1 (9.11) and thus 8
5 2− γ1 (jω + 2)
3
√ 1 +
H (ω ) = √
=
8
5 + 2 10
5(jω + 5 )
jω + 8
5 so that the optimal causal ﬁlter is
h(t) = 3
√
5 + 2 10 q δ (t) + (2 − 8
−t
)u(t)e
5 8
5 Finally, by (9.10) with T = 0, (9.11), and (9.1), the minimum mean square error is given by
∞ E [Xt − Xt 2 ] = RX (0) −
−∞ which is slightly larger than √1
10 2
γ1 dω
1 γ2
= − 1 ≈ 0.3246
1 + ω 2 2π
2
2 ≈ 0.3162, the MMSE found for the best noncausal estimator (see the example in Section 9.1), and slightly smaller than 1 , the MMSE for the best “instantaneous”
3
estimator of Xt given Yt , which is Xt .
3 Example 9.4.2 A special case of the causal ﬁltering problem formulated above is when the observed process Y is equal to X itself. This leads to the pure prediction problem. Let X be a
WSS mean zero random process and let T > 0. Then the optimal linear predictor of Xt+T given
(Xs : s ≤ t) corresponds to a linear timeinvariant system with transfer function H given by
−
−
+
+
(because SXY = SX , SY = SX , SY = SX , and SY = SX ):
H= 1
+ jωT
+Se
SX X (9.12) + To be more speciﬁc, suppose that SX (ω ) = ω41 . Observe that ω 4 + 4 = (ω 2 + 2j )(ω 2 − 2j ). Since
+4
2j = (1 + j )2 , we have (ω 2 + 2j ) = (ω + 1 + j )(ω − 1 − j ). Factoring the term (ω 2 − 2j ) in a similar
way, and rearranging terms as needed, yields that the factorization of SX is given by
SX (ω ) = 1
1
(jω + (1 + j ))(jω + (1 − j )) (−jω + (1 + j ))(−jω + (1 − j ))
−
SX (ω ) +
SX (ω ) so that
+
SX (ω ) = = 1
(jω + (1 + j ))(jω + (1 − j ))
γ1
γ2
+
jω + (1 + j ) jω + (1 − j ) where
γ1 =
γ2 = 1
jω + (1 − j )
1
jω + (1 + j )
267 =
j ω =−(1+j ) =
j ω =−1+j j
2 −j
2 +
yielding that the inverse Fourier transform of SX is given by
+
SX ↔ Hence j
j −(1+j )t
e
u(t) − e−(1−j )t u(t)
2
2
j −(1+j )(t+T )
2e +
SX (ω )ejωT ↔ so that
+
SX (ω )ejωT + = j
− 2 e−(1−j )(t+T ) t ≥ −T
0
else je−(1+j )T
je−(1−j )T
−
2(jω + (1 + j )) 2(jω + (1 − j )) The formula (9.12) for the optimal transfer function yields
je−(1+j )T (jω + (1 − j )) je−(1−j )T (jω + (1 + j ))
−
2
2
ejT (1 + j ) − e−jT (1 − j ) jω (ejT − e−jT )
+
= e−T
2j
2j H (ω ) = = e−T [cos(T ) + sin(T ) + jω sin(T )]
so that the optimal predictor for this example is given by
Xt+T t = Xt e−T (cos(T ) + sin(T )) + Xt e−T sin(T ) 9.5 Discrete time Wiener ﬁltering Causal Wiener ﬁltering for discretetime random processes can be handled in much the same way
that it is handled for continuous time random processes. An alternative approach can be based
on the use of whitening ﬁlters and linear innovations sequences. Both of these approaches will be
discussed in this section, but ﬁrst the topic of spectral factorization for discretetime processes is
discussed.
Spectral factorization for discrete time processes naturally involves z transforms. The z transform of a function (hk : k ∈ Z) is given by
∞ h(k )z −k H(z ) =
k=−∞ for z ∈ C. Setting z = ejω yields the Fourier transform: H (ω ) = H(ejω ) for 0 ≤ ω ≤ 2π . Thus, the
z transform H restricted to the unit circle in C is equivalent to the Fourier transform H on [0, 2π ],
and H(z) for other z ∈ C is an analytic continuation of its values on the unit circle.
Let h(k ) = h∗ (−k ) as before. Then the z transform of h is related to the z transform H of h as
follows:
∞ ∞ h(k )z
k=−∞ −k ∞
∗ h (−k )z =
k=−∞ −k ∗ ∞
∗ = ∗ −l l h (l )z =
l=−∞ 268 h(l)(1/z )
l=−∞ = H∗ (1/z ∗ ) The impulse response function h is called causal if h(k ) = 0 for k < 0. The z transform H is
said to be positive type if h is causal. Note that if H is positive type, then limz →∞ H(z ) = h(0).
The projection [H]+ is deﬁned as it was for Fourier transforms–it is the z transform of the function
u(k )h(k ), where u(k ) = I{k≥0} . (We will not need to deﬁne or use [ ]− for discrete time functions.)
If X is a discretetime WSS random process with correlation function RX , the z transform of
RX is denoted by SX . Similarly, if X and Y are jointly WSS then the z transform of RXY is
denoted by SXY . Recall that if Y is the output random process when X is passed through a linear
timeinvariant system with impulse response function h, then X and Y are jointly WSS and
RY X = h ∗ RX RXY = h ∗ RX RY = h ∗ h ∗ RX which in the z transform domain becomes:
SY X (z ) = H(z )SX (z ) SXY (z ) = H∗ (1/z ∗ )SX (z ) SY (z ) = H(z )H∗ (1/z ∗ )SX (z ) Example 9.5.1 Suppose Y is the output process when white noise W with RW (k ) = I{k=0} is
passed through a linear time invariant system with impulse response function h(k ) = ρk I{k≥0} ,
where ρ is a complex constant with ρ < 1. Let us ﬁnd H, SY , and RY . To begin,
∞ (ρ/z )k = H(z ) =
k=0 and the z transform of h is 1
1−ρ∗ z . 1
1 − ρ/z Note that the z transform for h converges absolutely for z  > ρ, whereas the z transform for h converges absolutely for z  < 1/ρ. Then
SY (z ) = H(z )H∗ (1/z ∗ )SX (z ) = 1
(1 − ρ/z )(1 − ρ∗ z ) The autocorrelation function RY can be found either in the time domain using RY = h ∗ h ∗ RW
or by inverting the z transform SY . Taking the later approach, factor out z and use the method of
partial fraction expansion to obtain
z
(z − ρ)(1 − ρ∗ z )
1
1
=z
+
2 )(z − ρ)
∗ ) − ρ)(1 − ρ∗ z )
(1 − ρ
((1/ρ
1
1
zρ∗
=
+
(1 − ρ2 ) 1 − ρ/z 1 − ρ∗ z SY (z ) = which is the z transform of
RY (k ) = ρk
1−ρ2
(ρ∗ )−k
1−ρ2 k≥0
k<0 The z transform SY of RY converges absolutely for ρ < z < 1/ρ.
269 Suppose that H(z ) is a rational function of z , meaning that it is a ratio of two polynomials
of z with complex coeﬃcients. We assume that the numerator and denominator have no zeros
in common, and that neither has a root on the unit circle. The function H is positive type (the
z transform of a causal function) if its poles (the zeros of its denominator polynomial) are inside
the unit circle in the complex plane. If H is positive type and if its zeros are also inside the unit
circle, then h and H are said to be minimum phase functions (in the time domain and z transform
domain, respectively). A positivetype, minimum phase function H has the property that both
H and its inverse 1/H are causal functions. Two linear timeinvariant systems in series, one with
transfer function H and one with transfer function 1/H, passes all signals. Thus if H is positive
type and minimum phase, we say that H is causal and causally invertible.
Assume that SY corresponds to a WSS random process Y and that SY is a rational function
with no poles or zeros on the unit circle in the complex plane. We shall investigate the symmetries
of SY , with an eye towards its factorization. First,
RY = RY so that ∗
SY (z ) = SY (1/z ∗ ) (9.13) ∗
Therefore, if z0 is a pole of SY with z0 = 0, then 1/z0 is also a pole. Similarly, if z0 is a zero of
∗
SY with z0 = 0, then 1/z0 is also a zero of SY . These observations imply that SY can be uniquely
factored as
−
+
SY (z ) = SY (z )SY (z ) such that for some constant β > 0:
+
• SY is a minimum phase, positive type z transform
−
+
• SY (z ) = (SY (1/z ∗ ))∗
+
• limz →∞ SY (z ) = β There is an additional symmetry if RY is realvalued:
∞ ∞ RY (k )z −k = SY (z ) =
k=−∞ ∗
(RY (k )(z ∗ )−k )∗ = SY (z ∗ ) (for realvalued RY ) (9.14) k=−∞ ∗
Therefore, if RY is real and if z0 is a nonzero pole of SY , then z0 is also a pole. Combining (9.13)
and (9.14) yields that if RY is real then the realvalued nonzero poles of SY come in pairs: z0 and
∗
∗
1/z0 , and the other nonzero poles of SY come in quadruples: z0 , z0 , 1/z0 , and 1/z0 . A similar
statement concerning the zeros of SY also holds true. Some example factorizations are as follows
(where ρ < 1 and β > 0): SY (z ) = β
β
1 − ρ/z 1 − ρ∗ z
+
SY (z ) SY (z ) = −
SY (z ) β (1 − .8/z )
β (1 − .8z )
(1 − .6/z )(1 .7/z ) (1 − .6z )(1 − .7z )
+
SY (z ) SY (z ) = −
SY (z ) β
β
(1 − ρ/z )(1 − ρ∗ /z ) (1 − ρz )(1 − ρ∗ z )
+
SY ( z ) 270 −
SY ( z ) An important application of spectral factorization is the generation of a discretetime WSS
random process with a speciﬁed correlation function RY . The idea is to start with a discretetime
white noise process W with RW (k ) = I{k=0} , or equivalently, with SW (z ) ≡ 1, and then pass it
through an appropriate linear, timeinvariant system. The appropriate ﬁlter is given by taking
+
H(z ) = SY (z ), for then the spectral density of the output is indeed given by
+
−
H(z )H∗ (1/z ∗ )SW (z ) = SY (z )SY (z ) = SY (z ) The spectral factorization can be used to solve the causal ﬁltering problem in discrete time.
Arguing just as in the continuous time case, we ﬁnd that if X and Y are jointly WSS random
processes, then the best estimator of Xn+T given (Yk : k ≤ n) having the form
∞ Yk h(n − k ) Xn+T n =
k=−∞ for a causal function h is the function h satisfying the WienerHopf equations (9.2) and (9.3), and
the z transform of the optimal h is given by
H= 1 z T SXY
+
−
SY
SY (9.15)
+ Finally, an alternative derivation of (9.15) is given, based on the use of a whitening ﬁlter. The
idea is the same as the idea of linear innovations sequence considered in Chapter 3. The ﬁrst step
is to notice that the causal estimation problem is particularly simple if the observation process is
white noise. Indeed, if the observed process Y is white noise with RY (k ) = I{k=0} then for each
k ≥ 0 the choice of h(k ) is simply made to minimize the mean square error when Xn+T is estimated
by the single term h(k )Yn−k . This gives h(k ) = RXY (T + k )I{k≥0} . Another way to get the same
result is to solve the WienerHopf equations (9.2) and (9.3) in discrete time in case RY (k ) = I{k=0} .
In general, of course, the observation process Y is not white, but the idea is to replace Y by an
equivalent observation process Z that is white.
Let Z be the result of passing Y through a ﬁlter with transfer function G (z ) = 1/S + (z ). Since
S + (z ) is a minimum phase function, G is a positive type function and the system is causal. Thus,
any random variable in the m.s. closure of the linear span of (Zk : k ≤ n) is also in the m.s. closure
of the linear span of (Yk : k ≤ n). Conversely, since Y can be recovered from Z by passing Z
through the causal linear timeinvariant system with transfer function S + (z ), any random variable
in the m.s. closure of the linear span of (Yk : k ≤ n) is also in the m.s. closure of the linear span
of (Zk : k ≤ n). Hence, the optimal causal linear estimator of Xn+T based on (Yk : k ≤ n) is equal
to the optimal causal linear estimator of Xn+T based on (Zk : k ≤ n). By the previous paragraph,
such estimator is obtained by passing Z through the linear timeinvariant system with impulse
response function RXZ (T + k )I{k≥0} , which has z transform [z T SXZ ]+ . See Figure 9.4.
The transfer function for two linear, timeinvariant systems in series is the product of their
z transforms. In addition,
SXZ (z ) = G ∗ (1/z ∗ )SXY (z ) =
271 SXY (z )
−
SY (z ) Y 1
+(z)
S Z Y ^
Xt+Tt [z TS (z)]+
XZ Figure 9.4: Optimal ﬁltering based on whitening ﬁrst.
Hence, the series system shown in Figure 9.4 is indeed equivalent to passing Y through the linear
time invariant system with H(z ) given by (9.15). Example 9.5.2 Suppose that X and N are discretetime mean zero WSS random processes such
that RXN = 0. Suppose SX (z ) = (1−ρ/z1 −ρz ) where 0 < ρ < 1, and suppose that N is a discrete)(1
time white noise with SN (z ) ≡ σ 2 and RN (k ) = σ 2 I{k=0} . Let the observed process Y be given
by Y = X + N . Let us ﬁnd the minimum mean square error linear estimator of Xn based on
(Yk : k ≤ n). We begin by factoring SY .
SY (z ) = SX (z ) + SN (z ) =
2 = z
+ σ2
(z − ρ)(1 − ρz ) −σ 2 ρ z 2 − ( 1+ρ +
ρ 1
)z
σ2 ρ +1 (z − ρ)(1 − ρz ) The quadratic expression in braces can be expressed as (z − z0 )(z − 1/z0 ), where z0 is the smaller
root of the expression in braces, yielding the factorization
SY (z ) = β (1 − z0 /z ) β (1 − z0 z )
(1 − ρ/z ) (1 − ρz ) where β2 = σ2ρ
z0 −
SY (z ) +
SY (z ) Using the fact SXY = SX , and appealing to a partial fraction expansion yields
SXY (z )
−
SY (z ) =
= 1
β (1 − ρ/z )(1 − z0 z )
1
z
+
β (1 − ρ/z )(1 − z0 ρ) β ((1/z0 ) − ρ)(1 − z0 z ) (9.16) The ﬁrst term in (9.16) is positive type, and the second term in (9.16) is the z transform of a
XY
.
function that is supported on the negative integers. Thus, the ﬁrst term is equal to SS −
Y +
Finally, dividing by SY yields that the z transform of the optimal ﬁlter is given by H(z ) = β 2 (1 or in the time domain
h(n) = 1
− z0 ρ)(1 − z0 /z )
n
z0 I{n≥0}
β 2 (1 − z0 ρ) 272 + 9.6 Problems 9.1 A quadratic predictor
Suppose X is a mean zero, stationary discretetime random process and that n is an integer with
n ≥ 1. Consider estimating Xn+1 by a nonlinear onestep predictor of the form
n Xn+1 = h0 + n j h1 (k )Xk + h2 (j, k )Xj Xk
j =1 k=1 k=1 (a) Find equations in term of the moments (second and higher, if needed) of X for the triple
(h0 , h1 , h2 ) to minimize the one step prediction error: E [(Xn+1 − Xn+1 )2 ].
(b) Explain how your answer to part (a) simpliﬁes if X is a Gaussian random process.
9.2 A smoothing problem
Suppose X and Y are mean zero, second order random processes in continuous time. Suppose the
MMSE estimator of X5 is to be found based on observation of (Yu : u ∈ [0, 3] ∪ [7, 10]). Assuming the
estimator takes the form of an integral, derive the optimality conditions that must be satisﬁed by
the kernal function (the function that Y is multiplied by before integrating). Use the orthogonality
principle.
9.3 A simple, noncausal estimation problem
Let X = (Xt : t ∈ R) be a real valued, stationary Gaussian process with mean zero and autocorrelation function RX (t) = A2 sinc(fo t), where A and fo are positive constants. Let N = (Nt : t ∈ R)
be a real valued Gaussian white noise process with RN (τ ) = σ 2 δ (τ ), which is independent of X .
∞
Deﬁne the random process Y = (Yt : t ∈ R) by Yt = Xt + Nt . Let Xt = −∞ h(t − s)Ys ds, where
2
the impulse response function h, which can be noncausal, is chosen to minimize E [Dt ] for each t,
where Dt = Xt − Xt . (a) Find h. (b) Identify the probability distribution of Dt , for t ﬁxed. (c)
Identify the conditional distribution of Dt given Yt , for t ﬁxed. (d) Identify the autocorrelation
function, RD , of the error process D, and the cross correlation function, RDY .
9.4 Interpolating a Gauss Markov process
Let X be a realvalued, mean zero stationary Gaussian process with RX (τ ) = e−τ  . Let a > 0.
Suppose X0 is estimated by X0 = c1 X−a + c2 Xa where the constants c1 and c2 are chosen to
minimize the mean square error (MSE).
(a) Use the orthogonality principle to ﬁnd c1 , c2 , and the resulting minimum MSE, E [(X0 − X0 )2 ].
(Your answers should depend only on a.)
(b) Use the orthogonality principle again to show that X0 as deﬁned above is the minimum MSE
estimator of X0 given (Xs : s ≥ a). (This implies that X has a twosided Markov property.)
9.5 Estimation of a ﬁltered narrowband random process in noise
Suppose X is a mean zero realvalued stationary Gaussian random process with the spectral density
shown.
S (2 π f)
X 8 Hz 1 8 Hz
f 10 4 Hz 10 273 4 Hz (a) Explain how X can be simulated on a computer using a pseudorandom number generator that
generates standard normal random variables. Try to use the minimum number per unit time. How
many normal random variables does your construction require per simulated unit time?
(b) Suppose X is passed through a linear timeinvariant system with approximate transfer function
H (2πf ) = 107 /(107 + f 2 ). Find an approximate numerical value for the power of the output.
(c) Let Zt = Xt + Wt where W is a Gaussian white noise random process, independent of X , with
RW (τ ) = δ (τ ). Find h to minimize the mean square error E [(Xt − Xt )2 ], where X = h ∗ Z .
(d) Find the mean square error for the estimator of part (c).
9.6 Proportional noise
Suppose X and N are second order, mean zero random processes such that RXN ≡ 0, and let
Y = X + N . Suppose the correlation functions RX and RN are known, and that RN = γ 2 RX
for some nonnegative constant γ 2 . Consider the problem of estimating Xt using a linear estimator
based on (Yu : a ≤ u ≤ b), where a, b, and t are given times with a < b.
(a) Use the orthogonality principle to show that if t ∈ [a, b], then the optimal estimator is given by
Xt = κYt for some constant κ, and identify the constant κ and the corresponding MSE.
(b) Suppose in addition that X and N are WSS and that Xt+T is to be estimated from (Ys : s ≤ t).
Show how the equation for the optimal causal ﬁlter reduces to your answer to part (a) in case
T ≤ 0.
(c) Continue under the assumptions of part (b), except consider T > 0. How is the optimal ﬁlter
for estimating Xt+T from (Ys : s ≤ t) related to the problem of predicting Xt+T from (Xs : s ≤ t)?
9.7 Predicting the future of a simple WSS process
Let X be a mean zero, WSS random process with power spectral density SX (ω ) = ω4 +131ω2 +36 .
+
+
(a) Find the positive type, minimum phase rational function SX such that SX (ω ) = SX (ω )2 .
(b) Let T be a ﬁxed known constant with T ≥ 0. Find Xt+T t , the MMSE linear estimator of Xt+T
given (Xs : s ≤ t). Be as explicit as possible. (Hint: Check that your answer is correct in case
T = 0 and in case T → ∞).
(c) Find the MSE for the optimal estimator of part (b).
9.8 Short answer ﬁltering questions
(a) Prove or disprove: If H is a positive type function then so is H 2 . (b) Prove or disprove: Suppose
X and Y are jointly WSS, mean zero random processes with continuous spectral densities such that
SX (2πf ) = 0 unless f  ∈[9012 MHz, 9015 MHz] and SY (2πf ) = 0 unless f  ∈[9022 MHz, 9025
MHz]. Then the best linear estimate of X0 given (Yt : t ∈ R) is 0. (c) Let H (2πf ) = sinc(f ). Find
[H ]+ .
9.9 On the MSE for causal estimation
Recall that if X and Y are jointly WSS and have power spectral densities, and if SY is rational with
a spectral factorization, then the mean square error for linear estimation of Xt+T using (Ys : s ≤ t)
is given by
∞ (MSE) = RX (0) −
−∞ ejωT SXY
−
SY 2 + dω
.
2π Evaluate and interpret the limits of this expression as T → −∞ and as T → ∞.
274 9.10 A singular estimation problem
Let Xt = Aej 2πfo t , where fo > 0 and A is a mean zero complex valued random variable with
2
2
E [A2 ] = 0 and E [A2 ] = σA . Let N be a white noise process with RN (τ ) = σN δ (τ ). Let
Yt = Xt + Nt . Let X denote the output process when Y is ﬁltered using the impulse response
function h(τ ) = αe−(α−j 2πfo )t I{t≥0} .
(a) Verify that X is a WSS periodic process, and ﬁnd its power spectral density (the power spectral
density only exists as a generalized function–i.e. there is a delta function in it).
(b) Give a simple expression for the output of the linear system when the input is X .
ˆ
(c) Find the mean square error, E [Xt − Xt 2 ]. How should the parameter α be chosen to approximately minimize the MSE?
9.11 Filtering a WSS signal plus noise
Suppose X and N are jointly WSS, mean zero, continuous time random processes with RXN ≡ 0.
The processes are the inputs to a system with the block diagram shown, for some transfer functions
K1 (ω ) and K2 (ω ):
X K1 + K2 Y=X out+Nout N Suppose that for every value of ω , Ki (ω ) = 0 for i = 1 and i = 2. Because the two subsystems are
linear, we can view the output process Y as the sum of two processes, Xout , due to the input X ,
plus Nout , due to the input N . Your answers to the ﬁrst four parts should be expressed in terms
of K1 , K2 , and the power spectral densities SX and SN .
(a) What is the power spectral density SY ?
(b) Find the signaltonoise ratio at the output (the power of Xout divided by the power of Nout ).
(c) Suppose Y is passed into a linear system with transfer function H , designed so that the output
at time t is Xt , the best linear estimator of Xt given (Ys : s ∈ R). Find H .
(d) Find the resulting minimum mean square error.
(e) The correct answer to part (d) (the minimum MSE) does not depend on the ﬁlter K2 . Why?
9.12 A prediction problem
Let X be a mean zero WSS random process with correlation function RX (τ ) = e−τ  . Using the
Wiener ﬁltering equations, ﬁnd the optimal linear MMSE estimator (i.e. predictor) of Xt+T based
on (Xs : s ≤ t), for a constant T > 0. Explain why your answer takes such a simple form.
9.13 Properties of a particular Gaussian process
Let X be a zeromean, widesense stationary Gaussian random process in continuous time with
autocorrelation function RX (τ ) = (1 + τ )e−τ  and power spectral density SX (ω ) = (2/(1 + ω 2 ))2 .
Answer the following questions, being sure to provide justiﬁcation.
(a) Is X mean ergodic in the m.s. sense?
(b) Is X a Markov process?
(c) Is X diﬀerentiable in the m.s. sense?
(d) Find the causal, minimum phase ﬁlter h (or its transform H ) such that if white noise with
autocorrelation function δ (τ ) is ﬁltered using h then the output autocorrelation function is RX .
(e) Express X as the solution of a stochastic diﬀerential equation driven by white noise.
275 9.14 Spectral decomposition and factorization
(a) Let x be the signal with Fourier transform given by x(2πf ) = sinc(100f )ej 2πf T + . Find the
energy of x for all real values of the constant T .
(b) Find the spectral factorization of the power spectral density S (ω ) = ω4 +161 2 +100 . (Hint: 1 + 3j
ω
is a pole of S .)
9.15 A continuoustime Wiener ﬁltering problem
Let (Xt ) and (Nt ) be uncorrelated, mean zero random processes with RX (t) = exp(−2t) and
SN (ω ) ≡ No /2 for a positive constant No . Suppose that Yt = Xt + Nt .
(a) Find the optimal (noncausal) ﬁlter for estimating Xt given (Ys : −∞ < s < +∞) and ﬁnd the
resulting mean square error. Comment on how the MMSE depends on No .
(b) Find the optimal causal ﬁlter with lead time T , that is, the Wiener ﬁlter for estimating Xt+T
given (Ys : −∞ < s ≤ t), and ﬁnd the corresponding MMSE. For simplicity you can assume that
T ≥ 0. Comment on the limiting value of the MMSE as T → ∞, as No → ∞, or as No → 0.
9.16 Estimation of a random signal, using the KL expansion
Suppose that X is a m.s. continuous, mean zero process over an interval [a, b], and suppose N is
a white noise process, with RXN ≡ 0 and RN (s, t) = σ 2 δ (s − t). Let (φk : k ≥ 1) be a complete
orthonormal basis for L2 [a, b] consisting of eigenfunctions of RX , and let (λk : k ≥ 1) denote the
corresponding eigenvalues. Suppose that Y = (Yt : a ≤ t ≤ b) is observed.
(a) Fix an index i. Express the MMSE estimator of (X, φi ) given Y in terms of the coordinates,
(Y, φ1 ), (Y, φ2 ), . . . of Y , and ﬁnd the corresponding mean square error.
(b) Now suppose f is a function in L2 [a, b]. Express the MMSE estimator of (X, f ) given Y in
terms of the coordinates ((f, φj ) : j ≥ 1) of f , the coordinates of Y , the λ’s, and σ . Also, ﬁnd the
mean square error.
9.17 Noiseless prediction of a baseband random process
Fix positive constants T and ωo , suppose X = (Xt : t ∈ R) is a baseband random process with
k
onesided frequency limit ωo , and let H (n) (ω ) = n=0 (jωT ) , which is a partial sum of the power
k
k!
(n) series of ejωT . Let Xt+T t denote the output at time t when X is passed through the linear time
(n) invariant system with transfer function H (n) . As the notation suggests, Xt+T t is an estimator (not
necessarily optimal) of Xt+T given (Xs : s ≤ t).
(n)
(a) Describe Xt+T t in terms of X in the time domain. Verify that the linear system is causal.
(b) Show that limn→∞ an = 0, where an = maxω≤ωo ejωT − H (n) (ω ). (This means that the power
series converges uniformly for ω ∈ [−ωo , ωo ].)
(c) Show that the mean square error can be made arbitrarily small by taking n suﬃciently large.
(n)
In other words, show that limn→∞ E [Xt+T − Xt+T t 2 ] = 0.
(d) Thus, the future of a narrowband random process X can be predicted perfectly from its past.
What is wrong with the following argument for general WSS processes? If X is an arbitrary WSS
random process, we could ﬁrst use a bank of (inﬁnitely many) narrowband ﬁlters to split X into
an equivalent set of narrowband random processes (call them “subprocesses”) which sum to X . By
the above, we can perfectly predict the future of each of the subprocesses from its past. So adding
together the predictions, would yield a perfect prediction of X from its past.
276 9.18 Linear innovations and spectral factorization
Suppose X is a discrete time WSS random process with mean zero. Suppose that the z transform
version of its power spectral density has the factorization as described in the notes: SX (z ) =
+
−
+
−
+
SX (z )SX (z ) such that SX (z ) is a minimum phase, positive type function, SX (z ) = (SX (1/z ∗ ))∗ ,
+
and limz →∞ SX (z ) = β for some β > 0. The linear innovations sequence of X is the sequence X
such that Xk = Xk − Xkk−1 , where Xkk−1 is the MMSE predictor of Xk given (Xl : l ≤ k − 1).
+
−
Note that there is no constant multiplying Xk in the deﬁnition of Xk . You should use SX (z ), SX (z ),
and/or β in giving your answers.
(a) Show that X can be obtained by passing X through a linear timeinvariant ﬁlter, and identify
the corresponding value of H.
(b) Identify the mean square prediction error, E [Xk − Xkk−1 2 ].
9.19 A singular nonlinear estimation problem
Suppose X is a standard Brownian motion with parameter σ 2 = 1 and suppose N is a Poisson
random process with rate λ = 10, which is independent of X . Let Y = (Yt : t ≥ 0) be deﬁned by
Yt = Xt + Nt .
(a) Find the optimal estimator of X1 among the estimators that are linear functions of (Yt : 0 ≤ t ≤
1) and the constants, and ﬁnd the corresponding mean square error. Your estimator can include a
constant plus a linear combination, or limits of linear combinations, of Yt : 0 ≤ t ≤ 1. (Hint: There
is a closely related problem elsewhere in this problem set.)
(b) Find the optimal possibly nonlinear estimator of X1 given (Yt : 0 ≤ t ≤ 1), and ﬁnd the
corresponding mean square error. (Hint: No computation is needed. Draw sample paths of the
processes.)
9.20 A discretetime Wiener ﬁltering problem
Extend the discretetime Wiener ﬁltering problem considered at the end of the notes to incorporate
a lead time T . Assume T to be integer valued. Identify the optimal ﬁlter in both the z transform
domain and in the time domain. (Hint: Treat the case T ≤ 0 separately. You need not identify the
covariance of error.)
9.21 Causal estimation of a channel input process
Let X = (Xt : t ∈ R) and N = (Nt : t ∈ R) denote WSS random processes with RX (τ ) = 3 e−τ 
2
and RN (τ ) = δ (τ ). Think of X as an input signal and N as noise, and suppose X and N are
orthogonal to each other. Let k denote the impulse response function given by k (τ ) = 2e−3τ I{τ ≥0} ,
and suppose an output process Y is generated according to the block diagram shown:
X k + Y N That is, Y = X ∗ k + N . Suppose Xt is to be estimated by passing Y through a causal ﬁlter with
impulse response function h, and transfer function H . Find the choice of H and h in order to
minimize the mean square error.
9.22 Estimation given a strongly correlated process
Suppose g and k are minimum phase causal functions in discretetime, with g (0) = k (0) = 1, and
z transforms G and K. Let W = (Wk : k ∈ Z) be a mean zero WSS process with SW (ω ) ≡ 1, let
277 Xn = ∞ −∞ g (n − i)Wi and Yn = ∞ −∞ k (n − i)Wi .
i=
i=
(a) Express RX , RY , RXY , SX , SY , and SXY in terms of g , k , G , K.
(b) Find h so that Xnn = ∞ −∞ Yi h(n − i) is the MMSE linear estimator of Xn given (Yi : i ≤ n).
i=
(c) Find the resulting mean square error. Give an intuitive reason for your answer.
9.23 Estimation of a process with raised cosine spectrum
Suppose Y = X + N, where X and N are independent, mean zero, WSS random processes with
SX (ω ) = (1 + cos( πω ))
ωo
2 I{ω≤ωo } and SN (ω ) = No
2 where No > 0 and ωo > 0. (a) Find the transfer function H for the ﬁlter such that if the input
process is Y , the output process, X , is such that X is the optimal linear estimator of Xt based on
(Ys : s ∈ R).
2
(b) Express the mean square error, σe = E [(Xt − Xt )2 ], as an integral in the frequency domain.
(You needn’t carry out the integration.)
(c) Describe the limits of your answers to (a) and (b) as No → 0.
(c) Describe the limits of your answers to (a) and (b) as No → ∞.
9.24 * Resolution of Wiener and Kalman ﬁltering
Consider the state and observation models:
Xn = F Xn−1 + Wn
Yn = H T Xn + Vn
where (Wn : −∞ < n < +∞) and (Vn : −∞ < n < +∞) are independent vectorvalued random
sequences of independent, identically distributed mean zero random variables. Let ΣW and ΣV
denote the respective covariance matrices of Wn and Vn . (F , H and the covariance matrices must
satisfy a stability condition. Can you ﬁnd it? ) (a) What are the autocorrelation function RX and
crosscorrelation function RXY ?
(b) Use the orthogonality principle to derive conditions for the causal ﬁlter h that minimizes
E [ Xn+1 − ∞ h(j )Yn−j 2 ]. (i.e. derive the basic equations for the WienerHopf method.)
j =0
(c) Write down and solve the equations for the Kalman predictor in steady state to derive an
expression for h, and verify that it satisﬁes the orthogonality conditions. 278 Chapter 10 Martingales
10.1 Conditional expectation revisited The general deﬁniton of a martingale requires the general deﬁnition of conditional expectation. We
begin by reviewing the deﬁnitions we have given so far. In Chapter 1 we reviewed the following
elementary deﬁnition of E [X Y ]. If X and Y are both discrete random variables,
E [X Y = i] = jP [X = j Y = i],
j which is well deﬁned if P {Y = i} > 0 and either the sum restricted to j > 0 or to j < 0 is
convergent. That is, E [X Y = i] is the mean of the conditional pmf of X given Y = i. Note that
g (i) = E [X Y = i] is a function of i. Let E [X Y ] be the random variable deﬁned by E [X Y ] = g (Y ).
Similarly, if X and Y have a joint pdf, E [X Y = y ] = xfX Y (xy )dx = g (y ) and E [X Y ] = g (Y ).
Chapter 3 showed that E [X Y ] could be deﬁned whenever E [X 2 ] < ∞, even if X and Y are
neither discrete random variables nor have a joint pdf. The deﬁnition is based on a projection,
characterized by the orthogonality principle. Speciﬁcally, if E [X 2 ] < ∞, then E [X Y  is the unique
random variable such that:
• it has the form g (Y ) for some (Borel measurable) function g such that E [(g (Y )2 ] < ∞, and
• E [(X − E [X Y ])f (Y )] = 0 for all (Borel measurable) functions f such that E [(f (Y ))2 ] < ∞.
That is, E [X Y ] is an unconstrained estimator based on Y, such that the error X − E [X Y ] is
orthogonal to all other unconstrained estimators based on Y. By the orthogonality principle, E [X Y ]
exists and is unique, if diﬀerences on a set of probability zero are ignored. This second deﬁnition
of E [X Y ] is more general than the elementary deﬁnition, because it doesn’t require X and Y to
be discrete or to have a joint pdf, but it is less general because it requires that E [X 2 ] < ∞.
The deﬁnition of E [X Y ] given next generalizes the previously given deﬁnition in two ways.
First, the deﬁnition applies as long as E [X ] < ∞, which is a weaker requirement than E [X 2 ] < ∞.
Second, the deﬁnition is based on having information represented by a σ algebra, rather than by a
random variable. Recall that, by deﬁnition, a σ algebra D for a set Ω is a set of subsets of Ω such
that:
(a) Ω ∈ D,
279 (b) if A ∈ D then Ac ∈ D,
(c) if A, B ∈ D then AB ∈ D, and more generally, if A1 , A2 , ... is such that Ai ∈ D for i ≥ 1, then
∪∞ Ai ∈ D.
i=1
In particular, the set of events, F , in a probability space (Ω, F , P ), is required to be a σ algebra.
The original motivation for introducing F in this context was a technical one, related to the
impossibility of extending P to be deﬁned on all subsets of Ω, for important examples such as
Ω = [0, 1] and P {(a, b)} = b − a for all intervals (a, b). However, σ algebras are also useful for
representing information available to an observer. We call D a subσ algebra of F if D is a σ algebra such that D ⊂ F . A random variable Z is said to be Dmeasurable if {Z ≤ c} ⊂ D for
all c. By deﬁnition, random variables are functions on Ω that are F measurable. The smaller the
σ algebra D is, the fewer the set of D measurable random variables. In practice, subσ algebras are
usually generated by collections of random variables:
Deﬁnition 10.1.1 The σ algebra generated by a collection of random variables (Yi : i ∈ I ), denoted
by σ (Yi : i ∈ I ), is the smallest σ algebra containing all sets of the form {Yi ≤ c}.1 The σ algebra
generated by a single random variable Y is denoted by σ (Y ), and sometimes as F Y .
An equivalent deﬁnition would be that σ (Yi : i ∈ I ) is the smallest σ algebra such that each Yi is
measurable with respect to it.
A subσ algebra of F represents knowledge about the probability experiment modeled by the
probability space (Ω, F , P ). In Chapter 3, the information gained from observing a random variable
Y was modeled by requiring estimators to be random variables of the form g (Y ), for a Borel
measurable function g . An equivalent condition would be to allow any estimator that is a σ (Y )measurable random variable. That is, as shown in a starred homework problem, if Y and Z are
random variables on the same probability space, then Z = g (Y ) for some Borel measurable function
g if and only if Z is σ (Y ) measurable. Using subσ algebras is more general, because some σ algebras
on some probability spaces are not generated by random variables. Using σ algebras to represent
information also works better when there is an uncountably inﬁnite number of observations, such
as observation of a continuous random process over an interval of time. But in engineering practice,
the main diﬀerence between the two ways to model information is simply a matter of notation.
Example 10.1.2 (The trivial σ algebra) Let (Ω, F , P ) be a probability space. Suppose X is a
random variable such that, for some constant co , X (ω ) = co for all ω ∈ Ω. Then X is measurable
with respect to the trivial σ algebra D deﬁned by D = {∅, Ω}. That is, constant random variables
are {∅, Ω}measurable.
Conversely, suppose Y is a {∅, Ω}measurable random variable. Select an arbitrary ωo ∈ Ω and
let co = Y (ωo ). On one hand, {ω : Y (ω ) ≤ c} can’t be empty for c ≥ co , so {ω : Y (ω ) ≤ c} = Ω for
c ≥ co . On the other hand, {ω : Y (ω ) ≤ co } doesn’t contain ωo for c < co , so {ω : Y (ω ) ≤ co } = ∅ for
c < co . Therefore, Y (ω ) = co for all ω. That is, {∅, Ω}measurable random variables are constant.
Deﬁnition 10.1.3 If X is a random variable on (Ω, F , P ) with ﬁnite mean and D is a subσ algebra of F , the conditional expectation of X given D, E [X D], is the unique (two versions equal
with probability one are considered to be the same) random variable on (Ω, F , P ) such that
1 The smallest one exists–it is equal to the intersection of all σ algebras which contain all sets of the form {Yi ≤ c}. 280 (i) E [X D] is Dmeasurable
(ii) E [(X − E [X D])ID ] = 0 for all D ∈ D. (Here ID is the indicator function of D).
We remark that a possible choice of D in property (ii) of the deﬁnition is D = Ω, so E [X D]
should satisfy E [X − E [X D]] = 0, or equivalently, since E [X ] is assumed to be ﬁnite, E [X ] =
E [E [X D]]. In particular, an implication of the deﬁnition is that E [X D] also has a ﬁnite mean.
Proposition 10.1.4 Deﬁnition 10.1.3 is well posed. Speciﬁcally, there exits a random variable
satisfying conditions (i) and (ii), and it is unique.
Proof. (Uniqueness) Suppose U and V are each Dmeasurable random variables such that
E [(X − U )ID ] = 0 and E [(X − V )ID ] = 0 for all D ∈ D. It follows that E [(U − V )ID ] = E [(X −
V )ID ] − E [(X − U )ID ] = 0 for any D ∈ D. A possible choice of D is {U > V }, so E [(U − V )I{U >V } ] =
0. Since (U − V )I{U >V } is nonnegative and is strictly positive on the event {U > V }, it must be
that P {U > V } = 0. Similarly, P {U < V } = 0. So P {U = V } = 1.
(Existence) Existence is ﬁrst proved under the added assumption that P {X ≥ 0} = 1. Let
L2 (D) be the space of Dmeasurable random variables with ﬁnite second moments. Then D is
a closed, linear subspace of L2 (Ω, F , P ), so the orthogonality principle can be applied. For any
n ≥ 0, the random variable X ∧ n is bounded and thus has a ﬁnite second moment. Let Xn be the
projection of X ∧ n onto L2 (D). Then by the orthogonality principle, X ∧ n − Xn is orthogonal
to any random variable in L2 (D). In particular, X ∧ n − Xn is orthogonal to ID for any D ∈ D.
Therefore, E [(X ∧ n − Xn )ID ] = 0 for all D ∈ D. Equivalently,
E [(X ∧ n)ID ] = E [Xn ID ]. (10.1) The next step is to take a limit as n → ∞. Since E [(X ∧ n)ID ] is nondecreasing in n for each
D ∈ D, the same is true of E [Xn ID ]. Thus, for any n ≥ 0, E [(Xn+1 − Xn )ID ] ≥ 0 for any D ∈ D.
Taking D = {Xn+1 − Xn < 0} implies that P {Xn+1 ≥ Xn } = 1. Therefore, the sequence (Xn )
converges a.s., and we denote the limit by X∞ . We show that X∞ satisﬁes the two properties,
(i) and (ii), required of E [X D]. First, X∞ is Dmeasurable because it is the limit of a sequence
of Dmeasurable random variables. Secondly, for any D ∈ D, the sequences or random variables
(X ∧ n)ID and Xn ID are a.s. nondecreasing and nonnegative, so by the monotone convergence
theorem (Theorem 11.6.6) and (10.1):
E [XID ] = lim E [(X ∧ n)ID ] = lim E [Xn ID ] = E [X∞ ID ].
n→∞ n→∞ So property (ii), E [(X − X∞ )ID ] = 0, is also satisﬁed. Existence is proved in case P {X ≥ 0} = 1.
For the general case, X can be represented as X = X+ − X− , where X+ and X− are nonnegative
with ﬁnite means. By the case already proved, E [X+ D] and E [X− D] exist, and, of course, they
satisfy conditions (i) and (ii) in Deﬁnition 10.1.3. Therefore, with E [X D] = E [X+ D] − E [X− D],
it is a simple matter to check that E ]X D] also satisﬁes conditions (i) and (ii), as required.
Proposition 10.1.5 Let X and Y be random variables on (Ω, F , P ) and let A and D be subσ algebras of F .
1. (Consistency with deﬁnition based on projection) If E [X 2 ] < ∞ and
V = {g (Y ) : g is Borel measurable such that E [g (Y )2 ] < ∞}, then E [X Y ], deﬁned as the
MMSE projection of X onto V (also written as ΠV (X )) is equal to E [X σ (Y )].
281 2. (Linearity) If E [X ] and E [Y ] are ﬁnite, then aE [X D] + bE [Y D] = E [aX + bY D].
3. (Tower property) If E [X ] is ﬁnite and A ⊂ D ⊂ F , then E [E [X D]A] = E [X A]. (In
particular, E [E [X D]] = E [X ].)
4. (Positivity preserving) If E [X ] is ﬁnite and X ≥ 0 a.s. then E [X D] ≥ 0 a.s.
5. (L1 contraction property) E [E X D]] ≤ E [X ].
6. (L1 continuity) If E [Xn ] is ﬁnite for all n and E [Xn − X∞ ] → 0, then
E [E [Xn D] − E [X∞ D]] → 0.
7. (Pull out property) If X is Dmeasurable and E [XY ] and E [Y ] are ﬁnite, then E [XY D] =
XE [Y D].
Proof. (Consistency with deﬁnition based on projection) Suppose X and V are as in part 1.
Then, by deﬁnition, E [X Y ] ∈ V and E [(X − E [X Y ])Z ] = 0 for any Z ∈ V . As mentioned above,
a random variable has the form g (Y ) if and only if it is σ (Y )measurable. In particular, V is simply
the set of σ (Y )measurable random variables Z such that E [Z 2 ] < ∞. Thus, E [X Y ] is σ (Y )
measurable, and E [(X − E [X Y ])Z ] = 0 for any σ (Y )measurable random variable Z such that
E [Z 2 ] < ∞. As a special case, E [(X − E [X Y ])ID ] = 0 for any D ∈ σ (Y ). Thus, E [X Y ] satisﬁes
conditions (i) and (ii) in Deﬁnition 10.1.3 of E [X σ (Y )]. So E [X Y ] = E [X σ (Y )].
(Linearity Property) (This is similar to the proof of linearity for projections, Proposition 3.2.2.)
It suﬃces to check that the linear combination aE [X D] + bE [Y D] satisﬁes the two conditions that
deﬁne E [aX + bY D]. First, E [X D] and E [Y D] are both D measurable, so their linear combination
is also Dmeasurable. Secondly, if D ∈ D, then E [(X − E [X D])ID ] = E [(Y − E [Y D])ID ] = 0,
from which if follows that
E [(aX + bY − E [aX + bY D]) ID ] = aE [(X − E [X D])ID ] + bE [(Y − E [Y D])ID ] = 0.
Therefore, aE [X D] + bE [Y D] = E [aX + bY D].
(Tower Property) (This is similar to the proof of Proposition 3.2.3, about projections onto nested
subspaces.) It suﬃces to check that E [E [X D]A] satisﬁes the two conditions that deﬁne E [X A].
First, E [E [X D]A] itself is a conditional expectation given A, so it is A measurable. Second, let
D ∈ A. Now X − E [E [X D] = (X − E [X D]) + (E [X D] − E [E [X D]), and (use the fact D ∈ D):
E [(X − E [X D])ID ] and E [(E [X D] − E [E [X D])ID ] = 0. Adding these last two equations yields
E [(X − E [E [X )D)ID ] = 0. Therefore, E [E [X D]A] = E [X A].
The proofs of linearity and tower property are nearly identical to the proofs of the same properties for projections (Propositions 3.2.2 and 3.2.3 ) and are left to the reader.
(Positivity preserving) Suppose E [X ] is ﬁnite and X ≥ 0 a.s. Let D = {E [X D] < 0}. Then
D ∈ D because E [X D] is Dmeasurable. So E [E [X D]ID ] = E [XID ] ≥ 0, while P {E [X D]ID ≤
0} = 1. Hence, P {E [X D]ID = 0} = 1, which is to say that E [X D] ≥ 0 a.s.
(L1 contraction property) (This property is a special case of the conditional version of Jensen’s
inequality, established in a starred homework problem. Here a diﬀerent proof is given.) The
variable X can be represented as X = X+ − X− , where X+ is the positive part of X and X− is
the negative part of X , given by X+ = X ∨ 0 and X− = (−X ) ∨ 0. Since X is assumed to have a
ﬁnite mean, the same is true of X± . Moreover, E [E [X± D]] = E [X± ], and by the linearity property,
E [X D] = E [X+ D] − E [X− D]. By the positivity preserving property, E [X+ D] and E [X− D] are
282 both nonnegative a.s., so E [X+ D] + E [X− D] ≥ E [X+ D] − E [X− D] a.s. (The inequality is strict
for ω such that both E [X+ D] and E [X− D] are strictly positive.) Therefore,
E [X ] = E [X+ ] + E [X− ]
= E [E [X+ D] + E [X− D]]
≥ E [E [X+ D] − E [X− D]]
= E [E [X D],
and the L1 contraction property is proved.
(L1 continuity) Since for any n, X∞  ≤ Xn  + Xn − X∞ , the hypotheses imply that X∞
has a ﬁnite mean. By linearity and the L1 contraction property, E [E [Xn D] − E [X∞ D]] =
E [E [Xn − X∞ D]] ≤ E [E [Xn − X∞ ]], which implies the L1 continuity property.
(Pull out property) The pull out property will be proved ﬁrst under the added assumption that
X and Y are nonnegative random variables. Clearly XE [Y D] is D measurable. Let D ∈ D. It
remains to show that
E [XY ID ] = E [XE [Y D]ID ].
(10.2)
If X has the form ID1 for D1 ∈ D then (10.2) becomes E [Y ID∩D1 ] = E [E [Y D]ID∩D1 ], which
holds by the deﬁnition of E [Y D] and the fact D ∩ D1 ∈ D. Equation (10.2) is thus also true if
X is a ﬁnite linear combination of random variables of the form ID1 , that is, if X is a simple Dmeasurable random variables. Then X is the a.s. limit of a nondecreasing sequence of nonnegative
simple random variables Xn . Now (10.2) holds for X replaced by Xn :
E [Xn Y ID ] = E [Xn E [Y D]ID ]. (10.3) Also, Xn Y ID is a nondecreasing sequence converging to XY ID a.s., and Xn E [Y D]ID is a nondecreasing sequence converging to XE [Y D]ID a.s. By the monotone convergence theorem, taking
n → ∞ on both sides of (10.3), yields (10.2). This proves the pull out property under the added
assumption that X and Y are nonnegative.
In the general case, X = X+ − X− , where X+ = X ∨ 0 and X− = (−X ) ∨ 0, and similarly Y =
Y+ − Y− . The hypotheses imply E [X± Y± ] and E [Y± ] are ﬁnite so that E [X± Y± D] = X± E [Y± D],
and therefore
E [X± Y± ID ] = E [X± E [Y± D]ID ],
(10.4)
where in these equations, the sign on both appearances of X should be the same, and the sign on
both appearances of Y should be the same. The left side of (10.2) can be expressed as a linear
combination of terms of the form E [X± Y± ID ]:
E [XY ID ] = E [X+ Y+ ID ] − E [X+ Y− ID ] − E [X− Y+ ID ] + E [X− Y− ID ].
Similarly, the right side of (10.2) can be expressed as a linear combination of terms of the form
E [X± E [Y± D]ID ]. Therefore, (10.2) follows from (10.4). 10.2 Martingales with respect to ﬁltrations A ﬁltration of a σ algebra F is a sequence of subσ algebras F = (Fn : n ≥ 0) of F , such that
Fn ⊂ Fn+1 for n ≥ 0. If Y = (Yn : n ≥ 0) or Y = (Yn : n ≥ 1) is a sequence of random variables on
283 Y
(Ω, F , P ), the ﬁltration generated by Y , often written as F Y = (Fn : n ≥ 0), is deﬁned by letting
Y
Y
Fn = σ (Yk : k ≤ n). (If there is no variable Y0 deﬁned, we take F0 to be the trivial σ algebra,
Y = {∅, Ω}, representing no observations.)
F0
In practice, a ﬁltration represents an sequence of observations or measurements. If the ﬁltration
is generated by a random process, then the information available at time n is represents observation
of the random process up to time n.
A random process (Xn : n ≥ 0) is adapted to a ﬁltration F if Xn is Fn measurable for each
n ≥ 0. Deﬁnition 10.2.1 Let (Ω, F , P ) be a probability space with a ﬁltration F = (Fn : n ≥ 0). Let
Y = (Yn : n ≥ 0) be a sequence of random variables adapted to F . Then Y is a martingale with
respect to F if for all n ≥ 0:
(0) Yn is Fn measurable (i.e. the process Y is adapted to F )
(i) E [Yn ] < ∞,
(ii) E [Yn+1 Fn ] = Yn a.s.
Similarly, Y is a submartingale relative to F if (i) and (ii) are true and E [Yn+1 Fn ] ≥ Yn , a.s.,
and Y is a supermartingale relative to F if (i) and (ii) are true and E [Yn+1 Fn ] ≤ Yn a.s.
Some comments are in order. Note the condition (ii) in the deﬁnition of a margingale implies
condition (0), because conditional expectations with respect to a σ algebra are random variables
measurable with respect to the σ algebra.
Note that if Y = (Yn : n ≥ 0) is a martingale with respect to a ﬁltration F = (Fn : n ≥ 0),
then Y is also a martingale with respect to the ﬁltration generated by Y itself. Indeed, for each n,
Y
Yn is Fn measurable, whereas Fn is the smallest σ algebra with respect to which Yn is measurable,
Y ⊂ F . Therefore, the tower property of conditional expectation, the fact Y is a margtingale
so Fn
n
Y
with respect to F , and the fact Yn is Fn measurable, imply
Y
Y
Y
E [Yn+1 Fn ] = E [E [Yn+1 Fn ]Fn ] = E [Yn Fn ] = Yn . Thus, in practice, if Y is said to be a martingale and no ﬁltration F is speciﬁed, at least Y is a
martingale with respect to the ﬁltration it generates.
Note that if Y is a martingale with respect to a ﬁltration F , then for any n, k ≥ 0,
E [Yn+k+1 Fn ] = E [E [Yn+k+1 Fn+k Fn ] = E [Yn+k Fn ]
Therefore, by induction on k for n ﬁxed:
E [Xn+k Fn ] = Xn , (10.5) for n, k ≥ 0.
Example 10.2.2 Suppose (Ui : i ≥ 1) is a collection of independent random variables, each with
mean zero. Let S0 = 0 and for n ≥ 1, Sn = n Ui . Let F = (Fn : n ≥ 0) denote the ﬁltration
i=1
generated by S : Fn = σ (S0 , . . . , Sn ). Equivalently, F is the ﬁltration generated by (Ui : i ≥ 1):
F0 = {∅, Ω} and Fn = σ (S0 , . . . , Sn ). for n ≥ 1. Then S = (Sn : n ≥ 0) is a martingale with respect
to F :
E [Sn+1 Fn ] = E [Un+1 Fn ] + E [Sn+1 Fn ] = 0 + Sn = Sn .
284 Example 10.2.3 Suppose S = (Sn : n ≥ 0) and F = (Fn : n ≥ 0) are deﬁned as in Example
10.2.2 in terms of a sequence of independent random variables U = (Ui : i ≥ 1). Suppose in
2
addition that Var(Ui ) = σ 2 for some ﬁnite constant σ 2 . Finally, let Mn = Sn − nσ 2 for n ≥ 0. Then
M = (Mn : n ≥ 0) is a martingale relative to F . Indeed, M is adapted to F . Since Sn+1 = Sn + Un ,
2
we have Mn+1 = Mn + 2Sn Un+1 + Un+1 − σ 2 so that
2
E [Mn+1 Fn ] = E [Mn Fn ] + 2Sn E [Un Fn ]] + E [Un − σ 2 Fn ]]
2
= Mn + 2Sn E [Un ] + E [Un − σ 2 ] = Mn Example 10.2.4 M (n) = eθSn /E [eθX1 ]n in case X ’s are iid. Example 10.2.5 (Branching process) Let Gn denote the number of individuals in the nth generation. Suppose the number of oﬀspring per individual be represented by a random variable X .
Select a > 0 so that E [aX ] = 1. Then, aGn is a martingale. Example 10.2.6 (Cumulative innovations process) Let Mn = Yn − n−1
k=0 E [Yk+1 −Yk Y0 , · · · , Yn−1 ]. Example 10.2.7 (Doob martingale) Yn = E [ΦFn ]. For example, n nodes, m edges in a graph,
Xi = I{ith edge exists} and the ﬁltration is generated by X0 , X1 , · · · . Deﬁnition 10.2.8 A martingale diﬀerence sequence (Dn : n ≥ 1) relative to a ﬁltration F = (Fn :
n ≥ 0) is a sequence of random variables (Dn : n ≥ 1) such that
(0) (Dn : n ≥ 1) is adapted to F (i.e. Dn is Fn measurable for each n ≥ 1)
(i) E [Dn ] < ∞ for n ≥ 1
(ii) E [Dn+1 Fn ] = 0 a.s. for all n ≥ 0.
Equivalently, (Dn : n ≥ 1) has the form Dn = Mn − Mn−1 for n ≥ 1, for some (Mn : n ≥ 0) which
is a martingale with respect to F .
Deﬁnition 10.2.9 A random process (Hn : n ≥ 1) is said to be predictable with respect to a
ﬁltration F = (Fn : n ≥ 0) if Hn is Fn−1 measurable for all n ≥ 1. (Sometimes this is called
“onestep” predictable, because Fn determines H one step ahead.)
285 Example 10.2.10 Suppose (Dn : n ≥ 1) is a martingale diﬀerence sequence and (Hk : k ≥ 1)
is predictable, both relative to a ﬁltration F = (Fn : n ≥ 0). We claim that the new process
D = (Dn : n ≥ 1) deﬁned by Dn = Hn Dn is also a martingale diﬀerence sequence with respect to
F . Indeed, it is adapted, has ﬁnite means, and
E [Hn+1 Dn+1 Fn ] = Hn+1 E [Dn+1 Fn ] = 0,
where we pulled out the Fn measurable random variable Hn+1 from the conditional expectation
given Fn . An interpretation is that Dn is the net gain to a gambler if one dollar is staked on the
outcome of a fair game in round n, and so Hn Dn is the net stake if Hn dollars are staked on round
n. The requirement that (Hk : k ≥ 1) be predictable means that the gambler must decide how
much to stake in round n based only on information available at the end of round n − 1. If would
be an unfair advantage if the gambler already knew Dn when deciding how much money to stake
in round n.
If the initial reserves of the gambler were some constant M0 , then the reserves of the gambler
after n rounds would be given by:
n Hk D k Mn = M0 +
k=1 Then (Mn : n ≥ 0) is a margingale with respect to F . The random variables are Hk Dk , 1 ≤ k ≤ n
2
2
are orthogonal. Also, E [(Hk Dk )2 ] = E [E [(Hk Dk )2 Fk−1 ]] = E [Hk E [Dk Fk−1 ]]. Therefore,
n
2
2
E [Hk E [Dk Fk−1 ]]. 2 E [(Mn − M0 ) ] =
k=1 10.3 AzumaHoeﬀding inequaltity One of the most simple inequalities for martingales is the AzumaHoeﬀding inequality. It is proven
in this section, and applications to prove concentration inequalities for some combinatorial problems
are given.2
Lemma 10.3.1 Suppose D is a random variable with E [D] = 0 and P {D − b ≤ d} = 1 for some
2
constant b. Then for any α ∈ R, E [eαD ] ≤ e(αd) /2 .
Proof. Since D has mean zero and D lies in the interval [b − d, b + d] with probability one, the
interval must contain zero, so b ≤ d. To avoid trivial cases we assume that b < d. Since eαx is
convex in x, the value of eαx for x ∈ [b − d, b + d] is bounded above by the linear function that is
equal to eαx at the endpoints, x = b ± d, of the interval:
eαx ≤
2 x − b + d α(b+d) b + d − x α(b−d)
e
+
e
.
2d
2d See McDiarmid survey paper 286 (10.6) Since D lies in that interval with probability one, (10.6) remains true if x is replaced by the random
variable D. Taking expectations on both sides and using E [D] = 0 yields
E [eαD ] ≤ d − b α(b+d) b + d α(b−d)
e
+
e
.
2d
2d (10.7)
2 The proof is completed by showing that the right side of (10.7) is less than or equal to e(αd) /2 for
2
any b < d. Letting u = αd and θ = b/d, the inequality to be proved becomes f (u) ≤ eu /2 , for
u ∈ R and θ < 1, where
f (u) = ln (1 − θ)eu(1+θ) + (1 + θ)eu(−1+θ)
2 . 2 v
Taylor’s formula implies that f (u) = f (0)+ f (0)u + f (2 )u for some v in the interval with endpoints
0 and u. Elementary, but somewhat tedious, calculations show that f (u) = (1 − θ2 )(eu − e−u )
(1 − θ)eu + (1 + θ)e−u and
f (u) =
= 4(1 − θ2 )
[(1 − θ)eu + (1 + θ)e−u ]2
1
,
2
cosh (u + β ) where β = 1 ln( 1−θ ). Note that f (0) = f (0) = 0, and f (u) ≤ 1 for all u ∈ R. Therefore,
2
1+θ
f (u) ≤ u2 /2 for all u ∈ R, as was to be shown.
Deﬁnition 10.3.2 A random process (Bn : n ≥ 0) is said to be predictable with respect to a
ﬁltration F = (Fn : n ≥ 0) if B0 is a constant and Bn is Fn−1 measurable for all n ≥ 1.
Proposition 10.3.3 (AzumaHoeﬀding inequality with centering) Let (Yn : n ≥ 0) be a martingale
and (Bn : n ≥ 0) be a predictable process, both with respect to a ﬁltration F = (Fn : n ≥ 0), such
that P [Yn+1 − Bn+1  ≤ dn ] = 1 for all n ≥ 0. Then
P {Yn − Y0  ≥ λ} ≤ 2 exp − λ2
n
2
i=1 di 2 . Proof. Let n ≥ 0. The idea is to write Yn = Yn − Yn−1 + Yn−1 , to use the tower property of
conditional expectation, and to apply Lemma 10.3.1 to the random variable Yn − Yn−1 for d = dn .
This yields:
E [eα(Yn −Y0 ) ] = E [E [eα(Yn −Yn−1 +Yn−1 −Y0 ) Fn−1 ]]
= E [eα(Yn−1 −Y0 ) E [eα(Yn −Yn−1 Fn−1 ]]
2 /2 ≤ E [eα(Yn−1 −Y0 ) ]e(αdn ) 287 . Thus, by induction on n,
E [eα(Yn −Y0 ) ] ≤ e(α 2 /2) Pn i=1 d2
i . The remainder of the proof is essentially the Chernoﬀ inequality:
P {Yn − Y0 ≥ λ} ≤ E [eα(Yn −Y0 −λ) ] ≤ e(α 2 /2) Pn Finally, taking α to make this bound as tight as possible, i.e. α =
P {Yn − Y0 ≥ λ} ≤ exp − λ2
2 n
2
i=1 di i=1 2 d2 −αλ
i λ
Pn i=1 d2
i . , yields . Similarly, P {Yn − Y0 ≤ −λ} satisﬁes the same bound because the previous bound applies for (Yn )
replaced by (−Yn ), yielding the proposition.
Deﬁnition 10.3.4 A function f of n variables x1 , . . . , xn is said to satisfy the Lipschitz condition
with constant c if f (x1 , . . . , xn ) − f (x1 , . . . , xn−1 , yi , xi+1 , . . . , xn ) ≤ c for any x1 , . . . , xn , i, and
yi .3
Proposition 10.3.5 (McDiarmid’s inequality) Suppose F = f (X1 , . . . , Xn ), where f satisﬁes the
Lipschitz condition with constant c, and X1 , . . . , Xn are independent random variables. Then
2
P {F − E [F ] ≥ λ} ≤ 2 exp(− 2λ2 ).
nc
X
Proof. Let (Zk : 0 ≤ k ≤ n) denote the Doob martingale deﬁned by Zk = E [F Fk ], where,
X
X
as usual, Fk = σ (Xk : 1 ≤ k ≤ n) is the ﬁltration generated by (Xk ). Note that F0 is the trivial
σ algebra {∅, Ω}, corresponding to no observations, so Z0 = E [F ]. Also, Zn = F . In words, Zk is
the expected value of F , given that the ﬁrst k X ’s are revealed.
For 0 ≤ k ≤ n − 1, let gk (x1 , . . . , xk , xk+1 ) = E [f (x1 , . . . , xk+1 , Xk+2 , . . . , Xn )].
Note that Zk+1 = gk (X1 , . . . , Xk+1 ). Since f satisﬁes the Lipschitz condition with constant c, the
same is true of gk . In particular, for x1 , . . . , xk ﬁxed, the set of possible values (i.e. range) of
gk (x1 , . . . , xk+1 ) as xk+1 varies, lies within some interval (depending on x1 , . . . , xk ) with length at
most c. We deﬁne mk (x1 , · · · , xk ) to be the midpoint of the smallest such interval:
m k (x1 , . . . , x k ) = supxk+1 gk (x1 , . . . , xk+1 ) + inf xk+1 gk (x1 , . . . , xk+1 )
2 c
and let Bk+1 = mk (X1 , . . . , Xk ). Then B is a predictable process and Zk+1 − Bk+1  ≤ 2 with
c
probability one. Thus, the AzumaHoeﬀding inequality with centering can be applied with di = 2
for all i, giving the desired result. 3 Equivalently, f (x) − f (y ) ≤ cdH (x, y ), where dH (x, y ) denotes the Hamming distance, which is the number of
coordinates in which x and y diﬀer. In the analysis of functions of a continuous variable, the Euclidean distance is
used instead of the Hamming distance. 288 Example 10.3.6 Let V = {v1 , . . . , vn } be a ﬁnite set of cardinality n ≥ 1. For each i, j with
1 ≤ i < j ≤ n, suppose that Zi,j is a Bernoulli random variable with parameter p, where 0 ≤ p ≤ 1.
Suppose that the Z ’s are mutually independent. Let G = (V, E ) be a random graph, such that for
i < j , there is an undirected edge between vertices vi and vj (i.e. vi and vj are neighbors) if and only
if Zi,j = 1. Equivalently, the set of edges is E = {{i, j } : i < j and Zi,j = 1}. An independent set in
the graph is a set of vertices, no two of which are neighbors. Let I = I (G) denote the maximum of
the cardinalities of all independent sets for G. Note that I is a random variable, because the graph
is random. We shall apply McDiarmid’s inequality to ﬁnd a concentration bound for I (G). Note
that I (G) = F ((Zi,j : 1 ≤ i < j ≤ n)), for an appropriate function F. We could write a computer
program for computing F, for example by cycling through all subsets of V , seeing which ones are
independent sets, and reporting the largest cardinality of the independent. However, there is no
need to be so explicit about what f is. Observe next that changing any one of the Z ’s would change
I (G) by at most one. In particular, if there is an independent set in a graph, and if one edge is
added to the graph, then at most one vertex would have to be removed from the independent set
for the original graph to obtain one for the new graph. Thus, F satisﬁes the Lipschitz condition
with constant c = 1. Thus, by McDiarmid’s inequality with c = 1 and m = n(n − 1)/2 variables,
P {I − E [I ] ≥ λ} ≤ 2 exp(− 4λ2
).
n(n − 1) More thought yields a tighter bound. For 1 ≤ i ≤ n, let Xi = (Zi,i+1 , Zi,i+2 , . . . , Zi,n ). In words, for
each i, Xi determines which vertices with index larger than i are neighbors of vertex vi . Of course
I is also determined by X1 , . . . , Xn . Moreover, if any one of the X ’s changes, I changes by at most
one. That is, I can be expressed as a function of the n variables X1 , . . . , Xn , such that the function
satisﬁes the Lipschitz condition with constant c = 1. Therefore, by McDiarmid’s inequality with
c = 1 and n variables,4
2λ2
P {I − E [I ] ≥ λ} ≤ 2 exp(−
).
n
√
For example, if λ = a n, we have
√
P {I − E [I ] ≥ a n} ≤ 2 exp(−2a2 )
whenever n ≥ 1, 0 ≤ p ≤ 1, and a > 0.
McDiarmid’s inequality can similiarly be applied to obtain concentration inequalities for many
other numbers associated with graphs, such as the size of a maximum matching (a matching is a
set of edges, no two of which have a node in common), chromatic index (number of colors needed
to color all edges so that all edges containing a single vertex are diﬀerent colors), chromatic number
(number of colors needed to color all vertices so that neighbors are diﬀerent colors), minimum
number of edges that need to be cut to break graph into two equal size components, and so on. 10.4 Stopping times and the optional sampling theorem Let X = (Xk : k ≥ 0) be a martingale with respect to a ﬁltration F = (Fk : k ≥ 0). Note that
E [Xk+1 ] = E [E [Xk+1 Fk ] = E [Xk ]. So, by induction on n, E [Xn ] = E [X0 ] for all n ≥ 0.
4 Since Xn is degenerate, we could use n − 1 instead of n, but it makes little diﬀerence. 289 A useful interpretation of a martingale X = (Xk : k ≥ 0) is that Xk is the reserve (amount of
money on hand) that a gambler playing a fair game at each time step, has after k time steps, if X0
is the initial reserve. (If the gambler is allowed to go into debt, the reserve can be negative.) The
condition E [Xk+1 Fk ] = Xk means that, given the knowledge that is observable up to time k , the
expected reserve after the next game is equal to the reserve at time k . The equality E [Xn ] = E [X0 ]
has the natural interpretation that the expected reserve of the gambler after n games have been
played, is equal to the inital reserve X0 .
What happens if the gambler stops after a random number, T , of games. Is it true that
E [XT ] = E [X0 ]?
Example 10.4.1 Suppose that Xn = W1 + · · · + Wn where P {Wk = 1} = P {Wk = −1} = 0.5 for
all k , and the W ’s are independent. Let T be the random time:
T= 3 if W1 + W2 + W3 = 1
0 else Then XT = 3 with probability 1/8, and XT = 0 otherwise. Hence, E [XT ] = 3/8.
Does example 10.4.1 give a realistic strategy for a gambler to obtain a strictly positive expected
payoﬀ from a fair game? To implement the strategy, the gambler should stop gambling after T
games. However, the event {T = 0} depends on the outcomes W1 , W2 , and W3 . Thus, at time
zero, the gambler is required to make a decision about whether to stop before any games are played
based on the outcomes of the ﬁrst thee games. Unless the gambler can somehow predict the future,
the gambler will be unable to implement the strategy of stopping play after T games.
Intuitively, a random time corresponds to an implementable stopping strategy if the gambler
has enough information after n games to tell whether to play future games. That type of condition
is captured by the notion of optional stopping time, deﬁned as follows.
Deﬁnition 10.4.2 An optional stopping time T relative to a ﬁltration F = (Fk : k ≥ 0) is a
random variable with values in Z+ such that for any n ≥ 0, {T ≤ n} ∈ Fn .
The intuitive interpretation of the condition {T ≤ n} ∈ Fn is that, the gambler should have enough
information by time n to know whether to stop by time n. Since σ algebras are closed under set
complements, the condition in the deﬁnition of an optional stopping time is equivalent to requiring
that, for any n ≥ 0, {T > n} ∈ Fn . This means that the gambler should have enough information
by time n to know whether to continue gambling strictly beyond time n.
Example 10.4.3 Let (Xn : n ≥ 0) be a random process adapted to a ﬁltration F = (Fn : n ≥ 0).
Let A be some ﬁxed (Borel measurable) set, and let T = min{n ≥ 0 : Xn ∈ A}. Then T is a
stopping time relative to F . Indeed, {T ≤ n} = {Xk ∈ A for some k with 0 ≤ k ≤ n}. So {T ≤ n}
is an event determined by (X0 , . . . , Xn ), which is in Fn because X is adapted to the ﬁltration. Example 10.4.4 Suppose W1 , W2 , . . . are independent Bernoulli random variables with p = 0.5,
modeling fair coin ﬂips. Suppose that if a gambler stakes some money at the beginning of the
nth round, then if Wn = 1, the gambler wins back the stake and an additional equal amount. If
290 Wn = 0, the gambler loses the money staked. Let Xn denote the reserve of the gambler after n
rounds. For simplicity, we assume that the gambler can borrow money as needed, and that the
initial reserve of the gambler is zero. So X0 = 0. Suppose the gambler adopts the following strategy.
The gambler continues playing until the ﬁrst win, and in each round until stopping, the gambler
stakes the amount of money needed to insure that the reserve at the time the gambler stops, is one
doller. For example, the gambler initially borrows one dollar, and stakes it on the ﬁrst outcome.
If W1 = 1 the gambler’s reserve (money in hand minus the amount borrowed) is one dollar, and
the gambler stops, so T = 1 and XT = 1. Since no money is staked after time T , Xk = XT for
all k ≥ T . If the gambler loses in the ﬁrst round (i.e. W1 = 0), then X1 = −1. In that case,
the gambler keeps playing, and, next, borrows two more dollars and stakes them on the second
outcome. If W2 = 1 the gambler’s reserve is one dollar, and the gambler stops. So T = 2 and
XT = 1. If the gambler loses in the second round (i.e. W2 = 0), then X2 = −3. In that case, the
gambler keeps playing, and, next, borrows four more dollars and stakes them on the third outcome,
and so on. The random process (Xn : n ≥ 0) is a martingale. For this strategy, the number of
rounds, T , that the gambler plays has the geometric distribution with parameter p = 0.5. Thus,
E [T ] = 2. In particular, T is ﬁnite with probability one. Thus, XT = 1 a.s., while X0 = 0. Thus,
E [XT ] = E [X0 ]. This strategy does not require the gambler to be able to predict the future, and
the gambler is always up one dollar after stopping.
But don’t run out and start playing this strategy, expecting to make money for sure. There is
a catch–the amount borrowed can be very large. Indeed, let us compute the expectation of B , the
total amount borrowed before the ﬁnal win. If T = 1 then B = 1 (only the dollar borrowed in the
ﬁrst round is counted). If T = 2 then B = 3 (the ﬁrst dollar in the ﬁrst round, and two more in
the second). In general, B = 2T − 1. Thus,
∞ E [B ] = ∞
n (2 − 1)P {T = n} =
n=1 ∞
n (2 − 1)2
n=1 −n (1 − 2−n ) = +∞ = n=1 That is, the expected amount of money the gambler will need to borrow is inﬁnite. Proposition 10.4.5 If X is a martingale and T is an optional stopping time, relative to (Ω, F , P ),
then E [XT ∧n ] = E [X0 ] for any n.
Proof. Note that
0
if T ≤ n
X ∧ (n + 1) − X ∧ n if T > n
= (X ∧ (n + 1) − X ∧ n)I{T >n} XT ∧(n+1) − XT ∧n = Using this and the tower property of conditional expectation yields
E [XT ∧(n+1) − XT ∧n ] = E [E [(X ∧ (n + 1) − X ∧ n)I{T >n} Fn ]]
= E [E [(X ∧ (n + 1) − X ∧ n)Fn ]I{T >n} ] = 0
because E [(X ∧ (n + 1) − X ∧ n)Fn ] = 0. Therefore, E [X ∧ (n + 1)] = E [X ∧ n] for all n ≥ 0. So
by induction on n, E [XT ∧n ] = E [X0 ] for all n ≥ 0.
The following corollary follows immediately from Proposition 10.4.5.
291 Corollary 10.4.6 If X is a martingale and T is an optional stopping time, relative to (Ω, F , P ),
then E [X0 ] = limn→∞ E [XT ∧n ]. In particular, if
lim E [XT ∧n ] = E [XT ] (10.8) n→∞ then E [XT ] = E [X0 ].
By Corollary 10.4.6, the trick to establishing E [XT ] = E [X0 ] comes down to proving (10.8). Note
a.s.
that XT ∧n → XT as n → ∞, so (10.8) is simply requiring the convergence of the means to the
mean of the limit, for an a.s. convergent sequence of random variables. There are several diﬀerent
suﬃcient conditions for this to happen, involving conditions on the martingale X , the stopping
time T , or both. For example:
Corollary 10.4.7 If X is a martingale and T is an optional stopping time, relative to (Ω, F , P ),
and if T is bounded (so P {T ≤ n} = 1 for some n) then E [XT ] = E [X0 ].
Proof. If P {T ≤ n} = 1 then T ∧ n = T with probability one, so E [XT ∧n = XT ]. Therefore, the
corollary follows from Proposition 10.4.5. Corollary 10.4.8 If X is a martingale and T is an optional stopping time, relative to (Ω, F , P ),
and if there is a random variable Z such that Xn  ≤ Z a.s. for all n, and E [Z ] < ∞, then
E [XT ] = E [X0 ].
Proof. Let > 0. Since E [Z ] < ∞, there exists δ > 0 so that if A is any set with P [A] < δ , then
p.
a.s.
E [ZIA ] < . Since XT ∧n → XT , we also have XT ∧n → XT . Therefore, if n is suﬃciently large,
P {XT ∧n − XT  ≥ } ≤ δ . For such n,
XT ∧n − XT  ≤
≤ + XT ∧n − XT I{XT ∧n −XT >
+ 2Z I{XT ∧n −XT > } } (10.9) Now E [Z I{XT ∧n −XT > } ] < by the choice of δ and n. So taking expectations of each side of (10.9)
yields E [XT ∧n − XT ] ≤ 3 . Both XT ∧n and XT have ﬁnite means, because both have absolute
values less than or equal to Z , so
E [XT ∧n ] − E [XT ] = E [XT ∧n − XT ] ≤ E [XT ∧n − XT ] < 3
Since was an arbitrary positive number, the corollary is proved. Corollary 10.4.9 Suppose (Xn : n ≥ 0) is a martingale relative to (Ω, F , P ). Suppose
(i) there is a constant c such that E [ Xn+1 − Xn  Fn ] ≤ c for n ≥ 0,
(ii) T is stopping time such that E [T ] < ∞.
Then E [XT ] = E [X0 ]. If, instead, (Xn : n ≥ 0) is a submartingale relative to (Ω, F , P ), satisfying
(i) and (ii), then E [XT ] ≥ E [X0 ].
292 Proof. Suppose (Xn : n ≥ 0) is a martingale relative to (Ω, F , P ), satisfying (i) and (ii). Looking
at the proof of Corollary 10.4.8, we see that it is enough to show that there is a random variable Z
such that E [Z ] < +∞ and XT ∧n  ≤ Z for all n ≥ 0. Let
Z = X0  + X1 − X0  + · · · + XT − XT −1 
Obviously, XT ∧n  ≤ Z for all n ≥ 0, so it remains to show that E [Z ] < ∞. But
∞ E [Z ] = E [X0 ] + E Xi − Xi−1 I{i≤T }
i=1
∞ = E [X0 ] + E E Xi − Xi−1 I{i≤T }  Fi−1
i=1
∞ = E [X0 ] + E I{i≤T } E Xi − Xi−1   Fi−1
i=1 ∞ = E [X0 ] + c P {i ≤ T }
i=1 = E [X0 ] + cE [T ] < ∞
The ﬁrst statement of the Corollary is proved. If instead X is a submartingale, then a minor
variation of Proposition 10.4.5 yields that E [XT ∧n ] ≥ E [X0 ]. The proof for the ﬁrst part of the
corollary, already given, shows that conditions (i) and (ii) imply that E [XT ∧n ] → E [XT ] as n → ∞.
Therefore, E [XT ] ≥ E [X0 ].
Martingale inequalities oﬀer a way to provide upper and lower bounds on the completion times
of algorithms. The following example shows how a lower bound can be found for a particular game.
Example 10.4.10 Consider the following game. There is an urn, initially with k1 red marbles
and k2 blue marbles. A player takes turns until the urn is empty, and the goal of the player is to
minimize the expected number of turns required. At the beginning of each turn, the player can
remove a set of marbles, and the set must be one of four types: one red, one blue, one red and
one blue, or two red and two blue. After removing the set of marbles, a fair coin is ﬂipped. If
tails appears, the turn is over. If heads appears, then some marbles are added back to the bag,
according to Table 10.1 Our goal will be to ﬁnd a lower bound on E [T ], where T is the number
Table 10.1: Rules of the marble game
Set removed
one red
one blue
two reds
two blues Set returned to bag on “heads”
one red and one blue
one red and one blue
three blues
three reds of turns needed by the player until the urn is empty. The bound should hold for any strategy the
293 player adopts. Let Xn denote the total number of marbles in the urn after n turns. If the player
elects to remove only one marble during a turn (either red or blue) then with probability one half,
two marbles are put back. Hence, for either set with one marble, the expected change in the total
number of marbles in the urn is zero. If the player elects to remove two reds or two blues, then
with probability one half, three marbles are put back into the urn. For these turns, the expected
change in the number of marbles in the urn is 0.5. Hence, for any choice of un (representing the
decision of the player for the n + 1th turn),
E [Xn+1 Xn , un ] ≥ Xn − 0.5 on {T > n} That is, the drift of Xn towards zero is at most 0.5 in magnitude, so we suspect that no strategy
can empty the urn in average time less than (k1 + k2 )/0.5. In fact, this result is true, and it is now
proved. Let Mn = Xn∧T + n∧T . By the observations above, M is a submartingale. Furthermore,
2
Mn+1 − Mn  ≤ 2. Either E [T ] = +∞ or E [T ] < ∞. If E [T ] = +∞ then the inequality to be
proved, E [T ] ≥ 2(k1 + k2 ), is trivially true, so suppose E [T ] < ∞. Then by Corollary 10.4.9,
E [MT ] ≥ E [M0 ] = k1 + k2 . Also, MT = T with probability one, so E [T ] ≥ 2(k1 + k2 ), as claimed.
2 10.5 Notes Material on AzumaHoeﬀding inequality and McDiarmid’s method can be found in McDiarmid’s
tutorial article [7]. 10.6 Problems 10.1 Two martingales associated with a simple branching process
Let Y = (Yn : n ≥ 0) denote a simple branching process. Thus, Yn is the number of individuals in
the nth generation, Y0 = 1, the numbers of oﬀspring of diﬀerent individuals are independent, and
each has the same distribution as a random variable X .
n
(a) Identify a constant θ so that Gn = Yn is a martingale.
θ
(b) Let E denote the event of eventual extinction, and let α = P {E}. Show that P [EY0 , . . . , Yn ] =
αYn . Thus, Mn = αYn is a martingale.
(c) Using the fact E [M1 ] = E [M0 ], ﬁnd an equation for α. (Note: Problem 4.29 shows that α is
the smallest positive solution to the equation, and α < 1 if and only if E [X ] > 1.)
10.2 A covering problem
Consider a linear array of n cells. Suppose that m base stations are randomly placed among the
cells, such that the locations of the base stations are independent, and uniformly distributed among
the n cell locations. Let r be a positive integer. Call a cell i covered if there is at least one base
station at some cell j with i − j  ≤ r − 1. Thus, each base station (unless those near the edge of
the array) covers 2r − 1 cells. Note that there can be more than one base station at a given cell,
and interference between base stations is ignored.
(a) Let F denote the number of cells covered. Apply the method of bounded diﬀerences based on
the AzumaHoeﬀding inequality to ﬁnd an upper bound on P [F − E [F ] ≥ γ ].
(b) (This part is related to the coupon collector problem and may not have anything to do with
martingales.) Rather than ﬁxing the number of base stations, m, let X denote the number of base
294 stations needed until all cells are covered. In case r = 1 we have seen that P [X ≥ n ln n + cn] →
exp(−e−c ) (the coupon collectors problem). For general r ≥ 1, ﬁnd g1 (r) and g2 (r) to so that for
any > 0, P [X ≥ (g2 (r) + )n ln n] → 0 and P [X ≤ (g1 (r) − )n ln n] → 0. (Ideally you can ﬁnd
g1 = g2 , but if not, it’d be nice if they are close.)
10.3 Doob decomposition
Suppose X = (Xk : k ≥ 0) is an integrable (meaning E [Xk ] < ∞ for each k ) sequence adapted to
a ﬁltration F = (Fk : k ≥ 1). (a) Show that there is sequence B = (Bk : k ≥ 0) which is predictable
relative to F (which means that B0 is a constant and Bk is Fk−1 measurable for k ≥ 1) and a mean
zero martingale M = (Mk : k ≥ 0), such that Xk = Bk + Mk for all k . (b) Are the sequences B
and M uniquely determined by X and F ?
10.4 On uniform integrability
(a) Show that if (Xi : i ∈ I ) and (Yi : i ∈ I ) are both uniformly integrable collections of random
variables with the same index set I , then (Zi : i ∈ I ), where Zi = Xi + Yi for all i, is also a
uniformly integrable collection. (b) Show that a collection of random variables (Xi : i ∈ I ) is
uniformly integrable if and only if there exists a convex increasing function ϕ : R+ → R+ with
limc→∞ ϕ(c) = +∞ and a constant K , such that E [ϕ(Xi )] ≤ K for all i ∈ I .
c
10.5 Stopping time properties
(a) Show that if S and T are stopping times for some ﬁltration F , then S ∧ T , S ∨ T , and S + T ,
are also stopping times.
(b) Show that if F is a ﬁltration and X = (Xk : k ≥ 0) is the random sequence deﬁned by
Xk = I{T ≤k} for some random time T with values in Z+ , then T is a stopping time if and only if
X is F adapted.
(c) If T is a stopping time for a ﬁltration F , recall that FT is the set of events A such that
A ∩ {T ≤ n} ∈ Fn for all n. (Or, for discrete time, the set of events A such that A ∩ {T = n} ∈ Fn
for all n.) Show that (i) FT is a σ algebra, (ii) T is FT measurable, and (iii) if X is an adapted
process then XT is FT measurable.
10.6 A stopped random walk
Let W1 , W2 , . . . be a sequence of independent, identically distributed mean zero random variables.
To avoid triviality, assume P {W1 = 0} = 0. Let S0 = 0 and Sn = W1 + . . . Wn for n ≥ 1. Fix
a constant c > 0 and let τ = min{n ≥ 0 : Sn  ≥ c}. The goal of this problem is to show that
E [Sτ ] = 0.
(a) Show that E [Sτ ] = 0 if there is a constant D so that P [Wi  > D] = 0. (Hint: Invoke a version
of the optional stopping theorem).
(b) In view of part (a), we need to address the case that the W ’s are not bounded. Let Wn = Wn if Wn  ≤ 2c
a if Wn > 2c
where the constants a and b are selected so that a ≥ 2c, b ≥ 2c, and −b if Wn < −2c
˜
E [Wi ] = 0. Note that if τ < n and if Wn = Wn , then τ = n. Thus, τ deﬁned above also
2 = Var(W ). Let S = W + . . . W for n ≥ 0 and let
satisﬁes τ = min{n ≥ 0 : Sn  ≥ c}. Let σ
i
n
1
n
2
Mn = Sn − nσ 2 . Show that M is a martingale. Hence, E [Mτ ∧n ] = 0 for all n. Conclude that
E [τ ] < ∞
295 (c) Show that E [Sτ ] = 0. (Hint: Use part (b) and invoke a version of the optional stopping
theorem).
10.7 Bounding the value of a game
Consider the following game. Initially a jar has ao red marbles and bo blue marbles. On each turn,
the player removes a set of marbles, consisting of either one or two marbles of the same color, and
then ﬂips a fair coin. If heads appears on the coin, then if one marble was removed, one of each
color is added to the jar, and if two marbles were removed, then three marbles of the other color
are added back to the jar. If tails appears, no marbles are added back to the jar. The turn is then
over. Play continues until the jar is empty after a turn, and then the game ends. Let τ be the
number of turns in the game. The goal of the player is to minimize E [τ ]. A strategy is a rule to
decide what set of marbles to remove at the beginning of each turn.
(a) Find a lower bound on E [τ ] that holds no matter what strategy the player selects.
(b) Suggest a strategy that approximately minimizes E [τ ], and for that strategy, ﬁnd an upper
bound on E [τ ].
10.8 On the size of a maximum matching in a random bipartite graph
Given 1 ≤ d < n, let U = {u1 , . . . , un } and V = {v1 , . . . , vn } be disjoint sets of cardinality n, and let
G be a bipartite random graph with vertex set U ∪ V , such that if Vi denotes the set of neighbors
of ui , then V1 , . . . , Vn are independent, and each is uniformly distributed over the set of all n
d
subsets of V of cardinality d. A matching for G is a subset of edges M such that no two edges in
M have a common vertex. Let Z denote the maximum of the cardinalities of the matchings for G.
(a) Find bounds a and b, with 0 < a ≤ b < n, so that a ≤ E [Z ] ≤ b.
√
(b) Give an upper bound on P {Z − E [Z ] ≥ γ n}, for γ > 0, showing that for ﬁxed d, the
distribution of Z is concentrated about its mean as n → ∞.
(c) Suggest a greedy algorithm for ﬁnding a large cardinality matching.
10.9 * Equivalence of having the form g (Y ) and being measurable relative to the sigma
algebra generated by Y .
Let Y and Z be random variables on the same probability space. The purpose of this problem is to
establish that Z = g (Y ) for some Borel measurable function g if and only if Z is σ (Y ) measurable.
(“only if” part) Suppose Z = g (Y ) for a Borel measurable function g , and let c ∈ R. It
must be shown that {Z ≤ c} ∈ σ (Y ). Since g is a Borel measurable function, by deﬁnition,
A = {y : g (y ) ≤ c} is a Borel subset of R. (a) Show that {Z ≤ c} = {Y ∈ A}. (b) Using the
deﬁnition of Borel sets, show that {Y ∈ A} ∈ σ (Y ) for any Borel set A. The “only if” part follows.
(“if” part) Suppose Z is σ (Y ) measurable. It must be shown that Z has the form g (Y ) for
some Borel measurable function g . (c) Prove this ﬁrst in the special case that Z has the form of
an indicator function: Z = IB , for some event B , which satisﬁes B ∈ σ (Y ). (Hint: Appeal to the
deﬁnition of σ (Y ).) (d) Prove the “if” part in general. (Hint: Z can be written as the supremum
of a countable set of random variables, with each being a constant times an indicator function:
Z = supn qn I{Z ≤qn } , where q1 , q2 , . . . is an enumeration of the set of rational numbers.)
10.10 * Regular conditional distributions
Let X be a random variable on (Ω, F , P ) and let D be a subσ algebra of F . A conditional probability such as P [X ≤ cD] for a ﬁxed constant c can sometimes have diﬀerent versions, but any
two such versions are equal with probability one. Roughly speaking, the idea of regular conditional
296 distributions, deﬁned next, is to select a version of P [X ≤ cD] for every real number c so that,
as a function of c for ω ﬁxed, the result is a valid CDF (i.e. nondecreasing, rightcontinuous, with
limit zero at −∞ and limit one at +∞.) The diﬃculty is that there are uncountably many choices
of c. Here is the deﬁnition. A regular conditional CDF of X given D, denoted by FX D (cω ), is a
function of (c, ω ) ∈ R × Ω such that:
(1) for each c ∈ R ﬁxed, FX D (cω ) is a D measurable function of ω ,
(2) for each ω ﬁxed, as a function of c, FX D (cω ) is a valid CDF,
(3) for any c ∈ R, FX D (cω ) is a version of P [X ≤ cD].
The purpose of this problem is to prove the existence of a regular conditional CDF. For each rational
number q , let Φ(q ) = P [X ≤ q D]. That is, for each rational number q , we pick Φ(q ) to be one
particular version of P [X ≤ q D]. Thus, Φ(q ) is a random variable, and so we can also write it at as
Φ(q, ω ) to make explicit the dependence on ω . By the positivity preserving property of conditional
expectations, P {Φ(q ) > Φ(q )} = 0 if q < q . Let {q1 , q2 , . . .} denote the set of rational numbers,
listed in some order. The event N deﬁned by
N = ∩n,m:qn <qm {Φ(qn ) > Φ(qm ).}
thus has probability zero. Modify Φ(q, ω ) for ω ∈ N by letting Φ(q, ω ) = Fo (q ) for ω ∈ N and all
rational q , where Fo is an arbitrary, ﬁxed CDF. Then for any c ∈ IR and ω ∈ Ω, let
Φ(c, ω ) = inf Φ(q, ω )
q ≥c Show that Φ so deﬁned is a regular, condtional CDF of X given D.
10.11 * An even more general deﬁnition of conditional expectation, and the conditional version of Jensen’s inequality
Let X be a random variable on (Ω, F , P ) and let D be a subσ algebra of F . Let FX D (cω ) be a
regular conditional CDF of X given D. Then for each ω , we can deﬁne E [X D] at ω to equal the
mean for the CDF FX D (cω ) : c ∈ R}, which is contained in the extended real line R ∪ {−∞, +∞}.
Symbollically: E [X D](ω ) = R cFX D (dcω ). Show that, in the special case that E [X ] < ∞, this
deﬁnition is consistent with the one given previously. As an application, the following conditional
version of Jensen’s inequality holds: If φ is a convex function on R, then E [φ(X )D] ≥ φ(E [X D])
a.s. The proof is given by applying the ordinary Jensen’s inequality for each ω ﬁxed, for the regular
conditional CDF of X given D evaluated at ω . 297 298 Chapter 11 Appendix
11.1 Some notation The following notational conventions are used in these notes. Ac = AB = complement of A
A∩B A ⊂ B ↔ any element of A is also an element of B
= AB c Ai = {a : a ∈ Ai for some i} Ai = {a : a ∈ Ai for all i} a ∨ b = max{a, b} = a∧b = a+ = IA (x) = A−B
∞
i=1
∞
i=1 a if a ≥ b
b if a < b
min{a, b} (a, b) = {x : a < x < b} a ∨ 0 = max{a, 0}
1 if x ∈ A
0
else
(a, b] = {x : a < x ≤ b} [a, b) = {x : a ≤ x < b} [a, b] = {x : a ≤ x ≤ b} Z − set of integers Z+ − set of nonnegative integers R − set of real numbers R+ − set of nonnegative real numbers C = set of complex numbers
299 A1 × · · · × An = {(a1 , . . . , an )T : ai ∈ Ai for 1 ≤ i ≤ n}
An = A × · · · × A
t = n times
greatest integer n such that n ≤ t t = least integer n such that n ≥ t A = expression − denotes that A is deﬁned by the expression
All the trigonometric identities required in these notes can be easily derived from the two
identities:
cos(a + b) = cos(a) cos(b) − sin(a) sin(b)
sin(a + b) = sin(a) cos(b) + cos(a) sin(b)
and the facts cos(−a) = cos(a) and sin(−b) = − sin(b).
A set of numbers is countably inﬁnite if the numbers in the set can be listed in a sequence
xi : i = 1, 2, . . .. For example, the set of rational numbers is countably inﬁnite, but the set of all
real numbers in any interval of positive length is not countably inﬁnite. 11.2 Convergence of sequences of numbers We begin with some basic deﬁnitions. Let (xn ) = (x1 , x2 , . . .) and (yn ) = (y1 , y2 , . . .) be sequences
of numbers and let x be a number. By deﬁnition, xn converges to x as n goes to inﬁnity if for each
> 0 there is an integer n so that  xn − x < for every n ≥ n . We write limn→∞ xn = x to
denote that xn converges to x.
+4
Example 11.2.1 Let xn = 2n+1 . Let us verify that limn→∞ xn = 0. The inequality  xn <
n2
holds if 2n + 4 ≤ (n2 + 1). Therefore it holds if 2n + 4 ≤ n2 . Therefore it holds if both 2n ≤ 2 n2 and 4 ≤ 2 n2 . So if n = max{ 4 , 8 } then n ≥ n implies that  xn < . So limn→∞ xn = 0. By deﬁnition, (xn ) converges to +∞ as n goes to inﬁnity if for every K > 0 there is an integer
nK so that xn ≥ K for every n ≥ nK . Convergence to −∞ is deﬁned in a similar way.1 For
example, n3 → ∞ as n → ∞ and n3 − 2n4 → −∞ as n → ∞.
Occasionally a twodimensional array of numbers (am,n : m ≥ 1, n ≥ 1) is considered. By
deﬁnition, am,n converges to a number a∗ as m and n jointly go to inﬁnity if for each > 0 there
is n > 0 so that  am,n − a∗ < for every m, n ≥ n . We write limm,n→∞ am,n = a to denote that
am,n converges to a as m and n jointly go to inﬁnity.
Theoretical Exercise Let am,n = 1 if m = n and am,n = 0 if m = n. Show that limn→∞ (limm→∞ am,n ) =
limm→∞ (limn→∞ amn ) = 0 but that limm,n→∞ am,n does not exist.
m+n (−1)
Theoretical Exercise Let am,n = min(m,n) . Show that limm→∞ am,n does not exist for any n
and limn→∞ am,n does not exist for any m, but limm,n→∞ am,n = 0.
1 Some authors reserve the word “convergence” for convergence to a ﬁnite limit. When we say a sequence converges
to +∞ some would say the sequence diverges to +∞. 300 Theoretical Exercise If limm,n→∞ amn = a∗ and limm→∞ amn = bn for each n, then limn→∞ bn =
a∗ .
The condition limm,n→∞ am,n = a∗ can be expressed in terms of convergence of sequences
depending on only one index (as can all the other limits discussed in these notes) as follows. Namely,
limm,n→∞ am,n = a∗ is equivalent to the following: limk→∞ amk ,nk = a∗ whenever ((mk , nk ) : k ≥ 1)
is a sequence of pairs of positive integers such that mk → ∞ and nk → ∞ as k → ∞. The condition
that the limit limm,n→∞ am,n exists, is equivalent to the condition that the limit limk→∞ amk ,nk
exists whenever ((mk , nk ) : k ≥ 1) is a sequence of pairs of positive integers such that mk → ∞
and nk → ∞ as k → ∞.2
A sequence a1 , a2 , . . . is said to be nondecreasing if ai ≤ aj for i < j . Similarly a function f
on the real line is nondecreasing if f (x) ≤ f (y ) whenever x < y . The sequence is called strictly
increasing if ai < aj for i < j and the function is called strictly increasing if f (x) < f (y ) whenever
x < y .3 A strictly increasing or strictly decreasing sequence is said to be strictly monotone, and a
nondecreasing or nonincreasing sequence is said to be monotone.
The sum of an inﬁnite sequence is deﬁned to be the limit of the partial sums. That is, by
deﬁnition,
∞ n yk = x means that k=1 lim n→∞ yk = x
k=1 Often we want to show that a sequence converges even if we don’t explicitly know the value of the
limit. A sequence (xn ) is bounded if there is a number L so that  xn ≤ L for all n. Any sequence
that is bounded and monotone converges to a ﬁnite number.
Example 11.2.2 Consider the sum ∞ k −α for a constant α > 1. For each n the nth partial
k=1
sum can be bounded by comparison to an integral, based on the fact that for k ≥ 2, the k th term
of the sum is less than the integral of x−α over the interval [k − 1, k ]:
n n k −α ≤ 1 +
k=1 x−α dx = 1 + 1 1 − n1−α
1
α
≤1+
=
(α − 1)
α−1
α−1 The partial sums are also monotone nondecreasing (in fact, strictly increasing). Therefore the sum
∞
−α exists and is ﬁnite.
k=1 k
A sequence (xn ) is a Cauchy sequence if limm,n→∞  xm − xn = 0. It is not hard to show that
if xn converges to a ﬁnite limit x then (xn ) is a Cauchy sequence. More useful is the converse
statement, called the Cauchy criteria for convergence, or the completeness property of R: If (xn )
is a Cauchy sequence then xn converges to a ﬁnite limit as n goes to inﬁnity. 2
We could add here the condition that the limit should be the same for all choices of sequences, but it is automatically true. If if two sequences were to yield diﬀerent limits of amk ,nk , a third sequence could be constructed by
interleaving the ﬁrst two, and amk ,nk wouldn’t be convergent for that sequence.
3
We avoid simply saying “increasing,” because for some authors it means strictly increasing and for other authors
it means nondecreasing. While inelegant, our approach is safer. 301 Example 11.2.3 Suppose (xn : n ≥ 1) is a sequence such that ∞ xi+1 − xi  < ∞. The Cauchy
i=1
criteria can be used to show that the sequence (xn : n ≥ 1) is convergent. Suppose 1 ≤ m < n.
Then by the triangle inequality for absolute values:
n−1 xn − xm  ≤ xi+1 − xi 
i=m or, equivalently,
n−1 xn − xm  ≤ m−1 xi+1 − xi  −
i=1 xi+1 − xi  . (11.1) i=1 Inequality (11.1) also holds if 1 ≤ n ≤ m. By the deﬁnition of the sum, ∞ xi+1 − xi , both sums
i=1
on the right side of (11.1) converge to ∞ xi+1 − xi  as m, n → ∞, so the right side of (11.1)
i=1
converges to zero as m, n → ∞. Thus, (xn ) is a Cauchy sequence, and it is hence convergent. Theoretical Exercise
1. Show that if limn→∞ xn = x and limn→∞ yn = y then limn→∞ xn yn = xy .
2. Find the limits and prove convergence as n → ∞ for the following sequences:
n2
1
n2
(a) xn = cos(+1) ,
(c) zn = n=2 k log k
(b) yn = log n
k
n2
The minimum of a set of numbers, A, written min A, is the smallest number in the set, if there
is one. For example, min{3, 5, 19, −2} = −2. Of course, min A is well deﬁned if A is ﬁnite (i.e. has
ﬁnite cardinality). Some sets fail to have a minimum, for example neither {1, 1/2, 1/3, 1/4, . . .} nor
{0, −1, −2, . . .} have a smallest number. The inﬁmum of a set of numbers A, written inf A, is the
greatest lower bound for A. If A is bounded below, then inf A = max{c : c ≤ a for all a ∈ A}. For
example, inf {1, 1/2, 1/3, 1/4, . . .} = 0. If there is no ﬁnite lower bound, the inﬁmum is −∞. For
example, inf {0, −1, −2, . . .} = −∞. By convention, the inﬁmum of the empty set is +∞. With
these conventions, if A ⊂ B then inf A ≥ inf B. The inﬁmum of any subset of R exists, and if min A
exists, then min A = inf A, so the notion of inﬁmum extends the notion of minimum to all subsets
of R.
Similarly, the maximum of a set of numbers A, written max A, is the largest number in the set,
if there is one. The supremum of a set of numbers A, written sup A, is the least upper bound for
A. We have sup A = − inf {−a : a ∈ A}. In particular, sup A = +∞ if A is not bounded above, and
sup ∅ = −∞. The supremum of any subset of R exists, and if max A exists, then max A = sup A,
so the notion of supremum extends the notion of maximum to all subsets of R.
The notions of inﬁmum and supremum of a set of numbers are useful because they exist for any
set of numbers. There is a pair of related notions that generalizes the notion of limit. Not every
sequence has a limit, but the following terminology is useful for describing the limiting behavior of
a sequence, whether or not the sequence has a limit.
Deﬁnition 11.2.4 The liminf (also called limit inferior) of a sequence (xn : n ≥ 1), is deﬁned by
lim inf xn = lim
n→∞ n→∞ [inf {xk : k ≥ n}] , 302 (11.2) and the limsup (also called limit superior) is deﬁned by
lim sup xn = lim
n→∞ n→∞ [sup{xk : k ≥ n}] , (11.3) The possible values of the liminf and limsup of a sequence are R ∪ {−∞, +∞}.
The limit on the right side of (11.2) exists because the inﬁmum inside the square brackets is
monotone nondecreasing in n. Similarly, the limit on the right side of (11.3) exists. So every
sequence of numbers has a liminf and limsup.
Deﬁnition 11.2.5 A subsequence of a sequence (xn : n ≥ 1) is a sequence of the form (xki : i ≥ 1),
where k1 , k2 , . . . is a strictly increasing sequence of integers. The set of limit points of a sequence is
the set of all limits of convergent subsequences. The values −∞ and +∞ are possible limit points.
Example 11.2.6 Suppose yn = 121 − 25n2 for n ≤ 100 and yn = 1/n for n ≥ 101. The liminf and
limsup of a sequence do not depend on any ﬁnite number of terms of the sequence, so the values
of yn for n ≤ 100 are irrelevant. For all n ≥ 101, inf {xk : k ≥ n} = inf {1/n, 1/(n + 1), . . .} = 0,
which trivially converges to 0 as n → ∞. So the liminf of (yn ) is zero. For all n ≥ 101, sup{xk :
1
k ≥ n} = sup{1/n, 1/(n + 1), . . .} = n , which converges also to 0 at n → ∞. So the limsup of (yn )
is also zero. Zero is also the only limit point of (yn ). Example 11.2.7 Consider the sequence of numbers (2, −3/2, 4/3, −5/4, 6/5, . . .), which we also
− n+1
write as (xn : n ≥ 1) such that xn = (n+1)(n 1) . The maximum (and supremum) of the sequence is
2, and the minimum (and inﬁmum) of the sequence is −3/2. But for large n, the sequence alternates
between numbers near one and numbers near minus one. More precisely, the subsequence of odd
numbered terms, (x2i−1 : i ≥ 1), converges to 1, and the subsequence of even numbered terms,
(x2i : i ≥ 1}, has limit +1. Thus, both 1 and 1 are limit points of the sequence, and there aren’t
any other limit points. The overall sequence itself does not converge (i.e. does not have a limit)
but lim inf n→∞ xn = −1 and lim supn→∞ xn = +1.
Some simple facts about the limit, liminf, limsup, and limit points of a sequence are collected
in the following proposition. The proof is left to the reader.
Proposition 11.2.8 Let (xn : n ≥ 1) denote a sequence of numbers.
1. The condition lim inf n→∞ xn = x∞ is equivalent to the following:
for any γ < x∞ , xn ≥ γ for all suﬃciently large n.
2. The condition lim supn→∞ xn = x∞ is equivalent to the following:
for any γ > x∞ , xn ≤ γ for all suﬃciently large n.
3. lim inf n→∞ xn ≤ lim supn→∞ xn .
4. limn→∞ xn exists if and only if the liminf equals the limsup, and if the limit exists, then the
limit, liminf, and limsup are equal.
303 5. limn→∞ xn exists if and only if the sequence has exactly one limit point, x∗ , and if the limit
exists, it is equal to that one limit point.
6. Both the liminf and limsup of the sequence are limit points. The liminf is the smallest limit
point and the limsup is the largest limit point (keep in mind that −∞ and +∞ are possible
values of the liminf, limsup, or a limit point). Theoretical Exercise
1. Prove Proposition 11.2.8
2. Here’s a more challenging one. Let r be an irrational constant, and let xn = nr − nr for
n ≥ 1. Show that every point in the interval [0, 1] is a limit point of (xn : n ≥ 1). (P. Bohl,
W. Sierpinski, and H. Weyl independently proved a stronger result in 19091910: namely,
the fraction of the ﬁrst n values falling into a subinterval converges to the length of the
subinterval.) 11.3 Continuity of functions Let f be a function on Rn for some n, and let xo ∈ Rn . The function has a limit y at xo , and such
situation is denoted by limx→xo f (x) = y , if the following is true. Given > 0, there exists δ > 0 so
that  f (x) − y ≤ whenever 0 < x − xo < δ . This convergence condition can also be expressed
in terms of convergence of sequences, as follows. The condition limx→xo f (x) = y is equivalent to
the condition f (xn ) → y for any sequence x1 , x2 , . . . from Rn − xo such that xn → xo .
The function f is said to be continuous at xo , or equivalently, xo is said to be a continuity
point of f , if limx→xo f (x) = f (xo ). In terms of sequences, f is continuous at xo if f (xn ) → f (xo )
whenever x1 , x2 , . . . is a sequence converging to xo . The function f is simply said to be continuous
if it is continuous at every point in Rn .
Let n = 1, so consider a function f on R, and let xo ∈ R. The function has a righthand limit
y at xo , and such situation is denoted by f (xo +) = y or limx xo f (x) = y, if the following is true.
Given > 0, there exists δ > 0 so that  f (x) − y ≤ whenever 0 < x − xo < δ . Equivalently,
f (xo +) = y if f (xn ) → y for any sequence x1 , x2 , . . . from (xo , +∞) such that xn → xo . The
lefthand limit f (xo −) = limx xo f (x) is deﬁned similarly. If f is monotone nondecreasing, then
the lefthand and righthand limits exist, and f (xo −) ≤ f (xo ) ≤ f (xo +) for all xo .
A function f is called rightcontinuous at xo if f (xo ) = f (xo +). A function f is simply called
rightcontinuous if it is rightcontinuous at all points.
Deﬁnition 11.3.1 A function f on a bounded interval (open, closed, or mixed) with endpoints
a < b is piecewise continuous, if there exist n ≥ 1 and a = t0 < t1 < · · · < tn = b, such that, for
1 ≤ k ≤ n: f is continuous over (tk−1 , tk ) and has ﬁnite limits at the endpoints of (tk−1 , tk ).
More generally, if T is all of R or an interval in R, f is piecewise continuous over T if it is
piecewise continuous over every bounded subinterval of T.
304 11.4 Derivatives of functions Let f be a function on R and let xo ∈ R. Then f is diﬀerentiable at xo if the following limit exists
and is ﬁnite:
f (x) − f (xo )
.
x − xo lim x→xo The value of the limit is the derivative of f at xo , written as f (xo ). In more detail, this condition
that f is diﬀerentiable at xo means there is a ﬁnite value f (xo ) so that, for any > 0, there exists
δ > 0, so that
f (x) − f (xo )
− f (xo ) ≤ δ
x − xo
whenever 0 < x − xo  < . Alternatively, in terms of convergence of sequences, it means there is a
ﬁnite value f (xo ) so that
f (xn ) − f (xo )
lim
= f (xo )
n→∞
xn − xo
whenever (xn : n ≥ 1) is a sequence with values in R − {xo } converging to xo . The function f is
diﬀerentiable if it is diﬀerentiable at all points.
The righthand derivative of f at a point xo , denoted by D+ f (xo ), is deﬁned the same way as
f (xo ), except the limit is taken using only x such that x > xo . The extra condition x > xo is
indicated by using a slanting arrow in the limit notation:
D+ f (x0 ) = lim
x xo f (x) − f (xo )
.
x − xo Similarly, the lefthand derivative of f at a point xo is D− f (x0 ) = limx xo f (x)−f (xo )
.
x− xo Theoretical Exercise
1. Suppose f is deﬁned on an open interval containing xo , then f (xo ) exists if and only if
D+ f (xo ) = D− f (x0 ). If f (xo ) exists then D+ f (xo ) = D− f (x0 ) = f (xo ).
We write f for the derivative of f . For an integer n ≥ 0 we write f (n) to denote the result of
diﬀerentiating f n times.
Theorem 11.4.1 (Mean value form of Taylor’s theorem) Let f be a function on an interval (a, b)
such that its nth derivative f (n) exists on (a, b). Then for a < x, x0 < b,
n−1 f (x) =
k=0 f (n) (y )(x − x0 )n
f (k) (x0 )
(x − x0 )k +
k!
n! for some y between x and x0 .
Clearly diﬀerentiable functions are continuous. But they can still have rather odd properties,
as indicated by the following example.
305 Example 11.4.2 Let f (t) = t2 sin(1/t2 ) for t = 0 and f (0) = 0. This function f is a classic
example of a diﬀerentiable function with a derivative function that is not continuous. To check the
derivative at zero, note that  f (s)−f (0)  ≤ s → 0 as s → 0, so f (0) = 0. The usual calculus can be
s
used to compute f (t) for t = 0, yielding
f (t) = 2t sin( t1 ) −
2
0 2 cos( 1 )
t2
t t=0
t=0 The derivative f is not even close to being continuous at zero. As t approaches zero, the cosine
term dominates, and f reaches both positive and negative values with arbitrarily large magnitude.
Even though the function f of Example 11.4.2 is diﬀerentiable, it does not satisfy the fundamental theorem of calculus (stated in the next section). One way to rule out the wild behavior
of Example 11.4.2, is to assume that f is continuously diﬀerentiable, which means that f is differentiable and its derivative function is continuous. For some applications, it is useful to work
with functions more general than continuously diﬀerentiable ones, but for which the fundamental
theorem of calculus still holds. A possible approach is to use the following condition.
Deﬁnition 11.4.3 A function f on a bounded interval (open, closed, or mixed) with endpoints
a < b is continuous and piecewise continuously diﬀerentiable, if f is continuous over the interval,
and if there exist n ≥ 1 and a = t0 < t1 < · · · < tn = b, such that, for 1 ≤ k ≤ n: f is continuously
diﬀerentiable over (tk−1 , tk ) and f has ﬁnite limits at the endpoints of (tk−1 , tk ).
More generally, if T is all of R or a subinterval of R, then a function f on T is continuous and
piecewise continuously diﬀerentiable if its restriction to any bounded interval is continuous and
piecewise continuously diﬀerentiable.
Example 11.4.4 Two examples of continuous, piecewise continuously diﬀerentiable functions on
R are: f (t) = min{t2 , 1} and g (t) =  sin(t). Example 11.4.5 The function given in Example 11.4.2 is not considered to be piecewise continuously diﬀerentiable because the derivative does not have ﬁnite limits at zero. Theoretical Exercise
1. Suppose f is a continuously diﬀerentiable function on an open bounded interval (a, b). Show
that if f has ﬁnite limits at the endpoints, then so does f .
2. Suppose f is a continuous function on a closed, bounded interval [a, b] such that f exists
and is continuous on the open subinterval (a, b). Show that if the righthand limit of the
derviative at a, f (a+) = limx a f (x), exists, then the righthand derivative at a, deﬁned by
f (x) − f (a)
a
x−a D+ f (a) = lim
x also exists, and the two limits are equal.
306 Let g be a function from Rn to Rm . Thus for each vector x ∈ Rn , g (x) is an m vector. The
∂g
∂gi
derivative matrix of g at a point x, ∂x (x), is the n × m matrix with ij th entry ∂xj (x). Sometimes for
brevity we write y = g (x) and think of y as a variable depending on x, and we write the derivative
∂y
matrix as ∂x (x).
∂y
Theorem 11.4.6 (Implicit function theorem) If m = n and if ∂x is continuous in a neighborhood
∂y
of x0 and if ∂x (x0 ) is nonsingular, then the inverse mapping x = g −1 (y ) is deﬁned in a neighborhood
of y0 = g (x0 ) and ∂x
(y0 ) =
∂y 11.5 −1 Integration 11.5.1 ∂y
(x0 )
∂x Riemann integration Let g be a bounded function on a bounded interval of the form (a, b]. Given:
• An partition of (a, b] of the form (t0 , t1 ], (t1 , t2 ], · · · , (tn−1 , tn ], where n ≥ 0 and
a = t0 ≤ t1 · · · < tn = b
• A sampling point from each subinterval, vk ∈ (tk−1 , tk ], for 1 ≤ k ≤ n,
the corresponding Riemann sum for g is deﬁned by
n g (vk )(tk − tk−1 ).
k=1
b The norm of the partition is deﬁned to be maxk tk − tk−1 . The Riemann integral a g (x)dx is
said to exist and its value is I if the following is true. Given any > 0, there is a δ > 0 so that
 n=1 g (vk )(tk − tk−1 ) − I  ≤ whenever the norm of the partition is less than or equal to δ .
k
This deﬁnition is equivalent to the following condition, expressed using convergence of sequences.
The Riemann integral exists and is equal to I , if for any sequence of partitions, speciﬁed by
m
m
((tm , tm , . . . , tm ) : m ≥ 1), with corresponding sampling points ((v1 , . . . , vnm ) : m ≥ 1), such that
nm
1
2
norm of the mth partition converges to zero as m → ∞, the corresponding sequence of Riemann
sums converges to I as m → ∞. The function g is said to be Reimann integrable over (a, b] if the
b
integral a g (x)dx exists and is ﬁnite.
Next, suppose g is deﬁned over the whole real line. If for every interval (a, b], g is bounded over
[a, b] and Riemann integrable over (a, b], then the Riemann integral of g over R is deﬁned by
∞ b g (x)dx =
−∞ lim a,b→∞ −a g (x)dx provided that the indicated limit exist as a, b jointly converge to +∞. The values +∞ or −∞ are
possible.
A function that is continuous, or just piecewise continuous, is Riemann integrable over any
bounded interval. Moreover, the following is true for Riemann integration:
307 Theorem 11.5.1 (Fundamental theorem of calculus) Let f be a continuously diﬀerentiable function
on R. Then for a < b,
b f (b) − f (a) = f (x)dx. (11.4) a More generally, if f is continuous and piecewise continuously diﬀerentiable, (11.4) holds with f (x)
replaced by the righthand derivative, D+ f (x). (Note that D+ f (x) = f (x) whenever f (x) is deﬁned.)
We will have occasion to use Riemann integrals in two dimensions. Let g be a bounded function
on a bounded rectangle of the form (a1 , b1 ] × (a2 , b2 ]. Given:
• A partition of (a1 , b1 ] × (a2 , b2 ] into n1 × n2 rectangles of the form (t1 , t1−1 ] × (t2 , t2 −1 ], where
jj
kk
ni ≥ 1 and ai = ti < ti < · · · < ti i = bi for i = 1, 2
n
0
1
1
2
• A sampling point (vjk , vjk ) inside (t1 , t1−1 ] × (t2 , t2 −1 ] for 1 ≤ j ≤ n1 and 1 ≤ k ≤ n2 ,
jj
kk the corresponding Riemann sum for g is
n1 n2
1
2
g (vj,k , vj,k )(t1 − t1−1 )(t2 − t2 −1 ).
j
j
k
k j =1 k=1 The norm of the partition is maxi∈{1,2} maxk  ti − ti −1 . As in the case of one dimension, g is said
k
k
to be Riemann integrable over (a1 , b1 ] × (a2 , b2 ], and
(a1 ,b1 ]×(a2 ,b2 ] g (x1 , x2 )dsdt = I , if the value
of the Riemann sum converges to I for any sequence of partitions and sampling point pairs, with
the norms of the partitions converging to zero.
The above deﬁnition of a Riemann sum allows the n1 × n2 sampling points to be selected
arbitrarily from the n1 × n2 rectangles. If, instead, the sampling points are restricted to have the
1
1
2
2
12
form (vj , vk ), for n1 + n2 numbers v1 , . . . , vn1 , v1 , . . . vn2 , we say the corresponding Riemann sum
uses aligned sampling. We deﬁne a function g on [a, b] × [a, b] to be Riemann integrable with aligned
sampling in the same way as we deﬁned g to be Riemann integrable, except the family of Riemann
sums used are the ones using aligned sampling. Since the set of sequences that must converge is
more restricted for aligned sampling, a function g on [a, b] × [a, b] that is Riemann integrable is also
Riemann integrable with aligned sampling.
Proposition 11.5.2 A suﬃcient condition for g to be Riemann integrable (and hence Riemann
integrable with aligned sampling) over (a1 , b1 ] × (a2 , b2 ] is that g be the restriction to (a1 , b1 ] ×
(a2 , b2 ] of a continuous function on [a1 , b1 ] × [a2 , b2 ]. More generally, g is Riemann integrable over
(a1 , b1 ] × (a2 , b2 ] if there is a partition of (a1 , b1 ] × (a2 , b2 ] into ﬁnitely many subrectangles of the form
(t1 , t1−1 ] × (t2 , t2 −1 ], such that g on (t1 , t1−1 ] × (t2 , t2 −1 ] is the restriction to (t1 , t1−1 ] × (t2 , t2 −1 ]
jj
jj
jj
kk
kk
kk
of a continuous function on [t1 , t1−1 ] × [t2 , t2 −1 ].
jj
kk
Proposition 11.5.2 is a standard result in real analysis. It’s proof uses the fact that continuous
functions on bounded, closed sets are uniformly continuous, from which if follows that, for any
> 0, there is a δ > 0 so that the Riemann sums for any two partitions with norm less than or
equal to δ diﬀer by most . The Cauchy criteria for convergence of sequences of numbers is also
used.
308 11.5.2 Lebesgue integration Lebesgue integration with respect to a probability measure is deﬁned in the section deﬁning the
expectation of a random variable X and is written as
X (ω )P (dω ) E [X ] =
Ω The idea is to ﬁrst deﬁne the expectation for simple random variables, then for nonnegative random
variables, and then for general random variables by E [X ] = E [X+ ] − E [X− ]. The same approach
can be used to deﬁne the Lebesgue integral
∞ g (ω )dω
−∞ for Borel measurable functions g on R. Such an integral is well deﬁned if either
∞
or −∞ g− (ω )dω < +∞. 11.5.3 ∞
−∞ g+ (ω )dω < +∞ RiemannStieltjes integration Let g be a bounded function on a closed interval [a, b] and let F be a nondecreasing function on
[a, b]. The RiemannStieltjes integral
b g (x)dF (x) (RiemannStieltjes) a is deﬁned the same way as the Riemann integral, except that the Riemann sums are changed to
n g (vk )(F (tk ) − F (tk−1 ))
k=1 Extension of the integral over the whole real line is done as it is for Riemann integration. An
∞
alternative deﬁnition of −∞ g (x)dF (x), preferred in the context of these notes, is given next. 11.5.4 LebesgueStieltjes integration ˜
Let F be a CDF. As seen in Section 1.3, there is a corresponding probability measure P on the
Borel subsets of R. Given a Borel measurable function g on R, the LebesgueStieltjes integral of g
˜
with respect to F is deﬁned to be the Lebesgue integral of g with respect to P :
∞ (LebesgueStieltjes) ∞ g (x)dF (x) =
−∞ ˜
g (x)P (dx) (Lebesgue) −∞ ∞ The same notation −∞ g (x)dF (x) is used for both RiemannStieltjes (RS) and LebesgueStieltjes
(LS) integration. If g is continuous and the LS integral is ﬁnite, then the integrals agree. In
309 particular,
integrals ∞
−∞ xdF (x) is identical as either an LS or RS integral. However, for equivalence of the
∞ g (x)dF (x), g (X (ω ))P (dω ) and
−∞ Ω even for continuous functions g , it is essential that the integral on the right be understood as an
LS integral. Hence, in these notes, only the LS interpretation is used, and RS integration is not
needed.
If F has a corresponding pdf f , then
∞ ∞ (LebesgueStieltjes) g (x)f (x)dx g (x)dF (x) = (Lebesgue) −∞ −∞ for any Borel measurable function g . 11.6 On the convergence of the mean
p. Suppose (Xn : n ≥ 1) is a sequence of random variables such that Xn → X∞ , for some random
variable X∞ . The theorems in this section address the question of whether E [Xn ] → E [X∞ ]. The
p.
hypothesis Xn → X∞ means that for any > 0 and δ > 0, P {Xn − X∞  ≤ } ≥ 1 − δ . Thus, the
event that Xn is close to X∞ has probability close to one. But the mean of Xn can diﬀer greatly
from the mean of X if, in the unlikely event that Xn − X∞  is not small, it is very, very large.
Example 11.6.1 Suppose U is a random variable with a ﬁnite mean, and suppose A1 , A2 , . . . is a
sequence of events, each with positive probability, but such that P [An ] → 0, and let b1 , b2 , · · · be a
sequence of nonzero numbers. Let Xn = U + bn IAn for n ≥ 1. Then for any > 0, P {Xn − U  ≥
p.
} ≤ P {Xn = U } = P [An ] → 0 as n → ∞, so Xn → U . However, E [Xn ] = E [U ] + bn P [An ]. Thus,
if the bn have very large magnitude, the mean E [Xn ] can be far larger or far smaller than E [U ],
for all large n.
The simplest way to rule out the very, very large values of Xn − X∞  is to require the sequence
(Xn ) to be bounded. That would rule out using constants bn with arbitrarily large magnitudes
in Example 11.6.1. The following result is a good start–it is generalized to yield the dominated
convergence theorem further below.
Theorem 11.6.2 (Bounded convergence theorem) Let X1 , X2 , . . . be a sequence of random varip.
ables such that for some ﬁnite L, P {Xn  ≤ L} = 1 for all n ≥ 1, and such that Xn → X as
n → ∞. Then E [Xn ] → E [X ].
Proof. For any > 0, P { X ≥ L + } ≤ P { X − Xn ≥ } → 0, so that P { X ≥ L + } = 0.
Since was arbitrary, P { X ≤ L} = 1. Therefore, P {X − Xn  ≤ 2L} = 1 for all n ≥ 1. Again let
> 0. Then
X − Xn  ≤ + 2LI{X −Xn ≥ } ,
(11.5)
so that E [X ] − E [Xn ] = E [X − Xn ] ≤ E [X − Xn ] ≤ + 2LP X − Xn  ≥ }. By the hypotheses,
P {X − Xn  ≥ } → 0 as n → ∞. Thus, for n large enough, E [X ] − E [Xn ] < 2 . Since is
arbitrary, E [Xn ] → E [X ].
310 Equation (11.5) is central to the proof just given. It bounds the diﬀerence X − Xn  by on
the event {X − Xn  < }, which has probability close to one for n large, and on the complement
of this event, the diﬀerence X − Xn  is still bounded so that its contribution is small for n large
enough.
The following lemma, used to establish the dominated convergence theorem, is similar to the
bounded convergence theorem, but the variables are assumed to be bounded only on one side:
speciﬁcally, the random variables are restricted to be greater than or equal to zero. The result is
that E [Xn ] for large n can still be much larger than E [X∞ ], but cannot be much smaller. The
restriction to nonnegative Xn ’s would rule out using negative constants bn with arbitrarily large
magnitudes in Example 11.6.1. The statement of the lemma uses “liminf,” which is deﬁned in
Appendix 11.2.
Lemma 11.6.3 (Fatou’s lemma) Suppose (Xn ) is a sequence of nonnegative random variables such
p.
that Xn → X∞ . Then lim inf n→∞ E [Xn ] ≥ E [X∞ ]. (Equivalently, for any γ < E [X∞ ], E [Xn ] ≥ γ
for all suﬃciently large n.)
Proof. We shall prove the equivalent form of the conclusion given in the lemma, so let γ be
any constant with γ < E [X∞ ]. By the deﬁnition of E [X∞ ], there is a simple random variable Z
with Z ≤ X∞ such that E [Z ] ≥ γ . Since Z = X∞ ∧ Z ,
p. Xn ∧ Z − Z  = Xn ∧ Z − X∞ ∧ Z  ≤ Xn − X∞  → 0,
p. so that Xn ∧ Z → Z . Therefore, by the bounded convergence theorem, limn→∞ E [Xn ∧ Z ] = E [Z ] >
γ. Since E [Xn ] ≥ E [Xn ∧ Zn ], it follows that E [Xn ] ≥ γ for all suﬃciently large n.
Theorem 11.6.4 (Dominated convergence theorem) If X1 , X2 , . . . is a sequence of random variables and X∞ and Y are random variables such that the following three conditions hold:
p. (i) Xn → X∞ as n → ∞
(ii) P {Xn  ≤ Y } = 1 for all n
(iii) E [Y ] < +∞
then E [Xn ] → E [X∞ ].
Proof. The hypotheses imply that (Xn + Y : n ≥ 1) is a sequence of nonnegative random variables
which converges in probability to X∞ + Y . So Fatou’s lemma implies that lim inf n→∞ E [Xn +
Y ] ≥ E [X∞ + Y ], or equivalently, subtracting E [Y ] from both sides, lim inf n→∞ E [Xn ] ≥ E [X∞ ].
Similarly, since (−Xn + Y : n ≥ 1) is a sequence of nonnegative random variables which converges
in probability to −X∞ + Y , Fatou’s lemma implies that lim inf n→∞ E [−Xn + Y ] ≥ E [−X∞ + Y ],
or equivalently, lim supn→∞ E [Xn ] ≤ E [X∞ ]. Summarizing,
lim sup E [Xn ] ≤ E [X∞ ] ≤ lim inf E [Xn ].
n→∞ n→∞ In general, the liminf of a sequence is less than or equal to the limsup, and if the liminf is equal to
the limsup, then the limit exists and is equal to both the liminf and limsup. Thus, E [Xn ] → E [X∞ ]. 311 Corollary 11.6.5 (A consequence of integrability) If Z has a ﬁnite mean, then given any
there exits a δ > 0 so that if P [A] < δ , then E [ZIA ] ≤ . > 0, Proof. If not, there would exist a sequence of events An with P {An } → 0 with E [ZIAn ] ≥ . But
p.
ZIAn → 0, and ZIAn is dominated by the integrable random variable Z for all n, so the dominated
convergence theorem implies that E [ZIAn ] → 0, which would result in a contradiction. Theoretical Exercise
1. Work backwards, and deduce the dominated convergence theorem from Corollary 11.6.5.
The following theorem is based on a diﬀerent way to control the diﬀerence between E [Xn ]
for large n and E [X∞ ]. Rather than a domination condition, it is assumed that the sequence is
monotone in n.
Theorem 11.6.6 (Monotone convergence theorem) Let X1 , X2 , . . . be a sequence of random variables such that E [X1 ] > −∞ and such that X1 (ω ) ≤ X2 (ω ) ≤ · · · . Then the limit X∞ given by
X∞ (ω ) = limn→∞ Xn (ω ) for all ω is an extended random variable (with possible value ∞) and
E [Xn ] → E [X∞ ] as n → ∞.
Proof. By adding min{0, −X1 } to all the random variables involved if necessary, we can assume
without loss of generality that X1 , X2 , . . . , and therefore also X , are nonnegative. Recall that E [X ]
is equal to the supremum of the expectation of simple random variables that are less than or equal
to X . So let γ be any number such that γ < E [X ]. Then, there is a simple random variable X
less than or equal to X with E [X ] ≥ γ . The simple random variable X takes only ﬁnitely many
possible values. Let L be the largest. Then X ≤ X ∧ L, so that E [X ∧ L] > γ . By the bounded
convergence theorem, E [Xn ∧ L] → E [X ∧ L]. Therefore, E [Xn ∧ L] > γ for all large enough n.
Since E [Xn ∧ L] ≤ E [Xn ] ≤ E [X ], if follows that γ < E [Xn ] ≤ E [X ] for all large enough n. Since
γ is an arbitrary constant with γ < E [X ], the desired conclusion, E [Xn ] → E [X ], follows. 11.7 Matrices An m × n matrix over the reals R has the form a11 a12 · · · a21 a22 · · · A=.
.
.
.
.
.
am1 am2 · · · a1n
a2n
.
.
. amn where aij ∈ R for all i, j . This matrix has m rows and n columns. A matrix over the complex
numbers C has the same form, with aij ∈ C for all i, j . The transpose of an m × n matrix A = (aij )
is the n × m matrix AT = (aji ). For example 12
T
103
= 0 1
211
31
312 The matrix A is symmetric if A = AT . Symmetry requires that the matrix A be square: m = n.
The diagonal of a matrix is comprised by the entries of the form aii . A square matrix A is called
diagonal if the entries oﬀ of the diagonal are zero. The n × n identity matrix is the n × n diagonal
matrix with ones on the diagonal. We write I to denote an identity matrix of some dimension n.
If A is an m × k matrix and B is a k × n matrix, then the product AB is the m × n matrix with
ij th element k=1 ail blj . A vector x is an m × 1 matrix, where m is the dimension of the vector.
l
Thus, vectors are written in column form: x1 x2 x=.
.
.
xm
The set of all dimension m vectors over R is the m dimensional Euclidean space Rm . The inner
product of two vectors x and y of the same dimension m is the number xT y , equal to m xi yi .
i=1
The vectors x and y are orthogonal if xT y = 0. The Euclidean length or norm of a vector x is given
1
by x = (xT x) 2 . A set of vectors ϕ1 , . . . , ϕn is orthonormal if the vectors are orthogonal to each
other and ϕi = 1 for all i.
A set of vectors v1 , . . . , vn in Rm is said to span Rm if any vector in Rm can be expressed as a
linear combination α1 v1 + α2 v2 + · · · + αn vn for some α1 , . . . , αn ∈ R. An orthonormal set of vectors
ϕ1 , . . . , ϕn in Rm spans Rm if and only if n = m. An orthonormal basis for Rm is an orthonormal
set of m vectors in Rm . An orthonormal basis ϕ1 , . . . , ϕm corresponds to a coordinate system for
Rm . Given a vector v in Rm , the coordinates of v relative to ϕ1 , . . . , ϕm are given by αi = ϕT v .
i
The coordinates α1 , . . . , αm are the unique numbers such that v = α1 ϕ1 + · · · + αm ϕm .
A square matrix U is called orthonormal if any of the following three equivalent conditions is
satisﬁed:
1. U T U = I
2. U U T = I
3. the columns of U form an orthonormal basis.
Given an m × m orthonormal matrix U and a vector v ∈ Rm , the coordinates of v relative to U are
given by the vector U T v . Given a square matrix A, a vector ϕ is an eigenvector of A and λ is an
eigenvalue of A if the eigen relation Aϕ = λϕ is satisﬁed.
A permutation π of the numbers 1, . . . , m is a onetoone mapping of {1, 2, . . . , m} onto itself.
That is (π (1), . . . , π (m)) is a reordering of (1, 2, . . . , m). Any permutation is either even or odd.
A permutation is even if it can be obtained by an even number of transpositions of two elements.
Otherwise a permutation is odd. We write
(−1)π = 1 if π is even
−1 if π is odd The determinant of a square matrix A, written det(A), is deﬁned by
m
π (−1) det(A) =
π aiπ(i)
i=1 The absolute value of the determinant of a matrix A is denoted by  A . Thus  A = det(A) .
Some important properties of determinants are the following. Let A and B be m × m matrices.
313 1. If B is obtained from A by multiplication of a row or column of A by a scaler constant c,
then det(B ) = c det(A).
2. If U is a subset of Rm and V is the image of U under the linear transformation determined
by A:
V = {Ax : x ∈ U}
then
(the volume of U ) =  A  × (the volume of V )
3. det(AB ) = det(A) det(B )
4. det(A) = det(AT )
5. U  = 1 if U is orthonormal.
6. The columns of A span Rn if and only if det(A) = 0.
7. The equation p(λ) = det(λI − A) deﬁnes a polynomial p of degree m called the characteristic
polynomial of A.
8. The zeros λ1 , λ2 , . . . , λm of the characteristic polynomial of A, repeated according to multiplicity, are the eigenvalues of A, and det(A) = n λi . The eigenvalues can be complex
i=1
valued with nonzero imaginary parts.
If K is a symmetric m × m matrix, then the eigenvalues λ1 , λ2 , . . . , λm , are realvalued (not
necessarily distinct) and there exists an orthonormal basis consisting of the corresponding eigenvectors ϕ1 , ϕ2 , . . . , ϕm . Let U be the orthonormal matrix with columns ϕ1 , . . . , ϕm and let Λ be
the diagonal matrix with diagonal entries given by the eigenvalues Λ= 0
∼ λ1
λ2
..
0
∼ .
λm Then the relations among the eigenvalues and eigenvectors may be written as KU = U Λ. Therefore
K = U ΛU T and Λ = U T KU . A symmetric m × m matrix A is positive semideﬁnite if αT Aα ≥ 0
for all mdimensional vectors α. A symmetric matrix is positive semideﬁnite if and only if its
eigenvalues are nonnegative.
The remainder of this section deals with matrices over C. The Hermitian transpose of a matrix
A is the matrix A∗ , obtained from AT by taking the complex conjugate of each element of AT . For
example, 1
2
∗
1 0 3 + 2j
0
−j =
2j
1
3 − 2j 1
314 The set of all dimension m vectors over C is the mcomplex dimensional space Cm . The inner
product of two vectors x and y of the same dimension m is the complex number y ∗ x, equal to
m
∗
∗
i=1 xi yi . The vectors x and y are orthogonal if x y = 0. The length or norm of a vector x is
1
given by x = (x∗ x) 2 . A set of vectors ϕ1 , . . . , ϕn is orthonormal if the vectors are orthogonal to
each other and ϕi = 1 for all i.
A set of vectors v1 , . . . , vn in Cm is said to span Cm if any vector in Cm can be expressed as a
linear combination α1 v1 + α2 v2 + · · · + αn vn for some α1 , . . . , αn ∈ C. An orthonormal set of vectors
ϕ1 , . . . , ϕn in Cm spans Cm if and only if n = m. An orthonormal basis for Cm is an orthonormal
set of m vectors in Cm . An orthonormal basis ϕ1 , . . . , ϕm corresponds to a coordinate system for
Cm . Given a vector v in Rm , the coordinates of v relative to ϕ1 , . . . , ϕm are given by αi = ϕ∗ v .
i
The coordinates α1 , . . . , αm are the unique numbers such that v = α1 ϕ1 + · · · + αm ϕm .
A square matrix U over C is called unitary (rather than orthonormal) if any of the following
three equivalent conditions is satisﬁed:
1. U ∗ U = I
2. U U ∗ = I
3. the columns of U form an orthonormal basis.
Given an m × m unitary matrix U and a vector v ∈ Cm , the coordinates of v relative to U are
given by the vector U ∗ v . Eigenvectors, eigenvalues, and determinants of square matrices over C
are deﬁned just as they are for matrices over R. The absolute value of the determinant of a matrix
A is denoted by  A . Thus  A = det(A) .
Some important properties of determinants of matrices over C are the following. Let A and B
by m × m matrices.
1. If B is obtained from A by multiplication of a row or column of A by a constant c ∈ C, then
det(B ) = c det(A).
2. If U is a subset of Cm and V is the image of U under the linear transformation determined
by A:
V = {Ax : x ∈ U}
then
(the volume of U ) =  A 2 × (the volume of V )
3. det(AB ) = det(A) det(B )
4. det∗ (A) = det(A∗ )
5.  U = 1 if U is unitary.
6. The columns of A span Cn if and only if det(A) = 0.
7. The equation p(λ) = det(λI − A) deﬁnes a polynomial p of degree m called the characteristic
polynomial of A.
315 8. The zeros λ1 , λ2 , . . . , λm of the characteristic polynomial of A, repeated according to multiplicity, are the eigenvalues of A, and det(A) = n λi . The eigenvalues can be complex
i=1
valued with nonzero imaginary parts.
A matrix K is called Hermitian symmetric if K = K ∗ . If K is a Hermitian symmetric m × m
matrix, then the eigenvalues λ1 , 2 , . . . , λm , are realvalued (not necessarily distinct) and there
exists an orthonormal basis consisting of the corresponding eigenvectors ϕ1 , ϕ2 , . . . , ϕm . Let U be
the unitary matrix with columns ϕ1 , . . . , ϕm and let Λ be the diagonal matrix with diagonal entries
given by the eigenvalues 0
λ1
∼ λ2 Λ= .. .
0
∼ λm Then the relations among the eigenvalues and eigenvectors may be written as KU = U Λ. Therefore
K = U ΛU ∗ and Λ = U ∗ KU . A Hermitian symmetric m × m matrix A is positive semideﬁnite if
α∗ Aα ≥ 0 for all α ∈ Cm . A Hermitian symmetric matrix is positive semideﬁnite if and only if its
eigenvalues are nonnegative.
Many questions about matrices over C can be addressed using matrices over R. If Z is an m × m
matrix over C, then Z can be expressed as Z = A + Bj , for some m × m matrices A and B over R.
Similarly, if x is a vector in Cm then it can be written as x = u + jv for vectors u, v ∈ Rm . Then
Zx = (Au − Bv ) + j (Bu + Av ). There is a onetoone and onto mapping from Cm to R2m deﬁned
by u + jv → u . Multiplication of x by the matrix Z is thus equivalent to multiplication of u by
v
v
A −B
. We will show that
Z=
BA
Z 2 = det(Z )
(11.6)
so that Property 2 for determinants of matrices over C follows from Property 2 for determinants
of matrices over R.
It remains to prove (11.6). Suppose that A−1 exists and examine the two 2m × 2m matrices
A −B
BA A
0
B A + BA−1 B and . (11.7) The second matrix is obtained from the ﬁrst by left multiplying each subblock in the right column of
the ﬁrst matrix by A−1 B, and adding the result to the left column. Equivalently, the second matrix
I A−1 B
I A−1 B
is obtained by right multiplying the ﬁrst matrix by
. But det
= 1, so
0
I
0
I
that the two matrices in (11.7) have the same determinant. Equating the determinants of the two
matrices in (11.7) yields det(Z ) = det(A) det(A + BA−1 B ). Similarly, the following four matrices
have the same determinant:
A + Bj
0 0
A − Bj A + Bj
0 A − Bj
A − Bj 2A
A − Bj A − Bj
A − Bj 2A
A − Bj 0
A+BA−1 B
2 (11.8) Equating the determinants of the ﬁrst and last of the matrices in (11.8) yields that Z 2 =
det(Z ) det∗ (Z ) = det(A + Bj ) det(A − Bj ) = det(A) det(A + BA−1 B ). Combining these observations yields that (11.6) holds if A−1 exists. Since each side of (11.6) is a continuous function
of A, (11.6) holds in general.
316 Chapter 12 Solutions to Problems
1.2 Independent vs. mutually exclusive
(a) If E is an event independent of itself, then P [E ] = P [E ∩ E ] = P [E ]P [E ]. This can happen if
P [E ] = 0. If P [E ] = 0 then cancelling a factor of P [E ] on each side yields P [E ] = 1. In summary,
either P [E ] = 0 or P [E ] = 1.
(b) In general, we have P [A ∪ B ] = P [A] + P [B ] − P [AB ]. If the events A and B are independent,
then P [A ∪ B ] = P [A] + P [B ] − P [A]P [B ] = 0.3 + 0.4 − (0.3)(0.4) = 0.58. On the other hand, if the
events A and B are mutually exclusive, then P [AB ] = 0 and therefore P ]A ∪ B ] = 0.3 + 0.4 = 0.7.
(c) If P [A] = 0.6 and P [B ] = 0.8, then the two events could be independent. However, if A and B
were mutually exclusive, then P [A] + P [B ] = P [A ∪ B ] ≤ 1, so it would not possible for A and B
to be mutually exclusive if P [A] = 0.6 and P [B ] = 0.8. 1.4 Frantic search
Let D,T ,B , and O denote the events that the glasses are in the drawer, on the table, in the briefcase, or in the oﬃce, respectively. These four events partition the probability space.
(a) Let E denote the event that the glasses were not found in the ﬁrst drawer search.
E
[E T ] [ ]
(1)(0.06)
P [T E ] = P [TE ] ] = P [E D]PPD]+PPETDc ]P [Dc ] = (0.1)(0.9)+(1)(0.1) = 0.06 ≈ 0.315
0.19
P[
[
[
(b) Let F denote the event that the glasses were not found after the ﬁrst drawer search and ﬁrst
[
BPB
table search. P [B F ] = PPBF ] = P [F D]P [D]+P [F T ]P [[F ]+]P [[F ]B ]P [B ]+P [F O]P [O]
[F ]
PT
(1)(0.03)
= (0.1)(0.9)+(0.1)(0.06)+(1)(0.03)+(1)(0.01) ≈ 0.22
(c) Let G denote the event that the glasses were not found after the two drawer searches, two table
searches, and one briefcase search.
OP O
P [OG] = PP[OG] = P [GD]P [D]+P [GT ]P [[G]+]P [[G]B ]P [B ]+P [GO]P [O]
[G ]
PT = (1)(0.01)
(0.1)2 (0.9)+(0.1)2 (0.06)+(0.1)(0.03)+(1)(0.01) ≈ 0.4225 1.6 Conditional probabilities–basic computations of iterative decoding
(a) Here is one of several approaches to this problem. Note that the n pairs (B1 , Y1 ), . . . , (Bn , Yn )
317 def are mutually independent, and λi (bi ) = P [Bi = bi Yi = yi ] =
P [B = 1Y1 = y1 , . . . , Yn = yn ] = qi (yi bi )
qi (yi 0)+qi (yi 1) . Therefore P [B1 = b1 , . . . , Bn = bn Y1 = y1 , . . . , Yn = yn ]
b1 ,...,bn :b1 ⊕···⊕bn =1
n = λi (bi ).
b1 ,...,bn :b1 ⊕···⊕bn =1 i=1 (b) Using the deﬁnitions,
P [B = 1Z1 = z1 , . . . , Zk = zk ] =
= = p(1, z1 , . . . , zk )
p(0, z1 , . . . , zk ) + p(1, z1 , . . . , zk )
k
j =1 rj (1zj )
k
k
1
j =1 rj (0zj ) + 2
j =1 rj (1zj )
1
2 1
2 η
1+η k where η =
j =1 rj (1zj )
.
rj (0zj ) 1.8 Blue corners
(a) There are 24 ways to color 5 corners so that at least one face has four blue corners (there are 6
choices of the face, and for each face there are four choices for which additional corner to color blue.)
Since there are 8 = 56 ways to select 5 out of 8 corners, P [B exactly 5 corners colored blue] =
5
24/56 = 3/7.
(b) By counting the number of ways that B can happen for diﬀerent numbers of blue corners we
ﬁnd P [B ] = 6p4 (1 − p)4 + 24p5 (1 − p)3 + 24p6 (1 − p)2 + 8p7 (1 − p) + p8 .
1.10 Recognizing cumulative distribution functions √
√
√
√
(a) Valid (draw a sketch) P [X 2 ≤ 5] = P [X ≤ − 5] + P [X ≥ 5] = F1 (− 5) + 1 − F1 ( 5) =
(b) Invalid. F (0) > 1. Another reason is that F is not nondecreasing
(c) Invalid, not right continuous at 0. e−5
2. 1.12 CDF and characteristic function of a mixed type random variable
(a) Range of X is [0, 0.5]. For 0 ≤ c ≤ 0.5, P [X ≤ c] = P [U ≤ c + 0.5] = c + 0.5 Thus, 0
c<0 c + 0.5 0 ≤ c ≤ 0.5
FX (c) = 1
c ≥ 0.5
(b) ΦX (u) = 0.5 + 0.5 jux
dx
0e = 0.5 + eju/2 −1
ju 1.14 Conditional expectation for uniform density over a triangular region
(a) The triangle has base and height one, so the area of the triangle is 0.5. Thus the joint pdf is 2
inside the triangle.
(b) x/2 2dy = x
if 0 < x < 1 ∞
0
x/2
fX (x) =
fXY (x, y )dy =
2dy = 2 − x if 1 < x < 2 x−1
−∞ 0
else
318 (c) In view of part (c), the conditional density fY X (y x) is not well deﬁned unless 0 < x < 2. In
general we have fY X (y x) = 2
x if
0
if
2
if
2−x
0
if
not deﬁned if 0<x≤1
0<x≤1
1<x<2
1<x<2
x ≤ 0 or and y ∈ [0, x ]
2
and y ∈ [0, x ]
2
and y ∈ [x − 1, x ]
2
and y ∈ [x − 1, x ]
2
x≥2 Thus, for 0 < x ≤ 1, the conditional distribution of Y is uniform over the interval [0, x ]. For
2
1 < x ≤ 2, the conditional distribution of Y is uniform over the interval [x − 1, x ].
2
(d) Finding the midpoints of the intervals that Y is conditionally uniformly distributed over, or
integrating x against the conditional density found in part (c), yields:
x
4
3x−2
4 if 0 < x ≤ 1
if 1 < x < 2
E [Y X = x] = not deﬁned if x ≤ 0 or x ≥ 2 1.16 Density of a function of a random variable
(a) P [X ≥ 0.4X ≤ 0.8] = P [0.4 ≤ X ≤ 0.8X ≤ 0.8] = (0.82 − 0.42 )/0.82 = 3 .
4
(b) The range of Y is the interval [0, +∞). For c ≥ 0,
1
P {− ln(X ) ≤ c} = P {ln(X ) ≥ −c} = P {X ≥ e−c } = e−c 2xdx = 1 − e−2c so fY (c) =
2 exp(−2c) c ≥ 0
That is, Y is an exponential random variable with parameter 2.
0
else
1.18 Functions of independent exponential random variables
(a) Z takes values in the positive real line. So let z ≥ 0.
P [Z ≤ z ] = P [min{X1 , X2 } ≤ z ] = P [X1 ≤ z or X2 ≤ z ]
= 1 − P [X1 > z and X2 > z ] = 1 − P [X1 > z ]P [X2 > z ] = 1 − e−λ1 z e−λ2 z = 1 − e−(λ1 +λ2 )z
Diﬀerentiating yields that
(λ1 + λ2 )e−(λ1 +λ2 )z , z ≥ 0
0,
z<0 fZ (z ) = That is, Z has the exponential distribution with parameter λ1 + λ2 .
(b) R takes values in the positive real line and by independence the joint pdf of X1 and X2 is the
product of their individual densities. So for r ≥ 0 ,
P [R ≤ r] = P [ X1
≤ r] = P [X1 ≤ rX2 ]
X2 ∞ r x2 =
0 = ∞ λ1 e−λ1 x1 λ2 e−λ2 x2 dx1 dx2 0 (1 − e−rλ1 x2 )λ2 e−λ2 x2 dx2 = 1 − 0 319 λ2
.
rλ1 + λ2 Diﬀerentiating yields that
fR (r) = λ1 λ2
(λ1 r+λ2 )2 0, r≥0
r<0 1.20 Gaussians and the Q function
(a) Cov(3X + 2Y, X + 5Y + 10) = 3Cov(X, X ) + 10Cov(Y, Y ) = 3Var(X ) + 10Var(Y ) = 13.
+4
(b) X + 4Y is N (0, 17), so P {X + 4Y ≥ 2} = P { X√17Y ≥ √2 } = Q( √2 ).
17
17 −
(c) X − Y is N (0, 2), so P {(X − Y )2 > 9} = P {(X − Y ) ≥ 3 orX − Y ≤ −3} = 2P { X√2Y ≥ 3
√}
2 = 3
2Q( √2 ). 1.22 Working with a joint density
(a) The density must integrate to one, so c = 4/19.
(b)
2
4
19 1 (1 + xy )dy =
fX (x) =
0
fY (y ) = 4
19 3
2 (1 + xy )dx =
0 4
19 [1 + 3x
2] 2≤x≤3
else 4
19 [1 + 5y
2] 1≤y≤2
else Therefore fX Y (xy ) is well deﬁned only if 1 ≤ y ≤ 2. For 1 ≤ y ≤ 2:
fX Y (xy ) = 1+xy
1+ 5 y
2 0 2≤x≤3
for other x 1.24 Density of a diﬀerence
(a) Method 1 The joint density is the product of the marginals, and for any c ≥ 0, the probability
P {X − Y  ≤ c} is the integral of the joint density over the region of the positive quadrant such
that {x − y  ≤ c}, which by symmetry is one minus twice the integral of the density over the region
∞
{y ≥ 0 and y ≤ y +c}. Thus, P {X −Y  ≤ c} = 1−2 0 exp(−λ(y +c))λ exp(−λy )dy = 1−exp(−λc).
λ exp(−λc) c ≥ 0
That is, Z has the exponential distribution with parameter
Thus, fZ (c) =
0
else
λ.
(Method 2 The problem can be solved without calculation by the memoryless property of the exponential distribution, as follows. Suppose X and Y are lifetimes of identical lightbulbs which are
turned on at the same time. One of them will burn out ﬁrst. At that time, the other lightbulb will
be the same as a new light bulb, and X − Y ] is equal to how much longer that lightbulb will last. 1.26 Some characteristic functions
(a) Diﬀerentiation is straightforward, yielding jEX = Φ (0) = 2j or EX = 2, and j 2 E [X 2 ] =
Φ (0) = −14, so Var(x) = 14 − 22 = 10. In fact, this is the characteristic function of a N (10, 22 )
random variable.
(b) Evaluation of the derivatives at zero requires l’Hospital’s rule, and is a little tedious. A simpler
way is to use the Taylor series expansion exp(ju) = 1 + (ju) + (ju)2 /2! + (ju)3 /3!... The result
320 is EX = 0.5 and Var(X ) = 1/12. In fact, this is the characteristic function of a U (0, 1) random
variable.
(c) Diﬀerentiation is straightforward, yielding EX = Var(X ) = λ. In fact, this is the characteristic
function of a P oi(λ) random variable.
1.28 A transformation of jointly continuous random variables
(a) We are using the mapping, from the square region {(u, v ) : 0 ≤ u, v ≤ 1} in the u − v plane to
the triangular region with corners (0,0), (3,0), and (3,1) in the x − y plane, given by
x = 3u
y = uv.
The mapping is onetoone, meaning that for any (x, y ) in the range we can recover (u, v ). Indeed,
the inverse mapping is given by
x
3
3y
.
x u=
v=
The Jacobian determinant of the transformation is
J (u, v ) = det ∂x
∂u
∂y
∂u ∂x
∂v
∂y
∂v = det 30 = 3u = 0, for all u, v ∈ (0, 1)2 . vu Therefore the required pdf is
fX,Y (x, y ) = fU,V (u, v )
9u2 v 2
9y 2
=
= 3uv 2 =
J (u, v )
3u
x within the triangle with corners (0,0), (3,0), and (3,1), and fX,Y (x, y ) = 0 elsewhere.
(b) Integrating out y from the joint pdf yields
x
3 fX (x) = 0 9y 2
x dy = x2
9 if 0 ≤ x ≤ 3
else 0 Therefore the conditional density fY X (y x) is well deﬁned only if 0 ≤ x ≤ 3. For 0 ≤ x ≤ 3,
81y 2
x3 fX,Y (x, y )
=
fY X (y x) =
fX (x) 1.30 Jointly distributed variables
∞
1
V2
(a) E [ 1+U ] = E [V 2 ]E [ 1+U ] = 0 v 2 λe−λv dv
(b) P {U ≤ V } = 1∞
−λv dvdu
0 u λe = 0 11
0 1+u du 1 −λu
du
0e 321 if 0 ≤ y ≤
else 2
= ( λ2 )(ln(2)) = = (1 − e−λ )/λ. x
3 2 ln 2
.
λ2 (c) The support of both fU V and fY Z is the strip [0, 1] × [0, ∞), and the mapping (u, v ) → (y, z )
1
deﬁned by y = u2 and z = uv is onetoone. Indeed, the inverse mapping is given by u = y 2 and
1
∂ (x,y
v = zy − 2 . The absolute value of the Jacobian determinant of the forward mapping is  ∂ (u,v)  =
)
2u 0
vu = 2u2 = 2y . Thus,
1 fY,Z (y, z ) = λ −λzy − 2
2y e 0 (y, z ) ∈ [0, 1] × [0, ∞)
otherwise. 2.2 The limit of the product is the product of the limits
(a) There exists n1 so large that yn − y  ≤ 1 for n ≥ n1 . Thus, yn  ≤ L for all n, where
L = max{y1 , y2 , . . . , yn1 −1 , y  + 1}..
(b) Given > 0, there exists n so large that xn − x ≤ 2L and yn − y  ≤ 2(x+1) . Thus, for n ≥ n ,
xn yn − xy  ≤ (xn − x)yn  + x(yn − y ) ≤ xn − xL + xyn − y  ≤ 2 + 2 ≤. So xn yn → xy as n → ∞.
2.4 Limits of some deterministic series
(a) Convergent. This is the power series expansion for ex , which is everywhere convergent, evaluated at x = 3. The value of the sum is thus e3 . Another way to show the series is convergent is to
n
3
3
notice that for n ≥ 3 the nth term can be bounded above by 3 ! = 3 3 3 · · · n ≤ (4.5)( 3 )n−3 . Thus,
n
3! 4 5
4
the sum is bounded by a constant plus a geometric series, so it is convergent.
(b) Convergent. Let 0 < η < 1. Then ln n < nη for all large enough n. Also, n + 2 ≤ 2n for all large
enough n, and n +5 ≥ n for all n. Therefore, the nth term in the series is bounded above, for all suﬃnη
ciently large n, by 2n·3 = 2nη−2 . Therefore, the sum in (b) is bounded above by ﬁnitely many terms
n
∞
of the sum, plus 2 ∞ nη−2 , which is ﬁnite, because, for α > 1, ∞ n−α < 1+ 1 x−α dx = αα 1 ,
n=1
n=1
−
as shown in an example in the appendix of the notes.
(c) Not convergent. Let 0 < η < 0.2. Then log(n + 1) ≤ nη for all n large enough, so for n large
enough the nth term in the series is greater than or equal to n−5η . The series is therefore divergent.
We used the fact that ∞ n−α is inﬁnite for any 0 ≤ α ≤ 1, because it is greater than or equal
n=1
∞
to the integral 1 x−α dx, which is inﬁnite for 0 ≤ α ≤ 1.
2.6 Convergence of sequences of random variables
(a) The distribution of Xn is the same for all n, so the sequence converges in distribution to any
random variable with the distribution of X1 . To check for mean square convergence, use the fact
1
cos(a) cos(b) = (cos(a + b)+cos(a − b))/2 to calculate that E [Xn Xm ] = 2 if n = m and E [Xn Xm ] = 0
if n = m. Therefore, limn,m→∞ E [Xn Xm ] does not exist, so the sequence (Xn ) does not satisfy
the Cauchy criteria for m.s. convergence, so it doesn’t converge in the m.s. sense. Since it is a
bounded sequence, it therefore does not converge in the p. sense either. (Because for bounded
sequences, convergence p. implies convergence m.s.) Therefore the sequence doesn’t converge in
the a.s. sense either. In summary, the sequence converges in distribution but not in the other three
senses. (Another approach is to note that the distribution of Xn − X2n is the same for all n, so
that the sequence doesn’t satisfy the Cauchy criteria for convergence in probability.)
322 (b) If ω is such that 0 < Θ(ω ) < 2π , then 1 − Θ(ω)  < 1 so that limn→∞ Yn (ω ) = 0 for such ω .
π
Since P [0 < Θ(ω ) < 2π ] = 1, it follows that (Yn ) converges to zero in the a.s. sense, and hence also
in the p. and d. senses. Since the sequence is bounded, it also converges to zero in the m.s. sense.
2.8 Convergence of random variables on (0,1], version 2
(a) (a.s, p., d., not m.s.) For any ω ∈ Ω ﬁxed, the deterministic sequence Xn (ω ) converges to zero.
So Xn → 0 a.s. The sequence thus also converges in p. and d. If the sequence converged in the
m.s. sense, the limit would also have to be zero, but
E [Xn − 02 ] = E [Xn 2 ] = 1 1
n2 0 1
dω = +∞ → 0.
ω The sequence thus does not converge in the m.s. sense.
(b) (a.s, p., d., not m.s.) For any ω ∈ Ω ﬁxed, except the single point 1 which has zero probability,
the deterministic sequence Xn (ω ) converges to zero. So Xn → 0 a.s. The sequence also converges
in p. and d. If the sequence converged in the m.s. sense, the limit would also have to be zero, but
1 ω 2n dω = E [Xn − 02 ] = E [Xn 2 ] = n2
0 n2
→ 0.
2n + 1 The sequence thus does not converge in the m.s. sense.
(c) (d. only) For ω ﬁxed and irrational, the sequence does not even come close to settling down,
so intuitively we expect the sequence does not converge in any of the three strongest senses: a.s.,
m.s., or p. To prove this, it suﬃces to prove that the sequence doesn’t converge in p. Since the
sequence is bounded, convergence in probability is equivalent to convergence in the m.s. sense, so it
also would suﬃce to prove the sequence does not converge in the m.s. sense. The Cauchy criteria
for m.s. convergence would be violated if E [(Xn − X2n )2 ] → 0 as n → ∞. By the double angle
formula, X2n (ω ) = 2ω sin(2πnω ) cos(2πnω ) so that
1 E [(Xn − X2n )2 ] = ω 2 (sin(2πnω ))2 (1 − 2 cos(2πnω ))2 dω
0 and this integral clearly does not converge to zero as n → ∞. In fact, following the heuristic
reasoning below, the limit can be shown to equal E [sin2 (Θ)(1 − 2 cos(Θ))2 ]/3, where Θ is uniformly
distributed over the interval [0, 2π ]. So the sequence (Xn ) does not converge in m.s., p., or a.s.
senses.
The sequence does converge in the distribution sense. We shall give a heuristic derivation of the
limiting CDF. Note that the CDF of Xn is given by
1 FXn (c) = I{f (ω) sin(2πnω)≤c} dω (12.1) 0 where f is the function deﬁned by f (ω ) = ω. As n → ∞, the integrand in (12.1) jumps between
zero and one more and more frequently. For any small > 0, we can imagine partitioning [0, 1] into
intervals of length . The number of oscillations of the integrand within each interval converges to
inﬁnity, and the factor f (ω ) is roughly constant over each interval. The fraction of a small interval
for which the integrand is one nearly converges to P {f (ω ) sin(Θ) ≤ c} , where Θ is a random
323 variable that is uniformly distributed over the interval [0, 2π ], and ω is a ﬁxed point in the small
interval. So the CDF of Xn converges for all constants c to:
1 P {f (ω ) sin(Θ) ≤ c} dω. (12.2) 0 (Note: The following observations can be used to make the above argument rigorous. The integrals
i
in (12.1) and (12.2) would be equal if f were constant within each interval of the form ( n , i+1 ). If f
n
is continuous on [0, 1], it can be approximated by such step functions with maximum approximation
error converging to zero as n → ∞. Details are left to the reader.)
2.10 Convergence of a sequence of discrete random variables
1
(a) The CDF of Xn is shown in Figure 12.1. Since Fn (x) = FX x − n it follows that limn→∞ Fn (x) =
1F X n 0
0 1 2 34 56 Figure 12.1: Fx
FX (x−) all x. So limn→∞ Fn (x) = FX (x) unless FX (x) = FX (x−) i.e., unless x = 1, 2, 3, 4, 5, or 6.
(b) FX is continuous at x unless x ∈ {1, 2, 3, 4, 5, 6}.
(c) Yes, limn→∞ Xn = X d. by deﬁnition.
2.12 Convergence of a minimum
(a) The sequence (Xn ) converges to zero in all four senses. Here is one proof, and there are others.
For any with 0 < < 1, P [Xn − 0 ≥ ] = P [U1 ≥ , . . . , Un ≥ ] = (1 − )n , which converges
to zero as n → ∞. Thus, by deﬁnition, Xn → 0 p. Thus, the sequence converges to zero in d.
sense and, since it is bounded, in the m.s. sense. For each ω , as a function of n, the sequence of
numbers X1 (ω ), X2 (ω ), . . . is a nonincreasing sequence of numbers bounded below by zero. Thus,
the sequence Xn converges in the a.s. sense to some limit random variable. If a limit of random
variables exists in diﬀerent senses, the limit random variable has to be the same, so the sequence
(Xn ) converges a.s. to zero.
(b) For n ﬁxed, the variable Yn is distributed over the interval [0, nθ ], so let c be a number in that
interval. Then P [Yn ≤ c] = P [Xn ≤ cn−θ ] = 1 − P [Xn > cn−θ ] = 1 − (1 − cn−θ )n . Thus, if θ = 1,
c
limn→∞ P [Yn ≤ c] = 1 − limn→∞ (1 − n )n = 1 − exp(−c) for any c ≥ 0. Therefore, if θ = 1, the
sequence (Yn ) converges in distribution, and the limit distribution is the exponential distribution
with parameter one. 324 2.14 Limits of functions of random variables
(a) Yes. Since g is a continuous function, if a sequence of numbers an converges to a limit a, then
g (an ) converges to g (a). Therefore, for any ω such that limn→∞ Xn (ω ) = X (ω ), it holds that
limn→∞ g (Xn (ω )) = g (X (ω )). If Xn → X a.s., then the set of all such ω has probability one, so
g (Xn ) → g (X ) a.s.
(b) Yes. A direct proof is to ﬁrst note that g (b) − g (a) ≤ b − a for any numbers a and b. So, if
Xn → X m.s., then E [g (Xn ) − g (X )2 ] ≤ E [X − Xn 2 ] → 0 as n → ∞. Therefore g (Xn ) → g (X )
m.s. A slightly more general proof would be to use the continuity of g (implying uniform continuity
on bounded intervals) to show that g (Xn ) → g (X ) p., and then, since g is bounded, use the fact
that convergence in probability for a bounded sequence implies convergence in the m.s. sense.)
(c) No. For a counter example, let Xn = (−1)n /n. Then Xn → 0 deterministically, and hence in
the a.s. sense. But h(Xn ) = (−1)n , which converges with probability zero, not with probability
one.
(d) No. For a counter example, let Xn = (−1)n /n. Then Xn → 0 deterministically, and hence in
the m.s. sense. But h(Xn ) = (−1)n does not converge in the m.s. sense. (For a proof, note that
E [h(Xm )h(Xn )] = (−1)m+n , which does not converge as m, n → ∞. Thus, h(Xn ) does not satisfy
the necessary Cauchy criteria for m.s. convergence.)
2.16 Sums of i.i.d. random variables, II
√
1
1
(a) ΦX1 (u) = 2 eju + 2 e−ju = cos(u), so ΦSn (u) = ΦX1 (u)n = (cos(u))n , and ΦVn (u) = ΦSn (u/ n) =
√n
cos(u/ n) .
(b) if u is an even multiple of π
1
does not exist if u is an odd multiple of π
lim ΦSn (u) =
n→∞ 0
if u is not a multiple of π . lim ΦVn (u) = n→∞ lim n→∞ 1
1−
2 u
√
n 2 +o u2
n n = e− u2
2 . (c) Sn does not converge in distribution, because, for example, limn→∞ ΦSn (π ) = limn→∞ (−1)n
does not exist. So Sn does not converge in the m.s., a.s. or p. sense either. The limit of ΦVn is
the characteristic function of the N (0, 1) distribution, so that (Vn ) converges in distribution and
the limit distribution is N (0, 1). It will next be proved that Vn does not converge in probability.
The intuitive idea is that if m is much larger than n, then most of the random variables in the sum
deﬁning Vm are independent of the variables deﬁning Vn . Hence, there is no reason for Vm to be
close to Vn with high probability. The proof below looks at the case m = 2n. Note that
V2n − Vn =
= X1 + · · · + X2n X1 + · · · + Xn
√
√
−
n
2n
√
2 − 2 X1 + · · · + Xn
1
Xn+1 + · · · + X2n
√
√
+√
2
n
n
2 The two terms within the two pairs of braces are independent, and by the central limit theorem, each converges in distribution to the N (0, 1) distribution. Thus limn→∞ d. V2n − Vn = W,
√
2
2
√
2−2
1
where W is a normal random variable with mean 0 and Var(W ) =
+ √2 = 2 − 2.
2
325 Thus, limn→∞ P (V2n − Vn  > ) = 0 so by the Cauchy criteria for convergence in probability, Vn
does not converge in probability. Hence Vn does not converge in the a.s. sense or m.s. sense either.
2.18 On the growth of the maximum of n independent exponentials
(a) Let n ≥ 2. Clearly FZn (c) = 0 for c ≤ 0. For c > 0,
FZn (c) = P {max{X1 , . . . , Xn } ≤ c ln n}
= P {X1 ≤ c ln n, X2 ≤ c ln n, . . . , Xn ≤ c ln n}
= P {X1 ≤ c ln n}P {X2 ≤ c ln n} · · · P {Xn ≤ c ln n}
= (1 − e−c ln n )n = (1 − n−c )n
(b) Or, FXn (c) = (1 + xn n
n) , where xn = −n1−c . Observe that as n → ∞, −∞ c < 1
−1 c = 1
xn → 0
c > 1, so by Lemma 2.3.1 (and the monotonicity of the function ex to extend to the case x = −∞), 0 c<1
e−1 c = 1
FZn (c) → 1 c > 1.
Therefore, if Z∞ is the random variable that is equal to one with probability one, then FZn (c) →
FZ∞ (c) at the continuity points (i.e. at c = 1) of FZ∞ . So the sequence (Zn ) converges to one in
distribution.
2.20 Limit behavior of a stochastic dynamical system
Due to the persistent noise, just as for the example following Theorem 2.1.5 in the notes, the
sequence does not converge to an ordinary random variables in the a.s., p., or m.s. senses. To gain
some insight, imagine (or simulate on a computer) a typical sample path of the process. A typical
sample sequence hovers around zero for a while, but eventually, since the Gaussian variables can
be arbitrarily large, some value of Xn will cross above any ﬁxed threshold with probability one.
After that, Xn would probably converge to inﬁnity quickly. For example, if Xn = 3 for some
n, and if the noise were ignored from that time forward, then X would go through the sequence
9, 81, 6561, 43046721, 1853020188851841, 2.43 × 1030 , . . ., and one suspects the noise terms would
not stop the growth. This suggests that Xn → +∞ in the a.s. sense (and hence in the p. and d.
senses as well. (Convergence to +∞ in the m.s. sense is not well deﬁned.) Of course, then, Xn does
not converge in any sense to an ordinary random variable.
We shall follow the above intuition to prove that Xn → ∞ a.s. If Wn−1 ≥ 3 for some n, then
Xn ≥ 3. Thus, the sequence Xn will eventually cross above the threshold 3. We say that X diverges
nicely from time n if the event En = {Xn+k ≥ 3 · 2k for all k ≥ 0} is true. Note that if Xn+k ≥ 3 · 2k
and Wn+k ≥ −3 · 2k , then Xn+k+1 ≥ (3 · 2k )2 − 3 · 2k = 3 · 2k (3 · 2k − 1) ≥ 3 · 2k+1 . Therefore,
En ⊃ {Xn ≥ 3 and Wn+k ≥ −3 · 2k for all k ≥ 0}. Thus, using a union bound and the bound
326 Q(u) ≤ 1
2 exp(−u2 /2) for u ≥ 0: P [En Xn ≥ 3] ≥ P {Wn+k ≥ −3 · 2k for all k ≥ 0}
= 1 − P ∪∞ {Wn+k ≤ −3 · 2k }
k=0
∞ ∞ P {Wn+k ≤ −3 · 2k } = 1 − ≥ 1− k=0
∞ ≥ 1− 1
2 exp(−(3 · 2k )2 ) ≥ 1 −
k=0 1
2 Q(3 · 2k · √ 2) k=0
∞ (e−9 )k+1 = 1 −
k=0 e−9
≥ 0.9999.
2(1 − e−9 ) The pieces are put together as follows. Let N1 be the smallest time such that XN1 ≥ 3. Then N1 is
ﬁnite with probability one, as explained above. Then X diverges nicely from time N1 with probability at least 0.9999. However, if X does not diverge nicely from time N1 , then there is some ﬁrst
time of the form N1 + k such that XN1 +k < 3 · 2k . Note that the future of the process beyond that
time has the same evolution as the original process. Let N2 be the ﬁrst time after that such that
XN2 ≥ 3. Then X again has chance at least 0.9999 to diverge nicely to inﬁnity. And so on. Thus,
X will have arbitrarily many chances to diverge nicely to inﬁnity, with each chance having probability at least 0.9999. The number of chances needed until success is a.s. ﬁnite (in fact it has the
geometric distribution), so that X diverges nicely to inﬁnity from some time, with probability one.
2.22 Convergence analysis of successive averaging
(b) The means µn of Xn for all n are determined by the recursion µ0 = 0, µ1 = 1, and, for n ≥ 1,
n
n
µn+1 = (µn + µn−1 )/2. This second order recursion has a solution of the form µn = Aθ1 + Bθ2 ,
2 = (1 + θ )/2. This yields µ = 2 (1 − (− 1 )n ).
where θ1 and θ2 are the solutions to the equation θ
n
3
2
(c) It is ﬁrst proved that limn→∞ Dn = 0 a.s.. Note that Dn = U1 · · · Un−1 . Since ln Dn =
1
ln(U1 ) + · · · ln(Un−1 ) and E [ln Ui ] = 0 ln(u)du = (x ln x − x)1 = −1, the strong law of large
0
Dn
numbers implies that limn→∞ ln −1 = −1 a.s., which in turn implies limn→∞ ln Dn = −∞ a.s., or
n
equivalently, limn→∞ Dn = 0 a.s., which was to be proved. By the hint, for each ω such that Dn (ω )
converges to zero, the sequence Xn (ω ) is a Cauchy sequence of numbers, and hence has a limit.
The set of such ω has probability one, so Xn converges a.s.
2.24 Mean square convergence of a random series
Let Yn = X1 + · · · + Xn . We are interested in determining whether limn→∞ Yn exists in the m.s.
sense. By Proposition 2.2.3, the m.s. limit exists if and only if the limit limm,n→∞ E [Ym Yn ] exists
∧m 2
2
and is ﬁnite. But E [Ym Yn ] = n=1 σk which converges to ∞ σk as n, m → ∞. Thus, (Yn )
k
k=1
∞
2
converges in the m.s. sense if and only if k=1 σk < ∞.
2.26 A large deviation
2
Since E [X1 ] = 2 > 1, Cram´r’s theorem implies that b = (2), which we now compute. Note, for
e
2 a > 0, ∞ −ax2
dx
−∞ e = x
∞ − 2( 1 )
e 2a dx
−∞ 2 = π
a. So
∞ M (θ) = ln E [eθx ] = ln
−∞ 1
1
21
√ e−x ( 2 −θ) dx = − ln(1 − 2θ)
2
2π
327 (a) = max θa +
θ 1
ln(1 − 2θ)
2 1
1
1−
2
a
1
1
(1 − )
θ∗ =
2
a
1
b = (2) = (1 − ln 2) = 0.1534
2
e−100b = 2.18 × 10−7 .
= 2.28 A rapprochement between the central limit theorem and large deviations
(a) Diﬀerentiating with respect to θ yields M (θ) = ( dE [exp(θX )] )/E [exp(θX )]. Diﬀerentiating again
dθ
d2 E [X exp(θX )]
E [exp(θX )] − ( dE [exp(θX )] )2 /E [exp(θX )]2 .
dθ
(dθ)2
dk E [exp(θX )]
k exp(θX )]. Therefore,
= E [X
expectation yields
(dθ)k yields M (θ) = Interchanging diﬀeren tiation and
M (θ) = E [X exp(θX )]/E [exp(θX )], which is the mean for the tilted distribution fθ , and
M (θ) = E [X 2 exp(θX )]E [exp(θX )] − E [X exp(θX )]2 /E [exp(θX )]2 , which is the second moment, minus the ﬁrst moment squared, or simply the variance, for the tilted density fθ .
(b) In particular, M (0) = 0 and M (0) = Var(X ) = σ 2 , so the second order Taylor’s approximation
2 σ2
2
for M near zero is M (θ) = θ2 σ 2 /2. Therefore, (a) for small a satisﬁes (a) ≈ maxθ (aθ − θ 2 ) = 2a 2 ,
σ
√
√
so as n → ∞, the large deviations upper bound behaves as P [Sn ≥ b n] ≤ exp(−n (b/ n)) ≈
2
2
b
exp(−n 2σ2 n ) = exp(− 2b 2 ). The exponent is the same as in the bound/approximation to the central
σ
limit approximation described in the problem statement. Thus, for moderately large b, the central
limit theorem approximation and large deviations bound/approximation are consistent with each
other.
2.30 Large deviations of a mixed sum
Modifying the derivation for iid random variables, we ﬁnd that for θ ≥ 0:
P Sn
≥a
n ≤ E [eθ(Sn −an) ]
= E [eθX1 ]nf E [eθY1 ]n(1−f ) e−nθa
= exp(−n[θa − f MX (θ) − (1 − f )MY (θ)]) where MX and MY are the log moment generating functions of X1 and Y1 respectively. Therefore,
l(f, a) = max θa − f MX (θ) − (1 − f )MY (θ)
θ where
MX (θ) = − ln(1 − θ) θ < 1
+∞
θ≥1 ∞ MY (θ) = ln
k=0 eθk e−1
θ
= ln(ee −1 ) = eθ − 1,
k! Note that l(a, 0) = a ln a + 1 − a (large deviations exponent for the P oi(1) distribution) and
l(a, 1) = a − 1 − ln(a) (large deviations exponent for the Exp(1) distribution). For 0 < f < 1 we
328 compute l(f, a) by numerical optimization. The result is
f
0
0+
1 /3
2/3
1
l(f, 4) 2.545 2.282 1.876 1.719 1.614
Note: l(4, f ) is discontinuous in f at f = 0. In fact, adding only one exponentially distributed
random variable to a sum of Poisson random variables can change the large deviations exponent.
2.32 The limit of a sum of cumulative products of a sequence of uniform random variables
m.s.
(a) Yes. E [(Bk − 0)2 ] = E [A2 ]k = ( 5 )k → 0 as k → ∞. Thus, Bk → 0.
1
8
(b) Yes. Each sample path of the sequence Bk is monotone nonincreasing and bounded below by
zero, and is hence convergent. Thus, limk→∞ Bk a.s. exists. (The limit has to be the same as the
m.s. limit, so Bk converges to zero almost surely.)
5
(c) If j ≤ k , then E [Bj Bk ] = E [A2 · · · A2 Aj +1 · · · Ak ] = ( 8 )j ( 3 )k−j . Therefore,
1
j
4
n E [Sn Sm ] = E [ m n Bj j =1
k=1
∞
∞ =2
j =1 k=j +1
∞∞ =2
j =1 l=1 ∞ 5
8 =
j =1 ∞ m 5
8 5
8 j 3
4
3
4 5
8 j =1
∞ 5
8 +
j =1 ∞ j ∞ + l 2
l=1 3
4 E [Bj Bk ] (12.3) j =1 k=1 j =1 k=1
k−j j ∞ E [Bj Bk ] → Bk ] = j j l (12.4) +1 5
35
(2 · 3 + 1) =
3
3 = A visual way to derive (12.4), is to note that (12.3) is the sum of all entries in the inﬁnite 2d array:
.
.
.
32
4
3
4 5
8
5
8
5
8
j .
.
. .
.
. 52 3
8
4
52
8
5
3
8
4 l 5
8
5
8 53
8
23
4
32
4 .·.
···
···
··· 5
Therefore, 8
2∞3
+ 1 is readily seen to be the sum of the j th term on the diagonal,
l=1 4
plus all terms directly above or directly to the right of that term.
(d) Mean square convergence implies convergence of the mean. Thus, the mean of the limit is
3
limn→∞ E [Sn ] = limn→∞ n=1 E [Bk ] = ∞ ( 4 )k = 3. The second moment of the limit is the
k
k=1
35
limit of the second moments, namely 3 , so the variance of the limit is 35 − 32 = 8 .
3
3
(e) Yes. Each sample path of the sequence Sn is monotone nondecreasing and is hence convergent.
Thus, limn→∞ Sn a.s. exists. The limit has to be the same as the m.s. limit. (To show that the limit
is ﬁnite with probability one, and hence an ordinary random variable, the monotone convergence 329 theorem could be applied, yielding E [limn→∞ S∞ ] = limn→∞ E [Sn ] = 1. )
3.2 Linear approximation of the cosine function over an interval
1π
E [Y Θ] = E [Y ] + Cov(Θ,Y ) (Θ − E [Θ]), where E [Y ] = π 0 cos(θ)dθ = 0, E [Θ] = π , Var(Θ) =
2
Var(Θ)
π π 2
E [ΘY ] = 0 θ cos(θ) dθ = θ sin(θ) π − 0 sin(θ) dθ = − π , and Cov(Θ, Y ) = E [ΘY ] − E [Θ]E [Y ] =
0
π
π
π
12
24
24
Therefore, E [Y Θ] = − π3 (Θ − π ), so the optimal choice is a = π2 and b = − π3 .
2 π2
12 ,
2
−π. 3.4 Valid covariance matrix
Set a = 1 to make K symmetric. Choose b so that the determinants of the following seven matrices
are nonnegative:
(2) (1) 21
11 (1) 2b
b1 10
01 K itself The ﬁfth matrix has determinant 2 − b2 and det(K ) = 2 − 1 − b2 = 1 − b2 . Hence K is a valid
covariance matrix (i.e. symmetric and positive semideﬁnite) if and only if a = 1 and −1 ≤ b ≤ 1.
3.6 Conditional probabilities with joint Gaussians II
1
1
(a) P [X − 1 ≥ 2] = P [X ≤ −1 or X ≥ 3] = P [ X ≤ − 2 ] + P [ X ≥ 3 ] = Φ(− 2 ) + 1 − Φ( 3 ).
2
2
2
2
Cov (X,Y )
(b) Given Y = 3, the conditional density of X is Gaussian with mean E [X ]+ Var(Y ) (3 − E [Y ]) = 1
2 2 6
and variance Var(X ) − Cov(X,Y)) = 4 − 18 = 2.
Var(Y
(c) The estimation error X − E [X Y ] is Gaussian, has mean zero and variance 2, and is independent of Y . (The variance of the error was calculated to be 2 in part (b)). Thus the probability is
1
1
1
1
Φ(− √2 ) + 1 − Φ( √2 ), which can also be written as 2Φ(− √2 ) or 2(1 − Φ( √2 )). 3.8 An MMSE estimation problem
1 1+x
5
(a) E [XY ] = 2 0 2x xydxdy = 12 . The other moments can be found in a similar way. Alternatively, note that the marginal densities are given by 0≤y≤1
y
2(1 − x) 0 ≤ x ≤ 1
2−y 1≤y ≤2
fY (y ) =
fX (x) =
0
else 0
else
so that EX = 1 , Var(X ) =
3
E [X  Y ] =
E [e2 ] = 1
18 , EY = 1, Var(Y ) = 1 , Cov(X, Y ) =
6 11
1
+ ( )−1 (Y − 1) =
3 12 6
1
11
1
− ( )( )−1 ( ) =
18
12 6
12 5
12 − 1
3 = 1
12 . So 1 Y −1
+
3
2
1
= the MMSE for E [X Y ]
72 Inspection of Figure 12.2 shows that for 0 ≤ y ≤ 2, the conditional distribution of X given Y = y is
the uniform distribution over the interval [0, y/2] if 0 ≤ y ≤ 1 and the over the interval [y − 1, y/2]
if 1 ≤ y ≤ 2. The conditional mean of X given Y = y is thus the midpoint of that interval, yielding:
E [X Y ] = Y
4
3Y −2
4 330 0≤Y ≤1
1≤Y ≤2 y 2 E[XY=y] 1 E[XY=y] x
0 1 Figure 12.2: Sketch of E [X Y = y ] and E [X Y = y ].
To ﬁnd the corresponding MSE, note that given Y , the conditional distribution of X is uniform
over some interval. Let L(Y ) denote the length of the interval. Then
1
L(Y )2 ].
12
1
y
1
1
y ( )2 dy =
12 0
2
96 E [e2 ] = E [E [e2 Y ]] = E [
=2 For this example, the MSE for the best estimator is 25% smaller than the MSE for the best linear
estimator.
(b)
∞
∞
12
1
y
2
2
√ e− 2 y dy =
EX =
y  √ e−y /2 dy = 2
and EY = 0,
π
2π
2π
−∞
0
20
2
+ Y≡
π1
π
That is, the best linear estimator is the constant EX . The corresponding MSE is Var(X ) =
2
2
E [X 2 ] − (EX )2 = E [Y 2 ] − π = 1 − π . Note that Y  is a function of Y with mean square error
E [(X − Y )2 ] = 0. Nothing can beat that, so Y  is the MMSE estimator of X given Y . So
Y  = E [X Y ]. The corresponding MSE is 0, or 100% smaller than the MSE for the best linear
estimator.
Var(Y ) = 1, Cov(X, Y ) = E [Y Y ] = 0 so E [X Y ] = 3.10 Conditional Gaussian comparison
X
(a) pa = P {X ≥ 2} = P { √10 ≥ √2 } = Q( √2 ) = Q(0.6324).
10
10
(b) By the theory of conditional distributions for jointly Gaussian random variables, the conditional
2
distribution of X given Y = y is Gaussian, with mean E [X Y = y ] and variance σe , which is the
MSE for estimation of X by E [X Y ]. Since X and Y are mean zero and Cov(X,Y ) = 0.8, we have
Var(Y )
2
2 = Var(X ) − Cov(X,Y ) = 3.6. Hence, given Y = y , the conditional
E [X Y = y ] = 0.8y, and σe
Var(Y )
331 √
distribution of X is N (0.8y, 3.6). Therefore, P [X ≥ 2Y = y ] = Q( 2−(0..8)y ). In particular, pb =
36
√
P [X ≥ 2Y = 3] = Q( 2−(0..8)3 ) = Q(−0.2108).
36
(c) Given the event {Y ≥ 3}, the conditional pdf of Y is obtained by setting the pdf of Y to zero on
the interval (−∞, 3), and then renormalizing by P {Y ≥ 3} = Q( √3 ) to make the density integrate
10
to one. We can write this as fY (y) = e−y2 /20
√
y≥3
1−FY (3)
Q( √3 ) 20π
fY Y ≥3 (y ) =
10 0
else. Using this density, by considering the possible values of Y , we have
∞ pc = P [X ≥ 2Y ≥ 3] = ∞ P [X ≥ 2, Y ∈ dy Y ≥ 3] =
3
∞ Q( = P [X ≥ 2Y = y ]P [Y ∈ dy Y ≥ 3]
3 3 2 − (0.8)y
√
)fY Y ≥3 (y )dy
3.6 (ALTERNATIVE) The same expression can be derived in a more conventional fashion as follows:
pe = P [X ≥ 2Y ≥ 3] =
∞ P {X ≥ 2, Y ≥ 3}
P {Y ≥ 3} ∞ fX Y (xy )dx fY (y )dy/P {Y ≥ 3} =
3 2
∞ = Q
3 2 − (0.8)y
√
3.6 ∞ fY (y )dy/(1 − FY (3)) = Q(
3 2 − (0.8)y
√
)fY Y ≥3 (y )dy
3.6 (d) We will show that pa < pb < pc . The inequality pa < pb follows from parts (a) and (b) and
√
the fact the function Q is decreasing. By part (c), pc is an average of Q( 2−(0..8)y ) with respect to y
36
√
over the region y ∈ [3, ∞) (using the pdf fY Y ≥3 ). But everywhere in that region, Q( 2−(0..8)y ) > pb ,
36
showing that pc > pb . 3.12 An estimator of an estimator
To show that E [X Y ] is the LMMSE estimator of E [X Y ], it suﬃces by the orthogonality principle
to note that E [X Y ] is linear in (1, Y ) and to prove that E [X Y ] − E [X Y ] is orthogonal to 1
and to Y . However E [X Y ] − E [X Y ] can be written as the diﬀerence of two random variables
(X − E [X Y ]) and (X − E [X Y ]), which are each orthogonal to 1 and to Y . Thus, E [X Y ] − E [X Y ]
is also orthogonal to 1 and to Y , and the result follows.
Here is a generalization, which can be proved in the same way. Suppose V0 and V1 are two
closed linear subspaces of random variables with ﬁnite second moments, such that V0 ⊃ V1 . Let
X be a random variable with ﬁnite second moment, and let Xi∗ be the variable in Vi with the
∗
minimum mean square distance to X , for i = 0 or i = 1. Then X1 is the variable in V1 with the
∗.
minimum mean square distance to X0
Another solution to the original problem can be obtained by using the formula for E [Z Y ]
applied to Z = E [X Y ]:
E [E [X Y ]Y ] = E [E [X Y ]] + Cov(Y, E [X Y ])Var(Y )−1 (Y − EY )
332 which can be simpliﬁed using E [E [X Y ]] = EX and
Cov(Y, E [X Y ]) = E [Y (E [X Y ] − EX )]
= E [Y E [X Y ]] − EY · EX
= E [E [XY Y ]] − EY · EX
= E [XY ] − EX · EY = Cov(X, Y )
to yield the desired result.
3.14 Some identities for estimators
(a) True. The random variable E [X Y ] cos(Y ) has the following two properties:
• It is a function of Y with ﬁnite second moments (because E [X Y ] is a function of Y with
ﬁnite second moment and cos(Y ) is a bounded function of Y )
• (X cos(Y ) − E [X Y ] cos(Y )) ⊥ g (Y ) for any g with E [g (Y )2 ] < ∞ (because for any such g ,
E [(X cos(Y )−E [X Y ] cos(Y ))g (Y )] = E [(X −E [X Y ])g (Y )] = 0, where g (Y ) = g (Y ) cos(Y ).)
Thus, by the orthogonality principle, E [X Y ] cos(Y ) is equal to E [X cos(Y )Y ].
(b) True. The left hand side is the projection of X onto the space {g (Y ) : E [g (Y )2 ] < ∞} and the
right hand side is the projection of X onto the space {f (Y 3 ) : E [f (Y 3 )2 ] < ∞}. But these two
spaces are the same, because for each function g there is the function f (u) = g (u1/3 ). The point is
that the function y 3 is an invertible function, so any function of Y can also be written as a function
of Y 3 .
(c) False. For example, let X be uniform on the interval [0, 1] and let Y be identically zero. Then
1
E [X 3 Y ] = E [X 3 ] = 4 and E [X Y ]3 = E [X ]3 = 1 .
8
(d) False. For example, let P {X = Y = 1} = P {X = Y = −1} = 0.5. Then E [X Y ] = Y while
E [X Y 2 ] = 0. The point is that the function y 2 is not invertible, so that not every function of Y
can be written as a function of Y 2 . Equivalently, Y 2 can give less information than Y .
(e) False. For example, let X be uniformly distributed on [−1, 1], and let Y = X . Then E [X Y ] = Y
3
4
while E [X Y 3 ] = E [X ] + Cov(X,Y ) ) (Y 3 − E [Y 3 ]) = E [X 6 ] Y 3 = 7 Y 3 .
5
E [X ]
Var(Y 3
3.16 Some simple examples
Of course there are many valid answers for this problem–we only give one.
(a) Let X denote the outcome of a roll of a fair die, and let Y = 1 if X is odd and Y = 2 if X is
even. Then E [X Y ] has to be linear. In fact, since Y has only two possible values, any function
of Y can be written in the form a + bY. That is, any function of Y is linear. (There is no need to
even calculate E [X Y ] here, but we note that is is given by E [Y X ] = X + 2.)
(b) Let X be a N(0,1) random variable, and let W be indpendent of X , with P {W = 1} = P {W =
−1} = 1 . Finally, let Y = XW . The conditional distribution of Y given W is N (0, 1), for either
2
possible value of W , so the unconditional value of Y is also N (0, 1). However, P {X − Y = 0} = 0.5,
so that X − Y is not a Gaussian random variable, so X and Y are not jointly Gaussian.
(c) Let the triplet (X, Y, Z ) take on the four values (0, 0, 0), (1, 1, 0), (1, 0, 1), (0, 1, 1) with equal
probability. Then any pair of these variables takes the values (0, 0), (0, 1), (1, 0), (1, 1) with equal
probability, indicating pairwise independence. But
1
P {(X, Y, Z } = (0, 0, 1)} = 0 = P {X = 0}P {Y = 0}P {Z = 1} = 8 . So the three random variables
333 are not independent.
3.18 Estimating a quadratic
(a) Recall the fact that E [Z 2 ] = E [Z ]2 + Var(Z ) for any second order random variable Z . The
idea is to apply the fact to the conditional distribution of X given Y . Given Y , the conditional
distribution of X is Gaussian with mean ρY and variance 1 − ρ2 . Thus, E [X 2 Y ] = (ρY )2 + 1 − ρ2 .
(b) The MSE=E [(X 2 )2 ]−E [(E [X 2 Y ])2 ] = E [X 4 ]−ρ4 E [Y 4 ]−2ρ2 E [Y 2 ](1−ρ2 )−(1−ρ2 )2 = 2(1−ρ4 )
(c) Since Cov(X 2 , Y ) = E [X 2 Y ] = 0, it follows that E [X 2 Y ] = E [X 2 ] = 1. That is, the best linear
estimator in this case is just the constant estimator equal to 1.
3.20 An innovations sequence and its application
2
2
e
2
(a) Y1 = Y1 . (Note: E [Y1 ] = 1), Y2 = Y2 − E [YfY1 ] Y1 = Y2 − 0.5Y1 (Note: E [Y2 ] = 0.75.)
2
E [ Y1 ] e
E [ Y3 Y1 ]
Y
f2 ] 1
E [ Y1 e
E [ Y3 Y 2 ]
Y
f2 ] 2
E [ Y2 Y3 = Y3 −
−
= Y3 − (0.5)Y1 − 1 Y2 = Y3 − 1 Y1 − 1 Y2 . Summarizing,
3
3
3 Y1
Y1
Y1
1
00 1 Y2 = A Y2 where A = − 2 1 0 Y2 .
−1 −1 1
Y3
Y3
Y3
3
3 Y1
Y1
1 0.5 0.5
(b) Since Cov Y2 = 0.5 1 0.5 and Cov(X, Y2 ) = (0 0.25 0.25),
0.5 0.5 1
Y3
Y3 Y1
1 0.5 0.5
100 it follows that Cov Y2 = A 0.5 1 0.5 AT = 0 3 0 ,
4
002
0.5 0.5 1
Y3
3 Y1 1
and that Cov(X, Y2 ) = (0 0.25 0.25)AT = (0 1 6 ).
4
Y3
e
e
e
Cov(X,Y1 ) = 0 b = Cov(X,Y2 ) = 1 c = Cov(X,Y3 ) = 1 .
(c) a =
e2
E [ Y1 ] e2
E [ Y2 ] 3 e2
E [ Y3 ] 4 3.22 A Kalman ﬁltering example
(a)
xk+1k = f xkk−1 + Kk (yk − xkk−1 )
2
2
22
2
σk+1 = f 2 (σk − σk (σk + 1)−1 σk ) + 1 = and 2
σk f 2
2 +1
1 + σk σ2 Kk = f ( 1+k 2 ).
σ
k 2
2
(b) Since σk ≤ 1 + f 2 for k ≥ 1, the sequence (σk ) is bounded for any value of f . 3.24 A variation of Kalman ﬁltering
Equations (3.25) and (3.24) are still true: xk+1k = f xkk−1 + Kk (yk − xkk−1 ) and
Kk = Cov(xk+1 − f xkk−1 , yk )Cov(˜k )−1 . The variable vk in (3.28) is replaced by wk to yield:
˜
y
Cov(xk+1 − f xkk−1 , yk ) = Cov(f (xk − xkk−1 ) + wk , xk − xkk−1 ) + wk
˜
2
= f σk + 1 334 2
As before, writing σk for Σkk−1 ,
2
Cov(˜k ) = Cov((xk − xkk−1 ) + wk ) = σk + 1
y So now
Kk = 2
f σk + 1
2
1 + σk and 2
2
σk+1 = (f 2 σk + 1) − = 2
(f σk + 1)2
2
1 + σk 2
(1 − f )2 σk
2
1 + σk 2
Note that if f = 1 then σk = 0 for all k ≥ 1, which makes sense because xk = yk−1 in that case. 3.26 An innovations problem
2
2
2
2
2
(a) E [Yn ] = E [U1 · · · Un ] = E [U1 ] · · · E [Un ] = 2−n and E [Yn ] = E [U1 · · · Un ] = E [U1 ] · · · E [Un ] =
−n , so Var(Y ) = 3−n − (2−n )2 = 3−n − 4−n .
3
n
(b) E [Yn Y0 , . . . , Yn−1 ] = E [Yn−1 Un Y0 , . . . , Yn−1 ] = Yn−1 E [Un Y0 , . . . , Yn−1 ] = Yn−1 E [Un ] = Yn−1 /2.
(c) Since the conditional expectation found in (b) is linear, it follows that E [Yn Y0 , . . . , Yn−1 ] =
E [Yn Y0 , . . . , Yn−1 ] = Yn−1 /2.
1
(d) Y0 = Y0 = 1, and Yn = Yn − Yn−1 /2 (also equal to U1 · · · Un−1 (Un − 2 )) for n ≥ 1.
2
2
(e) For n ≥ 1, Var(Yn ) = E [(Yn )2 ] = E [U1 · · · Un−1 (Un − 1 )2 ] = 3−(n−1) /12 and
2
1
Cov (XM , Yn ) = E [(U1 + · · · + UM )Yn ] = E [(U1 + · · · + UM )U1 · · · Un−1 (Un − 2 )]
1
= E [Un (U1 · · · Un−1 )(Un − 2 )] = 2−(n−1) Var(Un ) = 2−(n−1) /12. Since Y0 = 1 and all the other
innovations variables are mean zero, we have
E [XM Y0 , . . . , YM ] = = = M
+
2
M
+
2
M
+
2 M Cov (XM , Yn )Yn n=1
M
n=1
M Var(Yn )
2−n+1 /12
Yn
3−n+1 /12 3
( )n−1 Yn
2 n=1 3.28 Linear innovations and orthogonal polynomials for the uniform distribution
(a)
1n
1
u
un+1 1
n
n+1 n even
E [U ] =
du =
=
0
n odd
2(n + 1) −1
−1 2
(b) The formula for the linear innovations sequence yields:
1
U
Y1 = U , Y2 = U 2 − 3 , Y3 = U 3 − 35 , and
Y4 = U 4 − E [U 4 ·1]
E [12 ] ·1− 1
E [U 4 (U 2 − 3 )]
(U 2
1
E [(U 2 − 3 )2 ] 1
− 3) = U4 − 1 −
5 335 1
−1
7
5
1
− 2 +1
5
3 6
(U 2 − 1) = U 4 − 7 U 2 + 3
35 . Note: These mutually orthogonal (with respect to the uniform distribution on [1,1] ) polynomials 1, U ,
1
6
3
U 2 − 3 , U 3 − 3 U , U 4 − 7 U 2 + 35 are (up to constant multiples) known as the Legendre polynomials.
5
3.30 Kalman ﬁlter for a rotating state
(a) Given a nonzero vector xk , the vector F xk is obtained by rotating the vector xk one tenth
revolution counterclockwise about the origin, and then shrinking the vector towards zero by one
percent. Thus, successive iterates F k xo spiral in towards zero, with onetenth revolution per time
unit, and shrinking by about ten percent per revolution.
ac
T
(b) Suppose Σkk−1 =
. Note that Hk Σkk−1 Hk is simply the scaler a. The equations for
cb
Σk+1k and Kk can thus be written as
Σk+1k = F Σkk−1 −
Kk = F a
1+a
c
1+a a2
1+a
ac
1+a ac
1+a
c2
1+a FT + Q . 4.2 Correlation function of a product
RX (s, t) = E [Ys Zs Yt Zt ] = E [Ys Yt Zs Zt ] = E [Ys Yt ]E [Zs Zt ] = RY (s, t)RZ (s, t)
4.4 Another sinusoidal random process
(a) Since EX1 = EX2 = 0, EYt ≡ 0. The autocorrelation function is given by
2
2
RY (s, t) = E [X1 ] cos(2πs) cos(2πt) − 2E [X1 X2 ] cos(2πs) sin(2πt) + E [X2 ] sin(2πs) sin(2πt) = σ 2 (cos(2πs) cos(2πt) + sin(2πs) sin(2πt)]
= σ 2 cos(2π (s − t)) (a function of s − t only)
So (Yt : t ∈ R) is WSS.
(b) If X1 and X2 are each Gaussian random variables and are independent then (Yt : t ∈ R) is a
realvalued Gaussian WSS random process and is hence stationary.
(c) A simple solution to this problem is to take X1 and X2 to be independent, mean zero, variance
σ 2 random variables with diﬀerent distributions. For example, X1 could be N (0, σ 2 ) and X2 could
1
be discrete with P (X1 = σ ) = P (X1 = −σ ) = 2 . Then Y0 = X1 and Y3/4 = X2 , so Y0 and Y3/4 do
not have the same distribution, so that Y is not stationary.
4.6 Brownian motion: Ascension and smoothing
(a) Since the increments of W over nonoverlapping intervals are independent, mean zero Gaussian
random variables,
P {Wr ≤ Ws ≤ Wt } = P {Ws − Wr ≥ 0, Wt − Ws ≥ 0}
= P {Ws − Wr ≥ 0}P {Wt − Ws ≥ 0} = 1
11
·=.
22
4 (b) Since W is a Gaussian process, the three random variables Wr , Ws , Wt are jointly Gaussian.
336 They also all have mean zero, so that
E [Ws Wr , Wt ] = E [Ws Wr , Wt ]
= (Cov(Ws , Wr ), Cov(Ws , Wt )) Var(Xr )
Cov(Xr , Xt )
Cov(Xt , Xr )
Var(Xt ) −1 Wr
Wt −1 rr
Wr
rt
Wt
(t − s)Wr + (s − r)Wt
,
t−r = (r, s)
= −1 d −b
ab
1
. Note that as s varies from r to t,
= ad−bc
cd
−c a
E [Ws Wr , Wt ] is obtained by linearly interpolating between Wr and Wt . where we use the fact 4.8 Some Poisson process calculations
(a) (Method 1) Given N2 = 2, the distribution of the locations of the ﬁrst two jump points are as
if they are independently and uniformly distributed over the interval [0, 2]. Thus, each falls in the
second half of the interval with probability 1/2, and they both fall into the second interval with
1
probability ( 2 )2 = 1 . Thus, at least one of them falls in the ﬁrst half of the interval with probabilty
4
3
1 − 1 = 4.
4
(Method 2) Since N1 and N2 − N1 are independent, Poi(λ) random variables,
P {N1 ≥ 1, N2 = 2} = P {N1 = 1, N2 = 2} + P {N1 = 2, N2 = 2}
= P {N1 = 1, N2 − N1 = 1} + P {N1 = 2, N2 − N1 = 0}
λ2 e−λ −λ 3λ2 e−2λ
= (λe−λ )(λe−λ ) + (
)e =
2
2
2 −2λ 2 −2λ 2 −2λ and P {N2 } = (2λ) 2e . Thus, P [N2 = 2N1 ≥ 1] = 3λ e /( (2λ) 2e ) = 3 .
2
4
2 −2λ
(b) P {N1 ≥ 1} = 1 − e−λ . Therefore, P [N2 = 2N1 ≥ 1] = 3λ e /(1 − e−λ ).
2
(c) Yes. The process N is Markov (because N0 is constant and N has independent increments),
and since Xt and Nt are functions of each other for each t, the process X is also Markov. The state
space for X is S = {0, 1, 4, 9, . . .}. For i, j ∈ Z and t, τ ≥ 0,
2 2 pi2 ,j 2 (τ ) = P [Xt+τ = j Xt = i ] = (λτ )j −i e−λτ
(j −i)! 0 0≤i≤j
else 4.10 MMSE prediction for a Gaussian process based on two observations 5 0 −5
9
5
(a) Since RX (0) = 5, RX (1) = 0, and RX (2) = − 9 , the covariance matrix is 0 5 0 .
−5 0 5
9
(2)
(b) As the variables are all mean zero, E [X (4)X (2)] = Cov(X (4),X (2)) X (2) = − X9 .
Var(X (2)
(c) The variable X (3) is uncorrelated with (X (2), X (4))T . Since the variables are jointly Gaussian,
(2)
X (3) is also independent of (X (2), X (4))T . So E [X (4)X (2)] = E [X (4)X (2), X (3)] = − X9 . 337 4.12 Poisson process probabilities
(a) The numbers of arrivals in the disjoint intervals are independent, Poisson random variables with
mean λ. Thus, the probability is (λe−λ )3 = λ3 e−3λ .
(b) The event is the same as the event that the numbers of counts in the intervals [0,1], [1,2], and
2
2
2
[2,3] are 020, 111, or 202. The probability is thus e−λ ( λ e−λ )e−λ + (λe−λ )3 + ( λ e−λ )e−λ ( λ e−λ ) =
2
2
2
4
2
( λ + λ3 + λ )e−3λ .
2
4
(c) This is the same as the probability the counts are 020, divided by the answer to part (b), or
λ2
λ2
λ4
3
2
2
2 /( 2 + λ + 4 ) = 2λ /(2 + 4λ + λ ).
4.14 Adding jointly stationary Gaussian processes
X (t)+Y (t)
(a) RZ (s, t) = E [ X (s)+Y (s) = 1 [RX (s − t) + RY (s − t) + RXY (s − t) + RY X (s − t)].
2
2
4
So RZ (s, t) is a function of s − t. Also, RY X (s, t) = RXY (t, s). Thus,
−τ −3
−τ +3
1
RZ (τ ) = 4 [2e−τ  + e 2 + e 2 ].
(b) Yes, the mean function of Z is constant (µZ ≡ 0) and RZ (s, t) is a function of s − t only, so Z
is WSS. However, Z is obtained from the jointly Gaussian processes X and Y by linear operations,
so Z is a Gaussian process. Since Z is Gaussian and WSS, it is stationary.
1
1
(c) P [X (1) < 5Y (2) + 1] = P [ X (1)−5Y (2) ≤ σ ] = Φ( σ ), where
σ
−4
σ 2 = Var(X (1) − 5Y (2)) = RX (0) − 10RXY (1 − 2) + 25RY (0) = 1 − 10e + 25 = 26 − 5e−4 .
2
4.16 A linear evolution equation with random coeﬃcients
2
2
2
2
(a) Pk+1 = E [(Ak Xk + Bk )2 ] = E [A2 Xk ] + 2E [Ak Xk ]E [Bk ] + E [Bk ] = σA Pk + σB .
k
(b) Yes. Think of n as the present time. The future values Xn+1 , Xn+2 , . . . are all functions of Xn
and (Ak , Bk : k ≥ n). But the variables (Ak , Bk : k ≥ n) are independent of X0 , X1 , . . . Xn . Thus,
the future is conditionally independent of the past, given the present.
(c) No. For example, X1 − X0 = X1 = B1 , and X2 − X1 = A2 B1 + B2 , and clearly B1 and A2 B1 + B2
2
2
are not independent. (Given B1 = b, the conditional distribution of A2 B1 + B2 is N (0, σA b2 + σB ),
which depends on b.)
(d) Suppose s and t are integer times with s < t. Then RY (s, t) = E [Ys (At−1 Yt−1 + Bt−1 )] =
Pk if s = t = k
E [At−1 ]E [Ys Yt−1 ] + E [Ys ]E [Bt−1 ] = 0. Thus, RY (s, t) =
0 else.
(e) The variables Y1 , Y2 , . . . are already orthogonal by part (d) (and the fact the variables have
mean zero). Thus, Yk = Yk for all k ≥ 1.
4.18 A ﬂy on a cube
(a)(b) See the ﬁgures. For part (a), each twoheaded line represents a directed edge in each
direction and all directed edges have probability 1/3.
(a) 100 000 010 110 (b)
2/3 1 001 101 111 0 1/3 1 1/3
2 2/3 3
1 011 (c) Let ai be the mean time for Y to ﬁrst reach state zero starting in state i. Conditioning on the
338 2
ﬁrst time step yields a1 = 1 + 2 a2 , a2 = 1 + 3 a1 + 1 a3 , a3 = 1 + a2 . Using the ﬁrst and third
3
3
of these equations to eliminate a1 and a3 from the second equation yields a2 , and then a1 and a3 .
The solution is (a1 , a2 , a3 ) = (7, 9, 10). Therefore, E [τ ] = 1 + a1 = 8. 4.20 A random process created by interpolation
(a)
X n t n+1 t (b) Xt is the sum of two random variables, (1 − a)Ut , which is uniformly distributed on the interval
[0, 1 − a], and aUn+1 , which is uniformly distributed on the interval [0, a]. Thus, the density of Xt
is the convolution of the densities of these two variables:
1
a
1
1!a * 0 2 1!a = 0 a 1
1!a 0 a 1!a 1 2 (c) CX (t, t) = a +(1−a) for t = n + a. Since this depends on t, X is not WSS.
12
(d) P {max0≤t≤10 Xt ≤ 0.5} = P {Uk ≤ 0.5 for 0 ≤ k ≤ 10} = (0.5)11 .
4.22 Restoring samples
(a) Yes. The possible values of Xk are {1, . . . , k −1}. Given Xk , Xk+1 is equal to Xk with probability
Xk
Xk
k and is equal to Xk + 1 with probability 1 − k . Another way to say this is that the onestep
transition probabilities for the transition from Xk to Xk+1 are given by pij = i
k 1−
0 i
k for j = i
for j = i + 1
else (b) E [Xk+1 Xk ] = Xk ( Xk ) + (Xk + 1) 1 − Xk = Xk + 1 − Xk .
k
k
k
(c) The Markov property of X, the information equivalence of Xk and Mk , and part (b), inply that
1
E [Mk+1 M2 , . . . , Mk ] = E [Mk+1 Mk ] = k+1 (Xk + 1 − Xk ) = Mk , so that (Mk ) does not form a
k
martingale sequence.
(d) Using the transition probabilities mentioned in part (a) again, yields (with some tedious algebra
steps not shown)
2
E [Dk+1 Xk ] = =
= 1 2 Xk
Xk
Xk + 1 1 2 k − Xk
−
+
−
k+1 2
k
k+1
2
k
1
2
(4k − 8)Xk − (4k − 8)kXk + k (k − 1)2
4k (k + 1)2
1
1
2
k (k − 2)Dk +
(k + 1)2
4
339 2
2
(e) Since, by the tower property of conditional expectations, vk+1 = E [Dk+1 ] = E [E [Dk+1 Xk ]],
taking the expectation on each side of the equation found in part (d) yields vk+1 = 1
(k + 1)2 k (k − 2)vk + 1
4 and the initial condition v2 = 0 holds. The desired inequality, vk ≤ 41 , is thus true for k = 2. For
k
the purpose of proof by induction, suppose that vk ≤ 41 for some k ≥ 2. Then,
k
vk+1 ≤
= 1
1
1
k (k − 2) +
2
(k + 1)
4k 4
1
1
{k − 2 + 1} ≤
.
4(k + 1)2
4(k + 1) So the desired inequality is true for k +1. Therefore, by proof by induction, vk ≤ 41 for all k . Hence,
k
m.s.
vk → 0 as k → ∞. By deﬁnition, this means that Mk → 1 as k → ∞. (We could also note that,
2
since Mk is bounded, the convergence also holds in probability, and also it holds in distribution.)
4.24 An M/M/1/B queueing system (a) Q = . −λ
1
0
0
0 λ
0
0
0
−(1 + λ)
λ
0
0
1
−(1 + λ)
λ
0
0
1
−(1 + λ) λ
0
0
1
−1 (b) The equilibrium vector π = (π0 , π1 , . . . , πB ) solves πQ = 0. Thus, λπ0 = π1 . Also, λπ0 −
(1 + λ)π1 + π2 = 0, which with the ﬁrst equation yields λπ1 = π2 . Continuing this way yields
that πn = λπn−1 for 1 ≤ n ≤ B . Thus, πn = λn π0 . Since the probabilities must sum to one,
πn = λn /(1 + λ + · · · + λB ).
4.26 Identiﬁcation of special properties of two discretetime processes (version 2)
(a) (yes, yes, no). The process is Markov by its description. Think of a time k as the present
time. Given the number of cells alive at the present time k (i.e. given Xk ) the future evolution
does not depend on the past. To check for the martingale property in discrete time, it suﬃces to
check that E [Xk+1 X1 , . . . , Xk ] = Xk . But this equality is true because for each cell alive at time
k , the expected number of cells alive at time k + 1 is one (=0.5 × 0 + 0.5 × 2). The process does
not have independent increments, because, for example, P [X2 − X1 = 0X1 − X0 = −1] = 1 and
P [X2 − X1 = 0X1 − X0 = 1] = 1/2. So X2 − X1 is not independent of X1 − X0 .
(b) (yes, yes, no). Let k be the present time. Given Yk , the future values are all determined by
Yk , Uk+1 , Uk+2 , . . .. Since Uk+1 , Uk+2 , . . . is independent of Y0 , . . . , Yk , the future of Y is conditionally independent of the past, given the present value Yk . So Y is Markov. The process Y is a martingale because E [Yk+1 Y1 , . . . , Yk ] = E [Uk+1 Yk Y1 , . . . , Yk ] = Yk E [Uk+1 Y1 , . . . , Yk ] = Yk E [Uk+1 ] =
Yk . The process Y does not have independent increments, because, for example Y1 − Y0 = U1 − 1
is clearly not independent of Y2 − Y1 = U1 (U2 − 1). (To argue this further we could note that the
conditional density of Y2 − Y1 given Y1 − Y0 = y − 1 is the uniform distribution over the interval
[−y, y ], which depends on y .)
4.28 Identiﬁcation of special properties of two continuoustime processes (version 2)
(a) (yes,no,no) Z is Markov because W is Markov and the mapping from Wt to Zt is invertible. So
340 Wt and Zt have the same information. To see if W 3 is a martingale we suppose s ≤ t and use the
independent increment property of W to get:
E [Wt3 Wu , 0 ≤ u ≤ s] = E [Wt3 Ws ] = E [(Wt − Ws + Ws )3 Ws ] =
3
3
3
3E [(Wt − Ws )2 ]Ws + Ws = 3(t − s)Ws + Ws = Ws .
Therefore, W 3 is not a martingale. If the increments were independent, then since Ws is the increment Ws − W0 , it would have to be that E [(Wt − Ws + Ws )3 Ws ] doesn’t depend on Ws . But it
does. So the increments are not independent.
(b) (no, no, no) R is not Markov because knowing Rt for a ﬁxed t doesn’t quite determines Θ to
be one of two values. But for one of these values R has a positive derivative at t, and for the other
R has a negative derivative at t. If the past of R just before t were also known, then θ could be
completely determined, which would give more information about the future of R. So R is not
Markov. (ii)R is not a martingale. For example, observing R on a ﬁnite interval total determines
R. So E [Rt (Ru , 0 ≤ u ≤ s] = Rt , and if s − t is not an integer, Rs = Rt . (iii) R does not have
independent increments. For example the increments R(0.5) − R(0) and R(1.5) − R(1) are identical
random variables, not independent random variables.
4.30 Moving balls
(a) The states of the “relativeposition process” can be taken to be 111, 12, and 21. The state 111
means that the balls occupy three consecutive positions, the state 12 means that one ball is in the
left most occupied position and the other two balls are one position to the right of it, and the state
21 means there are two balls in the leftmost occupied position and one ball one position to the
right of them. With the states in the order 111, 12, 21, the onestep transition probability matrix 0.5 0.5 0
0 1 .
is given by P = 0
0.5 0.5 0
(b) The equilibrium distribution π of the process is the probability vector satisfying π = πP , from
11
which we ﬁnd π = ( 3 , 3 , 1 ). That is, all three states are equally likely in equilibrium. (c) Over a
3
long period of time, we expect the process to be in each of the states about a third of the time.
After each visit to states 111 or 12, the leftmost position of the conﬁguration advances one position to the right. After a visit to state 21, the next state will be 12, and the leftmost position of
the conﬁguration does not advance. Thus, after 2/3 of the slots there will be an advance. So the
longterm speed of the balls is 2/3. Another approach is to compute the mean distance the moved
ball travels in each slot, and divide by three.
(d) The same states can be used to track the relative positions of the balls as in discrete time. The −0.5 0.5 0
−1 1 . (Note that if the state is 111 and if the
generator matrix is given by Q = 0
0.5 0.5 −1
leftmost ball is moved to the rightmost position, the state of the relativeposition process is 111
the entire time. That is, the relativeposition process misses such jumps in the actual conﬁguration
process.) The equilibrium distribution can be determined by solving the equation πQ = 0, and
1
the solution is found to be π = ( 1 , 1 , 3 ) as before. When the relativeposition process is in states
33
111 or 12, the leftmost position of the actual conﬁguration advances one position to the right at
rate one, while when the relativeposition process is in state is 21, the rightmost position of the
actual conﬁguration cannot directly move right. The longterm average speed is thus 2/3, as in the
discretetime case. 341 4.32 Mean hitting time for a continuoustime, discretespace Markov process −1
1
0
1
Q = 10 −11
0
5 −5 50 5 1
,,
56 56 56 π= Consider Xh to get
a1 = h + (1 − h)a1 + ha2 + o(h)
a2 = h + 10a1 + (1 − 11h)a2 + o(h)
h
h
or equivalently 1 − a1 + a2 + o(h ) = 0 and 1 + 10a1 − 11a2 + o(h ) = 0. Let h → 0 to get 1 − a1 + a2 = 0
and 1 + 10a1 − 11a2 = 0, or a1 = 12 and a2 = 11. 4.34 Poisson splitting
This is basically the previous problem in reverse. This solution is based directly on the deﬁnition
of a Poisson process, but there are other valid approaches. Let X be Possion random variable, and
let each of X individuals be independently assigned a type, with type i having probability pi , for
some probability distribution p1 , . . . , pK . Let Xi denote the number assigned type i. Then,
P (X1 = i1 , X2 = i2 , · · · , XK = iK ) = P (X = i1 + · · · + iK )
K = (i1 + · · · + iK )! k1
p · · · p iK
K
i1 ! i2 ! · · · iK ! 1 i e−λj λjj j =1 ij ! where λi = λpi . Thus, independent splitting of a Poisson number of individuals yields that the
number of each type i is Poisson, with mean λi = λpi and they are independent of each other.
Now suppose that N is a rate λ Poisson process, and that Ni is the process of type i points,
given independent splitting of N with split distribution p1 , . . . , pK . By the deﬁnition of a Poisson
process, the following random variables are independent, with the ith having the P oi(λ(ti+1 − ti ))
distribution:
N (t1 ) − N (t0 ) N (t2 ) − N (t1 )
···
N (tp ) − N (tp−1 )
(12.5)
Suppose each column of the following array is obtained by independent splitting of the corresponding
variable in (12.5).
N1 (t1 ) − N1 (t0 )
N2 (t1 ) − N2 (t0 )
.
.
. N1 (t2 ) − N1 (t1 )
N2 (t2 ) − N2 (t1 )
.
.
. NK (t1 ) − NK (t0 ) NK (t2 ) − NK (t1 ) ···
···
···
··· N1 (tp ) − N1 (tp−1 )
N2 (tp ) − N2 (tp−1 )
.
.
. (12.6) NK (tp ) − NK (tp−1 ) Then by the splitting property of Poisson random variables described above, we get that all elements of the array (12.6) are independent, with the appropriate means. By deﬁnition, the ith
process Ni is a rate λpi random process for each i, and because of the independence of the rows of
the array, the K processes N1 , . . . , NK are mutually independent. 342 4.36 Some orthogonal martingales based on Brownian motion
Throughout the solution of this problem, let 0 < s < t, and let Y = Wt − Ws . Note that Y is
independent of Ws and it has the N (0, t − s) distribution.
2
Mt
Mt
Mt
Mt
(a) E [Mt Ws ] = Ms E [ Ms Ws ]. Now Ms = exp(θY − θ (t−s) ). Therefore, E [ Ms Ws ] = E [ Ms ] = 1.
2
Thus E [Mt Ws ] = Ms , so by the hint, M is a martingale.
2
(b) Wt2 − t = (Ws + Y )2 − s − (t − s) = Ws − s + 2Ws Y + Y 2 − (t − s), but
E [2Ws Y Ws ] = 2Ws E [Y Ws ] = 2Ws E [Y ] = 0, and E [Y 2 − (t − s)Ws ] = E [Y 2 − (t − s)] = 0. It
follows that E [2Ws Y + Y 2 − (t − s)Ws ] = 0, so the martingale property follows from the hint.
Similarly,
3
2
Wt3 − 3tWt = (Y + Ws )3 − 3(s + t − s)(Y + Ws ) = Ws − 3sWs +3Ws Y +3Ws (Y 2 − (t − s))+ Y 3 − 3tY .
Because Y is independent of Ws and because E [Y ] = E [Y 2 − (t − s)] = E [Y 3 ] = 0, it follows that
2
E [3Ws Y + 3Ws (Y 2 − (t − s)) + Y 3 − 3tY Ws ] = 0, so the martingale property follows from the hint.
(c) Fix distinct nonnegative integers m and n. Then
E [Mn (s)Mm (t)] = E [E [Mn (s)Mm (t)Ws ]] property of cond. expectation = E [Mn (s)E [Mm (t)Ws ]] property of cond. expectation = E [Mn (s)Mm (s)]
=0 martingale property orthogonality of variables at a ﬁxed time 5.2 A variance estimation problem with Poisson observation
(a)
2 (X 2 )n e−X
P {N = n} = E [P [N = nX ]] = E [
n!
∞ =
−∞ x2 x2n e−x e− 2σ2
√
dx
n!
2πσ 2
2 (b) To arrive at a simple answer, we could set the derivative of P {N = n} with respect to σ 2
equal to zero either before or after simplifying. Here we simplify ﬁrst, using the fact that if X
e2n (2n
1
1
is a N (0, σ 2 ) random variable, then E [X 2n ] = σ n!2n )! . Let σ 2 be such that 2˜ 2 = 1 + 2σ2 , or
˜
σ
σ2
equivalently, σ 2 = 1+2σ2 . Then the above integral can be written as follows:
−x2 P {N = n} =
= e
σ ∞ x2n e 2σ2
√
dx
σ −∞ n! 2π σ 2
c1 σ 2n+1
c1 σ 2n
=
2n+1 ,
σ
(1 + 2σ 2 ) 2 where the constant c1 depends on n but not on σ 2 . Taking the logarithm of P {N = n} and calculating the derivative with respect to σ 2 , we ﬁnd that P {N = n} is maximized at σ 2 = n. That is,
2
σM L (n) = n. 343 5.4 Transformation of estimators and estimators of transformations
(a) Yes, because the transformation is invertible.
(b) Yes, because the transformation is invertible.
(c) Yes, because the transformation is linear, the pdf of 3 + 5Θ is a scaled version of the pdf of Θ.
(d) No, because the transformation is not linear.
(e) Yes, because the MMSE estimator is given by the conditional expectation, which is linear. That
is, 3 + 5E [ΘY ] = E [3 + 5ΘY ].
(f) No. Typically E [Θ3 Y ] = E [ΘY ]3 .
5.6 Finding a most likely path
Finding the path z to maximize the posterior probability given the sequence 021201 is the same
as maximizing pcd (y, z θ). Due to the form of the parameter θ = (π, A, B ), for any path z =
(z1 , . . . , z6 ), pcd (y, z θ) has the form c6 ai for some i ≥ 0. Similarly, the variable δj (t) has the form
ct ai for some i ≥ 0. Since a < 1, larger values for pcd (y, z θ) and δj (t) correspond to smaller values
of i. Rather than keeping track of products, such as ai aj , we keep track of the exponents of the
products, which for ai aj would be i + j. Thus, the problem at hand is equivalent to ﬁnding a path
from left to right in trellis indicated in Figure 12.3(a) with minimum weight, where the weight of a
(a) Observations
0 2 1+1
0 1 1 3 1 3
3 State
1 3+2 2 3
3
1 1 1 1 2 3
3
1 1 1 1 3
3
1 3 0 3 3
3
1 1 2 1 2 t=1 3
t=6 Observations (b) 0
2
1+1
0 1 2
6
3 3
3 State
1 3+2
5
t=1 1
3
3 1 6 1
3
3 1 1 1
9
2 3
10 2
13
3 1 3
3
1 1
12 0
15
1 1 1
18
2 3
3
1 2
15 ! ’s
1 3
19
t=6 Figure 12.3: Trellis diagram for ﬁnding a MAP path.
path is the sum of all the numbers indicated on the vertices and edges of the graph. Figure 12.3(b)
shows the result of running the Viterbi algorithm. The value of δj (t) has the form ct ai , where for
i is indicated by the numbers in boxes. Of the two paths reaching the ﬁnal states of the trellis,
the upper one, namely the path 000000, has the smaller exponent, 18, and therefore, the larger
probability, namely c6 a18 . Therefore, 000000 is the MAP path. 344 5.8 Specialization of BaumWelch algorithm for no hidden data
(a) Suppose the sequence y = (y1 , . . . , yT ) is oberserved. If θ(0) = θ = (π, A, B ) is such that B is
the identity matrix, and all entries of π and A are nonzero, then directly by the deﬁnitions (without
using the α’s and β ’s):
γi (t) = P [Zt = iY1 = y1 , . . . , YT = yT , θ] = I{yt =i}
ξij (t) = P [Zt = i, Zt+1 = j Y1 = y1 , . . . , YT = yT , θ] = I{(yt ,yt+1 )=(i,j )}
Thus, (5.27)  (5.29) for the ﬁrst iteration, t = 0, become
(1) πi = I{y1 =i} i.e. the probability vector for S with all mass on y1 number of (i, j ) transitions observed
number of visits to i up to time T − 1
number of times the state is i and the observation is l
(1)
bil =
number of times the state is i
It is assumed that B is the identity matrix, so that each time the state is i the observation should
(1)
also be i. Thus, bil = I{i=l} for any state i that is visited. That is consistent with the assumption
that B is the identify matrix. (Alternatively, since B is ﬁxed to be the identity matrix, we could
just work with estimating π and A, and simply not consider B as part of the parameter to be estimated.) The next iteration will give the same values of π and A. Thus, the BaumWelch algorithm
converges in one iteration to the ﬁnal value θ(1) = (π (1) , A(1) , B (1) ) already described. Note that,
by Lemma 5.1.7, θ(1) is the ML estimate.
(1) ai,j = (b) In view of part (a), the ML estimates are π = (1, 0) and A = 2
3
1
3 1
3
2
3 . This estimator of A results from the fact that, of the ﬁrst 21 times, the sate was zero 12 times, and 8 of those 12 times
the next state was a zero. So a00 = 8/12 = 2/3 is the ML estimate. Similarly, the ML estimate of
a11 is 6/9, which simpliﬁes to 2/3.
5.10 BaumWelch saddlepoint
It turns out that π (k) = π (0) and A(k) = A(0) , for each k ≥ 0. Also, B (k) = B (1) for each k ≥ 1,
where B (1) is the matrix with identical rows, such that each row of B (1) is the empirical distribution
of the observation sequence. For example, if the observations are binary valued, and if there are
T = 100 observations, of which 37 observations are zero and 63 are 1, then each row of B (1) would
be (0.37, 0.63). Thus, the EM algorithm converges in one iteration, and unless θ(0) happens to
be a local maximum or local minimum, the EM algorithm converges to an inﬂection point of the
likelihood function.
One intuitive explanation for this assertion is that since all the rows of B (0) are the same, then
the observation sequence is initially believed to be independent of the state sequence, and the state
process is initially believed to be stationary. Hence, even if there is, for example, notable time
variation in the observed data sequence, there is no way to change beliefs in a particular direction
in order to increase the likelihood. In real computer experiments, the algorithm may still eventually
reach a near maximum likelihood estimate, due to roundoﬀ errors in the computations which allow
the algorithm to break away from the inﬂection point.
The assertion can be proved by use of the update equations for the BaumWelch algorithm. It
is enough to prove the assertion for the ﬁrst iteration only, for then it follows for all iterations by
induction.
345 (0) Since the rows of B (0) ) are all the same, we write bl to denote bil for an arbitrary value of i.
(0)
By induction on t, we ﬁnd αi (t) = by1 · · · byt πi and βj (t) = byt+1 · · · byT . In particular, βj (t) does
(0)
not depend on j . So the vector (αi βi : i ∈ S ) is proportional to π (0) , and therefore γi (t) = πi .
(0) (0)
Similarly, ξi,j (t) = P [Zt = i, Zt+1 = j y, θ(0) ] = πi ai,j . By (5.27), π (1) = π (0) , and by (5.28),
A(1) = A(0) . Finally, (5.29) gives
(1)
bi,l = T
t=1 πi I{yt =l} T πi = number of times l is observed
.
T 5.12 Constraining the BaumWelch algorithm
A quite simple way to deal with this problem is to take the initial parameter θ(0) = (π, A, B ) in
the BaumWelch algorithm to be such that aij > 0 if and only if aij = 1 and bil > 0 if and only if
bil = 1. (These constraints are added in addition to the usual constraints that π , A, and B have
the appropriate dimensions, with π and each row of A and b being probability vectors.) After all,
it makes sense for the initial parameter value to respect the constraint. And if it does, then the
same constraint will be satisﬁed after each iteration, and no changes are needed to the algorithm
itself.
6.2 A two station pipeline in continuous time
(a) S = {00, 01, 10, 11}
(b)
µ 2
01 00
! ! µ1
10 11
µ 2 −λ
0
λ
0
µ
−µ2 − λ
0
λ
.
(c) Q = 2
0
µ1
−µ1
0
0
0
µ2 −µ2
(d) η = (π00 + π01 )λ = (π01 + π11 )µ2 = π10 µ1 . If λ = µ1 = µ2 = 1.0 then π = (0.2, 0.2, 0.4, 0.2) and
η = 0.4.
(e) Let τ = min{t ≥ 0 : X (t) = 00}, and deﬁne hs = E [τ X (0) = s], for s ∈ S . We wish to ﬁnd h00 = 0
h00
0
λh11 h01 3 h01 = µ21 λ + µ2 h00 + µ2 +λ
+
µ2 +λ h11 .
For If λ = µ1 = µ2 = 1.0 this yields 1 h10 = 4 . Thus,
h10 = µ1 + h01
1
h11
5
h11 = µ2 + h10
h11 = 5 is the required answer. 346 6.4 A simple Poisson process calculation
Suppose 0 < s < t and 0 ≤ i ≤ k .
P [N (s) = i, N (t) = k ]
P [N (t) = k ] P [N (s) = iN (t) = k ] =
= e−λs (λs)i
i! = k
i s
t e−λ(t−s) (λ(t − s))k−i
(k − i)!
t−s
t i e−λt (λt)k
k! −1 k −1 That is, given N (t) = k , the conditional distribution of N (s) is binomial. This could have been
deduced with no calculation, using the fact that given N (t) = k , the locations of the k points are
uniformly and independently distributed on the interval [0, t].
6.6 A mean hitting time problem
(a)
2
0 1
1 2 1
2 2 3
πQ = 0 implies π = ( 2 , 2 , 7 ).
77
(b) Clearly a1 = 0. Condition on the ﬁrst step. The initial holding time in state i has mean − q1 and
ii the next state is j with probability pJ =
ij
a0
1
=
1.5
a2
(c) Clearly α2 (t) = 0 for all t.
Solving yields −qij
qii . Thus a0
a2 = − q1
00
− q1
22 + 0 pJ
02
pJ
0
20 a0
a2 . . α0 (t + h) = α0 (t)(1 + q00 h) + α1 (t)q10 h + o(h)
α1 (t + h) = α0 (t)q01 h + α1 (t)(1 + q11 h) + o(h)
q00 q01
with the
q10 q11
inital condition (α0 (0), α1 (0)) = (1, 0). (Note: the matrix involved here is the Q matrix with the
row and column for state 2 removed.)
(d) Similarly,
Subtract αi (t) from each side and let h → 0 to yield ( ∂α0 , ∂α1 ) = (α0 , α1 )
∂t
∂t β0 (t − h) = (1 + q00 h)β0 (t) + q01 hβ1 (t) + o(h)
β1 (t − h) = q10 hβ0 (t) + (1 + q11 h)β1 (t)) + o(h)
Subtract βi (t)’s, divide by h and let h → 0 to get:
β0 (tf )
β1 (tf ) = 1
1
347 − ∂β0
∂t
− ∂β1
∂t = q00 q01
q10 q11 β0
β1 with 6.8 Markov model for a link with resets
(a) Let S = {0, 1, 2, 3}, where the state is the number of packets passed since the last reset.
! ! 0 ! 1
µ 2 3 µ
µ (b) By the PASTA property, the dropping probability is π3 . We can ﬁnd the equilibrium distribution π by solving the equation πQ = 0. The balance equation for state 0 is λπ0 = µ(1 − π0 ) so that
µ
λ
π0 = λ+µ . The balance equation for state i ∈ {1, 2} is λπi−1 = (λ + µ)πi , so that π1 = π0 ( λ+µ )
λ
λ
λ
and π2 = π0 ( λ+µ )2 . Finally, λπ2 = µπ3 so that π3 = π0 ( λ+µ )2 µ = λ3
.
(λ+µ)3 The dropping prob λ3 ability is π3 = (λ+µ)3 . (This formula for π3 can be deduced with virtually no calculation from
the properties of merged Poisson processes. Fix a time t. Each event is a packet arrival with
λ
probability λ+µ and is a reset otherwise. The types of diﬀerent events are independent. Finally,
π3 (t) is the probability that the last three events before time t were arrivals. The formula follows.)
6.10 A queue with decreasing service rate
(a)
!
0 ! ! µ ! ... 1 K
µ µ !
K+1 µ/2 !
... K+2
µ/2 µ/2 X(t)
K t µ
(b) S2 = ∞ ( 2λ )k 2k∧K , where k ∧ K = min{k, K }. Thus, if λ < µ then S2 < +∞ and the
k=0
2
process is recurrent. S1 = ∞ ( 2λ )k 2−k∧K , so if λ < µ then S1 < +∞ and the process is positive
k=0 µ
2
recurrent. In this case, πk = ( 2λ )2−k∧K π0 , where
µ 1
1 − (λ/µ)K
(λ/µ)K
π0 =
=
+
S1
1 − (λ/µ)
1 − (2λ/µ) −1 . µ
(c) If λ = 23 , the queue appears to be stable until if ﬂuctuates above K . Eventually the queuelength will grow to inﬁnity at rate λ − µ = µ . See ﬁgure above.
2
6 6.12 An M/M/1 queue with impatient customers
(a)
!
0 ! µ µ+" ! !
2 1 3
µ+2" !
... 4
µ+3" µ+4" (b) The process is positive recurrent for all λ, µ if α > 0, and pk =
chosen so that the pk ’s sum to one.
348 cλk
µ(µ+α)···(µ+(k−1)α) where c is k k cλ
(c) If α = µ, pk = k!µk = cρ! . Therefore, (pk : k ≥ 0) is the Poisson distribution with mean ρ.
k
Furthermore, pD is the mean departure rate by defecting customers, divided by the mean arrival
rate λ. Thus, 1
pD =
λ ∞ pk (k − 1)α =
k=1 ρ − 1 + e−ρ
→
ρ 1 as ρ → ∞
0 as ρ → 0 where l’Hˆspital’s rule can be used to ﬁnd the limit as ρ → 0.
o
6.14 A queue with blocking
(a)
!
0 ! µ !
2 1
µ !
3 µ !
4 µ 5
µ k k ρ
πk = 1+ρ+ρ2 +ρ3 +ρ4 +ρ5 = ρ 1(1−ρ) for 0 ≤ k ≤ 5.
−ρ6
(b) pB = π5 by the PASTA property.
1
(c) W = NW /(λ(1 − pB )) where NW = 5 =1 (k − 1)πk . Alternatively, W = N /(λ(1 − pB )) − µ
k
(i.e. W is equal to the mean time in system minus the mean time in service)
1
1
(d) π0 =
=
Thereλ(mean cycle time for visits to state zero)
λ(1/λ+mean busy period duration)
ρ6
ρ5
11
fore, the mean busy period duration is given by λ [ π0 − 1] = λρ−−ρ) = µ1−−ρ)
(1
(1 6.16 On two distibutions seen by customers N(t) k+1
k
t As can be seen in the picture, between any two transtions from state k to k + 1 there is a transition
form state k + 1 to k , and vice versa. Thus, the number of transitions of one type is within one of
the number of transitions of the other type. This establishes that D(k, t) − R(k, t) ≤ 1 for all k .
(b)Observe that
D(k, t) R(k, t)
−
αt
δt ≤
≤
≤ Thus, D(k,t)
αt and R(k,t)
δt D(k, t) R(k, t)
R(k, t) R(k, t)
−
+
−
αt
αt
αt
δt
αt
1
R(k, t)
+
1−
αt
αt
δt
1
αt
→ 0 as t → ∞
+ 1−
αt
δt have the same limits, if the limits of either exists. 6.18 Positive recurrence of reﬂected random walk with negative drift
349 Let V (x) = 1 x2 . Then
2
(x + Bn + Ln )2
x2−
2
2
x2
(x + Bn )2−
≤ E[
2
2
B2
= xB +
2 P V (x) − V (x) = E [ Therefore, the conditions of the combined Foster stability criteria and moment bound corollary
B2
apply, yielding that X is positive recurrent, and X ≤ −2B . (This bound is somewhat weaker than
Kingman’s moment bound, disussed later in the notes: X ≤ Var(B ) .)
−2B 6.20 An inadequacy of a linear potential function
Suppose x is on the postive x2 axis (i.e. x1 = 0 and x2 > 0). Then, given X (t) = x, during
the slot, queue 1 will increase to 1 with probability a(1 − d1 ) = 0.42, and otherwise stay at zero.
Queue 2 will decrease by one with probability 0.4, and otherwise stay the same. Thus, the drift
of V , E [V (X (t + 1) − V (x)X (t) = x] is equal to 0.02. Therefore, the drift is strictly positive for
inﬁnitely many states, whereas the FosterLyapunov condition requires that the drift be negative
oﬀ of a ﬁnite set C . So, the linear choice for V does not work for this example.
6.22 Opportunistic scheduling
(a) The left hand side of (6.37) is the arrival rate to the set of queues in s, and the righthand side
is the probability that some queue in s is eligible for service in a given time slot. The condition is
necessary for the stability of the set of queues in s.
(b) Fix > 0 so that for all s ∈ E with s = ∅,
(ai + ) ≤
i∈ s w (B )
B :B ∩s=∅ Consider the ﬂow graph shown.
s
q w(s )
1
1 s a +!
1
q
a 2 +!
a .
.
.
a +!
N 2 .
.
. 2 w(s )
2
.
.
.
w(s )
k .
.
. q 1 sk N .
.
. .
.
.
s 350 N! .
.
. w(s )
N! b In addition to the source node a and sink node b, there are two columns of nodes in the graph. The
ﬁrst column of nodes corresponds to the N queues, and the second column of nodes corresponds
to the 2N subsets of E . There are three stages of links in the graph. The capacity of a link (a,qi )
in the ﬁrst stage is ai + , there is a link (qi , sj ) in the second stage if and only if qi ∈ sj , and each
such link has capacity greater than the sum of the capacities of all the links in the ﬁrst stage, and
the weight of a link (sk , t) in the third stage is w(sk ).
We claim that the minimum of the capacities of all a − b cuts is v ∗ = N (ai + ). Here is a
i=1
proof of the claim. The a − b cut ({a} : V − {a}) (here V is the set of nodes in the ﬂow network)
has capacity v ∗ , so to prove the claim, it suﬃces to show that any other a − b cut has capacity
greater than or equal to v ∗ . Fix any a − b cut (A : B ). Let A = A ∩ {q1 , . . . , qN }, or in words, A
is the set of nodes in the ﬁrst column of the graph (i.e. set of queues) that are in A. If qi ∈ A and
sj ∈ B such that (qi , sj ) is a link in the ﬂow graph, then the capacity of (A : B ) is greater than or
equal to the capacity of link (qi , sj ), which is greater than v ∗ , so the required inequality is proved
in that case. Thus, we can suppose that A contains all the nodes sj in the second column such
that sj ∩ A = ∅. Therefore,
C (A : B ) ≥ (ai + ) +
e
i∈{q1 ,...,qN }−A ≥ w(s)
e
s⊂E :s∩A=∅ (ai + ) = v ∗ , (ai + ) +
e
i∈{q1 ,...,qN }−A (12.7) e
i∈ A where the inequality in (12.7) follows from the choice of . The claim is proved.
Therefore there is an a − b ﬂow f which saturates all the links of the ﬁrst stage of the ﬂow graph.
Let u(i, s) = f (qi , s)/f (s, b) for all i, s such that f (s, b) > 0. That is, u(i, s) is the fraction of ﬂow
on link (s, b) which comes from link (qi , s). For those s such that f (s, b) = 0, deﬁne u(i, s) in some
arbitrary way, respecting the requirements u(i, s) ≥ 0, u(i, s) = 0 if i ∈ s, and i∈E u(i, s) = I{s=∅} .
Then ai + = f (a, qi ) = s f (qi , s) = s f (s, b)u(i, s) ≤ s w(s)u(i, s) = µi (u), as required.
1
(c) Let V (x) = 2 i∈E x2 . Let δ (t) denote the identity of the queue given a potential service at
i
time t, with δ (t) = 0 if no queue is given potential service. Then P [δ (t) = iS (t) = s] = u(i, s). The
dynamics of queue i are given by Xi (t + 1) = Xi (t) + Ai (t) − Ri (δ (t)) + Li (t), where Ri (δ ) = I{δ=i} .
Since i∈E (Ai (t) − Ri (δi (t)))2 ≤ i∈E (Ai (t))2 + (Ri (δi (t)))2 ≤ N + i∈E Ai (t)2 we have
P V (x) − V (x) ≤ xi (ai − µi (u)) +K (12.8) i∈ E ≤− xi +K (12.9) i∈ E where K = N + N Ki . Thus, under the necessary stability conditions we have that under the
i=1
2
vector of scheduling probabilities u, the system is positive recurrent, and
Xi ≤ K (12.10) i∈E (d) If u could be selected as a function of the state, x, then the right hand side of (12.8) would
be minimized by taking u(i, s) = 1 if i is the smallest index in s such that xi = maxj ∈s xj . This
351 suggests using the longest connected ﬁrst (LCF) policy, in which the longest connected queue is
served in each time slot. If P LCF denotes the onestep transition probability matrix for the LCF
policy, then (12.8) holds for any u, if P is replaced by P LCF . Therefore, under the necessary
condition and as in part (b), (12.9) also holds with P replaced by P LCF , and (12.10) holds for
the LCF policy.
6.24 Stability of two queues with transfers
(a) System is positive recurrent for some u if and only if λ1 < µ1 + ν, λ2 < µ2 , and λ1 + λ2 < µ1 + µ2 .
(b)
qxy (V (y ) − V (x)) QV (x) =
y :y =x = λ1
λ2
µ1
[(x1 + 1)2 − x2 ] + [(x2 + 1)2 − x2 ] + [(x1 − 1)2 − x2 ] +
1
2
+
1
2
2
2
uνI{x1 ≥1}
µ2
[(x2 − 1)2 − x2 ] +
[(x1 − 1)2 − x2 + (x2 + 1)2 − x2 ]
+
2
1
2
2
2 (12.11) (c) If the righthand side of (12.11) is changed by dropping the positive part symbols and dropping
the factor I{x1 ≥1} , then it is not increased, so that
QV (x) ≤ x1 (λ1 − µ1 − uν ) + x2 (λ2 + uν − µ2 ) + K
≤ −(x1 + x2 ) min{µ1 + uν − λ1 , µ2 − λ2 − uν } + K (12.12) where K = λ1 +λ2 +µ1 +µ2 +2ν . To get the best bound on X1 + X2 , we select u to maximize the min
2
−
term in (12.12), or u = u∗ , where u∗ is the point in [0, 1] nearest to µ1 +µ22νλ1 −λ2 . For u = u∗ , we
ﬁnd QV (x) ≤ − (x1 + x2 ) + K where = min{µ1 + ν − λ1 , µ2 − λ2 , µ1 +µ2 −λ1 −λ2 }. Which of the
2
three terms is smallest in the expression for corresponds to the three cases u∗ = 1, u∗ = 0, and
0 < u∗ < 1, respectively. It is easy to check that this same is the largest constant such that the
stability conditions (with strict inequality relaxed to less than or equal) hold with (λ1 , λ2 ) replaced
by (λ1 + , λ2 + ).
7.2 Lack of sample path continuity of a Poisson process
(a) The sample path of N is continuous over [0, T ] if and only if it has no jumps in the interval, equivalently, if and only if N (T ) = 0. So P [N is continuous over the interval [0,T] ] =
exp(−λT ). Since {N is continuous over [0, +∞)} = ∩∞ {N is continuous over [0, n]}, it follows
n=1
that P [N is continuous over [0, +∞)] = limn→∞ P [N is continuous over [0, n]] = limn→∞ e−λn =
0.
(b) Since P [N is continuous over [0, +∞)] = 1, N is not a.s. sample continuous. However N is
m.s. continuous. One proof is to simply note that the correlation function, given by RN (s, t) =
λ(s ∧ t) + λ2 st, is continuous. A more direct proof is to note that for ﬁxed t, E [Ns − Nt 2 ] =
λs − t + λ2 s − t2 → 0 as s → t.
7.4 Some statements related to the basic calculus of random processes
t
(a) False. limt→∞ 1 0 Xs ds = Z = E [Z ] (except in the degenerate case that Z has variance zero).
t
(b) False. One reason is that the function is continuous at zero, but not everywhere. For another,
we would have Var(X1 − X0 − X2 ) = 3RX (0) − 4RX (1) + 2RX (2) = 3 − 4 + 0 = −1.
352 (c) True. In general, RX X (τ ) = RX (τ ). Since RX is an even function, RX (0) = 0. Thus, for
any t, E [Xt Xt ] = RX X (0) = RX (0) = 0. Since the process X has mean zero, it follows that
Cov(Xt , Xt ) = 0 as well. Since X is a Gaussian process, and diﬀerentiation is a linear operation,
Xt and Xt are jointly Gaussian. Summarizing, for t ﬁxed, Xt and Xt are jointly Gaussian and
uncorrelated, so they are independent. (Note: Xs is not necessarily independent of Xt if s = t. )
7.6 Cross correlation between a process and its m.s. derivative
s
Fix t, u ∈ T . By assumption, lims→t X−− = Xt m.s. Therefore, by Corollary 2.2.4, E
st Xs −Xt
s−t Xu → E [Xt Xu ] as s → t. Equivalently, RX (s, u) − RX (t, u)
s−t Hence ∂1 RX (s, u) exists, and ∂1 RX (t, u) = RX → RX X (t, u) as s → t. X (t, u). 7.8 A windowed Poisson process
(a) The sample paths of X are piecewise constant, integer valued with initial value zero. They
jump by +1 at each jump of N , and jump by 1 one time unit after each jump of N .
(b) Method 1: If s − t ≥ 1 then Xs and Xt are increments of N over disjoint intervals, and are therefore independent, so CX (s, t) = 0. If s − t < 1, then there are three disjoint intervals, I0 , I1 , and I2 ,
with I0 = [s, s +1] ∪ [t, t +1], such that [s, s +1] = I0 ∪ I1 and [t, t +1] = I0 ∪ I2 . Thus, Xs = D0 + D1
and Xt = D0 + D2 , where Di is the increment of N over the interval Ii . The three increments
D1 , D2 , and D3 are independent, and D0 is a Poisson random variable with mean and variance equal
to λ times the length of I0 , which is 1 − s − t. Therefore, CX (s, t) = Cov(D0 + D1 , D0 + D2 ) =
λ(1 − s − t) if s − t < 1
Cov(D0 , D0 ) = λ(1 − s − t). Summarizing, CX (s, t) =
0
else
Method 2: CX (s, t) = Cov(Ns+1 − Ns , Nt+1 − Nt ) = λ[min(s + 1, t + 1) − min(s + 1, t) − min(s, t +
1) − min(s, t)]. This answer can be simpliﬁed to the one found by Method 1 by considering the
cases s − t > 1, t < s < t + 1, and s < t < s + 1 separately.
(c) No. X has a 1 jump one time unit after each +1 jump, so the value Xt for a “present” time t
tells less about the future, (Xs : s ≥ t), than the past, (Xs : 0 ≤ s ≤ t), tells about the future .
(d) Yes, recall that RX (s, t) = CX (s, t) − µX (s)µX (t). Since CX and µX are continuous functions,
so is RX , so that X is m.s. continuous.
(e) Yes. Using the facts that CX (s, t) is a function of s − t alone, and CX (s) → 0 as s → ∞, we
t
t
ﬁnd as in the section on ergodicity, Var( 1 0 Xs ds) = 2 0 (1 − s )CX (s)ds → 0 as t → ∞.
t
t
t
7.10 A singular integral with a Brownian motion
1 wt
(a) The integral
t dt exists in the m.s. sense for any > 0 because wt /t is m.s. continuous over
[ , 1]. To see if the limit exists we apply the correlation form of the Cauchy criteria (Proposition
2.2.2). Using diﬀerent letters as variables of integration and the fact Rw (s, t) = s ∧ t (the minimum
353 of s and t), yields that as ,
1 E → 0,
1 ws
ds
s wt
dt
t 1 1 s∧t
dsdt
st
1
1
s∧t
→
dsdt
st
0
0
1
t
1
t
s∧t
s
=2
dsdt = 2
dsdt
st
0
0
0
0 st
1
t
1
1
1dt = 2.
=2
dsdt = 2
0
0t
0
= Thus the m.s. limit deﬁning the integral exits. The integral has the N (0, 2) distribution.
(b) As a, b → ∞,
a E
1 ws
ds
s b
1 wt
dt
t a b s∧t
dsdt
st
1
1
∞
∞
s∧t
→
dsdt
st
1
1
∞
t
∞
t
s∧t
s
=2
dsdt = 2
dsdt
st
1
1
1
1 st
∞
t
∞
1
t−1
=2
dsdt = 2
dt = ∞,
t
1
1t
1
= so that the m.s. limit does not exist, and the integral is not well deﬁned.
7.12 Recognizing m.s. properties
(a) Yes m.s. continuous since RX is continuous. No not m.s. diﬀerentiable since RX (0) doesn’t
exist. Yes, m.s. integrable over ﬁnite intervals since m.s. continuous. Yes mean ergodic in m.s.
since RX (T ) → 0 as T  → ∞.
(b) Yes, no, yes, for the same reasons as in part (a). Since X is mean zero, RX (T ) = CX (T ) for all
T . Thus
lim CX (T ) = T →∞ lim RX (T ) = 1 T →∞ Since the limit of CX exists and is net zero, X is not mean ergodic in the m.s. sense.
(c) Yes, no, yes, yes, for the same reasons as in (a).
(d) No, not m.s. continuous since RX is not continuous. No, not m.s. diﬀerentiable since X is
not even m.s. continuous. Yes, m.s. integrable over ﬁnite intervals, because the Riemann integral
bb
a a RX (s, t)dsdt exists and is ﬁnite, for the region of integration is a simple bounded region and
the integrand is piecewise constant.
(e) Yes, m.s. continuous since RX is continuous. No, not m.s. diﬀerentiable. For example,
E Xt − X0
t 2 =
= 1
[RX (t, t) − RX (t, 0) − RX (0, t) + RX (0, 0)]
t2
1√
t − 0 − 0 + 0 → +∞ as t → 0.
t2
354 Yes, m.s. integrable over ﬁnite intervals since m.s. continuous.
7.14 A stationary Gaussian process
(a) No. All mean zero stationary, Gaussian Markov processes have autocorrelation functions of the
form RX (t) = Aρt , where A ≥ 0 and 0 ≤ ρ ≤ 1 for continuous time (or ρ ≤ 1 for discrete time).
RX (3)
(b) E [X3 X0 ] = E [X3 X0 ] = RX (0) X0 = X0 . The error is Gaussian with mean zero and variance
10
MSE = Var(X3 ) − Var( X0 ) = 1 − 0.01 = 0.99. So P {X3 − E [X3 X0 ] ≥ 10} = 2Q( √1099 ).
10
0.
2 2−
(c) RX (τ ) = −RX (τ ) = (1+6τ )3 . In particular, since −RX exists and is continuous, X is continuτ2
ously diﬀerentiable in the m.s. sense.
(d) The vector has a joint Gaussian distribution because X is a Gaussian process and diﬀer2τ
entiation is a linear operation. Cov(Xτ , X0 ) = RXX (τ ) = −RX (τ ) = (1+τ 2 )2 . In particular, Cov(X0 , X0 ) = 0 and Cov(X1 , X0 ) = 2 = 0.5. Also, Var(X0 ) = RX (0) = 2. So (X0 , X0 , X1 )T has 4
0
1
0 0 .5
2 0.5 distribution.
the N 0 , 0
0
0.5 0.5 1
7.16 Correlation ergodicity of Gaussian processes
Fix h and let Yt = Xt+h Xt . Clearly Y is stationary with mean µY = RX (h). Observe that
CY (τ ) = E [Yτ Y0 ] − µ2
Y
= E [Xτ +h Xτ Xh X0 ] − RX (h)2
= RX (h)2 + RX (τ )RX (τ ) + RX (τ + h)RX (τ − h) − RX (h)2
Therefore, CY (τ ) → 0 as τ  → ∞. Hence Y is mean ergodic, so X is correlation ergodic.
7.18 Gaussian review question
(a) Since X is Markovian, the best estimator of X2 given (X0 , X1 ) is a function of X1 alone.
Since X is Gaussian, such estimator is linear in X1 . Since X is mean zero, it is given by
Cov(X2 , X1 )Var(X1 )−1 X1 = e−1 X1 . Thus E [X2 X0 , X1 ] = e−1 X1 . No function of (X0 , X1 ) is
a better estimator! But e−1 X1 is equal to p(X0 , X1 ) for the polynomial p(x0 , x1 ) = x1 /e. This
is the optimal polynomial. The resulting mean square error is given by MMSE = Var(X2 ) −
(Cov(X1 X2 )2 )/Var(X1 ) = 9(1 − e−2 )
(b) Given (X0 = π, X1 = 3), X2 is N 3e−1 , 9(1 − e−2 ) so
P [X2 ≥ 4X0 = π, X1 = 3] = P X2 − 3e−1
9(1 − e−2 ) ≥ 4 − 3e−1
9(1 − e−2 ) =Q 4 − 3e−1
9(1 − e−2 ) 7.20 KL expansion of a simple random process
(a) Yes, because RX (τ ) is twice continuously diﬀerentiable.
t
t
(b) No. limt→∞ 2 0 ( t−τ )CX (τ )dτ = 50 + limt→∞ 100 0 ( t−τ ) cos(20πτ )dτ = 50 = 0. Thus, the
t
t
t
t
necessary and suﬃcient condition for mean ergodicity in the m.s. sense does not hold.
(c) APPROACH ONE Since RX (0) = RX (1), the process X is periodic with period one (actually,
with period 0.1). Thus, by the theory of WSS periodic processes, the eigenfunctions can be taken
to be φn (t) = e2πjnt for n ∈ Z. (Still have to identify the eigenvalues.)
355 APPROACH TWO The identity cos(θ) = 1 (ejθ + e−jθ ), yields
2
RX (s − t) = 50 + 25e20πj (s−t) + 25e−20πj (s−t) = 50 + 25e20πjs e−20πjt + 25e−20πjs e20πjt
= 50φ0 (s)φ∗ (t) + 25φ1 (s)φ∗ (t) + 25φ2 (s)φ∗ (t) for the choice φ0 (t) ≡ 1, φ1 (t) = e20πjt and φ2 =
0
1
2
e−20πjt . The eigenvalues are thus 50, 25, and 25. The other eigenfunctions can be selected to ﬁll
out an orthonormal basis, and the other eigenvalues are zero.
APPROACH THREE For s, t ∈ [0, 1] we have RX (s, t) = 50 + 50 cos(20π (s − t))
= 50 + 50 cos(20πs) cos(20πt) + 50 sin(20πs) sin(20πt) = 50φ0 (s)φ∗ (t) + 25φ1 (s)φ∗ (t) + 25φ2 (s)φ∗ (t)
0
1
2
√
√
for the choice φ0 (t) ≡ 1, φ1 (t) = 2 cos(20πt) and φ2 = 2 sin(20πt). The eigenvalues are thus
50, 25, and 25. The other eigenfunctions can be selected to ﬁll out an orthonormal basis, and the
other eigenvalues are zero.
(Note: the eigenspace for eigenvalue 25 is two dimensional, so the choice of eigen functions spanning
that space is not unique.)
7.22 KL expansion for derivative process
(a) Since φn (t) = (2πjn)φn (t), the derivative of each φn is a constant times φn itself. Therefore, the
equation given in the problem statement leads to: X (t) = n (X, φn )φn (t) = n [(2πjn)(X, φn )]φn (t),
which is a KL expansion, because the functions φn are orthonormal in L2 [0, 1] and the coordinates
are orthogonal random variables. Thus,
ψn (t) = φn (t), (X , ψn ) = (2πjn)(Xn , φn ), and µn = (2πn)2 λn for n ∈ Z (Recall that the eigenvalues are equal to the means of the squared magnitudes of the coordinates.)
(b) Note that φ1 = 0, φ2k (t) = −(2πk )φ2k+1 (t) and φ2k+1 (t) = (2πk )φ2k (t). This is similar to part
(a). The same basis functions can be used for X as for X, but the (2k )th and (2k + 1)th coordinates
of X come from the (2k + 1)th and (2k )th coordinates of X , respectively, for all k ≥ 1. Speciﬁcally,
we can take
ψn (t) = φn (t) for n ≥ 0, (X , ψ0 ) = 0 µ0 = 0,
(X , ψ2k ) = (2πk )(X, φ2k+1 ),
µ2k = (2πk )2 λ2k+1 ,
(X , ψ2k+1 ) = −(2πk )(X, φ2k ),
µ2k+1 = (2πk )2 λ2k , for k ≥ 1
(It would have been acceptable to not deﬁne ψ0 , because the corresponding eigenvalue is zero.)
√
(c) Note that φn (t) = (2n+1)π ψn (t), where ψn (t) = 2 cos( (2n+1)πt ), n ≥ 0. That is, ψn is the same
2
2
as φn , but with sin replaced by cos . Or equivalently, by the hint, we discover that ψn is obtained
from φn by timereversal: ψn (t) = φn (1 − t)(−1)n . Thus, the functions ψn are orthonormal. As
in part (a), we also have (X , ψn ) = (2n+1)π (X, φn ), and therefore, µn = ( (2n+1)π )2 λn . (The set
2
2
of eigenfunctions is not unique–for example, some could be multiplied by 1 to yield another valid
set.)
(d) Diﬀerentiating the KL expansion of X yields
√
√
Xt = (X, φ1 )φ1 (t) + (X, φ2 )φ2 (t) = (X, φ1 )c1 3 − (X, φ2 )c2 3.
That is, the random process X is constant in time. So its KL expansion involves only one nonzero
√
√
term, with the eigenfunction ψ1 (t) = 1 for 0 ≤ t ≤ 1. Then (X , ψ1 ) = (X, φ1 )c1 3 − (X, φ2 )c2 3,
and therefore µ1 = 3c2 λ1 + 3c2 λ2 .
1
2 356 7.24 Mean ergodicity of a periodic WSS random process
1
t t Xu du =
0 1
t where a0 = 1, and for n = 0, an,t  =  1
t
not important as t → ∞. Indeed, 2 Xn e2πjnu/T du =
0 n t 2πjnu/T
du
0e n∈Z,n=0 n∈Z,n=0 an,t Xn
n∈Z an,t 2 pX (n) ≤ an,t Xn = E t 2πjnt/T −
=  e 2πjnt/T 1  ≤ T2
π 2 t2 T
πnt . The n = 0 terms are pX (n) → 0 as t → ∞
n∈Z,n=0 t Therefore, 1 0 Xu du → X0 m.s. The limit has mean zero and variance pX (0). For mean ergodict
ity (in the m.s. sense), the limit should be zero with probability one, which is true if and only if
pX (0) = 0. That is, the process should have no zero frequency, or DC, component. (Note: More
generally, if X were not assumed to be mean zero, then X would be mean ergodic if and only if
Var(X0 ) = 0, or equivalently, pX (0) = µ2 , or equivalently, X0 is a constant a.s.)
X
8.2 On the cross spectral density
Follow the hint. Let U be the output if X is ﬁltered by H and V be the output if Y is ﬁltered by
H . The Schwarz inequality applied to random variables Ut and Vt for t ﬁxed yields RU V (0)2 ≤
RU (0)RV (0), or equivalently,
 SXY (ω )
J dω 2
≤
2π SX (ω )
J dω
2π SY (ω )
J dω
,
2π which implies that
 SXY (ωo ) + o( )2 ≤ ( SX (ωo ) + o( ))( SY (ωo ) + o( ))
Letting → 0 yields the desired conclusion. 8.4 Filtering a Gauss Markov process
(a) The process Y is the output when X is passed through the linear timeinvariant system with
impulse response function h(τ ) = e−τ I{τ ≥0} . Thus, X and Y are jointly WSS, and
1 −τ
τ ≥0
∞
∞
2e
RXY (τ ) = RX ∗ h(τ ) = t=−∞ RX (t)h(τ − t)dt = −∞ RX (t)h(t − τ )dt =
1
τ τ ≤0
( 2 − τ )e
(b) X5 and Y5 are jointly Gaussian, mean zero, with Var(X5 ) = RX (0) = 1, and Cov(Y5 , X5 ) =
RXY (0) = 1 , so E [Y5 X5 = 3] = (Cov(Y5 , X5 )/Var(X5 ))3 = 3/2.
2
(c) Yes, Y is Gaussian, because X is a Gaussian process and Y is obtained from X by linear operations.
(d) No, Y is not Markov. For example, we see that SY (ω ) = (1+2 2 )2 , which does not have the
ω
form required for a stationary mean zero Gaussian process to be Markov (namely α22Aω2 ). Another
+
explanation is that, if t is the present time, given Yt , the future of Y is determined by Yt and
(Xs : s ≥ t). The future could be better predicted by knowing something more about Xt than Yt
gives alone, which is provided by knowing the past of Y .
(Note: the R2 valued process ((Xt , Yt ) : t ∈ R) is Markov.) 357 8.6 A stationary twostate Markov process
πP = π implies π = ( 1 , 1 ) is the equilibrium distribution so P [Xn = 1] = P [Xn = −1] =
22
n. Thus µX = 0. For n ≥ 1 1
2 for all RX (n) = P [Xn = 1, X0 = 1] + P [Xn = −1, X0 = −1] − P [Xn = −1, X0 = 1] − P [Xn = 1, X0 = −1]
11 1
11 1
11 1
11 1
[ + (1 − 2p)n ] + [ + (1 − 2p)n ] − [ − (1 − 2p)n ] − [ − (1 − 2p)n ]
=
22 2
22 2
22 2
22 2
= (1 − 2p)n
So in general, RX (n) = (1 − 2p)n . The corresponding power spectral density is given by:
∞ ∞
n −jωn (1 − 2p) e SX (ω ) = ((1 − 2p)e = n=−∞ ∞
−jω n n=0 =
= ((1 − 2p)ejω )n − 1 )+ n=0 1
1
+
−1
−jω
1 − (1 − 2p)e
1 − (1 − 2p)ejω
1 − (1 − 2p)2
1 − 2(1 − 2p) cos(ω ) + (1 − 2p)2 8.8 A linear estimation problem
E [Xt − Zt 2 ] = E [(Xt − Zt )(Xt − Zt )∗ ]
= RX (0) + RZ (0) − RXZ (0) − RZX (0)
= RX (0) + h ∗ h ∗ RY (0) − 2Re(h ∗ RXY (0))
∞ = SX (ω ) + H (ω )2 SY (ω ) − 2Re(H ∗ (ω )SXY (ω )) −∞ The hint with σ 2 = SY (ω ), zo = S (XY (ω ), and z = H (ω ) implies Hopt (ω ) =
8.10 The accuracy of approximate diﬀerentiation
(a) SX (ω ) = SX (ω )H (ω )2 = ω 2 SX (ω ).
∞
(b) k (τ ) = 21 (δ (τ + a) − δ (τ − a)) and K (ω ) = −∞ k (τ )e−jωt dτ =
a 1
jωa
2a (e dω
2π SXY (ω )
SY (ω ) . − e−jωa ) = j sin(aω )
.
a By lima→0 jω cos(aω)
1 l’Hˆspital’s rule, lima→0 K (ω ) =
o
= jω.
(c) D is the output of the linear system with input X and transfer function H (ω ) − K (ω ). The
output thus has power spectral density SD (ω ) = SX (ω )H (ω ) − K (ω )2 = SX (ω )ω − sin(aω) 2 .
a
(d) Or, SD (ω ) = SX (ω )1 − sin(aω ) 2
aω  . √ Suppose 0 < a ≤ 0.6
0.77
ωo (≈ ωo ).
2
sin(aω )
≤ (aω)
aω
6 Then by the bound given
2 in the problem statement, if ω  ≤ ωo then 0 ≤ 1 −
≤ (aωo ) ≤ 0.1, so that
6
SD (ω ) ≤ (0.01)SX (ω ) for ω in the base band. Integrating this inequality over the band yields that
E [Dt 2 ] ≤ (0.01)E [Xt 2 ].
8.12 Filtering Poisson white noise
∞
(a) Since µN = λ, µX = λ −∞ h(t)dt. Also, CX = h ∗ h ∗ CN = λh ∗ h. (In particular, if
h(t) = I{0≤t<1} , then CX (τ ) = λ(1 − τ )+ , as already found in Problem 4.17.)
358 (b) In the special case, in between arrival times of N , X decreases exponentially, following the
equation X = −X. At each arrival time of N , X has an upward jump of size one. Formally, we can
write, X = −X + N . For a ﬁxed time to , which we think of as the present time, the process after
time to is the solution of the above diﬀerential equation, where the future of N is independent of
X up to time to . Thus, the future evolution of X depends only on the current value, and random
variables independent of the past. Hence, X is Markov.
8.14 Linear and nonlinear reconstruction from samples
(a) We ﬁrst ﬁnd the mean function and autocorrelation function of X. E [Xt ] =
U )]E [Bn ] = 0 because E [Bn ] = 0 for all n.
g (t − m − U )Bm g (s − n − U )Bn
m=−∞ n=−∞
∞
2 ∞ E [g (s − n − U )g (t − n − U )] = σ
n=−∞
∞ 1 2 g (s − n − u)g (t − n − u)du
n=−∞ 0 ∞ n+1 g (s − v )g (t − v )dv = σ 2 = σ2 g (s − v )g (t − v )dv
−∞ n=−∞ n
∞ = σ2 −n− ∞ ∞ RX (s, t) = E
=σ n E [g (t g (s − v )g (v − t)dv = σ 2 (g ∗ g )(s − t)
−∞ So X is WSS with mean zero and RX = σ 2 g ∗ g .
(b) By part (a), the power spectral density of X is σ 2 G(ω )2 . If g is a baseband signal, so that
G(ω )2  = 0 for ω ≥ ωo . then by the sampling theorem for WSS baseband random processes, X can
π
be recovered from the samples (X (nT ) : n ∈ Z) as long as T ≤ ωo .
2
(c) For this case, G(2πf ) = sinc (f ), which is not supported in any ﬁnite interval. So part (a) does
not apply. The sample paths of X are continuous and piecewise linear, and at least two sample
points fall within each linear portion of X. Either all pairs of samples of the form (Xn , Xn+0.5 ) fall
within linear regions (happens when 0.5 ≤ U ≤ 1), or all pairs of samples of the form (Xn+0.5 , Xn+1 )
fall within linear regions (happens when 0 ≤ U ≤ 0.5). We can try reconstructing X using both
cases. With probability one, only one of the cases will yield a reconstruction with change points
having spacing one. That must be the correct reconstruction of X. The algorithm is illustrated
in Figure 12.4. Figure 12.4(a) shows a sample path of B and a corresponding sample path of X ,
for U = 0.75. Thus, the breakpoints of X are at times of the form n + 0.75 for integers n. Figure
12.4(b) shows the corresponding samples, taken at integer multiples of T = 0.5. Figure 12.4(c)
shows the result of connecting pairs of the form (Xn , Xn+0.5 ), and Figure 12.4(d) shows the result
of connecting pairs of the form (Xn+0.5 , Xn+1 ). Of these two, only Figure 12.4(c) yields breakpoints
with unit spacing. Thus, the dashed lines in Figure 12.4(c) are connected to reconstruct X.
8.16 An approximation of white noise
(a) Since E [Bk Bl∗ ] = I{k=l} ,
K 1
2 E [ Nt dt ] = E [AT T
0 K
2 2 B k  ] = ( AT T ) E [ k=1
22 = (AT T ) σ K =
359 Bl∗ ] Bk
k=1 A2 T σ 2
T K
l=1 (a) 0 1 2 3 4 5 6 2 3 4 5 6 2 3 4 5 6 2 3 4 5 6 (b)
1
0
(c)
1
0
(d)
1
0 Figure 12.4: Nonlinear reconstruction of a signal from samples 1
(b) The choice of scaling constant AT such that A2 T ≡ 1 is AT = √T . Under this scaling the
T
process N approximates white noise with power spectral density σ 2 as T → 0.
1
(c) If the constant scaling AT = 1 is used, then E [ 0 Nt dt2 ] = T σ 2 → 0 as T → 0. 8.18 Filtering to maximize signal to noise ratio
∞
The problem is to select H to minimize σ 2 −∞ H (ω )2 dω , subject to the constraints (i) H (ω ) ≤ 1
2π
ωo
for all ω , and (ii) −ωo ω  H (ω )2 dω ≥ (the power of X )/2. First, H should be zero outside of
2π
the interval [−ωo , ωo ], for otherwise it would be contributing to the output noise power with no
contribution to the output signal power. Furthermore, if H (ω )2 > 0 over a small interval of length
2π contained in [ωo , ωo ], then the contribution to the output signal power is ω H (ω )2 , whereas
the contribution to the output noise is σ 2 H (ω )2 . The ratio of these contributions is ω /σ 2 , which
we would like to be as large as possible. Thus, the optimal H has the form H (ω ) = I{a≤ω≤ωo } ,
where a is selected to meet the signal power constraint with equality: (power of X )= (power of
√
X )/2. This yields a = ωo / 2. In summary, the optimal choice is H (ω )2 = I{ωo /√2≤ω≤ωo } .
8.20 Sampling a signal or process that is not band limited
(a) Evaluating the inverse transform of xo at nT , and using the fact that 2ωo T = 2π yields
∞
ωo
ωo
xo (nT ) = −∞ ejωnT xo (ω ) dω = −ωo ejωnT xo (ω ) dω = ∞=−∞ −ωo ejωnT x(ω + 2mωo ) dω
m
2π
2π
2π
(2m+1)ω ∞ = ∞=−∞ (2m−1)ωoo ejωnT x(ω ) dω = −∞ ejωnT x(ω ) dω = x(nT ).
m
2π
2π
(b) The observation follows from the sampling theorem for the narrowband signal xo and part (a).
o
(c) The fact RX (nT ) = RX (nT ) follows from part (a) with x(t) = RX (t).
o be a WSS baseband random process with autocorrelation function Ro . Then by the
(d) Let X
X
o
o
sampling theorem for baseband random processes, Xt = ∞ −∞ XnT sinc t−nT . But the discrete
n=
T
o
time processes (XnT : n ∈ Z) and (XnT : n ∈ Z) have the same autocorrelation function by part
o and Y have the same autocorrelation function. Also, Y has mean zero. So Y is WSS
(c). Thus X
o
with mean zero and autocorrelation function RX .
∞
o , S o (ω ) =
(e) For 0 ≤ ω ≤ ω
n=−∞ exp(−αω + 2nωo )
X
∞
∞
= n=0 exp(−α(ω + 2nωo )) + −=−1 exp(α(ω + 2nωo ))
n
360 = exp(−αω )+exp(α(ω −2ωo ))
1−exp(−2αωo ) = exp(−α(ω −ωo ))+exp(α(ω −ω0 ))
exp(αωo )−exp(−αωo ))
o
SX (ω ) = I{ω≤ωo } = cosh(α(ωo −ω ))
sinh(αωo ) . Thus, for any ω , cosh(α(ωo − ω ))
.
sinh(αωo ) 8.22 Another narrowband Gaussian process
(a)
∞ h(t)dt = µR H (0) = 0 µX = µR
SX (2πf ) = −∞
H (2πf )2 SR (2πf ) 4 = 10−2 e−f /10 I5000≤f ≤6000 (b)
∞ RX (0) = SX (2πf )df 2
102 = −∞ X25 ∼ N (0, 11.54) so 6000 4 e−f /10 df = (200)(e−0.5 − e−0.6 ) = 11.54 5000 P [X25 > 6] = Q √ 6
11.54 ≈ Q(1.76) ≈ 0.039 (c) For the narrowband representation about fc = 5500 (see Figure 12.5), SX SU=SV For fc= 5500 j SUV SU=SV For fc= 5000 SUV j Figure 12.5: Spectra of baseband equivalent signals for fc = 5500 and fc = 5000.
4 4 SU (2πf ) = SV (2πf ) = 10−2 e−(f +5500)/10 + e−(−f +5500)/10 If ≤500 e−.55
cosh(f /104 )If ≤500
50 = 4 4 SU V (2πf ) = 10−2 j e−(f +5500)/10 − je−(−f +5500)/10 If ≤500 = −je−.55
sinh(f /104 )If ≤500
50 (d) For the narrowband representation about fc = 5000 (see Figure 12.5),
4 SU (2πf ) = SV (2πf ) = 10−2 e0.5 e−f /10 If ≤1000
361 SU V (2πf ) = j sgn(f )SU (2πf ) 8.24 Declaring the center frequency for a given random process
(a) SU (ω ) = g (ω + ωc ) + g (−ω + ωc ) and SU V (ω ) = j (g (ω + ωc ) − g (−ω + ωc )).
(b) The integral becomes:
∞
∞
∞
dω
2 dω
2
2 dω
−∞ (g (ω + ωc )+ g (−ω + ωc )) 2π = 2 −∞ g (ω ) 2π +2 −∞ g (ω + ωc )g (−ω + ωc ) 2π = 2g  + g ∗ g (2ωc )
Thus, select ωc to maximize g ∗ g (2ωc ).
9.2 A smoothing problem
3
10
Write X5 = 0 g (s)Ys ds + 7 g (s)ys ds. The mean square error is minimized over all linear estimators if and only if (X5 − X5 ) ⊥ Yu for u ∈ [0, 3] ∪ [7, 10], or equivalently
3 RXY (5, u) = 10 g (s)RY (s, u)ds for u ∈ [0, 3] ∪ [7, 10]. g (s)RY (s, u)ds +
0 7 9.4 Interpolating a Gauss Markov process
(a) The constants must be selected so that X0 − X0 ⊥ Xa and X0 − X0 ⊥ X−a , or equivalently
e−a − [c1 e−2a + c2 ] = 0 and e−a − [c1 + c2 e−2a ] = 0. Solving for c1 and c2 (one could begin by
e−a
1
1
subtracting the two equations) yields c1 = c2 = c where c = 1+e−2a = ea +e−a = 2 cosh(a) . The corre2
2
sponding minimum MSE is given by E [X0 ] − E [X0 ] = 1 − c2 E [(X−a + Xa )2 ] = 1 − c2 (2 + 2e−2a ) =
a −e−a )(ea +e−a )
2a −e−2a
e
= (e (ea +e−a )2
= tanh(a).
(ea +e−a )2 (b) The claim is true if (X0 − X0 ) ⊥ Xu whenever u ≥ a. If u ≥ a then
1
E [(X0 − c(X−a + Xa ))Xu ] = e−u − ea +e−a (e−a−u + ea+u ) = 0. Similarly if u ≤ −a then
1
E [(X0 − c(X−a + Xa ))Xu ] = eu − ea +e−a (ea+u + e−a+u ) = 0. The orthogonality condition is thus
true whenever u ≥ a as required.
9.6 Proportional noise
(a) In order that κYt be the optimal estimator, by the orthogonality principle, it suﬃces to check
two things:
1. κYt must be in the linear span of (Yu : a ≤ u ≤ b). This is true since t ∈ [a, b] is assumed.
2. Orthogonality condition: (Xt − κYt ) ⊥ Yu for u ∈ [a, b]
It remains to show that κ can be chosen so that the orthogonality condition is true. The condition is
∗
equivalent to E [(Xt − κYt )Yu ] = 0 for u ∈ [a, b], or equivalently RXY (t, u) = κRY (t, u) for u ∈ [a, b].
The assumptions imply that RY = RX + RN = (1 + γ 2 )RX and RXY = RX , so the orthogonality
condition becomes RX (t, u) = κ(1 + γ 2 )RX (t, u) for u ∈ [a, b], which is true for κ = 1/(1 + γ 2 ).
The form of the estimator is proved. The MSE is given by E [Xt − Xt 2 ] = E [Xt 2 ] − E [Xt ]2 =
γ2
R (t, t).
1+γ 2 X
(b) Since SY is proportional to SX , the factors in the spectral factorization of SY are proportional
362 to the factors in the spectral factorization of X :
−
1 + γ 2 SX . +
1 + γ 2 SX SY = (1 + γ 2 )SX = −
SY +
SY That and the fact SXY = SX imply that
1 ejωT SXY
H (ω ) = +
−
SY
SY +
ejωT SX 1 = 1+ + +
γ 2 SX 1+ γ2 =
+ κ
+
SX (ω ) +
ejωT SX (ω ) + Therefore H is simply κ times the optimal ﬁlter for predicting Xt+T from (Xs : s ≤ t). In particular, if T < 0 then H (ω ) = κejωT , and the estimator of Xt+T is simply Xt+T t = κYt+T , which
agrees with part (a).
(c) As already observed, if T > 0 then the optimal ﬁlter is κ times the prediction ﬁlter for Xt+T
given (Xs : s ≤ t).
9.8 Short answer ﬁltering questions
(a) The convolution of a causal function h with itself is causal, and H 2 has transform h ∗ h. So if
H is a positive type function then H 2 is positive type.
(b) Since the intervals of support of SX and SY do not intersect, SX (2πf )SY (2πf ) ≡ 0. Since
SXY (2πf )2 ≤ SX (2πf )SY (2πf ) (by the ﬁrst problem in Chapter 6) it follows that SXY ≡ 0.
Hence the assertion is true.
(c) Since sinc(f ) is the Fourier transform of I[− 1 , 1 ] , it follows that
22 1
2 [H ]+ (2πf ) =
0 1
e−2πf jt dt = e−πjf /2 sinc
2 f
2 9.10 A singular estimation problem
(a) E [Xt ] = E [A]ej 2πfo t = 0, which does not depend on t.
2
RX (s, t) = E [Aej 2πfo s (Aej 2πfo t )∗ ] = σA ej 2πfo (s−t) is a function of s − t.
2
2
Thus, X is WSS with µX = 0 and RX (τ ) = σA ej 2πfo τ . Therefore, SX (2πf ) = σA δ (f − f0 ), or equiv2 δ (ω − ω ) (This makes R (τ ) = ∞ S (2πf )ej 2πf τ df = ∞ S (ω )ejωτ dω .)
alently, SX (ω ) = 2πσA
0
X
2π
−∞ X
−∞ X
∞
∞
∞
(b) (h ∗ X )t = −∞ h(τ )Xt−τ dτ = 0 αe−α−j 2πfo )τ Aej 2πfo (t−τ ) dτ = 0 αe−(ατ dτ Aej 2πfo t = Xt .
Another way to see this is to note that X is a pure tone sinusoid at frequency fo , and H (2πf0 ) = 1.
(c) In view of part (b), the mean square error is the power of the output due to the noise, or
∞ ∞ σ2 α 2
2
2
MSE=(h∗h∗RN )(0) = −∞ (h∗h)(t)RN (0−t)dt = σN h∗h(0) = σN h2 = σN 0 α2 e−2αt dt = N .
2
The MSE can be made arbitrarily small by taking α small enough. That is, the minimum mean
square error for estimation of Xt from (Ys : s ≤ t) is zero. Intuitively, the power of the signal X is
concentrated at a single frequency, while the noise power in a small interval around that frequency
is small, so that perfect estimation is possible. 9.12 A prediction problem
The optimal prediction ﬁlter is given by 1
+
SX +
ejωT SX . Since RX (τ ) = e−τ  , the spectral factoriza 363 tion of SX is given by √ √ 2
jω + 1 2
−jω + 1 +
SX SX (ω ) = −
SX +
+
so [ejωT SX ]+ = e−T SX (see Figure 12.6). Thus the optimal prediction ﬁlter is H (ω ) ≡ e−T , or in 2 e (t+T) T Figure 12.6: t 0 √ +
2ejωT SX in the time domain the time domain it is h(t) = e−T δ (t), so that XT +tt = e−T Xt . This simple form can be explained
and derived another way. Since linear estimation is being considered, only the means (assumed zero)
and correlation functions of the processes matter. We can therefore assume without loss of generality that X is a real valued Gaussian process. By the form of RX we recognize that X is Markov
so the best estimate of XT +t given (Xs : s ≤ t) is a function of Xt alone. Since X is Gaussian with
Cov(Xt+T ,Xt )Xt
mean zero, the optimal estimator of Xt+T given Xt is E [Xt+T Xt ] =
= e−T Xt .
Var(Xt )
9.14 Spectral decomposition and factorization
(a) Building up transform pairs by steps yields:
sinc(f ) ↔ I{− 1 ≤t≤ 1 }
2 −2 sinc(100f ) ↔ 10 2 I{− 1 ≤
2 sinc(100f )e2πjf T ↔ 10−2 I{− 1 ≤ t+T ≤ 1 }
2 sinc(100f )e −2 j 2πf T
+ t
≤1}
100
2 ↔ 10 100 2 I{−50−T ≤t≤50−T }∩{t≥0} so
2 −4 x = 10 length of ([−50 − T, 50 − T ] ∩ [0, +∞)) = 10−2
T ≤ −50
− T ) −50 ≤ T ≤ 50
0
T ≥ 50 10−4 (50 (b) By the hint, 1 + 3j is a pole of S . (Without the hint, the poles can be found by ﬁrst solving
for values of ω 2 for which the denominator of S is zero.) Since S is real valued, 1 − 3j must also
be a pole of S . Since S is an even function, i.e. S (ω ) = S (−ω ), −(1 + 3j ) and −(1 − 3j ) must also
be poles. Indeed, we ﬁnd
S (ω ) = 1
.
(ω − (1 + 3j ))(ω − (1 − 3j ))(ω + 1 + 3j )(ω + 1 − 3j )
364 or, multiplying each term by j (and using j 4 = 1) and rearranging terms:
S (ω ) = 1
1
(jω + 3 + j )(jω + 3 − j ) (−jω + 3 + j )(−jω + 3 − j )
S − (ω ) S + (ω ) or S + (ω ) =
constant. 1
.
(jω 2 )+6jω +10 The choice of S + is unique up to a multiplication by a unit magnitude 9.16 Estimation of a random signal, using the KL expansion
Note that (Y, φj ) = (X, φj ) + (N, φj ) for all j , where the variables (X, φj ), j ≥ 1 and (N, φj ), j ≥ 1
are all mutually orthogonal, with E [(X, φj )2 ] = λj and E [(N, φj )2 ] = σ 2 . Observation of the
process Y is linearly equivalent to observation of ((Y, φj ) : j ≥ 1). Since these random variables
are orthogonal and all random variables are mean zero, the MMSE estimator is the sum of the
projections onto the individual observations, (Y, φj ). But for ﬁxed i, only the ith observation,
(Y, φi ) = (X, φi )+(N, φi ), is not orthogonal to (X, φi ). Thus, the optimal linear estimator of (X, φi )
i ),
i
i
given Y is Cov((X,φY,φ(Y,φi )) (Y, φi ) = λλ(Y,φ2 ) . The mean square error is (using the orthogonality
i +σ
Var(( i ))
2
λ2 (λi +σ 2 )
i
i
iσ
principle): E [(X, φi )2 ] − E [ λλ(Y,φ2 ) 2 ] = λi − (iλi +σ2 )2 = λλ+σ2 .
i +σ
i
(b) Since f (t) = j (f, φj )φj (t), we have (X, f ) = j (f, φj )(X, φj ). That is, the random variable
to be estimated is the sum of the random variables of the form treated in part (a). Thus, the
best linear estimator of (X, f ) given Y can be written as the corresponding weighted sum of linear
estimators:
λi (Y, φi )(f, φi )
(MMSE estimator of (X, f ) given Y ) =
.
λi + σ 2
i The error in estimating (X, f ) is the sum of the errors for estimating the terms (f, φj )(X, φj ), and
those errors are orthogonal. Thus, the mean square error for (X, f ) is the sum of the mean square
errors of the individual terms:
λi σ 2 (f, φi )2
(MSE) =
.
λi + σ 2
i 9.18 Linear innovations and spectral factorization
First approach: The ﬁrst approach is motivated by the fact that
H (z ) = β
+
SX (z ) 1
+
SY is a whitening ﬁlter. Let and let Y be the output when X is passed through a linear timeinvariant system with z transform H(z ). We prove that Y is the innovations process for X . Since H is positive
type and limz →∞ H(z ) = 1, it follows that Yk = Xk + h(1)Xk−1 + h(2)Xk−2 + · · · Since SY (z ) =
H(z )H ∗ (1/z ∗ )SX (z ) ≡ β 2 , it follows that RY (k ) = β 2 I{k=0} . In particular,
Yk ⊥ linear span of {Yk−1 , Yk−2 , · · · }
Since H and 1/H both correspond to causal ﬁlters, the linear span of {Yk−1 , Yk−2 , · · · } is the same
as the linear span of {Xk−1 , Xk−2 , · · · }. Thus, the above orthogonality condition becomes,
Xk − (−h(1)Xk−1 − h(2)Xk−2 − · · · ) ⊥ linear span of {Xk−1 , Xk−2 , · · · }
365 Therefore −h(1)Xk−1 − h(2)Xk−2 − · · · must equal Xkk−1 , the one step predictor for Xk . Thus,
(Yk ) is the innovations sequence for (Xk ). The one step prediction error is E [Yk 2 ] = RY (0) = β 2 .
Second approah: The ﬁlter K for the optimal onestep linear predictor (Xk+1k ) is given by (take
T = 1 in the general formula):
K= 1
+
+ z SX
SX + . +
The z transform z SX corresponds to a function in the time domain with value β at time 1, and
zβ
+
+
value zero at all other negative times, so [z SX ]+ = z SX − zβ . Hence K(z ) = z − S + (z ) . If X is
X ﬁltered using K, the output at time k is Xk+1k . So if X is ﬁltered using 1 −
β
time k is Xkk−1 . So if X is ﬁltered using H(z ) = 1 − (1 − S + (z ) ) =
X β
+
SX (z ) β
+
SX (z ) , the output at then the output at time k is Xk − Xkk−1 = Xk , the innovations sequence. The output X has SX (z ) ≡ β 2 , so the prediction
e
error is RX (0) = β 2 .
e
9.20 A discretetime Wiener ﬁltering problem
To begin,
z T SXY (z )
−
SY (z ) = zT
z T +1
+
β (1 − ρ/z )(1 − zo ρ) β ( z1 − ρ)(1 − zo z )
o The right hand side corresponds in the time domain to the sum of an exponential function supported
on −T, −T + 1, −T + 2, . . . and an exponential function supported on −T − 1, −T − 2, . . .. If T ≥ 0
then only the ﬁrst term contributes to the positve part, yielding
z T SXY
−
SY
H (z ) = =
+ T
zo
β 2 (1 − zo ρ)(1 − zo /z ) T
zo
β (1 − ρ/z )(1 − zo ρ) and h(n) = T
zo
znI
.
β 2 (1 − zo ρ) o {n≥0} On the other hand if T ≤ 0 then
z T SXY
−
SY
so
H(z ) = =
+ T
zT
z (z T − z o )
+
β (1 − ρ/z )(1 − zo ρ) β ( z1 − ρ)(1 − zo z )
o T
zT
z (z T − zo )(1 − ρ/z )
+ 21
.
β 2 (1 − zo ρ)(1 − zo /z ) β ( z − ρ)(1 − zo z )(1 − zo /z )
o Inverting the z transforms and arranging terms yields that the impulse response function for the
optimal ﬁlter is given by
h(n) = 1
2
β 2 (1 − zo ) 
zon+T  − zo − ρ
1
zo − ρ n
zo +T I{n≥0} . (12.13) Graphically, h is the sum of a twosided symmetric exponential function, slid to the right by −T
and set to zero for negative times, minus a one sided exponential function on the nonnegative
366 integers. (This structure can be deduced by considering that the optimal casual estimator of Xt+T
is the optimal causal estimator of the optimal noncausal estimator of Xt+T .) Going back to the
z transform domain, we ﬁnd that H can be written as H (z ) = zT
β 2 (1 − zo /z )(1 − zo z ) −
+ T
zo (zo − ρ)
.
2
β 2 (1 − zo )( z1 − ρ)(1 − zo /z )
o (12.14) Although it is helpful to think of the cases T ≥ 0 and T ≤ 0 separately, interestingly enough, the
expressions (12.13) and (12.14) for the optimal h and H hold for any integer value of T .
9.22 Estimation given a strongly correlated process
(a) RX = g ∗ g ↔ SX (z ) = G (z )G ∗ (1/z ∗ ),
RY = k ∗ k ↔ SY (z ) = K(z )K∗ (1/z ∗ ),
RXY = g ∗ k ↔ SXY (z ) = G (z )K∗ (1/z ∗ ).
−
+
(b) Note that SY (z ) = K(z ) and SY (z ) = K∗ (1/z ∗ ). By the formula for the optimal causal
estimator,
H(z ) = 1 SXY
+
−
SY SY =
+ 1
G (z )K∗ (1/z ∗ )
K(z )
K∗ (1/z ∗ ) =
+ [G ]+
G (z )
=
K
K(z ) (c) The power spectral density of the estimator process X is given by H(z )H∗ (1/z ∗ )SY (z ) = SX (z ).
π
π
Therefore, M SE = RX (0) − RX (0) = −π SX (ejω ) dω − −π SX (ejω ) dω = 0. A simple explanation
b
b
2π
2π
1
1
for why the MSE is zero is the following. Using K inverts K, so that ﬁltering Y with K produces
the process W . Filtering that with G then yields X . That is, ﬁltering Y with H produces X , so
the estimation error is zero.
10.2 A covering problem
(a) Let Xi denote the location of the ith base station. Then F = f (X1 , . . . , Xm ), where f satisﬁes
the Lipschitz condition with constant (2r − 1). Thus, by the method of bounded diﬀerences based
2
on the AzumaHoeﬀding inequality, P {F − E [F ] ≥ γ } ≤ 2 exp(− m(2γ −1)2 ).
r
(b) Using the Possion method and associated bound technique, we compare to the case that the
number of stations has a Poisson distribution with mean m. Note that the mean number of stations
r
that cover cell i is m(2n−1) , unless cell i is near one of the boundaries. If cells 1 and n are covered,
then all the other cells within distance r of either boundary are covered. Thus,
P {X ≥ m} ≤ 2P {Poi(m) stations is not enough} 2ne−m(2r−1)/n + P {cell 1 or cell n is not covered}
(1 + )n ln n
→ 0 as n → ∞
if m =
2r − 1
≤ As to a bound going the other direction, note that if cells diﬀer by 2r − 1 or more then the events
367 that they are covered are independent. Hence,
P {X ≤ m} ≤ 2P {Poi(m) stations cover all cells} ≤ 2P {Poi(m) stations cover cells 1 + (2r − 1)j, 1 ≤ j ≤
− m(2r −1)
n ≤ 2 1−e ≤ 2 exp(−e− m(2r −1)
n → 0 as n → ∞ n−1
}
2r − 1 n−1
2r −1 n−1
)
2r − 1
(1 − )n ln n
if m =
2r − 1
· Thus, in conclusion, we can take g1 (r) = g2 (r) = 1
2r−1 . 10.4 On uniform integrability
(a) Use the small set characterization of u.i. First, supi (E [Xi ] + E [Yi ]) ≤ (supi E [Xi ]) +
(supi E [Yi ]) < ∞ Second, given > 0, there exists δ so small that E [Xi IA ] ≤ 2 and E [Yi IA ] ≤ 2
for all i, whenever A is an event with P {A} ≤ . Therefore E [(Xi + Yi IA ] ≤ E [Xi IA ]+ E [Yi IA ] ≤
for all i, whenever A is an event with P {A} ≤ . Thus, (Xi + Yi : i ∈ I ) is u.i.
(b) Suppose (Xi : i ∈ I ) is u.i., which by deﬁnition means that limc→∞ K (c) = 0, were K (c) =
supi∈I E [Xi I{Xi ≥c} ]. Let cn be a sequence of positive numbers monotonically converging to +∞
so that K (cn ) ≤ 2−n . Let ϕ(u) = ∞ (u − cn )+ . The sum is well deﬁned, because for any u, only
n=1
ﬁnitely many terms are nonzero. The function ϕ is convex and increasing because each term in the
sum is. The slope of ϕ(u) is at least n on the interval (cn , ∞) for all n, so that limu→∞ ϕ(u) = +∞.
u
Also, for any i ∈ I and c, E [(Xi  − c)+ ] = E [(Xi  − c)I{Xi ≥c} ] ≤ K (c).
Therefore, E [ϕ(Xi )] ≤ ∞ E [(Xi  − cn )+ ] ≤ ∞ 2−n ≤ 1. Thus, the function ϕ has the
n=1
n=1
required properties.
The converse is now proved. Suppose ϕ exists with the desired properties, and let > 0. Select
co so large that u ≥ co implies that u ≤ ϕ(u). Then, for all c ≥ co ,
E [Xi I{Xi ≥c} ] ≤ E [ϕ(Xi )I{Xi ≥c} ] ≤ E [ϕ(Xi )] ≤ K
Since this inequality holds for all i ∈ I , and is arbitrary, it follows that (Xi : i ∈ I ) is u.i. 10.6 A stopped random walk
(a) By the fact S0 = 0 and the choice of τ , Sn  ≤ c for 0 ≤ n ≤ τ − 1. Since the W ’s are bounded
by D, it follows that Sτ  ≤ c + D. Therefore, Sn∧τ is a bounded sequence of random variables,
so 0 = E [Sn∧τ ] → E [Sτ ] as n → ∞, by the dominated convergence theorem. (Or we could invoke
the ﬁrst or third form of the optional stopping theorem discussed in class. Note here that we are
taking it for granted that T < ∞ with probability one. If we don’t want to make that assumption,
we could appeal to part (b) or some other method. )
2
(b) Let Fn = σ (W1 , . . . , Wn ). Then Mn+1 − Mn = 2Wn+1 Sn + Wn+1 − σ 2 so that
2
E [Mn+1 − Mn Fn ] = 2E [Wn+1 Fn ]Sn + E [Wn+1 − σ 2 ] = 0, so that M is a martingale. Therefore,
2 ]/σ 2 . Arguing as in part a, we see that the sequence
E [Mτ ∧n ] = 0 for all n, or E [τ ∧ n] = E [Sn∧τ
a∧ 2
Sn∧τ  ≤ c + a ∧ b for all n. Thus, E [τ ∧ n] ≤ (c+σ2 b) for all n. But E [τ ∧ n] → E [τ ] as n → ∞ by
e the monotone convergence theorem, so that E [τ ] ≤
368 (c+a∧b)2
.
σ2
e (c) We have that E [Sn+1 − Sn Fn ] = C for all n and E [τ ] < ∞ by part (b), so by the third version
of the optional stopping theorem discussed in class, E [Sτ ] = E [S0 ] = 0.
10.8 On the size of a maximum matching in a random bipartite graph
(a) Many bounds are possible. Generally the tighter bounds are more complex. If d = 1, then
every vertex in V of degree one or more can be included in a single matching, and such a matching
1
is a maximum matching with expected size n(1 − (1 − n )n ) ≥ n(1 − e−1 ). The cardinality of a
maximum matching cannot decrease as more edges are added, so a = n(1 − e−1 ) is a lower bound
for any d and n. An upper bound is given by the expected number of vertices of V with degree at
d
least one, so b = n(1 − (1 − n )n ) ≈ n(1 − e−d ) works.
(b) The variable Z can be expressed as Z = F (V1 , . . . , Vn ), where Vi is the set of neighbors of
ui , and F satisﬁes the Lipschitz condition with constant 1. We are thus considering a martingale
associated with Z by exposing the vertices of U one at a time. The (improved version of the)
√
2
AzumaHoeﬀding inequality yields that P {Z − E [Z ] ≥ γ n} ≤ 2e−2γ .
(c) Process the vertices of U one at a time. Select an unprocessed vertex in U . If the vertex has
positive degree, select one of the edges associated with it and add it to the matching. Remove edges
from the graph that become ineligible for the matching, and reduce the degrees of the unprocessed
vertices correspondingly. (This algorithm can be improved by selecting the next vertex in U from
among those with smallest reduced degree, and also selecting outgoing edges with endpoint in V
having the smallest reduced degree in V , or both. If every matched edge has an endpoint with
reduced degree one at the time of matching, the resulting matching has maximum cardinality. ) 369 Index
autocorrelation function, see correlation function conjugate prior, 135
autocovariance function, see covariance function continuity
of a function, 302
baseband
of a function at a point, 302
random process, 238
of a random process, 190
signal, 237
of a random process at a point, 189
BaumWelch algorithm, 146
piecewise m.s., 192
Bayes’ formula, 6
convergence of sequences
Bernoulli distribution, 20
almost sure, 37
binomial distribution, 20
deterministic, 298
Borel sets, 3
in distribution, 42
BorelCantelli lemma, 7
in probability, 39
bounded convergence theorem, 308
mean square, 39, 209
bounded input bounded output (bibo) stability, convex function, 54
227
convolution, 227
Brownian motion, 101
correlation
coeﬃcient, 24
Cauchy
cross correlation matrix, 65
criterion for convergence, 47, 299
function, 96
criterion for m.s. convergence in correlation
matrix, 65
form, 50
count times, 102
sequence, 48, 299
countably inﬁnite, 298
central limit theorem, 53
counting process, 103
characteristic function
covariance
of a random variable, 19
cross covariance matrix, 65
of a random vector, 67
function, 96, 247
Chebychev inequality, 19
matrix, 65
circular symmetry, 246
pseudocovariance function, 247
joint, 246
pseudocovariance matrix, 246
completeness
Cram´r’s theorem, 56
e
of a probability space, 188
cumulative distribution function (CDF), 8, 23,
of an orthogonal basis, 212
96
of the real numbers, 299
conditional
derivative
expectation, 26, 71
righthand, 303
mean, see conditional expectation
diﬀerentiable, 303
pdf, 26
at a point, 303
probability, 5
continuous and piecewise continuously, 304
370 continuously, 304
m.s. at a point, 193
m.s. continuous and piecewise continuously,
197
m.s. continuously, 193
m.s. sense, 193
Dirichlet density, 135
discretetime random process, 95
dominated convergence theorem, 309
drift vector, 169, 176
energy spectral density, 230
Erlang B formula, 168
Erlang C formula, 167
expectation, 16
of a random vector, 65
expectationmaximization (EM) algorithm, 136
exponential distribution, 22
Fatou’s lemma, 309
forwardbackard algorithm, 144
Fourier transform, 229
inversion formula, 229
Parseval’s identity, 229
fundamental theorem of calculus, 306
gambler’s ruin problem, 99
gamma distribution, 22
Gaussian
distribution, 21
joint pdf, 77
random vector, 76
geometric distribution, 21
implicit function theorem, 305
impulse response function, 225
independence
events, 5
pairwise, 5
independent increment process, 100
inﬁmum, 300
information update, 84
inner product, 311, 313
integration
RiemannStieltjes, 307
Lebesgue, 307
LebesgueStieltjes, 307 m.s. Riemann, 197
Riemann, 305
intercount times, 102
Jacobian matrix, 26
Jensen’s inequality, 54
joint Gaussian distribution, 76
jointly Gaussian random variables, 76
Kalman ﬁlter, 82
law of total probability, 6
law of large numbers, 52
strong law, 52
weak law, 52
liminf, or limit inferior, 300, 301
limit points, 301
limsup, or limit superior, 301
linear innovations sequence, 82
Lipschitz condition, 286
Little’s law, 164
log moment generating function, 55
logsum inequality, 138
Markov inequality, 19
Markov process, 110
aperiodic, 153
birthdeath process, 157
ChapmanKolmogorov equations, 112
equilibrium distribution, 112
generator matrix, 115
holding times, 117
irreducible, 153
jump process, 117
Kolmogorov forward equations, 116
nonexplosive, 156
null recurrent, 154, 158
onestep transition probability matrix, 113
period of a state, 153
positive recurrent, 154, 158
purejump for a countable state space, 156
purejump for a ﬁnite state space, 114
spacetime structure, 117, 118
stationary, 112
time homogeneous, 112
transient, 154, 158
transition probabilities, 112 371 transition probability diagram, 113
transition rate diagram, 115
martingale, 100
matrices, 310
characteristic polynomial, 312, 313
determinant, 311
diagonal, 311
eigenvalue, 311, 312, 314
eigenvector, 311
Hermitian symmetric, 314
Hermitian transpose of, 312
identity matrix, 311
positive semideﬁnite, 312, 314
symmetric, 311
unitary, 313
maximum, 300
maximum a posteriori probability (MAP) estimator, 132
maximum likelihood (ML) estimator, 131
mean, see expectation
Mean ergodic, 203
mean function, 96
mean square closure, 255
memoryless property of the geometric distribution, 21
minimum, 300
monotone convergence theorem, 310
narrowband
random process, 242
signal, 240
norm
of a vector, 311
of an interval partition, 305
normal distribution, see Gaussian distribution
Nyquit sampling theorem, 237
orthogonal, 313
complex random variables, 209
random variables, 67
vectors, 311
orthogonality principle, 68
orthonormal, 311
basis, 311, 313
matrix, 311 periodic WSS random processes, 216
permutation, 311
piecewise continuous, 302
Poisson arrivals see time averages (PASTA), 166
Poisson distribution, 21
Poisson process, 103
posterior, or a posteriori, 132
power
of a random process, 230
spectral density, 230
prior, or a priori, 132
probability density function (pdf), 23
projection, 67
Rayleigh distribution, 23
Riemann sum, 305
sample path, 95
Schwarz’s inequality, 24
second order random process, 96
sinc function, 232
span, 311
spectral representation, 217
stationary, 105, 209
wide sense, 105, 247
strictly increasing, 299
subsequence, 301
supremum, 300
Taylor’s theorem, 303
time update, 84
timeinvariant linear system, 227
tower property, 74
transfer function, 229
triangle inequality, L2 , 24
uniform distribution, 22
uniform prior, 133
version
of a random process, 216
Viterbi algorithm, 145
wide sense sationary, 247
wide sense stationary, 105
Wiener process, see Brownian motion partition, 6
372 Bibliography
[1] S. Asmussen. Applied Probability and Queues. Springer, second edition, 2003.
[2] L.E. Baum, T. Petrie, G. Soules, and N. Weiss. A maximization technique occurring in the statistical
analysis of probabilisitic functions of markov chains. Ann. Math. Statist., 41:164–171, 1970.
[3] A.P. Dempster, N.M. Laird, and B.D. Rubin. Maximum likelihood from incomplete data via the EM
algorithm. J. Royal Statistical Society, 39(1):1–38, 1977.
[4] F.G. Foster. On the stochastic matrices associated with certain queueing processes. Ann. Math. Statist,
24:355–360, 1953.
[5] J.F.C. Kingman. Some inequalities for the queue GI/G/1. Biometrika, 49(3/4):315–324, 1962.
[6] P.R. Kumar and S.P. Meyn. Stability of queueing networks and scheduling policies. IEEE Trans. on
Automatic Control, 40(2):251–260, 1995.
[7] C. McDiarmid. On the method of bounded diﬀerences. Surveys in Combinatorics, 141:148–188, 1989.
[8] N. McKeown, A. Mekkittikul, V. Anantharam, and J. Walrand. Achieving 100% throughput in an
inputqueued switch. IEEE Trans. Communications, 47(8):1260–1267, 1999.
[9] S.P. Meyn and R.L. Tweedie. Markov chains and stochastic stability. SpringerVerlag London, 1993.
[10] J.R. Norris. Markov Chains. Cambridge University Press, 1997.
[11] L.R. Rabiner. A tutorial on hidden Markov models and selected applications in speech recognition.
Proceedings of the IEEE, 77(2):257–286, February 1989.
[12] L. Tassiulas. Scheduling and performance limits of networks with constantlychanging topology. IEEE
Trans. Information Theory, 43(3):1067–1073, 1997.
[13] L. Tassiulas and A. Ephremides. Stability properties of constrained queueing systems and schedulingpolicies for maximum throughput in multihop radio networks. IEEE Trans. on Automatic Control,
37(12):1936–1948, 1992.
[14] L. Tassiulas and A. Ephremides. Dynamic server allocation to parallel queues with randomly varyingconnectivity. IEEE Trans. Information Theory, 39(2):466–478, 1993.
[15] R.L. Tweedie. Existence of moments for stationary markov chains. Journal of Applied Probability,
20(1):191–196, 1983.
[16] C. Wu. On the convergence property of the EM algorithm. The Annals of Statistics, 11:95–103, 1983. 373 ...
View
Full
Document
This note was uploaded on 06/09/2010 for the course ECE ECE 534 taught by Professor Brucehajek during the Spring '10 term at Illinois College.
 Spring '10
 BruceHajek
 The Land

Click to edit the document details