Bruce Hajek Random Processes Notes

Bruce Hajek Random Processes Notes - Contents 1 Getting...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 5 8 11 17 21 25 26 27 28 31 ...... ...... variables ...... ...... ...... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 41 52 56 59 60 64 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 Discrete-time Kalman filtering . . . . . . . . . . . . . . . . . . . . . . . 3.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 73 75 80 80 82 83 85 91 92 96 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 . . . . . . . . . . . . . . . . . . . . . . . . iii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv CONTENTS 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 Discrete-state Markov processes . . . . . . . . . . . . . . 4.10 Space-time structure of discrete-state Markov processes 4.11 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 105 108 110 111 113 116 119 119 122 128 132 5 Inference for Markov Models 5.1 A bit of estimation theory . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 The expectation-maximization (EM) algorithm . . . . . . . . . . . . . . 5.3 Hidden Markov models . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Posterior state probabilities and the forward-backward algorithm 5.3.2 Most likely state sequence – Viterbi algorithm . . . . . . . . . . 5.3.3 The Baum-Welch algorithm, or EM algorithm for HMM . . . . . 5.4 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 143 149 154 155 158 159 161 161 . . . . . . . . . . . . 165 . 165 . 167 . 170 . 173 . 175 . 177 . 180 . 182 . 184 . 184 . 192 . 195 . . . . . 205 . 205 . 211 . 215 . 222 . 228 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Dynamics of Countable-State Markov Models 6.1 Examples with finite state space . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Classification and convergence of discrete-time Markov processes . . . . . . . 6.3 Classification and convergence of continuous-time Markov processes . . . . . 6.4 Classification of birth-death 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 birth-death processes 6.9 Foster-Lyapunov stability criterion and moment bounds . . . . . . . . . . . . 6.9.1 Stability criteria for discrete-time 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS 7.6 7.7 7.8 v The Karhunen-Lo`ve expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 e Periodic WSS random processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 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 Discrete-time processes in linear systems . . . . . . . . . . . . . . . 8.4 Baseband random processes . . . . . . . . . . . . . . . . . . . . . . 8.5 Narrowband random processes . . . . . . . . . . . . . . . . . . . . 8.6 Complexification, Part II . . . . . . . . . . . . . . . . . . . . . . . 8.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 246 249 256 258 261 267 269 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 ..... ..... 279 . 279 . 282 . 282 . 287 . 291 . 296 10 Martingales 10.1 Conditional expectation revisited . . . . . . . . . . 10.2 Martingales with respect to filtrations . . . . . . . 10.3 Azuma-Hoeffding inequaltity . . . . . . . . . . . . 10.4 Stopping times and the optional sampling theorem 10.5 Notes . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 . 325 . 326 . 330 . 331 . 333 . 333 . 335 . 336 . 336 . 337 . 339 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 Riemann-Stieltjes integration 11.5.4 Lebesgue-Stieltjes integration 11.6 On the convergence of the mean . . 11.7 Matrices . . . . . . . . . . . . . . . . 12 Solutions to Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 303 308 311 314 319 319 345 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, time-varying 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 time-varying 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 one-step 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 n-dimensional Gaussian vector into a Gaussian vector with n independent coordinates. In the new coordinates the joint probability distribution is the product of n one-dimensional distributions, representing a great vii viii CONTENTS reduction of complexity. Similarly, a random process on an interval of time, is diagonalized by the Karhunen-Lo`ve representation. A periodic random process is diagonalized by a Fourier e series representation. Stationary random processes are diagonalized by Fourier transforms. • 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. x CONTENTS 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 7-9 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 1-4 and 7-9 were covered in ECE 534 at Illinois. In Fall 2011, ECE 534 at Illinois Chapters 1-5, Sections 6.1-6.8, Chapter 7, Sections 8.1-8.4, and Section 9.1 are to be covered. A number of background topics are covered in the appendix, including basic notation. 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 1 2 CHAPTER 1. GETTING STARTED 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. 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 unit-interval 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 1.1. THE AXIOMS OF PROBABILITY THEORY 3 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. 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 → ∞. 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). 4 CHAPTER 1. GETTING STARTED 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 B1=D 1 D2 D3 ... Figure 1.1: A sequence of nested sets. j lim P [Bj ] j →∞ = (a) lim j →∞ P [Di ] i=1 ∞ = 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 can be written as a union of finitely many mutually exclusive rectangles, it follows that Tn ∈ F 1.2. INDEPENDENCE AND CONDITIONAL PROBABILITY 5 Tn 0 1 n 2 n 1 Figure 1.2: Approximation of a triangular region. ··· +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 6 CHAPTER 1. GETTING STARTED 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 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] 1.2. INDEPENDENCE AND CONDITIONAL PROBABILITY 7 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 Borel-Cantelli 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. Lemma 1.2.2 (Borel-Cantelli 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= n=1 pn < ∞, then ∞ limk→∞ n=k pn = 0, which implies part (a) of the lemma. (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 Therefore, P {An infinitely often} = limk→∞ P [∪n≥k An ] = 1. pn ) = 1 − exp(−∞) = 1. n=k 8 CHAPTER 1. GETTING STARTED 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 Borel-Cantelli lemma for independent events implies that P {An infinitely often} = 1. Example 1.2.4 Let (Ω, F , P ) be the standard unit-interval 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). 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 real-valued 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} = P {X ≤ c} (for short) (1.3) (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 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→∞ 1.3. RANDOM VARIABLES AND THEIR DISTRIBUTION FX 1 0 1 F 1 2 3 4 5 6 9 Y 0 1 Figure 1.4: Examples of CDFs. 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. 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 −1 0 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 n=1 = P [Ω] = 1. 10 CHAPTER 1. GETTING STARTED 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 ∞ 1 lim F (x + ) = lim P [An ] = P n→∞ 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.1-F.3. It must be shown that there exists a random variable with CDF F . Let Ω = R and let F be the set ˜ ˜ 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, FX (x) = pX (y ), (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 −∞ 1.4. FUNCTIONS OF A RANDOM VARIABLE 11 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. 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 ). 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. 2 Lebesgue integration is defined in Sections 1.5 and 11.5 12 CHAPTER 1. GETTING STARTED X g Ω X(ω) g(X(ω)) Figure 1.6: A function of a random variable as a composition of mappings. (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 0 √1 2π 2 exp(− s2 ) to obtain √ − + exp(−[ − √c6 2 ]2 )} 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 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 1.4. FUNCTIONS OF A RANDOM VARIABLE 13 B Θ a Figure 1.7: Direction of travel and a random direction. 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. fB −a 0 a Figure 1.8: The pdf of the effective speed in a uniformly distributed direction. 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 14 CHAPTER 1. GETTING STARTED Θ 0 Y Figure 1.9: A horizontal line, a fixed point at unit distance, and a line through the point with random direction. 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)} = tan−1 (c) + π π 2 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 ] 1.4. FUNCTIONS OF A RANDOM VARIABLE 15 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, lim fR (c) = λ→ 0 1 0≤c≤1 0 otherwise 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) 16 CHAPTER 1. GETTING STARTED !1 F (u) F(x) 1 u x 1 Figure 1.10: A CDF and its inverse. 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 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 <U ≤ = P {g (U ) = i} = P 6 6 6 so that g (U ) has the correct pmf, and hence the correct CDF. 1.5. EXPECTATION OF A RANDOM VARIABLE 1.5 17 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 E [X ] = X (ω )P [dω ] (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 18 CHAPTER 1. GETTING STARTED of X on Cj , then clearly E [X ] = xj P [Cj ] j 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 E [X ] = X (ω )dω (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 1.5. EXPECTATION OF A RANDOM VARIABLE 19 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 . 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 ∞ E [X ] = xPX (dx) (Lebesgue) (1.12) −∞ By definition, the integral (1.12) is the Lebesgue-Stieltjes integral of x with respect to FX , so that ∞ E [X ] = xdFX (x) (Lebesgue-Stieltjes) (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, g (X (ω ))P [dω ] E [g (X )] = (Lebesgue) Ω ∞ = g (x)dFX (x) (Lebesgue-Stieltjes) −∞ and in case X is a continuous type random variable ∞ E [g (X )] = g (x)fX (x)dx −∞ (Lebesgue) 20 CHAPTER 1. GETTING STARTED 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. 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 Lebesgue-Stieltjes 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 1.6. FREQUENTLY USED DISTRIBUTIONS 21 The Chebychev inequality follows by applying the Markov inequality with Y = |X − µ|2 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 ∞ exp(jux)fX (x)dx, ΦX (u) = −∞ 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 ] For a nonnegative integer-valued 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 real-valued, and n and i are integer-valued, except n can be noninteger-valued 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. 1 2 and 22 CHAPTER 1. GETTING STARTED Bi(n, p), n ≥ 1, 0 ≤ p ≤ 1 Binomial: 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. 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 integer-valued random variable with this property has a geometric distribution. 1.6. FREQUENTLY USED DISTRIBUTIONS Gaussian 23 (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 x2 1 √ e− 2 dx 2π 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, lim P n→∞ Exponential: X1 + · · · + Xn − nµ √ ≤c nσ 2 = 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. 3 As noted earlier, Φ is also used to denote characteristic functions. The meaning should be clear from the context. 24 Uniform: CHAPTER 1. GETTING STARTED 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: n n α2 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 exp − 2 r>0 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 1.7 25 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 . 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 (u1 ) = pX1 X2 (u1 , u2 ) u2 26 CHAPTER 1. GETTING STARTED 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 [XY ] | ≤ E [X 2 ]E [Y 2 ] (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 (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. 1.9. CONDITIONAL DENSITIES 27 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 ≤ m. Then E [Sm ] = mµ and Var(Sm ) = Cov(Sm , Sm ) = Var(Xi ) + i Cov(Xi , Xj ) i,j :i=j 2 = mσ . Therefore, 1.9 S√ −mµ m mσ 2 has mean zero and variance one. 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 −∞ 28 CHAPTER 1. GETTING STARTED 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 ∞ xfX |Y (x | y )dx E [X | Y = y ] = −∞ 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 one-to-one mapping from Rn to Rn . Think of g as mapping x-space (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 ) |. 1.10. TRANSFORMATION OF RANDOM VECTORS 29 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 √ fXY (x, y ) = y x+( √x −1) 2x 0 if (x, y ) ∈ A else 30 CHAPTER 1. GETTING STARTED 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 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 ∞ fX (x) = ∞ fU (x − y )fV (y )dy fXY (x, y )dy = −∞ −∞ 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 two-dimensional 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 1.11. PROBLEMS 31 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 ]. 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}. 32 CHAPTER 1. GETTING STARTED 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? 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 = 1|Y1 = 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 = 1|Z1 = 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. 1.11. PROBLEMS 33 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 > 8|X > 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 > 8|Y > 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. 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 34 CHAPTER 1. GETTING STARTED 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. 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 = i|Y < 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). 1.11. PROBLEMS 35 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 ≥ 0.4|X ≤ 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.) 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 Yi = 1 if a 2 shows on the ith roll 0 else 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 ?) 36 CHAPTER 1. GETTING STARTED (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 1 2 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. 1.11. PROBLEMS 37 (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 V2 (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, “one-dimensional 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 ). 38 CHAPTER 1. GETTING STARTED 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 sigma-algebras, 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. (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 real-valued random variable on a probability space (Ω, F , P ) is a real-valued 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). 1.11. PROBLEMS 39 (g) If X is a real-valued 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 ). 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 unit-interval 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. 41 42 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES X(ω ) 3 2 1 ω 1 4 0 1 2 3 4 1 Figure 2.1: A random variable on (Ω, F , P ). Example 2.1.2 Let (Xn : n ≥ 1) be the sequence of random variables on the standard unit-interval probability space defined by Xn (ω ) = ω n , illustrated in Figure 2.2. This sequence converges for all X1( ! ) X2( ! ) 1 X3( ! ) 1 ! 0 0 1 X4( ! ) 1 ! 0 0 1 1 ! 0 0 1 ! 0 0 1 Figure 2.2: Xn (ω ) = ω n on the standard unit-interval 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 unit-interval 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 X4 , X5 , X6 , and X7 are one on intervals of length 1 . In general, each n ≥ 1 can be written as 4 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. 2.1. FOUR DEFINITIONS OF CONVERGENCE OF RANDOM VARIABLES 43 X1( ω ) 1 ω 0 0 1 X2( ω ) X3( ω ) 1 1 ω 0 0 0 1 X4( ω ) ω 1 X5( ω ) 1 1 ω 0 0 ill lL I I II 0 1 0 X6( ω ) X7( ω ) 1 ω 0 1 1 ω 0 0 1 ω 0 0 1 Figure 2.3: A sequence of random variables on (Ω, F , P ). 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. 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: 44 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES 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 21 variables, (1.16), yields E [(X∞ )2 ] 2 ≤ E [(X∞ − Xno )2 ] 2 + E [Xno ] 2 < +∞. Example 2.1.7 (More moving, shrinking rectangles) This example is along the same lines as Example 2.1.3, using the standard unit-interval 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 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 − 0|2 ] = 2 a2 n n. m.s. Hence, Xn → 0 if and only if the sequence of constants (an ) is such that limn→∞ an = 0. For example, if an = ln(n) 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.) 2.1. FOUR DEFINITIONS OF CONVERGENCE OF RANDOM VARIABLES 45 Example 2.1.8 (Anchored, shrinking rectangles) Let (Xn : n ≥ 1) be a sequence of random variables defined on the standard unit-interval probability space, as indicated in Figure 2.5, where Xn( ! ) an ! 1 0 1/n Figure 2.5: A sequence of random variables corresponding to anchored, shrinking rectangles. 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 |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 unit-interval 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. 46 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES 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. 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 01 n F X n even n odd n −1 0 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 right-continuous 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−). 2.1. FOUR DEFINITIONS OF CONVERGENCE OF RANDOM VARIABLES 47 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 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. by th ted wi ina ble nt.) om varia ome m is d ce dom nd uen ran eco seq ingle nite s i (If a s af 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. 48 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES 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 |≥ } ≤ E [| X − Xn |2 ] (2.1) 2 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 2.1. FOUR DEFINITIONS OF CONVERGENCE OF RANDOM VARIABLES 49 IL 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. 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 right-continuitiy 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 50 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES 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) (2.3) 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. 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. 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 2.1. FOUR DEFINITIONS OF CONVERGENCE OF RANDOM VARIABLES Xk • • • • • •• • •• • • 51 k Figure 2.11: A typical sample sequence of X . 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 where σ∞ = 0.1 = 5.263. Therefore, the CDF of Xn converges everywhere to the CDF of any 19 d. 2 random variable X which has the N (0, σ∞ ) distribution. So Xn → X for any such X . 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 52 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES 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 2 where Φ−1 is the inverse of Φ. The first no − 1 terms of the sequence (σn ) are finite. Therefore, the 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 right-continuous, 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. 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 ). 2.2. CAUCHY CRITERIA FOR CONVERGENCE OF RANDOM VARIABLES 53 (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 (2.6) m.s. 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 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 21 E [Sn ] 2 ≤ 21 E [Xk1 ] 2 1 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 54 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES 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 Borel-Cantelli 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 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 ]. 2.2. CAUCHY CRITERIA FOR CONVERGENCE OF RANDOM VARIABLES 55 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 Cauchy-Schwarz 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 2 2 E [Yn ] → E [Y 2 ]. Since Xn Yn = ((Xn + Yn )2 − Xn − Yn )/2, the corollary follows. 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. 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 56 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES ∞ 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 + xn n n) → ex as n → ∞. 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 Letting s = xn n, 1 2 multiplying through by n, and applying the exponential function, yields that exp(xn − 1 if s ≥ − In fact, the sum is equal to is the main point here. π2 , 6 2x2 xn n n ) ≤ (1 + ) ≤ exp(xn ) n n if xn ≥ − n 2 but the technique of comparing the sum to an integral to show the sum is finite 2.3. LIMIT THEOREMS FOR SEQUENCES OF INDEPENDENT RANDOM VARIABLES 57 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 1 n2 Var(Xi ) ≤ i c n m.s. 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 ) = +∞. Here a second approach to proving (b) is given. The characteristic function of Xi is given by n E exp j uXi n u = E exp j ( )Xi n = ΦX 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) = n ΦX 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 58 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES u for some un between 0 and n for all n. Since Φ (un ) → jm as n → ∞, Lemma 2.3.1 yields un ΦX ( 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 we assume that EX1 = 0. Consider expanding Sn . There are n terms of the form Xi4 and 3n(n − 1) 2 X 2 with 1 ≤ i, j ≤ n and i = j . The other terms have the form X 3 X , X 2 X X terms of the form Xi j ij ijk 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 ] = Sn n E n=1 ∞ 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 → ∞. 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 22 u √ n u √ n = 1+ u2 Φ (un ) 2n for all n. Since Φ (un ) → −σ 2 as n → ∞, Lemma 2.3.1 yields σ u ΦX ( √n )n → exp(− u 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 2.4 59 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) + f (v )dvdu a 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 a, b, c with a ≥ 0, for λ constant, 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 ]). 60 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES 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 hence M (θ) itself, is well-defined 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 2.5. CHERNOFF BOUND AND LARGE DEVIATIONS THEORY 61 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 d ln M (θ) dθ E [X1 eθX1 ] E [eθX1 ] = θ=0 = 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) = sup θa − ln M (θ) (2.9) −∞<θ<∞ 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, the Chernoff bound gives the right exponent. e 62 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES Theorem 2.5.1 (Cram´r’s theorem) Suppose E [X1 ] is finite, and that E [X1 ] < a. Then for > 0 e there exists a number n such that P Sn ≥a n ≥ exp(−n(l(a) + )) (2.10) for all n ≥ n . Combining this bound with the Chernoff inequality yields 1 ln P n→∞ n Sn ≥a n lim = −l(a) In particular, if l(a) is finite (equivalently if P {X1 ≥ a} > 0) then P where ( n ) is a sequence with n Sn ≥a 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→∞ P Sn ∈ da n n n )) = 0. Informally, we can write for n large: ≈ e−nl(a) da (2.11) Proof. The lower bound (2.10) is proved here under the additional assumption that X1 is a bounded random variable: P {|X1 | ≤ C } = 1 for some constant C ; this assumption can be removed by a truncation argument covered in a homework problem. Also, to avoid trivialities, suppose P {X1 > a} > 0. The assumption that X1 is bounded and the monotone convergence theorem imply that the function M (θ) is finite and infinitely differentiable over θ ∈ R. Given θ ∈ R, let Pθ denote a new probability measure on the same probability space that X1 , X2 , . . . are defined on such that for any n and any event of the form {(X1 , . . . , Xn ) ∈ B }, Pθ {(X1 , . . . , Xn ) ∈ B } = E I{(X1 ,...,Xn )∈B } eθSn M (θ)n In particular, if Xi has pdf f for each i under the original probability measure P , then under (x θx the new probability measure Pθ , each Xi has pdf fθ defined by fθ (x) = f M )e ) , and the random (θ variables X1 , X2 , . . . are independent under Pθ . The pdf fθ is called the tilted version of f with parameter θ, and Pθ is similarly called the tilted version of P with parameter θ. It is not difficult to show that the mean and variance of the Xi ’s under Pθ are given by: E X1 eθX1 = (ln M (θ)) M (θ) 2 Varθ [X1 ] = Eθ [X1 ] − Eθ [X1 ]2 = (ln M (θ)) Eθ [X1 ] = 2.5. CHERNOFF BOUND AND LARGE DEVIATIONS THEORY 63 Under the assumptions we’ve made, X1 has strictly positive variance under Pθ for all θ, so that ln M (θ) is strictly convex. The assumption P {X1 > a} > 0 implies that (aθ − ln M (θ)) → −∞ as θ → ∞. Together with the fact that ln M (θ) is differentiable and strictly convex, there thus exists a unique value θ∗ of θ that maximizes aθ − ln M (θ). So l(a) = aθ∗ − ln M (θ∗ ). Also, the derivative of aθ − ln M (θ) at θ = θ∗ is zero, so that Eθ∗ [X ] = (ln M (θ)) = a. Observe that for any b with b > a, θ =θ ∗ P Sn ≥n n = 1 dP {ω :na≤Sn } ∗ M (θ∗ )n e−θ = ∗S n {ω :na≤Sn } = M (θ∗ )n e−θ ∗S n eθ Sn dP M (θ∗ )n dPθ∗ {ω :na≤Sn } ≥ M (θ∗ )n e−θ {ω :na≤Sn ≤nb} ∗ n −θ∗ nb M (θ ) e Pθ∗ {na ≤ ≥ ∗S n dPθ∗ Sn ≤ nb} ∗ Now M (θ∗ )n e−θ nb = exp(−n(l(a) + θ∗ (b − a)}), and by the central limit theorem, Pθ∗ {na ≤ Sn ≤ 1 nb} → 2 as n → ∞ so Pθ∗ {na ≤ Sn ≤ nb} ≥ 1/3 for n large enough. Therefore, for n large enough, P Sn ≥a n ≥ exp −n l(a) + θ∗ (b − a) + ln 3 n Taking b close enough to a, implies (2.10) for large enough n. 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 1 0 1 Figure 2.14: ln M (θ) and l(a) for an Exp(1) random variable. 64 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES Therefore, for any a ∈ R, l(a) = max{aθ − ln M (θ)} θ = max{aθ + ln(1 − θ)} θ<1 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 . 2.6. PROBLEMS 65 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∞ . 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 semi-open 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. 66 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES 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}. (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. 2.6. PROBLEMS 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 67 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? 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.) 68 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES 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 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 2.6. PROBLEMS 69 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 . (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 if u is moderately large. Thus, Q(b/σ ) ≈ 2 2 σ √ e−b /2σ b 2π if b/σ is moderately large. 70 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES √ √ (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 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 2.6. PROBLEMS 71 Let A1 , A2 , . . . be a sequence of independent random variables, with 1 P [Ai = 1] = P [Ai = 1 ] = 2 for all i. Let Bk = A1 · · · Ak . 2 (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, (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. 72 CHAPTER 2. CONVERGENCE OF A SEQUENCE OF RANDOM VARIABLES 2.35 * Completing the proof of Cramer’s theorem Prove Theorem 2.5.1 without the assumption that the random variables are bounded. To begin, select a large constant C and let Xi denote a random variable with the conditional distribution of Xi given that |Xi | ≤ C. Let Sn = X1 + · · · + Xn and let l denote the large deviations exponent for Xi . Then Sn Sn P ≥ n ≥ P {|X1 | ≤ C }n P ≥n n n One step is to show that l(a) converges to l(a) as C → ∞. It is equivalent to showing that if a pointwise monotonically increasing sequence of convex functions converges pointwise to a nonnegative convex function, then the minima of the convex functions converges to a nonnegative value. 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 EX1 EX2 EX = . . . 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 ]. 73 74CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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 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. 3.2. THE ORTHOGONALITY PRINCIPLE FOR MINIMUM MEAN SQUARE ERROR ESTIMATION75 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 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 . 76CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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. 3.2. THE ORTHOGONALITY PRINCIPLE FOR MINIMUM MEAN SQUARE ERROR ESTIMATION77 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. 78CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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 ], 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 ]. In particular, E [(X − ΠV2 (X ))2 ] ≥ E [(X − ΠV1 (X ))2 ]. (3.3) 3.2. THE ORTHOGONALITY PRINCIPLE FOR MINIMUM MEAN SQUARE ERROR ESTIMATION79 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 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. 80CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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. 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 ) 3.3. CONDITIONAL EXPECTATION AND LINEAR ESTIMATORS 81 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 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 g ∗ (Y ) = E [X | Y ] = E [X1 | Y ] E [X2 | 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 82CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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 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 ). (3.7) 3.3. CONDITIONAL EXPECTATION AND LINEAR ESTIMATORS 3.3.3 83 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 3.2.2-3.2.4 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 ]]. 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 (x|y ). ∞ fY (y ) = fXY (x, y )dx = −∞ 1 2 +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 2 + 3y . 3 + 6y Therefore, E [X | Y ] = . To find E [X | Y ] we compute EX = EY = 7 1 7 1 Cov(X, Y ) = − 144 so E [X | Y ] = 12 − 11 (Y − 12 ). 7 12 , Var(Y ) = 11 144 and 84CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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 1 x2 −x2 /2σ2 e dx = 2 σ 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 2π − 38 Var(Y ) = E [Y 2 ] − E [Y ]2 = E [X 2 ]E [U 2 ] − E [X ]2 E [U ]2 = σ 2 π 1 Cov(X, Y ) = E [U ]E [X 2 ] − E [U ]E [X ]2 = Var(X ) = σ 2 1 − 2 4 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 ∞ fY (y ) = fXY (x, y )dx = −∞ ∞ 1 −x2 /2σ 2 dx y σ2 e √ = 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 3.4. JOINT GAUSSIAN DISTRIBUTION AND GAUSSIAN RANDOM VECTORS 85 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 + b|Y ] = 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 − 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 . 86CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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. 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 − 1 exp (2π ) 2 |K | 2 (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. 3.4. JOINT GAUSSIAN DISTRIBUTION AND GAUSSIAN RANDOM VECTORS 87 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 right-hand 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. (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 88CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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. (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 m (2π ) 2 det(K ) exp − y T Λ−1 y 2 . 3.4. JOINT GAUSSIAN DISTRIBUTION AND GAUSSIAN RANDOM VECTORS 89 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 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 ]. 90CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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 (x|y ) = 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 random variables E [X 2 |Y ] and P [X ≥ c|Y ]. 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 ≥ c|Y = 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 ≥ c|Y ] = Q( c−√3/3) ). 3 3.5. LINEAR INNOVATIONS SEQUENCES 3.5 91 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 |Y1 , . . . , Yn ] = X + E [X − X |Yi ], (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 . 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 = X+ i=1 E [X Yi ]Yi E [(Yi )2 ] 92CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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 E [(Yi )2 ] i=1 k ≥ 2. The sequence Y1 , Y2 , . . . , Yn is called the linear innovations sequence for Y1 , Y2 , . . . , Yn . 3.6 Discrete-time Kalman filtering Kalman filtering is a state-space approach to the problem of estimating one random sequence from another. Recursive equations are found that are useful in many real-time 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 vk wk + xk+1 Delay xk T Hk + yk F k Figure 3.2: Block diagram of the state and observations equations. state sequence x0 , x1 , . . . is driven by the random vectors w0 , w1 , . . ., while the random vectors v0 , v1 , . . . , represent observation noise. 3.6. DISCRETE-TIME KALMAN FILTERING 93 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. 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 − xi|j ). The goal is to compute xk+1|k 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 xk|k−1 + Kk yk xk+1|k = (3.17) T = Fk xk|k−1 + Kk yk − Hk xk|k−1 with the initial condition x0|−1 = x0 , where the gain matrix Kk is given by T Kk = Fk Σk|k−1 Hk Hk Σk|k−1 Hk + Rk −1 (3.18) and the covariance of error matrices are recursively computed by T Σk+1|k = Fk Σk|k−1 − Σk|k−1 Hk Hk Σk|k−1 Hk + Rk −1 T T Hk Σk|k−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 (3.16) with xk|k−1 = xk , Σk|k−1 = Pk and Kk = 0. Taking Hk = 0 for all k is equivalent to having no observations available. 94CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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+1|k = (F − KH T )xk|k−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+1|k once xk|k−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+1|k = xk+1|k−1 + xk+1|k − xk+1|k−1 , (3.20) where the first term on the right of (3.20), namely xk+1|k−1 , represents the result of modifying xk|k−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+1|k−1 = E [Fk xk + wk |y k−1 ] = Fk E [xk |y k−1 ] + E [wk |y k−1 ] = Fk xk|k−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+1|k−1 , is given by Σk+1|k−1 = Cov(xk+1 − xk+1|k−1 ) = Cov(Fk (xk − xk|k−1 ) + wk ) T = Fk Σk|k−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 ˜ 3.6. DISCRETE-TIME KALMAN FILTERING 95 of incorporating the new observation we can pretend that yk is the new observation rather than yk . ˜ 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 xk|k−1 , T so yk = yk − Hk xk|k−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+1|k = E xk+1 |y k−1 , yk ˜ ˜ ˜ ˜ = xk+1 + E xk+1 − xk+1 |y k−1 + E [xk+1 − xk+1 |yk ] = xk+1|k−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+1|k = Fk xk|k−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+1|k = Σk+1|k−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 − xk|k−1 ) + vk ) ˜ T = Cov(Fk xk , Hk (xk − xk|k−1 )) T = Cov(Fk (xk − xk|k−1 ), Hk (xk − xk|k−1 )) = Fk Σk|k−1 Hk (3.27) 96CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION and T Cov(˜k ) = Cov(Hk (xk − xk|k−1 ) + vk ) y T = Cov(Hk (xk − xk|k−1 )) + Cov(vk ) T = Hk Σk|k−1 Hk + Rk To convert (3.26) into (3.19) use (3.22) and T Cov(Kk yk ) = Kk Cov(˜k )Kk ˜ y = Cov(xk+1 − Fk xk|k−1 )Cov(˜k )−1 Cov(xk+1 − Fk xk|k−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 real-valued random vector? 21b K = a 1 0 . b01 3.7. PROBLEMS 97 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 ≤ 1|Y ] in terms of ρ, Y , and the standard normal CDF, Φ. (b) Find E [(X − Y )2 |Y = y ] for real values of y . , 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 98CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION Cov(X, Y ) = 8. Express the following probabilities in terms of the Q function. (a) pa = P {X ≥ 2}. (b) pb = P [X ≥ 2|Y = 3]. (c) pc = P [X ≥ 2|Y ≥ 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 two-dimensional 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 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 ) 3.7. PROBLEMS 99 (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. (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 positive-semidefinite 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. 100CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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 [x|y ] is given by either x + Σx (Σx + Σn )−1 (y − (¯ + n)) ¯ x¯ 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, let f be a real constant, and let x0 denote a N (0, σ 2 ) random variable. 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 xk|k−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 Σk|k−1 ). (b) For what values of f is the sequence of error variances bounded? 3.23 Steady state gains for one-dimensional 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. 3.7. PROBLEMS 101 3.24 A variation of Kalman filtering (a) Let σ 2 > 0, let f be a real constant, and let x0 denote a N (0, σ 2 ) random variable. 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 xk|k−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 xk|k Suppose in a given application a Kalman filter has been implemented to recursively produce xk+1|k for k ≥ 0, as in class. Thus by time k , xk+1|k , Σk+1|k , xk|k−1 , and Σk|k−1 are already computed. Suppose that it is desired to also compute xk|k at time k . Give additional equations that can be used to compute xk|k . (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.) 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 . 102CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION 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.) 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 3.7. PROBLEMS 103 (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 (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 . 104CHAPTER 3. RANDOM VECTORS AND MINIMUM MEAN SQUARED ERROR ESTIMATION (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 ). 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 discrete-time random process. If T = R or if T is an interval of R, then X is called a continuous-time 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 discrete-time 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. 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 105 106 CHAPTER 4. RANDOM PROCESSES X (ω) W (ω) k ~ i k ~ ! •• ~ k k !! ••• Figure 4.1: Typical sample paths. 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 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 4.1. DEFINITION OF A RANDOM PROCESS 107 A(ω) −B(ω) 2 t A(ω)− B(ω) 4 2 Figure 4.2: Typical sample paths. 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. 108 CHAPTER 4. RANDOM PROCESSES 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 Uk Xt k t Figure 4.3: Typical sample paths. function and are given by RU (k, l) = 1 if k = l 0 else RX (s, t) = 1 if s = t 0 else . 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: 1 2 if t1 = t2 and either x1 = x2 = 1 or x1 = x2 = −1 1 if t1 = t2 and x1 , x2 ∈ {−1, 1} fX,2 (x1 , t1 ; x2 , t2 ) = 4 0 else. 4.2 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 . 4.2. RANDOM WALKS AND GAMBLER’S RUIN 109 Xn (ω) b k n Figure 4.4: A typical sample path. • 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. The so-called 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 . 110 CHAPTER 4. RANDOM PROCESSES 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 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. 4.4. BROWNIAN MOTION 111 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 1 process, and if p = 2 it is also a martingale. The following proposition is stated, without proof, to give an indication of some of the useful deductions that follow from the martingale property. 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. 112 CHAPTER 4. RANDOM PROCESSES X (ω ) t t Figure 4.5: A typical sample path of Brownian motion. A typical sample path of a Brownian motion is shown in Figure 4.5. A Brownian motion, being a 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 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 4.5 113 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 u2 t1 0 u3 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 . 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: 114 CHAPTER 4. RANDOM PROCESSES (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: f (t1 , . . . , tn |Nτ = n) = n! τn 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 n-dimensional 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 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). 4.5. COUNTING PROCESSES AND THE POISSON PROCESS 115 (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 n-dimensional 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. (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 ··· 116 CHAPTER 4. RANDOM PROCESSES 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 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 E [Xt ] = µX E [Xt1 Xt2 ] = RX (t1 − t2 ) t∈T t1 , t2 ∈ T. 4.6. STATIONARITY 117 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 with mean (µ, µ, . . . , µ)T and covariance matrix with ij th entry CX ((ti + s) − (tj + s)) = CX (ti − tj ). 118 CHAPTER 4. RANDOM PROCESSES 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) 4.7. JOINT PROPERTIES OF RANDOM PROCESSES 119 ˜ ˜ 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) 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 120 CHAPTER 4. RANDOM PROCESSES 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 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 (x|y )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 ). 4.8. CONDITIONAL INDEPENDENCE AND MARKOV PROCESSES 121 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 . 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 ∈ A|Y ]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. 122 CHAPTER 4. RANDOM 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 which, letting ρ(s, t) denote the correlation coefficient between Xs and Xt , is equivalent to the requirement ρ(t1 , tn+1 ) ρ(t1 , tn ) ρ(t2 , tn+1 )) ρ(t2 , tn ) = 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 discrete-time 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 ) = b|k| 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 ) = Ab|k| for some constants A and b with A > 0 and |b| ≤ 1. Similarly, if (Xt : t ∈ R) is a continuous-time 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(−α|τ |), for some constant α ≥ 0, or equivalently, if and only if CX has the form CX (τ ) = A exp(−α|τ |) for some constants A > 0 and α ≥ 0. 4.9 Discrete-state 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 4.9. DISCRETE-STATE MARKOV PROCESSES 123 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 real-valued 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 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 124 CHAPTER 4. RANDOM PROCESSES 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 s, τ, t ∈ T, s < τ < t. H (s, t) = H (s, τ )H (τ, t) (4.16) The relations (4.16) are known as the Chapman-Kolmogorov equations. A Markov process is time-homogeneous 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 time-homogeneous, 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 time-homogeneous 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 Chapman-Kolmogorov 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 P (k ) = H (k, k + 1) is the one-step 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 time-homogeneous then H (k ) = P k for all k , where P is the time independent one-step 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 two-stage 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 two-stage 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. 4.9. DISCRETE-STATE MARKOV PROCESSES a 125 ad 2 ad 2 01 00 d1 a ad 2 ad2 d1 10 11 d2 d 2 Figure 4.9: One-step transition probability diagram. 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 discrete-time 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 one-step 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. 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 126 CHAPTER 4. RANDOM PROCESSES Focus on the first of these three quantities to obtain η = P [an arrival at stage 1] = P [an arrival at stage 1|stage 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 continuous-time, finite-state 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 pure-jump 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 pure-jump Markov process is an S valued Markov process such that, with probability one, the sample functions are pure-jump functions. Let Q = (qij : i, j ∈ S ) be such that qij ≥ 0 qii = − j ∈S ,j =i qij 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 purejump, time-homogeneous 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) 4.9. DISCRETE-STATE MARKOV PROCESSES 127 0.5 2 1 1 1 0.5 1 3 Figure 4.10: Transition rate diagram for a continuous-time Markov process. 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 pure-jump, time-homogeneous 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 Chapman-Kolmogorov 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 (4.21) i∈S 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 two-state, continuous-time Markov process with the transition rate diagram shown in Figure 4.11 for some positive constants α and β . The generator matrix is given 128 CHAPTER 4. RANDOM PROCESSES α 1 2 β Figure 4.11: Transition rate diagram for a two-state continuous-time 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→∞ α β , 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. 4.10 Space-time structure of discrete-state Markov processes The previous section showed that the distribution of a time-homogeneous, discrete-state Markov process can be specified by an initial probability distribution, and either a one-step transition probability matrix P (for discrete-time processes) or a generator matrix Q (for continuous-time processes). Another way to describe these processes is to specify the space-time structure, which is simply the sequences of states visited and how long each state is visited. The space-time structure 4.10. SPACE-TIME STRUCTURE OF DISCRETE-STATE MARKOV PROCESSES 129 is discussed first for discrete-time processes, and then for continuous-time processes. One benefit is to show how little difference there is between discrete-time and continuous-time processes. Let (Xk : k ∈ Z+ ) be a time-homogeneous Markov process with one-step 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 !1 !2 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 time-homogeneous Markov process with onestep transition probability matrix P . (a) The jump process X J is itself a time-homogeneous Markov process, and its one-step 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 0 ≤ l ≤ n, k ≥ 1. 130 CHAPTER 4. RANDOM PROCESSES 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 − (pklljl 1 (1 − pjl jl )) j pJ 1 pJ1 j2 . . . pJn−1 jn ij j j l=0 This establishes the proposition. Next we consider the space-time structure of time-homogeneous continuous-time pure-jump Markov processes. Essentially the only difference between the discrete- and continuous-time Markov processes is that the holding times for the continuous-time 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 time-homogeneous, pure-jump Markov process with generator matrix Q. Then (a) The jump process X J is a discrete-time, time-homogeneous Markov process, and its one-step 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 ≥ c|X J (0) = j0 , . . . , X J (n) = jn ] = exp(cqjl jl ) 0 ≤ l ≤ n. 4.10. SPACE-TIME STRUCTURE OF DISCRETE-STATE MARKOV PROCESSES X(t) X(h)(1) 131 (h) X (2) (h) X (3) s 3 s2 s 1 t Figure 4.13: Illustration of sampling of a pure-jump function. 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 discrete-time Markov process with one-step 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) . The assumption that with probability one the sample paths of X are pure-jump 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 one-step transition probabilities pij (h), the formula for the jump process probabilities for discrete-time 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 one-step 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) = 132 CHAPTER 4. RANDOM PROCESSES θ + o(θ) by Taylor’s formula, we have for all c ≥ 0 that (h) > c|X J,h (0) = j0 , . . . , X J,h = jn ] c/h = (pjl jl (h)) = exp( c/h log(pjl jl (h))) = P [hTl 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. 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(−3|s − t|4 ). 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 wide-sense 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. 4.11. PROBLEMS 133 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? (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 ≥ 1|N2 = 2] in terms of λ. (b) Give a simple expression for P [N2 = 2|N1 ≥ 1] in terms of λ. (c) Let Xt = Nt2 . Is X = (Xt : t ≥ 0) a time-homogeneous 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 discrete-time 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 . 134 CHAPTER 4. RANDOM PROCESSES 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. 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 cross-correlation 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 for all k. Define a discrete-time random process Y by and let Bk have variance σB Y = (Yk : k ≥ 0), such that Y0 = 0 and Yk+1 = Ak Yk + Bk for k ≥ 0. 4.11. PROBLEMS 135 (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? (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 discrete-time 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 one-step 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 time-homogeneous Markov process. Indicate a suitable state space, and describe the one-step transition probabilities. (b) Find the distribution of Xk for k fixed. 136 CHAPTER 4. RANDOM PROCESSES (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? (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 one-step 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 4.11. PROBLEMS 137 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 space-time transformation of Brownian motion Suppose X = (Xt : t ≥ 0) is a real-valued, 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 continuous-time 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.) 4.25 Identification of special properties of two discrete-time 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 discrete-time 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. 138 CHAPTER 4. RANDOM PROCESSES (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 continuous-time 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 continuous-time 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 (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 4.11. PROBLEMS 139 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 finite-state 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 one-step 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 finite-state 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 discrete-time, discrete-state Markov process Let (Xk : k ≥ 0) be a time-homogeneous Markov process with the one-step transition probability diagram shown. 0.4 0.6 1 0.2 2 0.8 0.4 0.6 3 (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 continuous-time, discrete-space Markov process Let (Xt : t ≥ 0) be a time-homogeneous Markov process with the transition rate diagram shown. 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 . 140 CHAPTER 4. RANDOM PROCESSES 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). 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 4.11. PROBLEMS 141 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 tn/2 Hn ( Wt ), and Hn is the nth Hermite polynomial. The fact that M is a martinwhere Mn (t) = gale 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 time-homogeneous, discrete-time Markov process X = (Xk : k ≥ 0) with state space S = {1, 2, 3}, initial state X0 = 3, and one-step 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 one-step 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 one-step 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 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). 142 CHAPTER 4. RANDOM PROCESSES 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 expectation-maximization 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. 143 144 CHAPTER 5. INFERENCE FOR MARKOV MODELS 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 . 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 5.1. A BIT OF ESTIMATION THEORY 145 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 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: pΘ|Y (θ|y ) = 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 pΘ|Y (θ|y ) with respect to θ. For that purpose, the denominator in the right-hand side of (5.1) can be ignored, so that the MAP estimator is given by ΘM AP (y ) = arg max pΘ|Y (θ|y ) 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 fΘ|Y (θ|y ) 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 [Θ|Y = 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 [Θ|Y = y ]. Therefore, ΘM M SE (y ) = ΘM AP (y ) = E (Θ|Y = y ). It is interesting to compare 146 CHAPTER 5. INFERENCE FOR MARKOV MODELS 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. 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 5.1. A BIT OF ESTIMATION THEORY 147 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 148 CHAPTER 5. INFERENCE FOR MARKOV MODELS 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. 5.2. THE EXPECTATION-MAXIMIZATION (EM) ALGORITHM 5.2 149 The expectation-maximization (EM) algorithm The expectation-maximization 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 (x|θ), 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 (x|θ) (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 (Expectation-maximization (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(θ|θ(k) ) for all θ, where Q(θ|θ(k) ) = E [ log pcd (X |θ) | y, θ(k) ]. (5.5) (Maximization step) Compute θ(k+1) ∈ arg maxθ Q(θ|θ(k) ). In other words, find a value θ(k+1) of θ that maximizes Q(θ|θ(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 (x|θ), 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(θ|θ(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(θ|θ(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 150 CHAPTER 5. INFERENCE FOR MARKOV MODELS polynomial, such as the Gaussian or exponential distribution. In some cases there are closed form expressions for Q(θ|θ(k) ). In others, there may be an algorithm that generates samples of X with the desired pmf pcd (x|θ(k) ) using random number generators, and then log pcd (X |θ) is used as an approximation to Q(θ|θ(k) ). 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 one-dimensional 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, n|θ) = − log(2πθ) s2 log(2πσ 2 ) n2 − − − 2. 2 2θ 2 2σ For the estimation step, we find Q(θ|θ(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(θ|θ(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 5.2. THE EXPECTATION-MAXIMIZATION (EM) ALGORITHM 151 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 (5.6) Problem 5.3 shows that if θ(0) > 0, then θ(k) → θM L as k → ∞. 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 (x|θ) = p(y |θ)k (x|y, 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(p||q ), is defined by D(p||q ) = 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(p||q ) ≥ 0, with equality if and only if p = q (ii) D is a convex function of the pair (p, q ). 152 CHAPTER 5. INFERENCE FOR MARKOV MODELS 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 · qi ) = φ(1) = 0, D(p||q ) = φ( )qi ≥ φ( qi qi i i so (i) is proved. The proof of (ii) is based on the log-sum inequality, which is the fact that for nonnegative numbers a1 , . . . , an , b1 , . . . , bn : ai a ≥ a log , ai log (5.8) bi b i 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(a||b) ≥ 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 2 , q 2 ) are two pairs of probability distributions, and (p, q ) = α(p1 , q 1 ) + (1 − α)(p2 , q 2 ). For i fixed (p 1 i with 1 ≤ i ≤ n, the log-sum inequality (5.8) with (a1 , a2 , b1 , b2 ) = (αp1 , (1 − α)p2 , αqi , (1 − α)q2 ) i i yields αp1 log i p1 p2 i i + (1 − α)p2 log 2 i 1 qi qi = αp1 log i ≥ pi log αp1 (1 − α)p2 i i + (1 − α)p2 log i 1 2 αqi (1 − α)qi pi . qi Summing each side of this inequality over i yields αD(p1 ||q 1 ) + (1 − α)D(p2 ||q 2 ) ≥ D(p||q ), so that D(p||q ) is a convex function of the pair (p, q ). Proof of Proposition 5.2.3 Using the factorization pcd (x|θ) = p(y |θ)k (x|y, θ), Q(θ|θ(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) 5.2. THE EXPECTATION-MAXIMIZATION (EM) ALGORITHM 153 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 → ∞, (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 right-hand 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 Kuhn-Tucker 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 Kuhn-Tucker 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∗ . 154 5.3 CHAPTER 5. INFERENCE FOR MARKOV MODELS Hidden Markov models A popular model of one-dimensional 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 time-homogeneous Markov process, with one-step 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 state-space 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 = l|Z = 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 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 complete data, for a given choice of θ, is T −1 pcd (y, z |θ) = πz1 T azt ,zt+1 t=1 bzt ,yt . (5.14) t=1 The correspondence between the pmf and the graph shown in Figure 5.1 is that each term on the right-hand 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 forward-backward 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 Baum-Welch/EM algorithm). 5.3. HIDDEN MARKOV MODELS 155 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 forward-backward 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: Zt|tM AP = arg max P [Zt = i|Y1 = y1 , . . . , Yt = yt , θ] (5.15) Zt|T M AP = arg max P [Zt = i|Y1 = 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: “t|T ” 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 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 = i|θ], 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 = i|θ] = 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 right-hand side of (5.15) can be expressed in terms of the α’s as follows. P [Zt = i|Y1 = 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) 156 CHAPTER 5. INFERENCE FOR MARKOV MODELS 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 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 = i|Y1 = 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 right-hand side of (5.16), so that we have an efficient way to find the MAP smoothing estimator defined in (5.16). For later 5.3. HIDDEN MARKOV MODELS 157 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 state-transition 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 right-hand 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 forward-backward or α − β algorithm for computing the posterior distribution of the state or a transition is given by: Algorithm 5.3.1 (The forward-backward algorithm) The α’s can be recursively computed forward in time, and the β ’s recursively computed backward in time, using: αj (t + 1) = αi (t)aij bjyt+1 , with initial condition αi (1) = πi biy1 i∈S βi (t − 1) = aij bjyt βj (t), with initial condition βi (T ) = 1. j ∈S 158 CHAPTER 5. INFERENCE FOR MARKOV MODELS Then the posterior probabilities can be found: P [Zt = i|Y1 = y1 , . . . , Yt = yt , θ] = γi (t) = P [Zt = i|Y1 = 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 forward-backward 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]. The δ ’s can be computed by a recursion forward in time, using the initial values δi (1) = π (i)biy1 5.3. HIDDEN MARKOV MODELS 159 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 Baum-Welch 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(θ|θ(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(θ|θ(k) ) = E [log pcd (y, Z |θ)|y, θ(k) ] T −1 = γi (1) log πi + i∈S T ξij (t) log ai,j + t=1 i,j γi (t) log bi,yt , t=1 i∈S 160 CHAPTER 5. INFERENCE FOR MARKOV MODELS 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(θ|θ(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 forward-backward 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(θ|θ(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(θ|θ(k) ) to get T −1 Q(θ|θ (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) = γi (1) (k+1) = (k+1) = πi 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, π (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 right-hand side of (5.28) is the time-averaged 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 right-hand side of (5.29) is the time-averaged posterior conditional probability that, given the state is i at a typical time, the observation will be l. 5.4. NOTES 161 Algorithm 5.3.4 (Baum-Welch 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 forward-backward 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) ). 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 Baum-Welch 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, 162 F (θ ) = θ + CHAPTER 5. INFERENCE FOR MARKOV MODELS θ2 (y 2 −σ 2 ) σ4 + o(θ3 ). (Hint: For 0 < θ < σ 2 , 1 θ θ θ θ = σ2 (1 − σ2 + ( σ2 )2 σ 2 1+θ/σ 2 that if θ(0) > 0, then θ(k) → θM L . θ θ+σ 2 (c) Sketch F and argue, using the above properties of F, = − . . . ). 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 ? (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? 5.5. PROBLEMS 163 5.8 Specialization of Baum-Welch algorithm for no hidden data (a) Determine how the Baum-Welch 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 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(Q||BT ) + (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(Q||BT ) over Q in the same set. 5.10 Baum-Welch saddlepoint Suppose that the Baum-Welch 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 d-dimensional 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) ). 164 CHAPTER 5. INFERENCE FOR MARKOV MODELS (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 Baum-Welch algorithm The Baum-Welch 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 element-by-element ordering (for example, aij ≤ aij for all i, j.) Explain how the Baum-Welch 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 paramter θ = (π, A, B ), (b) To run the forward-backward algorithm and compute the α’s, β ’s, γ ’s, and ξ ’s , (c) To run the Baum-Welch algorithm. Experiment a bit and describe your results. For example, if T observations are generated, and then if the Baum-Welch algorithm is used to estimate the paramter, how large does T need to be to insure that the estimates of θ are pretty accurate. Chapter 6 Dynamics of Countable-State Markov Models Markov processes are useful for modeling a variety of dynamical systems. Often questions involving the long-time behavior of such systems are of interest, such as whether the process has a limiting distribution, or whether time-averages 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 discrete-time 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 countably-infinite state space, and propositions regarding convergence are presented. Example 6.1.1 Consider the discrete-time Markov process with the one-step 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 → ∞. 165 166 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS 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 one-step transition probability diagram with eight states. 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) the process is not aperiodic. Example 6.1.2 Consider the two-state, 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), O 1 ! "0 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 continuous-time processes. Example 6.1.3 Consider the continuous-time Markov process with the transition rate diagram in Figure 6.3. The Q matrix is the block-diagonal matrix given by −α α 0 0 β −β 0 0 Q= 0 0 −α α 0 0 β −β 6.2. CLASSIFICATION AND CONVERGENCE OF DISCRETE-TIME MARKOV PROCESSES167 ! ! 1 2 3 " 4 " Figure 6.3: A transition rate diagram with four states. 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 discrete-time Markov process with the transition probability diagram in Figure 6.4. The one-step transition probability matrix P is given by 1 2 1 1 1 3 Figure 6.4: A one-step 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 discrete-time Markov processes The following definition applies for either discrete time or continuous time. Definition 6.2.1 Let X be a time-homogeneous 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. 168 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS The next definition is relevant only for discrete-time 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. 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 1 Such as the Euclidean algorithm, Chinese remainder theorem, or Bezout theorem 6.2. CLASSIFICATION AND CONVERGENCE OF DISCRETE-TIME MARKOV PROCESSES169 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 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 finite-state Markov processes. Define δ (P ) for a one-step transition probability matrix P by pij ∧ pkj , δ (P ) = min 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 i,k Let µ for a vector µ denote the L1 norm: µ = |pij − pkj |. j i |µi |. 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 ˜ ˜ ˜ ˜ ˜ 170 1− CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS 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. 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 continuous-time Markov processes Chapter 4 discusses Markov processes in continuous time with a finite number of states. Here we extend the coverage of continuous-time 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 pure-jump 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 x(t) = si 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. 6.3. CLASSIFICATION AND CONVERGENCE OF CONTINUOUS-TIME MARKOV PROCESSES171 x(t) !0 !1 !2. . . !" t Figure 6.5: A pure-jump function with an explosion time. Definition 6.3.1 A pure-jump Markov process (Xt : t ≥ 0) is a Markov process such that, with probability one, its sample paths are pure-jump functions. Such a process is said to be nonexplosive if its sample paths are nonexplosive, with probability one. Generator matrices are defined for countable-state Markov processes just as they are for finitestate Markov processes. A pure-jump, time-homogeneous 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 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 space-time properties for continuous-time Markov processes with a countably infinite number of states are the same as for a finite number of states. There is a discrete-time 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 pure-jump, time-homogeneous Markov process with generator matrix Q and initial distribution π (0). The finite-dimensional distributions of the process are uniquely determined by π (0) and Q. The Chapman-Kolmogorov 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 (Birth-death processes) A useful class of countable-state Markov processes is the set of birth-death processes. A (continuous-time) birth-death process with parameters (λ0 , λ2 , . . .) and (µ1 , µ2 , . . .) (also set λ−1 = µ0 = 0) is a pure-jump 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 172 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS qij = 0 if |i − j | ≥ 2. The transition rate diagram is shown in Fig. 6.6. The space-time structure, as 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 birth-death processes are ∂πk (t) = λk−1 πk−1 (t) − (λ + µ)πk (t) + µk+1 k+1 (t) ∂t (6.5) Example 6.3.4 (Description of a Poisson process as a Markov process) Let λ > 0 and consider a birth-death process N with λk = λ and µk = 0 for all k , with initial state zero with probability one. The space-time 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. 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 discrete-time 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 discrete-time 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. !0 0 µ1 !1 1 µ2 !3 !2 2 µ3 3 µ4 Figure 6.6: Transition rate diagram of a birth-death process. 6.4. CLASSIFICATION OF BIRTH-DEATH PROCESSES 173 (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 birth-death processes The classification of birth-death processes, introduced in Example 6.3.3, is relatively simple. To avoid trivialities, consider a birth-death 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 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 ). 174 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS 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 space-time 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→∞ where ∞ S2 = i=0 (6.9) 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 reached|X (0) = 1] = 1/S2 . Consequently, X is recurrent if and only if S2 = ∞. In summary, the following proposition is proved. 6.5. TIME AVERAGES VS. STATISTICAL AVERAGES 175 Proposition 6.4.1 Suppose X is a continuous-time birth-death 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 = ∞. Discrete-time birth-death processes have a similar characterization. They are discrete-time, time-homogeneous 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 one-step 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 discrete-time process cannot be explosive.2 6.5 Time averages vs. statistical averages Let X be a positive recurrent, irreducible, time-homogeneous Markov process with equilibrium probability distribution π . To be definite, suppose X is a continuous-time process, with pure-jump sample paths and generator matrix Q. The results of this section apply with minor modifications to the discrete-time 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 t I{X (s)=i} ds 0 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 2 If in addition λi + µi = 1 for all i, then the discrete-time process has period 2. 176 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS 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 φ(j ) I{X (s)=j } ds = E 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. j ∈S φ− (j )πj < ∞, then 6.6. QUEUEING SYSTEMS, M/M/1 QUEUE AND LITTLE’S LAW 177 r------------------------~ , queue server , , ~?llllllllr~ , , , 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 first-come first-served (FCFS) in which customers are served in the order of their arrival, or last-come first-served (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, so-named 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. 178 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS If a potential departure occurs when the system is empty then the potential departure has no effect on the system. The number of customers in the system N can thus be expressed as t I{N (s−)≥1} dD(s) N (t) = N (0) + A(t) + 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 birth-death 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 birth-death 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 time-averaged 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 first-come firstserved order, the horizontal distance between the graphs gives the times in system for the customers. 6.6. QUEUEING SYSTEMS, M/M/1 QUEUE AND LITTLE’S LAW ! 179 s N(s) "s s t Figure 6.8: A single server queueing system. 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 time-averaged values of arrival rate and system content as λt = α(t)/t and Nt = 1 t t N (s)ds = γt /t 0 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 180 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS 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. 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 system-related 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 long-term 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 (Single-server, 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 6.7. MEAN ARRIVAL RATE, DISTRIBUTIONS SEEN BY ARRIVALS, AND PASTA 181 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 πk α 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 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 time-averages 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, positive-recurrent pure-jump 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 discrete-time 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 one-step 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 J . Furthermore, Y is positive recurrent and its equilibrium distribution is given distribution of X 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→∞ 182 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS with probability one. Therefore, if A ⊂ S × S , and if (i, j ) ∈ A, then lim n→∞ number of first n transitions of X that are type (i, j ) = number of first n transitions of X with type in A 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 birth-death 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). 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 birth-death 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 birth-death 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) 6.8. MORE EXAMPLES OF QUEUEING SYSTEMS MODELED AS MARKOV BIRTH-DEATH PROCESSES 1 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 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 (λ/µ)k πk = for 0 ≤ k ≤ m S1 k ! 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 birth-death 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 birth-death 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 184 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS 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. 6.9 Foster-Lyapunov 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 (1857-1918) 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 discrete-time 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 discrete-time processes Consider an irreducible discrete-time Markov process X on a countable state space S , with one-step 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 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. 6.9. FOSTER-LYAPUNOV STABILITY CRITERION AND MOMENT BOUNDS 185 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 well-defined, 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 (Foster-Lyapunov 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) In addition, suppose X is positive recurrent, so that the means, f = πf and g = πg are well-defined. Then f ≤ g . (In particular, if g is bounded, then g is finite, and therefore f is finite.) Corollary 6.9.3 (Combined Foster-Lyapunov 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 . 186 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS d queue 1 a 1 u u d queue 2 2 Figure 6.9: Two queues fed by a single arrival stream. 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. • 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 one-step 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) 6.9. FOSTER-LYAPUNOV STABILITY CRITERION AND MOMENT BOUNDS 187 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 For the system with u = u∗ , (6.23) yields 1 X1 + X2 ≤ . (6.24) We remark that, in fact, 2 (6.25) d1 + d2 − a 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. X1 + X2 ≤ 188 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS Example 6.9.5 (Route-to-shorter 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 one-step transition probability matrix when the route-to-shorter 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 route-to-shorter policy is less than or equal to the drift bound (6.22), for V for any choice of probabilistic splitting. In fact, route-to-shorter 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 route-to-shorter 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 route-to-shorter process. We remark that the stronger bound (6.25) is not always true for the route-to-shorter policy. The problem is that even if d1 − d2 < −a, the route-to-shorter 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 route-to-shorter 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. 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 discrete-time 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 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. 6.9. FOSTER-LYAPUNOV STABILITY CRITERION AND MOMENT BOUNDS input 1 1,1 189 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. 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 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 i∈ E (6.26) 190 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS 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 discrete-time 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∗ ) ≤ − and 1 2 E [(Aij (t) − Rij (σ (t)))2 |X (t) = x] ≤ i,j 1 2 E [(Aij (t))2 + (Rij (σ (t)))2 ] ≤ K i,j 6.9. FOSTER-LYAPUNOV STABILITY CRITERION AND MOMENT BOUNDS where K = 1 (N + 2 i,j 191 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 one-step transition probability matrix when the route-to-shorter 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 = xij Rij (σ ) ij xij µij (u) ij (6.30) 192 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS 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 Foster-Lyapunov stability criteria and the moment bounds. Suppose X is a time-homegeneous, irreducible, continuous-time 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 discrete-time process. Proposition 6.9.8 (Foster-Lyapunov 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. 6.9. FOSTER-LYAPUNOV STABILITY CRITERION AND MOMENT BOUNDS ! 1 queue 1 u 1 m ! 2 queue 2 3 queue 3 1 u 1 u 2 ! 193 m 2 u 2 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 off-diagonal 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 i=0 ii 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 well-defined. Then f ≤ g . Corollary 6.9.11 (Combined Foster-Lyapunov 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 194 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS necessary for stability: λ1 < m1 , λ3 < m2 , and λ1 + λ2 + λ3 < m1 + m2 (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 QV (x) = 1 2 3 ((xi + 1)2 − 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 6.10. PROBLEMS 195 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 u i=1 xi µi (u ) (6.36) i = max m1 (x1 u1 + x2 u1 ) + m2 (x2 u2 + x3 u2 ) u = m1 (x1 ∨ x2 ) + m2 (x2 ∨ x3 ) 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 one-step 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 i=1 − m1 (x1 ∨ x2 ) − m2 (x2 ∨ x3 ) + 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 discrete-time, time-homogeneous Markov chain with state space {0, 1, 2, 3} and one-step 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 continuous-time version of Example 4.9.1. Consider a pipeline consisting of two singlebuffer stages in series. Model the system as a continuous-time Markov process. Suppose new packets 196 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS 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 state-space 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 time-homogeneous 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, 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) = i|N (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 discrete-time Markov process with the nonzero one-step 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 time-homogeneous, pure-jump Markov process with state space {0, 1, 2} and Q matrix −4 2 2 Q = 1 −2 1 2 0 −2. 6.10. PROBLEMS 197 (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) = i|X (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 birth-death process with periodic rates Consider a single server queueing system in which the number in the system is modeled as a continuous time birth-death process with the transition rate diagram shown, where λa , λb , µa , and µb are strictly positive constants. !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 birth-death process Consider the birth-death process X with arrival rates λk = (p/(1 − p))k /ak and death rates µk = 198 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS (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 one-step transition probabilities for the jump-chain, 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 birth-death 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 discrete-time 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 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 one-way link in a store-and-forward 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 6.10. PROBLEMS 199 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. " 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. 200 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS 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. 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 routeto-shorter 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 discrete-time model, using the route to shorter policy, with ties broken in favor of queue 1, so u = I{x1 ≤x2 } ): 6.10. PROBLEMS 201 d queue 1 1 u a 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 Foster-Lyapunov 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: a a 1 2 queue 1 queue 2 channel . . . a N Fading 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 min-cut, max-flow theorem to an appropriate graph.) (c) Show that using the u found in part (b) that the process is positive recurrent. 202 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS (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 continuous-time 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 . (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 queue 1 1 0 ! server longer a 2 queue 2 2 6.10. PROBLEMS 203 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). 204 CHAPTER 6. DYNAMICS OF COUNTABLE-STATE MARKOV MODELS 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 Karhunen-Lo´ve expansion are discussed. In addition, e notation for complex-valued random processes is introduced. 7.1 Continuity of random processes The topic of this section is the definition of continuity of a continuous-time 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 205 206 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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. Similarly, and by essentially the same reasons, convergence for a continuous-time 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 continuous-time 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 )}. s→to n→∞ 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. 7.1. CONTINUITY OF RANDOM PROCESSES 207 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 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. 208 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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. sample-path 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. sample-path 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. sample-path 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. sample-path continuous, the set on the left-hand side of (7.1) contains an event F with P (F ) = 1 and F is also a subset of the set on the the right-hand 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. 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. sample-path continuous 7.1. CONTINUITY OF RANDOM PROCESSES 209 Example 7.1.8 (Shows a.s. sample-path 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 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. sample-path 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) 210 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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. 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. sample-path 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. 7.2. MEAN SQUARE DIFFERENTIATION OF RANDOM PROCESSES 211 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 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. 212 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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.) (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, (7.4) s−t 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 RX (s, t ) − RX (t, t ) = ∂1 RX (t, t ) s→t s−t X (t ) = lim 7.2. MEAN SQUARE DIFFERENTIATION OF RANDOM PROCESSES 213 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 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) 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 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 214 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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 ) and ∂2 ∂1 RX (s, t) = −RX (τ ). (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 =− 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 7.3. INTEGRATION OF RANDOM PROCESSES 215 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−α|τ | 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. 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: 216 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES • 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 norm of the mth partition converges to zero as m → ∞, the corresponding sequence of Riemann 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 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 two-dimensional 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 two-dimensional 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]. 7.3. INTEGRATION OF RANDOM PROCESSES 217 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 right-hand 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 two-dimensional 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]. 218 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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 right-hand 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 E Xt h(t)dt 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). 7.3. INTEGRATION OF RANDOM PROCESSES 219 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. 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 right-hand 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. 220 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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)). 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 RX (s, t) = E for t 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 RX (s, t) = 2 s t vdvdu + 0 = u t3 + 3 0 t2 (s − t) 2 vdvdu t = 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. 7.3. INTEGRATION OF RANDOM PROCESSES 221 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 ≥ t = 2Q t3 t3 3 3 t . 3 Note that P [|Xt | ≥ t] → 1 as 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 . 222 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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). (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. 7.4. ERGODICITY 223 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 Xu du = µX lim m.s. 0 A discrete time WSS random process X is similarly called mean ergodic (in the m.s. sense) if n 1 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 discrete-time version is true as well. Proposition 7.4.1 Let X be a real-valued, WSS, m.s. continuous random process. Then X is mean ergodic (in the m.s. sense) if and only if 2 t→∞ 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→∞ 1 t 2 t Xu du − µX 0 = 0. (7.32) 224 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES t 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 = = = = = t 1 t2 1 t2 2 t Xu du, 0 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 1 t2 1 t t 1 t = Cov 1 t (7.34) 0 t −t −τ CX (τ )dvdτ + 0 t 0 t − |τ | t −t t t−τ t 0 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 ! t t v !t Figure 7.4: Region of integration for (7.34) and (7.35). 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. 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 7.4. ERGODICITY 225 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 2ε t t − τ 2L (CX (0) + |c|) dτ + 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 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 226 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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 . 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 7.4. ERGODICITY 227 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 1 n = k=1 ((1 − S )Uk + SVk ) k=1 (1 − S ) = m.s. → n 1 n n Uk +S k=1 1 3 (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 1 1 1 5 3 + = . = 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 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 ). 228 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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, 1 (number of down crossings between times 1 and n + 1) = P [X1 > 0 ≥ X2 ]. (7.36) n→∞ n lim Equation (7.36) relates quantities that are quite different in nature. The left hand side of (7.36) is the long time-average 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 n-dimensional vector X can be written as X = U + jV , where U and V are each n-dimensional real vectors. As far as distributions are concerned, a random vector in n-dimensional 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 . 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 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. 7.5. COMPLEXIFICATION, PART I 229 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 pro- 230 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES cesses. By definition, X is stationary if and only if for any t1 , . . . , tn ∈ T, the joint distribution of (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 complex-valued 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 Karhunen-Lo`ve expansion e We’ve seen that under a change of coordinates, an n-dimensional 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 Karhunen-Lo`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 cn φn (t). f (t) = n=1 (7.37) ` 7.6. THE KARHUNEN-LOEVE EXPANSION 231 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 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 b a N cm φm (t) φ∗ (t)dt = n (f, φn ) = m=1 cm (φm , φn ) = cn . 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=1 232 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES or equivalently, N lim ||f − cn φn || = 0 N →∞ n=1 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) = sin( 2πkt ) for k ≥ 1. T 2 T πt sin( 2T ), (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 continuous-time 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 Karhunen-Lo`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 real-valued random variable with 0 < E [A2 ] < ∞, and Θ is a random variable uniformly distributed on [0, 2π ] and independent ` 7.6. THE KARHUNEN-LOEVE EXPANSION 233 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)] = −T E [A2 ]E [sin(2Θ)] = 0, this is indeed a KL expansion, with two terms. 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 Gramm-Schmidt 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 234 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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 λn φn (s)φ∗ (t)| = 0 n max |RX (s, t) − lim N →∞ a≤s,t≤b n=1 Theorem 7.6.6 ( Karhunen-Lo`ve expansion) Any process X satisfying the conditions of Mercer’s e theorem has a KL expansion, ∞ Xt = φn (t)(X, φn ), n=1 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 nth Thus, the mean of the 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 ∗ φ1 (t) λ1 φ∗ (t) λ2 2 . λ3 RX (s, t) = [φ1 (s)|φ2 (s)| · · · ] . . .. . ` 7.6. THE KARHUNEN-LOEVE EXPANSION 235 (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 |(f, φn )|2 . |f (t)| dt = (Energy of f ) = ||f || = a n=1 The mean total energy of (Xt : a < t < b) is thus given by b b |Xt |2 dt] = E[ RX (t, t)dt a a b∞ λn |φn (t)|2 dt = a n=1 ∞ = λn n=1 (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 These functions form a complete orthonormal basis for L2 [0, T ]. n≥0 236 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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 ∞ E −∞ ∗ ∞ f (s)Xs ds 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 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 7.7. PERIODIC WSS RANDOM PROCESSES 237 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 Karhunen-Lo`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 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 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 238 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES Therefore φn is an eigenfunction of RX with eigenvalue λn . The Karhunen-Lo`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 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): 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 ˆ ˆ E [|Xn e2πjnt/T |2 ] = E [|Xn |2 ] = pX ( 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. 7.8. PROBLEMS 7.8 239 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 is a valid autocorrelation function. (b) The function RX defined by RX (τ ) = 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. 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.) 240 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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 mean-square 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 2 , for some σ 2 > 0. (a) Find the autocorrelation function, R (s, t) for s, t ≥ 0. (b) Is X δ (τ )σ 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]. 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(−10|τ |), 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) 7.8. PROBLEMS 241 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, real-valued 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 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 242 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES and autocorrelation functions RY (τ ) = RZ (τ ) = e−|τ | . Let U be a real-valued 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 real-valued stationary Gauss-Markov 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 ≥ 4|X0 = π , X1 = 3]. You can express your answer using the Gaussian Q function 2 /2 ∞ Q(c) = c √1 π e−u du. (Hint: Think! Very little computation is needed.) 2 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 continuous-time analog of the discrete-time 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.) (b) Is X mean ergodic in the m.s. sense? (Justify your answer.) 7.8. PROBLEMS 243 (c) Describe a set of eigenfunctions and corresponding eigenvalues for the Karhunen-Lo`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 Karhunen-Lo`ve expansion of Z . That is, describe an orthornormal basis (φn : n ≥ 1), and e eigenvalues for the K-L 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 Xn e2πjnt/T . Xt = 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. 244 CHAPTER 7. BASIC CALCULUS OF RANDOM PROCESSES 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 signal. (a) What is the expected total energy of X ? (b) What are the mean and covariance functions of X ? (c) Describe the Karhunen-Lo´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 real-valued 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 Karhunen-Lo`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 Karhunen-Lo`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. 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 time-invariant 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 Karhunen-Lo`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 continuous-time random processes is all of R, while the set of frequencies relevant for discrete-time random processes is the interval [−π, π ]. For ease of notation we shall primarily concentrate on continuous-time 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 continuous-time 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 real-valued narrowband random processes. A related discussion of complex random processes is given in the last section of the chapter. 245 246CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS 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 (8.1) −∞ See Figure 8.1. For example, the linear system could be a simple integrator from time zero, defined X h Y Figure 8.1: A linear system with input X , impulse response function h, and output Y. by Ys = s 0 Xt dt 0 s≥0 s < 0, in which case the impulse response function is h(s, t) = 1 s≥t≥0 0 otherwise. 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 h µY Figure 8.2: A linear system with input µX and impulse response function h. 8.1. BASIC DEFINITIONS 247 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 ∗ ∞ h(u, τ )Xτ dτ RY (s, u) = E Ys −∞ ∞ = h∗ (u, τ )RY X (s, τ )dτ −∞ ∞ ∞ −∞ (8.5) −∞ = h∗ (u, τ )h(s, t)RX (t, τ )dtdτ (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 ∞ |y (t)| ≤ ∞ |h(t − s)|Lds = L −∞ |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 time-invariant 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, time-invariant 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 (s)g (t − s)ds f ∗ g (t) = −∞ 248CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS or equivalently ∞ f (t − s)g (s)ds f ∗ g (t) = −∞ 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 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 ∗ g ∗ k (a + b + c) = f (a + s + t)g (b − s)k (c − t)dsdt. 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. 8.2. FOURIER TRANSFORMS, TRANSFER FUNCTIONS AND POWER SPECTRAL DENSITIES249 To summarize, if X is WSS and if the linear system is time invariant, then X and Y are jointly WSS with ∞ h(t)dt RY X = h ∗ RX RY = h ∗ h ∗ RX . (8.10) µY = µX −∞ The convolution h ∗ h, equal to h ∗ h, can also be written as ∞ h(s)h(t − s)ds h ∗ h(t) = −∞ ∞ = 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 . 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 −∞ 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 (8.12) 250CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS 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 case-by-case 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 time-invariant 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 ). 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, ∞ Energy of x in frequency interval [a, b] = ||y ||2 = |H[a,b] (ω )|2 |x(ω )|2 −∞ dω = 2π b |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 8.2. FOURIER TRANSFORMS, TRANSFER FUNCTIONS AND POWER SPECTRAL DENSITIES251 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 real-valued. ∞ The Fourier inversion formula applied to SX yields that RX (τ ) = −∞ ejωτ SX (ω ) dω . In partic2π ular, ∞ dω SX (ω ) . E [|Xt |2 ] = RX (0) = (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 time-invariant 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 ) −∞ 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 = 1 T t Xs ds t−T Equivalently, Y is the output of the linear time-invariant 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 ) = 1 |τ | (1 − )+ T T 252CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS 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 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 1 u=0 u = 0. (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. Example 8.2.2 Consider two linear time-invariant systems in parallel as shown in Figure 8.5. The 8.2. FOURIER TRANSFORMS, TRANSFER FUNCTIONS AND POWER SPECTRAL DENSITIES253 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. X Y U h V k 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 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 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 and yt = qt C 254CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS R 1~' + x(t) q (t) C + − y(t) − I I Figure 8.6: An RC circuit modeled as a linear system. Therefore 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 real-valued, stationary Gaussian Markov process, so that its autocorrelation function has the form RX (τ ) = A2 e−α|τ | for some constants A2 and α > 0. Then 2A2 α SX (ω ) = 2 ω + α2 and SY (ω ) = SX (ω )|H (ω )|2 = 2A2 α (ω 2 + α2 )(1 + (RCω )2 ) Example 8.2.4 A random signal, modeled by the input random process X , is passed into a linear time-invariant system with feedback and with noise modeled by the random process N , as shown 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 signal-to-noise 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 8.2. FOURIER TRANSFORMS, TRANSFER FUNCTIONS AND POWER SPECTRAL DENSITIES255 Nt Xt + H (!) 1 + Yt H2(!) H3(!) Figure 8.7: A feedback system. Xt - H2(!)H1(!) 1!H (!)H ( )H2(!) 3 1! - ~ Xt ~~~ + 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 signal-to-noise 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 dω 2π dω 2π Example 8.2.5 Consider the linear time-invariant 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 ) 256CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS 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 ∞ SY (ω ) output power = −∞ 8.3 dω < ∞. 2π Discrete-time 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, discrete-time 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 time-invariant 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: 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 8.3. DISCRETE-TIME PROCESSES IN LINEAR SYSTEMS 257 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 discrete-time 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 Karhunen-Lo`ve expansion. The e functions in this basis all have norm 2π . Recall that when we considered the Karhunen-Lo`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 time-invariant 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, discrete-time 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 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 Power of X in frequency interval [a, b] = SX (ω ) 2π a 258CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS 8.4 Baseband random processes Deterministic baseband signals are considered first. Let x be a continuous-time 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 one-sided 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 ∞ t − nT T x(nT ) sinc x(t) = n=−∞ . (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 ω ∈ [−ωo , ωo ] (8.22) n=−∞ Plugging (8.22) into (8.19) yields ∞ x(t) = ωo x(nT )T n=−∞ −ωo ejωt e−jωnT dω . 2π (8.23) 8.4. BASEBAND RANDOM PROCESSES 259 The integral in (8.23) can be simplified using ωo T ejωτ −ωo dω τ = 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 one-sided band limit ωo if SX (ω ) = 0 for | ω |≥ ωo . Proposition 8.4.1 Suppose X is a WSS baseband random process with one-sided 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 1 ω 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 = −∞ ∞ jωnT e −∞ ωo ejωnT ejωt e−jωnT −ωo sinc t − nT T . dω 2π 260CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS 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 8.5 261 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 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 one-sided 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 real-valued baseband signals, each with one-sided 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 real-valued 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 jωc t . In a small neighborhood of a fixed time t, x is approximately a sinusoid complex sinusoid e 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 real-valued narrowband signal x results from modulating sinusoidal functions by a pair of real-valued baseband signals, or equivalently, modulating a complex sinusoidal 262CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS function by a complex-valued baseband signal. Does every real-valued narrowband signal have such a representation? The answer is yes, as we now show. Let x be a real-valued 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 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, time-invariant 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 real-valued. 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, time-invariant 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 real-valued baseband signals such that z (t) = u(t) + jv (t), and (8.32) becomes (8.30). In summary, any finite energy real-valued 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 8.5. NARROWBAND RANDOM PROCESSES 263 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 real-valued baseband random processes, and let X be defined by Xt = Ut cos(ωc t) − Vt sin(ωc t) (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 264CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS ∗ 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 real-valued 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 )I|ω|≤ωo +ωc , ˇ which has finite energy and it has a real-valued 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 real-valued random processes with mean zero and SU (ω ) = SV (ω ) = [SX (ω − ωc ) + SX (ω + ωc )] I|ω|≤ωo (8.36) and SU V (ω ) = j [SX (ω + ωc ) − SX (ω − ωc )] I|ω|≤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 . 8.5. NARROWBAND RANDOM PROCESSES 265 Proof To show that µX = 0, consider passing X through a linear, time-invariant 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 RX = RX ˇ Thus, for real numbers a and b, 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 ˇ X (a)ca + X (a)sa 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 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 266CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS SX /-:\ , SU=SV + = SUV , V xj + j = j Figure 8.10: A narrowband power spectral density and associated baseband spectral densities. 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 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 real-valued 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 discrete-time white noise sequence 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. 8.6. COMPLEXIFICATION, PART II 267 with unit variance through a discrete-time linear time-invariant system with real-valued 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 continuous-time baseband random processes with one-sided 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 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 268CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS 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 pseudo-covariance 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 pseudo-covariance 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 pseudo-covariance matrix are zero. 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 complex-valued random process and let U and V be the real-valued 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 ). 8.7. PROBLEMS 269 p The pseudo-covariance 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 real-valued 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 time-invariant 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 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 ||h||2 > 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 time-invariant 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 discrete-time mean-zero stationary random process with power E [X0 ] = 270CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS 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 time-invariant 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 8.6 A stationary two-state Markov process Let X = (Xk : k ∈ Z) be a stationary Markov process with state space S = {1, −1} and one-step transition probability matrix 1−p p P= , p 1−p 8.7. PROBLEMS 271 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 two-state 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, cross-correlation function, and associated spectral densities. Suppose Y is passed through a linear time-invariant 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 . 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? 272CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS (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 one-sided band limit of X . The process X is m.s. differentiable and X can be viewed as the output of a time-invariant 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 time-invariant 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 8.7. PROBLEMS 273 rise to a process N (a, b] as described in Problem 4.17. (The process N − λ is a white noise process, 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 time-invariant 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 one-sided band limit fo Hz. Thus, Xt = ∞ −∞ XnT sinc t−nT where T = 2fo . Let Yt = Xt for n= T each t. 274CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS (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. 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 discrete-time 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 ds|2 ] 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 continuous-time 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 real-valued 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 )| ≤ π|τ | 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 time-invariant 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 ? 8.7. PROBLEMS 275 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 .) 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 band-limited 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 (ω ) defined by S o (ω ) = I power spectral density SX {|ω |≤ωo } n=−∞ SX (ω + 2nωo ). Show that X 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(−α|ω |) for ω ∈ R. 8.21 A narrowband Gaussian process Let X be a real-valued 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 real-valued 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 real-valued 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 time-invariant 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 real-valued 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 time-invariant system with 2 276CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS transfer function H specified as follows, where f represents the frequency in gigahertz (GHz) and a gigahertz is 109 cycles per second. H (2πf ) = 1 19.12−|f | 0.01 0 19.10 ≤ |f | ≤ 19.11 19.11 ≤ |f | ≤ 19.12 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), 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 discrete-time random process and that g is a deterministic function. 8.7. PROBLEMS 277 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 zero-mean 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. 278CHAPTER 8. RANDOM PROCESSES IN LINEAR SYSTEMS AND SPECTRAL ANALYSIS 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 a 279 for u ∈ [a, b] 280 CHAPTER 9. WIENER FILTERING or equivalently b for u ∈ [a, b]. h(t, s)RY (s, u)ds RXY (t, u) = a 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 ) = SXY (ω ) SX (ω ) = SY (ω ) SX (ω ) + SN (ω ) 9.1. RETURN OF THE ORTHOGONALITY PRINCIPLE 281 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 ) − −∞ ∞ = −∞ 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−2|τ | 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 SX (ω ) = H (ω ) = SX (ω ) + SN (ω ) 1 1+ω 2 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−α|t| ↔ 2α , ω 2 + 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). 282 9.2 CHAPTER 9. WIENER FILTERING 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, time-invariant 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 time-invariant 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 ∞ h(t − s)RY (s − u)ds RXY (t + T − u) = 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 Wiener-Hopf 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: 9.3. CAUSAL FUNCTIONS AND SPECTRAL FACTORIZATION 283 • H = [H ]+ + [H ]− (because h = uh + (1 − u)h) • [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−α|t| 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 two-sided exponential function. 284 CHAPTER 9. WIENER FILTERING 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 = 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 − ω2 (jω + 3)(jω − 2) γ2 = 1 − ω2 (jω + 1)(jω − 2) γ3 = 1 − ω2 (jω + 1)(jω + 3) = 1 + (−1)2 1 =− (−1 + 3)(−1 − 2) 3 = 1 + 32 =1 (−3 + 1)(−3 − 2) j ω =−1 j ω =−3 = j ω =2 1 + 22 1 = (2 + 1)(2 + 3) 3 9.3. CAUSAL FUNCTIONS AND SPECTRAL FACTORIZATION 285 Consequently, [G]+ (ω ) = − 1 1 + 3(jω + 1) jω + 3 and [G]− (ω ) = 1 3(jω − 2) −jωT 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 e−α(t−T ) t ≥ T , 0 t<T g (t) = as pictured in Figure 9.3. Consider two cases. First, if T ≥ 0, then g is causal, G is positive type, g(t) T>0: t T T<0: g(t) t T Figure 9.3: Exponential function shifted by T. and therefore [G]+ = G and [G]− = 0. Second, if T ≤ 0 then g (t)u(t) = eαT e−αt t ≥ 0 0 t<0 −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 ) = e T α(T −t) −jωt e eαT −(α+jω)t dt = −(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 −∞ 286 CHAPTER 9. WIENER FILTERING 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 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 real-valued, 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 ∗ . 9.4. SOLUTION OF THE CAUSAL WIENER FILTERING PROBLEM FOR RATIONAL POWER SPECTRAL D 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 ) = √ 8 5) (jω + 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 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 Wiener-Hopf 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 288 CHAPTER 9. WIENER FILTERING 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) − −∞ 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 ) 9.4. SOLUTION OF THE CAUSAL WIENER FILTERING PROBLEM FOR RATIONAL POWER SPECTRAL D + − 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 = Therefore −jω + 2 √ =√ 8 5 ) j ω =−1 5(−jω + −jω + 2 √ 5(jω + 1) − =√ q j ω= 8 5 SXY (ω ) − SY (ω ) = + 3 √ 5+ 8 8 5 +2 √ 5+ 8 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 . 3 290 CHAPTER 9. WIENER FILTERING 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 time-invariant 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 1 (jω + (1 + j ))(jω + (1 − j )) γ1 γ2 + jω + (1 + j ) jω + (1 − j ) + SX (ω ) = = where γ1 = γ2 = 1 jω + (1 − j ) 1 jω + (1 + j ) = j ω =−(1+j ) = j ω =−1+j j 2 −j 2 + yielding that the inverse Fourier transform of SX is given by + SX ↔ j −(1+j )t j e u(t) − e−(1−j )t u(t) 2 2 Hence j −(1+j )(t+T ) 2e + SX (ω )ejωT ↔ j − 2 e−(1−j )(t+T ) t ≥ −T 0 else so that + SX (ω )ejωT + = je−(1+j )T je−(1−j )T − 2(jω + (1 + j )) 2(jω + (1 − j )) 9.5. DISCRETE TIME WIENER FILTERING 291 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 discrete-time 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 discrete-time 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=−∞ ∗ ∞ ∗ h(l)(1/z ) = H∗ (1/z ∗ ) l=−∞ 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 lim|z |→∞ 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 discrete-time 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 292 CHAPTER 9. WIENER FILTERING denoted by SXY . Recall that if Y is the output random process when X is passed through a linear time-invariant 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 − |ρ|2 ) 1 − ρ/z 1 − ρ∗ z SY (z ) = which is the z -transform of RY (k ) = ρk 1−|ρ|2 (ρ∗ )−k 1−|ρ|2 k≥0 k<0 The z -transform SY of RY converges absolutely for |ρ| < z < 1/|ρ|. 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 9.5. DISCRETE TIME WIENER FILTERING 293 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 positive-type, minimum phase function H has the property that both H and its inverse 1/H are causal functions. Two linear time-invariant 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 ∗ is also a zero of S . These observations imply that S can be uniquely SY with z0 = 0, then 1/z0 Y Y 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 ∗ ))∗ + • lim|z |→∞ SY (z ) = β There is an additional symmetry if RY is real-valued: ∞ ∞ ∗ (RY (k )(z ∗ )−k )∗ = SY (z ∗ ) RY (k )z −k = SY (z ) = k=−∞ (for real-valued 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 real-valued 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 ∗ /z ) (1 − ρz )(1 − ρ∗ z ) (1 − ρ/z )(1 − ρ + SY ( z ) − SY ( z ) 294 CHAPTER 9. WIENER FILTERING An important application of spectral factorization is the generation of a discrete-time WSS random process with a specified correlation function RY . The idea is to start with a discrete-time white noise process W with RW (k ) = I{k=0} , or equivalently, with SW (z ) ≡ 1, and then pass it through an appropriate linear, time-invariant 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 Wiener-Hopf 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 Wiener-Hopf 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 + (z ) is a minimum phase function, G is a positive type function and the system is causal. Thus, S 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 time-invariant 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 time-invariant system with impulse response function RXZ (T + k )I{k≥0} , which has z transform [z T SXZ ]+ . See Figure 9.4. 9.5. DISCRETE TIME WIENER FILTERING Y 1 S +(z) Z 295 [z TS XZ Y ^ Xt+T|t (z)]+ Figure 9.4: Optimal filtering based on whitening first. The transfer function for two linear, time-invariant systems in series is the product of their z -transforms. In addition, SXZ (z ) = G ∗ (1/z ∗ )SXY (z ) = SXY (z ) − SY (z ) 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 discrete-time 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 ) + SY (z ) where β2 = σ2ρ z0 − 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 ) z 1 + 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 + 296 CHAPTER 9. WIENER FILTERING + 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) = 9.6 1 − z0 ρ)(1 − z0 /z ) n z0 I{n≥0} β 2 (1 − z0 ρ) Problems 9.1 A quadratic predictor Suppose X is a mean zero, stationary discrete-time random process and that n is an integer with n ≥ 1. Consider estimating Xn+1 by a nonlinear one-step predictor of the form n Xn+1 = h0 + n j h1 (k )Xk + k=1 h2 (j, k )Xj Xk j =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 real-valued, 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 9.6. PROBLEMS 297 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 two-sided Markov property.) 9.5 Estimation of a filtered narrowband random process in noise Suppose X is a mean zero real-valued stationary Gaussian random process with the spectral density shown. S (2 π f) X 8 Hz 1 8 Hz f 10 4 Hz 10 4 Hz (a) Explain how X can be simulated on a computer using a pseudo-random 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 time-invariant 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). 298 CHAPTER 9. WIENER FILTERING 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 → ∞. 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 [|A|2 ] = 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 signal-to-noise 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 . 9.6. PROBLEMS 299 (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 zero-mean, wide-sense 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. 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 continuous-time Wiener filtering problem Let (Xt ) and (Nt ) be uncorrelated, mean zero random processes with RX (t) = exp(−2|t|) 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. 300 CHAPTER 9. WIENER FILTERING 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 one-sided 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 = max|ω|≤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. 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 lim|z |→∞ SX (z ) = β for some β > 0. The linear innovations sequence of X is the sequence X such that Xk = Xk − Xk|k−1 , where Xk|k−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 time-invariant filter, and identify the corresponding value of H. (b) Identify the mean square prediction error, E [|Xk − Xk|k−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.) 9.6. PROBLEMS 301 9.20 A discrete-time Wiener filtering problem Extend the discrete-time 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 discrete-time, 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 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 Xn|n = ∞ −∞ 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 (1 + cos( πω )) ωo No 2 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 → ∞. SX (ω ) = I{|ω|≤ωo } and SN (ω ) = 9.24 * Resolution of Wiener and Kalman filtering Consider the state and observation models: Xn = F Xn−1 + Wn Yn = H T Xn + Vn 302 CHAPTER 9. WIENER FILTERING where (Wn : −∞ < n < +∞) and (Vn : −∞ < n < +∞) are independent vector-valued 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 Wiener-Hopf 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. 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 (x|y )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: 303 304 CHAPTER 10. MARTINGALES (a) Ω ∈ D, (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 sub-σ -algebra of F if D is a σ algebra such that D ⊂ F . A random variable Z is said to be D-measurable 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, sub-σ -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 sub-σ -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 sub-σ -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. 1 The smallest one exists–it is equal to the intersection of all σ -algebras which contain all sets of the form {Yi ≤ c}. 10.1. CONDITIONAL EXPECTATION REVISITED 305 Definition 10.1.3 If X is a random variable on (Ω, F , P ) with finite mean and D is a sub-σ 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 (i) E [X |D] is D-measurable (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 D-measurable 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 D-measurable 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 D-measurable because it is the limit of a sequence of D-measurable 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. 306 CHAPTER 10. MARTINGALES 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 sub-σ 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 )]. 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 D-measurable 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 D-measurable. 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]. 10.1. CONDITIONAL EXPECTATION REVISITED 307 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 D-measurable. 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 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. 308 CHAPTER 10. MARTINGALES 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 sub-σ -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 Y (Ω, F , P ), the filtration generated by Y , often written as F Y = (Fn : n ≥ 0), is defined by letting Y = σ (Y : k ≤ n). (If there is no variable Y defined, we take F Y to be the trivial σ -algebra, Fn 0 k 0 Y F0 = {∅, Ω}, representing no observations.) 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, 10.2. MARTINGALES WITH RESPECT TO FILTRATIONS 309 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 . 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. 310 CHAPTER 10. MARTINGALES 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 [Φ|Fn ]. 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 “one-step” predictable, because Fn determines H one step ahead.) 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 Mn = M0 + Hk D k k=1 10.3. AZUMA-HOEFFDING INEQUALTITY 311 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 2 E [Hk E [Dk |Fk−1 ]]. E [(Mn − M0 ) ] = k=1 10.3 Azuma-Hoeffding inequaltity One of the most simple inequalities for martingales is the Azuma-Hoeffding 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 ≤ x − b + d α(b+d) b + d − x α(b−d) e + e . 2d 2d (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) = 2 See McDiarmid survey paper (1 − θ2 )(eu − e−u ) (1 − θ)eu + (1 + θ)e−u 312 CHAPTER 10. MARTINGALES 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 (Azuma-Hoeffding 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 ) . 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. 10.3. AZUMA-HOEFFDING INEQUALTITY 313 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 Azuma-Hoeffding inequality with centering can be applied with di = 2 for all i, giving the desired result. 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 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. 314 CHAPTER 10. MARTINGALES 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. 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 ]? 4 Since Xn is degenerate, we could use n − 1 instead of n, but it makes little difference. 10.4. STOPPING TIMES AND THE OPTIONAL SAMPLING THEOREM 315 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 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. 316 CHAPTER 10. MARTINGALES 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. 10.4. STOPPING TIMES AND THE OPTIONAL SAMPLING THEOREM 317 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 |> + 2|Z |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 ]. 318 CHAPTER 10. MARTINGALES 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 of turns needed by the player until the urn is empty. The bound should hold for any strategy the 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} 10.5. NOTES 319 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 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 Azuma-Hoeffding 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 [E|Y0 , . . . , 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. 320 CHAPTER 10. MARTINGALES (a) Let F denote the number of cells covered. Apply the method of bounded differences based on the Azuma-Hoeffding 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 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 = 10.6. PROBLEMS 321 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 satisfies τ = min{n ≥ 0 : |Sn | ≥ c}. Let σ 2 = Var(Wi ). Let Sn = W1 + . . . Wn for n ≥ 0 and let 2 Mn = Sn − nσ 2 . Show that M is a martingale. Hence, E [Mτ ∧n ] = 0 for all n. Conclude that E [τ ] < ∞ (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. 322 CHAPTER 10. MARTINGALES (“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 sub-σ -algebra of F . A conditional probability such as P [X ≤ c|D] 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 distributions, defined next, is to select a version of P [X ≤ c|D] for every real number c so that, as a function of c for ω fixed, the result is a valid CDF (i.e. nondecreasing, right-continuous, 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 (c|ω ), is a function of (c, ω ) ∈ R × Ω such that: (1) for each c ∈ R fixed, FX |D (c|ω ) is a D measurable function of ω , (2) for each ω fixed, as a function of c, FX |D (c|ω ) is a valid CDF, (3) for any c ∈ R, FX |D (c|ω ) is a version of P [X ≤ c|D]. 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 sub-σ -algebra of F . Let FX |D (c|ω ) 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 (c|ω ) : c ∈ R}, which is contained in the extended real line R ∪ {−∞, +∞}. Symbollically: E [X |D](ω ) = R cFX |D (dc|ω ). Show that, in the special case that E [|X |] < ∞, this 10.6. PROBLEMS 323 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 . 324 CHAPTER 10. MARTINGALES 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 325 326 CHAPTER 11. APPENDIX 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 two-dimensional 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. 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 +∞. 11.2. CONVERGENCE OF SEQUENCES OF NUMBERS 327 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. 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 1 x−α dx = 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 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. 328 CHAPTER 11. APPENDIX 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. 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 n2 1 (a) xn = cos(+1) , (b) yn = log n (c) zn = n=2 k log k 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. 11.2. CONVERGENCE OF SEQUENCES OF NUMBERS 329 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}] , (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. 330 CHAPTER 11. APPENDIX 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. 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 1909-1910: 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 right-hand 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 11.4. DERIVATIVES OF FUNCTIONS 331 left-hand limit f (xo −) = limx xo f (x) is defined similarly. If f is monotone nondecreasing, then the left-hand and right-hand limits exist, and f (xo −) ≤ f (xo ) ≤ f (xo +) for all xo . A function f is called right-continuous at xo if f (xo ) = f (xo +). A function f is simply called right-continuous if it is right-continuous 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. 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 right-hand 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 left-hand 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 ). 332 CHAPTER 11. APPENDIX 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. 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)|. 11.5. INTEGRATION 333 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 right-hand limit of the derviative at a, f (a+) = limx a f (x), exists, then the right-hand derivative at a, defined by D+ f (a) = lim x a f (x) − f (a) x−a also exists, and the two limits are equal. 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=1 334 CHAPTER 11. APPENDIX 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 th partition converges to zero as m → ∞, the corresponding sequence of Riemann norm of the m 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: 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 right-hand 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 11.5. INTEGRATION 335 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 12 1 1 2 2 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. 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 E [X ] = X (ω )P (dω ) 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 < +∞. ∞ −∞ g+ (ω )dω < +∞ 336 11.5.3 CHAPTER 11. APPENDIX Riemann-Stieltjes integration Let g be a bounded function on a closed interval [a, b] and let F be a nondecreasing function on [a, b]. The Riemann-Stieltjes integral b g (x)dF (x) (Riemann-Stieltjes) 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 Lebesgue-Stieltjes 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 Lebesgue-Stieltjes integral of g ˜ with respect to F is defined to be the Lebesgue integral of g with respect to P : ∞ (Lebesgue-Stieltjes) ∞ g (x)dF (x) = −∞ ˜ g (x)P (dx) (Lebesgue) −∞ ∞ The same notation −∞ g (x)dF (x) is used for both Riemann-Stieltjes (RS) and Lebesgue-Stieltjes (LS) integration. If g is continuous and the LS integral is finite, then the integrals agree. In ∞ particular, −∞ xdF (x) is identical as either an LS or RS integral. However, for equivalence of the integrals ∞ g (X (ω ))P (dω ) and g (x)dF (x), −∞ 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 ∞ (Lebesgue-Stieltjes) ∞ g (x)dF (x) = −∞ for any Borel measurable function g . g (x)f (x)dx −∞ (Lebesgue) 11.6. ON THE CONVERGENCE OF THE MEAN 11.6 337 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 ]. 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. 338 CHAPTER 11. APPENDIX 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∞ ]. 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. 11.7. MATRICES 339 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 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. 340 CHAPTER 11. APPENDIX 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 one-to-one 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 (−1)π 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. 11.7. MATRICES 341 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 real-valued (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 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 T and Λ = U T KU . A symmetric m × m matrix A is positive semidefinite if αT Aα ≥ 0 for all m-dimensional 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 1 342 CHAPTER 11. APPENDIX The set of all dimension m vectors over C is the m-complex 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. 11.7. MATRICES 343 7. The equation p(λ) = det(λI − A) defines a polynomial p of degree m called the characteristic polynomial of A. 8. The zeros 1 , 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 , λ2 , . . . , λm , are real-valued (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 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 ∗ 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 one-to-one 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 and A 0 B A + BA−1 B . (11.7) The second matrix is obtained from the first by left multiplying each sub-block 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 344 CHAPTER 11. APPENDIX 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. Index mean, see conditional expectation pdf, 28 probability, 5 baseband conjugate prior, 148 random process, 259 continuity signal, 258 of a function, 330 Baum-Welch algorithm, 161 of a function at a point, 330 Bayes’ formula, 6 of a random process, 208 Bernoulli distribution, 21 of a random process at a point, 207 binomial distribution, 22 piecewise m.s., 211 Borel sets, 3 convergence of sequences Borel-Cantelli lemma, 7 almost sure, 41 bounded convergence theorem, 337 deterministic, 326 bounded input bounded output (bibo) stability, in distribution, 47 247 in probability, 43 Brownian motion, 111 mean square, 43, 229 convex function, 59 Cauchy convolution, 247 criterion for convergence, 52, 328 criterion for m.s. convergence in correlation correlation coefficient, 26 form, 54 cross correlation matrix, 73 sequence, 53, 328 function, 105 central limit theorem, 58 matrix, 73 characteristic function count times, 113 of a random variable, 21 countably infinite, 326 of a random vector, 75 counting process, 113 Chebychev inequality, 20 covariance circular symmetry, 267 cross covariance matrix, 73 joint, 267 function, 105, 269 completeness matrix, 74 of a probability space, 206 pseudo-covariance function, 269 of an orthogonal basis, 232 pseudo-covariance matrix, 268 of the real numbers, 328 Cram´r’s theorem, 62 e conditional expectation, 28, 81 cumulative distribution function (CDF), 8, 25, autocorrelation function, see correlation function autocovariance function, see covariance function 401 402 INDEX events, 5 pairwise, 5 derivative independent increment process, 110 right-hand, 331 infimum, 328 differentiable, 331 information update, 94 at a point, 331 inner product, 340, 342 continuous and piecewise continuously, 332 integration continuously, 332 Riemann-Stieltjes, 336 m.s. at a point, 211 Lebesgue, 335 m.s. continuous and piecewise continuously, Lebesgue-Stieltjes, 336 215 m.s. Riemann, 216 m.s. continuously, 211 Riemann, 333 m.s. sense, 211 intercount times, 113 Dirichlet density, 148 Jacobian matrix, 28 discrete-time random process, 105 Jensen’s inequality, 59 dominated convergence theorem, 338 joint Gaussian distribution, 86 drift vector, 184, 192 jointly Gaussian random variables, 86 energy spectral density, 250 Kalman filter, 92 Erlang B formula, 183 Erlang C formula, 183 law of total probability, 6 expectation, 17 law of large numbers, 57 of a random vector, 73 strong law, 57 expectation-maximization (EM) algorithm, 149 weak law, 57 exponential distribution, 23 liminf, or limit inferior, 329, 330 limit points, 329, 330 Fatou’s lemma, 338 limsup, or limit superior, 329, 330 forward-backard algorithm, 157 linear innovations sequence, 92 Fourier transform, 249 Lipschitz condition, 313 inversion formula, 249 Little’s law, 179 Parseval’s identity, 249 log moment generating function, 60 fundamental theorem of calculus, 334 log-sum inequality, 152 gambler’s ruin problem, 109 Markov inequality, 20 gamma distribution, 24 Markov process, 121 Gaussian aperiodic, 168 distribution, 23 birth-death process, 171 joint pdf, 86 Chapman-Kolmogorov equations, 124 random vector, 86 equilibrium distribution, 124 geometric distribution, 22 generator matrix, 126 implicit function theorem, 333 holding times, 129 impulse response function, 246 irreducible, 167 jump process, 129 independence 105 INDEX Kolmogorov forward equations, 127 nonexplosive, 171 null recurrent, 168, 172 one-step transition probability matrix, 124 period of a state, 168 positive recurrent, 168, 172 pure-jump for a countable state space, 171 pure-jump for a finite state space, 126 space-time structure, 129, 130 stationary, 124 time homogeneous, 124 transient, 168, 172 transition probabilities, 123 transition probability diagram, 125 transition rate diagram, 126 martingale, 110 matrices, 339 characteristic polynomial, 341, 343 determinant, 340 diagonal, 339 eigenvalue, 340, 341, 343 eigenvector, 340 Hermitian symmetric, 343 Hermitian transpose of, 341 identity matrix, 339 positive semidefinite, 341, 343 symmetric, 339 unitary, 342 maximum, 328 maximum a posteriori probability (MAP) estimator, 144 maximum likelihood (ML) estimator, 143 mean, see expectation Mean ergodic, 223 mean function, 105 mean square closure, 279 memoryless property of the geometric distribution, 22 minimum, 328 monotone convergence theorem, 339 narrowband random process, 264 403 signal, 261 norm of a vector, 340 of an interval partition, 334 normal distribution, see Gaussian distribution Nyquit sampling theorem, 258 orthogonal, 342 complex random variables, 229 random variables, 75 vectors, 340 orthogonality principle, 77 orthonormal, 340 basis, 340, 342 matrix, 340 partition, 6 periodic WSS random processes, 236 permutation, 340 piecewise continuous, 331 Poisson arrivals see time averages (PASTA), 182 Poisson distribution, 22 Poisson process, 113 posterior, or a posteriori, 144 power of a random process, 251 spectral density, 250 prior, or a priori, 144 probability density function (pdf), 25 projection, 76 Rayleigh distribution, 24 Riemann sum, 333 sample path, 105 Schwarz’s inequality, 26 second order random process, 106 sinc function, 252 span, 340 spectral representation, 238 stationary, 116, 230 wide sense, 116, 269 strictly increasing, 327 subsequence, 329 404 supremum, 328 Taylor’s theorem, 332 time update, 94 time-invariant linear system, 247 tower property, 83 transfer function, 250 triangle inequality, L2 , 26 uniform distribution, 24 uniform prior, 145 version of a random process, 237 Viterbi algorithm, 159 wide sense sationary, 269 wide sense stationary, 116 Wiener process, see Brownian motion INDEX 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 input-queued switch. IEEE Trans. Communications, 47(8):1260–1267, 1999. [9] S.P. Meyn and R.L. Tweedie. Markov chains and stochastic stability. Springer-Verlag 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. 405 ...
View Full Document

This note was uploaded on 01/19/2012 for the course ECEN 646 taught by Professor Serapsavari during the Fall '11 term at Texas A&M.

Ask a homework question - tutors are online