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Unformatted text preview: Class Notes Econ 7800 Fall Semester 2003
Hans G. Ehrbar
Economics Department, University of Utah, 1645 Campus Center
Drive, Salt Lake City UT 841129300, U.S.A.
URL: www.econ.utah.edu/ehrbar/ecmet.pdf
Email address : ehrbar@econ.utah.edu Abstract. This is an attempt to make a carefully argued set of class notes
freely available. The source code for these notes can be downloaded from
www.econ.utah.edu/ehrbar/ecmetsources.zip Copyright Hans G. Ehrbar under the GNU Public License
The present version has those chapters relevant for Econ 7800. Contents
Chapter 1. Syllabus Econ 7800 Fall 2003 vii Chapter 2. Probability Fields
2.1. The Concept of Probability
2.2. Events as Sets
2.3. The Axioms of Probability
2.4. Objective and Subjective Interpretation of Probability
2.5. Counting Rules
2.6. Relationships Involving Binomial Coeﬃcients
2.7. Conditional Probability
2.8. Ratio of Probabilities as Strength of Evidence
2.9. Bayes Theorem
2.10. Independence of Events
2.11. How to Plot Frequency Vectors and Probability Vectors 1
1
5
8
10
11
12
13
18
19
20
22 Chapter 3. Random Variables
3.1. Notation
3.2. Digression about Inﬁnitesimals
3.3. Deﬁnition of a Random Variable
3.4. Characterization of Random Variables
3.5. Discrete and Absolutely Continuous Probability Measures
3.6. Transformation of a Scalar Density Function
3.7. Example: Binomial Variable
3.8. Pitfalls of Data Reduction: The Ecological Fallacy
3.9. Independence of Random Variables
3.10. Location Parameters and Dispersion Parameters of a Random Variable
3.11. Entropy 25
25
25
27
27
30
31
32
34
35
35
39 Chapter 4. Speciﬁc Random Variables
4.1. Binomial
4.2. The Hypergeometric Probability Distribution
4.3. The Poisson Distribution
4.4. The Exponential Distribution
4.5. The Gamma Distribution
4.6. The Uniform Distribution
4.7. The Beta Distribution
4.8. The Normal Distribution
4.9. The ChiSquare Distribution
4.10. The Lognormal Distribution
4.11. The Cauchy Distribution 49
49
52
52
55
56
59
59
60
62
63
63 iii iv CONTENTS Chapter 5. Chebyshev Inequality, Weak Law of Large Numbers, and Central
Limit Theorem
5.1. Chebyshev Inequality
5.2. The Probability Limit and the Law of Large Numbers
5.3. Central Limit Theorem 65
65
66
67 Chapter 6. Vector Random Variables
6.1. Expected Value, Variances, Covariances
6.2. Marginal Probability Laws
6.3. Conditional Probability Distribution and Conditional Mean
6.4. The Multinomial Distribution
6.5. Independent Random Vectors
6.6. Conditional Expectation and Variance
6.7. Expected Values as Predictors
6.8. Transformation of Vector Random Variables 69
70
73
74
75
76
77
79
83 Chapter 7. The Multivariate Normal Probability Distribution
7.1. More About the Univariate Case
7.2. Deﬁnition of Multivariate Normal
7.3. Special Case: Bivariate Normal
7.4. Multivariate Standard Normal in Higher Dimensions 87
87
88
88
97 Chapter 8. The Regression Fallacy 101 Chapter 9. A Simple Example of Estimation
9.1. Sample Mean as Estimator of the Location Parameter
9.2. Intuition of the Maximum Likelihood Estimator
9.3. Variance Estimation and Degrees of Freedom 109
109
110
112 Chapter 10. Estimation Principles and Classiﬁcation of Estimators
10.1. Asymptotic or LargeSample Properties of Estimators
10.2. Small Sample Properties
10.3. Comparison Unbiasedness Consistency
10.4. The CramerRao Lower Bound
10.5. Best Linear Unbiased Without Distribution Assumptions
10.6. Maximum Likelihood Estimation
10.7. Method of Moments Estimators
10.8. MEstimators
10.9. Suﬃcient Statistics and Estimation
10.10. The Likelihood Principle
10.11. Bayesian Inference 121
121
122
123
126
132
134
136
136
137
140
140 Chapter 11. 143 Chapter
12.1.
12.2.
12.3. Interval Estimation 12. Hypothesis Testing
Duality between Signiﬁcance Tests and Conﬁdence Regions
The Neyman Pearson Lemma and Likelihood Ratio Tests
The Wald, Likelihood Ratio, and Lagrange Multiplier Tests Chapter 13. 149
151
152
154 General Principles of Econometric Modelling 157 Chapter 14. MeanVariance Analysis in the Linear Model
14.1. Three Versions of the Linear Model
14.2. Ordinary Least Squares 159
159
160 CONTENTS 14.3. The Coeﬃcient of Determination
14.4. The Adjusted RSquare v 166
170 Chapter 15. Digression about Correlation Coeﬃcients
15.1. A Uniﬁed Deﬁnition of Correlation Coeﬃcients 173
173 Chapter
16.1.
16.2.
16.3.
16.4.
16.5. 177
177
184
189
190
190 16. Speciﬁc Datasets
Cobb Douglas Aggregate Production Function
Houthakker’s Data
Long Term Data about US Economy
Dougherty Data
Wage Data Chapter 17. The Mean Squared Error as an Initial Criterion of Precision
17.1. Comparison of Two Vector Estimators 203
203 Chapter
18.1.
18.2.
18.3.
18.4.
18.5. 207
207
209
210
218
219 18. Sampling Properties of the Least Squares Estimator
The Gauss Markov Theorem
Digression about Minimax Estimators
Miscellaneous Properties of the BLUE
Estimation of the Variance
Mallow’s CpStatistic as Estimator of the Mean Squared Error Chapter 19.
Chapter
20.1.
20.2.
20.3. Nonspherical Positive Deﬁnite Covariance Matrix 221 20. Best Linear Prediction
Minimum Mean Squared Error, Unbiasedness Not Required
The Associated Least Squares Problem
Prediction of Future Observations in the Regression Model 225
225
230
231 Chapter 21. Updating of Estimates When More Observations become Available 237
Chapter
22.1.
22.2.
22.3.
22.4.
22.5.
22.6.
22.7.
22.8.
22.9. 22. Constrained Least Squares
Building the Constraint into the Model
Conversion of an Arbitrary Constraint into a Zero Constraint
Lagrange Approach to Constrained Least Squares
Constrained Least Squares as the Nesting of Two Simpler Models
Solution by Quadratic Decomposition
Sampling Properties of Constrained Least Squares
Estimation of the Variance in Constrained OLS
Inequality Restrictions
Application: Biased Estimators and PreTest Estimators Chapter 23. Additional Regressors 241
241
242
243
245
246
247
248
251
251
253 Chapter
24.1.
24.2.
24.3. 24. Residuals: Standardized, Predictive, “Studentized”
Three Decisions about Plotting Residuals
Relationship between Ordinary and Predictive Residuals
Standardization 263
263
265
267 Chapter
25.1.
25.2.
25.3. 25. Regression Diagnostics
Missing Observations
Grouped Data
Inﬂuential Observations and Outliers 271
271
271
271 vi CONTENTS 25.4. Sensitivity of Estimates to Omission of One Observation 273 Chapter 26. Asymptotic Properties of the OLS Estimator
26.1. Consistency of the OLS estimator
26.2. Asymptotic Normality of the Least Squares Estimator 279
280
281 Chapter 27. 283 Least Squares as the Normal Maximum Likelihood Estimate Chapter 28. Random Regressors
28.1. Strongest Assumption: Error Term Well Behaved Conditionally on
Explanatory Variables
28.2. Contemporaneously Uncorrelated Disturbances
28.3. Disturbances Correlated with Regressors in Same Observation 289
290
291 Chapter 29. The Mahalanobis Distance
29.1. Deﬁnition of the Mahalanobis Distance 293
293 Chapter
30.1.
30.2.
30.3.
30.4. 30. Interval Estimation
A Basic Construction Principle for Conﬁdence Regions
Coverage Probability of the Conﬁdence Regions
Conventional Formulas for the Test Statistics
Interpretation in terms of Studentized Mahalanobis Distance 297
297
300
301
301 Chapter
31.1.
31.2.
31.3.
31.4.
31.5. 31. Three Principles for Testing a Linear Constraint
Mathematical Detail of the Three Approaches
Examples of Tests of Linear Hypotheses
The FTest Statistic is a Function of the Likelihood Ratio
Tests of Nonlinear Hypotheses
Choosing Between Nonnested Models 305
305
308
315
315
316 Chapter 32. Instrumental Variables 289 317 Appendix A. Matrix Formulas
A.1. A Fundamental Matrix Decomposition
A.2. The Spectral Norm of a Matrix
A.3. Inverses and gInverses of Matrices
A.4. Deﬁciency Matrices
A.5. Nonnegative Deﬁnite Symmetric Matrices
A.6. Projection Matrices
A.7. Determinants
A.8. More About Inverses
A.9. Eigenvalues and Singular Value Decomposition 321
321
321
322
323
326
329
331
332
335 Appendix B. Arrays of Higher Rank
B.1. Informal Survey of the Notation
B.2. Axiomatic Development of Array Operations
B.3. An Additional Notational Detail
B.4. Equality of Arrays and Extended Substitution
B.5. Vectorization and Kronecker Product 337
337
339
343
343
344 Appendix C. Matrix Diﬀerentiation
C.1. First Derivatives 353
353 Appendix. 359 Bibliography CHAPTER 1 Syllabus Econ 7800 Fall 2003
The class meets Tuesdays and Thursdays 12:25 to 1:45pm in BUC 207. First
class Thursday, August 21, 2003; last class Thursday, December 4.
Instructor: Assoc. Prof. Dr. Dr. Hans G. Ehrbar. Hans’s oﬃce is at 319 BUO,
Tel. 581 7797, email ehrbar@econ.utah.edu Oﬃce hours: Monday 10–10:45 am,
Thursday 5–5:45 pm or by appointment.
Textbook: There is no obligatory textbook in the Fall Quarter, but detailed
class notes are available at www.econ.utah.edu/ehrbar/ec7800.pdf, and you can
purchase a hardcopy containing the assigned chapters only at the University Copy
Center, 158 Union Bldg, tel. 581 8569 (ask for the class materials for Econ 7800).
Furthermore, the following optional texts will be available at the bookstore:
Peter Kennedy, A Guide to Econometrics (fourth edition), MIT Press, 1998 ISBN
0262611406.
The bookstore also has available William H. Greene’s Econometric Analysis, ﬁfth
edition, Prentice Hall 2003, ISBN 0130661899. This is the assigned text for Econ
7801 in the Spring semester 2004, and some of the introductory chapters are already
useful for the Fall semester 2003.
The following chapters in the class notes are assigned: 2, 3 (but not section 3.2),
4, 5, 6, 7 (but only until section 7.3), 8, 9, 10, 11, 12, 14, only section 15.1 in chapter
15, in chapter 16, we will perhaps do section 16.1 or 16.4, then in chapter 17 we do
section 17.1, then chapter 18 until and including 18.5, and in chapter 22 do sections
22.1, 22.3, 22.6, and 22.7. In chapter 29 only the ﬁrst section 29.1, ﬁnally chapters
30, and section 31.2 in chapter 31.
Summary of the Class: This is the ﬁrst semester in a twosemester Econometrics
ﬁeld, but it should also be useful for students taking the ﬁrst semester only as part
of their methodology requirement. The course description says: Probability, conditional probability, distributions, transformation of probability densities, suﬃcient
statistics, limit theorems, estimation principles, maximum likelihood estimation, interval estimation and hypothesis testing, least squares estimation, linear constraints.
This class has two focal points: maximum likelihood estimation, and the fundamental concepts of the linear model (regression).
If advanced mathematical concepts are necessary in these theoretical explorations, they will usually be reviewed very brieﬂy before we use them. The class
is structured in such a way that, if you allocate enough time, it should be possible
to refresh your math skills as you go along.
Here is an overview of the topics to be covered in the Fall Semester. They may
not come exactly in the order in which they are listed here
1. Probability ﬁelds: Events as sets, set operations, probability axioms, subjective vs. frequentist interpretation, ﬁnite sample spaces and counting rules (combinatorics), conditional probability, Bayes theorem, independence, conditional independence. vii viii 1. SYLLABUS ECON 7800 FALL 2003 2. Random Variables: Cumulative distribution function, density function;
location parameters (expected value, median) and dispersion parameters (variance).
3. Special Issues and Examples: Discussion of the “ecological fallacy”; entropy; moment generating function; examples (Binomial, Poisson, Gamma, Normal,
Chisquare); suﬃcient statistics.
4. Limit Theorems: Chebyshev inequality; law of large numbers; central limit
theorems.
The ﬁrst Midterm will already be on Thursday, September 18, 2003. It will be
closed book, but you are allowed to prepare one sheet with formulas etc. Most of
the midterm questions will be similar or identical to the homework questions in the
class notes assigned up to that time.
5. Jointly Distributed Random Variables: Joint, marginal, and conditional densities; conditional mean; transformations of random variables; covariance
and correlation; sums and linear combinations of random variables; jointly normal
variables.
6. Estimation Basics: Descriptive statistics; sample mean and variance; degrees of freedom; classiﬁcation of estimators.
7. Estimation Methods: Method of moments estimators; least squares estimators. Bayesian inference. Maximum likelihood estimators; large sample properties
of MLE; MLE and suﬃcient statistics; computational aspects of maximum likelihood.
8. Conﬁdence Intervals and Hypothesis Testing: Power functions; Neyman Pearson Lemma; likelihood ratio tests. As example of tests: the run test,
goodness of ﬁt test, contingency tables.
The second inclass Midterm will be on Thursday, October 16, 2003.
9. Basics of the “Linear Model.” We will discuss the case with nonrandom
regressors and a spherical covariance matrix: OLSBLUE duality, Maximum likelihood estimation, linear constraints, hypothesis testing, interval estimation (ttest,
F test, joint conﬁdence intervals).
The third Midterm will be a takehome exam. You will receive the questions on
Tuesday, November 25, 2003, and they are due back at the beginning of class on
Tuesday, December 2nd, 12:25 pm. The questions will be similar to questions which
you might have to answer in the Econometrics Field exam.
The Final Exam will be given according to the campuswide examination schedule, which is Wednesday December 10, 10:30–12:30 in the usual classroom. Closed
book, but again you are allowed to prepare one sheet of notes with the most important concepts and formulas. The exam will cover material after the second Midterm.
Grading: The three midterms and the ﬁnal exams will be counted equally. Every
week certain homework questions from among the questions in the class notes will
be assigned. It is recommended that you work through these homework questions
conscientiously. The answers provided in the class notes should help you if you get
stuck. If you have problems with these homeworks despite the answers in the class
notes, please write you answer down as far as you get and submit your answer to
me; I will look at them and help you out. A majority of the questions in the two
inclass midterms and the ﬁnal exam will be identical to these assigned homework
questions, but some questions will be diﬀerent.
Special circumstances: If there are special circumstances requiring an individualized course of study in your case, please see me about it in the ﬁrst week of
classes.
Hans G. Ehrbar CHAPTER 2 Probability Fields
2.1. The Concept of Probability
Probability theory and statistics are useful in dealing with the following types
of situations:
• Games of chance: throwing dice, shuﬄing cards, drawing balls out of urns.
• Quality control in production: you take a sample from a shipment, count
how many defectives.
• Actuarial Problems: the length of life anticipated for a person who has just
applied for life insurance.
• Scientiﬁc Eperiments: you count the number of mice which contract cancer
when a group of mice is exposed to cigarette smoke.
• Markets: the total personal income in New York State in a given month.
• Meteorology: the rainfall in a given month.
• Uncertainty: the exact date of Noah’s birth.
• Indeterminacy: The closing of the Dow Jones industrial average or the
temperature in New York City at 4 pm. on February 28, 2014.
• Chaotic determinacy: the relative frequency of the digit 3 in the decimal
representation of π .
• Quantum mechanics: the proportion of photons absorbed by a polarization
ﬁlter
• Statistical mechanics: the velocity distribution of molecules in a gas at a
given pressure and temperature.
In the probability theoretical literature the situations in which probability theory
applies are called “experiments,” see for instance [R´n70, p. 1]. We will not use this
e
terminology here, since probabilistic reasoning applies to several diﬀerent types of
situations, and not all these can be considered “experiments.”
Problem 1. (This question will not be asked on any exams) R´nyi says: “Obe
serving how long one has to wait for the departure of an airplane is an experiment.”
Comment.
Answer. R´ny commits the epistemic fallacy in order to justify his use of the word “expere
iment.” Not the observation of the departure but the departure itself is the event which can be
theorized probabilistically, and the word “experiment” is not appropriate here. What does the fact that probability theory is appropriate in the above situations
tell us about the world? Let us go through our list one by one:
• Games of chance: Games of chance are based on the sensitivity on initial
conditions: you tell someone to roll a pair of dice or shuﬄe a deck of cards,
and despite the fact that this person is doing exactly what he or she is asked
to do and produces an outcome which lies within a welldeﬁned universe
known beforehand (a number between 1 and 6, or a permutation of the
deck of cards), the question which number or which permutation is beyond
1 2 2. PROBABILITY FIELDS their control. The precise location and speed of the die or the precise order
of the cards varies, and these small variations in initial conditions give rise,
by the “butterﬂy eﬀect” of chaos theory, to unpredictable ﬁnal outcomes.
A critical realist recognizes here the openness and stratiﬁcation of the
world: If many diﬀerent inﬂuences come together, each of which is governed by laws, then their sum total is not determinate, as a naive hyperdeterminist would think, but indeterminate. This is not only a condition
for the possibility of science (in a hyperdeterministic world, one could not
know anything before one knew everything, and science would also not be
necessary because one could not do anything), but also for practical human
activity: the macro outcomes of human practice are largely independent of
micro detail (the postcard arrives whether the address is written in cursive
or in printed letters, etc.). Games of chance are situations which deliberately project this micro indeterminacy into the macro world: the micro
inﬂuences cancel each other out without one enduring inﬂuence taking over
(as would be the case if the die were not perfectly symmetric and balanced)
or deliberate human corrective activity stepping into the void (as a card
trickster might do if the cards being shuﬄed somehow were distinguishable
from the backside).
The experiment in which one draws balls from urns shows clearly another aspect of this paradigm: the set of diﬀerent possible outcomes is
ﬁxed beforehand, and the probability enters in the choice of one of these
predetermined outcomes. This is not the only way probability can arise;
it is an extensionalist example, in which the connection between success
and failure is external. The world is not a collection of externally related
outcomes collected in an urn. Success and failure are not determined by a
choice between diﬀerent spacially separated and individually inert balls (or
playing cards or faces on a die), but it is the outcome of development and
struggle that is internal to the individual unit.
• Quality control in production: you take a sample from a shipment, count
how many defectives. Why is statistics and probability useful in production? Because production is work, it is not spontaneous. Nature does not
voluntarily give us things in the form in which we need them. Production
is similar to a scientiﬁc experiment because it is the attempt to create local
closure. Such closure can never be complete, there are always leaks in it,
through which irregularity enters.
• Actuarial Problems: the length of life anticipated for a person who has
just applied for life insurance. Not only production, but also life itself is
a struggle with physical nature, it is emergence. And sometimes it fails:
sometimes the living organism is overwhelmed by the forces which it tries
to keep at bay and to subject to its own purposes.
• Scientiﬁc Eperiments: you count the number of mice which contract cancer
when a group of mice is exposed to cigarette smoke: There is local closure
regarding the conditions under which the mice live, but even if this closure were complete, individual mice would still react diﬀerently, because of
genetic diﬀerences. No two mice are exactly the same, and despite these
diﬀerences they are still mice. This is again the stratiﬁcation of reality. Two
mice are two diﬀerent individuals but they are both mice. Their reaction
to the smoke is not identical, since they are diﬀerent individuals, but it is
not completely capricious either, since both are mice. It can be predicted
probabilistically. Those mechanisms which make them mice react to the 2.1. THE CONCEPT OF PROBABILITY • • • • • • 3 smoke. The probabilistic regularity comes from the transfactual eﬃcacy of
the mouse organisms.
Meteorology: the rainfall in a given month. It is very fortunate for the
development of life on our planet that we have the chaotic alternation between cloud cover and clear sky, instead of a continuous cloud cover as in
Venus or a continuous clear sky. Butterﬂy eﬀect all over again, but it is
possible to make probabilistic predictions since the fundamentals remain
stable: the transfactual eﬃcacy of the energy received from the sun and
radiated back out into space.
Markets: the total personal income in New York State in a given month.
Market economies are a very much like the weather; planned economies
would be more like production or life.
Uncertainty: the exact date of Noah’s birth. This is epistemic uncertainty:
assuming that Noah was a real person, the date exists and we know a time
range in which it must have been, but we do not know the details. Probabilistic methods can be used to represent this kind of uncertain knowledge,
but other methods to represent this knowledge may be more appropriate.
Indeterminacy: The closing of the Dow Jones Industrial Average (DJIA)
or the temperature in New York City at 4 pm. on February 28, 2014: This
is ontological uncertainty, not only epistemological uncertainty. Not only
do we not know it, but it is objectively not yet decided what these data
will be. Probability theory has limited applicability for the DJIA since it
cannot be expected that the mechanisms determining the DJIA will be the
same at that time, therefore we cannot base ourselves on the transfactual
eﬃcacy of some stable mechanisms. It is not known which stocks will be
included in the DJIA at that time, or whether the US dollar will still be
the world reserve currency and the New York stock exchange the pinnacle
of international capital markets. Perhaps a diﬀerent stock market index
located somewhere else will at that time play the role the DJIA is playing
today. We would not even be able to ask questions about that alternative
index today.
Regarding the temperature, it is more defensible to assign a probability,
since the weather mechanisms have probably stayed the same, except for
changes in global warming (unless mankind has learned by that time to
manipulate the weather locally by cloud seeding etc.).
Chaotic determinacy: the relative frequency of the digit 3 in the decimal
representation of π : The laws by which the number π is deﬁned have very
little to do with the procedure by which numbers are expanded as decimals,
therefore the former has no systematic inﬂuence on the latter. (It has an
inﬂuence, but not a systematic one; it is the error of actualism to think that
every inﬂuence must be systematic.) But it is also known that laws can
have remote eﬀects: one of the most amazing theorems in mathematics is
the formula π = 1 − 1 + 1 − 1 + · · · which estalishes a connection between
4
3
5
4
the geometry of the circle and some simple arithmetics.
Quantum mechanics: the proportion of photons absorbed by a polarization
ﬁlter: If these photons are already polarized (but in a diﬀerent direction
than the ﬁlter) then this is not epistemic uncertainty but ontological indeterminacy, since the polarized photons form a pure state, which is atomic
in the algebra of events. In this case, the distinction between epistemic uncertainty and ontological indeterminacy is operational: the two alternatives
follow diﬀerent mathematics. 4 2. PROBABILITY FIELDS • Statistical mechanics: the velocity distribution of molecules in a gas at a
given pressure and temperature. Thermodynamics cannot be reduced to
the mechanics of molecules, since mechanics is reversible in time, while
thermodynamics is not. An additional element is needed, which can be
modeled using probability.
Problem 2. Not every kind of uncertainty can be formulated stochastically.
Which other methods are available if stochastic means are inappropriate?
Answer. Dialectics. Problem 3. How are the probabilities of rain in weather forecasts to be interpreted?
Answer. Renyi in [R´n70, pp. 33/4]: “By saying that the probability of rain tomorrow is
e
80% (or, what amounts to the same, 0.8) the meteorologist means that in a situation similar to that
observed on the given day, there is usually rain on the next day in about 8 out of 10 cases; thus,
while it is not certain that it will rain tomorrow, the degree of certainty of this event is 0.8.” Pure uncertainty is as hard to generate as pure certainty; it is needed for encryption and numerical methods.
Here is an encryption scheme which leads to a random looking sequence of numbers (see [Rao97, p. 13]): First a string of binary random digits is generated which is
known only to the sender and receiver. The sender converts his message into a string
of binary digits. He then places the message string below the key string and obtains
a coded string by changing every message bit to its alternative at all places where
the key bit is 1 and leaving the others unchanged. The coded string which appears
to be a random binary sequence is transmitted. The received message is decoded by
making the changes in the same way as in encrypting using the key string which is
known to the receiver.
Problem 4. Why is it important in the above encryption scheme that the key
string is purely random and does not have any regularities?
Problem 5. [Knu81, pp. 7, 452] Suppose you wish to obtain a decimal digit at
random, not using a computer. Which of the following methods would be suitable?
• a. Open a telephone directory to a random place (i.e., stick your ﬁnger in it
somewhere) and use the unit digit of the ﬁrst number found on the selected page.
Answer. This will often fail, since users select “round” numbers if possible. In some areas,
telephone numbers are perhaps assigned randomly. But it is a mistake in any case to try to get
several successive random numbers from the same page, since many telephone numbers are listed
several times in a sequence. • b. Same as a, but use the units digit of the page number.
Answer. But do you use the lefthand page or the righthand page? Say, use the lefthand
page, divide by 2, and use the units digit. • c. Roll a die which is in the shape of a regular icosahedron, whose twenty faces
have been labeled with the digits 0, 0, 1, 1,. . . , 9, 9. Use the digit which appears on
top, when the die comes to rest. (A felt table with a hard surface is recommended for
rolling dice.)
Answer. The markings on the face will slightly bias the die, but for practical purposes this
method is quite satisfactory. See Math. Comp. 15 (1961), 94–95, for further discussion of these
dice. 2.2. EVENTS AS SETS 5 • d. Expose a geiger counter to a source of radioactivity for one minute (shielding
yourself ) and use the unit digit of the resulting count. (Assume that the geiger
counter displays the number of counts in decimal notation, and that the count is
initially zero.)
Answer. This is a diﬃcult question thrown in purposely as a surprise. The number is not
uniformly distributed! One sees this best if one imagines the source of radioactivity is very low
level, so that only a few emissions can be expected during this minute. If the average number of
emissions per minute is λ, the probability that the counter registers k is e−λ λk /k! (the Poisson
∞
distribution). So the digit 0 is selected with probability e−λ
λ10k /(10k)!, etc.
k=0 • e. Glance at your wristwatch, and if the position of the secondhand is between
6n and 6(n + 1), choose the digit n.
Answer. Okay, provided that the time since the last digit selected in this way is random. A
bias may arise if borderline cases are not treated carefully. A better device seems to be to use a
stopwatch which has been started long ago, and which one stops arbitrarily, and then one has all
the time necessary to read the display. • f . Ask a friend to think of a random digit, and use the digit he names.
Answer. No, people usually think of certain digits (like 7) with higher probability. • g. Assume 10 horses are entered in a race and you know nothing whatever about
their qualiﬁcations. Assign to these horses the digits 0 to 9, in arbitrary fashion, and
after the race use the winner’s digit.
Answer. Okay; your assignment of numbers to the horses had probability 1/10 of assigning a
given digit to a winning horse. 2.2. Events as Sets
With every situation with uncertain outcome we associate its sample space U ,
which represents the set of all possible outcomes (described by the characteristics
which we are interested in).
Events are associated with subsets of the sample space, i.e., with bundles of
outcomes that are observable in the given experimental setup. The set of all events
we denote with F . (F is a set of subsets of U .)
Look at the example of rolling a die. U = {1, 2, 3, 4, 5, 6}. The events of getting
an even number is associated with the subset {2, 4, 6}; getting a six with {6}; not
getting a six with {1, 2, 3, 4, 5}, etc. Now look at the example of rolling two indistinguishable dice. Observable events may be: getting two ones, getting a one and a two,
etc. But we cannot distinguish between the ﬁrst die getting a one and the second a
two, and vice versa. I.e., if we deﬁne the sample set to be U = {1, . . . , 6}×{1, . . . , 6},
i.e., the set of all pairs of numbers between 1 and 6, then certain subsets are not
observable. {(1, 5)} is not observable (unless the dice are marked or have diﬀerent
colors etc.), only {(1, 5), (5, 1)} is observable.
If the experiment is measuring the height of a person in meters, and we make
the idealized assumption that the measuring instrument is inﬁnitely accurate, then
all possible outcomes are numbers between 0 and 3, say. Sets of outcomes one is
usually interested in are whether the height falls within a given interval; therefore
all intervals within the given range represent observable events.
If the sample space is ﬁnite or countably inﬁnite, very often all subsets are
observable events. If the sample set contains an uncountable continuum, it is not
desirable to consider all subsets as observable events. Mathematically one can deﬁne 6 2. PROBABILITY FIELDS quite crazy subsets which have no practical signiﬁcance and which cannot be meaningfully given probabilities. For the purposes of Econ 7800, it is enough to say that
all the subsets which we may reasonably deﬁne are candidates for observable events.
The “set of all possible outcomes” is well deﬁned in the case of rolling a die
and other games; but in social sciences, situations arise in which the outcome is
open and the range of possible outcomes cannot be known beforehand. If one uses
a probability theory based on the concept of a “set of possible outcomes” in such
a situation, one reduces a process which is open and evolutionary to an imaginary
predetermined and static “set.” Furthermore, in social theory, the mechanism by
which these uncertain outcomes are generated are often internal to the members of
the statistical population. The mathematical framework models these mechanisms
as an extraneous “picking an element out of a preexisting set.”
From given observable events we can derive new observable events by set theoretical operations. (All the operations below involve subsets of the same U .)
Mathematical Note: Notation of sets: there are two ways to denote a set: either
by giving a rule, or by listing the elements. (The order in which the elements are
listed, or the fact whether some elements are listed twice or not, is irrelevant.)
Here are the formal deﬁnitions of set theoretic operations. The letters A, B , etc.
denote subsets of a given set U (events), and I is an arbitrary index set. ω stands
for an element, and ω ∈ A means that ω is an element of A.
(2.2.1) A ⊂ B ⇐⇒ (ω ∈ A ⇒ ω ∈ B ) (2.2.2) A ∩ B = {ω : ω ∈ A and ω ∈ B } (A is contained in B )
(intersection of A and B ) Ai = {ω : ω ∈ Ai for all i ∈ I } (2.2.3)
i∈I (2.2.4) A ∪ B = {ω : ω ∈ A or ω ∈ B } (union of A and B ) Ai = {ω : there exists an i ∈ I such that ω ∈ Ai } (2.2.5)
i∈I Universal set: all ω we talk about are ∈ U . (2.2.6) U (2.2.7) A = {ω : ω ∈ A but ω ∈ U }
/ (2.2.8) ∅ = the empty set: ω ∈ ∅ for all ω .
/ These deﬁnitions can also be visualized by Venn diagrams; and for the purposes of
this class, demonstrations with the help of Venn diagrams will be admissible in lieu
of mathematical proofs.
Problem 6. For the following settheoretical exercises it is suﬃcient that you
draw the corresponding Venn diagrams and convince yourself by just looking at them
that the statement is true. For those who are interested in a precise mathematical
proof derived from the deﬁnitions of A ∪ B etc. given above, should remember that a
proof of the settheoretical identity A = B usually has the form: ﬁrst you show that
ω ∈ A implies ω ∈ B , and then you show the converse.
• a. Prove that A ∪ B = B ⇐⇒ A ∩ B = A.
Answer. If one draws the Venn diagrams, one can see that either side is true if and only
if A ⊂ B . If one wants a more precise proof, the following proof by contradiction seems most
illuminating: Assume the lefthand side does not hold, i.e., there exists a ω ∈ A but ω ∈ B . Then
/
ω ∈ A ∩ B , i.e., A ∩ B = A. Now assume the righthand side does not hold, i.e., there is a ω ∈ A
/
with ω ∈ B . This ω lies in A ∪ B but not in B , i.e., the lefthand side does not hold either.
/ • b. Prove that A ∪ (B ∩ C ) = (A ∪ B ) ∩ (A ∪ C ) 2.2. EVENTS AS SETS 7 Answer. If ω ∈ A then it is clearly always in the righthand side and in the lefthand side. If
there is therefore any diﬀerence between the righthand and the lefthand side, it must be for the
ω ∈ A: If ω ∈ A and it is still in the lefthand side then it must be in B ∩ C , therefore it is also in
/
/
the righthand side. If ω ∈ A and it is in the righthand side, then it must be both in B and in C ,
/
therefore it is in the lefthand side. • c. Prove that A ∩ (B ∪ C ) = (A ∩ B ) ∪ (A ∩ C ).
Answer. If ω ∈ A then it is clearly neither in the righthand side nor in the lefthand side. If
/
there is therefore any diﬀerence between the righthand and the lefthand side, it must be for the
ω ∈ A: If ω ∈ A and it is in the lefthand side then it must be in B ∪ C , i.e., in B or in C or in both,
therefore it is also in the righthand side. If ω ∈ A and it is in the righthand side, then it must be
in either B or C or both, therefore it is in the lefthand side. • d. Prove that A ∩ ∞
i=1 Bi = ∞
i=1 (A ∩ Bi ). Answer. Proof: If ω in lefthand side, then it is in A and in at least one of the Bi , say it is
in Bk . Therefore it is in A ∩ Bk , and therefore it is in the righthand side. Now assume, conversely,
that ω is in the righthand side; then it is at least in one of the A ∩ Bi , say it is in A ∩ Bk . Hence it
is in A and in Bk , i.e., in A and in
Bi , i.e., it is in the lefthand side. Problem 7. 3 points Draw a Venn Diagram which shows the validity of de
Morgan’s laws: (A ∪ B ) = A ∩ B and (A ∩ B ) = A ∪ B . If done right, the same
Venn diagram can be used for both proofs.
Answer. There is a proof in [HT83, p. 12]. Draw A and B inside a box which represents U ,
and shade A from the left (blue) and B from the right (yellow), so that A ∩ B is cross shaded
(green); then one can see these laws. Problem 8. 3 points [HT83, Exercise 1.213 on p. 14] Evaluate the following
unions and intersections of intervals. Use the notation (a, b) for open and [a, b] for
closed intervals, (a, b] or [a, b) for half open intervals, {a} for sets containing one
element only, and ∅ for the empty set.
∞ (2.2.9)
n=1
∞ (2.2.10)
n=1 ∞ 1
,2 =
n
1
,2 =
n 0, 1
=
n 0, 1 + 1
=
n n=1
∞ n=1 Answer.
∞ 1
,2
n (2.2.11)
n=1
∞ 1
, 2 = (0, 2]
n n=1
∞ 0, 1
n =∅ 0, 1 + 1
n = [0, 1] n=1 (2.2.12)
Explanation of n=1
none of the intervals. ∞ = (0, 2) 1
,2
n ∞ n=1 : for every α with 0 < α ≤ 2 there is a n with 1
n ≤ α, but 0 itself is in The set operations become logical operations if applied to events. Every experiment returns an element ω ∈U as outcome. Here ω is rendered green in the electronic
version of these notes (and in an upright font in the version for blackandwhite
printouts), because ω does not denote a speciﬁc element of U , but it depends on
chance which element is picked. I.e., the green color (or the unusual font) indicate
that ω is “alive.” We will also render the events themselves (as opposed to their
settheoretical counterparts) in green (or in an upright font).
• We say that the event A has occurred when ω ∈A. 8 2. PROBABILITY FIELDS • If A ⊂ B then event A implies event B , and we will write this directly in
terms of events as A ⊂ B .
• The set A ∩ B is associated with the event that both A and B occur (e.g.
an even number smaller than six), and considered as an event, not a set,
the event that both A and B occur will be written A ∩ B .
• Likewise, A ∪ B is the event that either A or B , or both, occur.
• A is the event that A does not occur.
• U the event that always occurs (as long as one performs the experiment).
• The empty set ∅ is associated with the impossible event ∅, because whatever
the value ω of the chance outcome ω of the experiment, it is always ω ∈ ∅.
/
If A ∩ B = ∅, the set theoretician calls A and B “disjoint,” and the probability
theoretician calls the events A and B “mutually exclusive.” If A ∪ B = U , then A
and B are called “collectively exhaustive.”
The set F of all observable events must be a σ algebra, i.e., it must satisfy:
∅∈F
A∈F ⇒A ∈F
A1 , A2 , . . . ∈ F ⇒ A1 ∪ A2 ∪ · · · ∈ F Ai ∈ F which can also be written as
i=1,2,... A1 , A2 , . . . ∈ F ⇒ A1 ∩ A2 ∩ · · · ∈ F Ai ∈ F . which can also be written as
i=1,2,... 2.3. The Axioms of Probability
A probability measure Pr : F → R is a mapping which assigns to every event a
number, the probability of this event. This assignment must be compatible with the
settheoretic operations between events in the following way:
Pr[U ] = 1 (2.3.1) Pr[A] ≥ 0 (2.3.2) ∞ (2.3.3) If Ai ∩ Aj = ∅ for all i, j with i = j then Pr[
i=1 for all events A ∞ Ai ] = Pr[Ai ]
i=1 Here an inﬁnite sum is mathematically deﬁned as the limit of partial sums. These
axioms make probability what mathematicians call a measure, like area or weight.
In a Venn diagram, one might therefore interpret the probability of the events as the
area of the bubble representing the event.
Problem 9. Prove that Pr[A ] = 1 − Pr[A].
Answer. Follows from the fact that A and A are disjoint and their union U has probability
1. Problem 10. 2 points Prove that Pr[A ∪ B ] = Pr[A] + Pr[B ] − Pr[A ∩ B ].
Answer. For Econ 7800 it is suﬃcient to argue it out intuitively: if one adds Pr[A] + Pr[B ]
then one counts Pr[A ∩ B ] twice and therefore has to subtract it again.
The brute force mathematical proof guided by this intuition is somewhat verbose: Deﬁne
D = A ∩ B , E = A ∩ B , and F = A ∩ B . D, E , and F satisfy
(2.3.4) D ∪ E = (A ∩ B ) ∪ (A ∩ B ) = A ∩ (B ∪ B ) = A ∩ U = A, (2.3.5) E ∪ F = B, (2.3.6) D ∪ E ∪ F = A ∪ B. 2.3. THE AXIOMS OF PROBABILITY 9 You may need some of the properties of unions and intersections in Problem 6. Next step is to
prove that D, E , and F are mutually exclusive. Therefore it is easy to take probabilities
(2.3.7) Pr[A] = Pr[D] + Pr[E ]; (2.3.8) Pr[B ] = Pr[E ] + Pr[F ];
Pr[A ∪ B ] = Pr[D] + Pr[E ] + Pr[F ]. (2.3.9) Take the sum of (2.3.7) and (2.3.8), and subtract (2.3.9):
(2.3.10) Pr[A] + Pr[B ] − Pr[A ∪ B ] = Pr[E ] = Pr[A ∩ B ]; A shorter but trickier alternative proof is the following. First note that A ∪ B = A ∪ (A ∩ B ) and
that this is a disjoint union, i.e., Pr[A∪B ] = Pr[A]+Pr[A ∩B ]. Then note that B = (A∩B )∪(A ∩B ),
and this is a disjoint union, therefore Pr[B ] = Pr[A∩B ]+Pr[A ∩B ], or Pr[A ∩B ] = Pr[B ]−Pr[A∩B ].
Putting this together gives the result. Problem 11. 1 point Show that for arbitrary events A and B , Pr[A ∪ B ] ≤
Pr[A] + Pr[B ].
Answer. From Problem 10 we know that Pr[A ∪ B ] = Pr[A] + Pr[B ] − Pr[A ∩ B ], and from
axiom (2.3.2) follows Pr[A ∩ B ] ≥ 0. Problem 12. 2 points (Bonferroni inequality) Let A and B be two events. Writing Pr[A] = 1 − α and Pr[B ] = 1 − β , show that Pr[A ∩ B ] ≥ 1 − (α + β ). You are
allowed to use that Pr[A ∪ B ] = Pr[A] + Pr[B ] − Pr[A ∩ B ] (Problem 10), and that
all probabilities are ≤ 1.
Answer.
(2.3.11)
(2.3.12) Pr[A ∪ B ] = Pr[A] + Pr[B ] − Pr[A ∩ B ] ≤ 1
Pr[A] + Pr[B ] ≤ 1 + Pr[A ∩ B ] (2.3.13) Pr[A] + Pr[B ] − 1 ≤ Pr[A ∩ B ] (2.3.14) 1 − α + 1 − β − 1 = 1 − α − β ≤ Pr[A ∩ B ] Problem 13. (Not eligible for inclass exams) Given a rising sequence of events
∞
B 1 ⊂ B 2 ⊂ B 3 · · · , deﬁne B = i=1 B i . Show that Pr[B ] = limi→∞ Pr[B i ].
Answer. Deﬁne C 1 = B 1 , C 2 = B 2 ∩ B 1 , C 3 = B 3 ∩ B 2 , etc. Then C i ∩ C j = ∅ for i = j ,
∞
n
and B n = i=1 C i and B = i=1 C i . In other words, now we have represented every B n and B
as a union of disjoint sets, and can therefore apply the third probability axiom (2.3.3): Pr[B ] =
n
∞
Pr[C i ], i.e.,
Pr[C i ]. The inﬁnite sum is merely a short way of writing Pr[B ] = limn→∞
i=1
i=1
n
the inﬁnite sum is the limit of the ﬁnite sums. But since these ﬁnite sums are exactly
Pr[C i ] =
i=1
n
Pr[ i=1 C i ] = Pr[B n ], the assertion follows. This proof, as it stands, is for our purposes entirely
acceptable. One can make some steps in this proof still more stringent. For instance, one might use
n
∞
induction to prove B n = i=1 C i . And how does one show that B = i=1 C i ? Well, one knows
∞
∞
that C i ⊂ B i , therefore i=1 C i ⊂ i=1 B i = B . Now take an ω ∈ B . Then it lies in at least one
of the B i , but it can be in many of them. Let k be the smallest k for which ω ∈ B k . If k = 1, then
ω ∈ C 1 = B 1 as well. Otherwise, ω ∈ B k−1 , and therefore ω ∈ C k . I.e., any element in B lies in
/
∞
at least one of the C k , therefore B ⊂ i=1 C i . Problem 14. (Not eligible for inclass exams) From problem 13 derive also
the following: if A1 ⊃ A2 ⊃ A3 · · · is a declining sequence, and A = i Ai , then
Pr[A] = lim Pr[Ai ].
Answer. If the Ai are declining, then their complements B i = Ai are rising: B 1 ⊂ B 2 ⊂
B 3 · · · are rising; therefore I know the probability of B =
B i . Since by de Morgan’s laws, B = A ,
this gives me also the probability of A. 10 2. PROBABILITY FIELDS The results regarding the probabilities of rising or declining sequences are equivalent to the third probability axiom. This third axiom can therefore be considered a
continuity condition for probabilities.
If U is ﬁnite or countably inﬁnite, then the probability measure is uniquely
determined if one knows the probability of every oneelement set. We will call
Pr[{ω }] = p(ω ) the probability mass function. Other terms used for it in the literature are probability function, or even probability density function (although it
is not a density, more about this below). If U has more than countably inﬁnite
elements, the probabilities of oneelement sets may not give enough information to
deﬁne the whole probability measure.
Mathematical Note: Not all inﬁnite sets are countable. Here is a proof, by
contradiction, that the real numbers between 0 and 1 are not countable: assume
there is an enumeration, i.e., a sequence a1 , a2 , . . . which contains them all. Write
them underneath each other in their (possibly inﬁnite) decimal representation, where
0.di1 di2 di3 . . . is the decimal representation of ai . Then any real number whose
decimal representation is such that the ﬁrst digit is not equal to d11 , the second digit
is not equal d22 , the third not equal d33 , etc., is a real number which is not contained
in this enumeration. That means, an enumeration which contains all real numbers
cannot exist.
On the real numbers between 0 and 1, the length measure (which assigns to each
interval its length, and to sets composed of several invervals the sums of the lengths,
etc.) is a probability measure. In this probability ﬁeld, every oneelement subset of
the sample set has zero probability.
This shows that events other than ∅ may have zero probability. In other words,
if an event has probability 0, this does not mean it is logically impossible. It may
well happen, but it happens so infrequently that in repeated experiments the average
number of occurrences converges toward zero.
2.4. Objective and Subjective Interpretation of Probability
The mathematical probability axioms apply to both objective and subjective
interpretation of probability.
The objective interpretation considers probability a quasi physical property of the
experiment. One cannot simply say: Pr[A] is the relative frequency of the occurrence
of A, because we know intuitively that this frequency does not necessarily converge.
E.g., even with a fair coin it is physically possible that one always gets head, or that
one gets some other sequence which does not converge towards 1 . The above axioms
2
resolve this dilemma, because they allow to derive the theorem that the relative
frequencies converges towards the probability with probability one.
Subjectivist interpretation (de Finetti: “probability does not exist”) deﬁnes probability in terms of people’s ignorance and willingness to take bets. Interesting for
economists because it uses money and utility, as in expected utility. Call “a lottery
on A” a lottery which pays $1 if A occurs, and which pays nothing if A does not
occur. If a person is willing to pay p dollars for a lottery on A and 1 − p dollars for
a lottery on A , then, according to a subjectivist deﬁnition of probability, he assigns
subjective probability p to A.
There is the presumption that his willingness to bet does not depend on the size
of the payoﬀ (i.e., the payoﬀs are considered to be small amounts).
Problem 15. Assume A, B , and C are a complete disjunction of events, i.e.,
they are mutually exclusive and A ∪ B ∪ C = U , the universal set. 2.5. COUNTING RULES 11 • a. 1 point Arnold assigns subjective probability p to A, q to B , and r to C .
Explain exactly what this means.
Answer. We know six diﬀerent bets which Arnold is always willing to make, not only on A,
B , and C , but also on their complements. • b. 1 point Assume that p + q + r > 1. Name three lotteries which Arnold would
be willing to buy, the net eﬀect of which would be that he loses with certainty.
Answer. Among those six we have to pick subsets that make him a sure loser. If p + q + r > 1,
then we sell him a bet on A, one on B , and one on C . The payoﬀ is always 1, and the cost is
p + q + r > 1. • c. 1 point Now assume that p + q + r < 1. Name three lotteries which Arnold
would be willing to buy, the net eﬀect of which would be that he loses with certainty.
Answer. If p + q + r < 1, then we sell him a bet on A , one on B , and one on C . The payoﬀ
is 2, and the cost is 1 − p + 1 − q + 1 − r > 2. • d. 1 point Arnold is therefore only coherent if Pr[A] + Pr[B ] + Pr[C ] = 1. Show
that the additivity of probability can be derived from coherence, i.e., show that any
subjective probability that satisﬁes the rule: whenever A, B , and C is a complete
disjunction of events, then the sum of their probabilities is 1, is additive, i.e., Pr[A ∪
B ] = Pr[A] + Pr[B ].
Answer. Since r is his subjective probability of C , 1 − r must be his subjective probability of
C = A ∪ B . Since p + q + r = 1, it follows 1 − r = p + q . This last problem indicates that the ﬁnite additivity axiom follows from the
requirement that the bets be consistent or, as subjectivists say, “coherent” with
each other. However, it is not possible to derive the additivity for countably inﬁnite
sequences of events from such an argument.
2.5. Counting Rules
In this section we will be working in a ﬁnite probability space, in which all atomic
events have equal probabilities. The acts of rolling dice or drawing balls from urns
can be modeled by such spaces. In order to compute the probability of a given event,
one must count the elements of the set which this event represents. In other words,
we count how many diﬀerent ways there are to achieve a certain outcome. This can
be tricky, and we will develop some general principles how to do it.
Problem 16. You throw two dice.
• a. 1 point What is the probability that the sum of the numbers shown is ﬁve or
less?
Answer. 11 12 13 14
21 22 23
,
31 32
41 i.e., 10 out of 36 possibilities, gives the probability 5
.
18 • b. 1 point What is the probability that both of the numbers shown are ﬁve or
less?
Answer. 11
21
31
41
51 12
22
32
42
52 13
23
33
43
53 14
24
34
44
54 15
25
35 ,
45
55 i.e., 25
.
36 • c. 2 points What is the probability that the maximum of the two numbers shown
is ﬁve? (As a clariﬁcation: if the ﬁrst die shows 4 and the second shows 3 then the
maximum of the numbers shown is 4.)
Answer. 15
25
35 ,
45
51 52 53 54 55 i.e., 1
.
4 12 2. PROBABILITY FIELDS In this and in similar questions to follow, the answer should be given as a fully
shortened fraction.
The multiplication principle is a basic aid in counting: If the ﬁrst operation can
be done n1 ways, and the second operation n2 ways, then the total can be done n1 n2
ways.
Deﬁnition: A permutation of a set is its arrangement in a certain order. It was
mentioned earlier that for a set it does not matter in which order the elements are
written down; the number of permutations is therefore the number of ways a given
set can be written down without repeating its elements. From the multiplication
principle follows: the number of permutations of a set of n elements is n(n − 1)(n −
2) · · · (2)(1) = n! (n factorial). By deﬁnition, 0! = 1.
If one does not arrange the whole set, but is interested in the number of k tuples made up of distinct elements of the set, then the number of possibilities is
n
n(n − 1)(n − 2) · · · (n − k + 2)(n − k + 1) = (n−!k)! . (Start with n and the number
of factors is k .) (k tuples are sometimes called ordered k tuples because the order in
n
which the elements are written down matters.) [Ame94, p. 8] uses the notation Pk
for this.
This leads us to the next question: how many k element subsets does a nelement
set have? We already know how many permutations into k elements it has; but always
k ! of these permutations represent the same subset; therefore we have to divide by
k !. The number of k element subsets of an nelement set is therefore
n
n(n − 1)(n − 2) · · · (n − k + 1)
n!
,
=
=
(2.5.1)
k
k !(n − k )!
(1)(2)(3) · · · k
It is pronounced as n choose k , and is also called a “binomial coeﬃcient.” It is
n
deﬁned for all 0 ≤ k ≤ n. [Ame94, p. 8] calls this number Ck .
Problem 17. 5 points Compute the probability of getting two of a kind and three
of a kind (a “full house”) when ﬁve dice are rolled. (It is not necessary to express it
as a decimal number; a fraction of integers is just ﬁne. But please explain what you
are doing.)
Answer. See [Ame94, example 2.3.3 on p. 9]. Sample space is all ordered 5tuples out of 6,
which has 65 elements. Number of full houses can be identiﬁed with number of all ordered pairs of
distinct elements out of 6, the ﬁrst element in the pair denoting the number which appears twice
6
and the second element that which appears three times, i.e., P2 = 6 · 5. Number of arrangements
5 = 5·4 (we have to specify the two places taken by the
of a given full house over the ﬁve dice is C2
1·2
6
5
twoofakind outcomes.) Solution is therefore P2 · C2 /65 = 50/64 = 0.03858. This approach uses
counting.
Alternative approach, using conditional probability: probability of getting 3 of one kind and
51
5
1
then two of a diﬀerent kind is 1 · 6 · 1 · 6 · 6 = 64 . Then multiply by 5 = 10, since this is the
6
2
number of arrangements of the 3 and 2 over the ﬁve cards. Problem 18. What is the probability of drawing the King of Hearts and the
1
Queen of Hearts if one draws two cards out of a 52 card game? Is it 522 ? Is it
52
1
2
(52)(51) ? Or is it 1
2 = (52)(51) ?
Answer. Of course the last; it is the probability of drawing one special subset. There are two
ways of drawing this subset: ﬁrst the King and then the Queen, or ﬁrst the Queen and then the
King. 2.6. Relationships Involving Binomial Coeﬃcients
Problem 19. Show that
be so. n
k = n
n−k . Give an intuitive argument why this must 2.7. CONDITIONAL PROBABILITY Answer. Because n
n−k 13 counts the complements of kelement sets. Assume U has n elements, one of which is ν ∈ U . How many k element subsets
of U have ν in them? There is a simple trick: Take all (k − 1)element subsets of the
set you get by removing ν from U , and add ν to each of these sets. I.e., the number
−1
is n−1 . Now how many k element subsets of U do not have ν in them? Simple; just
k
take the k element subsets of the set which one gets by removing ν from U ; i.e., it is
n−1
k . Adding those two kinds of subsets together one gets all k element subsets of
U:
n
k (2.6.1) = n−1
k−1 + n−1
k . This important formula is the basis of the Pascal triangle:
(2.6.2)
1
(0)
0
1
1
1
(0)
(1)
1
2
2
1
2
1
(1)
(0)
=
1
3
3
1
(3)
(3)
(3)
2
1
0
4
4
4
1
4
6
4
1
(2)
(1)
(0)
1
5
10
10
5
1 (5)
(5)
(5)
(5)
0
1
2
3 (2)
2
(3)
3
(4)
4 4
3 ()
5
4 () The binomial coeﬃcients also occur in the Binomial Theorem
n (2.6.3) (a + b)n = an + n
1 an−1 b + · · · + n
n−1 abn−1 + bn = n
k an−k bk k=0 Why? When the n factors a + b are multiplied out, each of the resulting terms selects
from each of the n original factors either a or b. The term an−k bk occurs therefore
n
n
n−k = k times.
As an application: If you set a = 1, b = 1, you simply get a sum of binomial
coeﬃcients, i.e., you get the number of subsets in a set with n elements: it is 2n
(always count the empty set as one of the subsets). The number of all subsets is
easily counted directly. You go through the set element by element and about every
element you ask: is it in the subset or not? I.e., for every element you have two
possibilities, therefore by the multiplication principle the total number of possibilities
is 2n .
2.7. Conditional Probability
The concept of conditional probability is arguably more fundamental than probability itself. Every probability is conditional, since we must know that the “experiment” has happened before we can speak of probabilities. [Ame94, p. 10] and
[R´n70] give axioms for conditional probability which take the place of the above
e
axioms (2.3.1), (2.3.2) and (2.3.3). However we will follow here the common procedure of deﬁning conditional probabilities in terms of the unconditional probabilities: (2.7.1) Pr[B A] = Pr[B ∩ A]
Pr[A] How can we motivate (2.7.1)? If we know that A has occurred, then of course the only
way that B occurs is when B ∩ A occurs. But we want to multiply all probabilities
of subsets of A with an appropriate proportionality factor so that the probability of
the event A itself becomes = 1. (5)
5 14 2. PROBABILITY FIELDS Problem 20. 3 points Let A be an event with nonzero probability. Show that
the probability conditionally on A, i.e., the mapping B → Pr[B A], satisﬁes all the
axioms of a probability measure:
Pr[U A] = 1 (2.7.2) Pr[B A] ≥ 0 (2.7.3)
(2.7.4) B i A] = Pr[ for all events B ∞ ∞ Pr[B i A] if B i ∩ B j = ∅ for all i, j with i = j . i=1 i=1 Answer. Pr[U A] = Pr[U ∩A]/ Pr[A] = 1. Pr[B A] = Pr[B ∩A]/ Pr[A] ≥ 0 because Pr[B ∩A] ≥
0 and Pr[A] > 0. Finally,
(2.7.5)
∞
∞
∞
∞
∞
Pr[( i=1 B i ) ∩ A]
Pr[ i=1 (B i ∩ A)]
1
B i A] =
=
=
Pr[B i ∩ A] =
Pr[B i A]
Pr[
Pr[A]
Pr[A]
Pr[A]
i=1 i=1 i=1 First equal sign is deﬁnition of conditional probability, second is distributivity of unions and intersections (Problem 6 d), third because the B i are disjoint and therefore the B i ∩ A are even more
disjoint: B i ∩ A ∩ B j ∩ A = B i ∩ B j ∩ A = ∅ ∩ A = ∅ for all i, j with i = j , and the last equal sign
again by the deﬁnition of conditional probability. Problem 21. You draw two balls without replacement from an urn which has 7
white and 14 black balls.
If both balls are white, you roll a die, and your payoﬀ is the number which the
die shows in dollars.
If one ball is black and one is white, you ﬂip a coin until you get your ﬁrst head,
and your payoﬀ will be the number of ﬂips it takes you to get a head, in dollars again.
If both balls are black, you draw from a deck of 52 cards, and you get the number
shown on the card in dollars. (Ace counts as one, J, Q, and K as 11, 12, 13, i.e.,
basically the deck contains every number between 1 and 13 four times.)
Show that the probability that you receive exactly two dollars in this game is 1/6.
Answer. You know a complete disjunction of events: U = {ww}∪{bb}∪{wb}, with Pr[{ww}] =
1
13
7
7
7 14
= 10 ; Pr[{bb}] = 14 13 = 30 ; Pr[{bw}] = 21 20 + 14 20 = 15 . Furthermore you know the con21 20
21
ditional probabilities of getting 2 dollars conditonally on each of these events: Pr[{2}{ww}] = 1 ;
6
1
Pr[{2}{bb}] = 13 ; Pr[{2}{wb}] = 1 . Now Pr[{2} ∩ {ww}] = Pr[{2}{ww}] Pr[{ww}] etc., therefore
4
76
21 20 (2.7.6)
(2.7.7)
(2.7.8) Pr[{2}] = Pr[{2} ∩ {ww}] + Pr[{2} ∩ {bw}] + Pr[{2} ∩ {bb}]
1
6
1
=
6
= 1 7 14
14 7
76
+
+
21 20
4 21 20
21 20
17
1 13
1
1
+
+
=
10
4 15
13 30
6 + 1 14 13
13 21 20 Problem 22. 2 points A and B are arbitrary events. Prove that the probability
of B can be written as:
(2.7.9) Pr[B ] = Pr[B A] Pr[A] + Pr[B A ] Pr[A ] This is the law of iterated expectations (6.6.2) in the case of discrete random variables: it might be written as Pr[B ] = E Pr[B A] .
Answer. B = B ∩ U = B ∩ (A ∪ A ) = (B ∩ A) ∪ (B ∩ A ) and this union is disjoint, i.e.,
(B ∩ A) ∩ (B ∩ A ) = B ∩ (A ∩ A ) = B ∩ ∅ = ∅. Therefore Pr[B ] = Pr[B ∩ A] + Pr[B ∩ A ].
Now apply deﬁnition of conditional probability to get Pr[B ∩ A] = Pr[B A] Pr[A] and Pr[B ∩ A ] =
Pr[B A ] Pr[A ]. Problem 23. 2 points Prove the following lemma: If Pr[B A1 ] = Pr[B A2 ] (call
it c) and A1 ∩ A2 = ∅ (i.e., A1 and A2 are disjoint), then also Pr[B A1 ∪ A2 ] = c. 2.7. CONDITIONAL PROBABILITY 15 Answer.
Pr[B ∩ (A1 ∪ A2 )]
Pr[(B ∩ A1 ) ∪ (B ∩ A2 )]
=
Pr[A1 ∪ A2 ]
Pr[A1 ∪ A2 ]
Pr[B ∩ A1 ] + Pr[B ∩ A2 ]
c Pr[A1 ] + c Pr[A2 ]
=
=
= c.
Pr[A1 ] + Pr[A2 ]
Pr[A1 ] + Pr[A2 ] Pr[B A1 ∪ A2 ] =
(2.7.10) Problem 24. Show by counterexample that the requirement A1 ∩ A2 = ∅ is
necessary for this result to hold. Hint: use the example in Problem 38 with A1 =
{HH, HT }, A2 = {HH, T H }, B = {HH, T T }.
Answer. Pr[B A1 ] = 1/2 and Pr[B A2 ] = 1/2, but Pr[B A1 ∪ A ] = 1/3. The conditional probability can be used for computing probabilities of intersections of events.
Problem 25. [Lar82, exercises 2.5.1 and 2.5.2 on p. 57, solutions on p. 597,
but no discussion]. Five white and three red balls are laid out in a row at random.
• a. 3 points What is the probability that both end balls are white? What is the
probability that one end ball is red and the other white?
is Answer. You can lay the ﬁrst ball ﬁrst and the last ball second: for white balls, the probability
5
5
= 14 ; for one white, one red it is 5 3 + 3 7 = 15 .
87
8
28 54
87 • b. 4 points What is the probability that all red balls are together? What is the
probability that all white balls are together?
Answer. All red balls together is the same as 3 reds ﬁrst, multiplied by 6, because you may
1
3
have between 0 and 5 white balls before the ﬁrst red. 3 2 6 · 6 = 28 . For the white balls you get
87
54321
1
· 4 = 14 .
87654
4
BTW, 3 reds ﬁrst is same probability as 3 reds last, ie., the 5 whites ﬁrst: 5 7 3 2 1 = 3 2 1 .
8 654
876 Problem 26. The ﬁrst three questions here are discussed in [Lar82, example
2.6.3 on p. 62]: There is an urn with 4 white and 8 black balls. You take two balls
out without replacement.
• a. 1 point What is the probability that the ﬁrst ball is white?
Answer. 1/3 • b. 1 point What is the probability that both balls are white?
Answer. It is Pr[second ball whiteﬁrst ball white] Pr[ﬁrst ball white] = 3
4
3+8 4+8 = 1
.
11 • c. 1 point What is the probability that the second ball is white?
Answer. It is Pr[ﬁrst ball white and second ball white]+Pr[ﬁrst ball black and second ball white] =
3
4
4
8
1
+
=.
3+84+8
7+48+4
3
This is the same as the probability that the ﬁrst ball is white. The probabilities are not dependent
on the order in which one takes the balls out. This property is called “exchangeability.” One can
see it also in this way: Assume you number the balls at random, from 1 to 12. Then the probability
for a white ball to have the number 2 assigned to it is obviously 1 .
3
(2.7.11) = • d. 1 point What is the probability that both of them are black?
Answer. 87
12 11 = 27
3 11 = 14
33 (or 56
).
132 • e. 1 point What is the probability that both of them have the same color?
Answer. The sum of the two above, 14
33 + 1
11 = 17
33 (or 68
).
132 16 2. PROBABILITY FIELDS Now you take three balls out without replacement.
• f . 2 points Compute the probability that at least two of the three balls are white.
13
Answer. It is 55 . The possibilities are wwb, wbw, bww, and www. Of the ﬁrst three, each
438
288
has probability 12 11 10 . Therefore the probability for exactly two being white is 1320 = 12 . The
55
4·3·2
24
1
312
13
probability for www is 12·11·10 = 1320 = 55 . Add this to get 1320 = 55 . More systematically, the answer is 4
2 8
1 + 4
3 12
3 . • g. 1 point Compute the probability that at least two of the three are black.
42
. For
55
42
. One
55 Answer. It is
Together
8
2 4
1 1008
1320
12
.
3 = 672
1320 exactly two: = 28
.
55 For three it is (8)(7)(6)
(12)(11)(10) = 336
1320 can also get is as: it is the complement of the last, or as =
8
3 14
.
55 + • h. 1 point Compute the probability that two of the three are of the same and
the third of a diﬀerent color.
Answer. It is 960
1320 = 40
55 = 8
,
11 or 4
1 8
2 + 4
2 8
1 12
3 . • i. 1 point Compute the probability that at least two of the three are of the same
color.
Answer. This probability is 1. You have 5 black socks and 5 white socks in your drawer.
There is a ﬁre at night and you must get out of your apartment in two minutes. There is no light.
You fumble in the dark for the drawer. How many socks do you have to take out so that you will
have at least 2 of the same color? The answer is 3 socks. Problem 27. If a poker hand of ﬁve cards is drawn from a deck, what is the probability that it will contain three aces? (How can the concept of conditional probability
help in answering this question?)
Answer. [Ame94, example 2.3.3 on p. 9] and [Ame94, example 2.5.1 on p. 13] give two
alternative ways to do it. The second answer uses conditional probability: Probability to draw
432
three aces in a row ﬁrst and then 2 nonaces is 52 51 50 48 47 Then multiply this by 5 = 5·4·3 = 10
49 48
3
1·2·3
This gives 0.0017, i.e., 0.17%. Problem 28. 2 points A friend tosses two coins. You ask: “did one of them
land heads?” Your friend answers, “yes.” What’s the probability that the other also
landed heads?
Answer. U = {HH, HT, T H, T T }; Probability is 13
/
44 = 1
.
3 Problem 29. (Not eligible for inclass exams) [Ame94, p. 5] What is the probability that a person will win a game in tennis if the probability of his or her winning
a point is p?
Answer.
(2.7.12) p4 1 + 4(1 − p) + 10(1 − p)2 + 20p(1 − p)3
1 − 2p(1 − p) How to derive this: {ssss} has probability p4 ; {sssf s}, {ssf ss}, {sf sss}, and {f ssss} have probability 4p4 (1 − p); {sssf f s} etc. (2 f and 3 s in the ﬁrst 5, and then an s, together 5 = 10
2
possibilities) have probability 10p4 (1 − p)2 . Now {sssf f f } and 6 = 20 other possibilities give
3
deuce at least once in the game, i.e., the probability of deuce is 20p3 (1 − p)3 . Now Pr[windeuce] =
p2 + 2p(1 − p)Pr[windeuce], because you win either if you score twice in a row (p2 ) or if you get
deuce again (probablity 2p(1 − p)) and then win. Solve this to get Pr[windeuce] = p2 / 1 − 2p(1 − p)
and then multiply this conditional probability with the probability of getting deuce at least once:
Pr[win after at least one deuce] = 20p3 (1 − p)3 p2 / 1 − 2p(1 − p) . This gives the last term in
(2.7.12). 2.7. CONDITIONAL PROBABILITY 17 Problem 30. (Not eligible for inclass exams) Andy, Bob, and Chris play the
following game: each of them draws a card without replacement from a deck of 52
cards. The one who has the highest card wins. If there is a tie (like: two kings and
no aces), then that person wins among those who drew this highest card whose name
comes ﬁrst in the alphabet. What is the probability for Andy to be the winner? For
Bob? For Chris? Does this probability depend on the order in which they draw their
cards out of the stack?
Answer. Let A be the event that Andy wins, B that Bob, and C that Chris wins.
One way to approach this problem is to ask: what are the chances for Andy to win when he
draws a king?, etc., i.e., compute it for all 13 diﬀerent cards. Then: what are the chances for Bob
to win when he draws a king, and also his chances for the other cards, and then for Chris.
It is computationally easier to make the following partitioning of all outcomes: Either all three
cards drawn are diﬀerent (call this event D), or all three cards are equal (event E ), or two of the
three cards are equal (T ). This third case will have to be split into T = H ∪ L, according to whether
the card that is diﬀerent is higher or lower.
If all three cards are diﬀerent, then Andy, Bob, and Chris have equal chances of winning; if all
three cards are equal, then Andy wins. What about the case that two cards are the same and the
third is diﬀerent? There are two possibilities. If the card that is diﬀerent is higher than the two
that are the same, then the chances of winning are evenly distributed; but if the two equal cards
are higher, then Andy has a 2 chance of winning (when the distribution of the cards Y (lower)
3
and Z (higher) among ABC is is ZZY and ZY Z ), and Bob has a 1 chance of winning (when
3
the distribution is Y ZZ ). What we just did was computing the conditional probabilities Pr[AD],
Pr[AE ], etc.
Now we need the probabilities of D, E , and T . What is the probability that all three cards
3
drawn are the same? The probability that the second card is the same as the ﬁrst is 51 ; and the
probability that the third is the same too is (3)(2)
2
6
; therefore the total probability is (51)(50) = 2550 .
50
48 44
2112
= 2550 . The probability that two are equal and
51 50 The probability that all three are unequal is
3
432
the third is diﬀerent is 3 51 48 = 2550 . Now in half of these cases, the card that is diﬀerent is higher,
50
and in half of the cases it is lower.
Putting this together one gets:
Uncond. Prob.
E
H
L
D
Sum all 3 equal
2 of 3 equal, 3rd higher
2 of 3 equal, 3rd lower
all 3 unequal 6/2550
216/2550
216/2550
2112/2550
2550/2550 Cond. Prob.
ABC
1
0
0
1
3
2
3
1
3 1
3
1
3
1
3 1
3 0
1
3 Prob. of intersection
A
B
C
6/2550
0
0
72/2550
72/2550
72/2550
144/2550
72/2550
0
704/2550 704/2550 704/2550
926/2550 848/2550 776/2550 I.e., the probability that A wins is 926/2550 = 463/1275 = .363, the probability that B wins is
848/2550 = 424/1275 = .3325, and the probability that C wins is 776/2550 = 338/1275 = .304.
Here we are using Pr[A] = Pr[AE ] Pr[E ] + Pr[AH ] Pr[H ] + Pr[AL] Pr[L] + Pr[AD] Pr[D]. Problem 31. 4 points You are the contestant in a game show. There are three
closed doors at the back of the stage. Behind one of the doors is a sports car, behind
the other two doors are goats. The game master knows which door has the sports car
behind it, but you don’t. You have to choose one of the doors; if it is the door with
the sports car, the car is yours.
After you make your choice, say door A, the game master says: “I want to show
you something.” He opens one of the two other doors, let us assume it is door B ,
and it has a goat behind it. Then the game master asks: “Do you still insist on door
A, or do you want to reconsider your choice?”
Can you improve your odds of winning by abandoning your previous choice and
instead selecting the door which the game master did not open? If so, by how much?
Answer. If you switch, you will lose the car if you had initially picked the right door, but you
will get the car if you were wrong before! Therefore you improve your chances of winning from 1/3
to 2/3. This is simulated on the web, see www.stat.sc.edu/∼west/javahtml/LetsMakeaDeal.html. 18 2. PROBABILITY FIELDS It is counterintuitive. You may think that one of the two other doors always has a goat behind
it, whatever your choice, therefore there is no reason to switch. But the game master not only shows
you that there is another door with a goat, he also shows you one of the other doors with a goat
behind it, i.e., he restricts your choice if you switch. This is valuable information. It is as if you
could bet on both other doors simultaneously, i.e., you get the car if it is behind one of the doors B
or C . I.e., if the quiz master had said: I give you the opportunity to switch to the following: you
get the car if it is behind B or C . Do you want to switch? The only doubt the contestant may have
about this is: had I not picked a door with the car behind it then I would not have been oﬀered
this opportunity to switch. 2.8. Ratio of Probabilities as Strength of Evidence
Pr1 and Pr2 are two probability measures deﬁned on the same set F of events.
Hypothesis H1 says Pr1 is the true probability, and H2 says Pr2 is the true probability.
Then the observation of an event A for which Pr1 [A] > Pr2 [A] is evidence in favor of
H1 as opposed to H2 . [Roy97] argues that the ratio of the probabilities (also called
“likelihood ratio”) is the right way to measure the strength of this evidence. Among
others, the following justiﬁcation is given [Roy97, p. 7]: If H2 is true, it is usually
not impossible to ﬁnd evidence favoring H1 , but it is unlikely ; and its probability is
bounded by the (reverse of) the ratio of probabilities.
This can be formulated mathematically as follows: Let S be the union of all
events A for which Pr1 [A] ≥ k Pr2 [A]. Then it can be shown that Pr2 [S ] ≤ 1/k , i.e.,
if H2 is true, the probability to ﬁnd evidence favoring H1 with strength k is never
greater than 1/k . Here is a proof in the case that there is only a ﬁnite number of possible outcomes U = {ω1 , . . . , ωn }: Renumber the outcomes such that for i = 1, . . . , m,
Pr1 [{ωi }] < k Pr2 [{ωi }], and for j = m + 1, . . . , n, Pr1 [{ωj }] ≥ k Pr2 [{ωj }]. Then
Pr [{ω }]
n
n
S = {ωm+1 , . . . , ωn }, therefore Pr2 [S ] = j =m+1 Pr2 [{ωj }] ≤ j =m+1 1 k j =
1
1
k Pr1 [S ] ≤ k as claimed. The last inequality holds because Pr1 [S ] ≤ 1, and the
equalsign before this is simply the deﬁnition of S .
With more mathematical eﬀort, see [Rob70], one can strengthen this simple inequality in a very satisfactory manner: Assume an unscrupulous researcher attempts
to ﬁnd evidence supporting his favorite but erroneous hypothesis H1 over his rival’s
H2 by a factor of at least k . He proceeds as follows: he observes an outcome of the
above experiment once, say the outcome is ωi(1) . If Pr1 [{ωi(1) }] ≥ k Pr2 [{ωi(1) }] he
publishes his result; if not, he makes a second independent observation of the experiment ωi(2) . If Pr1 [{ωi(1) }] Pr1 [{ωi(2) }] > k Pr2 [{ωi(1) }] Pr2 [{ωi(2) }] he publishes his
result; if not he makes a third observation and incorporates that in his publication as
well, etc. It can be shown that this strategy will not help: if his rival’s hypothesis is
true, then the probability that he will ever be able to publish results which seem to
show that his own hypothesis is true is still ≤ 1/k . I.e., the sequence of independent
observations ωi(2) , ωi(2) , . . . is such that
n n (2.8.1) Pr1 [{ωi(j ) }] ≥ k Pr2
j =1 Pr2 [{ωi(1) }]
j =1 for some n = 1, 2, . . . ≤ 1
k It is not possible to take advantage of the indeterminacy of a random outcome by
carrying on until chance places one ahead, and then to quit. If one fully discloses
all the evidence one is accumulating, then the probability that this accumulated
evidence supports one’s hypothesis cannot rise above 1/k .
Problem 32. It is usually not possible to assign probabilities to the hypotheses
H1 and H2 , but sometimes it is. Show that in this case, the likelihood ratio of event 2.9. BAYES THEOREM 19 A is the factor by which the ratio of the probabilities of H1 and H2 is changed by the
observation of A, i.e.,
Pr[H 1 A]
Pr[H 1 ] Pr[AH 1 ]
=
Pr[H 2 A]
Pr[H 2 ] Pr[AH 2 ] (2.8.2) Answer. Apply Bayes’s theorem (2.9.1) twice, once for the numerator, once for the denominator. A world in which probability theory applies is therefore a world in which the
transitive dimension must be distinguished from the intransitive dimension. Research
results are not determined by the goals of the researcher.
2.9. Bayes Theorem
In its simplest form Bayes’s theorem reads
Pr[AB ] = (2.9.1) Pr[B A] Pr[A]
.
Pr[B A] Pr[A] + Pr[B A ] Pr[A ] Problem 33. Prove Bayes theorem!
Answer. Obvious since numerator is Pr[B ∩ A] and denominator Pr[B ∩ A] + Pr[B ∩ A ] =
Pr[B ]. This theorem has its signiﬁcance in cases in which A can be interpreted as a
cause of B , and B an eﬀect of A. For instance, A is the event that a student who
was picked randomly from a class has learned for a certain exam, and B is the
event that he passed the exam. Then the righthand side expression contains that
information which you would know from the causeeﬀect relations: the unconditional
probability of the event which is the cause, and the conditional probabilities of the
eﬀect conditioned on whether or not the cause happened. From this, the formula
computes the conditional probability of the cause given that the eﬀect happened.
Bayes’s theorem tells us therefore: if we know that the eﬀect happened, how sure
can we be that the cause happened? Clearly, Bayes’s theorem has relevance for
statistical inference.
Let’s stay with the example with learning for the exam; assume Pr[A] = 60%,
Pr[B A] = .8, and Pr[B A ] = .5. Then the probability that a student who passed
.8)(.
the exam has learned for it is (.8)((6)+(6) .4) = .48 = .706. Look at these numbers:
.
.5)(
.68
The numerator is the average percentage of students who learned and passed, and
the denominator average percentage of students who passed.
Problem 34. AIDS diagnostic tests are usually over 99.9% accurate on those
who do not have AIDS (i.e., only 0.1% false positives) and 100% accurate on those
who have AIDS (i.e., no false negatives at all). (A test is called positive if it indicates
that the subject has AIDS.)
• a. 3 points Assuming that 0.5% of the population actually have AIDS, compute
the probability that a particular individual has AIDS, given that he or she has tested
positive.
Answer. A is the event that he or she has AIDS, and T the event that the test is positive.
Pr[T A] Pr[A]
1 · 0.005
=
=
Pr[T A] Pr[A] + Pr[T A ] Pr[A ]
1 · 0.005 + 0.001 · 0.995
1000 · 5
5000
1000
100 · 0.5
=
=
=
=
= 0.834028
100 · 0.5 + 0.1 · 99.5
1000 · 5 + 1 · 995
5995
1199 Pr[AT ] = Even after testing positive there is still a 16.6% chance that this person does not have AIDS. 20 2. PROBABILITY FIELDS • b. 1 point If one is young, healthy and not in one of the risk groups, then the
chances of having AIDS are not 0.5% but 0.1% (this is the proportion of the applicants
to the military who have AIDS). Recompute the probability with this alternative
number.
Answer.
1 · 0.001
100 · 0.1
1000 · 1
1000
1000
=
=
=
=
= 0.50025.
1 · 0.001 + 0.001 · 0.999
100 · 0.1 + 0.1 · 99.9
1000 · 1 + 1 · 999
1000 + 999
1999 2.10. Independence of Events
2.10.1. Deﬁnition of Independence. Heuristically, we want to say: event B
is independent of event A if Pr[B A] = Pr[B A ]. From this follows by Problem 23
that the conditional probability is equal to the unconditional probability Pr[B ], i.e.,
Pr[B ] = Pr[B ∩ A]/ Pr[A]. Therefore we will adopt as deﬁnition of independence the
socalled multiplication rule:
Deﬁnition: B and A are independent, notation B ⊥A, if Pr[B ∩ A] = Pr[B ] Pr[A].
This is a symmetric condition, i.e., if B is independent of A, then A is also
independent of B . This symmetry is not immediately obvious given the above deﬁnition of independence, and it also has the following nontrivial practical implication
(this example from [Daw79a, pp. 2/3]): A is the event that one is exposed to some
possibly carcinogenic agent, and B the event that one develops a certain kind of
cancer. In order to test whether B ⊥A, i.e., whether the exposure to the agent does
not increase the incidence of cancer, one often collects two groups of subjects, one
group which has cancer and one control group which does not, and checks whether
the exposure in these two groups to the carcinogenic agent is the same. I.e., the
experiment checks whether A⊥B , although the purpose of the experiment was to
determine whether B ⊥A.
Problem 35. 3 points Given that Pr[B ∩ A] = Pr[B ] · Pr[A] (i.e., B is independent of A), show that Pr[B ∩ A ] = Pr[B ] · Pr[A ] (i.e., B is also independent of
A ).
Answer. If one uses our heuristic deﬁnition of independence, i.e., B is independent of event
A if Pr[B A] = Pr[B A ], then it is immediately obvious since deﬁnition is symmetric in A and
A . However if we use the multiplication rule as the deﬁnition of independence, as the text of
this Problem suggests, we have to do a little more work: Since B is the disjoint union of (B ∩ A)
and (B ∩ A ), it follows Pr[B ] = Pr[B ∩ A] + Pr[B ∩ A ] or Pr[B ∩ A ] = Pr[B ] − Pr[B ∩ A] =
Pr[B ] − Pr[B ] Pr[A] = Pr[B ](1 − Pr[A]) = Pr[B ] Pr[A ]. Problem 36. 2 points A and B are two independent events with Pr[A] =
Pr[B ] = 1 . Compute Pr[A ∪ B ].
4
Answer. Pr[A ∪ B ] = Pr[A] + Pr[B ] − Pr[A ∩ B ] = Pr[A] + Pr[B ] − Pr[A] Pr[B ] =
1
.
2 1
3 1
3 and 1
+ 1 − 12 =
4 Problem 37. 3 points You have an urn with ﬁve white and ﬁve red balls. You
take two balls out without replacement. A is the event that the ﬁrst ball is white,
and B that the second ball is white. a. What is the probability that the ﬁrst ball
is white? b. What is the probability that the second ball is white? c. What is the
probability that both have the same color? d. Are these two events independent, i.e.,
is Pr[B A] = Pr[A]? e. Are these two events disjoint, i.e., is A ∩ B = ∅?
Answer. Clearly, Pr[A] = 1/2. Pr[B ] = Pr[B A] Pr[A] + Pr[B A ] Pr[A ] = (4/9)(1/2) +
5
(5/9)(1/2) = 1/2. The events are not independent: Pr[B A] = 4/9 = Pr[B ], or Pr[A ∩ B ] = 10 4 =
9 2.10. INDEPENDENCE OF EVENTS 21 2/9 = 1/4. They would be independent if the ﬁrst ball had been replaced. The events are also not
disjoint: it is possible that both balls are white. 2.10.2. Independence of More than Two Events. If there are more than
two events, we must require that all possible intersections of these events, not only
the pairwise intersections, follow the above multiplication rule. For instance,
Pr[A ∩ B ] = Pr[A] Pr[B ];
Pr[A ∩ C ] = Pr[A] Pr[C ]; (2.10.1) A, B , C mutually independent ⇐⇒ Pr[B ∩ C ] = Pr[B ] Pr[C ];
Pr[A ∩ B ∩ C ] = Pr[A] Pr[B ] Pr[C ]. This last condition is not implied by the other three. Here is an example. Draw a ball
at random from an urn containing four balls numbered 1, 2, 3, 4. Deﬁne A = {1, 4},
B = {2, 4}, and C = {3, 4}. These events are pairwise independent but not mutually
independent.
Problem 38. 2 points Flip a coin two times independently and deﬁne the following three events:
A = Head in ﬁrst ﬂip
B = Head in second ﬂip (2.10.2) C = Same face in both ﬂips.
Are these three events pairwise independent? Are they mutually independent?
Answer. U =
1
,
2 H H HT
TH TT
1
. They
2 . A = {HH, HT }, B = {HH, T H }, C = {HH, T T }. Pr[A] = 1
,
2 1
4 = Pr[B ] =
Pr[C ] =
are pairwise independent, but Pr[A ∩ B ∩ C ] = Pr[{HH }] =
Pr[A] Pr[B ] Pr[C ], therefore the events cannot be mutually independent. Problem 39. 3 points A, B , and C are pairwise independent events whose
probabilities are greater than zero and smaller than one, and A ∩ B ⊂ C . Can those
events be mutually independent?
Answer. No; from A ∩ B ⊂ C follows A ∩ B ∩ C = A ∩ B and therefore Pr[A ∩ B ∩ C ] =
Pr[A ∩ B ] Pr[C ] since Pr[C ] < 1 and Pr[A ∩ B ] > 0. If one takes unions, intersections, complements of diﬀerent mutually independent
events, one will still end up with mutually independent events. E.g., if A, B , C
mutually independent, then A , B , C are mutually independent as well, and A ∩ B
independent of C , and A ∪ B independent of C , etc. This is not the case if the events
are only pairwise independent. In Problem 39, A ∩ B is not independent of C .
'$
'$
'$
RST
UVW
&%
&%
X
&%
Figure 1. Generic Venn Diagram for 3 Events
2.10.3. Conditional Independence. If A and B are independent in the probability measure conditionally on C , i.e., if Pr[A ∩ B C ] = Pr[AC ] Pr[B C ], then they 22 2. PROBABILITY FIELDS are called conditionally independent given that C occurred, notation A⊥B C . In
formulas,
(2.10.3) Pr[A ∩ C ] Pr[B ∩ C ]
Pr[A ∩ B ∩ C ]
=
.
Pr[C ]
Pr[C ]
Pr[C ] Problem 40. 5 points Show that A⊥B C is equivalent to Pr[AB ∩C ] = Pr[AC ].
In other words: independence of A and B conditionally on C means: once we know
that C occurred, the additional knowledge whether B occurred or not will not help us
to sharpen our knowledge about A.
Literature about conditional independence (of random variables, not of events)
includes [Daw79a], [Daw79b], [Daw80].
2.10.4. Independent Repetition of an Experiment. If a given experiment
has sample space U , and we perform the experiment n times in a row, then this
repetition can be considered a single experiment with the sample space consisting of
ntuples of elements of U . This set is called the product set U n = U × U × · · · × U
(n terms).
If a probability measure Pr is given on F , then one can deﬁne in a unique way
a probability measure on the subsets of the product set so that events in diﬀerent
repetitions are always independent of each other.
The Bernoulli experiment is the simplest example of such an independent repetition. U = {s, f } (stands for success and failure). Assume Pr[{s}] = p, and that
the experimenter has several independent trials. For instance, U 5 has, among others,
the following possible outcomes:
If ω =(f, f, f, f, f )
(f, f, f, f, s) (1 − p)n−1 p (f, f, f, s, f ) (1 − p)n−1 p (f, f, f, s, s) (1 − p)n−2 p2 (f, f, s, f, f ) (2.10.4) then Pr[{ω }] = (1 − p)n (1 − p)n−1 p, etc. One sees, this is very cumbersome, and usually unnecessarily so. If we toss a coin
5 times, the only thing we usually want to know is how many successes there were.
As long as the experiments are independent, the question how the successes were
distributed over the n diﬀerent trials is far less important. This brings us to the
deﬁnition of a random variable, and to the concept of a suﬃcient statistic.
2.11. How to Plot Frequency Vectors and Probability Vectors
If there are only 3 possible outcomes, i.e., U = {ω1 , ω2 , ω3 }, then the set of all
probability measures is the set of nonnegative 3vectors whose components sum up to
1. Graphically, such vectors can be represented as points inside a trilateral triangle
with height 1: the three components of the vector are the distances of the point
to each of the sides of the triangle. The R/Splusfunction triplot in the ecmet
package, written by Jim Ramsay ramsay@ramsay2.psych.mcgill.ca, does this, with
optional rescaling if the rows of the data matrix do not have unit sums.
Problem 41. In an equilateral triangle, call a = the distance of the sides from
the center point, b = half the side length, and c = the distance of the corners from
√
the center point (as in Figure 2). Show that b = a 3 and c = 2a. 2.11. HOW TO PLOT FREQUENCY VECTORS AND PROBABILITY VECTORS 23 p
c
p p
p c
p b a
p p Figure 2. Geometry of an equilateral triangle
√
Answer. From (a + c)2 + b2 = 4b2 , i.e., (a + c)2 = 3b2 , follows a + c = b 3. But we
2 + b2 = c2 . Therefore a2 + 2ac + c2 = 3b2 = 3c2 − 3a2 , or 4a2 + 2ac − 2c2 = 0
also have a
or 2a2 + ac − √2 = (2a − √ a + c) = 0. The positive solution is therefore c = 2a. This gives
c
c)(
a + c = 3a = b 3, or b = a 3. And the function quadplot, also written by Jim Ramsey, does quadrilinear plots,
meaning that proportions for four categories are plotted within a regular tetrahedron. Quadplot displays the probability tetrahedron and its points using XGobi.
Each vertex of the triangle or tetrahedron corresponds to the degenerate probability distribution in which one of the events has probability 1 and the others have
probability 0. The labels of these vertices indicate which event has probability 1.
The script kai is an example visualizing data from [Mor65]; it can be run using
the command ecmet.script(kai).
Example: Statistical linguistics.
In the study of ancient literature, the authorship of texts is a perplexing problem.
When books were written and reproduced by hand, the rights of authorship were
limited and what would now be considered forgery was common. The names of
reputable authors were borrowed in order to sell books, get attention for books, or the
writings of disciples and collaborators were published under the name of the master,
or anonymous old manuscripts were optimistically attributed to famous authors. In
the absence of conclusive evidence of authorship, the attribution of ancient texts
must be based on the texts themselves, for instance, by statistical analysis of literary
style. Here it is necessary to ﬁnd stylistic criteria which vary from author to author,
but are independent of the subject matter of the text. An early suggestion was to use
the probability distribution of word length, but this was never acted upon, because
it is too dependent on the subject matter. Sentencelength distributions, on the
other hand, have proved highly reliable. [Mor65, p. 184] says that sentencelength
is “periodic rather than random,” therefore the sample should have at least about
100 sentences. “Sentencelength distributions are not suited to dialogue, they cannot
be used on commentaries written on one author by another, nor are they reliable on
such texts as the fragmentary books of the historian Diodorus Siculus.”
Problem 42. According to [Mor65, p. 184], sentencelength is “periodic rather
than random.” What does this mean?
Answer. In a text, passages with long sentences alternate with passages with shorter sentences. This is why one needs at least 100 sentences to get a representative distribution of sentences, and this is why fragments and drafts and commentaries on others’ writings do not exhibit
an average sentence length distribution: they do not have the melody of the ﬁnished text. Besides the length of sentences, also the number of common words which express
a general relation (“and”, “in”, “but”, “I”, “to be”) is random with the same distribution at least among the same genre. By contrast, the occurrence of the deﬁnite 24 2. PROBABILITY FIELDS article “the” cannot be modeled by simple probabilistic laws because the number of
nouns with deﬁnite article depends on the subject matter.
Table 1 has data about the epistles of St. Paul. Abbreviations: Rom Romans; Co1
1st Corinthians; Co2 2nd Corinthians; Gal Galatians; Phi Philippians; Col Colossians; Th1 1st Thessalonians; Ti1 1st Timothy; Ti2 2nd Timothy; Heb Hebrews. 2nd
Thessalonians, Titus, and Philemon were excluded because they were too short to
give reliable samples. From an analysis of these and other data [Mor65, p. 224] the
ﬁrst 4 epistles (Romans, 1st Corinthians, 2nd Corinthians, and Galatians) form a
consistent group, and all the other epistles lie more than 2 standard deviations from
the mean of this group (using χ2 statistics). If Paul is deﬁned as being the author of
Galatians, then he also wrote Romans and 1st and 2nd Corinthians. The remaining
epistles come from at least six hands.
Table 1. Number of Sentences in Paul’s Epistles with 0, 1, 2, and
≥ 3 occurrences of kai no kai
one
two
3 or more Rom Co1
386 424
141 152
34
35
17
16 Co2
192
86
28
13 Gal Phi Col Th1 Ti1 Ti2 Heb
128
42 23
34
49
45 155
48
29 32
23
38
28
94
5
19
17
8
9
11
37
6
12
9
16
10
4
24 Problem 43. Enter the data from Table 1 into xgobi and brush the four epistles
which are, according to Morton, written by Paul himself. 3 of those points are almost
on top of each other, and one is a little apart. Which one is this?
Answer. In R, issue the commands library(xgobi) then data(PaulKAI) then quadplot(PaulKAI,
normalize = TRUE). If you have xgobi but not R, this dataset is one of the default datasets coming
with xgobi. CHAPTER 3 Random Variables
3.1. Notation
Throughout these class notes, lower case bold letters will be used for vectors
and upper case bold letters for matrices, and letters that are not bold for scalars.
The (i, j ) element of the matrix A is aij , and the ith element of a vector b is bi ;
the arithmetic mean of all elements is ¯. All vectors are column vectors; if a row
b
vector is needed, it will be written in the form b . Furthermore, the online version
of these notes uses green symbols for random variables, and the corresponding black
symbols for the values taken by these variables. If a blackandwhite printout of
the online version is made, then the symbols used for random variables and those
used for speciﬁc values taken by these random variables can only be distinguished
by their grey scale or cannot be distinguished at all; therefore a special monochrome
version is available which should be used for the blackandwhite printouts. It uses
an upright math font, called “Euler,” for the random variables, and the same letter
in the usual slanted italic font for the values of these random variables.
Example: If y is a random vector, then y denotes a particular value, for instance
an observation, of the whole vector; y i denotes the ith element of y (a random scalar),
and yi is a particular value taken by that element (a nonrandom scalar).
With realvalued random variables, the powerful tools of calculus become available to us. Therefore we will begin the chapter about random variables with a
digression about inﬁnitesimals 3.2. Digression about Inﬁnitesimals
In the following pages we will recapitulate some basic facts from calculus. But
it will diﬀer in two respects from the usual calculus classes. (1) everything will be
given its probabilitytheoretic interpretation, and (2) we will make explicit use of
inﬁnitesimals. This last point bears some explanation.
You may say inﬁnitesimals do not exist. Do you know the story with Achilles and
the turtle? They are racing, the turtle starts 1 km ahead of Achilles, and Achilles
runs ten times as fast as the turtle. So when Achilles arrives at the place the turtle
started, the turtle has run 100 meters; and when Achilles has run those 100 meters,
the turtle has run 10 meters, and when Achilles has run the 10 meters, then the turtle
has run 1 meter, etc. The Greeks were actually arguing whether Achilles would ever
reach the turtle.
This may sound like a joke, but in some respects, modern mathematics never
went beyond the level of the Greek philosophers. If a modern mathematicien sees
something like
(3.2.1) 1
= 0,
i→∞ i
lim n or lim n→∞
25 i=0 1
10
=
,
i
10
9 26 3. RANDOM VARIABLES then he will probably say that the lefthand term in each equation never really reaches
the number written on the right, all he will say is that the term on the left comes
arbitrarily close to it.
This is like saying: I know that Achilles will get as close as 1 cm or 1 mm to the
turtle, he will get closer than any distance, however small, to the turtle, instead of
simply saying that Achilles reaches the turtle. Modern mathematical proofs are full
of races between Achilles and the turtle of the kind: give me an ε, and I will prove to
you that the thing will come at least as close as ε to its goal (socalled epsilontism),
but never speaking about the moment when the thing will reach its goal.
Of course, it “works,” but it makes things terribly cumbersome, and it may have
prevented people from seeing connections.
Abraham Robinson in [Rob74] is one of the mathematicians who tried to remedy
it. He did it by adding more numbers, inﬁnite numbers and inﬁnitesimal numbers.
Robinson showed that one can use inﬁnitesimals without getting into contradictions,
and he demonstrated that mathematics becomes much more intuitive this way, not
only its elementary proofs, but especially the deeper results. One of the elemrntary
books based on his calculus is [HK79].
The wellknow logician Kurt G¨del said about Robinson’s work: “I think, in
o
coming years it will be considered a great oddity in the history of mathematics that
the ﬁrst exact theory of inﬁnitesimals was developed 300 years after the invention of
the diﬀerential calculus.”
G¨del called Robinson’s theory the ﬁrst theory. I would like to add here the folo
lowing speculation: perhaps Robinson shares the following error with the “standard”
mathematicians whom he criticizes: they consider numbers only in a static way, without allowing them to move. It would be beneﬁcial to expand on the intuition of the
inventors of diﬀerential calculus, who talked about “ﬂuxions,” i.e., quantities in ﬂux,
in motion. Modern mathematicians even use arrows in their symbol for limits, but
they are not calculating with moving quantities, only with static quantities.
This perspective makes the categorytheoretical approach to inﬁnitesimals taken
in [MR91] especially promising. Category theory considers objects on the same
footing with their transformations (and uses lots of arrows).
Maybe a few years from now mathematics will be done right. We should not let
this temporary backwardness of mathematics allow to hold us back in our intuition.
∆y
The equation ∆x = 2x does not hold exactly on a parabola for any pair of given
(static) ∆x and ∆y ; but if you take a pair (∆x, ∆y ) which is moving towards zero
then this equation holds in the moment when they reach zero, i.e., when they vanish.
Writing dy and dx means therefore: we are looking at magnitudes which are in the
process of vanishing. If one applies a function to a moving quantity one again gets a
moving quantity, and the derivative of this function compares the speed with which
the transformed quantity moves with the speed of the original quantity. Likewise,
n
the equation i=1 21 = 1 holds in the moment when n reaches inﬁnity. From this
n
point of view, the axiom of σ additivity in probability theory (in its equivalent form
of rising or declining sequences of events) indicates that the probability of a vanishing
event vanishes.
Whenever we talk about inﬁnitesimals, therefore, we really mean magnitudes
which are moving, and which are in the process of vanishing. dVx,y is therefore not,
as one might think from what will be said below, a static but small volume element
located close to the point (x, y ), but it is a volume element which is vanishing into
the point (x, y ). The probability density function therefore signiﬁes the speed with
which the probability of a vanishing element vanishes. 3.4. CHARACTERIZATION OF RANDOM VARIABLES 27 3.3. Deﬁnition of a Random Variable
The best intuition of a random variable would be to view it as a numerical
variable whose values are not determinate but follow a statistical pattern, and call
it x, while possible values of x are called x.
In order to make this a mathematically sound deﬁnition, one says: A mapping x :
U → R of the set U of all possible outcomes into the real numbers R is called a random
variable. (Again, mathematicians are able to construct pathological mappings that
cannot be used as random variables, but we let that be their problem, not ours.) The
green x is then deﬁned as x = x(ω ). I.e., all the randomness is shunted oﬀ into the
process of selecting an element of U . Instead of being an indeterminate function, it
is deﬁned as a determinate function of the random ω . It is written here as x(ω ) and
not as x(ω ) because the function itself is determinate, only its argument is random.
Whenever one has a mapping x : U → R between sets, one can construct from it
in a natural way an “inverse image” mapping between subsets of these sets. Let F ,
as usual, denote the set of subsets of U , and let B denote the set of subsets of R. We
will deﬁne a mapping x−1 : B → F in the following way: For any B ⊂ R, we deﬁne
x−1 (B ) = {ω ∈ U : x(ω ) ∈ B }. (This is not the usual inverse of a mapping, which
does not always exist. The inverseimage mapping always exists, but the inverse
image of a oneelement set is no longer necessarily a oneelement set; it may have
more than one element or may be the empty set.)
This “inverse image” mapping is well behaved with respect to unions and intersections, etc. In other words, we have identities x−1 (A ∩ B ) = x−1 (A) ∩ x−1 (B ) and
x−1 (A ∪ B ) = x−1 (A) ∪ x−1 (B ), etc.
Problem 44. Prove the above two identities.
Answer. These are a very subtle proofs. x−1 (A ∩ B ) = {ω ∈ U : x(ω ) ∈ A ∩ B } = {ω ∈
U : x(ω ) ∈ A and x(ω ) ∈ B = {ω ∈ U : x(ω ) ∈ A} ∩ {ω ∈ U : x(ω ) ∈ B } = x−1 (A) ∩ x−1 (B ). The
other identity has a similar proof. Problem 45. Show, on the other hand, by a counterexample, that the “direct
image” mapping deﬁned by x(E ) = {r ∈ R : there exists ω ∈ E with x(ω ) = r} no
longer satisﬁes x(E ∩ F ) = x(E ) ∩ x(F ).
By taking inverse images under a random variable x, the probability measure
on F is transplanted into a probability measure on the subsets of R by the simple
prescription Pr[B ] = Pr x−1 (B ) . Here, B is a subset of R and x−1 (B ) one of U , the
Pr on the right side is the given probability measure on U , while the Pr on the left is
the new probability measure on R induced by x. This induced probability measure
is called the probability law or probability distribution of the random variable.
Every random variable induces therefore a probability measure on R, and this
probability measure, not the mapping itself, is the most important ingredient of
a random variable. That is why Amemiya’s ﬁrst deﬁnition of a random variable
(deﬁnition 3.1.1 on p. 18) is: “A random variable is a variable that takes values
acording to a certain distribution.” In other words, it is the outcome of an experiment
whose set of possible outcomes is R.
3.4. Characterization of Random Variables
We will begin our systematic investigation of random variables with an overview
over all possible probability measures on R.
The simplest way to get such an overview is to look at the cumulative distribution
functions. Every probability measure on R has a cumulative distribution function, 28 3. RANDOM VARIABLES but we will follow the common usage of assigning the cumulative distribution not
to a probability measure but to the random variable which induces this probability
measure on R.
Given a random variable x : U ω → x(ω ) ∈ R. Then the cumulative distribution function of x is the function Fx : R → R deﬁned by:
(3.4.1) Fx (a) = Pr[{ω ∈ U : x(ω ) ≤ a}] = Pr[x≤a]. This function uniquely deﬁnes the probability measure which x induces on R.
Properties of cumulative distribution functions: a function F : R → R is a cumulative distribution function if and only if
a ≤ b ⇒ F (a) ≤ F (b) (3.4.2)
(3.4.3) lim F (a) = 0 a→−∞ (3.4.4)
(3.4.5) lim F (a) = 1 a→∞ lim ε→0,ε>0 F (a + ε) = F (a) Equation (3.4.5) is the deﬁnition of continuity from the right (because the limit
holds only for ε ≥ 0). Why is a cumulative distribution function continuous from
the right? For every nonnegative sequence ε1 , ε2 , . . . ≥ 0 converging to zero which
also satisﬁes ε1 ≥ ε2 ≥ . . . follows {x ≤ a} = i {x ≤ a + εi }; for these sequences,
therefore, the statement follows from what Problem 14 above said about the probability of the intersection of a declining set sequence. And a converging sequence of
nonnegative εi which is not declining has a declining subsequence.
A cumulative distribution function need not be continuous from the left. If
limε→0,ε>0 F (x − ε) = F (x), then x is a jump point, and the height of the jump is
the probability that x = x.
It is a matter of convention whether we are working with right continuous or
left continuous functions here. If the distribution function were deﬁned as Pr[x < a]
(some authors do this, compare [Ame94, p. 43]), then it would be continuous from
the left but not from the right.
Problem 46. 6 points Assume Fx (x) is the cumulative distribution function of
the random variable x (whose distribution is not necessarily continuous). Which of
the following formulas are correct? Give proofs or verbal justiﬁcations.
(3.4.6) Pr[x = x] = (3.4.7) Pr[x = x] = Fx (x) − (3.4.8) Pr[x = x] = lim ε>0; ε→0 lim ε>0; ε→0 Fx (x + ε) − Fx (x)
lim δ>0; δ →0 Fx (x − δ ) Fx (x + ε) − lim δ>0; δ →0 Fx (x − δ ) Answer. (3.4.6) does not hold generally, since its rhs is always = 0; the other two equations
always hold. Problem 47. 4 points Assume the distribution of z is symmetric about zero,
i.e., Pr[z < −z ] = Pr[z >z ] for all z . Call its cumulative distribution function Fz (z ).
Show that the cumulative distribution function of the random variable q = z 2 is
√
Fq (q ) = 2Fz ( q ) − 1 for q ≥ 0, and 0 for q < 0. 3.4. CHARACTERIZATION OF RANDOM VARIABLES 29 Answer. If q ≥ 0 then
(3.4.9)
(3.4.10)
(3.4.11)
(3.4.12)
(3.4.13) √
√
Fq (q ) = Pr[z 2 ≤q ] = Pr[− q ≤z ≤ q ]
√
√
= Pr[z ≤ q ] − Pr[z < − q ]
√
√
= Pr[z ≤ q ] − Pr[z > q ]
√
√
= Fz ( q ) − (1 − Fz ( q ))
√
= 2Fz ( q ) − 1. Instead of the cumulative distribution function Fy one can also use the quan−
tile function Fy 1 to characterize a probability measure. As the notation suggests,
the quantile function can be considered some kind of “inverse” of the cumulative
distribution function. The quantile function is the function (0, 1) → R deﬁned by
(3.4.14) −
Fy 1 (p) = inf {u : Fy (u) ≥ p} or, plugging the deﬁnition of Fy into (3.4.14),
(3.4.15) −
Fy 1 (p) = inf {u : Pr[y ≤u] ≥ p}. The quantile function is only deﬁned on the open unit interval, not on the endpoints
0 and 1, because it would often assume the values −∞ and +∞ on these endpoints,
and the information given by these values is redundant. The quantile function is
continuous from the left, i.e., from the other side than the cumulative distribution
function. If F is continuous and strictly increasing, then the quantile function is
the inverse of the distribution function in the usual sense, i.e., F −1 (F (t)) = t for
all t ∈ R, and F (F −1 ((p)) = p for all p ∈ (0, 1). But even if F is ﬂat on certain
intervals, and/or F has jump points, i.e., F does not have an inverse function, the
following important identity holds for every y ∈ R and p ∈ (0, 1):
(3.4.16) p ≤ Fy (y ) −
iﬀ Fy 1 (p) ≤ y Problem 48. 3 points Prove equation (3.4.16).
Answer. ⇒ is trivial: if F (y ) ≥ p then of course y ≥ inf {u : F (u) ≥ p}. ⇐: y ≥ inf {u :
F (u) ≥ p} means that every z > y satisﬁes F (z ) ≥ p; therefore, since F is continuous from the
right, also F (y ) ≥ p. This proof is from [Rei89, p. 318]. Problem 49. You throw a pair of dice and your random variable x is the sum
of the points shown.
• a. Draw the cumulative distribution function of x.
Answer. This is Figure 1: the cdf is 0 in (−∞, 2), 1/36 in [2,3), 3/36 in [3,4), 6/36 in [4,5),
10/36 in [5,6), 15/36 in [6,7), 21/36 in [7,8), 26/36 on [8,9), 30/36 in [9,10), 33/36 in [10,11), 35/36
on [11,12), and 1 in [12, +∞). • b. Draw the quantile function of x.
Answer. This is Figure 2: the quantile function is 2 in (0, 1/36], 3 in (1/36,3/36], 4 in
(3/36,6/36], 5 in (6/36,10/36], 6 in (10/36,15/36], 7 in (15/36,21/36], 8 in (21/36,26/36], 9 in
(26/36,30/36], 10 in (30/36,33/36], 11 in (33/36,35/36], and 12 in (35/36,1]. Problem 50. 1 point Give the formula of the cumulative distribution function
of a random variable which is uniformly distributed between 0 and b.
Answer. 0 for x ≤ 0, x/b for 0 ≤ x ≤ b, and 1 for x ≥ b. 30 3. RANDOM VARIABLES q qq q
q
q
q
q
q
q q Figure 1. Cumulative Distribution Function of Discrete Variable
Empirical Cumulative Distribution Function:
Besides the cumulative distribution function of a random variable or of a probability measure, one can also deﬁne the empirical cumulative distribution function of
a sample. Empirical cumulative distribution functions are zero for all values below
the lowest observation, then 1/n for everything below the second lowest, etc. They
are step functions. If two observations assume the same value, then the step at
that value is twice as high, etc. The empirical cumulative distribution function can
be considered an estimate of the cumulative distribution function of the probability
distribution underlying the sample. [Rei89, p. 12] writes it as a sum of indicator
functions:
1
(3.4.17)
F=
1[xi ,+∞)
ni
3.5. Discrete and Absolutely Continuous Probability Measures
One can deﬁne two main classes of probability measures on R:
One kind is concentrated in countably many points. Its probability distribution
can be deﬁned in terms of the probability mass function.
Problem 51. Show that a distribution function can only have countably many
jump points. q q q q q q q q q q Figure 2. Quantile Function of Discrete Variable 3.6. TRANSFORMATION OF A SCALAR DENSITY FUNCTION ≥ 1
,
4 Answer. Proof: There are at most two with jump height ≥
etc. 1
,
2 31 at most four with jump height Among the other probability measures we are only interested in those which can
be represented by a density function (absolutely continuous). A density function is a
nonnegative integrable function which, integrated over the whole line, gives 1. Given
b
such a density function, called fx (x), the probability Pr[x∈(a, b)] = a fx (x)dx. The
density function is therefore an alternate way to characterize a probability measure.
But not all probability measures have density functions.
Those who are not familiar with integrals should read up on them at this point.
Start with derivatives, then: the indeﬁnite integral of a function is a function whose
derivative is the given function. Then it is an important theorem that the area under
the curve is the diﬀerence of the values of the indeﬁnite integral at the end points.
This is called the deﬁnite integral. (The area is considered negative when the curve
is below the xaxis.)
The intuition of a density function comes out more clearly in terms of inﬁnitesimals. If fx (x) is the value of the density function at the point x, then the probability
that the outcome of x lies in an interval of inﬁnitesimal length located near the point
x is the length of this interval, multiplied by fx (x). In formulas, for an inﬁnitesimal
dx follows
(3.5.1) Pr x∈[x, x + dx] = fx (x) dx . The name “density function” is therefore appropriate: it indicates how densely the
probability is spread out over the line. It is, so to say, the quotient between the
probability measure induced by the variable, and the length measure on the real
numbers.
If the cumulative distribution function has everywhere a derivative, this derivative is the density function.
3.6. Transformation of a Scalar Density Function
Assume x is a random variable with values in the region A ⊂ R, i.e., Pr[x∈A] = 0,
/
and t is a onetoone mapping A → R. Onetoone (as opposed to manytoone)
means: if a, b ∈ A and t(a) = t(b), then already a = b. We also assume that t has a
continuous nonnegative ﬁrst derivative t ≥ 0 everywhere in A. Deﬁne the random
variable y by y = t(x). We know the density function of y , and we want to get that of
x. (I.e., t expresses the old variable, that whose density function we know, in terms
of the new variable, whose density function we want to know.)
Since t is onetoone, it follows for all a, b ∈ A that a = b ⇐⇒ t(a) = t(b). And
recall the deﬁnition of a derivative in terms of inﬁnitesimals dx: t (x) = t(x+dx)−t(x) .
dx
In order to compute fx (x) we will use the following identities valid for all x ∈ A:
(3.6.1)
(3.6.2) fx (x) dx = Pr x∈[x, x + dx] = Pr t(x)∈[t(x), t(x + dx)]
= Pr t(x)∈[t(x), t(x) + t (x) dx] = fy (t(x)) t (x)dx Absolute values are multiplicative, i.e., t (x)dx = t (x) dx; divide by dx to get
(3.6.3) fx (x) = fy t(x) t (x) . This is the transformation formula how to get the density of x from that of y . This
formula is valid for all x ∈ A; the density of x is 0 for all x ∈ A.
/
dy
Heuristically one can get this transformation as follows: write t (x) = dx , then

one gets it from fx (x) dx = fy (t(x)) dy  by just dividing both sides by dx. 32 3. RANDOM VARIABLES In other words, this transformation rule consists of 4 steps: (1) Determine A,
the range of the new variable; (2) obtain the transformation t which expresses the
old variable in terms of the new variable, and check that it is onetoone on A; (3)
plug expression (2) into the old density; (4) multiply this pluggedin density by the
absolute value of the derivative of expression (2). This gives the density inside A; it
is 0 outside A.
An alternative proof is conceptually simpler but cannot be generalized to the
multivariate case: First assume t is monotonically increasing. Then Fx (x) = Pr[x ≤
x] = Pr[t(x) ≤ t(i)] = Fy (t(x)). Now diﬀerentiate and use the chain rule. Then
also do the monotonically decresing case. This is how [Ame94, theorem 3.6.1 on
pp. 48] does it. [Ame94, pp. 52/3] has an extension of this formula to manytoone
functions.
Problem 52. 4 points [Lar82, example 3.5.4 on p. 148] Suppose y has density
function
fy (y ) = (3.6.4) 1
0 for 0 < y < 1
otherwise. Obtain the density fx (x) of the random variable x = − log y .
Answer. (1) Since y takes values only between 0 and 1, its logarithm takes values between
−∞ and 0, the negative logarithm therefore takes values between 0 and +∞, i.e., A = {x : 0 < x}.
(2) Express y in terms of x: y = e−x . This is onetoone on the whole line, therefore also on A.
(3) Plugging y = e−x into the density function gives the number 1, since the density function does
not depend on the precise value of y , as long is we know that 0 < y < 1 (which we do). (4) The
derivative of y = e−x is −e−x . As a last step one has to multiply the number 1 by the absolute
value of the derivative to get the density inside A. Therefore fx (x) = e−x for x > 0 and 0 otherwise. Problem 53. 6 points [Dhr86, p. 1574] Assume the random variable z has
the exponential distribution with parameter λ, i.e., its density function is fz (z ) =
λ exp(−λz ) for z > 0 and 0 for z ≤ 0. Deﬁne u = − log z . Show that the density
function of u is fu (u) = exp µ − u − exp(µ − u) where µ = log λ. This density will
be used in Problem 140.
Answer. (1) Since z only has values in (0, ∞), its log is well deﬁned, and A = R. (2) Express
old variable in terms of new: −u = log z therefore z = e−u ; this is onetoone everywhere. (3)
plugging in (since e−u > 0 for all u, we must plug it into λ exp(−λz )) gives . . . . (4) the derivative of
z = e−u is −e−u , taking absolute values gives the Jacobian factor e−u . Plugging in and multiplying
−u
gives the density of u: fu (u) = λ exp(−λe−u )e−u = λe−u−λe , and using λ exp(−u) = exp(µ − u)
this simpliﬁes to the formula above.
Alternative without transformation rule for densities: Fu (u) = Pr[u≤u] = Pr[− log z ≤u] =
−u
+∞
Pr[log z ≥ − u] = Pr[z ≥e−u ] = −u λe−λz dz = −e−λz +∞ = e−λe , now diﬀerentiate.
e−u
e Problem 54. 4 points Assume the random variable z has the exponential distribution with λ = 1, i.e., its density function is fz (z ) = exp(−z ) for z ≥ 0 and 0
√
for z < 0. Deﬁne u = z . Compute the density function of u.
√
Answer. (1) A = {u : u ≥ 0} since always denotes the nonnegative square root; (2) Express
2 , this is onetoone on A (but not onetoone on all of R);
old variable in terms of new: z = u
(3) then the derivative is 2u, which is nonnegative as well, no absolute values are necessary; (4)
multiplying gives the density of u: fu (u) = 2u exp(−u2 ) if u ≥ 0 and 0 elsewhere. 3.7. EXAMPLE: BINOMIAL VARIABLE 33 3.7. Example: Binomial Variable
Go back to our Bernoulli trial with parameters p and n, and deﬁne a random
variable x which represents the number of successes. Then the probability mass
function of x is
(3.7.1) px (k ) = Pr[x=k ] = nk
p (1 − p)(n−k)
k k = 0, 1, 2, . . . , n Proof is simple, every subset of k elements represents one possibility of spreading
out the k successes.
We will call any observed random variable a statistic. And we call a statistic t
suﬃcient for a parameter θ if and only if for any event A and for any possible value
t of t, the conditional probability Pr[At≤t] does not involve θ. This means: after
observing t no additional information can be obtained about θ from the outcome of
the experiment.
Problem 55. Show that x, the number of successes in the Bernoulli trial with
parameters p and n, is a suﬃcient statistic for the parameter p (the probability of
success), with n, the number of trials, a known ﬁxed number.
Answer. Since the distribution of x is discrete, it is suﬃcient to show that for any given k,
Pr[Ax=k] does not involve p whatever the event A in the Bernoulli trial. Furthermore, since the
Bernoulli trial with n tries is ﬁnite, we only have to show it if A is an elementary event in F , i.e.,
an event consisting of one element. Such an elementary event would be that the outcome of the
trial has a certain given sequence of successes and failures. A general A is the ﬁnite disjoint union
of all elementary events contained in it, and if the probability of each of these elementary events
does not depend on p, then their sum does not either.
Now start with the deﬁnition of conditional probability
(3.7.2) Pr[Ax=k] = Pr[A ∩ {x=k}]
.
Pr[x=k] If A is an elementary event whose number of sucesses is not k, then A ∩ {x=k} = ∅, therefore its
probability is 0, which does not involve p. If A is an elementary event which has k successes, then
A ∩ {x=k} = A, which has probability pk (1 − p)n−k . Since Pr[{x=k}] = n pk (1 − p)n−k , the
k
terms in formula (3.7.2) that depend on p cancel out, one gets Pr[Ax=k] = 1/
no p in that formula. n
k . Again there is Problem 56. You perform a Bernoulli experiment, i.e., an experiment which
can only have two outcomes, success s and failure f . The probability of success is p.
• a. 3 points You make 4 independent trials. Show that the probability that the
ﬁrst trial is successful, given that the total number of successes in the 4 trials is 3,
is 3/4.
Answer. Let B = {sf f f, sf f s, sf sf, sf ss, ssf f, ssf s, sssf, ssss} be the event that the ﬁrst
trial is successful, and let {x=3} = {f sss, sf ss, ssf s, sssf } be the event that there are 3 successes,
it has 4 = 4 elements. Then
3
(3.7.3) Pr[B x=3] = Pr[B ∩ {x=3}]
Pr[x=3] Now B ∩ {x=3} = {sf ss, ssf s, sssf }, which has 3 elements. Therefore we get
(3.7.4) Pr[B x=3] = • b. 2 points Discuss this result. 3 · p3 (1 − p)
3
=.
4 · p3 (1 − p)
4 34 3. RANDOM VARIABLES Answer. It is signiﬁcant that this probability is independent of p. I.e., once we know how
many successes there were in the 4 trials, knowing the true p does not help us computing the
probability of the event. From this also follows that the outcome of the event has no information
about p. The value 3/4 is the same as the unconditional probability if p = 3/4. I.e., whether we
know that the true frequency, the one that holds in the long run, is 3/4, or whether we know that
the actual frequency in this sample is 3/4, both will lead us to the same predictions regarding the
ﬁrst throw. But not all conditional probabilities are equal to their unconditional counterparts: the
conditional probability to get 3 successes in the ﬁrst 4 trials is 1, but the unconditional probability
is of course not 1. 3.8. Pitfalls of Data Reduction: The Ecological Fallacy
The nineteenthcentury sociologist Emile Durkheim collected data on the frequency of suicides and the religious makeup of many contiguous provinces in Western Europe. He found that, on the average, provinces with greater proportions of
Protestants had higher suicide rates and those with greater proportions of Catholics
lower suicide rates. Durkheim concluded from this that Protestants are more likely
to commit suicide than Catholics. But this is not a compelling conclusion. It may
have been that Catholics in predominantly Protestant provinces were taking their
own lives. The oversight of this logical possibility is called the “Ecological Fallacy”
[Sel58].
This seems like a farfetched example, but arguments like this have been used to
discredit data establishing connections between alcoholism and unemployment etc.
as long as the unit of investigation is not the individual but some aggregate.
One study [RZ78] found a positive correlation between driver education and
the incidence of fatal automobile accidents involving teenagers. Closer analysis
showed that the net eﬀect of driver education was to put more teenagers on the
road and therefore to increase rather than decrease the number of fatal crashes involving teenagers.
Problem 57. 4 points Assume your data show that counties with high rates of
unemployment also have high rates of heart attacks. Can one conclude from this that
the unemployed have a higher risk of heart attack? Discuss, besides the “ecological
fallacy,” also other objections which one might make against such a conclusion.
Answer. Ecological fallacy says that such a conclusion is only legitimate if one has individual
data. Perhaps a rise in unemployment is associated with increased pressure and increased workloads
among the employed, therefore it is the employed, not the unemployed, who get the heart attacks.
Even if one has individual data one can still raise the following objection: perhaps unemployment
and heart attacks are both consequences of a third variable (both unemployment and heart attacks
depend on age or education, or freezing weather in a farming community causes unemployment for
workers and heart attacks for the elderly). But it is also possible to commit the opposite error and rely too much on individual data and not enough on “neighborhood eﬀects.” In a relationship between
health and income, it is much more detrimental for your health if you are poor in a
poor neighborhood, than if you are poor in a rich neighborhood; and even wealthy
people in a poor neighborhood do not escape some of the health and safety risks
associated with this neighborhood.
Another pitfall of data reduction is Simpson’s paradox. According to table 1,
the new drug was better than the standard drug both in urban and rural areas. But
if you aggregate over urban and rural areas, then it looks like the standard drug was
better than the new drug. This is an artiﬁcial example from [Spr98, p. 360]. 3.10. LOCATION AND DISPERSION PARAMETERS 35 Responses in Urban and Rural Areas to Each of Two Drugs
Standard Drug
New Drug
Urban Rural Urban
Rural
No Eﬀect
500
350
1050
120
Cure
100
350
359
180
Table 1. Disaggregated Results of a New Drug
Response to Two Drugs
Standard Drug New Drug
No Eﬀect
850
1170
Cure
450
530
Table 2. Aggregated Version of Table 1 3.9. Independence of Random Variables
The concept of independence can be extended to random variables: x and y are
independent if all events that can be deﬁned in terms of x are independent of all
events that can be deﬁned in terms of y , i.e., all events of the form {ω ∈ U : x(ω ) ∈
C } are independent of all events of the form {ω ∈ U : y (ω ) ∈ D} with arbitrary
(measurable) subsets C, D ⊂ R. Equivalent to this is that all events of the sort x≤a
are independent of all events of the sort y ≤b.
Problem 58. 3 points The simplest random variables are indicator functions,
i.e., functions which can only take the values 0 and 1. Assume x is indicator function
of the event A and y indicator function of the event B , i.e., x takes the value 1 if A
occurs, and the value 0 otherwise, and similarly with y and B . Show that according
to the above deﬁnition of independence, x and y are independent if and only if the
events A and B are independent. (Hint: which are the only two events, other than
the certain event U and the null event ∅, that can be deﬁned in terms of x)?
Answer. Only A and A . Therefore we merely need the fact, shown in Problem 35, that if A
and B are independent, then also A and B are independent. By the same argument, also A and
B are independent, and A and B are independent. This is all one needs, except the observation
that every event is independent of the certain event and the null event. 3.10. Location Parameters and Dispersion Parameters of a Random
Variable
3.10.1. Measures of Location. A location parameter of random variables is
a parameter which increases by c if one adds the constant c to the random variable.
The expected value is the most important location parameter. To motivate it,
assume x is a discrete random variable, i.e., it takes the values x1 , . . . , xr with probr
abilities p1 , . . . , pr which sum up to one:
i=1 pi = 1. x is observed n times independently. What can we expect the average value of x to be? For this we ﬁrst need
a formula for this average: if ki is the number of times that x assumed the value
xi (i = 1, . . . , r) then
ki = n, and the average is k1 x1 + · · · + kn xn . With an
n
n
appropriate deﬁnition of convergence, the relative frequencies ki converge towards
n
pi . Therefore the average converges towards p1 x1 + · · · + pn xn . This limit is the
expected value of x, written as
(3.10.1) E[x] = p1 x1 + · · · + pn xn . Problem 59. Why can one not use the usual concept of convergence here? 36 3. RANDOM VARIABLES Answer. Because there is no guarantee that the sample frequencies converge. It is not physically impossible (although it is highly unlikely) that certain outcome will never be realized. Note the diﬀerence between the sample mean, i.e., the average measured in a
given sample, and the “population mean” or expected value. The former is a random
variable, the latter is a parameter. I.e., the former takes on a diﬀerent value every
time the experiment is performed, the latter does not.
Note that the expected value of the number of dots on a die is 3.5, which is not
one of the possible outcomes when one rolls a die.
Expected value can be visualized as the center of gravity of the probability mass.
If one of the tails has its weight so far out that there is no ﬁnite balancing point then
the expected value is inﬁnite of minus inﬁnite. If both tails have their weights so far
out that neither one has a ﬁnite balancing point, then the expected value does not
exist.
It is trivial to show that for a function g (x) (which only needs to be deﬁned for
those values which x can assume with nonzero probability), E[g (x)] = p1 g (x1 ) + · · · +
pn g (xn ).
Example of a countable probability mass distribution which has an inﬁnite ex∞1
a
pected value: Pr[x = x] = x2 for x = 1, 2, . . .. (a is the constant 1
i=1 i2 .) The
∞ expected value of x would be i=1 a , which is inﬁnite. But if the random variable
i
is bounded, then its expected value exists.
The expected value of a continuous random variable is deﬁned in terms of its
density function:
+∞ (3.10.2) E[x] = xfx (x) dx
−∞ It can be shown that for any function g (x) deﬁned for all those x for which fx (x) = 0
follows:
(3.10.3) E[g (x)] = g (x)fx (x) dx
fx (x)=0 Here the integral is taken over all the points which have nonzero density, instead of
the whole line, because we did not require that the function g is deﬁned at the points
where the density is zero.
Problem 60. Let the random variable x have the Cauchy distribution, i.e., its
density function is
(3.10.4) fx (x) = 1
π (1 + x2 ) Show that x does not have an expected value.
Answer.
(3.10.5) x dx
1
=
π (1 + x2 )
2π 2x dx
1
=
1 + x2
2π d(x2 )
1
=
ln(1 + x2 )
1 + x2
2π Rules about how to calculate with expected values (as long as they exist):
(3.10.6)
(3.10.7)
(3.10.8) E[c] = c if c is a constant
E[ch] = c E[h]
E[h + j ] = E[h] + E[j ] 3.10. LOCATION AND DISPERSION PARAMETERS 37 and if the random variables h and j are independent, then also
(3.10.9) E[hj ] = E[h] E[j ]. Problem 61. 2 points You make two independent trials of a Bernoulli experiment with success probability θ, and you observe t, the number of successes. Compute
the expected value of t3 . (Compare also Problem 169.)
Answer. Pr[t = 0] = (1 − θ)2 ; Pr[t = 1] = 2θ(1 − θ); Pr[t = 2] = θ2 . Therefore an application
of (3.10.1) gives E[t3 ] = 03 · (1 − θ)2 + 13 · 2θ(1 − θ) + 23 · θ2 = 2θ + 6θ2 . Theorem 3.10.1. Jensen’s Inequality: Let g : R → R be a function which is
convex on an interval B ⊂ R, which means
(3.10.10) g (λa + (1 − λ)b) ≤ λg (a) + (1 − λ)g (b) for all a, b ∈ B . Furthermore let x : R → R be a random variable so that Pr[x ∈
B ] = 1. Then g (E[x]) ≤ E[g (x)].
Proof. The Jensen inequality holds with equality if h(x) is a linear function (with a constant term), i.e., in this case, E[h(x)] = h(E[x]). (2) Therefore
Jensen’s inequality is proved if we can ﬁnd a linear function h with the two properties h(E[x]) = g (E[x]), and h(x) ≤ g (x) for all other x—because with such a
h, E[g (x)] ≥ E[h(x)] = h(E[x]). (3) The existence of such a h follows from convexity. Since g is convex, for every point a ∈ B there is a number β so that
g (x) ≥ g (a) + β (x − a). This β is the slope of g if g is diﬀerentiable, and otherwise it is some number between the left and the right derivative (which both always
exist for a convex function). We need this for a = E[x].
This existence is the deepest part of this proof. We will not prove it here, for a
proof see [Rao73, pp. 57, 58]. One can view it as a special case of the separating
hyperplane theorem.
Problem 62. Use Jensen’s inequality to show that (E[x])2 ≤ E[x2 ]. You are
allowed to use, without proof, the fact that a function is convex on B if the second
derivative exists on B and is nonnegative.
Problem 63. Show that the expected value of the empirical distribution of a
sample is the sample mean.
Other measures of locaction: The median is that number m for which there is
as much probability mass to the left of m as to the right, i.e.,
(3.10.11) Pr[x≤m] = 1
2 or, equivalently, Fx (m) = 1
.
2 It is much more robust with respect to outliers than the mean. If there is more than
one m satisfying (3.10.11), then some authors choose the smallest (in which case the
median is a special case of the quantile function m = F −1 (1/2)), and others the
average between the biggest and smallest. If there is no m with property (3.10.11),
i.e., if the cumulative distribution function jumps from a value that is less than 1 to
2
a value that is greater than 1 , then the median is this jump point.
2
The mode is the point where the probability mass function or the probability
density function is highest. 38 3. RANDOM VARIABLES 3.10.2. Measures of Dispersion. Here we will discuss variance, standard deviation, and quantiles and percentiles: The variance is deﬁned as
var[x] = E[(x − E[x])2 ], (3.10.12)
but the formula var[x] = E[x2 ] − (E[x])2 (3.10.13) is usually more convenient.
How to calculate with variance?
(3.10.14) var[ax] = a2 var[x] (3.10.15) var[x + c] = var[x] if c is a constant (3.10.16) var[x + y ] = var[x] + var[y ] if x and y are independent. Note that the variance is additive only when x and y are independent; the expected
value is always additive.
Problem 64. Here we make the simple step from the deﬁnition of the variance
to the usually more convenient formula (3.10.13).
• a. 2 points Derive the formula var[x] = E[x2 ] − (E[x])2 from the deﬁnition of a
variance, which is var[x] = E[(x − E[x])2 ]. Hint: it is convenient to deﬁne µ = E[x].
Write it down carefully, you will lose points for missing or unbalanced parentheses
or brackets.
Answer. Here it is side by side with and without the notation E[x] = µ:
var[x] = E[(x − E[x])2 ] var[x] = E[(x − µ)2 ] = E[x2 − 2x(E[x]) + (E[x])2 ]
(3.10.17) 2 2 = E[x ] − 2(E[x]) + (E[x])
= E[x2 ] − (E[x])2 . 2 = E[x2 − 2xµ + µ2 ]
= E[x2 ] − 2µ2 + µ2
= E[x2 ] − µ2 . • b. 1 point Assume var[x] = 3, var[y ] = 2, x and y are independent. Compute
var[−x], var[3y + 5], and var[x − y ].
Answer. 3, 18, and 5. Problem 65. If all y i are independent with same variance σ 2 , then show that y
¯
has variance σ 2 /n.
The standard deviation is the square root of the variance. Often preferred because has same scale as x. The variance, on the other hand, has the advantage of a
simple addition rule.
Standardization: if the random variable x has expected value µ and standard
deviation σ , then z = x−µ has expected value zero and variance one.
σ
An αth quantile or a 100αth percentile of a random variable x was already
deﬁned previously to be the smallest number x so that Pr[x≤x] ≥ α.
3.10.3. MeanVariance Calculations. If one knows mean and variance of a
random variable, one does not by any means know the whole distribution, but one
has already some information. For instance, one can compute E[y 2 ] from it, too.
Problem 66. 4 points Consumer M has an expected utility function for money
income u(x) = 12x − x2 . The meaning of an expected utility function is very simple:
if he owns an asset that generates some random income y , then the utility he derives
from this asset is the expected value E[u(y )]. He is contemplating acquiring two 3.10. LOCATION AND DISPERSION PARAMETERS 39 assets. One asset yields an income of 4 dollars with certainty. The other yields an
expected income of 5 dollars with standard deviation 2 dollars. Does he prefer the
certain or the uncertain asset?
Answer. E[u(y )] = 12 E[y ] − E[y 2 ] = 12 E[y ] − var[y ] − (E[y ])2 . Therefore the certain asset
gives him utility 48 − 0 − 16 = 32, and the uncertain one 60 − 4 − 25 = 31. He prefers the certain
asset. 3.10.4. Moment Generating Function and Characteristic Function. Here
we will use the exponential function ex , also often written exp(x), which has the two
2
3
x
properties: ex = limn→∞ (1 + n )n (Euler’s limit), and ex = 1 + x + x + x + · · · .
2!
3!
Many (but not all) random variables x have a moment generating function mx (t)
for certain values of t. If they do for t in an open interval around zero, then their
distribution is uniquely determined by it. The deﬁnition is
mx (t) = E[etx ] (3.10.18) It is a powerful computational device.
The moment generating function is in many cases a more convenient characterization of the random variable than the density function. It has the following
uses:
1. One obtains the moments of x by the simple formula
E[xk ] = (3.10.19) dk
mx (t)
dtk .
t=0 Proof:
t3 x3
t2 x2
+
+ ···
2!
3!
t2
t3
(3.10.21)
mx (t) = E[etx ] = 1 + t E[x] + E[x2 ] + E[x3 ] + · · ·
2!
3!
d
t2
(3.10.22)
mx (t) = E[x] + t E[x2 ] + E[x3 ] + · · ·
dt
2!
d2
(3.10.23)
mx (t) = E[x2 ] + t E[x3 ] + · · · etc.
dt2
2. The moment generating function is also good for determining the probability
distribution of linear combinations of independent random variables.
a. it is easy to get the m.g.f. of λx from the one of x:
etx = 1 + tx + (3.10.20) (3.10.24) mλx (t) = mx (λt) because both sides are E[eλtx ].
b. If x, y independent, then
(3.10.25) mx+y (t) = mx (t)my (t). The proof is simple:
(3.10.26) E[et(x+y) ] = E[etx ety ] = E[etx ] E[ety ] due to independence.
√
The characteristic function is deﬁned as ψx (t) = E[eitx ], where i = −1. It has
the disadvantage that it involves complex numbers, but it has the advantage that it
always exists, since exp(ix) = cos x + i sin x. Since cos and sin are both bounded,
they always have an expected value.
And, as its name says, the characteristic function characterizes the probability
distribution. Analytically, many of its properties are similar to those of the moment
generating function. 40 3. RANDOM VARIABLES 3.11. Entropy
3.11.1. Deﬁnition of Information. Entropy is the average information gained
by the performance of the experiment. The actual information yielded by an event
A with probabbility Pr[A] = p = 0 is deﬁned as follows:
(3.11.1) I [A] = log2 1
Pr[A] This is simply a transformation of the probability, and it has the dual interpretation
of either how unexpected the event was, or the informaton yielded by the occurrense
of event A. It is characterized by the following properties [AD75, pp. 3–5]:
• I [A] only depends on the probability of A, in other words, the information
content of a message is independent of how the information is coded.
• I [A] ≥ 0 (nonnegativity), i.e., after knowing whether A occurred we are no
more ignorant than before.
• If A and B are independent then I [A ∩ B ] = I [A] + I [B ] (additivity for
independent events). This is the most important property.
• Finally the (inessential) normalization that if Pr[A] = 1/2 then I [A] = 1,
i.e., a yesorno decision with equal probability (coin ﬂip) is one unit of
information.
Note that the information yielded by occurrence of the certain event is 0, and that
yielded by occurrence of the impossible event is ∞.
But the important informationtheoretic results refer to average, not actual,
information, therefore let us deﬁne now entropy:
3.11.2. Deﬁnition of Entropy. The entropy of a probability ﬁeld (experiment) is a measure of the uncertainty prevailing before the experiment is performed,
or of the average information yielded by the performance of this experiment. If the
set U of possible outcomes of the experiment has only a ﬁnite number of diﬀerent elements, say their number is n, and the probabilities of these outcomes are p1 , . . . , pn ,
then the Shannon entropy H[F ] of this experiment is deﬁned as
(3.11.2) H[F ]
=
bits n pk log2
k=1 1
pk This formula uses log2 , logarithm with base 2, which can easily be computed from the
natural logarithms, log2 x = log x/ log 2. The choice of base 2 is convenient because
in this way the most informative Bernoulli experiment, that with success probability
p = 1/2 (coin ﬂip), has entropy 1. This is why one says: “the entropy is measured
in bits.” If one goes over to logarithms of a diﬀerent base, this simply means that
one measures entropy in diﬀerent units. In order to indicate this dependence on the
measuring unit, equation (3.11.2) was written as the deﬁnition H[F ] instead of H[F ]
bits
itself, i.e., this is the number one gets if one measures the entropy in bits. If one uses
natural logarithms, then the entropy is measured in “nats.”
Entropy can be characterized axiomatically by the following axioms [Khi57]:
• The uncertainty associated with a ﬁnite complete scheme takes its largest
value if all events are equally likely, i.e., H(p1 , . . . , pn ) ≤ H(1/n, . . . , 1/n).
• The addition of an impossible event to a scheme does not change the amount
of uncertainty.
• Composition Law: If the possible outcomes are arbitrarily combined into
m groups W 1 = X 11 ∪ · · · ∪ X 1k1 , W 2 = X 21 ∪ · · · ∪ X 2k2 , . . . , W m = 3.11. ENTROPY 41 X m1 ∪ · · · ∪ X mkm , with corresponding probabilities w1 = p11 + · · · + p1k1 ,
w2 = p21 + · · · + p2k2 , . . . , wm = pm1 + · · · + pmkm , then
H(p1 , . . . , pn ) = H(w1 , . . . , wn ) +
+ w1 H (p11 /w1 + · · · + p1k1 /w1 ) +
+ w2 H (p21 /w2 + · · · + p2k2 /w2 ) + · · · +
+ wm H (pm1 /wm + · · · + pmkm /wm ).
Since pij /wj = Pr[X ij Wj ], the composition law means: if you ﬁrst learn half the
outcome of the experiment, and then the other half, you will in the average get as
much information as if you had been told the total outcome all at once.
The entropy of a random variable x is simply the entropy of the probability
ﬁeld induced by x on R. It does not depend on the values x takes but only on the
probabilities. For discretely distributed random variables it can be obtained by the
following “eerily selfreferential” prescription: plug the random variable into its own
probability mass function and compute the expected value of the negative logarithm
of this, i.e.,
H[x]
= E[− log2 px (x)]
bits
One interpretation of the entropy is: it is the average number of yesorno questions necessary to describe the outcome of the experiment. For instance, consider an
experiment which has 32 diﬀerent outcomes occurring with equal probabilities. The
entropy is (3.11.3) (3.11.4) H
=
bits 32 i=1 1
log2 32 = log2 32 = 5
32 i.e., H = 5 bits which agrees with the number of bits necessary to describe the outcome.
Problem 67. Design a questioning scheme to ﬁnd out the value of an integer
between 1 and 32, and compute the expected number of questions in your scheme if
all numbers are equally likely.
Answer. In binary digits one needs a number of length 5 to describe a number between 0 and
31, therefore the 5 questions might be: write down the binary expansion of your number minus 1.
Is the ﬁrst binary digit in this expansion a zero, then: is the second binary digit in this expansion a
zero, etc. Formulated without the use of binary digits these same questions would be: is the number
between 1 and 16?, then: is it between 1 and 8 or 17 and 24?, then, is it between 1 and 4 or 9 and
12 or 17 and 20 or 25 and 28?, etc., the last question being whether it is odd. Of course, you can
formulate those questions conditionally: First: between 1 and 16? if no, then second: between 17
and 24? if yes, then second: between 1 and 8? Etc. Each of these questions gives you exactly the
entropy of 1 bit. Problem 68. [CT91, example 1.1.2 on p. 5] Assume there is a horse race
with eight horses taking part. The probabilities for winning for the eight horses are
1
1
1
1
1111
2 , 4 , 8 , 16 , 64 , 64 , 64 , 64 .
• a. 1 point Show that the entropy of the horse race is 2 bits.
Answer.
H
1
1
1
1
4
= log2 2 + log2 4 + log2 8 +
log2 16 +
log2 64 =
bits
2
4
8
16
64
1
1
3
1
3
4+4+3+2+3
=++++=
=2
2
2
8
4
8
8 42 3. RANDOM VARIABLES • b. 1 point Suppose you want to send a binary message to another person
indicating which horse won the race. One alternative is to assign the bit strings 000,
001, 010, 011, 100, 101, 110, 111 to the eight horses. This description requires 3 bits
for any of the horses. But since the win probabilities are not uniform, it makes sense
to use shorter descriptions for the horses more likely to win, so that we achieve
a lower expected value of the description length. For instance, we could use the
following set of bit strings for the eight horses: 0, 10, 110, 1110, 111100, 111101,
111110, 111111. Show that the the expected length of the message you send to your
friend is 2 bits, as opposed to 3 bits for the uniform code. Note that in this case the
expected value of the description length is equal to the entropy.
Answer. The math is the same as in the ﬁrst part of the question:
1
1
1
1
1
1
1
3
1
3
4+4+3+2+3
·1+ ·2+ ·3+
·4+4·
·6= + + + + =
=2
2
4
8
16
64
2
2
8
4
8
8 Problem 69. [CT91, example 2.1.2 on pp. 14/15]: The experiment has four
possible outcomes; outcome x=a occurs with probability 1/2, x=b with probability
1/4, x=c with probability 1/8, and x=d with probability 1/8.
• a. 2 points The entropy of this experiment (in bits) is one of the following
three numbers: 11/8, 7/4, 2. Which is it?
• b. 2 points Suppose we wish to determine the outcome of this experiment with
the minimum number of questions. An eﬃcient ﬁrst question is “Is x=a?” This
splits the probability in half. If the answer to the ﬁrst question is no, then the second
question can be “Is x=b?” The third question, if it is necessary, can then be: “Is
x=c?” Compute the expected number of binary questions required.
c. 2 points Show that the entropy gained by each question is 1 bit.
• d. 3 points Assume we know about the ﬁrst outcome that x=a. What is the
entropy of the remaining experiment (i.e., under the conditional probability)?
• e. 5 points Show in this example that the composition law for entropy holds.
Problem 70. 2 points In terms of natural logarithms equation (3.11.4) deﬁning
entropy reads
H
1
=
bits
ln 2 (3.11.5) n pk ln
k=1 1
.
pk Compute the entropy of (i.e., the average informaton gained by) a roll of an unbiased
die.
Answer. Same as the actual information gained, since each outcome is equally likely:
(3.11.6) H
1
=
bits
ln 2 1
1
ln 6 + · · · + ln 6
6
6 = ln 6
= 2.585
ln 2 • a. 3 points How many questions does one need in the average to determine the
outcome of the roll of an unbiased die? In other words, pick a certain questioning
scheme (try to make it eﬃcient) and compute the average number of questions if
this scheme is followed. Note that this average cannot be smaller than the entropy
H /bits, and if one chooses the questions optimally, it is smaller than H /bits + 1. 3.11. ENTROPY 43 Answer. First question: is it bigger than 3? Second question: is it even? Third question (if
necessary): is it a multiple of 3? In this scheme, the number of questions for the six faces of the
4
die are 3, 2, 3, 3, 2, 3, therefore the average is 6 · 3 + 2 · 2 = 2 2 . Also optimal: (1) is it bigger than
6
3
2? (2) is it odd? (3) is it bigger than 4? Gives 2, 2, 3, 3, 3, 3. Also optimal: 1st question: is it 1 or
2? If anser is no, then second question is: is it 3 or 4?; otherwise go directly to the third question:
is it odd or even? The steamroller approach: Is it 1? Is it 2? etc. gives 1, 2, 3, 4, 5, 5 with expected
number 3 1 . Even this is here < 1 + H /bits.
3 Problem 71.
• a. 1 point Compute the entropy of a roll of two unbiased dice if they are
distinguishable.
Answer. Just twice the entropy from Problem 70.
1
1
1
H
=
ln 36 + · · · +
ln 36
(3.11.7)
bits
ln 2 36
36 = ln 36
= 5.170
ln 2 • b. Would you expect the entropy to be greater or less in the more usual case
that the dice are indistinguishable? Check your answer by computing it.
Answer. If the dice are indistinguishable, then one gets less information, therefore the experiment has less entropy. One has six like pairs with probability 1/36 and 6 · 5/2 = 15 unlike pairs
with probability 2/36 = 1/18 each. Therefore the average information gained is
(3.11.8) 1
1
1
H
=
6·
ln 36 + 15 ·
ln 18
bits
ln 2
36
18 = 1
ln 2 1
5
ln 36 + ln 18
6
6 = 4.337 • c. 3 points Note that the diﬀerence between these two entropies is 5/6 = 0.833.
How can this be explained?
Answer. This is the composition law (??) in action. Assume you roll two dice which you ﬁrst
consider indistinguishable and afterwards someone tells you which is which. How much information
do you gain? Well, if the numbers are the same, then telling you which die is which does not give
you any information, since the outcomes of the experiment are deﬁned as: which number has the
ﬁrst die, which number has the second die, regardless of where on the table the dice land. But if
the numbers are diﬀerent, then telling you which is which allows you to discriminate between two
outcomes both of which have conditional probability 1/2 given the outcome you already know; in
this case the information you gain is therefore 1 bit. Since the probability of getting two diﬀerent
numbers is 5/6, the expected value of the information gained explains the diﬀerence in entropy.
1
All these deﬁnitions use the convention 0 log 0 = 0, which can be justiﬁed by the
following continuity argument: Deﬁne the function, graphed in Figure 3: (3.11.9) η (w) = w log
0 1
w if w > 0
if w = 0. η is continuous for all w ≥ 0, even at the boundary point w = 0. Diﬀerentiation gives
η (w) = −(1 + log w), and η (w) = −w−1 . The function starts out at the origin with
a vertical tangent, and since the second derivative is negative, it is strictly concave
for all w > 0. The deﬁnition of strict concavity is η (w) < η (v ) + (w − v )η (v ) for
w = v , i.e., the function lies below all its tangents. Substituting η (v ) = −(1 + log v )
and simplifying gives w − w log w ≤ v − w log v for v, w > 0. One veriﬁes that this
inequality also holds for v, w ≥ 0.
Problem 72. Make a complete proof, discussing all possible cases, that for
v, w ≥ 0 follows
(3.11.10) w − w log w ≤ v − w log v 44 3. RANDOM VARIABLES Answer. We already know it for v, w > 0. Now if v = 0 and w = 0 then the equation reads
0 ≤ 0; if v > 0 and w = 0 the equation reads 0 ≤ v , and if w > 0 and v = 0 then the equation reads
w − w log w ≤ +∞. 3.11.3. How to Keep Forecasters Honest. This mathematical result allows
an interesting alternative mathematical characterization of entropy. Assume Anita
performs a Bernoulli experiment whose success probability she does not know but
wants to know. Clarence knows this probability but is not on very good terms with
Anita; therefore Anita is unsure that he will tell the truth if she asks him.
Anita knows “how to keep forecasters honest.” She proposes the following deal
to Clarence: “you tell me the probability q , and after performing my experiment I
pay you the amount log2 (q ) if the experiment is a success, and log2 (1 − q ) if it is a
failure. If Clarence agrees to this deal, then telling Anita that value q which is the
true success probability of the Bernoulli experiment maximizes the expected value of
his payoﬀ. And the maximum expected value of this payoﬀ is exactly the negative
of the entropy of the experiment.
Proof: Assume the correct value of the probability is p, and the number Clarence
tells Tina is q . For every p, q between 0 and 1 we have to show:
p log p + (1 − p) log(1 − p) ≥ p log q + (1 − p) log(1 − q ). (3.11.11) For this, plug w = p and v = q as well as w = 1 − p and v = 1 − q into equation
(3.11.10) and add.
w log 1
w . 1
e d d
.
.
. ... .......
............................... ...
.................................... d
..
.....
....
.....
.....
....
...
....d
....
...
... ....
....
...
...
....
..
....
..
..
....
....
.
..
....
d..... ....
..
..
..
...
...
..
...
...
..
d......... .
.
..
.
...
.
...
.
..
.
..
.
d...........
. .
.
.
.
..
..
..
.
.
.
.
.
d......... .
.
T 1
e Figure 3. η : w → w log 1 1
w E ..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
. w is continuous at 0, and concave everywhere 3.11.4. The Inverse Problem. Now let us go over to the inverse problem:
computing those probability ﬁelds which have maximum entropy subject to the information you have.
If you know that the experiment has n diﬀerent outcomes, and you do not know
the probabilities of these outcomes, then the maximum entropy approach amounts
to assigning equal probability 1/n to each outcome.
Problem 73. (Not eligible for inclass exams) You are playing a slot machine.
Feeding one dollar to this machine leads to one of four diﬀerent outcomes: E 1 :
machine returns nothing, i.e., you lose $1. E 2 : machine returns $1, i.e., you lose 3.11. ENTROPY 45 nothing and win nothing. E 3 : machine returns $2, i.e., you win $1. E 4 : machine
returns $10, i.e., you win $9. Events E i occurs with probability pi , but these probabilities are unknown. But due to a new “TruthinGambling Act” you ﬁnd a sticker
on the side of the machine which says that in the long run the machine pays out only
$0.90 for every dollar put in. Show that those values of p1 , p2 , p3 , and p4 which
maximize the entropy (and therefore make the machine most interesting) subject to
the constraint that the expected payoﬀ per dollar put in is $0.90, are p1 = 0.4473,
p2 = 0.3158, p3 = 0.2231, p4 = 0.0138.
Answer. Solution is derived in [Rie85, pp. 68/9 and 74/5], and he refers to [Rie77]. You
have to maximize −
pn log pn subject to
pn = 1 and
cn pn = d. In our case c1 = 0, c2 = 1,
c3 = 2, and c4 = 10, and d = 0.9, but the treatment below goes through for arbitrary ci as long as
not all of them are equal. This case is discussed in detail in the answer to Problem 74. • a. Diﬃcult: Does the maximum entropy approach also give us some guidelines
how to select these probabilities if all we know is that the expected value of the payout
rate is smaller than 1?
Answer. As shown in [Rie85, pp. 68/9 and 74/5], one can give the minimum value of the
entropy for all distributions with payoﬀ smaller than 1: H < 1.6590, and one can also give some
bounds for the probabilities: p1 > 0.4272, p2 < 0.3167, p3 < 0.2347, p4 < 0.0214. • b. What if you also know that the entropy of this experiment is 1.5?
Answer. This was the purpose of the paper [Rie85]. Problem 74. (Not eligible for inclass exams) Let p1 , p2 , . . . , pn ( pi = 1) be
the proportions of the population of a city living in n residential colonies. The cost of
living in colony i, which includes cost of travel from the colony to the central business
district, the cost of the time this travel consumes, the rent or mortgage payments,
and other costs associated with living in colony i, is represented by the monetary
amount ci . Without loss of generality we will assume that the ci are numbered in
such a way that c1 ≤ c2 ≤ · · · ≤ cn . We will also assume that the ci are not all
equal. We assume that the ci are known and that also the average expenditures on
travel etc. in the population is known; its value is d. One approach to modelling the
population distribution is to maximize the entropy subject to the average expenditures,
1
i.e., to choose p1 , p2 , . . . pn such that H = pi log pi is maximized subject to the two
constraints
pi = 1 and
pi ci = d. This would give the greatest uncertainty about
where someone lives.
• a. 3 points Set up the Lagrange function and show that
(3.11.12) pi = exp(−λci )
exp(−λci ) where the Lagrange multiplier λ must be chosen such that pi ci = d. Answer. The Lagrange function is
(3.11.13) L=− pn log pn − κ( pn − 1) − λ( cn pn − d) Partial diﬀerentiation with respect to pi gives the ﬁrst order conditions
(3.11.14) − log pi − 1 − κ − λci = 0. Therefore pi = exp(−κ − 1) exp(−λci ). Plugging this into the ﬁrst constraint gives 1 =
pi =
1
exp(−κ − 1)
. This constraint therefore deﬁnes κ
exp(−λci ) or exp(−κ − 1) =
exp(−λci ) uniquely, and we can eliminate κ from the formula for pi :
(3.11.15) pi = exp(−λci )
exp(−λci ) 46 3. RANDOM VARIABLES Now all the pi depend on the same unknown λ, and this λ must be chosen such that the second
constraint holds. This is the MaxwellBoltzmann distribution if µ = kT where k is the Boltzmann
constant and T the temperature. • b. 2 points Here is a mathematical lemma needed for the next part: Prove that
for ai ≥ 0 and ci arbitrary follows
ai ai c2 ≥ ( ai ci )2 , and if all ai > 0 and
i
not all ci equal, then this inequality is strict.
Answer. By choosing the same subscripts in the second sum as in the ﬁrst we pair elements
of the ﬁrst sum with elements of the second sum:
(3.11.16)
c2 a j −
j ai
i ci a i
i j (c2 − ci cj ) a i a j
j cj a j =
j i,j but if we interchange i and j on the rhs we get
(3.11.17) (c2 − ci cj ) a i a j
i ( c 2 − cj c i ) a j a i =
i = i,j j,i Now add the righthand sides to get
(3.11.18)
2 c2 a j −
j ai
i j ci a i
i ( c 2 + c 2 − 2 ci c j ) a i a j =
i
j cj a j =
j i,j (ci − cj )2 a i a j ≥ 0
i,j • c. 3 points It is not possible to solve equations (3.11.12) analytically for λ, but
the following can be shown [Kap89, p. 310/11]: the function f deﬁned by
(3.11.19) f (λ) = ci exp(−λci )
exp(−λci ) is a strictly decreasing function which decreases from cn to c1 as λ goes from −∞
to ∞, and f (0) = c where c = (1/n) ci . We need that λ for which f (λ) = d, and
¯
¯
this equation has no real root if d < c1 or d > cn , it has a unique positive root if
c1 < d < c it has the unique root 0 for d = c, and it has a unique negative root for
¯
¯
c < d < cn . From this follows: as long as d lies between the lowest and highest cost,
¯
and as long as the cost numbers are not all equal, the pi are uniquely determined by
the above entropy maximization problem.
Answer. Here is the derivative; it is negative because of the mathematical lemma just shown: (3.11.20) f (λ) = u v − uv
=−
v2 exp(−λci ) c2 exp(−λci ) −
i
exp(−λci ) ci exp(−λci )
2 2 <0 Since c1 ≤ c2 ≤ · · · ≤ cn , it follows
(3.11.21) c1 = c1 exp(−λci )
exp(−λci ) ≤ ci exp(−λci )
exp(−λci ) ≤ cn exp(−λci )
exp(−λci ) = cn Now the statement about the limit can be shown if not all cj are equal, say c1 < ck+1 but c1 = ck .
The fraction can be written as
(3.11.22) kc1 exp(−λc1 ) +
k exp(−λc1 ) + n−k
c
exp(−λck+i )
i=1 k+i
i = 1n−k exp(−λck+i ) = kc1 + Since ck+i − c1 > 0, this converges towards c1 for λ → ∞. k+ n−k
c
exp(−λ(ck+i − c1 ))
i=1 k+i
n−k
exp(−λ(ck+i − c1 ))
i=1 3.11. ENTROPY 47 • d. 3 points Show that the maximum attained entropy is H = λd + k (λ) where
(3.11.23) exp(−λcj ) . k (λ) = log Although λ depends on d, show that ∂ H = λ, i.e., it is the same as if λ did not
∂d
depend on d. This is an example of the “envelope theorem,” and it also gives an
interpretation of λ.
Answer. We have to plug the optimal pi = exp(−λci )
exp(−λci ) into the formula for H = − pi log pi . For this note that − log pi = λci + k(λ) where k(λ) = log( exp(−λcj )) does not depend on i.
Therefore H =
pi (λci + k(λ)) = λ
pi ci + k(λ)
pi = λd + k(λ), and ∂ H = λ + d ∂λ + k (λ) ∂λ .
∂d
∂d
∂d
Now we need the derivative of k(λ), and we discover that k (λ) = −f (λ) where f (λ) was deﬁned in
(3.11.19). Therefore ∂ H = λ + (d − f (λ)) ∂λ = λ.
∂d
∂d • e. 5 points Now assume d is not known (but the ci are still known), i.e., we
know that (3.11.12) holds for some λ but we don’t know which. We want to estimate
this λ (and therefore all pi ) by taking a random sample of m people from that metropolitan area and asking them what their regional living expenditures are and where
they live. Assume xi people in this sample live in colony i. One way to estimate this
xi
λ would be to use the average consumption expenditure of the sample,
m ci , as an
estimate of the missing d in the above procedure, i.e., choose that λ which satisﬁes
xi
f (λ) =
m ci . Another procedure, which seems to make a better use of the information given by the sample, would be to compute the maximum likelihood estimator
of λ based on all xi . Show that these two estimation procedures are identical.
Answer. The xi have the multinomial distribution. Therefore, given that the proportion pi
of the population lives in colony i, and you are talking a random sample of size m from the whole
population, then the probability to get the outcome x1 , . . . , xn is
m!
px1 px2 · · · pxn
(3.11.24)
L=
n
x1 ! · · · xn ! 1 2
This is what we have to maximize, subject to the condition that the pi are an entropy maximizing
population distribution. Let’s take logs for computational simplicity:
log L = log m! − (3.11.25) log xj ! + xi log pi j All we know about the pi is that they must be some entropy maximizing probabilities, but we don’t
know yet which ones, i.e., they depend on the unknown λ. Therefore we need the formula again
− log pi = λci + k(λ) where k(λ) = log( exp(−λcj )) does not depend on i. This gives
(3.11.26) log L = log m!− log xj !−
j (for this last term remember that xi (λci +k(λ)) = log m!− log xj !−λ xi ci +k(λ)m j xi = m. Therefore the derivative is 1∂
xi
log L =
ci − f (λ)
(3.11.27)
m ∂λ
m
I.e., using the obvious estimate for d is the same as maximum likelihood under the assumption of
maximum entropy. This is a powerful estimation strategy. An article with sensational image reconstitutions using maximum entropy algorithms is [SG85, pp. 111, 112, 115, 116].
And [GJM96] applies maximum entropy methods to illposed or underdetermined
problems in econometrics! CHAPTER 4 Speciﬁc Random Variables
4.1. Binomial
We will begin with mean and variance of the binomial variable, i.e., the number
of successes in n independent repetitions of a Bernoulli trial (3.7.1). The binomial
variable has the two parameters n and p. Let us look ﬁrst at the case n = 1, in which
the binomial variable is also called indicator variable: If the event A has probability
p, then its complement A has the probability q = 1 − p. The indicator variable of
A, which assumes the value 1 if A occurs, and 0 if it doesn’t, has expected value p
and variance pq . For the binomial variable with n observations, which is the sum of
n independent indicator variables, the expected value (mean) is np and the variance
is npq .
Problem 75. The random variable x assumes the value a with probability p and
the value b with probability q = 1 − p. Show that var[x] = pq (a − b)2 .
Answer. E[x] = pa + qb; var[x] = E[x2 ] − (E[x])2 = pa2 + qb2 − (pa + qb)2 = (p − p2 )a2 −
2pqab + (q − q 2 )b2 = pq (a − b)2 . For this last equality we need p − p2 = p(1 − p) = pq . The Negative Binomial Variable is, like the binomial variable, derived from the
Bernoulli experiment; but one reverses the question. Instead of asking how many
successes one gets in a given number of trials, one asks, how many trials one must
make to get a given number of successes, say, r successes.
First look at r = 1. Let t denote the number of the trial at which the ﬁrst success
occurs. Then
Pr[t=n] = pq n−1 (4.1.1) (n = 1, 2, . . .). This is called the geometric probability.
Is the probability derived in this way σ additive? The sum of a geometrically
declining sequence is easily computed:
(4.1.2) 1 + q + q 2 + q 3 + · · · = s Now multiply by q : (4.1.3) q + q 2 + q 3 + · · · = qs (4.1.4) 1 = ps Now subtract and write 1 − q = p: Equation (4.1.4) means 1 = p + pq + pq 2 + · · · , i.e., the sum of all probabilities is
indeed 1.
Now what is the expected value of a geometric variable? Use deﬁnition of ex∞
pected value of a discrete variable: E[t] = p k=1 kq k−1 . To evaluate the inﬁnite
sum, solve (4.1.4) for s:
(4.1.5) s= 1
p ∞ or 1 + q + q2 + q3 + q4 · · · = qk =
k=0 49 1
1−q 50 4. SPECIFIC RANDOM VARIABLES and diﬀerentiate both sides with respect to q :
∞ (4.1.6) kq k−1 = 1 + 2q + 3q 2 + 4q 3 + · · · =
k=1 1
1
= 2.
(1 − q )2
p The expected value of the geometric variable is therefore E[t] = p
p2 1
= p. Problem 76. Assume t is a geometric random variable with parameter p, i.e.,
it has the values k = 1, 2, . . . with probabilities
(4.1.7) pt (k ) = pq k−1 , where q = 1 − p. The geometric variable denotes the number of times one has to perform a Bernoulli
experiment with success probability p to get the ﬁrst success.
• a. 1 point Given a positive integer n. What is Pr[t>n]? (Easy with a simple
trick!)
Answer. t>n means, the ﬁrst n trials must result in failures, i.e., Pr[t>n] = q n . Since
{t > n} = {t = n + 1} ∪ {t = n + 2} ∪ · · · , one can also get the same result in a more tedious way:
It is pq n + pq n+1 + pq n+2 + · · · = s, say. Therefore qs = pq n+1 + pq n+2 + · · · , and (1 − q )s = pq n ;
since p = 1 − q , it follows s = q n . • b. 2 points Let m and n be two positive integers with m < n. Show that
Pr[t=nt>m] = Pr[t=n − m].
Answer. Pr[t=nt>m] = Pr[t=n]
Pr[t>m] = pq n−1
qm = pq n−m−1 = Pr[t=n − m]. • c. 1 point Why is this property called the memoryless property of the geometric
random variable?
Answer. If you have already waited for m periods without success, the probability that success
will come in the nth period is the same as the probability that it comes in n − m periods if you
start now. Obvious if you remember that geometric random variable is time you have to wait until
1st success in Bernoulli trial. Problem 77. t is a geometric random variable as in the preceding problem. In
order to compute var[t] it is most convenient to make a detour via E[t(t − 1)]. Here
are the steps:
• a. Express E[t(t − 1)] as an inﬁnite sum.
Answer. Just write it down according to the deﬁnition of expected values:
∞
=
k(k − 1)pq k−1 .
k=2 1)pq k−1 ∞
k=0 k (k − • b. Derive the formula
∞ k (k − 1)q k−2 = (4.1.8)
k=2 2
(1 − q )3 by the same trick by which we derived a similar formula in class. Note that the sum
starts at k = 2.
Answer. This is just a second time diﬀerentiating the geometric series, i.e., ﬁrst time diﬀerentiating (4.1.6). • c. Use a. and b. to derive
(4.1.9) E[t(t − 1)] = 2q
p2 4.1. BINOMIAL 51 Answer.
∞ ∞ k(k − 1)pq k−1 = pq (4.1.10)
k=2 k(k − 1)q k−2 = pq 2
2q
= 2.
(1 − q )3
p k=2 • d. Use c. and the fact that E[t] = 1/p to derive
q
(4.1.11)
var[t] = 2 .
p
Answer.
(4.1.12) var[t] = E[t2 ] − (E[t])2 = E[t(t − 1)] + E[t] − (E[t])2 = 1
1
q
2q
+ − 2 = 2.
p2
p
p
p Now let us look at the negative binomial with arbitrary r. What is the probability
that it takes n trials to get r successes? (That means, with n − 1 trials we did not yet
have r successes.) The probability that the nth trial is a success is p. The probability
−1
that there are r − 1 successes in the ﬁrst n − 1 trials is n−1 pr−1 q n−r . Multiply
r
those to get:
(4.1.13) Pr[t=n] = n − 1 r n−r
pq
.
r−1 This is the negative binomial, also called the Pascal probability distribution with
parameters r and p.
One easily gets the mean and variance, because due to the memoryless property
it is the sum of r independent geometric variables:
r
rq
(4.1.14)
E[t] =
var[t] = 2
p
p
Some authors deﬁne the negative binomial as the number of failures before the
rth success. Their formulas will look slightly diﬀerent than ours.
Problem 78. 3 points A fair coin is ﬂipped until heads appear 10 times, and x
is the number of times tails appear before the 10th appearance of heads. Show that
the expected value E[x] = 10.
Answer. Let t be the number of the throw which gives the 10th head. t is a negative binomial
with r = 10 and p = 1/2, therefore E[t] = 20. Since x = t − 10, it follows E[x] = 10. Problem 79. (Banach’s matchbox problem) (Not eligible for inclass exams)
There are two restaurants in town serving hamburgers. In the morning each of them
obtains a shipment of n raw hamburgers. Every time someone in that town wants
to eat a hamburger, he or she selects one of the two restaurants at random. What is
the probability that the (n + k )th customer will have to be turned away because the
restaurant selected has run out of hamburgers?
Answer. For each restaurant it is the negative binomial probability distribution in disguise:
if a restaurant runs out of hamburgers this is like having n successes in n + k tries.
But one can also reason it out: Assume one of the restaurantes must turn customers away
after the n + kth customer. Write down all the n + k decisions made: write a 1 if the customer
goes to the ﬁrst restaurant, and a 2 if he goes to the second. I.e., write down n + k ones and twos.
Under what conditions will such a sequence result in the n + kth move eating the last hamburgerthe
ﬁrst restaurant? Exactly if it has n ones and k twos, a n + kth move is a one. As in the reasoning
k
for the negative binomial probability distribution, there are n+−−1 p ossibilities, each of which
n1
has probability 2−n−k . Emptying the second restaurant has the same probability. Together the
k
probability is therefore n+−−1 21−n−k .
n1 52 4. SPECIFIC RANDOM VARIABLES 4.2. The Hypergeometric Probability Distribution
Until now we had independent events, such as, repeated throwing of coins or
dice, sampling with replacement from ﬁnite populations, ar sampling from inﬁnite
populations. If we sample without replacement from a ﬁnite population, the probability of the second element of the sample depends on what the ﬁrst element was.
Here the hypergeometric probability distribution applies.
Assume we have an urn with w white and n − w black balls in it, and we take a
sample of m balls. What is the probability that y of them are white?
We are not interested in the order in which these balls are taken out; we may
therefore assume that they are taken out simultaneously, therefore the set U of
outcomes is the set of subsets containing m of the n balls. The total number of such
n
subsets is m . How many of them have y white balls in them? Imagine you ﬁrst
pick y white balls from the set of all white balls (there are w possibilities to do
y
that), and then you pick m − y black balls from the set of all black balls, which can
n
be done in m−w diﬀerent ways. Every union of such a set of white balls with a set
−y
of black balls gives a set of m elements with exactly y white balls, as desired. There
n
are therefore w m−w diﬀerent such sets, and the probability of picking such a set
y
−y
is
(4.2.1) Pr[Sample of m elements has exactly y white balls] = w
y n−w
m−y
n
m . Problem 80. You have an urn with w white and n − w black balls in it, and you
take a sample of m balls with replacement, i.e., after pulling each ball out you put it
back in before you pull out the next ball. What is the probability that y of these balls
are white? I.e., we are asking here for the counterpart of formula (4.2.1) if sampling
is done with replacement.
Answer.
(4.2.2) w
n y n−w
n m−y m
y Without proof we will state here that the expected value of y , the number of
white balls in the sample, is E[y ] = m w , which is the same as if one would select the
n
balls with replacement.
Also without proof, the variance of y is
(4.2.3) var[y ] = m w (n − w) (n − m)
.
n
n
(n − 1) This is smaller than the variance if one would choose with replacement, which is
represented by the above formula without the last term n−m . This last term is
n−1
called the ﬁnite population correction. More about all this is in [Lar82, p. 176–183].
4.3. The Poisson Distribution
The Poisson distribution counts the number of events in a given time interval.
This number has the Poisson distribution if each event is the cumulative result of a
large number of independent possibilities, each of which has only a small chance of
occurring (law of rare events). The expected number of occurrences is proportional
to time with a proportionality factor λ, and in a short time span only zero or one
event can occur, i.e., for inﬁnitesimal time intervals it becomes a Bernoulli trial. 4.3. THE POISSON DISTRIBUTION 53 t
Approximate it by dividing the time from 0 to t into n intervals of length n ; then
the occurrences are approximately n independent Bernoulli trials with probability of
success λt . (This is an approximation since some of these intervals may have more
n
than one occurrence; but if the intervals become very short the probability of having
two occurrences in the same interval becomes negligible.)
In this discrete approximation, the probability to have k successes in time t is
n λt k
λt (n−k)
Pr[x=k ] =
1−
(4.3.1)
k
n
n
λt n
λt −k
1 n(n − 1) · · · (n − k + 1)
(4.3.2)
(λt)k 1 −
1−
=
k!
nk
n
n
k
(λt) −λt
→
e
(4.3.3)
for n → ∞ while k remains constant
k!
(4.3.3) is the limit because the second and the last term in (4.3.2) → 1. The sum
k
∞
of all probabilities is 1 since k=0 (λt!) = eλt . The expected value is (note that we
k
can have the sum start at k = 1):
∞ E[x] = e−λt (4.3.4) k
k=1 (λt)k
= λte−λt
k! ∞ k=1 (λt)k−1
= λt.
(k − 1)! This is the same as the expected value of the discrete approximations.
Problem 81. x follows a Poisson distribution, i.e.,
(λt)k −λt
e
k!
• a. 2 points Show that E[x] = λt. (4.3.5) Pr[x=k ] = for k = 0, 1, . . .. Answer. See (4.3.4). • b. 4 points Compute E[x(x − 1)] and show that var[x] = λt.
Answer. For E[x(x − 1)] we can have the sum start at k = 2:
∞ (4.3.6) E[x(x − 1)] = e−λt k(k − 1)
k=2 (λt)k
= (λt)2 e−λt
k! ∞ (λt)k−2
= (λt)2 .
(k − 2)! k=2 From this follows
(4.3.7) var[x] = E[x2 ] − (E[x])2 = E[x(x − 1)] + E[x] − (E[x])2 = (λt)2 + λt − (λt)2 = λt. The Poisson distribution can be used as an approximation to the Binomial distribution when n large, p small, and np moderate.
Problem 82. Which value of λ would one need to approximate a given Binomial
with n and p?
Answer. That which gives the right expected value, i.e., λ = np. Problem 83. Two researchers counted cars coming down a road, which obey a
Poisson distribution with unknown parameter λ. In other words, in an interval of
length t one will have k cars with probability
(λt)k −λt
e.
k!
Their assignment was to count how many cars came in the ﬁrst half hour, and how
many cars came in the second half hour. However they forgot to keep track of the
time when the ﬁrst half hour was over, and therefore wound up only with one count,
(4.3.8) 54 4. SPECIFIC RANDOM VARIABLES namely, they knew that 213 cars had come down the road during this hour. They
were afraid they would get ﬁred if they came back with one number only, so they
applied the following remedy: they threw a coin 213 times and counted the number of
heads. This number, they pretended, was the number of cars in the ﬁrst half hour.
• a. 6 points Did the probability distribution of the number gained in this way
diﬀer from the distribution of actually counting the number of cars in the ﬁrst half
hour?
Answer. First a few deﬁnitions: x is the total number of occurrences in the interval [0, 1]. y
1
is the number of occurrences in the interval [0, t] (for a ﬁxed t; in the problem it was t = 2 , but we
will do it for general t, which will make the notation clearer and more compact. Then we want to
compute Pr[y =mx=n]. By deﬁnition of conditional probability:
(4.3.9) Pr[y =mx=n] = Pr[y =m and x=n]
.
Pr[x=n] How can we compute the probability of the intersection Pr[y =m and x=n]? Use a trick: express
this intersection as the intersection of independent events. For this deﬁne z as the number of
events in the interval (t, 1]. Then {y =m and x=n} = {y =m and z =n − m}; therefore Pr[y =m and
x=n] = Pr[y =m] Pr[z =n − m]; use this to get
(4.3.10)
Pr[y =mx=n] = Pr[y =m] Pr[z =n − m]
=
Pr[x=n] λm tm −λt λ
e
m! k n−m (1−t)n−m −λ(1−t)
e
(n−m)!
λn −λ
e
n!
(λt)k = nm
t (1−t)n−m ,
m (1−λ)k tk e−(1−λ)t .
Here we use the fact that Pr[x=k] = t ! e−t , Pr[y =k] = k! e−λt , Pr[z =k] =
k
k!
One sees that a. Pr[y =mx=n] does not depend on λ, and b. it is exactly the probability of having m
successes and n − m failures in a Bernoulli trial with success probability t. Therefore the procedure
with the coins gave the two researchers a result which had the same probability distribution as if
they had counted the number of cars in each half hour separately. • b. 2 points Explain what it means that the probability distribution of the number
for the ﬁrst half hour gained by throwing the coins does not diﬀer from the one gained
by actually counting the cars. Which condition is absolutely necessary for this to
hold?
Answer. The supervisor would never be able to ﬁnd out through statistical analysis of the
data they delivered, even if they did it repeatedly. All estimation results based on the faked statistic
would be as accurate regarding λ as the true statistics. All this is only true under the assumption
that the cars really obey a Poisson distribution and that the coin is fair.
The fact that the Poisson as well as the binomial distributions are memoryless has nothing to
do with them having a suﬃcient statistic. Problem 84. 8 points x is the number of customers arriving at a service counter
in one hour. x follows a Poisson distribution with parameter λ = 2, i.e.,
(4.3.11) Pr[x=j ] = 2j −2
e.
j! • a. Compute the probability that only one customer shows up at the service
counter during the hour, the probability that two show up, and the probability that no
one shows up.
• b. Despite the small number of customers, two employees are assigned to the
service counter. They are hiding in the back, and whenever a customer steps up to
the counter and rings the bell, they toss a coin. If the coin shows head, Herbert serves 4.4. THE EXPONENTIAL DISTRIBUTION 55 the customer, and if it shows tails, Karl does. Compute the probability that Herbert
has to serve exactly one customer during the hour. Hint:
1
1
1
(4.3.12)
e = 1 + 1 + + + + ··· .
2! 3! 4!
• c. For any integer k ≥ 0, compute the probability that Herbert has to serve
exactly k customers during the hour.
Problem 85. 3 points Compute the moment generating function of a Poisson
k
variable observed over a unit time interval, i.e., x satisﬁes Pr[x=k ] = λ ! e−λ and
k
you want E[etx ] for all t.
k
∞
etk λ ! e−λ
k
k=0 Answer. E[etx ] = ∞
(λet )k −λ
e
k!
k=0 = t t = eλe e−λ = eλ(e −1) . 4.4. The Exponential Distribution
Now we will discuss random variables which are related to the Poisson distribution. At time t = 0 you start observing a Poisson process, and the random
variable t denotes the time you have to wait until the ﬁrst occurrence. t can have
any nonnegative real number as value. One can derive its cumulative distribution
as follows. t>t if and only if there are no occurrences in the interval [0, t]. There0
fore Pr[t>t] = (λt) e−λt = e−λt , and hence the cumulative distribution function
0!
Ft (t) = Pr[t≤t] = 1 − e−λt when t ≥ 0, and Ft (t) = 0 for t < 0. The density function
is therefore ft (t) = λe−λt for t ≥ 0, and 0 otherwise. This is called the exponential
density function (its discrete analog is the geometric random variable). It can also
be called a Gamma variable with parameters r = 1 and λ.
Problem 86. 2 points An exponential random variable t with parameter λ > 0
has the density ft (t) = λe−λt for t ≥ 0, and 0 for t < 0. Use this density to compute
the expected value of t.
Answer. E[t] = ∞
0 λte−λt dt = ∞ uv dt = uv 0 can also use the more abbreviated notation =
Either way one obtains E[t] = ∞
−te−λt
0 + ∞ ∞
0 u dv = uv 0
∞ −λt
e
dt
0 =0 ∞ u=t v =λe−λt
. One
u =1 v =−e−λt
∞
u=t dv =λe−λt dt
−
.
v du, where
0
du =dt
v =−e−λt
0
1 −λt ∞
1
− λe
0 = λ . − 0
∞ u v dt, where Problem 87. 4 points An exponential random variable t with parameter λ > 0
has the density ft (t) = λe−λt for t ≥ 0, and 0 for t < 0. Use this density to compute
the expected value of t2 .
∞ Answer. One can use that Γ(r) =
2/λ2 . Or all from scratch: u = t2
u = 2t E[t2 ] = 0
∞
0 λr tr−1 e−λt dt for r = 3 to get: E[t2 ] = (1/λ2 )Γ(3) = λt2 e−λt dt = v = λe−λt
. Therefore E[t2 ] = −t2 e−λt
v = −e−λt the second do it again: ∞
0 2te−λt dt = ∞
0 ∞
0 + uv dt = uv Therefore the second term becomes 2(t/λ)e−λt ∞
0 +2 ∞
0
∞
0 ∞
0 − ∞
0 uv dt = uv 2te−λt
∞ 0 ∞
0 − ∞
0 u v dt, where dt. The ﬁrst term vanishes, for u v dt, where u=t
u =1 v = e−λt
.
v = −(1/λ)e−λt (1/λ)e−λt dt = 2/λ2 . Problem 88. 2 points Does the exponential random variable with parameter
λ > 0, whose cumulative distribution function is Ft (t) = 1 − e−λt for t ≥ 0, and
0 otherwise, have a memoryless property? Compare Problem 76. Formulate this
memoryless property and then verify whether it holds or not.
Answer. Here is the formulation: for s<t follows Pr[t>tt>s] = Pr[t>t − s]. This does indeed
hold. Proof: lhs = Pr[t>t and t>s]
Pr[t>s] = Pr[t>t]
Pr[t>s] = e−λt
e−λs = e−λ(t−s) . 56 4. SPECIFIC RANDOM VARIABLES Problem 89. The random variable t denotes the duration of an unemployment
spell. It has the exponential distribution, which can be deﬁned by: Pr[t>t] = e−λt for
t ≥ 0 (t cannot assume negative values).
• a. 1 point Use this formula to compute the cumulative distribution function
Ft (t) and the density function ft (t)
Answer. Ft (t) = Pr[t≤t] = 1 − Pr[t>t] = 1 − e−λt for t ≥ 0, zero otherwise. Taking the
derivative gives ft (t) = λe−λt for t ≥ 0, zero otherwise. • b. 2 points What is the probability that an unemployment spell ends after time
t + h, given that it has not yet ended at time t? Show that this is the same as the
unconditional probability that an unemployment spell ends after time h (memoryless
property).
Answer.
Pr[t>t + ht>t] = (4.4.1) Pr[t>t + h]
Pr[t>t] = e−λ(t+h)
= e−λh
e−λt • c. 3 points Let h be a small number. What is the probability that an unemployment spell ends at or before t + h, given that it has not yet ended at time t? Hint:
for small h, one can write approximately
Pr[t < t≤t + h] = hft (t). (4.4.2)
Answer. Pr[t≤t + ht>t] =
(4.4.3) = Pr[t≤t + h and t>t]
=
Pr[t>t]
h ft (t)
h λe−λt
=
= h λ.
1 − Ft (t)
e−λt 4.5. The Gamma Distribution
The time until the second occurrence of a Poisson event is a random variable
which we will call t(2) . Its cumulative distribution function is Ft(2) (t) = Pr[t(2) ≤t] =
1 − Pr[t(2) >t]. But t(2) >t means: there are either zero or one occurrences in the time
between 0 and t; therefore Pr[t(2) >t] = Pr[x=0]+Pr[x=1] = e−λt + λte−λt . Putting it
all together gives Ft(2) (t) = 1 − e−λt − λte−λt . In order to diﬀerentiate the cumulative
distribution function we need the product rule of diﬀerentiation: (uv ) = u v + uv .
This gives
ft(2) (t) = λe−λt − λe−λt + λ2 te−λt = λ2 te−λt . (4.5.1) Problem 90. 3 points Compute the density function of t(3) , the time of the third
occurrence of a Poisson variable.
Answer.
(4.5.2)
(4.5.3)
(4.5.4) Pr[t(3) >t] = Pr[x=0] + Pr[x=1] + Pr[x=2]
λ2 2 −λt
t )e
2
λ3 2 −λt
∂
λ2 2
ft(3) (t) =
Ft(3) (t) = − −λ(1 + λt +
t ) + (λ + λ2 t) e−λt =
te
.
∂t
2
2
Ft(3) (t) = Pr[t(3) ≤t] = 1 − (1 + λt + 4.5. THE GAMMA DISTRIBUTION 57 If one asks for the rth occurrence, again all but the last term cancel in the
diﬀerentiation, and one gets
λr
tr−1 e−λt .
(r − 1)! ft(r) (t) = (4.5.5) This density is called the Gamma density with parameters λ and r.
The following deﬁnite integral, which is deﬁned for all r > 0 and all λ > 0 is
called the Gamma function:
∞ (4.5.6) λr tr−1 e−λt dt. Γ(r) =
0 Although this integral cannot be expressed in a closed form, it is an important
function in mathematics. It is a well behaved function interpolating the factorials in
the sense that Γ(r) = (r − 1)!.
Problem 91. Show that Γ(r) as deﬁned in (4.5.6) is independent of λ, i.e.,
instead of (4.5.6) one can also use the simpler equation
∞ (4.5.7) tr−1 e−t dt. Γ(r) =
0 Problem 92. 3 points Show by partial integration that the Gamma function
satisﬁes Γ(r + 1) = rΓ(r).
Answer. Start with
∞ (4.5.8) λr+1 tr e−λt dt Γ(r + 1) =
0 and integrate by parts:
and v = rλr tr−1 :
(4.5.9) uv dt with u = λe−λt and v = λr tr , therefore u = −e−λt u v dt = uv − Γ(r + 1) = −λr tr e−λt ∞ ∞ rλr tr−1 e−λt dt = 0 + rΓ(r ). +
0 0 Problem 93. Show that Γ(r) = (r − 1)! for all natural numbers r = 1, 2, . . ..
Answer. Proof by induction. First verify that it holds for r = 1, i.e., that Γ(1) = 1:
∞ (4.5.10) λe−λt dt = −e−λt Γ(1) =
0 ∞
0 =1 and then, assuming that Γ(r) = (r − 1)! Problem 92 says that Γ(r + 1) = rΓ(r ) = r (r − 1)! = r!. √
1
Without proof: Γ( 2 ) = π . This will be shown in Problem 141.
Therefore the following deﬁnes a density function, called the Gamma density
with parameter r and λ, for all r > 0 and λ > 0:
(4.5.11) f (x) = λr r−1 −λx
xe
Γ(r) for x ≥ 0, 0 otherwise. The only application we have for it right now is: this is the distribution of the time
one has to wait until the rth occurrence of a Poisson distribution with intensity λ.
Later we will have other applications in which r is not an integer.
Problem 94. 4 points Compute the moment generating function of the Gamma
distribution. 58 4. SPECIFIC RANDOM VARIABLES Answer.
∞ (4.5.12) mx (t) = E[etx ] = etx
0 = (4.5.14) λ
λ−t ∞ λr
(λ − t)r = (4.5.13) λr r−1 −λx
x
e
dx
Γ(r ) 0 (λ − t)r xr−1 −(λ−t)x
e
dx
Γ(r ) r since the integrand in (4.5.12) is the density function of a Gamma distribution with parameters r
and λ − t. Problem 95. 2 points The density and moment generating functions of a Gamma
variable x with parameters r > 0 and λ > 0 are
λr r−1 −λx
(4.5.15)
fx (x) =
xe
for x ≥ 0, 0 otherwise.
Γ(r)
r
λ
.
λ−t
Show the following: If x has a Gamma distribution with parameters r and 1, then v =
x/λ has a Gamma distribution with parameters r and λ. You can prove this either
using the transformation theorem for densities, or the momentgenerating function. mx (t) = (4.5.16) Answer. Solution using density function: The random variable whose density we know is x;
1
its density is Γ(r) xr−1 e−x . If x = λv , then dx = λ, and the absolute value is also λ. Therefore the
dv
density of v is
(4.5.17) (4.5.18) λr
v r−1 e−λv .
Γ(r ) Solution using the mgf:
mx (t) = E[etx ] = 1
1−t r 1
1 − (t/λ) mv (t) E[etv ] = E[e(t/λ)x ] = r = λ
λ−t r but this last expression can be recognized to be the mgf of a Gamma with r and λ. Problem 96. 2 points It x has a Gamma distribution with parameters r and
λ, and y one with parameters p and λ, and both are independent, show that x + y
has a Gamma distribution with parameters r + p and λ (reproductive property of the
Gamma distribution.) You may use equation (4.5.14) without proof
Answer.
(4.5.19) λ
λ−t r λ
λ−t p = λ
λ−t r +p . Problem 97. Show that a Gamma variable x with parameters r and λ has
expected value E[x] = r/λ and variance var[x] = r/λ2 .
Answer. Proof with moment generating function:
(4.5.20) d
dt λ
λ−t r = r
λ λ
λ−t r +1 , r
therefore E[x] = λ , and by diﬀerentiating twice (apply the same formula again), E[x2 ] =
r
therefore var[x] = λ2 . Proof using density function: For the expected value one gets E[t] =
∞ r r +1 −λt
r
1
r Γ(r +1)
r
tλ
e
dt = λ · Γ(r+1) = λ . Using
λ Γ(r +1) 0
∞ λr+2 r +1 −λt
r (r +1)
r (r +1)
t
e
dt = λ2 .
λ2
0 Γ(r +2)
2 ] − (E[t])2 = r/λ2 .
Therefore var[t] = E[t the same tricks E[t2 ] = ∞ 0 r (r +1)
,
λ2 λr r −1 −λt
t
e
dt =
Γ(r )
∞ 2 λr r −1 −λt
t · Γ(r) t
e
dt =
0 t· 4.7. THE BETA DISTRIBUTION 59 4.6. The Uniform Distribution
Problem 98. Let x be uniformly distributed in the interval [a, b], i.e., the density
function of x is a constant for a ≤ x ≤ b, and zero otherwise.
• a. 1 point What is the value of this constant?
Answer. It is 1
b−a • b. 2 points Compute E[x]
Answer. E[x] = bx
a b−a dx = 1 b2 −a2
b−a
2 b 2 a+b
2 since b2 − a2 = (b + a)(b − a). a2 +ab+b2
.
3 • c. 2 points Show that E[x2 ] =
x
Answer. E[x2 ] =
dx =
a b−a
(check it by multiplying out). = 1 b3 −a3
.
b−a
3 Now use the identity b3 − a3 = (b − a)(b2 + ab + a2 ) • d. 2 points Show that var[x] = (b−a)2
12 . Answer. var[x] = E[x2 ] − (E[x])2 = a2 +ab+b2
3 (b−a)
12 2 − (a+b)2
4 4a2 +4ab+4b2
12 = − 3a2 +6ab+3b2
12 = . 4.7. The Beta Distribution
Assume you have two independent variables, both distributed uniformly over
the interval [0, 1], and you want to know the distribution of their maximum. Or of
their minimum. Or you have three and you want the distribution of the one in the
middle. Then the densities have their maximum to the right, or to the left, or in the
middle. The distribution of the rth highest out of n independent uniform variables
is an example of the Beta density function. Can also be done and is probabilitytheoretically meaningful for arbitrary real r and n.
Problem 99. x and y are two independent random variables distributed uniformly over the interval [0, 1]. Let u be their minimum u = min(x, y ) (i.e., u
takes the value of x when x is smaller, and the value of y when y is smaller), and
v = max(x, y ).
• a. 2 points Given two numbers q and r between 0 and 1. Draw the events u≤q
and v ≤r into the unit square and compute their probabilities.
• b. 2 points Compute the density functions fu (u) and fv (v ).
• c. 2 points Compute the expected values of u and v .
Answer. For u: Pr[u ≤ q ] = 1 − Pr[u > q ] = 1 − (1 − q )2 = 2q − q 2 . fv (v ) = 2v Therefore
f u (u) = 2 − 2 u
1 (4.7.1) (2 − 2u)u du = E[u] = u2 − 0 1 2 u3
3 =
0 1
.
3 For v it is: Pr[v ≤ r] = r2 ; this is at the same time the cumulative distribution function. Therefore
the density function is fv (v ) = 2v for 0 ≤ v ≤ 1 and 0 elsewhere.
1 (4.7.2) E[v ] = v 2v dv =
0 2v 3
3 1 =
0 2
.
3 60 4. SPECIFIC RANDOM VARIABLES 4.8. The Normal Distribution
By deﬁnition, y is normally distributed with mean µ and variance σ 2 , in symbols,
y ∼ N (µ, σ 2 ), if it has the density function
(4.8.1) fy (y ) = √ 1
2πσ 2 e− (y −µ)2
2σ 2 . It will be shown a little later that this is indeed a density function. This distribution
has the highest entropy among all distributions with a given mean and variance
[Kap89, p. 47].
If y ∼ N (µ, σ 2 ), then z = (y − µ)/σ ∼ N (0, 1), which is called the standard
Normal distribution.
Problem 100. 2 points Compare [Gre97, p. 68]: Assume x ∼ N (3, 4) (mean
is 3 and variance 4). Determine with the help of a table of the Standard Normal
Distribution function Pr[2<x≤5].
Answer. Pr[2 < x≤5] = Pr[2 − 3 < x − 3 ≤ 5 − 3] = Pr[ 2−3 < x−3 ≤ 5−3 ] = Pr[− 1 < x−3 ≤
2
2
2
2
2
1
1] = Φ(1) − Φ(− 2 ) = Φ(1) − (1 − Φ( 1 )) = Φ(1) + Φ( 1 ) − 1 = 0.8413 + 0.6915 − 1 = 0.5328. Some
2
2
tables (Greene) give the area between 0 and all positive values; in this case it is 0.3413 + 0.1915. The moment generating function of a standard normal z ∼ N (0, 1) is the following integral:
+∞ (4.8.2) mz (t) = E[etz ] =
−∞ 1 −z 2
etz √ e 2 dz.
2π To solve this integral, complete the square in the exponent:
(4.8.3) tz − t2
1
z2
=
− (z − t)2 ;
2
2
2
t2 2 Note that the ﬁrst summand, t2 , no longer depends on z ; therefore the factor e 2
can be written in front of the integral:
(4.8.4) +∞ t2 mz (t) = e 2 −∞ 2
1
t2
1
√ e− 2 (z−t) dz = e 2 ,
2π because now the integrand is simply the density function of a N (t, 1).
A general univariate normal x ∼ N (µ, σ 2 ) can be written as x = µ + σ z with
z ∼ N (0, 1), therefore
(4.8.5) mx (t) = E[e(µ+σz)t ] = eµt E[eσzt ] = e(µt+σ 22 t /2) . 2
Problem 101. Given two independent normal variables x ∼ N (µx , σx ) and
2
y ∼ N (µy , σy ). Using the moment generating function, show that (4.8.6) 2
2
αx + β y ∼ N (αµx + βµy , α2 σx + β 2 σy ). Answer. Because of independence, the moment generating function of αx + β y is the product
of the m.g.f. of αx and the one of β y :
(4.8.7) 2 mαx+β y (t) = eµx αt+σx α 2 22
t /2 µy βt+σy β t /2 22 e which is the moment generating function of a N (αµx + 2 = e(µx α+µy β )t+(σx α 2
βµy , α2 σx + 2 2
+σy β 2 )t2 /2 , 2
β 2 σy ). We will say more about the univariate normal later when we discuss the multivariate normal distribution. 4.8. THE NORMAL DISTRIBUTION 61 Sometimes it is also necessary to use the truncated normal distributions. If z is
standard normal, then
(4.8.8) E[z z >z ] = fz (z )
,
1 − Fz (z ) var[z z >z ] = 1 − µ(µ − z ), where µ = E[z z >z ]. This expected value is therefore the ordinate of the density function at point z divided
by the tail area of the tail over which z is known to vary. (This rule is only valid for
the normal density function, not in general!) These kinds of results can be found in
[JK70, pp. 81–83] or in the original paper [Coh50]
Problem 102. Every customer entering a car dealership in a certain location
can be thought of as having a reservation price y in his or her mind: if the car will be
oﬀered at or below this reservation price, then he or she will buy the car, otherwise
there will be no sale. (Assume for the sake of the argument all cars are equal.)
Assume this reservation price is Normally distributed with mean $6000 and standard
deviation $1000 (if you randomly pick a customer and ask his or her reservation
price). If a sale is made, a person’s consumer surplus is the diﬀerence between the
reservation price and the price actually paid, otherwise it is zero. For this question
you will need the table for the standard normal cumulative distribution function.
• a. 2 points A customer is oﬀered a car at a price of $5800. The probability
that he or she will take the car is . Answer. We need Pr[y ≥5800. If y =5800 then z = y−6000 = −0.2; Pr[z ≥ − 0.2] = 1 − Pr[z ≤ −
1000
0.2] = 1 − 0.4207 = 0.5793. • b. 3 points Since it is the 63rd birthday of the owner of the dealership, all
cars in the dealership are sold for the price of $6300. You pick at random one of
the people coming out of the dealership. The probability that this person bought a car
and his or her consumer surplus was more than $500 is . Answer. This is the unconditional probability that the reservation price was higher than
$6300 + $500 = $6800. i.e., Pr[y ≥6800. Deﬁne z = (y − $6000)/$1000. It is a standard normal, and
y ≤$6800 ⇐⇒ z ≤.8, Therefore p = 1 − Pr[z ≤.8] = .2119. • c. 4 points Here is an alternative scenario: Since it is the 63rd birthday of
the owner of the dealership, all cars in the dealership are sold for the “birthday
special” price of $6300. You pick at random one of the people who bought one of
these “birthday specials” priced $6300. The probability that this person’s consumer
surplus was more than $500 is . The important part of this question is: it depends on the outcome of the experiment whether or not someone is included in the sample sample selection bias.
Answer. Here we need the conditional probability:
(4.8.9) p = Pr[y >$6800y >$6300] = 1 − Pr[y ≤$6800]
Pr[y >$6800]
=
.
Pr[y >$6300]
1 − Pr[y ≤$6300] Again use the standard normal z = (y − $6000)/$1000. As before, y ≤$6800 ⇐⇒ z ≤.8, and
y ≤$6300 ⇐⇒ z ≤.3. Therefore
(4.8.10) p= 1 − Pr[z ≤.8]
.2119
=
= .5546.
1 − Pr[z ≤.3]
.3821 It depends on the layout of the normal distribution table how this should be looked up. 62 4. SPECIFIC RANDOM VARIABLES • d. 5 points We are still picking out customers that have bought the birthday
specials. Compute the median value m of such a customer’s consumer surplus. It is
deﬁned by
(4.8.11) Pr[y >$6300 + my >$6300] = Pr[y ≤$6300 + my >$6300] = 1/2. Answer. Obviously, m ≥ $0. Therefore
(4.8.12) Pr[y >$6300 + my >$6300] = 1
Pr[y >$6300 + m]
=,
Pr[y >$6300]
2 or Pr[y >$6300 + m] = (1/2) Pr[y >$6300] = (1/2).3821 = .1910. I.e., Pr[ y−6000 > 6300−6000+m =
1000
1000
m
300
m
300
+ 1000 ] = .1910. For this we ﬁnd in the table 1000 + 1000 = 0.875, therefore 300 + m = 875,
1000
or m = $575. • e. 3 points Is the expected value of the consumer surplus of all customers that
have bought a birthday special larger or smaller than the median? Fill in your answer
here: . Proof is not required, as long as the answer is correct. Answer. The mean is larger because it is more heavily inﬂuenced by outliers.
(4.8.13) E[y − 6300y ≥6300] = E[6000 + 1000z − 63006000 + 1000z ≥6300] (4.8.14) = E[1000z − 3001000z ≥300] (4.8.15) = E[1000z z ≥0.3] − 300 (4.8.16) = 1000 E[z z ≥0.3] − 300 (4.8.17) = 1000 f (0.3)
− 300 = 698 > 575.
1 − Ψ(0.3) 4.9. The ChiSquare Distribution
A χ2 with one degree of freedom is deﬁned to be the distribution of the square
q = z 2 of a univariate standard normal variable.
Call the cumulative distribution function of a standard normal Fz (z ). Then the
cumulative distribution function of the χ2 variable q = z 2 is, according to Problem
√
47, Fq (q ) = 2Fz ( q ) − 1. To get the density of q take the derivative of Fq (q ) with
respect to q . For this we need the chain rule, ﬁrst taking the derivative with respect
√
to z = q and multiply by dz :
dq
d
d
√
2Fz ( q ) − 1 =
2Fz (z ) − 1
dq
dq
2
dFz
dz
2
1
(4.9.2)
=2
(z )
= √ e−z /2 √
dz
dq
2q
2π
1
(4.9.3)
=√
e−q/2 .
2πq
√
Now remember the Gamma function. Since Γ(1/2) = π (Proof in Problem 141),
one can rewrite (4.9.3) as
(4.9.1) (4.9.4) fq (q ) = fq (q ) = (1/2)1/2 q −1/2 e−q/2
,
Γ(1/2) i.e., it is a Gamma density with parameters r = 1/2, λ = 1/2.
A χ2 with p degrees of freedom is deﬁned as the sum of p independent univariate
2
χ variables. By the reproductive property of the Gamma distribution (Problem 96) 4.11. THE CAUCHY DISTRIBUTION 63 this gives a Gamma variable with parameters r = p/2 and λ = 1/2.
If q ∼ χ2
p (4.9.5) then E[q ] = p and var[q ] = 2p We will say that a random variable q is distributed as a σ 2 χ2 iﬀ q /σ 2 is a χ2 . This
p
p
is the distribution of a sum of p independent N (0, σ 2 ) variables.
4.10. The Lognormal Distribution
This is a random variable whose log has a normal distribution. See [Gre97, p.
71]. Parametrized by the µ and σ 2 of its log. Density is
2
1
√
e−(ln x−µ/σ )/2
2
x 2πσ
[Cow77, pp. 82–87] has an excellent discussion of the properties of the lognormal
for income distributions. (4.10.1) 4.11. The Cauchy Distribution
Problem 103. 6 points [JK70, pp. 155/6] An example of a distribution without
mean and variance is the Cauchy distribution, whose density looks much like the
normal density, but has much thicker tails. The density and characteristic functions
are (I am not asking you to compute the characteristic function)
(4.11.1) fx (x) = 1
π (1 + x2 ) E[eitx ] = exp(− t). √
Here i = −1, but you should not be afraid of it, in most respects, i behaves like any
real number. The characteristic function has properties very similar to the moment
generating function, with the added advantage that it always exists. Using the characteristic functions show that if x and y are independent Cauchy distributions, then
(x + y )/2 has the same distribution as x or y .
Answer.
(4.11.2) E exp it x+y
2 t
t
= E exp i x exp i y
2
2 = exp(− t
t
) exp(−
) = exp(− t).
2
2 It has taken a historical learning process to distinguish signiﬁcant from insignificant events. The order in which the birds sit down on a tree is insigniﬁcant, but
the constellation of stars on the night sky is highly signiﬁcant for the seasons etc.
The confusion between signiﬁcant and insigniﬁcant events can explain how astrology arose: after it was discovered that the constellation of stars was signiﬁcant, but
without knowledge of the mechanism through which the constellation of stars was
signiﬁcant, people experimented to ﬁnd evidence of causality between those aspects
of the night sky that were changing, like the locations of the planets, and events on
earth, like the births of babies. Romans thought the constellation of birds in the sky
was signiﬁcant.
Freud discovered that human error may be signiﬁcant. Modern political consciousness still underestimates the extent to which the actions of states are signiﬁcant: If a welfare recipient is faced with an intractable labyrinth of regulations and
a multitude of agencies, then this is not the unintended result of bureaucracy gone
wild, but it is deliberate: this bureaucratic nightmare deters people from using welfare, but it creates the illusion that welfare exists and it does give relief in some
blatant cases. 64 4. SPECIFIC RANDOM VARIABLES Also “mistakes” like the bombing of the Chinese embassy are not mistakes but
are signiﬁcant.
In statistics the common consensus is that the averages are signiﬁcant and the
deviations from the averages are insigniﬁcant. By taking averages one distills the
signiﬁcant, systematic part of the date from the insigniﬁcant part. Usually this is
justiﬁed by the “law of large numbers.” I.e., people think that this is something
about reality which can be derived and proved mathematically. However this is an
irrealist position: how can math tell us which events are signiﬁcant?
Here the Cauchy distribution is an interesting counterexample: it is a probability
distribution for which it does not make sense to take averages. If one takes the
average of n observations, then this average does not have less randomness than
each individual observation, but it has exactly the same distribution as one single
observation. (The law of large numbers does not apply here because the Cauchy
distribution does not have an expected value.)
In a world in which random outcomes are Cauchydistributed, taking averages
is not be a good way to learn from one’s experiences. People who try to keep track
of things by taking averages (or by running regressions, which is a natural extension
of taking averages) would have the same status in that world as astrologers have in
our world. Taking medians and other quantiles would be considered scientiﬁc, but
taking averages would be considered superstition.
The lesson of this is: even a scientiﬁc procedure as innocuous as that of taking
averages cannot be justiﬁed on purely epistemological grounds. Although it is widely
assumed that the law of large numbers is such a justiﬁcation, it is not. The law of
large numbers does not always hold; it only holds if the random variable under
consideration has an expected value.
The transcendental realist can therefore say: since it apparently does make sense
to take averages in our world, we can deduce transcendentally that many random
variables which we are dealing with do have ﬁnite expected values.
This is perhaps the simplest case of a transcendental conclusion. But this simplest case also vindicates another one of Bhaskar’s assumptions: these transcendental
conclusions cannot be arrived at in a nontranscendental way, by staying in the science itself. It is impossible to decide, using statistical means alone, whether one’s
data come from a distribution which has ﬁnite expected values or not. The reason
is that one always has only ﬁnite datasets, and the empirical distribution of a ﬁnite
sample always has ﬁnite expected values, even if the sample comes from a population
which does not have ﬁnite expected values. CHAPTER 5 Chebyshev Inequality, Weak Law of Large
Numbers, and Central Limit Theorem
5.1. Chebyshev Inequality
If the random variable y has ﬁnite expected value µ and standard deviation σ ,
and k is some positive number, then the Chebyshev Inequality says
1
(5.1.1)
Pr y − µ≥kσ ≤ 2 .
k
In words, the probability that a given random variable y diﬀers from its expected
value by more than k standard deviations is less than 1/k 2 . (Here “more than”
and “less than” are short forms for “more than or equal to” and “less than or equal
to.”) One does not need to know the full distribution of y for that, only its expected
value and standard deviation. We will give here a proof only if y has a discrete
distribution, but the inequality is valid in general. Going over to the standardized
1
variable z = y−µ we have to show Pr[z ≥k ] ≤ k2 . Assuming z assumes the values
σ
z1 , z2 ,. . . with probabilities p(z1 ), p(z2 ),. . . , then
(5.1.2) Pr[z ≥k ] = p(zi ).
i : zi ≥k Now multiply by k 2 :
(5.1.3) k 2 Pr[z ≥k ] = k 2 p(zi )
i : zi ≥k
2
zi p(zi ) ≤ (5.1.4) i : zi ≥k
2
zi p(zi ) = var[z ] = 1. ≤ (5.1.5) all i The Chebyshev inequality is sharp for all k ≥ 1. Proof: the random variable
1
which takes the value −k with probability 2k2 and the value +k with probability
1
1
2k2 , and 0 with probability 1 − k2 , has expected value 0 and variance 1 and the
≤sign in (5.1.1) becomes an equal sign.
Problem 104. [HT83, p. 316] Let y be the number of successes in n trials of a
Bernoulli experiment with success probability p. Show that
y
1
(5.1.6)
Pr
− p <ε ≥ 1 −
.
n
4nε2
Hint: ﬁrst compute what Chebyshev will tell you about the lefthand side, and then
you will need still another inequality.
Answer. E[y /n] = p and var[y /n] = pq/n (where q = 1 − p). Chebyshev says therefore
(5.1.7) Pr y
− p ≥k
n
65 pq
n ≤ 1
.
k2 5.
66 CHEBYSHEV INEQUALITY, WEAK LAW OF LARGE NUMBERS, AND CENTRAL LIMIT THEOREM Setting ε = k pq/n, therefore 1/k2 = pq/nε2 one can rewerite (5.1.7) as (5.1.8) Pr y
− p ≥ε
n ≤ pq
.
nε2 Now note that pq ≤ 1/4 whatever their values are. Problem 105. 2 points For a standard normal variable, Pr[z ≥1] is approximately 1/3, please look up the precise value in a table. What does the Chebyshev
inequality says about this probability? Also, Pr[z ≥2] is approximately 5%, again
look up the precise value. What does Chebyshev say?
Answer. Pr[z ≥1] = 0.3174, the Chebyshev inequality says that Pr[z ≥1] ≤ 1.
Pr[z ≥2] = 0.0456, while Chebyshev says it is ≤ 0.25. Also, 5.2. The Probability Limit and the Law of Large Numbers
Let y 1 , y 2 , y 3 , . . . be a sequence of independent random variables all of which
n
1
have the same expected value µ and variance σ 2 . Then y n = n i=1 y i has expected
¯
2
value µ and variance σ . I.e., its probability mass is clustered much more closely
n
around the value µ than the individual y i . To make this statement more precise we
need a concept of convergence of random variables. It is not possible to deﬁne it in
the “obvious” way that the sequence of random variables y n converges toward y if
every realization of them converges, since it is possible, although extremely unlikely,
that e.g. all throws of a coin show heads ad inﬁnitum, or follow another sequence
for which the average number of heads does not converge towards 1/2. Therefore we
will use the following deﬁnition:
The sequence of random variables y 1 , y 2 , . . . converges in probability to another
random variable y if and only if for every δ > 0
lim Pr y n − y  ≥δ = 0. (5.2.1) n→∞ One can also say that the probability limit of y n is y , in formulas
(5.2.2) plim y n = y . n→∞ In many applications, the limiting variable y is a degenerate random variable, i.e., it
is a constant.
The Weak Law of Large Numbers says that, if the expected value exists, then the
probability limit of the sample means of an ever increasing sample is the expected
value, i.e., plimn→∞ y n = µ.
¯
Problem 106. 5 points Assuming that not only the expected value but also the
variance exists, derive the Weak Law of Large Numbers, which can be written as
(5.2.3) lim Pr y n − E[y ]≥δ = 0 for all δ > 0,
¯ n→∞ from the Chebyshev inequality
(5.2.4) Pr[x − µ≥kσ ] ≤ 1
k2 where µ = E[x] and σ 2 = var[x] Answer. From nonnegativity of probability and the Chebyshev inequality for x = y follows
¯
√
kσ
1
σ2
0 ≤ Pr[y − µ≥ √n ] ≤ k2 for all k. Set k = δ σ n to get 0 ≤ Pr[y n − µ≥δ ] ≤ nδ2 . For any ﬁxed
¯
¯
δ > 0, the upper bound converges towards zero as n → ∞, and the lower bound is zero, therefore
the probability iself also converges towards zero. 5.3. CENTRAL LIMIT THEOREM 67 Problem 107. 4 points Let y 1 , . . . , y n be a sample from some unknown probn
1
ability distribution, with sample mean y = n i=1 y i and sample variance s2 =
¯
n
1
¯2
i=1 (y i − y ) . Show that the data satisfy the following “sample equivalent” of
n
the Chebyshev inequality: if k is any ﬁxed positive number, and m is the number of
¯
observations y j which satisfy y j − y ≥k s, then m ≤ n/k 2 . In symbols,
n
(5.2.5)
#{y i : y i − y  ≥k s} ≤ 2 .
¯
k
Hint: apply the usual Chebyshev inequality to the socalled empirical distribution of
the sample. The empirical distribution is a discrete probability distribution deﬁned
by Pr[y =y i ] = k/n, when the number y i appears k times in the sample. (If all y i are
diﬀerent, then all probabilities are 1/n). The empirical distribution corresponds to
the experiment of randomly picking one observation out of the given sample.
Answer. The only thing to note is: the sample mean is the expected value in that empirical
distribution, the sample variance is the variance, and the relative number m/n is the probability.
#{y i : y i ∈ S } = n Pr[S ] (5.2.6) • a. 3 points What happens to this result when the distribution from which the
y i are taken does not have an expected value or a variance?
Answer. The result still holds but y and s2 do not converge as the number of observations
¯
increases. 5.3. Central Limit Theorem
Assume all y i are independent and have the same distribution with mean µ,
variance σ 2 , and also a moment generating function. Again, let y n be the sample
¯
mean of the ﬁrst n observations. The central limit theorem says that the probability
distribution for
¯
yn − µ
√
(5.3.1)
σ/ n
converges to a N (0, 1). This is a diﬀerent concept of convergence than the probability
limit, it is convergence in distribution.
Problem 108. 1 point Construct a sequence of random variables y 1 , y 2 . . . with
the following property: their cumulative distribution functions converge to the cumulative distribution function of a standard normal, but the random variables themselves
do not converge in probability. (This is easy!)
Answer. One example would be: all y i are independent standard normal variables.
yn −
¯
Why do we have the funny expression σ/√µ ? Because this is the standardized
n
version of y n . We know from the law of large numbers that the distribution of
¯
y n becomes more and more concentrated around µ. If we standardize the sample
¯
averages y n , we compensate for this concentration. The central limit theorem tells
¯
us therefore what happens to the shape of the cumulative distribution function of y n .
¯
If we disregard the fact that it becomes more and more concentrated (by multiplying
it by a factor which is chosen such that the variance remains constant), then we see
that its geometric shape comes closer and closer to a normal distribution.
Proof of the Central Limit Theorem: By Problem 109, (5.3.2) yn − µ
¯
1
√ =√
σ/ n
n n i=1 yi − µ
1
=√
σ
n n zi
i=1 where z i = yi − µ
.
σ 5.
68 CHEBYSHEV INEQUALITY, WEAK LAW OF LARGE NUMBERS, AND CENTRAL LIMIT THEOREM Let m3 , m4 , etc., be the third, fourth, etc., moments of z i ; then the m.g.f. of z i is
(5.3.3)
Therefore the m.g.f. of
(5.3.4) t2
m3 t3
m4 t4
+
+
+ ···
2!
3!
4!
√
n
i=1 z i is (multiply and substitute t/ n for t): mzi (t) = 1 + 1+ 1
√
n t2
m3 t3
m4 t4
+ ···
+√+
2!n 3! n3
4!n2 n = 1+ wn
n n where
(5.3.5) wn = t2
m3 t 3
m4 t4
+√+
+ ··· .
2! 3! n
4!n Now use Euler’s limit, this time in the form: if wn → w for n → ∞, then 1+ wn
n n → t2
2 2 ew . Since our wn → t2 , the m.g.f. of the standardized y n converges toward e , which
¯
is that of a standard normal distribution.
The Central Limit theorem is an example of emergence: independently of the
distributions of the individual summands, the distribution of the sum has a very
speciﬁc shape, the Gaussian bell curve. The signals turn into white noise. Here
emergence is the emergence of homogenity and indeterminacy. In capitalism, much
more speciﬁc outcomes emerge: whether one quits the job or not, whether one sells
the stock or not, whether one gets a divorce or not, the outcome for society is to
perpetuate the system. Not many activities don’t have this outcome.
Problem 109. Show in detail that
Answer. Lhs =
µ √
n
σ 1
n n
i=1 y n −µ
¯
√
σ/ n
√ y i −µ = n
σ = 1
√
n
1
n n
y i −µ
i=1 σ .
n
i=1 yi − 1
n n
i=1 µ = √
n1
σn = rhs. Problem 110. 3 points Explain verbally clearly what the law of large numbers
means, what the Central Limit Theorem means, and what their diﬀerence is.
Problem 111. (For this problem, a table is needed.) [Lar82, exercise 5.6.1,
p. 301] If you roll a pair of dice 180 times, what is the approximate probability that
the sum seven appears 25 or more times? Hint: use the Central Limit Theorem (but
don’t worry about the continuity correction, which is beyond the scope of this class).
Answer. Let xi be the random variable that equals one if the ith roll is a seven, and zero
otherwise. Since 7 can be obtained in six ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1), the probability
to get a 7 (which is at the same time the expected value of xi ) is 6/36=1/6. Since x2 = xi ,
i
180 1
1
5
var[xi ] = E[xi ] − (E[xi ])2 = 6 − 36 = 36 . Deﬁne x =
x . We need Pr[x≥25]. Since x
i=1 i
is the sum of many independent identically distributed random variables, the CLT says that x is
asympotically normal. Which normal? That which has the same expected value and variance as
x. E[x] = 180 · (1/6) = 30 and var[x] = 180 · (5/36) = 25. Therefore deﬁne y ∼ N (30, 25). The
CLT says that Pr[x≥25] ≈ Pr[y ≥25]. Now y ≥25 ⇐⇒ y − 30≥ − 5 ⇐⇒ y − 30≤ + 5 ⇐⇒
(y − 30)/5≤1. But z = (y − 30)/5 is a standard Normal, therefore Pr[(y − 30)/5≤1] = Fz (1), i.e.,
the cumulative distribution of the standard Normal evaluated at +1. One can look this up in a
table, the probability asked for is .8413. Larson uses the continuity correction: x is discrete, and
Pr[x≥25] = Pr[x>24]. Therefore Pr[y ≥25] and Pr[y >24] are two alternative good approximations;
but the best is Pr[y ≥24.5] = .8643. This is the continuity correction. n
i=1 yi − CHAPTER 6 Vector Random Variables
In this chapter we will look at two random variables x and y deﬁned on the same
sample space U , i.e.,
(6.0.6) x: U ω → x(ω ) ∈ R and y: U ω → y (ω ) ∈ R. As we said before, x and y are called independent if all events of the form x ≤ x
are independent of any event of the form y ≤ y . But now let us assume they are
not independent. In this case, we do not have all the information about them if we
merely know the distribution of each.
The following example from [Lar82, example 5.1.7. on p. 233] illustrates the
issues involved. This example involves two random variables that have only two
possible outcomes each. Suppose you are told that a coin is to be ﬂipped two times
and that the probability of a head is .5 for each ﬂip. This information is not enough
to determine the probability of the second ﬂip giving a head conditionally on the
ﬁrst ﬂip giving a head.
For instance, the above two probabilities can be achieved by the following experimental setup: a person has one fair coin and ﬂips it twice in a row. Then the
two ﬂips are independent.
But the probabilities of 1/2 for heads and 1/2 for tails can also be achieved as
follows: The person has two coins in his or her pocket. One has two heads, and one
has two tails. If at random one of these two coins is picked and ﬂipped twice, then
the second ﬂip has the same outcome as the ﬁrst ﬂip.
What do we need to get the full picture? We must consider the two variables not
separately but jointly, as a totality. In order to do this, we combine x and y into one
x
entity, a vector
∈ R2 . Consequently we need to know the probability measure
y
x(ω )
∈ R2 .
induced by the mapping U ω →
y (ω )
It is not suﬃcient to look at random variables individually; one must look at
them as a totality.
Therefore let us ﬁrst get an overview over all possible probability measures on the
plane R2 . In strict analogy with the onedimensional case, these probability measures
can be represented by the joint cumulative distribution function. It is deﬁned as
(6.0.7) Fx,y (x, y ) = Pr[ x
x
≤ = Pr[x ≤ x and y ≤ y ].
y
y For discrete random variables, for which the cumulative distribution function is
a step function, the joint probability mass function provides the same information:
(6.0.8) px,y (x, y ) = Pr[ x
x
= = Pr[x=x and y =y ].
y
y Problem 112. Write down the joint probability mass functions for the two versions of the two coin ﬂips discussed above.
69 70 6. VECTOR RANDOM VARIABLES Answer. Here are the probability mass functions for these two cases: (6.0.9) First
Flip Second Flip
H
T
H
.25
.25
T
.25
.25
sum .50
.50 sum
.50
.50
1.00 First
Flip Second Flip
H
T
H
.50
.00
T
.00
.50
sum .50
.50 sum
.50
.50
1.00 The most important case is that with a diﬀerentiable cumulative distribution
function. Then the joint density function fx,y (x, y ) can be used to deﬁne the probability measure. One obtains it from the cumulative distribution function by taking
derivatives:
∂2
(6.0.10)
fx,y (x, y ) =
Fx,y (x, y ).
∂x ∂y
Probabilities can be obtained back from the density function either by the integral condition, or by the inﬁnitesimal condition. I.e., either one says for a subset
B ⊂ R2 :
x
Pr[
(6.0.11)
∈ B] =
f (x, y ) dx dy,
y
B
or one says, for a inﬁnitesimal twodimensional volume element dVx,y located at [ x ],
y
which has the twodimensional volume (i.e., area) dV ,
(6.0.12) Pr[ x
∈ dVx,y ] = f (x, y ) dV .
y The vertical bars here do not mean the absolute value but the volume of the argument
inside.
6.1. Expected Value, Variances, Covariances
To get the expected value of a function of x and y , one simply has to put this
function together with the density function into the integral, i.e., the formula is
(6.1.1) E[g (x, y )] = g (x, y )fx,y (x, y ) dx dy.
R2 Problem 113. Assume there are two transportation choices available: bus and
car. If you pick at random a neoclassical individual ω and ask which utility this
person derives from using bus or car, the answer will be two numbers that can be
u(ω )
written as a vector
(u for bus and v for car).
v (ω )
• a. 3 points Assuming u
has a uniform density in the rectangle with corners
v 66
66
71
71
,
,
, and
, compute the probability that the bus will be preferred.
68
72
68
72
Answer. The probability is 9/40. u and v have a joint density function that is uniform in
the rectangle below and zero outside (u, the preference for buses, is on the horizontal, and v , the
preference for cars, on the vertical axis). The probability is the fraction of this rectangle below the
diagonal.
72
71 70
69
68 66 67 68 69 70 71 6.1. EXPECTED VALUE, VARIANCES, COVARIANCES 71 • b. 2 points How would you criticize an econometric study which argued along
the above lines?
Answer. The preferences are not for a bus or a car, but for a whole transportation systems.
And these preferences are not formed independently and individualistically, but they depend on
which other infrastructures are in place, whether there is suburban sprawl or concentrated walkable
cities, etc. This is again the error of detotalization (which favors the status quo). Jointly distributed random variables should be written as random vectors. Iny
stead of
we will also write x (bold face). Vectors are always considered to be
z
column vectors. The expected value of a random vector is a vector of constants,
notation E[x1 ]
.
(6.1.2)
E [x] = . .
E[xn ]
For two random variables x and y , their covariance is deﬁned as
(6.1.3) cov[x, y ] = E (x − E[x])(y − E[y ]) Computation rules with covariances are
(6.1.4) cov[x, z ] = cov[z , x] cov[x, x] = var[x] (6.1.5) cov[x + y , z ] = cov[x, z ] + cov[y , z ] cov[x, α] = 0 cov[αx, y ] = α cov[x, y ] Problem 114. 3 points Using deﬁnition (6.1.3) prove the following formula:
(6.1.6) cov[x, y ] = E[xy ] − E[x] E[y ]. Write it down carefully, you will lose points for unbalanced or missing parantheses
and brackets.
Answer. Here it is side by side with and without the notation E[x] = µ and E[y ] = ν :
cov[x, y ] = E (x − E[x])(y − E[y ]) cov[x, y ] = E[(x − µ)(y − ν )] = E xy − x E[y ] − E[x]y + E[x] E[y ] = E[xy − xν − µy + µν ] = E[xy ] − E[x] E[y ] − E[x] E[y ] + E[x] E[y ] = E[xy ] − µν − µν + µν = E[xy ] − E[x] E[y ]. (6.1.7) = E[xy ] − µν. Problem 115. 1 point Using (6.1.6) prove the ﬁve computation rules with covariances (6.1.4) and (6.1.5).
Problem 116. Using the computation rules with covariances, show that
(6.1.8) var[x + y ] = var[x] + 2 cov[x, y ] + var[y ]. If one deals with random vectors, the expected value becomes a vector, and the
variance becomes a matrix, which is called dispersion matrix or variancecovariance
matrix or simply covariance matrix. We will write it V [x]. Its formal deﬁnition is
(6.1.9) V [x] = E (x − E [x])(x − E [x]) , 72 6. VECTOR RANDOM VARIABLES but we can look at it simply as the matrix of all variances and covariances, for
example
x
var[x]
cov[x, y ]
.
V [ y ] = cov[y , x]
var[y ] (6.1.10) An important computation rule for the covariance matrix is
V [x] = Ψ ⇒ V [Ax] = AΨA . (6.1.11) Problem 117. 4 points Let x = y
z be a vector consisting of two random variables, with covariance matrix V [x] = Ψ, and let A = ab
be an arbitrary
cd 2 × 2 matrix. Prove that
V [Ax] = AΨA . (6.1.12) Hint: You need to multiply matrices, and to use the following computation rules for
covariances:
(6.1.13)
cov[x + y , z ] = cov[x, z ] + cov[y , z ] cov[αx, y ] = α cov[x, y ] cov[x, x] = var[x].
Answer. V [Ax] =
V[ a
c b
d y
z = V[ On the other hand, AΨA
a
c b
d var[y ]
cov[y , z ] ay + bz
var[ay + bz ]=
cy + dz
cov[cy + dz , ay + bz ] cov[ay + bz , cy + dz ]
var[cy + dz ] = cov[y , z ]
var[z ] a
b c
a var[y ] + b cov[y , z ]
=
d
c var[y ] + d cov[y , z ] a cov[y , z ] + b var[z ]
c cov[y , z ] + d var[z ] a
b c
d Multiply out and show that it is the same thing. Since the variances are nonnegative, one can see from equation (6.1.11) that
covariance matrices are nonnegative deﬁnite (which is in econometrics is often also
called positive semideﬁnite ). By deﬁnition, a symmetric matrix Σ is nonnegative definite if for all vectors a follows a Σ a ≥ 0. It is positive deﬁnite if it is nonnegativbe
deﬁnite, and a Σ a = 0 holds only if a = o.
Problem 118. 1 point A symmetric matrix Ω is nonnegative deﬁnite if and only
if a Ω a ≥ 0 for every vector a. Using this criterion, show that if Σ is symmetric and
nonnegative deﬁnite, and if R is an arbitrary matrix, then R ΣR is also nonnegative
deﬁnite.
One can also deﬁne a covariance matrix between diﬀerent vectors, C [x, y ]; its
i, j element is cov[xi , y j ].
The correlation coeﬃcient of two scalar random variables is deﬁned as
(6.1.14) corr[x, y ] = cov[x, y ]
var[x] var[y ] . The advantage of the correlation coeﬃcient over the covariance is that it is always
between −1 and +1. This follows from the CauchySchwartz inequality
(6.1.15) (cov[x, y ])2 ≤ var[x] var[y ]. Problem 119. 4 points Given two random variables y and z with var[y ] = 0,
compute that constant a for which var[ay − z ] is the minimum. Then derive the
CauchySchwartz inequality from the fact that the minimum variance is nonnegative. 6.2. MARGINAL PROBABILITY LAWS 73 Answer.
(6.1.16)
(6.1.17) var[ay − z ] = a2 var[y ] − 2a cov[y , z ] + var[z ]
First order condition: 0 = 2a var[y ] − 2 cov[y , z ] Therefore the minimum value is a∗ = cov[y , z ]/ var[y ], for which the cross product term is −2 times
the ﬁrst item:
(cov[y , z ])2
2(cov[y , z ])2
−
+ var[z ]
var[y ]
var[y ] (6.1.18) 0 ≤ var[a∗ y − z ] = (6.1.19) 0 ≤ −(cov[y , z ])2 + var[y ] var[z ]. This proves (6.1.15) for the case var[y ] = 0. If var[y ] = 0, then y is a constant, therefore cov[y , z ] = 0
and (6.1.15) holds trivially. 6.2. Marginal Probability Laws
The marginal probability distribution of x (or y ) is simply the probability distribution of x (or y ). The word “marginal” merely indicates that it is derived from
the joint probability distribution of x and y .
If the probability distribution is characterized by a probability mass function,
we can compute the marginal probability mass functions by writing down the joint
probability mass function in a rectangular scheme and summing up the rows or
columns:
(6.2.1) px (x) = px,y (x, y ).
y : p(x,y )=0 For density functions, the following argument can be given:
(6.2.2) Pr[x ∈ dVx ] = Pr[ By the deﬁnition of a product set:
many small disjoint intervals, R =
(6.2.3) x
∈ dVx × R].
y x
∈ A × B ⇔ x ∈ A and y ∈ B . Split R into
y
i dVyi , then Pr[x ∈ dVx ] = x
∈ dVx × dVyi
y Pr
i (6.2.4) fx,y (x, yi )dVx dVyi  =
i (6.2.5) = dVx  fx,y (x, yi )dVyi .
i Therefore
i fx,y (x, y )dVyi  is the density function we are looking for. Now the
dVyi  are usually written as dy , and the sum is usually written as an integral (i.e.,
an inﬁnite sum each summand of which is inﬁnitesimal), therefore we get
y =+∞ (6.2.6) fx (x) = fx,y (x, y ) dy.
y =−∞ In other words, one has to “integrate out” the variable which one is not interested
in. 74 6. VECTOR RANDOM VARIABLES 6.3. Conditional Probability Distribution and Conditional Mean
The conditional probability distribution of y given x=x is the probability distribution of y if we count only those experiments in which the outcome of x is x. If the
distribution is deﬁned by a probability mass function, then this is no problem:
(6.3.1) Pr[y =y and x=x]
px,y (x, y )
=
.
Pr[x=x]
px (x) pyx (y, x) = Pr[y =y x=x] = For a density function there is the problem that Pr[x=x] = 0, i.e., the conditional
probability is strictly speaking not deﬁned. Therefore take an inﬁnitesimal volume
element dVx located at x and condition on x ∈ dVx :
(6.3.2) Pr[y ∈ dVy and x ∈ dVx ]
Pr[x ∈ dVx ]
fx,y (x, y )dVx dVy 
=
fx (x)dVx 
fx,y (x, y )
=
dVy .
fx (x) Pr[y ∈ dVy x ∈ dVx ] = (6.3.3)
(6.3.4) This no longer depends on dVx , only on its location x. The conditional density is
therefore
fx,y (x, y )
(6.3.5)
fyx (y, x) =
.
fx (x)
As y varies, the conditional density is proportional to the joint density function, but
for every given value of x the joint density is multiplied by an appropriate factor so
that its integral with respect to y is 1. From (6.3.5) follows also that the joint density
function is the product of the conditional times the marginal density functions.
Problem 120. 2 points The conditional density is the joint divided by the marginal:
(6.3.6) fyx (y, x) = fx,y (x, y )
.
fx (x) Show that this density integrates out to 1.
Answer. The conditional is a density in y with x as parameter. Therefore its integral with
respect to y must be = 1. Indeed,
+∞ +∞ (6.3.7) fyx=x (y, x) dy = y =−∞ fx,y (x, y ) dy f x ( x) y =−∞ = f x ( x)
=1
f x ( x) because of the formula for the marginal:
+∞ (6.3.8) f x ( x) = fx,y (x, y ) dy
y =−∞ You see that formula (6.3.6) divides the joint density exactly by the right number which makes the
integral equal to 1. Problem 121. [BD77, example 1.1.4 on p. 7]. x and y are two independent
random variables uniformly distributed over [0, 1]. Deﬁne u = min(x, y ) and v =
max(x, y ).
• a. Draw in the x, y plane the event {max(x, y ) ≤ 0.5 and min(x, y ) > 0.4} and
compute its probability.
Answer. The event is the square between 0.4 and 0.5, and its probability is 0.01. 6.4. THE MULTINOMIAL DISTRIBUTION 75 • b. Compute the probability of the event {max(x, y ) ≤ 0.5 and min(x, y ) ≤ 0.4}.
Answer. It is Pr[max(x, y ) ≤ 0.5] − Pr[max(x, y ) ≤ 0.5 and min(x, y ) > 0.4], i.e., the area of
the square from 0 to 0.5 minus the square we just had, i.e., 0.24. • c. Compute Pr[max(x, y ) ≤ 0.5 min(x, y ) ≤ 0.4].
Answer.
(6.3.9) 0.24
0.24
3
Pr[max(x, y ) ≤ 0.5 and min(x, y ) ≤ 0.4]
=
=
=.
Pr[min(x, y ) ≤ 0.4]
1 − 0.36
0.64
8 • d. Compute the joint cumulative distribution function of u and v .
Answer. One good way is to do it geometrically: for arbitrary 0 ≤ u, v ≤ 1 draw the area
{u ≤ u and v ≤ v } and then derive its size. If u ≤ v then Pr[u ≤ u and v ≤ v ] = Pr[v ≤ v ] − Pr[u ≤ u
and v > v ] = v 2 − (v − u)2 = 2uv − u2 . If u ≥ v then Pr[u ≤ u and v ≤ v ] = Pr[v ≤ v ] = v 2 . • e. Compute the joint density function of u and v . Note: this joint density is
discontinuous. The values at the breakpoints themselves do not matter, but it is very
important to give the limits within this is a nontrivial function and where it is zero.
Answer. One can see from the way the cumulative distribution function was constructed that
the density function must be
(6.3.10) 2
0 fu,v (u, v ) = if 0 ≤ u ≤ v ≤ 1
otherwise I.e., it is uniform in the abovediagonal part of the square. This is also what one gets from diﬀerentiating 2vu − u2 once with respect to u and once with respect to v . • f . Compute the marginal density function of u.
Answer. Integrate v out: the marginal density of u is
1 (6.3.11) f u (u) = 1 = 2 − 2u 2 dv = 2v
v =u if 0 ≤ u ≤ 1, and 0 otherwise. u • g. Compute the conditional density of v given u = u.
Answer. Conditional density is easy to get too; it is the joint divided by the marginal, i.e., it
is uniform:
(6.3.12) fvu=u (v ) = 1
1−u for 0 ≤ u ≤ v ≤ 1 0 otherwise. 6.4. The Multinomial Distribution
Assume you have an experiment with r diﬀerent possible outcomes, with outcome
i having probability pi (i = 1, . . . , r). You are repeating the experiment n diﬀerent
times, and you count how many times the ith outcome occurred. Therefore you get
a random vector with r diﬀerent components xi , indicating how often the ith event
occurred. The probability to get the frequencies x1 , . . . , xr is
m!
(6.4.1)
Pr[x1 = x1 , . . . , xr = xr ] =
px1 px2 · · · pxr
r
x1 ! · · · xr ! 1 2
This can be explained as follows: The probability that the ﬁrst x1 experiments
yield outcome 1, the next x2 outcome 2, etc., is px1 px2 · · · pxr . Now every other
r
12
sequence of experiments which yields the same number of outcomes of the diﬀerent
categories is simply a permutation of this. But multiplying this probability by n! 76 6. VECTOR RANDOM VARIABLES may count certain sequences of outcomes more than once. Therefore we have to
divide by the number of permutations of the whole n element set which yield the
same original sequence. This is x1 ! · · · xr !, because this must be a permutation which
permutes the ﬁrst x1 elements amongst themselves, etc. Therefore the relevant count
n!
of permutations is x1 !···xr ! .
Problem 122. You have an experiment with r diﬀerent outcomes, the ith outcome occurring with probability pi . You make n independent trials, and the ith outcome occurred xi times. The joint distribution of the x1 , . . . , xr is called a multinomial distribution with parameters n and p1 , . . . , pr .
• a. 3 points Prove that their mean vector and covariance matrix are
(6.4.2) p1
p1 − p2 −p1 p2 · · · −p1 pr
1
x1
x1
p2 −p2 p1 p2 − p2 · · · −p2 pr 2
.
. µ = E [ . ] = n . and Ψ = V [ . ] = n .
.
. .
..
.
.
.
.
.
.
.
.
.
.
.
xr
xr
pr
−pr p1
−pr p2 · · · pr − p2
r
Hint: use the fact that the multinomial distribution with parameters n and p1 , . . . , pr
is the independent sum of n multinomial distributions with parameters 1 and p1 , . . . , pr .
Answer. In one trial, x2 = xi , from which follows the formula for the variance, and for i = j ,
i
xi xj = 0, since only one of them can occur. Therefore cov[xi , xj ] = 0 − E[xi ] E[xj ]. For several
independent trials, just add this. • b. 1 point How can you show that this covariance matrix is singular?
Answer. Since x1 + · · · + xr = n with zero variance, we should expect (6.4.3) p1 − p2
1 −p2 p1
.
n
.
.
−pr p1 −p1 p2
p2 − p2
2
.
.
.
−pr p2 ···
···
..
.
··· −p1 pr
1
0
−p2 pr 1 0 . = .
. . .
.
.
.
.
2
pr − pr
1
0 6.5. Independent Random Vectors
The same deﬁnition of independence, which we already encountered with scalar
random variables, also applies to vector random variables: the vector random variables x : U → Rm and y : U → Rn are called independent if all events that can be
deﬁned in terms of x are independent of all events that can be deﬁned in terms of
y , i.e., all events of the form {x(ω ) ∈ C } are independent of all events of the form
{y (ω ) ∈ D} with arbitrary (measurable) subsets C ⊂ Rm and D ⊂ Rn .
For this it is suﬃcient that for all x ∈ Rm and y ∈ Rn , the event {x ≤ x}
is independent of the event {y ≤ y }, i.e., that the joint cumulative distribution
function is the product of the marginal ones.
Since the joint cumulative distribution function of independent variables is equal
to the product of the univariate cumulative distribution functions, the same is true
for the joint density function and the joint probability mass function.
Only under this strong deﬁnition of independence is it true that any functions
of independent random variables are independent.
Problem 123. 4 points Prove that, if x and y are independent, then E[xy ] =
E[x] E[y ] and therefore cov[x, y ] = 0. (You may assume x and y have density functions). Give a counterexample where the covariance is zero but the variables are
nevertheless dependent. 6.6. CONDITIONAL EXPECTATION AND VARIANCE 77 Answer. Just use that the joint density function is the product of the marginals. It can also be
done as follows: E[xy ] = E E[xy x] = E x E[y x] = now independence is needed = E x E[y ] =
E[x] E[y ]. A counterexample is given in Problem 139. Problem 124. 3 points Prove the following: If the scalar random variables x
and y are indicator variables (i.e., if each of them can only assume the values 0 and
1), and if cov[x, y ] = 0, then x and y are independent. (I.e., in this respect indicator
variables have similar properties as jointly normal random variables.)
Answer. Deﬁne the events A = {ω ∈ U : x(ω ) = 1} and B = {ω ∈ U : y (ω ) = 1}, i.e.,
x = iA (the indicator variable of the event A) and y = iB . Then xy = iA∩B . If cov[x, y ] =
E[xy ] − E[x] E[y ] = Pr[A ∩ B ] − Pr[A] Pr[B ] = 0, then A and B are independent. Problem 125. If the vector random variables x and y have the property that
xi is independent of every y j for all i and j , does that make x and y independent
random vectors? Interestingly, the answer is no. Give a counterexample that this
fact does not even hold for indicator variables. I.e., construct two random vectors x
and y , consisting of indicator variables, with the property that each component of x
is independent of each component of y , but x and y are not independent as vector
random variables. Hint: Such an example can be constructed in the simplest possible
case that x has two components and y has one component; i.e., you merely have to
ﬁnd three indicator variables x1 , x2 , and y with the property that x1 is independent
x1
of y , and x2 is independent of y , but the vector
is not independent of y . For
x2
these three variables, you should use three events which are pairwise independent but
not mutually independent.
Answer. Go back to throwing a coin twice independently and deﬁne A = {HH, HT }; B =
{T H, HH }, and C = {HH, T T }, and x1 = IA , x2 = IB , and y = IC . They are pairwise independent, but A ∩ B ∩ C = A ∩ B , i.e., x1 x2 y = x1 x2 , therefore E[x1 x2 y ] = E[x1 x2 ] E[y ] therefore they
are not independent. Problem 126. 4 points Prove that, if x and y are independent, then var[xy ] =
(E[x])2 var[y ] + (E[y ])2 var[x] + var[x] var[y ].
Answer. Start with result and replace all occurrences of var[z ] with E[z 2 ]−E[z ]2 , then multiply
out: E[x]2 (E[y 2 ] − E[y ]2 ) + E[y ]2 (E[x2 ] − E[x]2 ) + (E[x2 ] − E[x]2 )(E[y 2 ] − E[y ]2 ) = E[x2 ] E[y 2 ] −
E[x]2 E[y ]2 = E[(xy )2 ] − E[xy ]2 . 6.6. Conditional Expectation and Variance
The conditional expectation of y is the expected value of y under the conditional
density. If joint densities exist, it follows
(6.6.1) E[y x=x] = y fx,y (x, y ) dy
=: g (x).
fx (x) This is not a random variable but a constant which depends on x, i.e., a function of
x, which is called here g (x). But often one uses the term E[y x] without specifying
x. This is, by deﬁnition, the random variable g (x) which one gets by plugging x into
g ; it assigns to every outcome ω ∈ U the conditional expectation of y given x=x(ω ).
Since E[y x] is a random variable, it is possible to take its expected value. The
law of iterated expectations is extremely important here. It says that you will get
the same result as if you had taken the expected value of y :
(6.6.2) E E[y x] = E[y ]. 78 6. VECTOR RANDOM VARIABLES Proof (for the case that the densities exist):
E E[y x] = E[g (x)] =
(6.6.3)
= y fx,y (x, y ) dy
fx (x) dx
fx (x)
y fx,y (x, y ) dy dx = E[y ]. Problem 127. Let x and y be two jointly distributed variables. For every ﬁxed
value x, var[y x = x] is the variance of y under the conditional distribution, and
var[y x] is this variance as a random variable, namely, as a function of x.
• a. 1 point Prove that
(6.6.4) var[y x] = E[y 2 x] − (E[y x])2 . This is a very simple proof. Explain exactly what, if anything, needs to be done to
prove it.
Answer. For every ﬁxed value x, it is an instance of the law
var[y ] = E[y 2 ] − (E[y ])2 (6.6.5) applied to the conditional density given x = x. And since it is true for every ﬁxed x, it is also true
after plugging in the random variable x. • b. 3 points Prove that
(6.6.6) var[y ] = var E[y x] + E var[y x] , i.e., the variance consists of two components: the variance of the conditional mean
and the mean of the conditional variances. This decomposition of the variance is
given e.g. in [Rao73, p. 97] or [Ame94, theorem 4.4.2 on p. 78].
Answer. The ﬁrst term on the rhs is E[(E[y x])2 ] − (E[E[y x]])2 , and the second term, due
to (6.6.4), becomes E[E[y 2 x]] − E[(E[y x])2 ]. If one adds, the two E[(E[y x])2 ] cancel out, and the
other two terms can be simpliﬁed by the law of iterated expectations to give E[y 2 ] − (E[y ])2 . • c. 2 points [Coo98, p. 23] The conditional expected value is sometimes called
the population regression function. In graphical data analysis, the sample equivalent
of the variance ratio
(6.6.7) E var[y x]
var E[y x] can be used to determine whether the regression function E[y x] appears to be visually welldetermined or not. Does a small or a big variance ratio indicate a welldetermined regression function?
Answer. For a welldetermined regression function the variance ratio should be small. [Coo98,
p. 23] writes: “This ratio is reminiscent of a oneway analysis of variance, with the numerator representing the average within group (slice) variance, and the denominator representing the varince
between group (slice) means.” Now some general questions:
Problem 128. The ﬁgure on page 79 shows 250 independent observations of the
random vector [ x ].
y
• a. 2 points Draw in by hand the approximate location of E [[ x ]] and the graph
y
of E[y x]. Draw into the second diagram the approximate marginal density of x. 6.7. EXPECTED VALUES AS PREDICTORS 79 • b. 2 points Is there a law that the graph of the conditional expectation E[y x]
always goes through the point E [[ x ]]—for arbitrary probability distributions for which
y
these expectations exist, or perhaps for an important special case? Indicate how this
could be proved or otherwise give (maybe geometrically) a simple counterexample.
Answer. This is not the law of iterated expectations. It is true for jointly normal variables,
not in general. It is also true if x and y are independent; then the graph of E[y x] is a horizontal line
at the height of the unconditional expectation E[y ]. A distribution with Ushaped unconditional
distribution has the unconditional mean in the center of the U, i.e., here the unconditional mean
does not lie on the curve drawn out by the conditional mean. • c. 2 points Do you have any ideas how the strangelooking cluster of points in
the ﬁgure on page 79 was generated?
Problem 129. 2 points Given two independent random variables x and y with
density functions fx (x) and gy (y ). Write down their joint, marginal, and conditional
densities.
Answer. Joint density: fx,y (x, (y ) = fx (x)gy (y ).
∞
∞
Marginal density of x is
fx (x)gy (y ) dy = fx (x)
gy (y ) dy = fx (x), and that of y is
−∞
−∞
gy (y ). The text of the question should have been: “Given two independent random variables x
and y with marginal density functions fx (x) and gy (y )”; by just calling them “density functions”
without specifying “marginal” it committed the error of detotalization, i.e., it treated elements of
a totality, i.e., of an ensemble in which each depends on everything else, as if they could be deﬁned
independently of each other.
Conditional density functions: fxy=y (x; y ) = fx (x) (i.e., it does not depend on y ); and
gyx=x (y ; x) = gy (y ). You can see this by dividing the joint by the marginal. 6.7. Expected Values as Predictors
`
Expected values and conditional expected values have optimal properties as predictors.
`
Problem 130. 3 points What is the best predictor of a random variable y by a
constant a, if the loss function is the “mean squared error” (MSE) E[(y − a)2 ]?
`
`
`` T `
T `
`
` `` `` ``
`` ` ` ` ``` ` `
` ` ` `` `
` ` ` `` ` ` ` ` ` `` ` `` `` ` `
`
` ` ` ` ` ` ```` ```` ` ` ````` `` ` `` ` ` ```` `` `` ` `` ` ` ` `
` `` ` ` ` ``` ``` `` ` ` ` ` ` ` `
`
` `` ` ` `` ` ` `` ``` ``` `` ` ` ``` ``` ` `` `` ` `
` ` `` ` ` ` `` ` ` `
`
` `` ` ` ` ` `` ` `` ` ` `
` ``
`
` ``
``
` `` `
`
`` ` `
`
`` `` E ` E 80 6. VECTOR RANDOM VARIABLES Answer. Write E[y ] = µ; then
(y − a )2 = ( y − µ) − ( a − µ) 2 = (y − µ)2 − 2(y − µ)(a − µ) + (a − µ)2 ; (6.7.1)
therefore E[(y − a)2 ] = E[(y − µ)2 ] − 0 + (a − µ)2 This is minimized by a = µ. The expected value of y is therefore that constant which, as predictor of y , has
smallest MSE.
What if we want to predict y not by a constant but by a function of the random
vector x, call it h(x)?
Problem 131. 2 points Assume the vector x = [x1 , . . . xj ] and the scalar y
are jointly distributed random variables, and assume conditional means exist. x is
observed, but y is not observed. The joint distribution of x and y is known. Show
that the conditional expectation E[y x] is the minimum MSE predictor of y given x,
i.e., show that for any other function of x, call it h(x), the following inequality holds:
2 2 E[ y − h(x) ] ≥ E[ y − E[y x] ]. (6.7.2) For this proof and the proofs required in Problems 132 and 133, you may use (1)
the theorem of iterated expectations E E[y x] = E[y ], (2) the additivity E[g (y ) +
h(y )x] = E[g (y )x]+ E[h(y )x], and (3) the fact that E[g (x)h(y )x] = g (x)E[h(y )x].
Be very speciﬁc about which rules you are applying at every step. You must show
that you understand what you are writing down.
Answer.
(6.7.3)
E[ y − h(x) 2
=E y − E[y x] − (h(x) − E[y x]) 2 = E[(y − E[y x])2 ] − 2 E[(y − E[y x])(h(x) − E[y x])] + E[(h(x) − E[y x])2 ].
Here the cross product term E[(y − E[y x])(h(x) − E[y x])] is zero. In order to see this, ﬁrst use the
law of iterated expectations
(6.7.4) E[(y − E[y x])(h(x) − E[y x])] = E E[(y − E[y x])(h(x) − E[y x])x] and then look at the inner term, not yet doing the outer expectation:
E[(y − E[y x])(h(x) − E[y x])x] = (h(x) − E[y x]) =
E[(y − E[y x])x] = (h(x) − E[y x])(E[y x] − E[y x]) == (h(x) − E[y x]) · 0 = 0
Plugging this into (6.7.4) gives E[(y − E[y x])(h(x) − E[y x])] = E 0 = 0. This is one of the few clear cut results in probability theory where a best estimator/predictor exists. In this case, however, all parameters of the distribution are
known, the only uncertainty comes from the fact that some random variables are
unobserved.
Problem 132. Assume the vector x = [x1 , . . . xj ] and the scalar y are jointly
distributed random variables, and assume conditional means exist. Deﬁne ε = y −
E[y x]. 6.7. EXPECTED VALUES AS PREDICTORS 81 • a. 5 points Demonstrate the following identities:
(6.7.5) E[εx] = 0 (6.7.6) E[ε] = 0 (6.7.7) E[xi εx] = 0 for all i, 1 ≤ i ≤ j (6.7.8) E[xi ε] = 0 for all i, 1 ≤ i ≤ j (6.7.9) cov[xi , ε] = 0 for all i, 1 ≤ i ≤ j . Interpretation of (6.7.9): ε is the error in the best prediction of y based on x. If this
error were correlated with one of the components xi , then this correlation could be
used to construct a better prediction of y .
Answer. (6.7.5): E[εx] = E[y x] − E E[y x]x = 0 since E[y x] is a function of x and therefore
equal to its own expectation conditionally on x. (This is not the law of iterated expectations but
the law that the expected value of a constant is a constant.)
(6.7.6) follows from (6.7.5) (i.e., (6.7.5) is stronger than (6.7.6)): if an expectation is zero conditionally on every possible outcome of x then it is zero altogether. In formulas, E[ε] = E E[εx] =
E[0] = 0. It is also easy to show it in one swoop, without using (6.7.5): E[ε] = E[y − E[y x]] = 0.
Either way you need the law of iterated expectations for this.
(6.7.7): E[xi εx] = xi E[εx] = 0.
(6.7.8): E[xi ε] = E E[xi εx] = E[0] = 0; or in one swoop: E[xi ε] = E xi y − xi E[y x] =
E xi y − E[xi y x] = E[xi y ] − E[xi y ] = 0. The following “proof” is not correct: E[xi ε] = E[xi ] E[ε] =
E[xi ] · 0 = 0. xi and ε are generally not independent, therefore the multiplication rule E[xi ε] =
E[xi ] E[ε] cannot be used. Of course, the following “proof” does not work either: E[xi ε] = xi E[ε] =
xi · 0 = 0. xi is a random variable and E[xi ε] is a constant; therefore E[xi ε] = xi E[ε] cannot hold.
(6.7.9): cov[xi , ε] = E[xi ε] − E[xi ] E[ε] = 0 − E[xi ] · 0 = 0. • b. 2 points This part can only be done after discussing the multivariate normal
distribution:If x and y are jointly normal, show that x and ε are independent, and
that the variance of ε does not depend on x. (This is why one can consider it an
error term.)
Answer. If x and y are jointly normal, then x and ε are jointly normal as well, and independence follows from the fact that their covariance is zero. The variance is constant because in the
Normal case, the conditional variance is constant, i.e., E[ε2 ] = E E[ε2 x] = constant (does not
depend on x). Problem 133. 5 points Under the permanent income hypothesis, the assumption
is made that consumers’ lifetime utility is highest if the same amount is consumed
every year. The utilitymaximizing level of consumption c for a given consumer
depends on the actual state of the economy in each of the n years of the consumer’s
life c = f (y 1 , . . . , y n ). Since c depends on future states of the economy, which are
not known, it is impossible for the consumer to know this optimal c in advance; but
it is assumed that the function f and the joint distribution of y 1 , . . . , y n are known to
him. Therefore in period t, when he only knows the values of y 1 , . . . , y t , but not yet
the future values, the consumer decides to consume the amount ct = E[cy 1 , . . . , y t ],
which is the best possible prediction of c given the information available to him. Show
that in this situation, ct+1 − ct is uncorrelated with all y 1 , . . . , y t . This implication of
the permanent income hypothesis can be tested empirically, see [Hal78]. Hint: you
are allowed to use without proof the following extension of the theorem of iterated
expectations:
(6.7.10) E E[xy , z ] y = E[xy ]. Here is an explanation of (6.7.10): E[xy ] is the best predictor of x based on information set y . E[xy , z ] is the best predictor of x based on the extended information 82 6. VECTOR RANDOM VARIABLES set consisting of y and z . E E[xy , z ] y is therefore my prediction, based on y only,
how I will reﬁne my prediction when z becomes available as well. Its equality with
E[xy ], i.e., (6.7.10) says therefore that I cannot predict how I will change my mind
after better information becomes available.
Answer. In (6.7.10) set x = c = f (y 1 , . . . , y t , y t+1 , . . . , y n ), y = [y 1 , . . . , y t ] , and z = y t+1
to get
E E[cy 1 , . . . , y t+1 ] y 1 , . . . , y t = E[cy 1 , . . . , y t ]. (6.7.11) Writing ct for E[cy 1 , . . . , y t ], this becomes E[ct+1 y 1 , . . . , y t ] = ct , i.e., ct is not only the best
predictor of c, but also that of ct+1 . The change in consumption ct+1 − ct is therefore the prediction
error, which is uncorrelated with the conditioning variables, as shown in Problem 132. Problem 134. 3 points Show that for any two random variables x and y whose
covariance exists, the following equation holds:
cov[x, y ] = cov x, E[y x] (6.7.12) Note: Since E[y x] is the best predictor of y based on the observation of x, (6.7.12)
can also be written as
cov x, (y − E[y x]) = 0, (6.7.13) i.e., x is uncorrelated with the prediction error of the best prediction of y given x.
(Nothing to prove for this Note.)
Answer. Apply (6.1.6) to the righthand side of (6.7.12):
(6.7.14)
cov x, E[y x] = E xE[y x] −E[x] E E[y x] = E E[xy x] −E[x] E[y ] = E[xy ]−E[x] E[y ] = cov[x, y ].
The tricky part here is to see that xE[y x] = E[xy x]. Problem 135. Assume x and y have a joint density function fx,y (x, y ) which
is symmetric about the xaxis, i.e.,
fx,y (x, y ) = fx,y (x, −y ).
Also assume that variances and covariances exist. Show that cov[x, y ] = 0. Hint:
one way to do it is to look at E[y x].
Answer. We know that cov[x, y ] = cov x, E[y x] . Furthermore, from symmetry follows
E[y x] = 0. Therefore cov[x, y ] = cov[x, 0] = 0. Here is a detailed proof of E[y x] = 0: E[y x=x] =
∞ f (x,y ) ,
y xfy (x) dy . Now substitute z = −y , then also dz = −dy , and the boundaries of integration
x
are reversed: −∞ −∞ (6.7.15) E[y x=x] = z
∞ fx,y (x, −z )
dz =
f x ( x) −∞ z
∞ fx,y (x, z )
dz = − E[y x=x].
f x ( x) One can also prove directly under this presupposition cov[x, y ] = cov[x, −y ] and therefore it must
be zero. Problem 136. [Wit85, footnote on p. 241] Let p be the logarithm of the price
level, m the logarithm of the money supply, and x a variable representing real inﬂuences on the price level (for instance productivity). We will work in a model of the
economy in which p = m + γ x, where γ is a nonrandom parameter, and m and x are
2
2
independent normal with expected values µm , µx , and variances σm , σx . According
to the rational expectations assumption, the economic agents know the probability
distribution of the economy they live in, i.e., they know the expected values and variances of m and x and the value of γ . But they are unable to observe m and x, they 6.8. TRANSFORMATION OF VECTOR RANDOM VARIABLES 83 can only observe p. Then the best predictor of x using p is the conditional expectation
E[xp].
• a. Assume you are one of these agents and you observe p = p. How great
would you predict x to be, i.e., what is the value of E[xp = p]?
Answer. It is, according to formula (7.3.18), E[xp = p] = µx +
E[p] = µm + γµx , cov[x, p] = cov[x, m] + γ cov[x, x] = 2
γσx , and var(p) = cov(x,p)
(p − E[p]). Now
var(p)
2
2
σm + γ 2 σx . Therefore 2
γσx
E[xp = p] = µx + 2
(p − µm − γµx ).
2
σm + γ 2 σx (6.7.16) • b. Deﬁne the prediction error ε = x − E[xp]. Compute expected value and
variance of ε.
Answer.
(6.7.17) ε = x − µx − 2
γσx
(p − µm − γµx ).
2
2
σm + γ 2 σx This has zero expected value, and its variance is
(6.7.18) var[ε] = var[x] + (6.7.19) 2
= σx + (6.7.20) = 2
γ σx
2
2
σm + γ 2 σx 2 var[p] − 2 2
γ σx
cov[x, p] =
2
2
σm + γ 2 σ x 2
2
γ 2 (σ x ) 2
γ 2 ( σx )2
−2 2
2 σ2
2
γx
σ m + γ 2 σx 2
σm +
2 σ2
σx m
2
2
σm + γ 2 σx = 1+ 2
σx
.
2 σ 2 /σ 2
γxm 2
• c. In an attempt to ﬁne tune the economy, the central bank increases σm . Does
that increase or decrease var(ε)? Answer. From (6.7.20) follows that it increases the variance. 6.8. Transformation of Vector Random Variables
In order to obtain the density or probability mass function of a onetoone transformation of random variables, we have to follow the same 4 steps described in
Section 3.6 for a scalar random variable. (1) Determine A, the range of the new
variable, whose density we want to compute; (2) express the old variable, the one
whose density/mass function is known, in terms of the new variable, the one whose
x
x
density or mass function is needed. If that of
is known, set
= t(u, v ). Here
y
y
q (u, v )
t is a vectorvalued function, (i.e., it could be written t(u, v ) =
, but we will
r(u, v )
use one symbol t for this whole transformation), and you have to check that it is
onetoone on A, i.e., t(u, v ) = t(u1 , v1 ) implies u = u1 and v = v1 for all (u, v ) and
u1 , v1 ) in A. (A function for which two diﬀerent arguments (u, v ) and u1 , v1 ) give
the same function value is called manytoone.)
If the joint probability distribution of x and y is described by a probability mass
function, then the joint probability mass function of u and v can simply be obtained
by substituting t into the joint probability mass function of x and y (and it is zero
for any values which are not in A):
(6.8.1)
u
u
x
pu,v (u, v ) = Pr
=
= Pr t(u, v ) = t(u, v ) = Pr
= t(u, v ) = px,y t(u, v ) .
v
v
y 84 6. VECTOR RANDOM VARIABLES The second equal sign is where the condition enters that t : R2 → R2 is onetoone.
If one works with the density function instead of a mass function, one must
perform an additional step besides substituting t. Since t is onetoone, it follows
u
∈ dVu,v } = {t(u, v ) ∈ t(dV )x,y }.
v { (6.8.2)
Therefore
(6.8.3)
fu,v (u, v )dVu,v  = Pr[ u
∈ dVu,v ] = Pr[t(u, v ) ∈ t(dV )x,y ] = fx,y (t(u, v ))t(dV )x,y  =
v
= fx,y (t(u, v )) (6.8.4)
t(dV ) t(dV )x,y 
dVu,v .
dVu,v   x,y
The term dVu,v  is the local magniﬁcation factor of the transformation t;
analytically it is the absolute value J  of the Jacobian determinant (6.8.5) J= ∂x
∂u
∂y
∂u ∂x
∂v
∂y
∂v = ∂q
∂u (u, v )
∂r
∂u (u, v ) ∂q
∂v (u, v )
∂r
∂v (u, v ) . Remember, u, v are the new and x, y the old variables. To compute J one
has to express the old in terms of the new variables. If one expresses the new in
terms of the old, one has to take the inverse of the corresponding determinant! The
transformation rule for density functions can therefore be summarized as:
(x, y ) = t(u, v ) ⇒ fu,v (u, v ) = fx,y t(u, v ) J  where onetoone J= ∂x
∂u
∂y
∂u ∂x
∂v
∂y
∂v Problem 137. Let x and y be two random variables with joint density function
fx,y (x, y ).
• a. 3 points Deﬁne u = x + y . Derive the joint density function of u and y .
Answer. You have to express the “old” x and y as functions of the “new” u and y :
x=u−y
y=y or x
y = −1
1 1
0 u
y therefore J= ∂x
∂u
∂y
∂u ∂x
∂y
∂y
∂y = 1
0 −1
= 1.
1 Therefore
fu,y (u, y ) = fx,y (u − y, y ). (6.8.6) • b. 1 point Derive from this the following formula computing the density function fu (u) of the sum u = x + y from the joint density function fx,y (x, y ) of x and
y.
y =∞ (6.8.7) fx,y (u − y, y )dy. fu (u) =
y =−∞ Answer. Write down the joint density of u and y and then integrate y out, i.e., take its integral
over y from −∞ to +∞:
y =∞ (6.8.8) f u (u) = y =∞ y =−∞ i.e., one integrates over all x
y fx,y (u − y, y )dy. fu,y (u, y )dy = with x + y = u. y =−∞ . 6.8. TRANSFORMATION OF VECTOR RANDOM VARIABLES 85 Problem 138. 6 points Let x and y be independent and uniformly distributed
over the interval [0, 1]. Compute the density function of u = x + y and draw its
graph. Hint: you may use formula (6.8.7) for the density of the sum of two jointly
distributed random variables. An alternative approach would be to ﬁrst compute the
cumulative distribution function Pr[x + y ≤ u] for all u.
Answer. Using equation (6.8.7): (6.8.9) fx,y (u − y, y ) dy = fx+y (u) = Tq d
E
q
dq
for 1 ≤ u ≤ 2 −∞ u for 0 ≤ u ≤ 1 ∞ otherwise. 2−u
0 To help evaluate this integral, here is the area in u, y plane (u = x + y on the horizontal and y on
the vertical axis) in which fx,y (u − v, v ) has the value 1: q
Tq E
q
q
This is the area between (0,0), (1,1), (2,1), and (1,0).
One can also show it this way: fx,y (x, y ) = 1 iﬀ 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. Now take any ﬁxed
u. It must be between 0 and 2. First assume 0 ≤ u ≤ 1: then fx,y (u − y, y ) = 1 iﬀ 0 ≤ u − y ≤ 1
and 0 ≤ y ≤ 1 iﬀ 0 ≤ y ≤ u. Now assume 1 ≤ u ≤ 2: then fx,y (u − y, y ) = 1 iﬀ u − 1 ≤ y ≤ 1. Problem 139. Assume [ x ] is uniformly distributed on a round disk around the
y
origin with radius 10.
• a. 4 points Derive the joint density, the marginal density of x, and the conditional density of y given x=x.
• b. 3 points Now let us go over to polar coordinates r and φ, which satisfy
(6.8.10) x = r cos φ
y = r sin φ , i.e., the vector transformation t is t( r
r cos φ
)=
.
φ
r sin φ r
Which region in φ space is necessary to cover x space? Compute the Jacobian
y
determinant of this transformation. Give an intuitive explanation in terms of local
magniﬁcation factor of the formula you get. Finally compute the transformed density
function. • c. 1 point Compute cov[x, y ].
• d. 2 points Compute the conditional variance var[y x=x].
• e. 2 points Are x and y independent?
Problem 140. [Ame85, pp. 296–7] Assume three transportation choices are
available: bus, train, and car. If you pick at random a neoclassical individual ω
and ask him or her which utility this person derives from using bus, train, and car,
the answer will be three numbers u1 (ω ), u2 (ω ), u3 (ω ). Here u1 , u2 , and u3 are assumed to be independent random variables with the following cumulative distribution
functions:
(6.8.11) Pr[ui ≤ u] = Fi (u) = exp − exp(µi − u) , i = 1, 2, 3. I.e., the functional form is the same for all three transportation choices (exp indicates the exponential function); the Fi only diﬀer by the parameters µi . These
probability distributions are called Type I extreme value distributions, or log Weibull
distributions. 86 6. VECTOR RANDOM VARIABLES Often these kinds of models are set up in such a way that these µi to depend on
the income etc. of the individual, but we assume for this exercise that this distribution
applies to the population as a whole.
• a. 1 point Show that the Fi are indeed cumulative distribution functions, and
derive the density functions fi (u).
Individual ω likes cars best if and only if his utilities satisfy u3 (ω ) ≥ u1 (ω ) and
u3 (ω ) ≥ u2 (ω ). Let I be a function of three arguments such that I (u1 , u2 , u3 ) is the
indicator function of the event that one randomly chooses an individual ω who likes
cars best, i.e.,
(6.8.12) I (u1 , u2 , u3 ) = 1
0 if u1 ≤ u3 and u2 ≤ u3
otherwise. Then Pr[car] = E[I (u1 , u2 , u3 )]. The following steps have the purpose to compute
this probability:
• b. 2 points For any ﬁxed number u, deﬁne g (u) = E[I (u1 , u2 , u3 )u3 = u].
Show that
g (u) = exp − exp(µ1 − u) − exp(µ2 − u) . (6.8.13)
• c. 2 points This here is merely the evaluation of an integral. Show that +∞ exp − exp(µ1 − u) − exp(µ2 − u) − exp(µ3 − u) exp(µ3 − u) du =
−∞ exp µ3
.
exp µ1 + exp µ2 + exp µ3
Hint: use substitution rule with y = − exp(µ1 − u) − exp(µ2 − u) − exp(µ3 − u).
= • d. 1 point Use b and c to show that
(6.8.14) Pr[car] = exp µ3
.
exp µ1 + exp µ2 + exp µ3 CHAPTER 7 The Multivariate Normal Probability Distribution
7.1. More About the Univariate Case
By deﬁnition, z is a standard normal variable, in symbols, z ∼ N (0, 1), if it has
the density function
z2
1
fz (z ) = √ e− 2 .
2π (7.1.1) To verify that this is a density function we have to check two conditions. (1) It is
everywhere nonnegative. (2) Its integral from −∞ to ∞ is 1. In order to evaluate this
integral, it is easier to work with the independent product of two standard normal
x2 +y 2 variables x and y ; their joint density function is fx,y (x, y ) = 21 e− 2 . In order to
π
see that this joint density integrates to 1, go over to polar coordinates x = r cos φ,
y = r sin φ, i.e., compute the joint distribution of r and φ from that of x and y : the
absolute value of the Jacobian determinant is r, i.e., dx dy = r dr dφ, therefore
y =∞ x=∞ y =−∞ x=−∞ (7.1.2) 1 − x2 +y2
2
e
dx dy =
2π 2π ∞ φ=0 r =0 1 − r2
e 2 r dr dφ.
2π
∞ By substituting t = r2 /2, therefore dt = r dr, the inner integral becomes − 21 e−t 0 =
π
1
2π ; therefore the whole integral is 1. Therefore the product of the integrals of the
marginal densities is 1, and since each such marginal integral is positive and they are
equal, each of the marginal integrals is 1 too.
∞ Problem 141. 6 points The Gamma function can be deﬁned as Γ(r) = 0 xr−1 e−x dx.
√
Show that Γ( 1 ) = π . (Hint: after substituting r = 1/2, apply the variable transfor2
mation x = z 2 /2 for nonnegative x and z only, and then reduce the resulting integral
to the integral over the normal density function.)
Answer. Then dx = z dz ,
normal density:
∞ (7.1.3)
0 dx
√
x = dz √
1
√ e−x dx = 2
x √ 2. Therefore one can reduce it to the integral over the ∞ e−z
0 2 /2 1
dz = √
2 ∞ e−z
−∞ 2 /2 √
√
2π
dz = √ = π.
2 A univariate normal variable with mean µ and variance σ 2 is a variable x whose
standardized version z = x−µ ∼ N (0, 1). In this transformation from x to z , the
σ
dz
1
Jacobian determinant is dx = σ ; therefore the density function of x ∼ N (µ, σ 2 ) is
(two notations, the second is perhaps more modern:)
(7.1.4) fx (x) = √ 1
2πσ 2 e− (x−µ)2
2σ 2 = (2πσ 2 )−1/2 exp −(x − µ)2 /2σ 2 .
87 88 7. MULTIVARIATE NORMAL Problem 142. 3 points Given n independent observations of a Normally distributed variable y ∼ N (µ, 1). Show that the sample mean y is a suﬃcient statis¯
tic for µ. Here is a formulation of the factorization theorem for suﬃcient statistics, which you will need for this question: Given a family of probability densities
fy (y1 , . . . , yn ; θ) deﬁned on Rn , which depend on a parameter θ ∈ Θ. The statistic
T : Rn → R, y1 , . . . , yn → T (y1 , . . . , yn ) is suﬃcient for parameter θ if and only if
there exists a function of two variables g : R × Θ → R, t, θ → g (t; θ), and a function
of n variables h : Rn → R, y1 , . . . , yn → h(y1 , . . . , yn ) so that
fy (y1 , . . . , yn ; θ) = g T (y1 , . . . , yn ); θ · h(y1 , . . . , yn ). (7.1.5) Answer. The joint density function can be written (factorization indicated by ·):
(7.1.6)
(2π )−n/2 exp − 1
2 n (yi −µ)2 = (2π )−n/2 exp −
i=1 1
2 n n
y
y
(yi −y )2 ·exp − (¯−µ)2 = h(y1 , . . . , yn )·g (¯; µ).
¯
2 i=1 7.2. Deﬁnition of Multivariate Normal
The multivariate normal distribution is an important family of distributions with
very nice properties. But one must be a little careful how to deﬁne it. One might
naively think a multivariate Normal is a vector random variable each component
of which is univariate Normal. But this is not the right deﬁnition. Normality of
the components is a necessary but not suﬃcient condition for a multivariate normal
x
vector. If u =
with both x and y multivariate normal, u is not necessarily
y
multivariate normal.
Here is a recursive deﬁnition from which one gets all multivariate normal distributions:
(1) The univariate standard normal z , considered as a vector with one component, is multivariate normal.
x
(2) If x and y are multivariate normal and they are independent, then u =
y
is multivariate normal.
(3) If y is multivariate normal, and A a matrix of constants (which need not
be square and is allowed to be singular), and b a vector of constants, then Ay + b
is multivariate normal. In words: A vector consisting of linear combinations of the
same set of multivariate normal variables is again multivariate normal.
For simplicity we will go over now to the bivariate Normal distribution.
7.3. Special Case: Bivariate Normal
The following two simple rules allow to obtain all bivariate Normal random
variables:
(1) If x and y are independent and each of them has a (univariate) normal
distribution with mean 0 and the same variance σ 2 , then they are bivariate normal.
(They would be bivariate normal even if their variances were diﬀerent and their
means not zero, but for the calculations below we will use only this special case, which
together with principle (2) is suﬃcient to get all bivariate normal distributions.)
x
(2) If x =
is bivariate normal and P is a 2 × 2 nonrandom matrix and µ
y
a nonrandom column vector with two elements, then P x + µ is bivariate normal as
well. 7.3. BIVARIATE NORMAL 89 All other properties of bivariate Normal variables can be derived from this.
First let us derive the density function of a bivariate Normal distribution. Write
x
x=
. x and y are independent N (0, σ 2 ). Therefore by principle (1) above the
y
vector x is bivariate normal. Take any nonsingular 2 × 2 matrix P and a 2 vector
u
µ
= u = P x + µ. We need nonsingularity because otherwise
µ=
, and deﬁne
v
ν
the resulting variable would not have a bivariate density; its probability mass would
be concentrated on one straight line in the twodimensional plane. What is the
joint density function of u? Since P is nonsingular, the transformation is ontoone,
therefore we can apply the transformation theorem for densities. Let us ﬁrst write
down the density function of x which we know:
1
1
exp − 2 (x2 + y 2 ) .
2πσ 2
2σ
For the next step, remember that we have to express the old variable in terms
of the new one: x = P −1 (u − µ). The Jacobian determinant is therefore J =
x
u−µ
det(P −1 ). Also notice that, after the substitution
= P −1
, the expoy
v−ν
fx,y (x, y ) = (7.3.1) 1
1
nent in the joint density function of x and y is − 2σ2 (x2 + y 2 ) = − 2σ2 x
y x
=
y u−µ
u−µ
P −1 P −1
. Therefore the transformation theorem of density
v−ν
v−ν
functions gives
1
− 2σ 2 (7.3.2) fu,v (u, v ) = 1
1 u−µ
det(P −1 ) exp − 2
2
2πσ
2σ v − ν P −1 P −1 u−µ
v−ν . This expression can be made nicer. Note that the covariance matrix of the
u = σ 2 P P = σ 2 Ψ, say. Since P −1 P −1 P P = I ,
transformed variables is V [
v
it follows P −1 P −1 = Ψ−1 and det(P −1 ) = 1/ det(Ψ), therefore
(7.3.3) fu,v (u, v ) = 1
2πσ 2 1
det(Ψ) exp − 1 u−µ
2σ 2 v − ν Ψ−1 u−µ
v−ν . This is the general formula for the density function of a bivariate normal with nonsingular covariance matrix σ 2 Ψ and mean vector µ. One can also use the following
notation which is valid for the multivariate Normal variable with n dimensions, with
mean vector µ and nonsingular covariance matrix σ 2 Ψ:
(7.3.4) fx (x) = (2πσ 2 )−n/2 (det Ψ)−1/2 exp − 1
(x − µ) Ψ−1 (x − µ) .
2σ 2 Problem 143. 1 point Show that the matrix product of (P −1 ) P −1 and P P
is the identity matrix.
Problem 144. 3 points All vectors in this question are n × 1 column vectors.
Let y = α + ε , where α is a vector of constants and ε is jointly normal with E [ε ] = o.
Often, the covariance matrix V [ε ] is not given directly, but a n × n nonsingular matrix
T is known which has the property that the covariance matrix of T ε is σ 2 times the
n × n unit matrix, i.e.,
(7.3.5) 2
V [T ε ] = σ I n . 90 7. MULTIVARIATE NORMAL Show that in this case the density function of y is
1
(7.3.6)
fy (y ) = (2πσ 2 )−n/2 det(T ) exp − 2 T (y − α) T (y − α) .
2σ
Hint: deﬁne z = T ε , write down the density function of z , and make a transformation between z and y .
Answer. Since E [z ] = o and V [z ] = σ 2 I n , its density function is (2πσ 2 )−n/2 exp(−z z /2σ 2 ).
Now express z , whose density we know, as a function of y , whose density function we want to know.
z = T (y − α) or
(7.3.7) z1 = t11 (y1 − α1 ) + t12 (y2 − α2 ) + · · · + t1n (yn − αn )
.
.
. (7.3.8)
(7.3.9) zn = tn1 (y1 − α1 ) + tn2 (y1 − α2 ) + · · · + tnn (yn − αn ) therefore the Jacobian determinant is det(T ). This gives the result. 7.3.1. Most Natural Form of Bivariate Normal Density.
Problem 145. In this exercise we will write the bivariate normal density in its
most natural form. For this we set the multiplicative “nuisance parameter” σ 2 = 1,
i.e., write the covariance matrix as Ψ instead of σ 2 Ψ.
u
• a. 1 point Write the covariance matrix Ψ = V [ in terms of the standard
v
deviations σu and σv and the correlation coeﬃcient ρ.
• b. 1 point Show that the inverse of a 2 × 2 matrix has the following form:
ab
cd (7.3.10) −1 = 1
d −b
.
ad − bc −c a • c. 2 points Show that
(7.3.11)
(7.3.12) q 2 = u − µ v − ν Ψ−1
= u−µ
v−ν u−µv−ν
1
(u − µ)2
(v − ν )2
− 2ρ
+
.
2
2
2
1−ρ
σu
σu
σv
σv • d. 2 points Show the following quadratic decomposition:
(7.3.13) q2 = (u − µ)2
1
σv
+
v − ν − ρ (u − µ)
2
2
σu
(1 − ρ2 )σv
σu 2 . • e. 1 point Show that (7.3.13) can also be written in the form
2
(u − µ)2
σ2
σ uv
+ 22 u
v − ν − 2 (u − µ) .
2
2
σu
σu σv − (σuv )
σu
√
• f . 1 point Show that d = det Ψ can be split up, not additively but multiplicatively, as follows: d = σu · σv 1 − ρ2 . (7.3.14) q2 = • g. 1 point Using these decompositions of d and q 2 , show that the density function fu,v (u, v ) reads
(7.3.15)
2
σv
(v − ν ) − ρ σu (u − µ)
1
(u − µ)2
1
exp −
·
exp −
.
2
2
2
2
2σ u
2(1 − ρ2 )σv
2πσu
2πσv 1 − ρ2 7.3. BIVARIATE NORMAL 91 σv
2
The second factor in (7.3.15) is the density of a N (ρ σu u, (1 − ρ2 )σv ) evaluated
at v , and the ﬁrst factor does not depend on v . Therefore if I integrate v out to
get the marginal density of u, this simply gives me the ﬁrst factor. The conditional
density of v given u = u is the joint divided by the marginal, i.e., it is the second
factor. In other words, by completing the square we wrote the joint density function
in its natural form as the product of a marginal and a conditional density function:
fu,v (u, v ) = fu (u) · fvu (v ; u).
From this decomposition one can draw the following conclusions:
2
• u ∼ N (0, σu ) is normal and, by symmetry, v is normal as well. Note that u
(or v ) can be chosen to be any nonzero linear combination of x and y . Any
nonzero linear transformation of independent standard normal variables is
therefore univariate normal.
• If ρ = 0 then the joint density function is the product of two independent
univariate normal density functions. In other words, if the variables are
normal, then they are independent whenever they are uncorrelated. For
general distributions only the reverse is true.
• The conditional density of v conditionally on u = u is the second term on
the rhs of (7.3.15), i.e., it is normal too.
• The conditional mean is
σv
(7.3.16)
E[v u = u] = ρ u,
σu i.e., it is a linear function of u. If the (unconditional) means are not zero,
then the conditional mean is
σv
(7.3.17)
E[v u = u] = µv + ρ (u − µu ).
σu
Since ρ =
(7.3.18) cov[u,v ]
σu σv , (7.3.17) can als be written as follows: E[v u = u] = E[v ] + cov[u, v ]
(u − E[u])
var[u] • The conditional variance is the same whatever value of u was chosen: its
value is
(7.3.19) 2
var[v u = u] = σv (1 − ρ2 ), which can also be written as
(7.3.20) var[v u = u] = var[v ] − (cov[u, v ])2
.
var[u] We did this in such detail because any bivariate normal with zero mean has this
form. A multivariate normal distribution is determined by its means and variances
and covariances (or correlations coeﬃcients). If the means are not zero, then the
densities merely diﬀer from the above by an additive constant in the arguments, i.e.,
if one needs formulas for nonzero mean, one has to replace u and v in the above
equations by u − µu and v − µv . du and dv remain the same, because the Jacobian
of the translation u → u − µu , v → v − µv is 1. While the univariate normal was
determined by mean and standard deviation, the bivariate normal is determined by
the two means µu and µv , the two standard deviations σu and σv , and the correlation
coeﬃcient ρ. 92 7. MULTIVARIATE NORMAL 7.3.2. Level Lines of the Normal Density.
Problem 146. 8 points Deﬁne the angle δ = arccos(ρ), i.e, ρ = cos δ . In terms
of δ , the covariance matrix (??) has the form
(7.3.21) Ψ= 2
σu
σu σv cos δ σu σv cos δ
2
σv Show that for all φ, the vector
(7.3.22) r σu cos φ
r σv cos(φ + δ ) x= satisﬁes x Ψ−1 x = r2 . The opposite holds too, all vectors x satisfying x Ψ−1 x =
r2 can be written in the form (7.3.22) for some φ, but I am not asking to prove this.
This formula can be used to draw level lines of the bivariate Normal density and
conﬁdence ellipses, more details in (??).
Problem 147. The ellipse in Figure 1 contains all the points x, y for which
(7.3.23) x−1 y−1 0.5
−0.25
−0.25
1 −1 x−1
≤6
y−1 • a. 3 points Compute the probability that a random variable
1
0.5
−0.25
,
1
−0.25
1 x
∼N
y (7.3.24) falls into this ellipse. Hint: you should apply equation (7.4.9). Then you will have
to look up the values of a χ2 distribution in a table, or use your statistics software
to get it.
• b. 1 point Compute the standard deviations of x and y , and the correlation
coeﬃcient corr(x, y )
• c.√ points The vertical tangents to the ellipse in Figure 1 are at the locations
2
x = 1 ± 3. What is the probability that [ x ] falls between these two vertical tangents?
y
√
• d. 1 point The horizontal tangents are at the locations y = 1 ± 6. What is
the probability that [ x ] falls between the horizontal tangents?
y
• e. 1 point Now take an arbitrary linear combination u = ax + by . Write down
its mean and its standard deviation.
√
• f . 1 point Show that the set of realizations x, y for which u lies less than 6
standard deviation away from its mean is
√
(7.3.25)
a(x − 1) + b(y − 1) ≤ 6 a2 var[x] + 2ab cov[x, y ] + b2 var[y ].
The set of all these points forms a band limited by two parallel lines. What is the
probability that [ x ] falls between these two lines?
y
• g. 1 point It is our purpose to show that this band is again tangent to the
ellipse. This is easiest if we use matrix notation. Deﬁne
(7.3.26) x= x
y µ= 1
1 Ψ= 0.5
−0.25
−0.25
1 a= a
b Equation (7.3.23) in matrix notation says: the ellipse contains all the points for
which
(7.3.27) (x − µ) Ψ−1 (x − µ) ≤ 6. 7.3. BIVARIATE NORMAL −2 −1 x=0 1 93 2 3 4 4 4
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....................... 3 2 1 y=0 −1 3 2 1 y=0 −1 −2 −2
−2 −1 x=0 1 2 3 Figure 1. Level Line for Normal Density
Show that the band deﬁned by inequality (7.3.25) contains all the points for which
a (x − µ)
a Ψa (7.3.28) 2 ≤ 6. • h. 2 points Inequality (7.3.28) can also be written as:
(x − µ) a(a Ψa)−1 a (x − µ) ≤ 6 (7.3.29)
or alternatively
(7.3.30) x−1 y−1 a
b a b Ψ−1 a
b −1 Show that the matrix
(7.3.31) Ω = Ψ−1 − a(a Ψa)−1 a x−1
y−1 a b ≤ 6. 4 94 7. MULTIVARIATE NORMAL satisﬁes Ω ΨΩ = Ω . Derive from this that Ω is nonnegative deﬁnite. Hint: you may
use, without proof, that any symmetric matrix is nonnegative deﬁnite if and only if
it can be written in the form RR .
• i. 1 point As an aside: Show that Ω Ψa = o and derive from this that Ω is not
positive deﬁnite but only nonnegative deﬁnite.
• j. 1 point Show that the following inequality holds for all x − µ,
(x − µ) Ψ−1 (x − µ) ≥ (x − µ) a(a Ψa)−1 a (x − µ). (7.3.32) In other words, if x lies in the ellipse then it also lies in each band. I.e., the ellipse
is contained in the intersection of all the bands.
• k. 1 point Show: If x − µ = Ψaα with some arbitrary scalar α, then (7.3.32)
is an equality, and if α = ± 6/a Ψa, then both sides in (7.3.32) have the value 6.
I.e., the boundary of the ellipse and the boundary lines of the band intersect. Since
the ellipse is completely inside the band, this can only be the case if the boundary
lines of the band are tangent to the ellipse.
• l. 2 points The vertical lines in Figure 1 which are not tangent to the ellipse
delimit a band which, if extended to inﬁnity, has as much probability mass as the
ellipse itself. Compute the xcoordinates of these two lines.
7.3.3. Miscellaneous Exercises.
Problem 148. Figure 2 shows the level line for a bivariate Normal density which
contains 95% of the probability mass.
..................
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−1
0
1
3
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..... −1 0 1 2 3 2 1 0 −1
3 Figure 2. Level Line of Bivariate Normal Density, see Problem 148 7.3. BIVARIATE NORMAL 95 x
. Ψ1 =
y
0.62 −0.56
1.85 1.67
0.62 0.56
1.85 −1.67
, Ψ2 =
, Ψ3 =
, Ψ4 =
,
−0.56 1.04
1.67 3.12
0.56 1.04
1.67 3.12
3.12 −1.67
1.04 0.56
3.12 1.67
0.62 0.81
Ψ5 =
, Ψ6 =
, Ψ7 =
, Ψ8 =
,
−1.67 1.85
0.56 0.62
1.67 1.85
0.81 1.04
3.12 1.67
0.56 0.62
Ψ9 =
, Ψ10 =
. Which is it? Remember that for a uni2.67 1.85
0.62 −1.04
variate Normal, 95% of the probability mass lie within ±2 standard deviations from
the mean. If you are not sure, cross out as many of these covariance matrices as
possible and write down why you think they should be crossed out.
• a. 3 points One of the following matrices is the covariance matrix of Answer. Covariance matrix must be symmetric, therefore we can cross out 4 and 9. It must
also be nonnegative deﬁnite (i.e., it must have nonnegative elements in the diagonal), therefore
cross out 10, and a nonnegative determinant, therefore cross out 8. Covariance must be positive, so
cross out 1 and 5. Variance in xdirection is smaller than in ydirection, therefore cross out 6 and
7. Remains 2 and 3.
Of these it is number 3. By comparison with Figure 1 one can say that the vertical band
between 0.4 and 2.6 and the horizontal band between 3 and 1 roughly have the same probability
as the ellipse, namely 95%. Since a univariate Normal has 95% of its probability mass in an
interval centered around the mean which is 4 standard deviations long, standard deviations must
be approximately 0.8 in the horizontal and 1 in the vertical directions.
Ψ1 is negatively correlated; Ψ2 has the right correlation but is scaled too big; Ψ3 this is it; Ψ4
not symmetric; Ψ5 negatively correlated, and x has larger variance than y ; Ψ6 x has larger variance
than y ; Ψ7 too large, x has larger variance than y ; Ψ8 not positive deﬁnite; Ψ9 not symmetric;
Ψ10 not positive deﬁnite. The next Problem constructs a counterexample which shows that a bivariate distribution, which is not bivariate Normal, can nevertheless have two marginal densities
which are univariate Normal.
Problem 149. Let x and y be two independent standard normal random vari2
2
ables, and let u and v be bivariate normal with mean zero, variances σu = σv = 1,
and correlation coeﬃcient ρ = 0. Let fx,y and fu,v be the corresponding density
functions, i.e.,
fx,y (a, b) = 1
a2 + b2
exp(−
) fu,v (a, b) =
2π
2
2π 1
1 − ρ2 exp(−a2 + b2 − 2ρa b
).
2(1 − ρ2 ) Assume the random variables a and b are deﬁned by the following experiment: You
ﬂip a fair coin; if it shows head, then you observe x and y and give a the value
observed on x, and b the value observed of y . If the coin shows tails, then you
observe u and v and give a the value of u, and b the value of v .
• a. Prove that the joint density of a and b is
1
1
(7.3.33)
fa,b (a, b) = fx,y (a, b) + fu,v (a, b).
2
2
Hint: ﬁrst show the corresponding equation for the cumulative distribution functions.
Answer. Following this hint:
(7.3.34)
(7.3.35) Fa,b (a, b) = Pr[a ≤ a and b ≤ b] =
= Pr[a ≤ a and b ≤ bhead] Pr[head] + Pr[a ≤ a and b ≤ btail] Pr[tail] 1
1
+ Fu,v (a, b) .
2
2
The density function is the function which, if integrated, gives the above cumulative distribution
function.
(7.3.36) = Fx,y (a, b) 96 7. MULTIVARIATE NORMAL • b. Show that the marginal distribution of a and b each is normal.
Answer. You can either argue it out: each of the above marginal distributions is standard
normal, but you can also say integrate b out; for this it is better to use form (7.3.15) for fu,v , i.e.,
write
a2
1
exp −
fu,v (a, b) = √
2
2π (7.3.37) ·√ 1 exp − 1 − ρ2 2π (b − ρa)2
.
2(1 − ρ2 ) Then you can see that the marginal is standard normal. Therefore you get a mixture of two
distributions each of which is standard normal, therefore it is not really a mixture any more. • c. Compute the density of b conditionally on a = 0. What are its mean and
variance? Is it a normal density?
√ Answer. Fba (b; a) = fa,b (a,b)
.
fa (a) We don’t need it for every a, only for a = 0. Since fa (0) = 1/ 2π , therefore
(7.3.38) fba=0 (b) = √ 2πfa,b (0, b) = 1
11
−b2
1
−b2
exp
+√
exp
.
√
2 2π
2
2 2 π 1 − ρ2
2(1 − ρ2 ) It is not normal, it is a mixture of normals with diﬀerent variances. This has mean zero and variance
1
(1 + (1 − ρ2 )) = 1 − 1 ρ2 .
2
2 • d. Are a and b jointly normal?
Answer. Since the conditional distribution is not normal, they cannot be jointly normal. Problem 150. This is [HT83, 4.86 on p. 263] with variance σ 2 instead of 1:
Let x and y be independent normal with mean 0 and variance σ 2 . Go over to polar
coordinates r and φ, which satisfy
x = r cos φ (7.3.39) y = r sin φ. • a. 1 point Compute the Jacobian determinant.
Answer. Express the variables whose density you know in terms of those whose density you
want to know. The Jacobian determinant is
(7.3.40) J= ∂x
∂r
∂y
∂r ∂x
∂φ
∂y
∂φ = cos φ
sin φ −r sin φ
= (cos φ)2 + (sin φ)2 r = r.
r cos φ • b. 2 points Find the joint probability density function of r and φ. Also indicate
the area in (r, φ) space in which it is nonzero.
Answer. fx,y (x, y ) =
∞ and 0 ≤ φ < 2π . 2
2
2
1
e−(x +y )/2σ ;
2πσ 2 therefore fr,φ (r, φ) = 2
2
1
re−r /2σ
2πσ 2 for 0 ≤ r < • c. 3 points Find the marginal distributions of r and φ. Hint: for one of the
integrals it is convenient to make the substitution q = r2 /2σ 2 .
2
2
1
re−r /2σ for 0
σ2
2
2
∞
re−r /2σ dr = 21 , set
π
0 Answer. fr (r ) =
1
we need 2πσ2
∞ −q
1
e dq .
2π 0 ≤ r < ∞, and fφ (φ) =
q= r 2 /2σ 2 , then dq = 1
for 0 ≤ φ < 2π . For the latter
2π
1
2 r dr , and the integral becomes
σ • d. 1 point Are r and φ independent?
Answer. Yes, because joint density function is the product of the marginals. 7.4. MULTIVARIATE STANDARD NORMAL IN HIGHER DIMENSIONS 97 7.4. Multivariate Standard Normal in Higher Dimensions
Here is an important fact about the multivariate normal, which one cannot see in
x
two dimensions: if the partitioned vector
is jointly normal, and every component
y
of x is independent of every component of y , then the vectors x and y are already
independent. Not surprised? You should be, see Problem 125.
Let’s go back to the construction scheme at the beginning of this chapter. First
we will introduce the multivariate standard normal, which one obtains by applying
only operations (1) and (2), i.e., it is a vector composed of independent univariate
standard normals, and give some properties of it. Then we will go over to the
multivariate normal with arbitrary covariance matrix, which is simply an arbitrary
linear transformation of the multivariate standard normal. We will always carry the
“nuisance parameter” σ 2 along.
Definition 7.4.1. The random vector z is said to have a multivariate standard
normal distribution with variance σ 2 , written as z ∼ N (o, σ 2 I ), if each element z i is
a standard normal with same variance σ 2 , and all elements are mutually independent
of each other. (Note that this deﬁnition of the standard normal is a little broader
than the usual one; the usual one requires that σ 2 = 1.)
The density function of a multivariate standard normal z is therefore the product
of the univariate densities, which gives fx (z ) = (2πσ 2 )−n/2 exp(−z z /2σ 2 ).
The following property of the multivariate standard normal distributions is basic:
Theorem 7.4.2. Let z be multivariate standard normal pvector with variance
σ 2 , and let P be a m × p matrix with P P = I . Then x = P z is a multivariate
standard normal mvector with the same variance σ 2 , and z z − x x ∼ σ 2 χ2−m
p
independent of x.
Proof. P P = I means all rows are orthonormal. If P is not square, it
must therefore have more columns than rows, and one can add more rows to get an
P
orthogonal square matrix, call it T =
. Deﬁne y = T z , i.e., z = T y . Then
Q
z z = y T T y = y y , and the Jacobian of the transformation from y to z has absolute value one. Therefore the density function of y is (2πσ 2 )−n/2 exp(−y y /2σ 2 ),
which means y is standard normal as well. In other words, every y i is univariate standard normal with same variance σ 2 and y i is independent of y j for i = j . Therefore
also any subvector of y , such as x, is standard normal. Since z z −x x = y y −x x
is the sum of the squares of those elements of y which are not in x, it follows that it
is an independent σ 2 χ2−m .
p
Problem 151. Show that the moment generating function of a multivariate standard normal with variance σ 2 is mz (t) = E [exp(t z )] = exp(σ 2 t t/2).
Answer. Proof: The moment generating function is deﬁned as
(7.4.1) mz (t) = E[exp(t z )] (7.4.2) = (2πσ 2 )n/2 ··· exp(− 1
z z ) exp(t z ) dz1 · · · dzn
2σ 2 (7.4.3) = (2πσ 2 )n/2 ··· exp(− 1
σ2
(z − σ 2 t ) (z − σ 2 t ) +
t t) dz1 · · · dzn
2σ 2
2 (7.4.4) = exp( σ2
t t)
2 since ﬁrst part of integrand is density function. 98 7. MULTIVARIATE NORMAL Theorem 7.4.3. Let z ∼ N (o, σ 2 I ), and P symmetric and of rank r. A necessary and suﬃcient condition for q = z P z to have a σ 2 χ2 distribution is P 2 = P .
In this case, the χ2 has r degrees of freedom.
Proof of suﬃciency: If P 2 = P with rank r, then a matrix T exists with P =
T T and T T = I . Deﬁne x = T z ; it is standard normal by theorem 7.4.2.
r
Therefore q = z T T z = i=1 x2 .
Proof of necessity by construction of the moment generating function of q =
z P z for arbitrary symmetric P with rank r. Since P is symmetric, there exists a
T with T T = I r and P = T ΛT where Λ is a nonsingular diagonal matrix, write
r
it Λ = diag(λ1 , . . . , λr ). Therefore q = z T ΛT z = x Λx = i=1 λi x2 where
i
2
x = T z ∼ N (o, σ I r ). Therefore the moment generating function
r (7.4.5) λi x2 )]
i E[exp(q t)] = E[exp(t
i=1 (7.4.6) = E[exp(tλ1 x2 )] · · · E[exp(tλr x2 )]
1
r (7.4.7) = (1 − 2λ1 σ 2 t)−1/2 · · · (1 − 2λr σ 2 t)−1/2 . By assumption this is equal to (1 − 2σ 2 t)−k/2 with some integer k ≥ 1. Taking
squares and inverses one obtains
(7.4.8) (1 − 2λ1 σ 2 t) · · · (1 − 2λr σ 2 t) = (1 − 2σ 2 t)k . Since the λi = 0, one obtains λi = 1 by uniqueness of the polynomial roots. Furthermore, this also implies r = k .
From Theorem 7.4.3 one can derive a characterization of all the quadratic forms
of multivariate normal variables with arbitrary covariance matrices that are χ2 ’s.
Assume y is a multivariate normal vector random variable with mean vector µ and
covariance matrix σ 2 Ψ, and Ω is a symmetric nonnegative deﬁnite matrix. Then
(y − µ) Ω (y − µ) ∼ σ 2 χ2 iﬀ
k
(7.4.9) ΨΩ ΨΩ Ψ = ΨΩ Ψ, and k is the rank of ΨΩ.
Here are the three best known special cases (with examples):
• Ψ = I (the identity matrix) and Ω 2 = Ω , i.e., the case of theorem 7.4.3.
This is the reason why the minimum value of the SSE has a σ 2 χ2 distribution, see (27.0.10).
• Ψ nonsingular and Ω = Ψ−1 . The quadratic form in the exponent of
the normal density function is therefore a χ2 ; one needs therefore the χ2
to compute the probability that the realization of a Normal is in a given
equidensityellipse (Problem 147).
• Ψ singular and Ω = Ψ− , its ginverse. The multinomial distribution has a
singular covariance matrix, and equation (??) gives a convenient ginverse
which enters the equation for Pearson’s goodness of ﬁt test.
Here are, without proof, two more useful theorems about the standard normal:
Theorem 7.4.4. Let x a multivariate standard normal. Then x P x is independent of x Qx if and only if P Q = O .
This is called Craig’s theorem, although Craig’s proof in [Cra43] is incorrect.
Kshirsagar [Ksh19, p. 41] describes the correct proof; he and Seber [Seb77] give
Lancaster’s book [Lan69] as basic reference. Seber [Seb77] gives a proof which is
only valid if the two quadratic forms are χ2 . 7.4. MULTIVARIATE STANDARD NORMAL IN HIGHER DIMENSIONS 99 The next theorem is known as James’s theorem, it is a stronger version of
Cochrane’s theorem. It is from Kshirsagar [Ksh19, p. 41].
Theorem 7.4.5. Let x be pvariate standard normal with variance σ 2 , and
k
x x=
i=1 x P i x. Then for the quadratic forms x P i x to be independently
distributed as σ 2 χ2 , any one of the following three equivalent conditions is necessary
and suﬃcient:
P2 = Pi
i (7.4.10)
(7.4.11) P iP j = O
k (7.4.12) rank(P i ) = p
i=1 for all i
i=j CHAPTER 8 The Regression Fallacy
Only for the sake of this exercise we will assume that “intelligence” is an innate
property of individuals and can be represented by a real number z . If one picks at
random a student entering the U of U, the intelligence of this student is a random
variable which we assume to be normally distributed with mean µ and standard
deviation σ . Also assume every student has to take two intelligence tests, the ﬁrst
at the beginning of his or her studies, the other half a year later. The outcomes of
these tests are x and y . x and y measure the intelligence z (which is assumed to be
the same in both tests) plus a random error ε and δ , i.e.,
(8.0.13) x=z+ε (8.0.14) y =z+δ Here z ∼ N (µ, τ 2 ), ε ∼ N (0, σ 2 ), and δ ∼ N (0, σ 2 ) (i.e., we assume that both errors
have the same variance). The three variables ε, δ , and z are independent of each
other. Therefore x and y are jointly normal. var[x] = τ 2 + σ 2 , var[y ] = τ 2 + σ 2 ,
2
cov[x, y ] = cov[z + ε, z + δ ] = τ 2 + 0 + 0 + 0 = τ 2 . Therefore ρ = τ 2τ σ2 . The contour
+
lines of the joint density are ellipses with center (µ, µ) whose main axes are the lines
y = x and y = −x in the x, y plane.
Now what is the conditional mean? Since var[x] = var[y ], (7.3.17) gives the
line E[y x=x] = µ + ρ(x − µ), i.e., it is a line which goes through the center of the
ellipses but which is ﬂatter than the line x = y representing the real underlying linear
relationship if there are no errors. Geometrically one can get it as the line which
intersects each ellipse exactly where the ellipse is vertical.
Therefore, the parameters of the best prediction of y on the basis of x are not
the parameters of the underlying relationship. Why not? Because not only y but
also x is subject to errors. Assume you pick an individual by random, and it turns
out that his or her ﬁrst test result is very much higher than the average. Then it is
more likely that this is an individual which was lucky in the ﬁrst exam, and his or
her true IQ is lower than the one measured, than that the individual is an Einstein
who had a bad day. This is simply because z is normally distributed, i.e., among the
students entering a given University, there are more individuals with lower IQ’s than
Einsteins. In order to make a good prediction of the result of the second test one
must make allowance for the fact that the individual’s IQ is most likely lower than
his ﬁrst score indicated, therefore one will predict the second score to be lower than
the ﬁrst score. The converse is true for individuals who scored lower than average,
i.e., in your prediction you will do as if a “regression towards the mean” had taken
place.
The next important point to note here is: the “true regression line,” i.e., the
prediction line, is uniquely determined by the joint distribution of x and y . However
the line representing the underlying relationship can only be determined if one has
information in addition to the joint density, i.e., in addition to the observations.
E.g., assume the two tests have diﬀerent standard deviations, which may be the case
101 102 8. THE REGRESSION FALLACY simply because the second test has more questions and is therefore more accurate.
Then the underlying 45◦ line is no longer one of the main axes of the ellipse! To be
more precise, the underlying line can only be identiﬁed if one knows the ratio of the
variances, or if one knows one of the two variances. Without any knowledge of the
variances, the only thing one can say about the underlying line is that it lies between
the line predicting y on the basis of x and the line predicting x on the basis of y .
The name “regression” stems from a confusion between the prediction line and
the real underlying relationship. Francis Galton, the cousin of the famous Darwin,
measured the height of fathers and sons, and concluded from his evidence that the
heights of sons tended to be closer to the average height than the height of the
fathers, a purported law of “regression towards the mean.” Problem 152 illustrates
this:
Problem 152. The evaluation of two intelligence tests, one at the beginning
of the semester, one at the end, gives the following disturbing outcome: While the
underlying intelligence during the ﬁrst test was z ∼ N (100, 20), it changed between
the ﬁrst and second test due to the learning experience at the university. If w is the
intelligence of each student at the second test, it is connected to his intelligence z
at the ﬁrst test by the formula w = 0.5z + 50, i.e., those students with intelligence
below 100 gained, but those students with intelligence above 100 lost. (The errors
of both intelligence tests are normally distributed with expected value zero, and the
variance of the ﬁrst intelligence test was 5, and that of the second test, which had
more questions, was 4. As usual, the errors are independent of each other and of the
actual intelligence.)
80 90 100 110 120 110 110
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..... 100 100 90 90 80 90 100 110 120 Figure 1. Ellipse containing 95% of the probability mass of test
results x and y 8. THE REGRESSION FALLACY 103 • a. 3 points If x and y are the outcomes of the ﬁrst and second intelligence
test, compute E[x], E[y ], var[x], var[y ], and the correlation coeﬃcient ρ = corr[x, y ].
Figure 1 shows an equidensity line of their joint distribution; 95% of the probability
mass of the test results are inside this ellipse. Draw the line w = 0.5z + 50 into
Figure 1.
Answer. We know z ∼ N (100, 20); w = 0.5z + 50; x = z + ε; ε ∼ N (0, 4); y = w + δ ;
δ ∼ N (0, 5); therefore E[x] = 100; E[y ] = 100; var[x] = 20 + 5 = 25; var[y ] = 5 + 4 = 9;
cov[x, y ] = 10; corr[x, y ] = 10/15 = 2/3. In matrix notation
x
∼N
y (8.0.15) 100
25
,
100
10 10
9 The line y = 50 + 0.5x goes through the points (80, 90) and (120, 110). • b. 4 points Compute E[y x=x] and E[xy =y ]. The ﬁrst is a linear function of
x and the second a linear function of y . Draw the two lines representing these linear
functions into Figure 1. Use (7.3.18) for this.
Answer.
2
10
(x − 100) = 60 + x
25
5
10
100
10
E[xy =y ] = 100 +
(y − 100) = −
+
y.
9
9
9 E[y x=x] = 100 + (8.0.16)
(8.0.17) The line y = E[y x=x] goes through the points (80, 92) and (120, 108) at the edge of Figure 1; it
intersects the ellipse where it is vertical. The line x = E[xy =y ] goes through the points (80, 82) and
(120, 118), which are the corner points of Figure 1; it intersects the ellipse where it is horizontal.
The two lines intersect in the center of the ellipse, i.e., at the point (100, 100). 6
• c. 2 points Another researcher says that w = 10 z + 40, z ∼ N (100, 100 ),
6
50
ε ∼ N (0, 6 ), δ ∼ N (0, 3). Is this compatible with the data?
6
100+40 = 100;
10
6
cov[x, y ] = 10 var[z ] = Answer. Yes, it is compatible: E[x] = E[z ]+E[ε] = 100; E[y ] = E[w]+E[δ ] =
var[x] =
10. 100
6 + 50
6 = 25; var[y ] = 62
10 var[z ] + var[δ ] = 63 100
100 6 + 3 = 9; • d. 4 points A third researcher asserts that the IQ of the students really did not
change. He says w = z , z ∼ N (100, 5), ε ∼ N (0, 20), δ ∼ N (0, 4). Is this compatible
with the data? Is there unambiguous evidence in the data that the IQ declined?
Answer. This is not compatible. This scenario gets everything right except the covariance:
E[x] = E[z ] + E[ε] = 100; E[y ] = E[z ] + E[δ ] = 100; var[x] = 5 + 20 = 25; var[y ] = 5 + 4 = 9;
cov[x, y ] = 5. A scenario in which both tests have same underlying intelligence cannot be found.
Since the two conditional expectations are on the same side of the diagonal, the hypothesis that
the intelligence did not change between the two tests is not consistent with the joint distribution
of x and y . The diagonal goes through the points (82, 82) and (118, 118), i.e., it intersects the two
horizontal boundaries of Figure 1. We just showed that the parameters of the true underlying relationship cannot
be inferred from the data alone if there are errors in both variables. We also showed
that this lack of identiﬁcation is not complete, because one can specify an interval
which in the plim contains the true parameter value.
Chapter ?? has a much more detailed discussion of all this. There we will see
that this lack of identiﬁcation can be removed if more information is available, i.e., if
one knows that the two error variances are equal, or if one knows that the regression
has zero intercept, etc. Question 153 shows that in this latter case, the OLS estimate
is not consistent, but other estimates exist that are consistent. 104 8. THE REGRESSION FALLACY Problem 153. [Fri57, chapter 3] According to Friedman’s permanent income
hypothesis, drawing at random families in a given country and asking them about
their income y and consumption c can be modeled as the independent observations of
two random variables which satisfy
y = yp + yt , (8.0.18)
(8.0.19) c = cp + ct , (8.0.20) p c = β yp . Here y p and cp are the permanent and y t and ct the transitory components of income
and consumption. These components are not observed separately, only their sums y
and c are observed. We assume that the permanent income y p is random, with
2
E[y p ] = µ = 0 and var[y p ] = τy . The transitory components y t and ct are assumed
2
to be independent of each other and of y p , and E[y t ] = 0, var[y t ] = σy , E[ct ] = 0,
t
2
and var[c ] = σc . Finally, it is assumed that all variables are normally distributed.
• a. 2 points Given the above information, write down the vector of expected values E [ y ] and the covariance matrix V [ y ] in terms of the ﬁve unknown parameters
c
c
2
2
2
of the model µ, β , τy , σy , and σc .
Answer.
(8.0.21) E y
c = µ
βµ and V y
c = 2
2
τy + σy
2
βτy 2
βτy
2
2.
β 2 τy + σc • b. 3 points Assume that you know the true parameter values and you observe a
family’s actual income y . Show that your best guess (minimum mean squared error)
of this family’s permanent income y p is
(8.0.22) yp ∗ = 2
2
σy
τy
µ+ 2
y.
2
2
2
τy + σ y
τy + σ y Note: here we are guessing income, not yet consumption! Use (7.3.17) for this!
Answer. This answer also does the math for part c. The best guess is the conditional mean
cov[y p , y ]
(22,000 − E[y ])
var[y ]
16,000,000
(22,000 − 12,000) = 20,000
= 12,000 +
20,000,000 E[y p y = 22,000] = E[y p ] + or equivalently
E[y p y = 22,000] = µ +
= 2
τy
2
σy 2
τy
2
+ σy 2
2
τy + σy µ+ (22,000 − µ)
2
τy
2
2
τy + σy 22,000 = (0.2)(12,000) + (0.8)(22,000) = 20,000. • c. 3 points To make things more concrete, assume the parameters are
(8.0.23) β = 0.7 (8.0.24) σy = 2,000 (8.0.25) σc = 1,000 (8.0.26) µ = 12,000 (8.0.27) τy = 4,000. 8. THE REGRESSION FALLACY 105 If a family’s income is y = 22,000, what is your best guess of this family’s permanent
income y p ? Give an intuitive explanation why this best guess is smaller than 22,000.
Answer. Since the observed income of 22,000 is above the average of 12,000, chances are
greater that it is someone with a positive transitory income than someone with a negative one. • d. 2 points If a family’s income is y , show that your best guess about this
family’s consumption is
c∗ = β (8.0.28) 2
2
σy
τy
y.
µ+ 2
2
2
2
τy + σ y
τy + σ y Instead of an exact mathematical proof you may also reason out how it can be obtained
from (8.0.22). Give the numbers for a family whose actual income is 22,000.
Answer. This is 0.7 times the best guess about the family’s permanent income, since the
transitory consumption is uncorrelated with everything else and therefore must be predicted by 0.
This is an acceptable answer, but one can also derive it from scratch:
(8.0.29)
E[cy = 22,000] = E[c] + cov[c, y ]
(22,000 − E[y ])
var[y ]
2
βτy (22,000 − µ) = 8,400 + 0.7 16,000,000
(22,000 − 12,000) = 14,000
20,000,000 (8.0.30) = βµ + (8.0.31) or =β (8.0.32) = 0.7 (0.2)(12,000) + (0.8)(22,000) = (0.7)(20,000) = 14,000. 2
τy
2
σy 2
+ σy µ+
2 2
τy + σy 2
τy
2
2
τy + σy 22,000 The remainder of this Problem uses material that comes later in these Notes:
• e. 4 points From now on we will assume that the true values of the parameters
are not known, but two vectors y and c of independent observations are available.
We will show that it is not correct in this situation to estimate β by regressing c on
y with the intercept suppressed. This would give the estimator
ci y i
ˆ
(8.0.33)
β=
y2
i
Show that the plim of this estimator is
E[cy ]
ˆ
(8.0.34)
plim[β ] =
E[y 2 ]
ˆ
Which theorems do you need for this proof ? Show that β is an inconsistent estimator
of β , which yields too small values for β .
ˆ
Answer. First rewrite the formula for β in such a way that numerator and denominator each
has a plim: by the weak law of large numbers the plim of the average is the expected value, therefore
we have to divide both numerator and denominator by n. Then we can use the Slutsky theorem
that the plim of the fraction is the fraction of the plims.
ˆ
β= 1
n
1
n ci y i
y2
i ; ˆ
plim[β ] = 2
2
µβµ + βτy
µ2 + τy
E[cy ]
E[c] E[y ] + cov[c, y ]
=
=2
=β 2
.
2
2
2
2
E[y 2 ]
(E[y ])2 + var[y ]
µ + τy + σy
µ + τy + σy • f . 4 points Give the formulas of the method of moments estimators of the ﬁve
2
2
2
paramaters of this model: µ, β , τy , σy , and σp . (For this you have to express these
ﬁve parameters in terms of the ﬁve moments E[y ], E[c], var[y ], var[c], and cov[y , c],
and then simply replace the population moments by the sample moments.) Are these
consistent estimators? 106 8. THE REGRESSION FALLACY
E[c]
. This together with cov[y , c] =
E[y ]
2 + σ 2 gives σ 2 = var[y ] − τ 2 =
This together with var[y ] = τy
y
y
y
cov[y ,c] E[c]
2
2
2
equation var[c] = β 2 τy + σc one get σc = var[c] −
.
E[y ] Answer. From (8.0.21) follows E[c] = β E[y ], therefore β =
cov[y ,c]
cov[y ,c] E[y ]
=
.
β
E[c]
cov[y ,c] E[y ]
. And from the last
E[c] 2
2
βτy gives τy = var[y ] −
All these are consistent estimators, as long as E[y ] = 0 and β = 0. • g. 4 points Now assume you are not interested in estimating β itself, but in
addition to the two nvectors y and c you have an observation of y n+1 and you want
to predict the corresponding cn+1 . One obvious way to do this would be to plug the
methodof moments estimators of the unknown parameters into formula (8.0.28) for
the best linear predictor. Show that this is equivalent to using the ordinary least
ˆ
ˆˆ
ˆ
squares predictor c∗ = α + β y n+1 where α and β are intercept and slope in the simple
regression of c on y , i.e.,
c
(y i − y )(ci − ¯)
¯
(y i − y )2
¯
α = ¯ − βy
ˆ c ˆ¯
ˆ
β= (8.0.35)
(8.0.36) Note that we are regressing c on y with an intercept, although the original model
does not have an intercept.
Answer. Here I am writing population moments where I should be writing sample moments.
2
2
First substitute the method of moments estimators in the denominator in (8.0.28): τy + σy = var[y ].
Therefore the ﬁrst summand becomes
E[c]
1
cov[y , c] E[y ]
1
cov[y , c] E[y ]
cov[y , c] E[y ]
2
=
var[y ]−
E[y ]
= E[c] 1−
= E[c]−
βσy µ
var[y ]
E[y ]
E[c]
var[y ]
var[y ] E[c]
var[y ]
But since
to show: cov[y ,c]
var[y ] ˆ
= β and α + β E[y ] = E[c] this expression is simply α. The second term is easier
ˆˆ
ˆ
β 2
τy var[y ] y= cov[y , c]
ˆ
y = βy
var[y ] • h. 2 points What is the “Iron Law of Econometrics,” and how does the above
relate to it?
Answer. The Iron Law says that all eﬀects are underestimated because of errors in the independent variable. Friedman says Keynesians obtain their low marginal propensity to consume due
to the “Iron Law of Econometrics”: they ignore that actual income is a measurement with error of
the true underlying variable, permanent income. Problem 154. This question follows the original article [SW76] much more
closely than [HVdP02] does. Sargent and Wallace ﬁrst reproduce the usual argument
why “activist” policy rules, in which the Fed “looks at many things” and “leans
against the wind,” are superior to policy rules without feedback as promoted by the
monetarists.
They work with a very stylized model in which national income is represented by
the following time series:
(8.0.37) y t = α + λy t−1 + β mt + ut Here y t is GNP, measured as its deviation from “potential” GNP or as unemployment
rate, and mt is the rate of growth of the money supply. The random disturbance ut
is assumed independent of y t−1 , it has zero expected value, and its variance var[ut ]
is constant over time, we will call it var[u] (no time subscript).
• a. 4 points First assume that the Fed tries to maintain a constant money
supply, i.e., mt = g0 + εt where g0 is a constant, and εt is a random disturbance
since the Fed does not have full control over the money supply. The εt have zero 8. THE REGRESSION FALLACY 107 expected value; they are serially uncorrelated, and they are independent of the ut .
This constant money supply rule does not necessarily make y t a stationary time
series (i.e., a time series where mean, variance, and covariances do not depend on
t), but if λ < 1 then y t converges towards a stationary time series, i.e., any initial
deviations from the “steady state” die out over time. You are not required here to
prove that the time series converges towards a stationary time series, but you are
asked to compute E[y t ] in this stationary time series.
• b. 8 points Now assume the policy makers want to steer the economy towards
a desired steady state, call it y ∗ , which they think makes the best tradeoﬀ between
unemployment and inﬂation, by setting mt according to a rule with feedback:
(8.0.38) mt = g0 + g1 y t−1 + εt Show that the following values of g0 and g1
(8.0.39) g0 = (y ∗ − α)/β g1 = −λ/β represent an optimal monetary policy, since they bring the expected value of the steady
state E[y t ] to y ∗ and minimize the steady state variance var[y t ].
• c. 3 points This is the conventional reasoning which comes to the result that a
policy rule with feedback, i.e., a policy rule in which g1 = 0, is better than a policy rule
without feedback. Sargent and Wallace argue that there is a ﬂaw in this reasoning.
Which ﬂaw?
• d. 5 points A possible system of structural equations from which (8.0.37) can
be derived are equations (8.0.40)–(8.0.42) below. Equation (8.0.40) indicates that
unanticipated increases in the growth rate of the money supply increase output, while
anticipated ones do not. This is a typical assumption of the rational expectations
school (Lucas supply curve).
(8.0.40) y t = ξ0 + ξ1 (mt − Et−1 mt ) + ξ2 y t−1 + ut The Fed uses the policy rule
(8.0.41) mt = g0 + g1 y t−1 + εt and the agents know this policy rule, therefore
(8.0.42) Et−1 mt = g0 + g1 y t−1 . Show that in this system, the parameters g0 and g1 have no inﬂuence on the time
path of y .
• e. 4 points On the other hand, the econometric estimations which the policy
makers are running seem to show that these coeﬃcients have an impact. During a
certain period during which a constant policy rule g0 , g1 is followed, the econometricians regress y t on y t−1 and mt in order to estimate the coeﬃcients in (8.0.37).
Which values of α, λ, and β will such a regression yield? CHAPTER 9 A Simple Example of Estimation
We will discuss here a simple estimation problem, which can be considered the
prototype of all least squares estimation. Assume we have n independent observations
y1 , . . . , yn of a Normally distributed random variable y ∼ N (µ, σ 2 ) with unknown
location parameter µ and dispersion parameter σ 2 . Our goal is to estimate the
location parameter and also estimate some measure of the precision of this estimator. 9.1. Sample Mean as Estimator of the Location Parameter
The obvious (and in many cases also the best) estimate of the location parameter
n
1
of a distribution is the sample mean y = n i=1 yi . Why is this a reasonable
¯
estimate?
1. The location parameter of the Normal distribution is its expected value, and
by the weak law of large numbers, the probability limit for n → ∞ of the sample
mean is the expected value.
2. The expected value µ is sometimes called the “population mean,” while y is
¯
the sample mean. This terminology indicates that there is a correspondence between
population quantities and sample quantities, which is often used for estimation. This
is the principle of estimating the unknown distribution of the population by the
empirical distribution of the sample. Compare Problem 63.
3. This estimator is also unbiased. By deﬁnition, an estimator t of the parameter
θ is unbiased if E[t] = θ. y is an unbiased estimator of µ, since E[¯] = µ.
¯
y
4. Given n observations y1 , . . . , yn , the sample mean is the number a = y which
¯
minimizes (y1 − a)2 + (y2 − a)2 + · · · + (yn − a)2 . One can say it is the number whose
squared distance to the given sample numbers is smallest. This idea is generalized
in the least squares principle of estimation. It follows from the following frequently
used fact:
5. In the case of normality the sample mean is also the maximum likelihood
estimate.
Problem 155. 4 points Let y1 , . . . , yn be an arbitrary vector and α an arbitrary
n
number. As usual, y = n i=1 yi . Show that
¯1 n n (yi − α)2 = (9.1.1)
i=1 (yi − y )2 + n(¯ − α)2
¯
y
i=1
109 110 9. A SIMPLE EXAMPLE OF ESTIMATION ...
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.
..........
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............. ...................... .....................
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..............
.... .
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...
.....
.
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......
.......
.........................................................
..........................................................
..........................................................
........................................................... q
µ2 µ1 µ3 µ4 Figure 1. Possible Density Functions for y
Answer.
n n (yi − α )2 = (9.1.2)
i=1 (9.1.3) (yi − y ) + (¯ − α)
¯
y
i=1
n n (yi − y )2 + 2
¯ =
i=1
n (9.1.4) 2 n i=1 i=1
n (yi − y )2 + 2(¯ − α)
¯
y = (¯ − α)2
y (yi − y )(¯ − α) +
¯y i=1 (yi − y ) + n(¯ − α)2
¯
y
i=1 Since the middle term is zero, (9.1.1) follows. Problem 156. 2 points Let y be a nvector. (It may be a vector of observations
of a random variable y , but it does not matter how the yi were obtained.) Prove that
the scalar α which minimizes the sum
(9.1.5) (y1 − α)2 + (y2 − α)2 + · · · + (yn − α)2 = (yi − α)2 is the arithmetic mean α = y .
¯
Answer. Use (9.1.1). Problem 157. Give an example of a distribution in which the sample mean is
not a good estimate of the location parameter. Which other estimate (or estimates)
would be preferable in that situation?
9.2. Intuition of the Maximum Likelihood Estimator
In order to make intuitively clear what is involved in maximum likelihood estimation, look at the simplest case y = µ + ε, ε ∼ N (0, 1), where µ is an unknown
parameter. In other words: we know that one of the functions shown in Figure 1 is
the density function of y , but we do not know which:
Assume we have only one observation y . What is then the MLE of µ? It is that
µ for which the value of the likelihood function, evaluated at y , is greatest. I.e., you
˜
look at all possible density functions and pick the one which is highest at point y ,
and use the µ which belongs this density as your estimate.
2) Now assume two independent observations of y are given, y1 and y2 . The
family of density functions is still the same. Which of these density functions do we
choose now? The one for which the product of the ordinates over y1 and y2 gives
the highest value. For this the peak of the density function must be exactly in the
middle between the two observations.
3) Assume again that we made two independent observations y1 and y2 of y , but
this time not only the expected value but also the variance of y is unknown, call it
σ 2 . This gives a larger family of density functions to choose from: they do not only
diﬀer by location, but some are low and fat and others tall and skinny. 9.2. INTUITION OF THE MAXIMUM LIKELIHOOD ESTIMATOR 111 ...... ... ...
..... ... ...
....................................
............................................................ ..............................................
.....................................
............................................................ ...............................................
... ...
... ...
.
.
....... ..............
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... ..................................... ................... ......
.................
................
.........
....
..........
...........
.......................................................... ............................................
........................................................... ............................................. q µ1 q µ2 µ3 µ4 Figure 2. Two observations, σ 2 = 1
Figure 3. Two observations, σ 2 unknown
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.....................................................................................................................................................................................................................................................................................................................................
........ ......................................................................................
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.................................................................................... .........
... . . .
.. . Figure 4. Only those centered over the two observations need to be considered
Figure 5. Many Observations
For which density function is the product of the ordinates over y1 and y2 the
largest again? Before even knowing our estimate of σ 2 we can already tell what µ is:
˜
it must again be (y1 + y2 )/2. Then among those density functions which are centered
over (y1 + y2 )/2, there is one which is highest over y1 and y2 . Figure 4 shows the
densities for standard deviations 0.01, 0.05, 0.1, 0.5, 1, and 5. All curves, except
the last one, are truncated at the point where the resolution of TEX can no longer
distinguish between their level and zero. For the last curve this point would only be
reached at the coordinates ±25.
4) If we have many observations, then the density pattern of the observations,
as indicated by the histogram below, approximates the actual density function of y
itself. That likelihood function must be chosen which has a high value where the
points are dense, and which has a low value where the points are not so dense.
9.2.1. Precision of the Estimator. How good is y as estimate of µ? To an¯
swer this question we need some criterion how to measure “goodness.” Assume your
business depends on the precision of the estimate µ of µ. It incurs a penalty (extra
ˆ
cost) amounting to (ˆ − µ)2 . You don’t know what this error will be beforehand,
µ
but the expected value of this “loss function” may be an indication how good the
estimate is. Generally, the expected value of a loss function is called the “risk,” and
for the quadratic loss function E[(ˆ − µ)2 ] it has the name “mean squared error of
µ
µ as an estimate of µ,” write it MSE[ˆ; µ]. What is the mean squared error of y ?
ˆ
µ
¯
2
Since E[¯] = µ, it is E[(¯ − E[¯])2 ] = var[¯] = σ .
y
y
y
y
n 112 9. A SIMPLE EXAMPLE OF ESTIMATION Note that the MSE of y as an estimate of µ does not depend on µ. This is
¯
convenient, since usually the MSE depends on unknown parameters, and therefore
one usually does not know how good the estimator is. But it has more important
y
y
y
advantages. For any estimator y of µ follows MSE[˜; µ] = var[˜] + (E[˜] − µ)2 . If
˜
y is linear (perhaps with a constant term), then var[˜] is a constant which does
y
˜
not depend on µ, therefore the MSE is a constant if y is unbiased and a quadratic
˜
function of µ (parabola) if y is biased. Since a parabola is an unbounded function,
˜
a biased linear estimator has therefore the disadvantage that for certain values of µ
its MSE may be very high. Some estimators are very good when µ is in one area,
and very bad when µ is in another area. Since our unbiased estimator y has bounded
¯
MSE, it will not let us down, wherever nature has hidden the µ.
On the other hand, the MSE does depend on the unknown σ 2 . So we have to
estimate σ 2 . 9.3. Variance Estimation and Degrees of Freedom
It is not so clear what the best estimator of σ 2 is. At least two possibilities are
in common use:
s2 =
m 1
n s2 =
u (9.3.1) 1
n−1 (y i − y )2
¯ or
(9.3.2) (y i − y )2 .
¯ Let us compute the expected value of our two estimators. Equation (9.1.1) with
α = E[y ] allows us to simplify the sum of squared errors so that it becomes easy to
take expected values:
n (9.3.3) n (y i − y )2 ] =
¯ E[
i=1 y
E[(y i − µ)2 ] − n E[(¯ − µ)2 ]
i=1
n (9.3.4) σ2 − n =
i=1 σ2
= (n − 1)σ 2 .
n because E[(y i − µ)2 ] = var[y i ] = σ 2 and E[(¯ − µ)2 ] = var[¯] =
y
y
use as estimator of σ 2 the quantity
(9.3.5) s2 =
u 1
n−1 σ2
n. Therefore, if we n (y i − y )2
¯
i=1 then this is an unbiased estimate.
Problem 158. 4 points Show that
(9.3.6) s2
u 1
=
n−1 n (y i − y )2
¯
i=1 is an unbiased estimator of the variance. List the assumptions which have to be made
about y i so that this proof goes through. Do you need Normality of the individual
observations y i to prove this? 9.3. VARIANCE ESTIMATION AND DEGREES OF FREEDOM 113 Answer. Use equation (9.1.1) with α = E[y ]:
n (9.3.7) n ( y i − y )2 ] =
¯ E[
i=1 E[(y i − µ)2 ] − n E[(¯ − µ)2 ]
y
i=1
n (9.3.8) σ2 − n = σ2
= (n − 1)σ 2 .
n i=1 You do not need Normality for this. For testing, conﬁdence intervals, etc., one also needs to know the probability
distribution of s2 . For this look up once more Section 4.9 about the ChiSquare
u
distribution. There we introduced the terminology that a random variable q is distributed as a σ 2 χ2 iﬀ q /σ 2 is a χ2 . In our model with n independent normal variables
¯
y i with same mean and variance, the variable (y i − y )2 is a σ 2 χ2 −1 . Problem 159
n
gives a proof of this in the simplest case n = 2, and Problem 160 looks at the case
2
n = 3. But it is valid for higher n too. Therefore s2 is a nσ 1 χ2 −1 . This is reu
n
−
markable: the distribution of s2 does not depend on µ. Now use (4.9.5) to get the
u
2σ 4
variance of s2 : it is n−1 .
u
Problem 159. Let y 1 and y 2 be two independent Normally distributed variables
with mean µ and variance σ 2 , and let y be their arithmetic mean.
¯
• a. 2 points Show that
2 (9.3.9) (y i − y )2 ∼ σ 2 χ2
¯
1 SSE =
i−1 Hint: Find a Normally distributed random variable z with expected value 0 and variance 1 such that SSE = σ 2 z 2 .
Answer.
(9.3.10)
(9.3.11) y1 − y
¯ (9.3.12) ¯
y2 − y (9.3.13) (y 1 − y )2 + ( y 2 − y )2
¯
¯ (9.3.14) y1 + y2
2
y − y2
=1
2
y − y2
=− 1
2
(y 1 − y 2 )2
(y − y 2 )2
=
+1
4
4
2
2 y1 − y2
=σ
,
√
2σ 2 y=
¯ = (y 1 − y 2 )2
2 √
and since z = (y 1 − y 2 )/ 2σ 2 ∼ N (0, 1), its square is a χ2 .
1 • b. 4 points Write down the covariance matrix of the vector
y1 − y
¯
y2 − y
¯ (9.3.15)
and show that it is singular.
Answer. (9.3.11) and (9.3.12) give
(9.3.16)
and V [D y ] = D V [y ]D 1
y1 − y
¯
2
=
y2 − y
¯
−1
2 −1
2
1
2 y1
y2 = Dy = σ 2 D because V [y ] = σ 2 I and D = idempotent. D is singular because its determinant is zero. 1
2 −1
2 1
−2 1
2 is symmetric and 114 9. A SIMPLE EXAMPLE OF ESTIMATION • c. 1 point The joint distribution of y 1 and y 2 is bivariate normal, why did we
then get a χ2 with one, instead of two, degrees of freedom?
Answer. Because y 1 − y and y 2 − y are not independent; one is exactly the negative of the
¯
¯
other; therefore summing their squares is really only the square of one univariate normal. Problem 160. Assume y 1 , y 2 , and y 3 are independent N (µ, σ 2 ). Deﬁne three
new variables z 1 , z 2 , and z 3 as follows: z 1 is that multiple of y which has variance
¯
σ 2 . z 2 is that linear combination of z 1 and y 2 which has zero covariance with z 1
and has variance σ 2 . z 3 is that linear combination of z 1 , z 2 , and y 3 which has zero
covariance with both z 1 and z 2 and has again variance σ 2 . These properties deﬁne
z 1 , z 2 , and z 3 uniquely up factors ±1, i.e., if z 1 satisﬁes the above conditions, then
−z 1 does too, and these are the only two solutions.
• a. 2 points Write z 1 and z 2 (not yet z 3 ) as linear combinations of y 1 , y 2 , and
y3 .
• b. 1 point To make the computation of z 3 less tedious, ﬁrst show the following:
if z 3 has zero covariance with z 1 and z 2 , it also has zero covariance with y 2 .
• c. 1 point Therefore z 3 is a linear combination of y 1 and y 3 only. Compute
its coeﬃcients.
• d. 1 point How does the joint distribution of z 1 , z 2 , and z 3 diﬀer from that of
y 1 , y 2 , and y 3 ? Since they are jointly normal, you merely have to look at the expected
values, variances, and covariances.
• e. 2 points Show that z 2 + z 2 + z 2 = y 2 + y 2 + y 2 . Is this a surprise?
1
2
3
1
2
3
• f . 1 point Show further that s2 = 1
u
2
simple trick!) Conclude from this that s2 ∼
u 3
12
2
¯2
i=1 (y i − y ) = 2 (z 2 + z 3 ).
2
σ
2
¯
2 χ2 , independent of y . (There is a For a matrixinterpretation of what is happening, see equation (7.4.9) together
with Problem 161.
1
Problem 161. 3 points Verify that the matrix D = I − n ιι is symmetric and
idempotent, and that the sample covariance of two vectors of observations x and y
can be written in matrix notation as
1
1
(9.3.17)
sample covariance(x, y ) =
(xi − x)(yi − y ) = x D y
¯
¯
n
n In general, one can always ﬁnd n − 1 normal variables with variance σ 2 , independent of each other and of y , whose sum of squares is equal to (y i − y )2 . Simply
¯
¯
√
start with y n and generate n − 1 linear combinations of the y i which are pairwise
¯
uncorrelated and have √
variances σ 2 . You are simply building an orthonormal coordinate system with y n as its ﬁrst vector; there are many diﬀerent ways to do
¯
this.
Next let us show that y and s2 are statistically independent. This is an ad¯
u
vantage. Assume, hypothetically, y and s2 were negatively correlated. Then, if the
¯
u
observed value of y is too high, chances are that the one of s2 is too low, and a look
¯
u
at s2 will not reveal how far oﬀ the mark y may be. To prove independence, we will
¯
u
ﬁrst show that y and y i − y are uncorrelated:
¯
¯
(9.3.18)
(9.3.19) y
¯
y
y
cov[¯, y i − y ] = cov[¯, y i ] − var[¯]
1
σ2
= cov[ (y 1 + · · · + y i + · · · + y n ), y i ] −
=0
n
n 9.3. VARIANCE ESTIMATION AND DEGREES OF FREEDOM 115 ¯
By normality, y is therefore independent of y i − y for all i. Since all variables in¯
volved are jointly normal, it follows from this that y is independent of the vector
¯
y 1 − y · · · y n − y ; therefore it is also independent of any function of this vec¯
¯
tor, such as s2 .
u
The above calculations explain why the parameter of the χ2 distribution has
the colorful name “degrees of freedom.” This term is sometimes used in a very
broad sense, referring to estimation in general, and sometimes in a narrower sense,
in conjunction with the linear model. Here is ﬁrst an interpretation of the general use
of the term. A “statistic” is deﬁned to be a function of the observations and of other
known parameters of the problem, but not of the unknown parameters. Estimators
are statistics. If one has n observations, then one can ﬁnd at most n mathematically
independent statistics; any other statistic is then a function of these n. If therefore
a model has k independent unknown parameters, then one must have at least k
observations to be able to estimate all parameters of the model. The number n − k ,
i.e., the number of observations not “used up” for estimation, is called the number
of “degrees of freedom.”
There are at least three reasons why one does not want to make the model such
that it uses up too many degrees of freedom. (1) the estimators become too inaccurate
if one does; (2) if there are no degrees of freedom left, it is no longer possible to make
any “diagnostic” tests whether the model really ﬁts the data, because it always gives
a perfect ﬁt whatever the given set of data; (3) if there are no degrees of freedom left,
then one can usually also no longer make estimates of the precision of the estimates.
Speciﬁcally in our linear estimation problem, the number of degrees of freedom
is n − 1, since one observation has been used up for estimating the mean. If one
runs a regression, the number of degrees of freedom is n − k , where k is the number
of regression coeﬃcients. In the linear model, the number of degrees of freedom
becomes immediately relevant for the estimation of σ 2 . If k observations are used
up for estimating the slope parameters, then the other n − k observations can be
combined into a n − k variate Normal whose expected value does not depend on the
slope parameter at all but is zero, which allows one to estimate the variance.
If we assume that the original observations are normally distributed, i.e., y i ∼
2
NID(µ, σ 2 ), then we know that s2 ∼ nσ 1 χ2 −1 . Therefore E[s2 ] = σ 2 and var[s2 ] =
u
u
u
n
−
2σ 4 /(n − 1). This estimate of σ 2 therefore not only gives us an estimate of the
precision of y , but it has an estimate of its own precision built in.
¯
(y −y )2
¯ i
Interestingly, the MSE of the alternative estimator s2 =
is smaller
m
n
2
2
2
than that of su , although sm is a biased estimator and su an unbiased estimator of
σ 2 . For every estimator t, MSE[t; θ] = var[t] + (E[t − θ])2 , i.e., it is variance plus
2σ 4
squared bias. The MSE of s2 is therefore equal to its variance, which is n−1 . The
u
4 2 4 (
/n
alternative s2 = n−1 s2 has bias − σ and variance 2σ nn−1) . Its MSE is (2−1n )σ .
2
m
u
n
n
2
Comparing that with the formula for the MSE of su one sees that the numerator is
smaller and the denominator is bigger, therefore s2 has smaller MSE.
m Problem 162. 4 points Assume y i ∼ NID(µ, σ 2 ). Show that the socalled Theil
Schweitzer estimator [TS61]
(9.3.20) s2 =
t 1
n+1 (y i − y )2
¯ has even smaller MSE than s2 and s2 as an estimator of σ 2 .
u
m 116 9. A SIMPLE EXAMPLE OF ESTIMATION
........
.....................
..... .. .
........................
.. ..
........
... ......
.........
..
.
... ....
.........
... ...
..
..........
.. ...
...........
.. ...
..
............
.............
... .
.. .
................
..... .
.......
...............
.
.........................
.
...........................
..
..
.....
.................
.................
0
1
2
3
4 5 6 Figure 6. Densities of Unbiased and Theil Schweitzer Estimators
Answer. s2 =
t n−1 2
s;
n+1 u 2 2
therefore its bias is − nσ and its variance is
+1 4 2
MSE is nσ . That this is smaller than the MSE of s2 means
m
+1
(2n − 1)(n + 1) = 2n2 + n − 1 > 2n2 for n > 1. 2n−1
n2 ≥ 2
,
n+1 2(n−1)σ 4
,
(n+1)2 and the which follows from Problem 163. 3 points Computer assignment: Given 20 independent observations of a random variable y ∼ N (µ, σ 2 ). Assume you know that σ 2 = 2. Plot
the density function of s2 . Hint: In R, the command dchisq(x,df=25) returns the
u
density of a Chisquare distribution with 25 degrees of freedom evaluated at x. But
the number 25 was only taken as an example, this is not the number of degrees of
freedom you need here. You also do not need the density of a ChiSquare but that
of a certain multiple of a Chisquare. (Use the transformation theorem for density
functions!)
2
Answer. s2 ∼ 19 χ2 . To express the density of the variable whose density is known by that
u
19
whose density one wants to know, say 19 s2 ∼ χ2 . Therefore
19
2u f s 2 ( x) = (9.3.21) u 19
19
f 2 ( x) .
2 χ19 2 • a. 2 points In the same plot, plot the density function of the TheilSchweitzer
estimate s2 deﬁned in equation (9.3.20). This gives a plot as in Figure 6. Can one see
t
from the comparison of these density functions that the TheilSchweitzer estimator
has a better MSE?
Answer. Start with plotting the TheilSchweitzer plot, because it is higher, and therefore it
will give the right dimensions of the plot. You can run this by giving the command ecmetscript(theilsch).
The two areas between the densities have equal size, but the area where the TheilSchweitzer density
is higher is overall closer to the true value than the area where the unbiased density is higher. Problem 164. 4 points The following problem illustrates the general fact that
if one starts with an unbiased estimator and “shrinks” it a little, one will end up
with a better MSE. Assume E[y ] = µ, var(y ) = σ 2 , and you make n independent
observations y i .
The best linear unbiased estimator of µ on the basis of these
observations is the sample mean y . Show that, whenever α satisﬁes
¯
nµ2 − σ 2
<α<1
nµ2 + σ 2 (9.3.22) then MSE[αy ; µ] < MSE[¯; µ]. Unfortunately, this condition depends on µ and σ 2
¯
y
and can therefore not be used to improve the estimate.
Answer. Here is the mathematical relationship:
(9.3.23) MSE[αy ; µ] = E (αy − µ)2 = E (αy − αµ + αµ − µ)2 < MSE[¯; µ] = var[¯]
¯
¯
¯
y
y (9.3.24) α2 σ 2 /n + (1 − α)2 µ2 < σ 2 /n Now simplify it:
(9.3.25) (1 − α)2 µ2 < (1 − α2 )σ 2 /n = (1 − α)(1 + α)σ 2 /n 9.3. VARIANCE ESTIMATION AND DEGREES OF FREEDOM 117 This cannot be true for α ≥ 1, because for α = 1 one has equality, and for α > 1, the righthand side
is negative. Therefore we are allowed to assume α < 1, and can divide by 1 − α without disturbing
the inequality:
(9.3.26) (1 − α)µ2 < (1 + α)σ 2 /n (9.3.27) µ2 − σ 2 /n < α(µ2 + σ 2 /n) The answer is therefore
(9.3.28) nµ2 − σ 2
< α < 1.
nµ2 + σ 2 This the range. Note that nµ2 − σ 2 < 0 may be negative. The best value is in the middle of this
range, see Problem 165. Problem 165. [KS79, example 17.14 on p. 22] The mathematics in the following
problem is easier than it looks. If you can’t prove a., assume it and derive b. from
it, etc.
• a. 2 points Let t be an estimator of the nonrandom scalar parameter θ. E[t − θ]
is called the bias of t, and E (t − θ)2 is called the mean squared error of t as an
estimator of θ, written MSE[t; θ]. Show that the MSE is the variance plus the squared
bias, i.e., that
(9.3.29) 2 MSE[t; θ] = var[t] + E[t − θ] . Answer. The most elegant proof, which also indicates what to do when θ is random, is:
(9.3.30) MSE[t; θ] = E (t − θ)2 = var[t − θ] + (E[t − θ])2 = var[t] + (E[t − θ])2 . • b. 2 points For the rest of this problem assume that t is an unbiased estimator
of θ with var[t] > 0. We will investigate whether one can get a better MSE if one
estimates θ by a constant multiple at instead of t. Show that
(9.3.31) MSE[at; θ] = a2 var[t] + (a − 1)2 θ2 . Answer. var[at] = a2 var[t] and the bias of at is E[at − θ] = (a − 1)θ. Now apply (9.3.30). • c. 1 point Show that, whenever a > 1, then MSE[at; θ] > MSE[t; θ]. If one
wants to decrease the MSE, one should therefore not choose a > 1.
Answer. MSE[at; θ] − MSE[t; θ] = (a2 − 1) var[t]+(a − 1)2 θ2 > 0 since a > 1 and var[t] > 0. • d. 2 points Show that
(9.3.32) d
MSE[at; θ]
da > 0.
a=1 From this follows that the MSE of at is smaller than the MSE of t, as long as a < 1
and close enough to 1.
Answer. The derivative of (9.3.31) is
d
MSE[at; θ] = 2a var[t] + 2(a − 1)θ2
da
Plug a = 1 into this to get 2 var[t] > 0. (9.3.33) • e. 2 points By solving the ﬁrst order condition show that the factor a which
gives smallest MSE is
(9.3.34) a= θ2
.
var[t] + θ2 Answer. Rewrite (9.3.33) as 2a(var[t] + θ2 ) − 2θ2 and set it zero. 118 9. A SIMPLE EXAMPLE OF ESTIMATION • f . 1 point Assume t has an exponential distribution with parameter λ > 0, i.e.,
ft (t) = λ exp(−λt), (9.3.35) t≥0 and ft (t) = 0 otherwise. Check that ft (t) is indeed a density function.
∞ Answer. Since λ > 0, ft (t) > 0 for all t ≥ 0. To evaluate
λ exp(−λt) dt, substitute
0
s = −λt, therefore ds = −λdt, and the upper integration limit changes from +∞ to −∞, therefore
−∞
the integral is −
exp(s) ds = 1.
0 • g. 4 points Using this density function (and no other knowledge about the
exponential distribution) prove that t is an unbiased estimator of 1/λ, with var[t] =
1/λ2 .
Answer. To evaluate ∞
0 λt exp(−λt) dt, use partial integration uv dt = uv − u v dt with u = t, u = 1, v = − exp(−λt), v = λ exp(−λt). Therefore the integral is −t exp(−λt)
∞
0 exp(−λt) dt = 1/λ, since we just saw that
To evaluate ∞
0 ∞
0 0 + λ exp(−λt) dt = 1. λt2 exp(−λt) dt, use partial integration with u = t2 , u = 2t, v = − exp(−λt), v = λ exp(−λt). Therefore the integral is −t2 exp(−λt)
2/λ2 . ∞ Therefore var[t] = E[t2 ] − (E[t])2 = 2/λ2 − 1/λ2 ∞
0 = +2 ∞ 0
1/λ2 . t exp(−λt) dt = 2
λ ∞
0 λt exp(−λt) dt = • h. 2 points Which multiple of t has the lowest MSE as an estimator of 1/λ?
Answer. It is t/2. Just plug θ = 1/λ into (9.3.34).
(9.3.36) a= 1/λ2
1/λ2
1
=
=.
var[t] + 1/λ2
1/λ2 + 1/λ2
2 • i. 2 points Assume t1 , . . . , tn are independently distributed, and each of them
has the exponential distribution with the same parameter λ. Which multiple of the
n
1
sample mean ¯ = n i=1 ti has best MSE as estimator of 1/λ?
t
Answer. ¯ has expected value 1/λ and variance 1/nλ2 . Therefore
t
(9.3.37) a= i.e., for the best estimator ˜ =
t 1/λ2
n
1/λ2
=
=
,
2
var[t] + 1/λ
1/nλ2 + 1/λ2
n+1
1
n+1 ti divide the sum by n + 1 instead of n. 1
• j. 3 points Assume q ∼ σ 2 χ2 (in other words, σ2 q ∼ χ2 , a Chisquare distrim
m
bution with m degrees of freedom). Using the fact that E[χ2 ] = m and var[χ2 ] = 2m,
m
m
compute that multiple of q that has minimum MSE as estimator of σ 2 . Answer. This is a trick question since q itself is not an unbiased estimator of σ 2 . E[q ] = mσ 2 ,
therefore q /m is the unbiased estimator. Since var[q /m] = 2σ 4 /m, it follows from (9.3.34) that
qm
a = m/(m + 2), therefore the minimum MSE multiple of q is m m+2 = mq . I.e., divide q by m + 2
+2
instead of m. • k. 3 points Assume you have n independent observations of a Normally distributed random variable y with unknown mean µ and standard deviation σ 2 . The
1
best unbiased estimator of σ 2 is n−1 (y i − y )2 , and the maximum likelihood extima¯
1
2
tor is n (y i − y ) . What are the implications of the above for the question whether
¯
one should use the ﬁrst or the second or still some other multiple of (y i − y )2 ?
¯
Answer. Taking that multiple of the sum of squared errors which makes the estimator unbiased is not necessarily a good choice. In terms of MSE, the best multiple of
(y i − y )2 is
¯
1
(y i − y )2 .
¯
n+1 9.3. VARIANCE ESTIMATION AND DEGREES OF FREEDOM 119 • l. 3 points We are still in the model deﬁned in k. Which multiple of the sample
mean y has smallest MSE as estimator of µ? How does this example diﬀer from the
¯
ones given above? Can this formula have practical signiﬁcance?
2 µ
Answer. Here the optimal a = µ2 +(σ2 /n) . Unlike in the earlier examples, this a depends on
the unknown parameters. One can “operationalize” it by estimating the parameters from the data,
but the noise introduced by this estimation can easily make the estimator worse than the simple y .
¯
Indeed, y is admissible, i.e., it cannot be uniformly improved upon. On the other hand, the Stein
¯
rule, which can be considered an operationalization of a very similar formula (the only diﬀerence
being that one estimates the mean vector of a vector with at least 3 elements), by estimating µ2
1
and µ2 + n σ 2 from the data, shows that such an operationalization is sometimes successful. We will discuss here one more property of y and s2 : They together form suﬃcient
¯
u
statistics for µ and σ 2 . I.e., any estimator of µ and σ 2 which is not a function of y
¯
and s2 is less eﬃcient than it could be. Since the factorization theorem for suﬃcient
u
statistics holds even if the parameter θ and its estimate t are vectors, we have to
write the joint density of the observation vector y as a product of two functions, one
depending on the parameters and the suﬃcient statistics, and the other depending
on the value taken by y , but not on the parameters. Indeed, it will turn out that
this second function can just be taken to be h(y ) = 1, since the density function can
be rearranged as
n (9.3.38) fy (y1 , . . . , yn ; µ, σ 2 ) = (2πσ 2 )−n/2 exp − (yi − µ)2 /2σ 2 =
i=1 n (9.3.39) = (2πσ 2 )−n/2 exp − (yi − y )2 − n(¯ − µ)2 /2σ 2 =
¯
y
i=1 (9.3.40) = (2πσ 2 )−n/2 exp − (n − 1)s2 − n(¯ + µ)2
y
u
.
2
2σ CHAPTER 10 Estimation Principles and Classiﬁcation of
Estimators
10.1. Asymptotic or LargeSample Properties of Estimators
We will discuss asymptotic properties ﬁrst, because the idea of estimation is to
get more certainty by increasing the sample size.
Strictly speaking, asymptotic properties do not refer to individual estimators
but to sequences of estimators, one for each sample size n. And strictly speaking, if
one alters the ﬁrst 10 estimators or the ﬁrst million estimators and leaves the others
unchanged, one still gets a sequence with the same asymptotic properties. The results
that follow should therefore be used with caution. The asymptotic properties may
say very little about the concrete estimator at hand.
The most basic asymptotic property is (weak) consistency. An estimator tn
(where n is the sample size) of the parameter θ is consistent iﬀ
(10.1.1) plim tn = θ. n→∞ Roughly, a consistent estimation procedure is one which gives the correct parameter
values if the sample is large enough. There are only very few exceptional situations
in which an estimator is acceptable which is not consistent, i.e., which does not
converge in the plim to the true parameter value.
Problem 166. Can you think of a situation where an estimator which is not
consistent is acceptable?
Answer. If additional data no longer give information, like when estimating the initial state
of a timeseries, or in prediction. And if there is no identiﬁcation but the value can be conﬁned to
an interval. This is also inconsistency. The following is an important property of consistent estimators:
Slutsky theorem: If t is a consistent estimator for θ, and the function g is continuous at the true value of θ, then g (t) is consistent for g (θ).
For the proof of the Slutsky theorem remember the deﬁnition of a continuous
function. g is continuous at θ iﬀ for all ε > 0 there exists a δ > 0 with the property
that for all θ1 with θ1 − θ < δ follows g (θ1 ) − g (θ) < ε. To prove consistency of
g (t) we have to show that for all ε > 0, Pr[g (t) − g (θ) ≥ ε] → 0. Choose for the
given ε a δ as above, then g (t) − g (θ) ≥ ε implies t − θ ≥ δ , because all those
values of t for with t − θ < δ lead to a g (t) with g (t) − g (θ) < ε. This logical
implication means that
(10.1.2) Pr[g (t) − g (θ) ≥ ε] ≤ Pr[t − θ ≥ δ ]. Since the probability on the righthand side converges to zero, the one on the lefthand
side converges too.
Diﬀerent consistent estimators can have quite diﬀerent speeds of convergence.
Are there estimators which have optimal asymptotic properties among all consistent
121 122 10. ESTIMATION PRINCIPLES estimators? Yes, if one limits oneself to a fairly reasonable subclass of consistent
estimators.
Here are the details: Most consistent estimators we will encounter are asymptotically normal, i.e., the “shape” of their distribution function converges towards
the normal distribution, as we had it for the sample mean in the central limit theorem. In order to be able to use this asymptotic distribution for signiﬁcance tests
and conﬁdence intervals, however, one needs more than asymptotic normality (and
many textbooks are not aware of this): one needs the convergence to normality to
be uniform in compact intervals [Rao73, p. 346–351]. Such estimators are called
consistent uniformly asymptotically normal estimators (CUAN estimators)
If one limits oneself to CUAN estimators it can be shown that there are asymptotically “best” CUAN estimators. Since the distribution is asymptotically normal,
there is no problem to deﬁne what it means to be asymptotically best: those estimators are asymptotically best whose asymptotic MSE = asymptotic variance is
smallest. CUAN estimators whose MSE is asymptotically no larger than that of
any other CUAN estimator, are called asymptotically eﬃcient. Rao has shown that
for CUAN estimators the lower bound for this asymptotic variance is the asymptotic
limit of the Cramer Rao lower bound (CRLB). (More about the CRLB below). Maximum likelihood estimators are therefore usually eﬃcient CUAN estimators. In this
sense one can think of maximum likelihood estimators to be something like asymptotically best consistent estimators, compare a statement to this eﬀect in [Ame94, p.
144]. And one can think of asymptotically eﬃcient CUAN estimators as estimators
who are in large samples as good as maximum likelihood estimators.
All these are large sample properties. Among the asymptotically eﬃcient estimators there are still wide diﬀerences regarding the small sample properties. Asymptotic
eﬃciency should therefore again be considered a minimum requirement: there must
be very good reasons not to be working with an asymptotically eﬃcient estimator.
Problem 167. Can you think of situations in which an estimator is acceptable
which is not asymptotically eﬃcient?
Answer. If robustness matters then the median may be preferable to the mean, although it
is less eﬃcient. 10.2. Small Sample Properties
In order to judge how good an estimator is for small samples, one has two
dilemmas: (1) there are many diﬀerent criteria for an estimator to be “good”; (2)
even if one has decided on one criterion, a given estimator may be good for some
values of the unknown parameters and not so good for others.
If x and y are two estimators of the parameter θ, then each of the following
conditions can be interpreted to mean that x is better than y :
(10.2.1)
(10.2.2) Pr[x − θ ≤ y − θ] = 1
E[g (x − θ)] ≤ E[g (y − θ)] for every continuous function g which is and nonincreasing for x < 0 and nondecreasing for x > 0
(10.2.3) E[g (x − θ)] ≤ E[g (y − θ)] 10.3. COMPARISON UNBIASEDNESS CONSISTENCY 123 for every continuous and nondecreasing function g
(10.2.4) Pr[{x − θ > ε}] ≤ Pr[{y − θ > ε}]
2 for every ε 2 (10.2.5) E[(x − θ) ] ≤ E[(y − θ) ] (10.2.6) Pr[x − θ < y − θ] ≥ Pr[x − θ > y − θ] This list is from [Ame94, pp. 118–122]. But we will simply use the MSE.
Therefore we are left with dilemma (2). There is no single estimator that has
uniformly the smallest MSE in the sense that its MSE is better than the MSE of
any other estimator whatever the value of the parameter value. To see this, simply
think of the following estimator t of θ: t = 10; i.e., whatever the outcome of the
experiments, t always takes the value 10. This estimator has zero MSE when θ
happens to be 10, but is a bad estimator when θ is far away from 10. If an estimator
existed which had uniformly best MSE, then it had to be better than all the constant
estimators, i.e., have zero MSE whatever the value of the parameter, and this is only
possible if the parameter itself is observed.
Although the MSE criterion cannot be used to pick one best estimator, it can be
used to rule out estimators which are unnecessarily bad in the sense that other estimators exist which are never worse but sometimes better in terms of MSE whatever
the true parameter values. Estimators which are dominated in this sense are called
inadmissible.
But how can one choose between two admissible estimators? [Ame94, p. 124]
gives two reasonable strategies. One is to integrate the MSE out over a distribution
of the likely values of the parameter. This is in the spirit of the Bayesians, although
Bayesians would still do it diﬀerently. The other strategy is to choose a minimax
strategy. Amemiya seems to consider this an alright strategy, but it is really too
defensive. Here is a third strategy, which is often used but less well founded theoretically: Since there are no estimators which have minimum MSE among all estimators,
one often looks for estimators which have minimum MSE among all estimators with
a certain property. And the “certain property” which is most often used is unbiasedness. The MSE of an unbiased estimator is its variance; and an estimator which has
minimum variance in the class of all unbiased estimators is called “eﬃcient.”
The class of unbiased estimators has a highsounding name, and the results
related with CramerRao and Least Squares seem to conﬁrm that it is an important
class of estimators. However I will argue in these class notes that unbiasedness itself
is not a desirable property.
10.3. Comparison Unbiasedness Consistency
Let us compare consistency with unbiasedness. If the estimator is unbiased,
then its expected value for any sample size, whether large or small, is equal to the
true parameter value. By the law of large numbers this can be translated into a
statement about large samples: The mean of many independent replications of the
estimate, even if each replication only uses a small number of observations, gives
the true parameter value. Unbiasedness says therefore something about the small
sample properties of the estimator, while consistency does not.
The following thought experiment may clarify the diﬀerence between unbiasedness and consistency. Imagine you are conducting an experiment which gives you
every ten seconds an independent measurement, i.e., a measurement whose value is
not inﬂuenced by the outcome of previous measurements. Imagine further that the
experimental setup is connected to a computer which estimates certain parameters of
that experiment, recalculating its estimate every time twenty new observation have 124 10. ESTIMATION PRINCIPLES become available, and which displays the current values of the estimate on a screen.
And assume that the estimation procedure used by the computer is consistent, but
biased for any ﬁnite number of observations.
Consistency means: after a suﬃciently long time, the digits of the parameter
estimate displayed by the computer will be correct. That the estimator is biased,
means: if the computer were to use every batch of 20 observations to form a new
estimate of the parameter, without utilizing prior observations, and then would use
the average of all these independent estimates as its updated estimate, it would end
up displaying a wrong parameter value on the screen.
A biased extimator gives, even in the limit, an incorrect result as long as one’s
updating procedure is the simple taking the averages of all previous estimates. If
an estimator is biased but consistent, then a better updating method is available,
which will end up in the correct parameter value. A biased estimator therefore is not
necessarily one which gives incorrect information about the parameter value; but it
is one which one cannot update by simply taking averages. But there is no reason to
limit oneself to such a crude method of updating. Obviously the question whether
the estimate is biased is of little relevance, as long as it is consistent. The moral of
the story is: If one looks for desirable estimators, by no means should one restrict
one’s search to unbiased estimators! The highsounding name “unbiased” for the
technical property E[t] = θ has created a lot of confusion.
Besides having no advantages, the category of unbiasedness even has some inconvenient properties: In some cases, in which consistent estimators exist, there are
no unbiased estimators. And if an estimator t is an unbiased estimate for the parameter θ, then the estimator g (t) is usually no longer an unbiased estimator for
g (θ). It depends on the way a certain quantity is measured whether the estimator is
unbiased or not. However consistency carries over.
Unbiasedness is not the only possible criterion which ensures that the values of
the estimator are centered over the value it estimates. Here is another plausible
deﬁnition:
ˆ
Definition 10.3.1. An estimator θ of the scalar θ is called median unbiased for
all θ ∈ Θ iﬀ
1
ˆ
ˆ
(10.3.1)
Pr[θ < θ] = Pr[θ > θ] =
2
This concept is always applicable, even for estimators whose expected value does
not exist.
Problem 168. 6 points (Not eligible for inclass exams) The purpose of the following problem is to show how restrictive the requirement of unbiasedness is. Sometimes no unbiased estimators exist, and sometimes, as in the example here, unbiasedness leads to absurd estimators. Assume the random variable x has the geometric
distribution with parameter p, where 0 ≤ p ≤ 1. In other words, it can only assume
the integer values 1, 2, 3, . . ., with probabilities
(10.3.2) Pr[x = r] = (1 − p)r−1 p. Show that the unique unbiased estimator of p on the basis of one observation of x is
the random variable f (x) deﬁned by f (x) = 1 if x = 1 and 0 otherwise. Hint: Use
the mathematical fact that a function φ(q ) that can be expressed as a power series
∞
φ(q ) = j =0 aj q j , and which takes the values φ(q ) = 1 for all q in some interval of
nonzero length, is the power series with a0 = 1 and aj = 0 for j = 0. (You will need
the hint at the end of your answer, don’t try to start with the hint!) 10.3. COMPARISON UNBIASEDNESS CONSISTENCY 125 ∞ Answer. Unbiasedness means that E[f (x)] =
f (r )(1 − p)r−1 p = p for all p in the unit
r =1
∞
interval, therefore
f (r )(1 − p)r−1 = 1. This is a power series in q = 1 − p, which must be
r =1
identically equal to 1 for all values of q between 0 and 1. An application of the hint shows that
the constant term in this power series, corresponding to the value r − 1 = 0, must be = 1, and all
other f (r) = 0. Here older formulation: An application of the hint with q = 1 − p, j = r − 1, and
aj = f (j + 1) gives f (1) = 1 and all other f (r ) = 0. This estimator is absurd since it lies on the
boundary of the range of possible values for q . Problem 169. As in Question 61, you make two independent trials of a Bernoulli
experiment with success probability θ, and you observe t, the number of successes.
• a. Give an unbiased estimator of θ based on t (i.e., which is a function of t).
• b. Give an unbiased estimator of θ2 .
• c. Show that there is no unbiased estimator of θ3 .
Hint: Since t can only take the three values 0, 1, and 2, any estimator u which
is a function of t is determined by the values it takes when t is 0, 1, or 2, call them
u0 , u1 , and u2 . Express E[u] as a function of u0 , u1 , and u2 .
Answer. E[u] = u0 (1 − θ)2 + 2u1 θ(1 − θ) + u2 θ2 = u0 + (2u1 − 2u0 )θ + (u0 − 2u1 + u2 )θ2 . This
is always a second degree polynomial in θ, therefore whatever is not a second degree polynomial in θ
cannot be the expected value of any function of t. For E[u] = θ we need u0 = 0, 2u1 − 2u0 = 2u1 = 1,
therefore u1 = 0.5, and u0 − 2u1 + u2 = −1 + u2 = 0, i.e. u2 = 1. This is, in other words, u = t/2.
For E[u] = θ2 we need u0 = 0, 2u1 − 2u0 = 2u1 = 0, therefore u1 = 0, and u0 − 2u1 + u2 = u2 = 1,
This is, in other words, u = t(t − 1)/2. From this equation one also sees that θ3 and higher powers,
or things like 1/θ, cannot be the expected values of any estimators. • d. Compute the moment generating function of t.
Answer.
(10.3.3) E[eλt ] = e0 · (1 − θ)2 + eλ · 2θ(1 − θ) + e2λ · θ2 = 1 − θ + θeλ 2 Problem 170. This is [KS79, Question 17.11 on p. 34], originally [Fis, p. 700].
• a. 1 point Assume t and u are two unbiased estimators of the same unknown
scalar nonrandom parameter θ. t and u have ﬁnite variances and satisfy var[u − t] =
0. Show that a linear combination of t and u, i.e., an estimator of θ which can be
written in the form αt + β u, is unbiased if and only if α = 1 − β . In other words,
any unbiased estimator which is a linear combination of t and u can be written in
the form
t + β (u − t). (10.3.4) • b. 2 points By solving the ﬁrst order condition show that the unbiased linear
combination of t and u which has lowest MSE is
cov[t, u − t]
ˆ
(10.3.5)
θ =t−
(u − t)
var[u − t]
Hint: your arithmetic will be simplest if you start with (10.3.4).
• c. 1 point If ρ2 is the squared correlation coeﬃcient between t and u − t, i.e.,
(10.3.6) ρ2 = (cov[t, u − t])2
var[t] var[u − t] ˆ
show that var[θ] = var[t](1 − ρ2 ).
• d. 1 point Show that cov[t, u − t] = 0 implies var[u − t] = 0. 126 10. ESTIMATION PRINCIPLES • e. 2 points Use (10.3.5) to show that if t is the minimum MSE unbiased
estimator of θ, and u another unbiased estimator of θ, then
cov[t, u − t] = 0. (10.3.7) • f . 1 point Use (10.3.5) to show also the opposite: if t is an unbiased estimator
of θ with the property that cov[t, u − t] = 0 for every other unbiased estimator u of
θ, then t has minimum MSE among all unbiased estimators of θ.
There are estimators which are consistent but their bias does not converge to
zero:
ˆ
θn = (10.3.8)
ˆ
Then Pr( θn − θ ≥ ε) ≤
θ + 1 = 0. 1
n, θ
n with probability 1 −
1
with probability n 1
n ˆ
i.e., the estimator is consistent, but E[θ] = θ n−1 + 1 →
n Problem 171. 4 points Is it possible to have a consistent estimator whose bias
becomes unbounded as the sample size increases? Either prove that it is not possible
or give an example.
Answer. Yes, this can be achieved by making the rare outliers even wilder than in (10.3.8),
say
ˆ
θn = (10.3.9)
ˆ
Here Pr( θn − θ ≥ ε) ≤ 1
,
n θ
n2 with probability 1 −
1
with probability n 1
n ˆ
i.e., the estimator is consistent, but E[θ] = θ n−1 + n → θ + n.
n And of course there are estimators which are unbiased but not consistent: simply take the ﬁrst observation x1 as an estimator if E[x] and ignore all the other
observations.
10.4. The CramerRao Lower Bound
Take a scalar random variable y with density function fy . The entropy of y , if it
exists, is H[y ] = − E[log(fy (y ))]. This is the continuous equivalent of (3.11.2). The
entropy is the measure of the amount of randomness in this variable. If there is little
information and much noise in this variable, the entropy is high.
Now let y → g (y ) be the density function of a diﬀerent random variable x. In
+∞
other words, g is some function which satisﬁes g (y ) ≥ 0 for all y , and −∞ g (y ) dy = 1.
Equation (3.11.10) with v = g (y ) and w = fy (y ) gives
(10.4.1) fy (y ) − fy (y ) log fy (y ) ≤ g (y ) − fy (y ) log g (y ). This holds for every value y , and integrating over y gives 1 − E[log fy (y )] ≤ 1 −
E[log g (y )] or
(10.4.2) E[log fy (y )] ≥ E[log g (y )]. This is an important extremal value property which distinguishes the density function
fy (y ) of y from all other density functions: That density function g which maximizes
E[log g (y )] is g = fy , the true density function of y .
This optimality property lies at the basis of the CramerRao inequality, and it
is also the reason why maximum likelihood estimation is so good. The diﬀerence
between the left and right hand side in (10.4.2) is called the KullbackLeibler discrepancy between the random variables y and x (where x is a random variable whose
density is g ). 10.4. THE CRAMERRAO LOWER BOUND 127 The Cramer Rao inequality gives a lower bound for the MSE of an unbiased
estimator of the parameter of a probability distribution (which has to satisfy certain regularity conditions). This allows one to determine whether a given unbiased
estimator has a MSE as low as any other unbiased estimator (i.e., whether it is
“eﬃcient.”)
Problem 172. Assume the density function of y depends on a parameter θ,
write it fy (y ; θ), and θ◦ is the true value of θ. In this problem we will compare the
expected value of y and of functions of y with what would be their expected value if the
true parameter value were not θ◦ but would take some other value θ. If the random
variable t is a function of y , we write Eθ [t] for what would be the expected value of t
if the true value of the parameter were θ instead of θ◦ . Occasionally, we will use the
subscript ◦ as in E◦ to indicate that we are dealing here with the usual case in which
the expected value is taken with respect to the true parameter value θ◦ . Instead of E◦
one usually simply writes E, since it is usually selfunderstood that one has to plug
the right parameter values into the density function if one takes expected values. The
subscript ◦ is necessary here only because in the present problem, we sometimes take
expected values with respect to the “wrong” parameter values. The same notational
convention also applies to variances, covariances, and the MSE.
Throughout this problem we assume that the following regularity conditions hold:
(a) the range of y is independent of θ, and (b) the derivative of the density function
with respect to θ is a continuous diﬀerentiable function of θ. These regularity conditions ensure that one can diﬀerentiate under the integral sign, i.e., for all function
t(y ) follows
∞ (10.4.3) ∂
∂
fy (y ; θ)t(y ) dy =
∂θ
∂θ −∞
∞
2 (10.4.4)
−∞ ∞ fy (y ; θ)t(y ) dy =
−∞ ∂
∂2
fy (y ; θ)t(y ) dy =
(∂θ)2
(∂θ)2 ∂
Eθ [t(y )]
∂θ ∞ fy (y ; θ)t(y ) dy =
−∞ ∂2
Eθ [t(y )].
(∂θ)2 • a. 1 point The score is deﬁned as the random variable
∂
log fy (y ; θ).
∂θ
In other words, we do three things to the density function: take its logarithm, then
take the derivative of this logarithm with respect to the parameter, and then plug the
random variable into it. This gives us a random variable which also depends on the
nonrandom parameter θ. Show that the score can also be written as
(10.4.5) (10.4.6) q (y ; θ) = q (y ; θ) = 1
∂fy (y ; θ)
fy (y ; θ)
∂θ Answer. This is the chain rule for diﬀerentiation: for any diﬀerentiable function g (θ), ∂
∂θ log g (θ) = 1 ∂g (θ )
.
g (θ ) ∂θ • b. 1 point If the density function is member of an exponential dispersion family
(??), show that the score function has the form
(10.4.7) q (y ; θ) = y − ∂b(θ)
∂θ
a(ψ ) Answer. This is a simple substitution: if
(10.4.8) fy (y ; θ, ψ ) = exp y θ − b(θ)
+ c(y, ψ ) ,
a (ψ ) 128 10. ESTIMATION PRINCIPLES then
∂b(θ ) y − ∂θ
∂ log fy (y ; θ, ψ )
=
∂θ
a(ψ ) (10.4.9) • c. 3 points If fy (y ; θ◦ ) is the true density function of y , then we know from
(10.4.2) that E◦ [log fy (y ; θ◦ )] ≥ E◦ [log f (y ; θ)] for all θ. This explains why the score
is so important: it is the derivative of that function whose expected value is maximized
if the true parameter is plugged into the density function. The ﬁrstorder conditions
in this situation read: the expected value of this derivative must be zero for the true
parameter value. This is the next thing you are asked to show: If θ◦ is the true
parameter value, show that E◦ [q (y ; θ◦ )] = 0.
Answer. First write for general θ
∞ (10.4.10) ∞ q (y ; θ)fy (y ; θ◦ ) dy = E◦ [q (y ; θ)] =
−∞ −∞ 1
∂fy (y ; θ)
fy (y ; θ◦ ) dy.
f y (y ; θ )
∂θ For θ = θ◦ this simpliﬁes:
∞ (10.4.11) E◦ [q (y ; θ◦ )] =
−∞ Here I am writing ∂fy (y ;θ )
∂θ ∂fy (y ; θ)
∂θ dy =
θ =θ ◦ ∂
∂θ ∞ fy (y ; θ) dy
∞ =
θ =θ ◦ ∂
1 = 0.
∂θ ∂fy (y ;θ ◦ )
, in order to emphasize
∂θ
◦ into that derivative.
one plugs θ instead of the simpler notation
θ =θ ◦ that one ﬁrst has to take a derivative with respect to θ and then • d. Show that, in the case of the exponential dispersion family,
(10.4.12) E◦ [y ] = ∂b(θ)
∂θ θ =θ ◦ Answer. Follows from the fact that the score function of the exponential family (10.4.7) has
zero expected value. • e. 5 points If we diﬀerentiate the score, we obtain the Hessian
(10.4.13) h(θ) = ∂2
log fy (y ; θ).
(∂θ)2 From now on we will write the score function as q (θ) instead of q (y ; θ); i.e., we will
no longer make it explicit that q is a function of y but write it as a random variable
which depends on the parameter θ. We also suppress the dependence of h on y ; our
notation h(θ) is short for h(y ; θ). Since there is only one parameter in the density
function, score and Hessian are scalars; but in the general case, the score is a vector
and the Hessian a matrix. Show that, for the true parameter value θ◦ , the negative
of the expected value of the Hessian equals the variance of the score, i.e., the expected
value of the square of the score:
E◦ [h(θ◦ )] = − E◦ [q 2 (θ◦ )]. (10.4.14) Answer. Start with the deﬁnition of the score
∂
1
∂
q (y ; θ ) =
log fy (y ; θ) =
f y (y ; θ ),
∂θ
fy (y ; θ) ∂θ (10.4.15) and diﬀerentiate the rightmost expression one more time:
(10.4.16)
(10.4.17) h (y ; θ ) = ∂
1
q (y ; θ ) = − 2
(∂θ)
f y (y ; θ ) = − q 2 (y ; θ ) + ∂
f y (y ; θ )
∂θ 1
∂2
f y (y ; θ )
fy (y ; θ) ∂θ2 2 + 1
∂2
f y (y ; θ )
fy (y ; θ) ∂θ2 10.4. THE CRAMERRAO LOWER BOUND 129 Taking expectations we get
+∞ (10.4.18) E◦ [h(y ; θ)] = − E◦ [q 2 (y ; θ)] +
−∞ 1
f y (y ; θ ) ∂2
fy (y ; θ) fy (y ; θ◦ ) dy
∂θ2 Again, for θ = θ◦ , we can simplify the integrand and diﬀerentiate under the integral sign:
+∞ (10.4.19)
−∞ ∂2
∂2
fy (y ; θ) dy =
∂θ2
∂θ2 +∞ fy (y ; θ) dy =
−∞ ∂2
1 = 0.
∂θ2 • f . Derive from (10.4.14) that, for the exponential dispersion family (??),
(10.4.20) var◦ [y ] = ∂ 2 b(θ)
a(φ)
∂θ2 Answer. Diﬀerentiation of (10.4.7) gives h(θ) = −
equal to its own expected value. (10.4.14) says therefore
(10.4.21) ∂ 2 b(θ)
∂θ2 θ =θ ◦ θ =θ ◦ 2 ∂ b(θ ) 1
.
∂θ 2 a(φ) 1
= E◦ [q 2 (θ◦ )] =
a(φ) 1
a(φ) 2 This is constant and therefore var◦ [y ] from which (10.4.20) follows. Problem 173.
• a. Use the results from question 172 to derive the following strange and interesting result: for any random variable t which is a function of y , i.e., t = t(y ),
∂
follows cov◦ [q (θ◦ ), t] = ∂θ Eθ [t] θ=θ◦ .
Answer. The following equation holds for all θ:
∞ (10.4.22) E◦ [q (θ)t] =
−∞ 1
∂fy (y ; θ)
t(y )fy (y ; θ◦ ) dy
f y (y ; θ )
∂θ If the θ in q (θ) is the right parameter value θ◦ one can simplify:
∞ (10.4.23) E◦ [q (θ◦ )t] =
−∞ (10.4.24)
(10.4.25) =
= ∂
∂θ ∂fy (y ; θ)
∂θ t(y ) dy
θ =θ ◦ ∞ fy (y ; θ)t(y ) dy
−∞ ∂
Eθ [t]
∂θ θ =θ ◦ θ =θ ◦ This is at the same time the covariance: cov◦ [q (θ◦ ), t] = E◦ [q (θ◦ )t] − E◦ [q (θ◦ )] E◦ [t] = E◦ [q (θ◦ )t],
since E◦ [q (θ◦ )] = 0. Explanation, nothing to prove here: Now if t is an unbiased estimator of θ,
∂
whatever the value of θ, then it follows cov◦ [q (θ◦ ), t] = ∂θ θ = 1. From this fol◦
lows by CauchySchwartz var◦ [t] var◦ [q (θ )] ≥ 1, or var◦ [t] ≥ 1/ var◦ [q (θ◦ )]. Since
E◦ [q (θ◦ )] = 0, we know var◦ [q (θ◦ )] = E◦ [q 2 (θ◦ )], and since t is unbiased, we know
var◦ [t] = MSE◦ [t; θ◦ ]. Therefore the CauchySchwartz inequality reads
(10.4.26) MSE◦ [t; θ◦ ] ≥ 1/ E◦ [q 2 (θ◦ )]. This is the CramerRao inequality. The inverse of the variance of q (θ◦ ), 1/ var◦ [q (θ◦ )] =
1/ E◦ [q 2 (θ◦ )], is called the Fisher information, written I (θ◦ ). It is a lower bound for
the MSE of any unbiased estimator of θ. Because of (10.4.14), the Cramer Rao
inequality can also be written in the form
(10.4.27) MSE[t; θ◦ ] ≥ −1/ E◦ [h(θ◦ )]. 130 10. ESTIMATION PRINCIPLES (10.4.26) and (10.4.27) are usually written in the following form: Assume y has
density function fy (y ; θ) which depends on the unknown parameter θ, and and let
t(y ) be any unbiased estimator of θ. Then
(10.4.28) 1 var[t] ≥
E[ 2 ∂
∂θ = log fy (y ; θ) ] ∂2
E[ ∂θ2 −1
.
log fy (y ; θ)] (Sometimes the ﬁrst and sometimes the second expression is easier to evaluate.)
If one has a whole vector of observations then the CramerRao inequality involves
the joint density function:
(10.4.29) 1 var[t] ≥
E[ ∂
∂θ 2 = log fy (y ; θ) ] ∂2
E[ ∂θ2 −1
.
log fy (y ; θ)] This inequality also holds if y is discrete and one uses its probability mass function
instead of the density function. In small samples, this lower bound is not always
attainable; in some cases there is no unbiased estimator with a variance as low as
the Cramer Rao lower bound.
Problem 174. 4 points Assume n independent observations of a variable y ∼
N (µ, σ 2 ) are available, where σ 2 is known. Show that the sample mean y attains the
¯
CramerRao lower bound for µ.
Answer. The density function of each y i is
fyi (y ) = (2πσ 2 )−1/2 exp − (10.4.30) (y − µ )2
2σ 2 therefore the log likelihood function of the whole vector is
n (10.4.31) log fyi (y i ) = − ( y ; µ) = n
n
1
log(2π ) − log σ 2 −
2
2
2σ 2 i=1 ( y i − µ )2
i=1 n ∂
1
( y ; µ) = 2
∂µ
σ (10.4.32) n (y i − µ)
i=1 In order to apply (10.4.29) you can either square this and take the expected value
(10.4.33) E[ ∂
(y ; µ)
∂µ 2
= 1
σ4 E[(y i − µ)2 ] = n/σ 2 alternatively one may take one more derivative from (10.4.32) to get
(10.4.34) n
∂2
( y ; µ) = − 2
∂µ2
σ This is constant, therefore equal to its expected value. Therefore the CramerRao Lower Bound
says that var[¯] ≥ σ 2 /n. This holds with equality.
y Problem 175. Assume y i ∼ NID(0, σ 2 ) (i.e., normally independently distributed)
1
with unknown σ 2 . The obvious estimate of σ 2 is s2 = n
y2 .
i
2 • a. 2 points Show that s2 is an unbiased estimator of σ 2 , is distributed ∼ σ χ2 ,
nn
and has variance 2σ 4 /n. You are allowed to use the fact that a χ2 has variance 2n,
n
which is equation (4.9.5). 10.4. THE CRAMERRAO LOWER BOUND 131 Answer.
2
E[yi ] = var[yi ] + (E[yi ])2 = σ 2 + 0 = σ 2
yi
zi =
∼ NID(0, 1)
σ
yi = σzi (10.4.35)
(10.4.36)
(10.4.37) 2
2
yi = σ 2 zi (10.4.38) n n 1
n (10.4.40) (10.4.41) 2
zi ∼ σ 2 χ2
n 2
yi = σ 2 (10.4.39) var 1
n i=1
n
2
yi =
i=1
n
2
yi = σ2
n i=1
n
2
zi ∼ σ2 2
χ
nn i=1 σ4
n2 var[χ2 ] =
n σ4
2σ 4
2n =
2
n
n i=1 • b. 4 points Show that this variance is at the same time the Cramer Rao lower
bound.
Answer.
(10.4.42) 1
1
y2
log 2π − log σ 2 −
2
2
2σ 2
1
y2
y2 − σ2
∂ log fy
(y ; σ 2 ) = − 2 +
=
∂σ 2
2σ
2σ 4
2σ 4 (y, σ 2 ) = log fy (y ; σ 2 ) = − (10.4.43) Since y2 − σ2
has zero mean, it follows
2σ 4 (10.4.44) E[ ∂ log fy
(y ; σ 2 )
∂σ 2 2
= var[y 2 ]
1
=
.
4σ 8
2σ 4 Alternatively, one can diﬀerentiate one more time:
∂ 2 log fy
y2
1
(y ; σ 2 ) = − 6 +
(∂σ 2 )2
σ
2σ 4 (10.4.45)
(10.4.46) E[ 1
1
σ2
∂ 2 log fy
(y ; σ 2 )] = − 6 +
=
(∂σ 2 )2
σ
2σ 4
2σ 4 (10.4.47)
This makes the Cramer Rao lower bound 2σ 4 /n. Problem 176. 4 points Assume x1 , . . . , xn is a random sample of independent
observations of a Poisson distribution with parameter λ, i.e., each of the xi has
probability mass function
(10.4.48) pxi (x) = Pr[xi = x] = λx −λ
e
x! x = 0, 1, 2, . . . . A Poisson variable with parameter λ has expected value λ and variance λ. (You
are not required to prove this here.) Is there an unbiased estimator of λ with lower
variance than the sample mean x?
¯
Here is a formulation of the Cramer Rao Inequality for probability mass functions, as you need it for Question 176. Assume y 1 , . . . , y n are n independent observations of a random variable y whose probability mass function depends on the
unknown parameter θ and satisﬁes certain regularity conditions. Write the univariate probability mass function of each of the y i as py (y ; θ) and let t be any unbiased 132 10. ESTIMATION PRINCIPLES estimator of θ. Then
(10.4.49) 1 var[t] ≥
n E[ ∂
∂θ 2 = ln py (y ; θ) ] ∂2
n E[ ∂θ2 −1
.
ln py (y ; θ)] Answer. The Cramer Rao lower bound says no.
log px (x; λ) = x log λ − log x! − λ (10.4.50)
(10.4.51)
(10.4.52) x
x−λ
∂ log px
(x; λ) = − 1 =
∂λ
λ
λ
2
var[x]
1
∂ log px
(x − λ)2
(x; λ) ] = E[=
=.
E[
∂λ
λ2
λ2
λ Or alternatively, after (10.4.51) do
(10.4.53)
(10.4.54) ∂ 2 log px
x
(x; λ) = − 2
∂λ2
λ
E[x]
1
∂ 2 log px
(x; λ) ] = 2 = .
− E[
∂λ2
λ
λ Therefore the Cramer Rao lower bound is λ
,
n which is the variance of the sample mean. If the density function depends on more than one unknown parameter, i.e., if
it has the form fy (y ; θ1 , . . . , θk ), the Cramer Rao Inequality involves the following
steps: (1) deﬁne (y ; θ1 , · · · , θk ) = log fy (y ; θ1 , . . . , θk ), (2) form the following matrix
which is called the information matrix :
(10.4.55) 2
2
∂2
∂
∂
−n E[ ∂ θ2 ]
· · · −n E[ ∂ θ∂ ∂θk ]
n E[ ∂ θ1 ] · · · n E[ ∂ θ1 ∂∂θk ]
1
1 .
.
.
.
..
..
=
,
.
.
.
.
I=
.
.
.
.
.
. 2
∂2
∂2
∂
∂
∂
−n E[ ∂ θ2 ]
−n E[ ∂ θk ∂θ1 ] · · ·
n E[ ∂ θk ∂ θ1 ] · · · n E[ ∂ θk ]
k t1
.
−1
and (3) form the matrix inverse I . If the vector random variable t = . .
tn θ1
.
is an unbiased estimator of the parameter vector θ = . , then the inverse of
. θn
the information matrix I −1 is a lower bound for the covariance matrix V [t] in the
following sense: the diﬀerence matrix V [t] − I −1 is always nonnegative deﬁnite.
From this follows in particular: if iii is the ith diagonal element of I −1 , then
var[ti ] ≥ iii .
10.5. Best Linear Unbiased Without Distribution Assumptions
If the xi are Normal with unknown expected value and variance, their sample
mean has lowest MSE among all unbiased estimators of µ. If one does not assume
Normality, then the sample mean has lowest MSE in the class of all linear unbiased
estimators of µ. This is true not only for the sample mean but also for all least squares
estimates. This result needs remarkably weak assumptions: nothing is assumed about
the distribution of the xi other than the existence of mean and variance. Problem
177 shows that in some situations one can even dispense with the independence of
the observations.
Problem 177. 5 points [Lar82, example 5.4.1 on p 266] Let y 1 and y 2 be two
random variables with same mean µ and variance σ 2 , but we do not assume that they 10.5. BEST LINEAR UNBIASED WITHOUT DISTRIBUTION ASSUMPTIONS 133 are uncorrelated; their correlation coeﬃcient is ρ, which can take any value ρ ≤ 1.
Show that y = (y 1 + y 2 )/2 has lowest mean squared error among all linear unbiased
¯
estimators of µ, and compute its MSE. (An estimator µ of µ is linear iﬀ it can be
˜
written in the form µ = α1 y 1 + α2 y 2 with some constant numbers α1 and α2 .)
˜
Answer.
(10.5.1)
(10.5.2) y = α1 y 1 + α2 y 2
˜
var y = α2 var[y 1 ] + α2 var[y 2 ] + 2α1 α2 cov[y 1 , y 2 ]
˜
1
2
= σ 2 (α2 + α2 + 2α1 α2 ρ).
1
2 (10.5.3) Here we used (6.1.14). Unbiasedness means α2 = 1 − α1 , therefore we call α1 = α and α2 = 1 − α:
(10.5.4) y
var[˜]/σ 2 = α2 + (1 − α)2 + 2α(1 − α)ρ Now sort by the powers of α:
(10.5.5) = 2α2 (1 − ρ) − 2α(1 − ρ) + 1 (10.5.6) = 2(α2 − α)(1 − ρ) + 1. This takes its minimum value where the derivative
α1 = α2 − 1/2 into (10.5.3) to get σ2
2 ∂
(α 2
∂α − α) = 2α − 1 = 0. For the MSE plug 1+ρ . Problem 178. You have two unbiased measurements with errors of the same
quantity µ (which may or may not be random). The ﬁrst measurement y 1 has mean
squared error E[(y 1 − µ)2 ] = σ 2 , the other measurement y 2 has E[(y 1 − µ)2 ] =
τ 2 . The measurement errors y 1 − µ and y 2 − µ have zero expected values (i.e., the
measurements are unbiased) and are independent of each other.
• a. 2 points Show that the linear unbiased estimators of µ based on these two
measurements are simply the weighted averages of these measurements, i.e., they can
be written in the form µ = αy 1 + (1 − α)y 2 , and that the MSE of such an estimator
˜
is α2 σ 2 + (1 − α)2 τ 2 . Note: we are using the word “estimator” here even if µ is
random. An estimator or predictor µ is unbiased if E[˜ − µ] = 0. Since we allow µ
˜
µ
to be random, the proof in the class notes has to be modiﬁed.
Answer. The estimator µ is linear (more precisely: aﬃne) if it can written in the form
˜
µ = α1 y 1 + α2 y 2 + γ
˜ (10.5.7) The measurements themselves are unbiased, i.e., E[y i − µ] = 0, therefore
(10.5.8) E[˜ − µ] = (α1 + α2 − 1) E[µ] + γ = 0
µ for all possible values of E[µ]; therefore γ = 0 and α2 = 1 − α1 . To simplify notation, we will call
from now on α1 = α, α2 = 1 − α. Due to unbiasedness, the MSE is the variance of the estimation
error
(10.5.9) var[˜ − µ] = α2 σ 2 + (1 − α)2 τ 2
µ • b. 4 points Deﬁne ω 2 by
1
1
σ2 τ 2
1
= 2+ 2
which can be solved to give
ω2 = 2
.
ω2
σ
τ
σ + τ2
Show that the Best (i.e., minimum MSE) linear unbiased estimator (BLUE) of µ
based on these two measurements is
ω2
ω2
(10.5.11)
y = 2 y1 + 2 y2
ˆ
σ
τ
i.e., it is the weighted average of y 1 and y 2 where the weights are proportional to the
inverses of the variances.
(10.5.10) 134 10. ESTIMATION PRINCIPLES Answer. The variance (10.5.9) takes its minimum value where its derivative with respect of
α is zero, i.e., where (10.5.13) ∂
α2 σ 2 + (1 − α)2 τ 2 = 2ασ 2 − 2(1 − α)τ 2 = 0
∂α
ασ 2 = τ 2 − ατ 2 (10.5.14) α= (10.5.12) In terms of ω one can write
ω2
τ2
(10.5.15)
=2
α= 2
2
σ +τ
σ σ2 τ2
+ τ2 and 1−α= σ2 σ2
ω2
= 2.
2
+τ
τ • c. 2 points Show: the MSE of the BLUE ω 2 satisﬁes the following equation:
1
1
1
(10.5.16)
= 2+ 2
ω2
σ
τ
Answer. We already have introduced the notation ω 2 for the quantity deﬁned by (10.5.16);
therefore all we have to show is that the MSE or, equivalently, the variance of the estimation error
is equal to this ω 2 :
(10.5.17) var[˜ − µ] =
µ ω2
σ2 22 σ+ ω2
τ2 22 τ = ω4 1
1
+2
σ2
τ = ω4 1
= ω2
ω2 Examples of other classes of estimators for which a best estimator exists are: if
one requires the estimator to be translation invariant, then the least squares estimators are best in the class of all translation invariant estimators. But there is no best
linear estimator in the linear model. (Theil)
10.6. Maximum Likelihood Estimation
This is an excellent and very widely applicable estimation principle. Its main
drawback is its computational complexity, but with modern computing power it
becomes more and more manageable. Another drawback is that it requires a full
speciﬁcation of the distribution.
Problem 179. 2 points What are the two greatest disadvantages of Maximum
Likelihood Estimation?
Answer. Its high information requirements (the functional form of the density function must
be known), and computational complexity. In our discussion of entropy in Section 3.11 we derived an extremal value property
which distinguishes the actual density function fy (y ) of a given random variable y
from all other possible density functions of y , i.e., from all other functions g ≥ 0
+∞
with −∞ g (y ) dy = 1. The true density function of y is the one which maximizes
E[log g (y )]. We showed that this principle can be used to design a payoﬀ scheme
by which it is in the best interest of a forecaster to tell the truth. Now we will see
that this principle can also be used to design a good estimator. Say you have n
independent observations of y . You know the density of y belongs to a given family
F of density functions, but you don’t know which member of F it is. Then form
the arithmetic mean of log f (yi ) for all f ∈ F . It converges towards E[log f (y )]. For
the true density function, this expected value is higher than for all the other density
functions. If one does not know which the true density function is, then it is a good
strategy to select that density function f for which the sample mean of the log f (yi )
is largest. This is the maximum likelihood estimator. 10.6. MAXIMUM LIKELIHOOD ESTIMATION 135 Let us interject here a short note about the deﬁnitional diﬀerence between density
function and likelihood function. If we know µ = µ0 , we can write down the density
function as
(y −µ0 )2
1
(10.6.1)
fy (y ; µ0 ) = √ e− 2 .
2π
It is a function of y , the possible values assumed by y , and the letter µ0 symbolizes
a constant, the true parameter value. The same function considered as a function of
the variable µ, representing all possible values assumable by the true mean, with y
being ﬁxed at the actually observed value, becomes the likelihood function.
In the same way one can also turn probability mass functions px (x) into likelihood
functions.
Now let us compute some examples of the MLE. You make n independent
observations y 1 , . . . , y n from a N (µ, σ 2 ) distribution. Write the likelihood function
as
n (10.6.2) L(µ, σ 2 ; y 1 , . . . , y n ) = fy (y i ) = √
i=1 1
2πσ 2 n 1 e− 2σ2 (y i −µ)2 . Its logarithm is more convenient to maximize:
n
n
1
= ln L(µ, σ 2 ; y 1 , . . . , y n ) = − ln 2π − ln σ 2 − 2
(y i − µ)2 .
(10.6.3)
2
2
2σ
To compute the maximum we need the partial derivatives:
1
∂
=2
(y i − µ)
(10.6.4)
∂µ
σ
n
1
∂
(10.6.5)
(y i − µ)2 .
=− 2 + 4
∂ σ2
2σ
2σ
The maximum likelihood estimators are those values µ and σ 2 which set these two
ˆ
ˆ
partials zero. I.e., at the same time at which we set the partials zero we must put
the hats on µ and σ 2 . As long as σ 2 = 0 (which is the case with probability one),
ˆ
1
the ﬁrst equation determines µ:
ˆ
y i − nµ = 0, i.e., µ = n
ˆ
ˆ
y i = y . (This would
¯
2
be the MLE of µ even if σ were known). Now plug this µ into the second equation
ˆ
1
1
¯
ˆ
¯
to get n = 2ˆ 2 (y i − y )2 , or σ 2 = n (y i − y )2 .
2
σ
Here is another example: t1 , . . . , tn are independent and follow an exponential
distribution, i.e.,
(10.6.6)
(10.6.7)
(10.6.8) ft (t; λ) = λe−λt (t > 0) n −λ(t1 +···+tn ) L(t1 , . . . , tn ; λ) = λ e (t1 , . . . , tn λ) = n ln λ − λ(t1 + · · · + tn ) ∂
n
= − (t1 + · · · + tn ).
∂λ
λ
n
ˆ instead of λ to get λ =
ˆ
¯
Set this zero, and write λ
t1 +···+tn = 1/t.
Usually the MLE is asymptotically unbiased and asymptotically normal. Therefore it is important to have an estimate of its asymptotic variance. Here we can use
the fact that asymptotically the Cramer Rao Lower Bound is not merely a lower
bound for this variance but is equal to its variance. (From this follows that the maximum likelihood estimator is asymptotically eﬃcient.) The Cramer Rao lower bound
itself depends on unknown parameters. In order to get a consistent estimate of the
Cramer Rao lower bound, do the following: (1) Replace the unknown parameters
in the second derivative of the log likelihood function by their maximum likelihood
estimates. (2) Instead of taking expected values over the observed values xi you may
(10.6.9) 136 10. ESTIMATION PRINCIPLES simply insert the sample values of the xi into these maximum likelihood estimates,
and (3) then invert this estimate of the information matrix.
ˆ
MLE obeys an important functional invariance principle: if θ is the MLE of θ,
ˆ) is the MLE of g (θ). E.g., µ = 1 is the expected value of the exponential
then g (θ
λ
variable, and its MLE is x.
¯
Problem 180. x1 , . . . , xm is a sample from a N (µx , σ 2 ), and y 1 , . . . , y n from a
N (µy , σ 2 ) with diﬀerent mean but same σ 2 . All observations are independent of each
other.
• a. 2 points Show that the MLE of µx , based on the combined sample, is x. (By
¯
symmetry it follows that the MLE of µy is y .)
¯
Answer.
(10.6.10) (µx , µ y , σ 2 ) = − m m
m
1
ln 2π −
ln σ 2 −
2
2
2σ 2 ( xi − µ x ) 2
i=1
n n
n
1
− ln 2π − ln σ 2 −
2
2
2σ 2 ( y j − µ y )2
j =1 (10.6.11) 1
∂
=− 2
∂ µx
2σ −2(xi − µx ) =0 for µx = x
¯ • b. 2 points Derive the MLE of σ 2 , based on the combined samples.
Answer.
(10.6.12) ∂
m+n
1
=−
+
∂ σ2
2σ 2
2σ 4 m n ( xi − µx ) 2 +
i=1 (10.6.13) σ2 =
ˆ 1
m+n m j =1
n ( xi − x) 2 +
¯
i=1 ( y j − µ y )2 (y i − y ) 2 .
¯
j =1 10.7. Method of Moments Estimators
Method of moments estimators use the sample moments as estimates of the
population moments. I.e., the estimate of µ is x, the estimate of the variance σ 2 is
¯
1
(xi − x)2 , etc. If the parameters are a given function of the population moments,
¯
n
use the same function of the sample moments (using the lowest moments which do
the job).
The advantage of method of moments estimators is their computational simplicity. Many of the estimators discussed above are method of moments estimators.
However if the moments do not exist, then method of moments estimators are inconsistent, and in general method of moments estimators are not as good as maximum
likelihood estimators.
10.8. MEstimators
The class of M estimators maximizes something other than a likelihood function: it includes nonlinear least squares, generalized method of moments, minimum
distance and minimum chisquared estimators. The purpose is to get a “robust”
estimator which is good for a wide variety of likelihood functions. Many of these are
asymptotically eﬃcient; but their smallsample properties may vary greatly. 10.9. SUFFICIENT STATISTICS AND ESTIMATION 137 10.9. Suﬃcient Statistics and Estimation
Weak Suﬃciency Principle: If x has a p.d.f. fx (x; θ) and if a suﬃcient statistic
s(x) exists for θ, then identical conclusions should be drawn from data x1 and x2
which have same value s(x1 ) = s(x2 ).
Why? Suﬃciency means: after knowing s(x), the rest of the data x can be
regarded generated by a random mechanism not dependent on θ, and are therefore
uninformative about θ.
This principle can be used to improve on given estimators. Without proof we
will state here
Rao Blackwell Theorem: Let t(x) be an estimator of θ and s(x) a suﬃcient
statistic for θ. Then one can get an estimator t∗ (x) of θ which has no worse a
MSE than t(x) by taking expectations conditionally on the suﬃcient statistic, i.e.,
t∗ (x) = E[t(x)s(x)].
To recapitulate: t∗ (x) is obtained by the following two steps: (1) Compute the
conditional expectation t∗∗ (s) = E[t(x)s(x) = s], and (2) plug s(x) into t∗∗ , i.e.,
t∗ (x) = t∗∗ (s(x)).
A statistic s is said to be complete, if the only realvalued function g deﬁned on
the range of s, which satisﬁes E[g (s)] = 0 whatever the value of θ, is the function
which is identically zero. If a statistic s is complete and suﬃcient, then every function
g (s) is the minimum MSE unbiased estimator of its expected value E[g (s)].
If a complete and suﬃcient statistic exists, this gives a systematic approach to
minimum MSE unbiased estimators (Lehmann Scheﬀ´ Theorem ): if t is an unbiased
e
estimator of θ and s is complete and suﬃcient, then t∗ (x) = E[t(x)s(x)] has lowest
MSE in the class of all unbiased estimators of θ. Problem 181 steps you through the
proof.
Problem 181. [BD77, Problem 4.2.6 on p. 144] If a statistic s is complete
and suﬃcient, then every function g (s) is the minimum MSE unbiased estimator
of E[g (s)] ( LehmannScheﬀ´ theorem). This gives a systematic approach to ﬁnding
e
minimum MSE unbiased estimators. Here are the deﬁnitions: s is suﬃcient for θ
if for any event E and any value s, the conditional probability Pr[E s ≤ s] does not
involve θ. s is complete for θ if the only function g (s) of s, which has zero expected
value whatever the value of θ, is the function which is identically zero, i.e., g (s) = 0
for all s.
• a. 3 points Given an unknown parameter θ, and a complete suﬃcient statistic
s, how can one ﬁnd that function of s whose expected value is θ? There is an easy
trick: start with any statistic p with E[p] = θ, and use the conditional expectation
E[ps]. Argue why this conditional expectation does not depend on the unknown
parameter θ, is an unbiased estimator of θ, and why this leads to the same estimate
regardless which p one starts with.
Answer. You need suﬃciency for the ﬁrst part of the problem, the law of iterated expectations
for the second, and completeness for the third.
Set E = {p ≤ p} in the deﬁnition of suﬃciency given at the beginning of the Problem to see
that the cdf of p conditionally on s being in any interval does not involve θ, therefore also E[ps]
does not involve θ.
Unbiasedness follows from the theorem of iterated expectations E E[ps] = E[p] = θ.
The independence on the choice of p can be shown as follows: Since the conditional expectation
conditionally on s is a function of s, we can use the notation E[ps] = g1 (s) and E[q s] = g2 (s).
From E[p] = E[q ] follows by the law of iterated expectations E[g1 (s) − g2 (s)] = 0, therefore by
completeness g1 (s) − g2 (s) ≡ 0. 138 10. ESTIMATION PRINCIPLES • b. 2 points Assume y i ∼ NID(µ, 1) (i = 1, . . . , n), i.e., they are independent
and normally distributed with mean µ and variance 1. Without proof you are allowed
to use the fact that in this case, the sample mean y is a complete suﬃcient statistic
¯
for µ. What is the minimum MSE unbiased estimate of µ, and what is that of µ2 ?
Answer. We have to ﬁnd functions of y with the desired parameters as expected values.
¯
Clearly, y is that of µ, and y 2 − 1/n is that of µ2 .
¯
¯ • c. 1 point For a given j , let π be the probability that the j th observation is
nonnegative, i.e., π = Pr[y j ≥ 0]. Show that π = Φ(µ) where Φ is the cumulative
distribution function of the standard normal. The purpose of the remainder of this
Problem is to ﬁnd a minimum MSE unbiased estimator of π .
Answer.
(10.9.1) π = Pr[y i ≥ 0] = Pr[y i − µ ≥ −µ] = Pr[y i − µ ≤ µ] = Φ(µ) because y i − µ ∼ N (0, 1). We needed symmetry of the distribution to ﬂip the sign. • d. 1 point As a ﬁrst step we have to ﬁnd an unbiased estimator of π . It does
not have to be a good one, any ubiased estimator will do. And such an estimator is
indeed implicit in the deﬁnition of π . Let q be the “indicator function” for nonnegative
values, satisfying q (y ) = 1 if y ≥ 0 and 0 otherwise. We will be working with the
random variable which one obtains by inserting the j th observation y j into q , i.e.,
with q = q (y j ). Show that q is an unbiased estimator of π .
Answer. q (y j ) has a discrete distribution and Pr[q (y j ) = 1] = Pr[y j ≥ 0] = π by (10.9.1) and
therefore Pr[q (y j ) = 0] = 1 − π
The expected value is E[q (y j )] = (1 − π ) · 0 + π · 1 = π . • e. 2 points Given q we can apply the LehmannScheﬀ´ theorem: E[q (y j )y ] is
e
¯
the best unbiased estimator of π . We will compute E[q (y j )y ] in four steps which build
¯
on each other. First step: since for every indicator function follows E[q (y j )y ] =
¯
Pr[y j ≥ 0y ], we need for every given value y , the conditional distribution of y j
¯
¯
conditionally on y = y . (Not just the conditional mean but the whole conditional
¯
¯
distribution.) In order to construct this, we ﬁrst have to specify exactly the joint
distribution of y j and y :
¯
Answer. They are jointly normal:
(10.9.2) yj
y
¯ ∼N µ
1
,
µ
1/n 1/n
1/n • f . 2 points Second step: From this joint distribution derive the conditional
distribution of y j conditionally on y = y . (Not just the conditional mean but the
¯
¯
whole conditional distribution.) For this you will need formula (7.3.18) and (7.3.20).
Answer. Here are these two formulas: if u and v are jointly normal, then the conditional
distribution of v conditionally on u = u is Normal with mean
(10.9.3) E[v u = u] = E[v ] + cov[u, v ]
(u − E[u])
var[u] and variance
(10.9.4) var[v u = u] = var[v ] − (cov[u, v ])2
.
var[u] 10.9. SUFFICIENT STATISTICS AND ESTIMATION 139 Plugging u = y and v = y j into (7.3.18) and (7.3.20) gives: the conditional distribution of y j
¯
conditionally on y = y has mean
¯
¯
(10.9.5) E[y j y = y ] = E[y j ] +
¯
¯
=µ+ (10.9.6) y
cov[¯, y j ]
var[¯]
y (¯ − E[¯])
y
y 1/n
(¯ − µ) = y
y
¯
1/n and variance
(10.9.7) var[y j y = y ] = var[y j ] −
¯
¯ y
(cov[¯, y j ])2
var[¯]
y 1
(1/n)2
=1− .
=1−
1/n
n (10.9.8) y
Therefore the conditional distribution of y j conditional on y is N (¯, (n − 1)/n). How can this
¯
be motivated? if we know the actual arithmetic mean of the variables, then our best estimate is
that each variable is equal to this arithmetic mean. And this additional knowledge cuts down the
variance by 1/n. • g. 2 points The variance decomposition (6.6.6) gives a decomposition of var[y j ]:
give it here:
Answer.
(10.9.9) var[y j ] = var E[y j y ] + E var[y j y ]
¯
¯ (10.9.10) = var[¯] + E
y n−1
n = n−1
1
+
n
n • h. Compare the conditional with the unconditional distribution.
Answer. Conditional distribution does not depend on unknown parameters, and it has smaller
variance! • i. 2 points Third step: Compute the probability, conditionally on y = y , that
¯
¯
y j ≥ 0.
Answer. If x ∼ N (¯, (n − 1)/n) (I call it x here instead of y j since we use it not with its
y
familiar unconditional distribution N (µ, 1) but with a conditional distribution), then Pr[x ≥ 0] =
Pr[x − y ≥ −y ] = Pr[x − y ≤ y ] = Pr (x − y ) n/(n − 1) ≤ y n/(n − 1) = Φ(¯ n/(n − 1))
¯
¯
¯
¯
¯
¯
y
¯
¯
because (x − y ) n/(n − 1) ∼ N (0, 1) conditionally on y . Again we needed symmetry of the
distribution to ﬂip the sign. • j. 1 point Finally, put all the pieces together and write down E[q (y j )y ], the
¯
¯
e
conditional expectation of q (y j ) conditionally on y , which by the LehmannScheﬀ´
theorem is the minimum MSE unbiased estimator of π . The formula you should
come up with is
(10.9.11) π = Φ(¯
ˆ
y n/(n − 1)), where Φ is the standard normal cumulative distribution function.
Answer. The conditional expectation of q (y j ) conditionally on y = y is, by part d, simply
¯
¯
the probability that y j ≥ 0 under this conditional distribution. In part i this was computed as
Φ(¯
y n/(n − 1)). Therefore all we have to do is replace y by y to get the minimum MSE unbiased
¯
¯ estimator of π as Φ(¯
y n/(n − 1)). Remark: this particular example did not give any brand new estimators, but it can
rather be considered a proof that certain obvious estimators are unbiased and eﬃcient.
But often this same procedure gives new estimators which one would not have been
able to guess. Already when the variance is unknown, the above example becomes
quite a bit more complicated, see [Rao73, p. 322, example 2]. When the variables 140 10. ESTIMATION PRINCIPLES have an exponential distribution then this example (probability of early failure) is
discussed in [BD77, example 4.2.4 on pp. 124/5].
10.10. The Likelihood Principle
Consider two experiments whose likelihood functions depend on the same parameter vector θ . Suppose that for particular realizations of the data y 1 and y 2 ,
the respective likelihood functions are proportional to each other, i.e., 1 (θ ; y 1 ) =
α 2 (θ ; y 2 ) where α does not depend on θ although it may depend on y 1 and y 2 .
Then the likelihood principle states that identical conclusions should be drawn from
these two experiments about θ .
The likelihood principle is equivalent to the combination of two simpler principles: the weak suﬃciency principle, and the following principle, which seems very
plausible:
Weak Conditonality Principle: Given two possible experiments A and B . A
mixed experiment is one in which one throws a coin and performs A if the coin
shows head and B if it shows tails. The weak conditionality principle states: suppose it is known that the coin shows tails. Then the evidence of the mixed experiment
is equivalent to the evidence gained had one not thrown the coin but performed B
without the possible alternative of A. This principle says therefore that an experiment which one did not do but which one could have performed does not alter the
information gained from the experiment actually performed.
As an application of the likelihood principle look at the following situation:
Problem 182. 3 points You have a Bernoulli experiment with unknown parameter θ, 0 ≤ θ ≤ 1. Person A was originally planning to perform this experiment
12 times, which she does. She obtains 9 successes and 3 failures. Person B was
originally planning to perform the experiment until he has reached 9 successes, and
it took him 12 trials to do this. Should both experimenters draw identical conclusions
from these two experiments or not?
Answer. The probability mass function in the ﬁrst is by (3.7.1)
second it is by (4.1.13)
matter! 11
8 θ9 (1 − θ)3 . 12
9 θ9 (1 − θ)3 , and in the They are proportional, the stopping rule therefore does not 10.11. Bayesian Inference
Reallife estimation usually implies the choice between competing estimation
methods all of which have their advantages and disadvantages. Bayesian inference
removes some of this arbitrariness.
Bayesians claim that “any inferential or decision process that does not follow from
some likelihood function and some set of priors has objectively veriﬁable deﬁciencies”
[Cor69, p. 617]. The “prior information” used by Bayesians is a formalization of
the notion that the information about the parameter values never comes from the
experiment alone. The Bayesian approach to estimation forces the researcher to cast
his or her prior knowledge (and also the loss function for estimation errors) in a
mathematical form, because in this way, unambiguous mathematical prescriptions
can be derived as to how the information of an experiment should be evaluated.
To the objection that these are large information requirements which are often
not satisﬁed, one might answer that it is less important whether these assumptions
are actually the right ones. The formulation of prior density merely ensures that the
researcher proceeds from a coherent set of beliefs. 10.11. BAYESIAN INFERENCE 141 The mathematics which the Bayesians do is based on a “ﬁnal” instead of an “initial” criterion of precision. In other words, not an estimation procedure is evaluated
which will be good in hypothetical repetitions of the experiment in the average, but
one which is good for the given set of data and the given set of priors. Data which
could have been observed but were not observed are not taken into consideration.
Both Bayesians and nonBayesians deﬁne the probabilistic properties of an experiment by the density function (likelihood function) of the observations, which may
depend on one or several unknown parameters. The nonBayesian considers these
parameters ﬁxed but unknown, while the Bayesian considers the parameters random,
i.e., he symbolizes his prior information about the parameters by a prior probability
distribution.
An excellent example in which this prior probability distribution is discrete is
given in [Ame94, pp. 168–172]. In the more usual case that the prior distribution
has a density function, a Bayesian is working with the joint density function of the
parameter values and the data. Like all joint density function, it can be written
as the product of a marginal and conditional density. The marginal density of the
parameter value represents the beliefs the experimenter holds about the parameters
before the experiment (prior density), and the likelihood function of the experiment
is the conditional density of the data given the parameters. After the experiment has
been conducted, the experimenter’s belief about the parameter values is represented
by their conditional density given the data, called the posterior density.
Let y denote the observations, θ the unknown parameters, and f (y , θ ) their
joint density. Then
(10.11.1)
(10.11.2) f (y , θ ) = f (θ )f (y θ )
= f (y )f (θ y ). Therefore
(10.11.3) f (θ y ) = f (θ )f (y θ )
.
f (y ) In this formula, the value of f (y ) is irrelevant. It only depends on y but not on
θ , but y is ﬁxed, i.e., it is a constant. If one knows the posterior density function
of θ up to a constant, one knows it altogether, since the constant is determined by
the requirement that the area under the density function is 1. Therefore (10.11.3) is
usually written as (∝ means “proportional to”)
(10.11.4) f (θ y ) ∝ f (θ )f (y θ ); here the lefthand side contains the posterior density function of the parameter, the
righthand side the prior density function and the likelihood function representing the
probability distribution of the experimental data.
The Bayesian procedure does not yield a point estimate or an interval estimate,
but a whole probability distribution for the unknown parameters (which represents
our information about these parameters) containing the “prior” information “updated” by the information yielded by the sample outcome.
Of course, such probability distributions can be summarized by various measures
of location (mean, median), which can then be considered Bayesian point estimates.
Such summary measures for a whole probability distribution are rather arbitrary.
But if a loss function is given, then this process of distilling point estimates from
the posterior distribution can once more be systematized. For a concrete decision it
tells us that parameter value which minimizes the expected loss function under the 142 10. ESTIMATION PRINCIPLES posterior density function, the socalled “Bayes risk.” This can be considered the
Bayesian analog of a point estimate.
For instance, if the loss function is quadratic, then the posterior mean is the
parameter value which minimizes expected loss.
There is a diﬀerence between Bayes risk and the notion of risk we applied previously. The frequentist minimizes expected loss in a large number of repetitions of the
trial. This risk is dependent on the unknown parameters, and therefore usually no
estimators exist which give minimum risk in all situations. The Bayesian conditions
on the data (ﬁnal criterion!) and minimizes the expected loss where the expectation
is taken over the posterior density of the parameter vector.
The irreducibility of absence to presences: the absence of knowledge (or also
the absence of regularity itself) cannot be represented by a probability distribution.
Proof: if I give a certain random variable a neutral prior, then functions of this
random variable have nonneutral priors. This argument is made in [Roy97, p. 174].
Many good Bayesians drift away from the subjective point of view and talk about
a stratiﬁed world: their center of attention is no longer the world out there versus
our knowledge of it, but the empirical world versus the underlying systematic forces
that shape it.
Bayesians say that frequentists use subjective elements too; their outcomes depend on what the experimenter planned to do, even if he never did it. This again
comes from [Roy97, p. ??]. Nature does not know about the experimenter’s plans,
and any evidence should be evaluated in a way independent of this. CHAPTER 11 Interval Estimation
Look at our simplest example of an estimator, the sample mean of an independent
sample from a normally distributed variable. Since the population mean of a normal
variable is at the same time its median, the sample mean will in 50 percent of the
cases be larger than the population mean, and in 50 percent of the cases it will be
smaller. This is a statement about the procedure how the sample mean was obtained,
not about any given observed value of the sample mean. Say in one particular sample
the observed sample mean was 3.5. This number 3.5 is either larger or smaller than
the true mean, there is no probability involved. But if one were to compute sample
means of many diﬀerent independent samples, then these means would in 50% of the
cases lie above and in 50% of the cases below the population mean. This is why one
can, from knowing how this one given number was obtained, derive the “conﬁdence”
of 50% that the actual mean lies above 3.5, and the same with below. The sample
mean can therefore be considered a onesided conﬁdence bound, although one usually
wants higher conﬁdence levels than 50%. (I am 95% conﬁdent that φ is greater or
equal than a certain value computed from the sample.) The concept of “conﬁdence”
is nothing but the usual concept of probability if one uses an initial criterion of
precision.
The following thought experiment illustrates what is involved. Assume you
bought a widget and want to know whether it is defective or not. The obvious
way (which would correspond to a “ﬁnal” criterion of precision) would be to open
it up and look if it is defective or not. Now assume we cannot do it: there is no
way telling by just looking at it whether it will work. Then another strategy would
be to go by an “initial” criterion of precision: we visit the widget factory and look
how they make them, how much quality control there is and such. And if we ﬁnd
out that 95% of all widgets coming out of the same factory have no defects, then we
have the “conﬁdence” of 95% that our particular widget is not defective either.
The matter becomes only slightly more mystiﬁed if one talks about intervals.
Again, one should not forget that conﬁdence intervals are random intervals. Besides
conﬁdence intervals and onesided conﬁdence bounds one can, if one regards several
parameters simultaneously, also construct conﬁdence rectangles, ellipsoids and more
complicated shapes. Therefore we will deﬁne in all generality:
Let y be a random vector whose distribution depends on some vector of unknown
parameters φ ∈ Ω. A conﬁdence region is a prescription which assigns to every
possible value y of y a subset R(y ) ⊂ Ω of parameter space, so that the probability
that this subset covers the true value of φ is at least a given conﬁdence level 1 − α,
i.e.,
(11.0.5) Pr R(y ) φ0 φ = φ0 ≥ 1 − α for all φ0 ∈ Ω. The important thing to remember about this deﬁnition is that these regions R(y )
are random regions; every time one performs the experiment one obtains a diﬀerent
region.
143 144 11. INTERVAL ESTIMATION Now let us go to the speciﬁc case of constructing an interval estimate for the
parameter µ when we have n independent observations from a normally distributed
population ∼ N (µ, σ 2 ) in which neither µ nor σ 2 are known. The vector of observations is therefore distributed as y ∼ N (ιµ, σ 2 I ), where ιµ is the vector every
component of which is µ.
I will give you now what I consider to be the cleanest argument deriving the
socalled tinterval. It generalizes directly to the F test in linear regression. It is not
the same derivation which you will usually ﬁnd, and I will bring the usual derivation
below for comparison. Recall the observation made earlier, based on (9.1.1), that the
sample mean y is that number y = a which minimizes the sum of squared deviations
¯
¯
(yi − a)2 . (In other words, y is the “least squares estimate” in this situation.) This
¯
least squares principle also naturally leads to interval estimates for µ: we will say
that a lies in the interval for µ if and only if
(yi − a)2
≤c
(yi − y )2
¯ (11.0.6) for some number c ≥ 1. Of course, the value of c depends on the conﬁdence level,
but the beauty of this criterion here is that the value of c can be determined by the
conﬁdence level alone without knowledge of the true values of µ or σ 2 .
To show this, note ﬁrst that (11.0.6) is equivalent to
(11.0.7) (yi − a)2 − (yi − y )2
¯
≤c−1
(yi − y )2
¯ and then apply the identity (yi − a)2 = (yi − y )2 + n(¯ − a)2 to the numerator
¯
y
to get the following equivalent formulation of (11.0.6):
n(¯ − a)2
y
≤c−1
(yi − y )2
¯ (11.0.8) The conﬁdence level of this interval is the probability that the true µ lies in an
interval randomly generated using this principle. In other words, it is
(11.0.9) Pr n(¯ − µ)2
y
≤c−1
(y i − y )2
¯ Although for every known a, the probability that a lies in the conﬁdence interval
depends on the unknown µ and σ 2 , we will show now that the probability that the
unknown µ lies in the conﬁdence interval does not depend on any unknown parameters. First look at the distribution of the numerator: Since y ∼ N (µ, σ 2 /n), it follows
¯
(¯ − µ)2 ∼ (σ 2 /n)χ2 . We also know the distribution of the denominator. Earlier we
y
1
have shown that the variable (y i − y )2 is a σ 2 χ2 −1 . It is not enough to know the
¯
n
distribution of numerator and denominator separately, we also need their joint distribution. For this go back to our earlier discussion of variance estimation again; there
¯
¯
we also showed that y is independent of the vector y 1 − y · · · y n − y ; there¯
fore any function of y is also independent of any function of this vector, from which
¯
follows that numerator and denominator in our fraction are independent. Therefore
this fraction is distributed as an σ 2 χ2 over an independent σ 2 χ2 −1 , and since the
1
n
σ 2 ’s cancel out, this is the same as a χ2 over an independent χ2 −1 . In other words,
1
n
this distribution does not depend on any unknown parameters!
The deﬁnition of a F distribution with k and m degrees of freedom is the distribution of a ratio of a χ2 /k divided by a χ2 /m; therefore if we divide the sum of
m
k 11. INTERVAL ESTIMATION 145 squares in the numerator by n − 1 we get a F distribution with 1 and n − 1 d.f.:
(¯ − µ)2
y
∼ F 1,n−1
(y i − y )2
¯ (11.0.10) 11
n n−1 If one does not take the square in the numerator, i.e., works with y − µ instead of
¯
y
(¯ − µ)2 , and takes square root in the denominator, one obtains a tdistribution:
y−µ
¯
(11.0.11)
∼ tn−1
1
1
(y i − y )2
¯
n
n−1
The left hand side of this last formula has a suggestive form. It can be written as
(¯ − µ)/sy , where sy is an estimate of the standard deviation of y (it is the square
y
¯
¯
¯
root of the unbiased estimate of the variance of y ). In other words, this tstatistic
¯
can be considered an estimate of the number of standard deviations the observed
value of y is away from µ.
¯
Now we will give, as promised, the usual derivation of the tconﬁdence intervals,
which is based on this interpretation. This usual derivation involves the following
two steps:
(1) First assume that σ 2 is known. Then it is obvious what to do; for every
observation y of y construct the following interval:
R(y ) = {u ∈ R : u − y  ≤ N(α/2) σy }.
¯
¯ (11.0.12) What do these symbols mean? The interval R (as in region) has y as an argument,
i.e.. it is denoted R(y ), because it depends on the observed value y . R is the set of real
numbers. N(α/2) is the upper α/2quantile of the Normal distribution, i.e., it is that
number c for which a standard Normal random variable z satisﬁes Pr[z ≥ c] = α/2.
Since by the symmetry of the Normal distribution, Pr[z ≤ −c] = α/2 as well, one
obtains for a twosided test:
Pr[z  ≥ N(α/2) ] = α. (11.0.13) From this follows the coverage probability:
(11.0.14)
(11.0.15) Pr[R(y ) µ] = Pr[µ − y  ≤ N(α/2) σy ]
¯
¯
= Pr[(µ − y )/σy  ≤ N(α/2) ] = Pr[−z  ≤ N(α/2) ] = 1 − α
¯¯ since z = (¯ − µ)/σy is a standard Normal. I.e., R(y ) is a conﬁdence interval for µ
y
¯
with conﬁdence level 1 − α.
(2) Second part: what if σ 2 is not known? Here a seemingly adhoc way out
would be to replace σ 2 by its unbiased estimate s2 . Of course, then the Normal
distribution no longer applies. However if one replaces the normal critical values by
those of the tn−1 distribution, one still gets, by miraculous coincidence, a conﬁdence
level which is independent of any unknown parameters.
Problem 183. If y i ∼ NID(µ, σ 2 ) (normally independently distributed) with µ
and σ 2 unknown, then the conﬁdence interval for µ has the form
(11.0.16) R(y ) = {u ∈ R : u − y  ≤ t(n−1;α/2) sy }.
¯
¯ Here t(n−q;α/2) is the upper α/2quantile of the t distribution with n − 1 degrees
of freedom, i.e., it is that number c for which a random variable t which has a t
distribution with n − 1 degrees of freedom satisﬁes Pr[t ≥ c] = α/2. And sy is
¯
obtained as follows: write down the standard deviation of y and replace σ by s. One
¯
s
can also say sy = σy σ where σy is an abbreviated notation for std. dev[y ] = var[y ].
¯
¯
¯
• a. 1 point Write down the formula for sy .
¯ 146 11. INTERVAL ESTIMATION Table 1. Percentiles of Student’s t Distribution. Table entry x
satisﬁes Pr[tn ≤ x] = p. n
1
2
3
4
5 .750
1.000
0.817
0.765
0.741
0.727 p=
.950
.975
.990
.995
6.314 12.706 31.821 63.657
2.920
4.303
6.965
9.925
2.354
3.182
4.541
5.841
2.132
2.776
3.747
4.604
2.015
2.571
3.365
4.032 .900
3.078
1.886
1.638
1.533
1.476 2
Answer. Start with σy = var[¯] =
y
¯ σ2
,
n √
therefore σy = σ/ n, and
¯ √
sy = s/ n =
¯ (11.0.17) (y i − y )2
¯
n(n − 1) • b. 2 points Compute the coverage probability of the interval (11.0.16).
Answer. The coverage probability is
(11.0.18) Pr[R(y ) µ] = Pr[ µ − y ≤ t(n−1;α/2) sy ]
¯
¯ (11.0.19)
(11.0.20)
(11.0.21)
(11.0.22) µ−y
¯
≤ t(n−1;α/2) ]
sy
¯
(µ − y )/σy
¯
¯
= Pr[
≤ t(n−1;α/2) ]
sy /σy
¯
¯
(y − µ)/σy
¯
¯
= Pr[
≤ t(n−1;α/2) ]
s/σ
= 1 − α,
= Pr[ because the expression in the numerator is a standard normal, and the expression in the denominator
is the square root of an independent χ2 −1 divided by n − 1. The random variable between the
n
absolute signs has therefore a tdistribution, and (11.0.22) follows from (30.4.8). • c. 2 points Four independent observations are available of a normal √
random
√
variable with unknown mean µ and variance σ 2 : the values are −2, − 2, + 2, and
+2. (These are not the kind of numbers you are usually reading oﬀ a measurement
instrument, but they make the calculation easy). Give a 95% conﬁdence interval for
µ. Table 1 gives the percentiles of the tdistribution.
Answer. In our situation
(11.0.23) ¯
x−µ
√ ∼ t3
s/ n According to table 1, for b = 3.182 follows
(11.0.24) Pr[t3 ≤ b] = 0.975 therefore
(11.0.25) Pr[t3 > b] = 0.025 and by symmetry of the tdistribution
(11.0.26) Pr[t3 < −b] = 0.025 Now subtract (11.0.26) from (11.0.24) to get
(11.0.27) Pr[−b ≤ t3 ≤ b] = 0.95 11. INTERVAL ESTIMATION 147 or
Pr[t3  ≤ b] = 0.95 (11.0.28)
or, plugging in the formula for t3 , ¯
x−µ
≤ b = .95
√
s/ n
√
Pr[x − µ ≤ bs/ n] = .95
¯
(11.0.30)
√
√
(11.0.31)
¯
Pr[−bs/ n ≤ µ − x ≤ bs/ n] = .95
√
√
Pr[¯ − bs/ n ≤ µ ≤ x + bs/ n] = .95
(11.0.32)
x
¯
√
√
the conﬁdence interval is therefore x − bs/ n, x + bs/ n . In our sample, x = 0, s2 = 12 = 4,
¯
¯
¯
3
2 /n = 1, therefore also s/√n = 1. So the sample value of the conﬁdence interval
n = 4, therefore s
is [−3.182, +3.182].
(11.0.29) Pr Problem 184. Using R, construct 20 samples of 12 observation each from a
N (0, 1) distribution, construct the 95% conﬁdence tintervals for the mean based on
these 20 samples, plot these intervals, and count how many intervals contain the true
mean.
Here are the commands: stdnorms<matrix(rnorm(240),nrow=12,ncol=20 gives
a 12 × 20 matrix containing 240 independent random normals. You get the vector
containing the midpoints of the conﬁdence intervals by the assignment midpts <apply(stdnorms,2,mean). About apply see [BCW96, p. 130]. The vector containing the half width of each conﬁdence interval can be obtained by another use of apply:
halfwidth < (qt(0.975,11)/sqrt(12)) * sqrt(apply(stdnorms,2,var)); To
print the values on the screen you may simply issue the command cbind(midptshalfwidth,midpts+halfwidth).
But it is much better to plot them. Since such a plot does not have one of the usual
formats, we have to put it together with some lowlevel commands. See [BCW96,
page 325]. At the very minimum we need the following: frame() starts a new plot.
par(usr = c(1,20, range(c(midptshalfwidth,midpts+halfwidth)) sets a coordinate system which accommodates all intervals. The 20 conﬁdence intervals are
constructed by segments(1:20, midptshalfwidth, 1:20, midpts+halfwidth).
Finally, abline(0,0) adds a horizontal line, so that you can see how many intervals
contain the true mean.
The ecmet package has a function confint.segments which draws such plots
automatically. Choose how many observations in each experiment (the argument
n), and how many conﬁdence intervals (the argument rep), and the conﬁdence level
level (the default is here 95%), and then issue, e.g. the command confint.segments(n=50,rep=100,level=.9).
Here is the transcript of the function:
confint.segments < function(n, rep, level = 95/100)
{
stdnormals < matrix(rnorm(n * rep), nrow = n, ncol = rep)
midpts < apply(stdnormals, 2, mean)
halfwidth < qt(p=(1 + level)/2, df= n  1) * sqrt(1/n)* sqrt(apply(stdnormals, 2, var))
frame()
x < c(1:rep, 1:rep)
y < c(midpts + halfwidth, midpts  halfwidth)
par(usr = c(1, rep, range(y)))
segments(1:rep, midpts  halfwidth, 1:rep, midpts + halfwidth)
abline(0, 0)
invisible(cbind(x,y))
} 148 11. INTERVAL ESTIMATION This function draws the plot as a “side eﬀect,” but it also returns a matrix with
the coordinates of the endpoints of the plots (without printing them on the screen).
This matrix can be used as input for the identify function. If you do for instance
iddata<confint.segments(12,20) and then identify(iddata,labels=iddata[,2],
then the following happens: if you move the mouse cursor on the graph near one of
the endpoints of one of the intervals, and click the left button, then it will print on
the graph the coordinate of the bounday of this interval. Clicking any other button of
the mouse gets you out of the identify function. CHAPTER 12 Hypothesis Testing
Imagine you are a business person considering a major investment in order to
launch a new product. The sales prospects of this product are not known with
certainty. You have to rely on the outcome of n marketing surveys that measure
the demand for the product once it is oﬀered. If µ is the actual (unknown) rate of
return on the investment, each of these surveys here will be modeled as a random
variable, which has a Normal distribution with this mean µ and known variance 1.
Let y1 , y2 , . . . , yn be the observed survey results. How would you decide whether to
build the plant?
The intuitively reasonable thing to do is to go ahead with the investment if
the sample mean of the observations is greater than a given value c, and not to do
it otherwise. This is indeed an optimal decision rule, and we will discuss in what
respect it is, and how c should be picked.
Your decision can be the wrong decision in two diﬀerent ways: either you decide
to go ahead with the investment although there will be no demand for the product,
or you fail to invest although there would have been demand. There is no decision
rule which eliminates both errors at once; the ﬁrst error would be minimized by the
rule never to produce, and the second by the rule always to produce. In order to
determine the right tradeoﬀ between these errors, it is important to be aware of their
asymmetry. The error to go ahead with production although there is no demand has
potentially disastrous consequences (loss of a lot of money), while the other error
may cause you to miss a proﬁt opportunity, but there is no actual loss involved, and
presumably you can ﬁnd other opportunities to invest your money.
To express this asymmetry, the error with the potentially disastrous consequences
is called “error of type one,” and the other “error of type two.” The distinction
between type one and type two errors can also be made in other cases. Locking up
an innocent person is an error of type one, while letting a criminal go unpunished
is an error of type two; publishing a paper with false results is an error of type one,
while foregoing an opportunity to publish is an error of type two (at least this is
what it ought to be).
Such an asymmetric situation calls for an asymmetric decision rule. One needs
strict safeguards against committing an error of type one, and if there are several
decision rules which are equally safe with respect to errors of type one, then one will
select among those that decision rule which minimizes the error of type two.
Let us look here at decision rules of the form: make the investment if y > c.
¯
An error of type one occurs if the decision rule advises you to make the investment
while there is no demand for the product. This will be the case if y > c but µ ≤ 0.
¯
The probability of this error depends on the unknown parameter µ, but it is at most
α = Pr[¯ > c  µ = 0]. This maximum value of the type one error probability is called
y
the signiﬁcance level, and you, as the director of the ﬁrm, will have to decide on α
depending on how tolerable it is to lose money on this venture, which presumably
depends on the chances to lose money on alternative investments. It is a serious
149 150 12. HYPOTHESIS TESTING
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3
2
1
0
1
2
3 Figure 1. Eventually this Figure will show the Power function of
a onesided normal test, i.e., the probability of error of type one as
a function of µ; right now this is simply the cdf of a Standard Normal
shortcoming of the classical theory of hypothesis testing that it does not provide
good guidelines how α should be chosen, and how it should change with sample size.
Instead, there is the tradition to choose α to be either 5% or 1% or 0.1%. Given α,
a table of the cumulative standard normal distribution function allows you to ﬁnd
y
that c for which Pr[¯ > c  µ = 0] = α.
Problem 185. 2 points Assume each y i ∼ N (µ, 1), n = 400 and α = 0.05, and
y
diﬀerent y i are independent. Compute the value c which satisﬁes Pr[¯ > c  µ = 0] =
α. You shoule either look it up in a table and include a xerox copy of the table with
the entry circled and the complete bibliographic reference written on the xerox copy,
or do it on a computer, writing exactly which commands you used. In R, the function
qnorm does what you need, ﬁnd out about it by typing help(qnorm).
Answer. In the case n = 400, y has variance 1/400 and therefore standard deviation 1/20 =
¯
0.05. Therefore 20¯ is a standard normal: from Pr[¯ > c  µ = 0] = 0.05 follows Pr[20¯ > 20c  µ =
y
y
y
0] = 0.05. Therefore 20c = 1.645 can be looked up in a table, perhaps use [JHG+ 88, p. 986], the
row for ∞ d.f.
Let us do this in R. The p“quantile” of the distribution of the random variable y is deﬁned
as that value q for which Pr[y ≤ q ] = p. If y is normally distributed, this quantile is computed
by the Rfunction qnorm(p, mean=0, sd=1, lower.tail=TRUE). In the present case we need either
qnorm(p=10.05, mean=0, sd=0.05) or qnorm(p=0.05, mean=0, sd=0.05, lower.tail=FALSE) which
gives the value 0.08224268. Choosing a decision which makes a loss unlikely is not enough; your decision
must also give you a chance of success. E.g., the decision rule to build the plant if
−0.06 ≤ y ≤ −0.05 and not to build it otherwise is completely perverse, although
¯
the signiﬁcance level of this decision rule is approximately 4% (if n = 100). In other
words, the signiﬁcance level is not enough information for evaluating the performance
of the test. You also need the “power function,” which gives you the probability
with which the test advises you to make the “critical” decision, as a function of
the true parameter values. (Here the “critical” decision is that decision which might
potentially lead to an error of type one.) By the deﬁnition of the signiﬁcance level, the
power function does not exceed the signiﬁcance level for those parameter values for
which going ahead would lead to a type 1 error. But only those tests are “powerful”
whose power function is high for those parameter values for which it would be correct
to go ahead. In our case, the power function must be below 0.05 when µ ≤ 0, and
we want it as high as possible when µ > 0. Figure 1 shows the power function for
the decision rule to go ahead whenever y ≥ c, where c is chosen in such a way that
¯
the signiﬁcance level is 5%, for n = 100.
The hypothesis whose rejection, although it is true, constitutes an error of type
one, is called the null hypothesis, and its alternative the alternative hypothesis. (In the
examples the null hypotheses were: the return on the investment is zero or negative,
the defendant is innocent, or the results about which one wants to publish a research
paper are wrong.) The null hypothesis is therefore the hypothesis that nothing is 12.1. DUALITY BETWEEN SIGNIFICANCE TESTS AND CONFIDENCE REGIONS 151 the case. The test tests whether this hypothesis should be rejected, will safeguard
against the hypothesis one wants to reject but one is afraid to reject erroneously. If
you reject the null hypothesis, you don’t want to regret it.
Mathematically, every test can be identiﬁed with its null hypothesis, which is
a region in parameter space (often consisting of one point only), and its “critical
region,” which is the event that the test comes out in favor of the “critical decision,”
i.e., rejects the null hypothesis. The critical region is usually an event of the form
that the value of a certain random variable, the “test statistic,” is within a given
range, usually that it is too high. The power function of the test is the probability
of the critical region as a function of the unknown parameters, and the signiﬁcance
level is the maximum (or, if this maximum depends on unknown parameters, any
upper bound) of the power function over the null hypothesis.
Problem 186. Mr. Jones is on trial for counterfeiting Picasso paintings, and
you are an expert witness who has developed foolproof statistical signiﬁcance tests
for identifying the painter of a given painting.
• a. 2 points There are two ways you can set up your test.
a: You can either say: The null hypothesis is that the painting was done by
Picasso, and the alternative hypothesis that it was done by Mr. Jones.
b: Alternatively, you might say: The null hypothesis is that the painting was
done by Mr. Jones, and the alternative hypothesis that it was done by Picasso.
Does it matter which way you do the test, and if so, which way is the correct one.
Give a reason to your answer, i.e., say what would be the consequences of testing in
the incorrect way.
Answer. The determination of what the null and what the alternative hypothesis is depends
on what is considered to be the catastrophic error which is to be guarded against. On a trial, Mr.
Jones is considered innocent until proven guilty. Mr. Jones should not be convicted unless he can be
proven guilty beyond “reasonable doubt.” Therefore the test must be set up in such a way that the
hypothesis that the painting is by Picasso will only be rejected if the chance that it is actually by
Picasso is very small. The error of type one is that the painting is considered counterfeited although
it is really by Picasso. Since the error of type one is always the error to reject the null hypothesis
although it is true, solution a. is the correct one. You are not proving, you are testing. • b. 2 points After the trial a customer calls you who is in the process of acquiring
a very expensive alleged Picasso painting, and who wants to be sure that this painting
is not one of Jones’s falsiﬁcations. Would you now set up your test in the same way
as in the trial or in the opposite way?
Answer. It is worse to spend money on a counterfeit painting than to forego purchasing a
true Picasso. Therefore the null hypothesis would be that the painting was done by Mr. Jones, i.e.,
it is the opposite way. 12.1. Duality between Signiﬁcance Tests and Conﬁdence Regions
There is a duality between conﬁdence regions with conﬁdence level 1 − α and
certain signiﬁcance tests. Let us look at a family of signiﬁcance tests, which all have
a signiﬁcance level ≤ α, and which deﬁne for every possible value of the parameter
φ0 ∈ Ω a critical region C (φ0 ) for rejecting the simple null hypothesis that the true
parameter is equal to φ0 . The condition that all signiﬁcance levels are ≤ α means
mathematically
(12.1.1) Pr C (φ0 )φ = φ0 ≤ α for all φ0 ∈ Ω. 152 12. HYPOTHESIS TESTING Mathematically, conﬁdence regions and such families of tests are one and the
same thing: if one has a conﬁdence region R(y ), one can deﬁne a test of the null
hypothesis φ = φ0 as follows: for an observed outcome y reject the null hypothesis
if and only if φ0 is not contained in R(y ). On the other hand, given a family of tests,
one can build a conﬁdence region by the prescription: R(y ) is the set of all those
parameter values which would not be rejected by a test based on observation y .
Problem 187. Show that with these deﬁnitions, equations (11.0.5) and (12.1.1)
are equivalent.
Answer. Since φ0 ∈ R(y ) iﬀ y ∈ C (φ0 ) (the complement of the critical region rejecting that
the parameter value is φ0 ), it follows Pr[R(y ) ∈ φ0 φ = φ0 ] = 1 − Pr[C (φ0 )φ = φ0 ] ≥ 1 − α. This duality is discussed in [BD77, pp. 177–182].
12.2. The Neyman Pearson Lemma and Likelihood Ratio Tests
Look one more time at the example with the fertilizer. Why are we considering
only regions of the form y ≥ µ0 , why not one of the form µ1 ≤ y ≤ µ2 , or maybe not
¯
¯
use the mean but decide to build if y 1 ≥ µ3 ? Here the µ1 , µ2 , and µ3 can be chosen
such that the probability of committing an error of type one is still α.
It seems intuitively clear that these alternative decision rules are not reasonable.
The Neyman Pearson lemma proves this intuition right. It says that the critical
regions of the form y ≥ µ0 are uniformly most powerful, in the sense that every
¯
other critical region with same probability of type one error has equal or higher
probability of committing error of type two, regardless of the true value of µ.
Here are formulation and proof of the Neyman Pearson lemma, ﬁrst for the
case that both null hypothesis and alternative hypothesis are simple: H0 : θ = θ0 ,
HA : θ = θ1 . In other words, we want to determine on the basis of the observations of
the random variables y 1 , . . . , y n whether the true θ was θ0 or θ1 , and a determination
θ = θ1 when in fact θ = θ0 is an error of type one. The critical region C is the set of
all outcomes that lead us to conclude that the parameter has value θ1 .
The Neyman Pearson lemma says that a uniformly most powerful test exists in
this situation. It is a socalled likelihoodratio test, which has the following critical
region:
(12.2.1) C = {y1 , . . . , yn : L(y1 , . . . , yn ; θ1 ) ≥ kL(y1 , . . . , yn ; θ0 )}. C consists of those outcomes for which θ1 is at least k times as likely as θ0 (where k
is chosen such that Pr[C θ0 ] = α).
To prove that this decision rule is uniformly most powerful, assume D is the critical region of a diﬀerent test with same signiﬁcance level α, i.e., if the null hypothesis
is correct, then C and D reject (and therefore commit an error of type one) with
equally low probabilities α. In formulas, Pr[C θ0 ] = Pr[Dθ0 ] = α. Look at ﬁgure 2
with C = U ∪ V and D = V ∪ W . Since C and D have the same signiﬁcance level,
it follows
(12.2.2) Pr[U θ0 ] = Pr[W θ0 ]. Also
(12.2.3) Pr[U θ1 ] ≥ k Pr[U θ0 ], 12.2. THE NEYMAN PEARSON LEMMA AND LIKELIHOOD RATIO TESTS 153 Figure 2. Venn Diagram for Proof of Neyman Pearson Lemma ec660.1005
since U ⊂ C and C were chosen such that the likelihood (density) function of the
alternative hypothesis is high relatively to that of the null hypothesis. Since W lies
outside C , the same argument gives
Pr[W θ1 ] ≤ k Pr[W θ0 ]. (12.2.4) Linking those two inequalities and the equality gives
(12.2.5) Pr[W θ1 ] ≤ k Pr[W θ0 ] = k Pr[U θ0 ] ≤ Pr[U θ1 ], hence Pr[Dθ1 ] ≤ Pr[C θ1 ]. In other words, if θ1 is the correct parameter value, then
C will discover this and reject at least as often as D. Therefore C is at least as
powerful as D, or the type two error probability of C is at least as small as that of
D.
Back to our fertilizer example. To make both null and alternative hypotheses
simple, assume that either µ = 0 (fertilizer is ineﬀective) or µ = t for some ﬁxed
t > 0. Then the likelihood ratio critical region has the form
(12.2.6)
C = {y1 , . . . , yn : 1
√
2π n 1 e− 2 ((y1 −t) 2 +···+(yn −t)2 ) 1
≥k √
2π n 1 2 2 e− 2 (y1 +···+yn ) } (12.2.7)
12
1
2
= {y1 , . . . , yn : − ((y1 − t)2 + · · · + (yn − t)2 ) ≥ ln k − (y1 + · · · + yn )}
2
2
(12.2.8)
t2 n
= {y1 , . . . , yn : t(y1 + · · · + yn ) −
≥ ln k }
2
(12.2.9)
t
ln k
+}
= {y1 , . . . , yn : y ≥
¯
nt
2
i.e., C has the form y ≥ some constant. The dependence of this constant on k is not
¯
relevant, since this constant is usually chosen such that the maximum probability of
error of type one is equal to the given signiﬁcance level.
Problem 188. 8 points You have four independent observations y1 , . . . , y4 from
an N (µ, 1), and you are testing the null hypothesis µ = 0 against the alternative
hypothesis µ = 1. For your test you are using the likelihood ratio test with critical
region
(12.2.10) C = {y1 , . . . , y4 : L(y1 , . . . , y4 ; µ = 1) ≥ 3.633 · L(y1 , . . . , y4 ; µ = 0)}. Compute the signiﬁcance level of this test. (According to the NeymanPearson
lemma, this is the uniformly most powerful test for this signiﬁcance level.) Hints: 154 12. HYPOTHESIS TESTING In order to show this you need to know that ln 3.633 = 1.29, everything else can be
done without a calculator. Along the way you may want to show that C can also be
written in the form C = {y1 , . . . , y4 : y1 + · · · + y4 ≥ 3.290}.
Answer. Here is the equation which determines when y1 , . . . , y4 lie in C :
(12.2.11)
(12.2.12)
(12.2.13) 1
12
2
(y1 − 1)2 + · · · + (y4 − 1)2 ≥ 3.633 · (2π )−2 exp −
y + · · · + y4
2
21
1
12
2
−
(y1 − 1)2 + · · · + (y4 − 1)2 ≥ ln(3.633) −
y + · · · + y4
2
21
y1 + · · · + y4 − 2 ≥ 1.290 (2π )−2 exp − Since Pr[y 1 + · · · + y 4 ≥ 3.290] = Pr[z = (y 1 + · · · + y 4 )/2 ≥ 1.645] and z is a standard normal, one
obtains the signiﬁcance level of 5% from the standard normal table or the ttable. Note that due to the properties of the Normal distribution, this critical region,
for a given signiﬁcance level, does not depend at all on the value of t. Therefore this
test is uniformly most powerful against the composite hypothesis µ > 0.
One can als write the null hypothesis as the composite hypothesis µ ≤ 0, because
the highest probability of type one error will still be attained when µ = 0. This
completes the proof that the test given in the original fertilizer example is uniformly
most powerful.
Most other distributions discussed here are equally well behaved, therefore uniformly most powerful onesided tests exist not only for the mean of a normal with
known variance, but also the variance of a normal with known mean, or the parameters of a Bernoulli and Poisson distribution.
However the given onesided hypothesis is the only situation in which a uniformly
most powerful test exists. In other situations, the generalized likelihood ratio test has
good properties even though it is no longer uniformly most powerful. Many known
tests (e.g., the F test) are generalized likelihood ratio tests.
Assume you want to test the composite null hypothesis H0 : θ ∈ ω , where ω is
a subset of the parameter space, against the alternative HA : θ ∈ Ω, where Ω ⊃ ω
is a more comprehensive subset of the parameter space. ω and Ω are deﬁned by
functions with continuous ﬁrstorder derivatives. The generalized likelihood ratio
critical region has the form
(12.2.14) C = {x1 , . . . , xn : supθ∈Ω L(x1 , . . . , xn ; θ)
≥ k}
supθ∈ω L(x1 , . . . , xn ; θ) where k is chosen such that the probability of the critical region when the null
hypothesis is true has as its maximum the desired signiﬁcance level. It can be shown
that twice the log of this quotient is asymptotically distributed as a χ2−s , where q
q
is the dimension of Ω and s the dimension of ω . (Sometimes the likelihood ratio
is deﬁned as the inverse of this ratio, but whenever possible we will deﬁne our test
statistics so that the null hypothjesis is rejected if the value of the test statistic is
too large.)
In order to perform a likelihood ratio test, the following steps are necessary:
First construct the MLE’s for θ ∈ Ω and θ ∈ ω , then take twice the diﬀerence of the
attained levels of the log likelihoodfunctions, and compare with the χ2 tables.
12.3. The Wald, Likelihood Ratio, and Lagrange Multiplier Tests
˜
Let us start with the generalized Wald test. Assume θ is an asymptotically
normal estimator of θ , whose asymptotic distribution is N (θ , Ψ). Assume furtherˆ
more that Ψ is a consistent estimate of Ψ. Then the following statistic is called the 12.3. WALD, LIKELIHOOD RATIO, LAGRANGE MULTIPLIER TESTS 155 generalized Wald statistic. It can be used for an asymtotic test of the hypothesis
h(θ ) = o, where h is a q vectorvalued diﬀerentiable function:
−1
∂h ˆ ∂h
˜
Ψ
h(θ )
˜
˜
∂θ θ ∂θ θ
Under the null hypothesis, this test statistic is asymptotically distributed as a χ2 . To
q
˜ , h(θ ) h(θ ) + ∂ h (θ − θ ). Taking
˜
˜
understand this, note that for all θ close to θ (12.3.1) ˜
G.W. = h(θ ) ∂θ covariances ˜
θ ∂h ˆ ∂h
Ψ
˜
˜
∂θ θ ∂θ θ
˜
˜
is an estimate of the covariance matrix of h(θ ). I.e., one takes h(θ ) twice and
“divides” it by its covariance matrix.
Now let us make more stringent assumptions. Assume the density fx (x; θ ) of
x depends on the parameter vector θ . We are assuming that the conditions are
ˆ
satisﬁed which ensure asymptotic normality of the maximum likelihood estimator θ
¯
and also of θ , the constrained maximum likelihood estimator subject to the constraint
h(θ ) = o.
There are three famous tests to test this hypothesis, which asymptotically are
all distributed like χ2 . The likehihoodratio test is
q
(12.3.2) (12.3.3) LRT = −2 log maxh(θ)=o fy (y ; θ )
¯
ˆ
= 2(log fy (y , θ ) − log fy (y , θ ))
maxθ fy (y ; θ ) It rejects if imposing the constraint reduces the attained level of the likelihood function too much.
The Wald test has the form
−1 ∂ h
−1
∂h
∂ 2 log f (y ; θ )
ˆ
ˆ
(12.3.4)
Wald = −h(θ )
h(θ )
ˆ
ˆ
ˆ
∂θ θ
θ
∂θ θ
∂ θ∂ θ
2 −1 log (
To understand this formula, note that − E ∂ ∂ θ∂fθ y;θ)
is the Cramer Rao
lower bound, and since all maximum likelihood estimators asymptotically attain the
ˆ
CRLB, it is the asymptotic covariance matrix of θ . If one does not take the expected
ˆ into these partial derivatives of the log likelihood function, one
value but plugs θ
gets a consistent estimate of the asymtotic covariance matrix. Therefore the Wald
test is a special case of the generalized Wald test.
Finally the score test has the form
−1 ∂ log f (y ; θ )
∂ log f (y ; θ )
∂ 2 log f (y ; θ )
¯
¯
¯
∂θ
θ
θ
θ
∂θ
∂ θ∂ θ
This test tests whether the score, i.e., the gradient of the unconstrained log likelihood
function, evaluated at the constrained maximum likelihood estimator, is too far away
from zero. To understand this formula, remember that we showed in the proof of
the CramerRao lower bound that the negative of the expected value of the Hessian
2
log (
− E ∂ ∂ θ∂fθ y;θ) is the covariance matrix of the score, i.e., here we take the score
twice and divide it by its estimated covariance matrix. (12.3.5) Score = − CHAPTER 13 General Principles of Econometric Modelling
[Gre97, 6.1 on p. 220] says: “An econometric study begins with a set of propositions about some aspect of the economy. The theory speciﬁes a set of precise,
deterministic relationships among variables. Familiar examples are demand equations, production functions, and macroeconomic models. The empirical investigation
provides estimates of unknown parameters in the model, such as elasticities or the
marginal propensity to consume, and usually attempts to measure the validity of the
theory against the behavior of the observable data.”
[Hen95, p. 6] distinguishes between two extremes: “‘Theorydriven’ approaches,
in which the model is derived from a priori theory and calibrated from data evidence.
They suﬀer from theory dependence in that their credibility depends on the credibility of the theory from which they arose—when that theory is discarded, so is the
associated evidence.” The other extreme is “‘Datadriven’ approaches, where models
are developed to closely describe the data . . . These suﬀer from sample dependence in
that accidental and transient data features are embodied as tightly in the model as
permanent aspects, so that extension of the data set often reveal predictive failure.”
Hendry proposes the following useful distinction of 4 levels of knowledge:
A Consider the situation where we know the complete structure of the process
which gernerates economic data and the values of all its parameters. This is the
equivalent of a probability theory course (example: rolling a perfect die), but involves
economic theory and econometric concepts.
B consider a known economic structure with unknown values of the parameters.
Equivalent to an estimation and inference course in statistics (example: independent
rolls of an imperfect die and estimating the probabilities of the diﬀerent faces) but
focusing on econometrically relevant aspects.
C is “the empirically relevant situation where neither the form of the datagenerating process nor its parameter values are known. (Here one does not know
whether the rolls of the die are independent, or whether the probabilities of the
diﬀerent faces remain constant.) Model discovery, evaluation, data mining, modelsearch procedures, and associated methodological issues.
D Forecasting the future when the data outcomes are unknown. (Model of money
demand under ﬁnancial innovation).
The example of Keynes’s consumption function in [Gre97, pp. 221/22] sounds at
the beginning as if it was close to B , but in the further discussion Greene goes more
and more over to C . It is remarkable here that economic theory usually does not yield
functional forms. Greene then says: the most common functional form is the linear
one c = α + β x with α > 0 and 0 < β < 1. He does not mention the aggregation
problem hidden in this. Then he says: “But the linear function is only approximate;
in fact, it is unlikely that consumption and income can be connected by any simple
relationship. The deterministic relationship is clearly inadequate.” Here Greene
uses a random relationship to model a relationship which is quantitatively “fuzzy.”
This is an interesting and relevant application of randomness.
157 158 13. GENERAL PRINCIPLES OF ECONOMETRIC MODELLING A sentence later Green backtracks from this insight and says: “We are not so
ambitious as to attempt to capture every inﬂuence in the relationship, but only those
that are substantial enough to model directly.” The “fuzziness” is not due to a lack
of ambition of the researcher, but the world is inherently quantiatively fuzzy. It is
not that we don’t know the law, but there is no law; not everything that happens in
an economy is driven by economic laws. Greene’s own example, in Figure 6.2, that
during the war years consumption was below the trend line, shows this.
Greene’s next example is the relationship between income and education. This
illustrates multiple instead of simple regression: one must also include age, and then
also the square of age, even if one is not interested in the eﬀect which age has, but
in order to “control” for this eﬀect, so that the eﬀects of education and age will not
be confounded.
Problem 189. Why should a regression of income on education include not only
age but also the square of age?
Answer. Because the eﬀect of age becomes smaller with increases in age. Critical Realist approaches are [Ron02] and [Mor02]. CHAPTER 14 MeanVariance Analysis in the Linear Model
In the present chapter, the only distributional assumptions are that means and
variances exist. (From this follows that also the covariances exist).
14.1. Three Versions of the Linear Model
As background reading please read [CD97, Chapter 1].
Following [JHG+ 88, Chapter 5], we will start with three diﬀerent linear statistical models. Model 1 is the simplest estimation problem already familiar from chapter
9, with n independent observations from the same distribution, call them y 1 , . . . , y n .
The only thing known about the distribution is that mean and variance exist, call
them µ and σ 2 . In order to write this as a special case of the “linear model,” deﬁne
εi = y i −µ, and deﬁne the vectors y = y 1 y 2 · · · y n , ε = ε1 ε2 · · · εn ,
and ι = 1 1 ··· 1 . Then one can write the model in the form
ε ∼ (o, σ 2 I ) y = ιµ + ε (14.1.1) The notation ε ∼ (o, σ 2 I ) is shorthand for E [ε ] = o (the null vector) and V [ε ] = σ 2 I
(σ 2 times the identity matrix, which has 1’s in the diagonal and 0’s elsewhere). µ is
the deterministic part of all the y i , and εi is the random part.
Model 2 is “simple regression” in which the deterministic part µ is not constant
but is a function of the nonrandom variable x. The assumption here is that this
function is diﬀerentiable and can, in the range of the variation of the data, be approximated by a linear function [Tin51, pp. 19–20]. I.e., each element of y is a
constant α plus a constant multiple of the corresponding element of the nonrandom
vector x plus a random error term: y t = α + xt β + εt , t = 1, . . . , n. This can be
written as x1
1
1 x1
y1
ε1
ε1
. . . . .
. α + . .
(14.1.2)
. . = . α + . β + . = .
.
.
.
.
.
.
.
β
xn
1
1 xn
yn
εn
εn
or
ε ∼ (o, σ 2 I ) y = Xβ + ε (14.1.3) Problem 190. 1 point Compute the matrix product 40
1 2 5
2 1
031
38
Answer.
1
0 2
3 5
1 4
2
3 0
1
8 = 1·4+2·2+5·3
0·4+3·2+1·3 159 1·0+2·1+5·8
23
=
0·0+3·1+1·8
9 42
11 160 14. MEANVARIANCE ANALYSIS IN THE LINEAR MODEL If the systematic part of y depends on more than one variable, then one needs
multiple regression, model 3. Mathematically, multiple regression has the same form
(14.1.3), but this time X is arbitrary (except for the restriction that all its columns
are linearly independent). Model 3 has Models 1 and 2 as special cases.
Multiple regression is also used to “correct for” disturbing inﬂuences. Let me
explain. A functional relationship, which makes the systematic part of y dependent
on some other variable x will usually only hold if other relevant inﬂuences are kept
constant. If those other inﬂuences vary, then they may aﬀect the form of this functional relation. For instance, the marginal propensity to consume may be aﬀected
by the interest rate, or the unemployment rate. This is why some econometricians
(Hendry) advocate that one should start with an “encompassing” model with many
explanatory variables and then narrow the speciﬁcation down by hypothesis tests.
Milton Friedman, by contrast, is very suspicious about multiple regressions, and
argues in [FS91, pp. 48/9] against the encompassing approach.
Friedman does not give a theoretical argument but argues by an example from
Chemistry. Perhaps one can say that the variations in the other inﬂuences may have
more serious implications than just modifying the form of the functional relation:
they may destroy this functional relation altogether, i.e., prevent any systematic or
predictable behavior.
observed unobserved
random
y
ε
nonrandom
X
β, σ2
14.2. Ordinary Least Squares
ˆ
In the model y = Xβ + ε , where ε ∼ (o, σ 2 I ), the OLSestimate β is deﬁned to
ˆ which minimizes
be that value β = β
(14.2.1) SSE = (y − Xβ ) (y − Xβ ) = y y − 2y X β + β X X β . Problem 156 shows that in model 1, this principle yields the arithmetic mean.
Problem 191. 2 points Prove that, if one predicts a random variable y by a
constant a, the constant which gives the best MSE is a = E[y ], and the best MSE one
can get is var[y ].
Answer. E[(y − a)2 ] = E[y 2 ] − 2a E[y ] + a2 . Diﬀerentiate with respect to a and set zero to
get a = E[y ]. One can also diﬀerentiate ﬁrst and then take expected value: E[2(y − a)] = 0. We will solve this minimization problem using the ﬁrstorder conditions in vector
notation. As a preparation, you should read the beginning of Appendix C about
matrix diﬀerentiation and the connection between matrix diﬀerentiation and the
Jacobian matrix of a vector function. All you need at this point is the two equations
(C.1.6) and (C.1.7). The chain rule (C.1.23) is enlightening but not strictly necessary
for the present derivation.
The matrix diﬀerentiation rules (C.1.6) and (C.1.7) allow us to diﬀerentiate
(14.2.1) to get
(14.2.2) ∂ SSE /∂ β = −2y X + 2β X X . Transpose it (because it is notationally simpler to have a relationship between column
ˆ
vectors), set it zero while at the same time replacing β by β , and divide by 2, to get
the “normal equation”
ˆ
(14.2.3)
X y = X X β. 14.2. ORDINARY LEAST SQUARES 161 Due to our assumption that all columns of X are linearly independent, X X has
an inverse and one can premultiply both sides of (14.2.3) by (X X )−1 :
ˆ
β = (X X )−1 X y .
(14.2.4)
If the columns of X are not linearly independent, then (14.2.3) has more than one
solution, and the normal equation is also in this case a necessary and suﬃcient
ˆ
condition for β to minimize the SSE (proof in Problem 194).
Problem 192. 4 points Using the matrix diﬀerentiation rules
(14.2.5)
(14.2.6) ∂ w x/∂ x = w
∂ x M x/∂ x = 2x M ˆ
for symmetric M , compute the leastsquares estimate β which minimizes
(14.2.7) SSE = (y − Xβ ) (y − Xβ ) You are allowed to assume that X X has an inverse.
Answer. First you have to multiply out
(14.2.8) (y − Xβ ) (y − Xβ ) = y y − 2y X β + β X X β . The matrix diﬀerentiation rules (14.2.5) and (14.2.6) allow us to diﬀerentiate (14.2.8) to get
(14.2.9) ∂ SSE /∂ β = −2y X + 2β X X . Transpose it (because it is notationally simpler to have a relationship between column vectors), set
ˆ
it zero while at the same time replacing β by β , and divide by 2, to get the “normal equation”
ˆ
(14.2.10)
X y = X X β.
Since X X has an inverse, one can premultiply both sides of (14.2.10) by (X X )−1 :
(14.2.11) ˆ
β = (X X )−1 X y . Problem 193. 2 points Show the following: if the columns of X are linearly
independent, then X X has an inverse. (X itself is not necessarily square.) In your
proof you may use the following criteria: the columns of X are linearly independent
(this is also called: X has full column rank) if and only if Xa = o implies a = o.
And a square matrix has an inverse if and only if its columns are linearly independent.
Answer. We have to show that any a which satisﬁes X X a = o is itself the null vector.
From X X a = o follows a X X a = 0 which can also be written X a 2 = 0. Therefore Xa = o,
and since the columns of X are linearly independent, this implies a = o. Problem 194. 3 points In this Problem we do not assume that X has full column
rank, it may be arbitrary.
• a. The normal equation (14.2.3) has always at least one solution. Hint: you
are allowed to use, without proof, equation (A.3.3) in the mathematical appendix.
ˆ
Answer. With this hint it is easy: β = (X X )− X y is a solution. ˆ
• b. If β satisﬁes the normal equation and β is an arbitrary vector, then
ˆ
ˆ
ˆ
(14.2.12) (y − Xβ ) (y − Xβ ) = (y − X β ) (y − X β ) + (β − β ) X X (β − β ).
Answer. This is true even if X has deﬁcient rank, and it will be shown here in this general
ˆ
ˆ
ˆ
ˆ
(y − X β ) − X (β − β ) ;
case. To prove (14.2.12), write (14.2.1) as SSE = (y − X β ) − X (β − β )
ˆ satisﬁes (14.2.3), the cross product terms disappear.
since β • c. Conclude from this that the normal equation is a necessary and suﬃcient
ˆ
condition characterizing the values β minimizing the sum of squared errors (14.2.12). 162 14. MEANVARIANCE ANALYSIS IN THE LINEAR MODEL Answer. (14.2.12) shows that the normal equations are suﬃcient. For necessity of the normal
ˆ
equations let β be an arbitrary solution of the normal equation, we have seen that there is always
ˆ
at least one. Given β , it follows from (14.2.12) that for any solution β ∗ of the minimization,
ˆ
ˆ
X X (β ∗ − β ) = o. Use (14.2.3) to replace (X X )β by X y to get X X β ∗ = X y . ˆˆ
It is customary to use the notation X β = y for the socalled ﬁtted values, which
ˆ
are the estimates of the vector of means η = Xβ . Geometrically, y is the orthogonal
projection of y on the space spanned by the columns of X . See Theorem A.6.1 about
projection matrices.
The vector of diﬀerences between the actual and the ﬁtted values is called the
ˆ
ˆ
vector of “residuals” ε = y − y . The residuals are “predictors” of the actual (but
unobserved) values of the disturbance vector ε . An estimator of a random magnitude
is usually called a “predictor,” but in the linear model estimation and prediction are
treated on the same footing, therefore it is not necessary to distinguish between the
two.
You should understand the diﬀerence between disturbances and residuals, and
between the two decompositions
ˆˆ
y = Xβ + ε = X β + ε
(14.2.13)
Problem 195. 2 points Assume that X has full column rank. Show that ε = M y
ˆ
where M = I − X (X X )−1 X . Show that M is symmetric and idempotent.
ˆ
Answer. By deﬁnition, ε = y − X β = y − X (X X )−1 X y = I − X (X X )−1 X y . Idemˆ
potent, i.e. M M = M :
(14.2.14)
M M = I − X (X X )−1 X I − X (X X )−1 X = I − X (X X )−1 X − X (X X )−1 X + X (X X )−1 X X (X X )−1 Problem 196. Assume X has full column rank. Deﬁne M = I −X (X X )−1 X .
• a. 1 point Show that the space M projects on is the space orthogonal to all
columns in X , i.e., M q = q if and only if X q = o.
Answer. X q = o clearly implies M q = q . Conversely, M q = q implies X (X X )−1 X q =
o. Premultiply this by X to get X q = o. • b. 1 point Show that a vector q lies in the range space of X , i.e., the space
spanned by the columns of X , if and only if M q = o. In other words, {q : q = Xa
for some a} = {q : M q = o}.
Answer. First assume M q = o. This means q = X (X X )−1 X q = Xa with a =
(X X )−1 X q . Conversely, if q = Xa then M q = M Xa = Oa = o. Problem 197. In 2dimensional space, write down the projection matrix on the
diagonal line y = x (call it E ), and compute E z for the three vectors a = [ 2 ],
1
b = [ 2 ], and c = [ 3 ]. Draw these vectors and their projections.
2
2
Assume we have a dependent variable y and two regressors x1 and x2 , each with
15 observations. Then one can visualize the data either as 15 points in 3dimensional
space (a 3dimensional scatter plot), or 3 points in 15dimensional space. In the
ﬁrst case, each point corresponds to an observation, in the second case, each point
corresponds to a variable. In this latter case the points are usually represented
as vectors. You only have 3 vectors, but each of these vectors is a vector in 15dimensional space. But you do not have to draw a 15dimensional space to draw
ˆ
these vectors; these 3 vectors span a 3dimensional subspace, and y is the projection
of the vector y on the space spanned by the two regressors not only in the original 14.2. ORDINARY LEAST SQUARES 163 15dimensional space, but already in this 3dimensional subspace. In other words,
[DM93, Figure 1.3] is valid in all dimensions! In the 15dimensional space, each
dimension represents one observation. In the 3dimensional subspace, this is no
longer true.
Problem 198. “Simple regression” is regression with an intercept and one explanatory variable only, i.e.,
(14.2.15) y t = α + βxt + εt Here X = ι x and β = α
ˆ
for β = α β :
ˆˆ β . Evaluate (14.2.4) to get the following formulas x2 y t − xt xt y t
t
n x2 − ( xt )2
t
n xt y t − xt y t
ˆ
β=
n x2 − ( xt )2
t (14.2.16) α=
ˆ (14.2.17)
Answer.
(14.2.18) X X= (14.2.19) ι
x ι X X −1 = (14.2.20) x= 1
x2 − (
t n X y= ιι
xι ιx
=
xx
x2
t
xt − xt ) 2 ιy
=
xy n
xt xt
x2
t − xt
n yt
xi y t Therefore (X X )−1 X y gives equations (14.2.16) and (14.2.17). Problem 199. Show that
n n (xt − x)(y t − y ) =
¯
¯ (14.2.21)
t=1 xt y t − nxy
¯¯
t=1 (Note, as explained in [DM93, pp. 27/8] or [Gre97, Section 5.4.1], that the left
hand side is computationally much more stable than the right.)
Answer. Simply multiply out. Problem 200. Show that (14.2.17) and (14.2.16) can also be written as follows:
(14.2.22)
(14.2.23) Answer. Using (xt − x)(y t − y )
¯
¯
(xt − x)2
¯
α = y − βx
ˆ ¯ ˆ¯
ˆ
β= xi = nx and
¯ y i = ny in (14.2.17), it can be written as
¯
ˆ
β= (14.2.24) xt y t − nxy
¯¯
x2 − nx2
¯
t Now apply Problem 199 to the numerator of (14.2.24), and Problem 199 with y = x to the denominator, to get (14.2.22).
To prove equation (14.2.23) for α, let us work backwards and plug (14.2.24) into the righthand
ˆ
side of (14.2.23):
(14.2.25) y − xβ =
¯ ¯ˆ y
¯ x2 − y nx2 − x
¯¯
¯
t
x2
t − xt y t + nxxy
¯ ¯¯
nx2
¯ 164 14. MEANVARIANCE ANALYSIS IN THE LINEAR MODEL The second and the fourth term in the numerator cancel out, and what remains can be shown to
be equal to (14.2.16). Problem 201. 3 points Show that in the simple regression model, the ﬁtted
regression line can be written in the form
y t = y + β (xt − x).
ˆ
¯
¯ˆ (14.2.26) From this follows in particular that the ﬁtted regression line always goes through the
point x, y .
¯¯
ˆ
β xt . Answer. Follows immediately if one plugs (14.2.23) into the deﬁning equation y t = α +
ˆ
ˆ Formulas (14.2.22) and (14.2.23) are interesting because they express the regression coeﬃcients in terms of the sample means and covariances. Problem 202 derives
the properties of the population equivalents of these formulas:
Problem 202. Given two random variables x and y with ﬁnite variances, and
var[x] > 0. You know the expected values, variances and covariance of x and y , and
you observe x, but y is unobserved. This question explores the properties of the Best
Linear Unbiased Predictor (BLUP) of y in this situation.
• a. 4 points Give a direct proof of the following, which is a special case of theorem
20.1.1: If you want to predict y by an aﬃne expression of the form a + bx, you will get
the lowest mean squared error MSE with b = cov[x, y ]/ var[x] and a = E[y ] − b E[x].
Answer. The MSE is variance plus squared bias (see e.g. problem 165), therefore
(14.2.27) MSE[a + bx; y ] = var[a + bx − y ] + (E[a + bx − y ])2 = var[bx − y ] + (a − E[y ] + b E[x])2 .
Therefore we choose a so that the second term is zero, and then you only have to minimize the ﬁrst
term with respect to b. Since
var[bx − y ] = b2 var[x] − 2b cov[x, y ] + var[y ] (14.2.28)
the ﬁrst order condition is 2b var[x] − 2 cov[x, y ] = 0 (14.2.29) ∂
∂a • b. 2 points For the ﬁrstorder conditions you needed the partial derivatives
∂
E[(y − a − bx)2 ] and ∂b E[(y − a − bx)2 ]. It is also possible, and probably shorter, to interchange taking expected value and partial derivative, i.e., to compute E
2 ∂
∂b (y a − bx) and E
alternative fashion.
Answer. E 2 − a − bx) ∂
(y − a − bx)2
∂a the formula for a. Now E ∂
(y
∂b − and set those zero. Do the above proof in this = −2 E[y − a − bx] = −2(E[y ] − a − b E[x]). Setting this zero gives − a − bx)2 = −2 E[x(y − a − bx)] = −2(E[xy ] − a E[x] − b E[x2 ]). Setting this zero gives E[xy ] − a E[x] − b E[x2 ] = 0. Plug in formula for a and solve for b:
(14.2.30) ∂
∂a (y b= E[xy ] − E[x] E[y ]
cov[x, y ]
=
.
E[x2 ] − (E[x])2
var[x] • c. 2 points Compute the MSE of this predictor. 14.2. ORDINARY LEAST SQUARES 165 Answer. If one plugs the optimal a into (14.2.27), this just annulls the last term of (14.2.27)
so that the MSE is given by (14.2.28). If one plugs the optimal b = cov[x, y ]/ var[x] into (14.2.28),
one gets
(14.2.31)
(14.2.32) MSE = cov[x, y ]
var[x] = var[y ] − 2 var[x] − 2 (cov[x, y ])
cov[x, y ] + var[x]
var[x] (cov[x, y ])2
.
var[x] • d. 2 points Show that the prediction error is uncorrelated with the observed x.
Answer.
(14.2.33) cov[x, y − a − bx] = cov[x, y ] − a cov[x, x] = 0 • e. 4 points If var[x] = 0, the quotient cov[x, y ]/ var[x] can no longer be formed,
but if you replace the inverse by the ginverse, so that the above formula becomes
(14.2.34) b = cov[x, y ](var[x])− then it always gives the minimum MSE predictor, whether or not var[x] = 0, and
regardless of which ginverse you use (in case there are more than one). To prove this,
you need to answer the following four questions: (a) what is the BLUP if var[x] = 0?
(b) what is the ginverse of a nonzero scalar? (c) what is the ginverse of the scalar
number 0? (d) if var[x] = 0, what do we know about cov[x, y ]?
Answer. (a) If var[x] = 0 then x = µ almost surely, therefore the observation of x does not
give us any new information. The BLUP of y is ν in this case, i.e., the above formula holds with
b = 0.
(b) The ginverse of a nonzero scalar is simply its inverse.
(c) Every scalar is a ginverse of the scalar 0.
(d) if var[x] = 0, then cov[x, y ] = 0.
Therefore pick a ginverse 0, an arbitrary number will do, call it c. Then formula (14.2.34)
says b = 0 · c = 0. Problem 203. 3 points Carefully state the speciﬁcations of the random variables
involved in the linear regression model. How does the model in Problem 202 diﬀer
from the linear regression model? What do they have in common?
Answer. In the regression model, you have several observations, in the other model only one.
In the regression model, the xi are nonrandom, only the y i are random, in the other model both
x and y are random. In the regression model, the expected value of the y i are not fully known,
in the other model the expected values of both x and y are fully known. Both models have in
common that the second moments are known only up to an unknown factor. Both models have in
common that only ﬁrst and second moments need to be known, and that they restrict themselves
to linear estimators, and that the criterion function is the MSE (the regression model minimaxes
it, but the other model minimizes it since there is no unknown parameter whose value one has to
minimax over. But this I cannot say right now, for this we need the GaussMarkov theorem. Also
the GaussMarkov is valid in both cases!) Problem 204. 2 points We are in the multiple regression model y = Xβ + ε
with intercept, i.e., X is such that there is a vector a with ι = Xa. Deﬁne the
1
¯
row vector x = n ι X , i.e., it has as its j th component the sample mean of the
ˆ
j th independent variable. Using the normal equations X y = X X β , show that
¯ˆ
y = x β (i.e., the regression plane goes through the center of gravity of all data
¯
points).
Answer. Premultiply the normal equation by a
1/n to get the result. ˆ
to get ι y − ι X β = 0. Premultiply by 166 14. MEANVARIANCE ANALYSIS IN THE LINEAR MODEL Problem 205. The ﬁtted values y and the residuals ε are “orthogonal” in two
ˆ
ˆ
diﬀerent ways.
• a. 2 points Show that the inner product y ε = 0. Why should you expect this
ˆˆ
from the geometric intuition of Least Squares?
Answer. Use ε = M y and y = (I − M )y : y ε = y (I − M )M y = 0 because M (I − M ) = O .
ˆ
ˆ
ˆˆ
This is a consequence of the more general result given in problem ??. • b. 2 points Sometimes two random variables are called “orthogonal” if their
covariance is zero. Show that y and ε are orthogonal also in this sense, i.e., show
ˆ
ˆ
that for every i and j , cov[y i , εj ] = 0. In matrix notation this can also be written
ˆˆ
ˆˆ
C [y , ε] = O .
Answer. C [y , ε] = C [(I − M )y , M y ] = (I − M ) V [y ]M = (I − M )(σ 2 I )M = σ 2 (I − M )M =
ˆˆ
O . This is a consequence of the more general result given in question 246. 14.3. The Coeﬃcient of Determination
Among the criteria which are often used to judge whether the model is appro¯
priate, we will look at the “coeﬃcient of determination” R2 , the “adjusted” R2 , and
later also at Mallow’s Cp statistic. Mallow’s Cp comes later because it is not a ﬁnal
but an initial criterion, i.e., it does not measure the ﬁt of the model to the given
data, but it estimates its MSE. Let us ﬁrst look at R2 .
A value of R2 always is based (explicitly or implicitly) on a comparison of two
models, usually nested in the sense that the model with fewer parameters can be
viewed as a specialization of the model with more parameters. The value of R2 is
then 1 minus the ratio of the smaller to the larger sum of squared residuals.
Thus, there is no such thing as the R2 from a single ﬁtted model—one must
always think about what model (perhaps an implicit “null” model) is held out as a
standard of comparison. Once that is determined, the calculation is straightforward,
based on the sums of squared residuals from the two models. This is particularly
appropriate for nls(), which minimizes a sum of squares.
The treatment which follows here is a little more complete than most. Some
textbooks, such as [DM93], never even give the leftmost term in formula (14.3.6)
according to which R2 is the sample correlation coeﬃcient. Other textbooks, such
that [JHG+ 88] and [Gre97], do give this formula, but it remains a surprise: there
is no explanation why the same quantity R2 can be expressed mathematically in
two quite diﬀerent ways, each of which has a diﬀerent interpretation. The present
treatment explains this.
ˆ
If the regression has a constant term, then the OLS estimate β has a third
optimality property (in addition to minimizing the SSE and being the BLUE): no
other linear combination of the explanatory variables has a higher squared sample
ˆ
correlation with y than y = X β .
ˆ
In the proof of this optimality property we will use the symmetric and idempotent
1
¯
projection matrix D = I − n ιι . Applied to any vector z , D gives Dz = z − ιz ,
which is z with the mean taken out. Taking out the mean is therefore a projection,
on the space orthogonal to ι. See Problem 161.
Problem 206. In the reggeom visualization, see Problem 293, in which x1 is
the vector of ones, which are the vectors Dx2 and D y ?
Answer. Dx2 is og , the dark blue line starting at the origin, and D y is cy , the red line
starting on x1 and going up to the peak. 14.3. THE COEFFICIENT OF DETERMINATION 167 As an additional mathematical tool we will need the CauchySchwartz inequality
for the vector product:
(u v )2 ≤ (u u)(v v ) (14.3.1) Problem 207. If Q is any nonnegative deﬁnite matrix, show that also
(u Qv )2 ≤ (u Qu)(v Qv ). (14.3.2) Answer. This follows from the fact that any nnd matrix Q can be written in the form Q =
R R. In order to prove that y has the highest squared sample correlation, take any
ˆ
˜
vector c and look at y = Xc. We will show that the sample correlation of y with
˜
y cannot be higher than that of y with y . For this let us ﬁrst compute the sample
ˆ
˜
covariance. By (9.3.17), n times the sample covariance between y and y is
(14.3.3) ˜
n times sample covariance(˜ , y ) = y D y = c X D (y + ε ).
y
ˆˆ ˆ
ˆ
ˆ
ˆ
By Problem 208, Dε = ε , hence X Dε = X ε = o (this last equality is
˜
equivalent to the Normal Equation (14.2.3)), therefore (14.3.3) becomes y D y =
˜
y D y . Together with (14.3.2) this gives
ˆ
(14.3.4) n times sample covariance(˜ , y )
y 2 ˜ˆ
= (˜ D y )2 ≤ (˜ D y )(y D y )
y
ˆ
y
ˆ In order to get from n2 times the squared sample covariance to the squared
sample correlation coeﬃcient we have to divide it by n2 times the sample variances
˜
of y and of y :
(14.3.5)
¯
y Dy
ˆ
ˆ
(yj − y )2
ˆ
ˆ
(yj − y )2
ˆ
¯
(˜ D y )2
y
2
≤
=
=
.
sample correlation(˜ , y ) =
y
y Dy
(yj − y )2
¯
(yj − y )2
¯
˜ )(y D y )
(˜ D y
y
For the rightmost equal sign in (14.3.5) we need Problem 209.
˜
If y = y , inequality (14.3.4) becomes an equality, and therefore also (14.3.5)
ˆ
becomes an equality throughout. This completes the proof that y has the highest
ˆ
possible squared sample correlation with y , and gives at the same time two diﬀerent
formulas for the same entity
(14.3.6) R2 = 2
¯
(yj − y )(yj − y )
ˆ
ˆ
¯
=
¯
(yj − y )2 (yj − y )2
ˆ
ˆ
¯ (yj − y )2
ˆ
¯
.
(yj − y )2
¯ ˆˆ
Problem 208. 1 point Show that, if X contains a constant term, then Dε = ε .
ˆ
ε = o, which is equivalent to the normal
You are allowed to use the fact that X
equation (14.2.3).
ˆ
Answer. Since X has a constant term, a vector a exists such that Xa = ι, therefore ι ε =
ˆ
ˆ
ˆˆ
a X ε = a o = 0. From ι ε = 0 follows Dε = ε . ¯¯
Problem 209. 1 point Show that, if X has a constant term, then y = y
ˆ
ˆ
Answer. Follows from 0 = ι ε = ι y − ι y . In the visualization, this is equivalent with the
ˆ
fact that both ocb and ocy are right angles. Problem 210. Instead of (14.3.6) one often sees the formula
2 (14.3.7) (yj − y )(yj − y )
ˆ
¯
¯
=
2
(yj − y )
ˆ
¯
(yj − y )2
¯ (yj − y )2
ˆ
¯
.
(yj − y )2
¯ Prove that they are equivalent. Which equation is better? 168 14. MEANVARIANCE ANALYSIS IN THE LINEAR MODEL The denominator in the righthand side expression of (14.3.6),
(yj − y )2 , is
¯
usually called “SST ,” the total (corrected) sum of squares. The numerator (yj −
ˆ
y )2 is usually called “SSR,” the sum of squares “explained” by the regression. In
¯
order to understand SSR better, we will show next the famous “Analysis of Variance”
identity SST = SSR + SSE .
Problem 211. In the reggeom visualization, again with x1 representing the
vector of ones, show that SST = SSR + SSE , and show that R2 = cos2 α where α
is the angle between two lines in this visualization. Which lines?
Answer. ε is the by , the green line going up to the peak, and SSE is the squared length of
ˆ
¯
¯
it. SST is the squared length of y − ιy . Sincer ιy is the projection of y on x1 , i.e., it is oc, the part
of x1 that is red, one sees that SST is the squared length of cy . SSR is the squared length of cb.
The analysis of variance identity follows because cby is a right angle. R2 = cos2 α where α is the
angle between bcy in this same triangle. Since the regression has a constant term, the decomposition
y = (y − y ) + (y − ιy ) + ιy
ˆ
ˆ
¯
¯ (14.3.8) is an orthogonal decomposition (all three vectors on the righthand side are orthogonal
to each other), therefore in particular
(y − y ) (y − ιy ) = 0.
ˆ
ˆ
¯ (14.3.9) Geometrically this follows from the fact that y − y is orthogonal to the column space
ˆ
of X , while y − ιy lies in that column space.
ˆ
¯
Problem 212. Show the decomposition 14.3.8 in the reggeomvisualization.
Answer. From y take the green line down to b, then the light blue line to c, then the red line
to the origin. This orthogonality can also be explained in terms of sequential projections: instead of projecting y on x1 directly I can ﬁrst project it on the plane spanned by x1
and x2 , and then project this projection on x1 .
From (14.3.9) follows (now the same identity written in three diﬀerent notations):
(14.3.10) (y − ιy ) (y − ιy ) = (y − y ) (y − y ) + (y − ιy ) (y − ιy )
¯
¯
ˆ
ˆ
ˆ
¯
ˆ
¯
(yt − y )2 =
¯ (14.3.11)
t (yt − yt )2 +
ˆ
t (14.3.12) (ˆt − y )2
y
¯
t SST = SSE + SSR Problem 213. 5 points Show that the “analysis of variance” identity SST =
SSE + SSR holds in a regression with intercept, i.e., prove one of the two following
equations:
(14.3.13) (y − ιy ) (y − ιy ) = (y − y ) (y − y ) + (y − ιy ) (y − ιy )
¯
¯
ˆ
ˆ
ˆ
¯
ˆ
¯
(yt − y )2 =
¯ (14.3.14)
t (yt − yt )2 +
ˆ
t (ˆt − y )2
y
¯
t Answer. Start with
(14.3.15) SST = (yt − y )2 =
¯ (yt − yt + yt − y ) 2
ˆ
ˆ
¯ ˆ1
ˆ
and then show that the cross product term
(yt −yt )(ˆt −y ) =
ˆy¯
εt (ˆt −y ) = ε (X β −ι n ι y ) = 0
ˆy ¯
ˆ
ˆ
ε X = o and in particular, since a constant term is included, ε ι = 0.
since 14.3. THE COEFFICIENT OF DETERMINATION 169 From the socalled “analysis of variance” identity (14.3.12), together with (14.3.6),
one obtains the following three alternative expressions for the maximum possible correlation, which is called R2 and which is routinely used as a measure of the “ﬁt” of
the regression:
(14.3.16) 2
¯
(yj − y )(yj − y )
ˆ
ˆ
¯
SSR
SST − SSE
=
=
¯
SST
SST
¯
(yj − y )2 (yj − y )2
ˆ
ˆ 2 R= The ﬁrst of these three expressions is the squared sample correlation coeﬃcient between y and y , hence the notation R2 . The usual interpretation of the middle
ˆ
expression is the following: SST can be decomposed into a part SSR which is “explained” by the regression, and a part SSE which remains “unexplained,” and R2
measures that fraction of SST which can be “explained” by the regression. [Gre97,
pp. 250–253] and also [JHG+ 88, pp. 211/212] try to make this notion plausible.
Instead of using the vague notions “explained” and “unexplained,” I prefer the following reading, which is based on the third expression for R2 in (14.3.16): ιy is the
¯
vector of ﬁtted values if one regresses y on a constant term only, and SST is the SSE
in this “restricted” regression. R2 measures therefore the proportionate reduction in
the SSE if one adds the nonconstant regressors to the regression. From this latter
formula one can also see that R2 = cos2 α where α is the angle between y − ιy and
¯
y − ιy .
ˆ
¯
Problem 214. Given two data series x and y . Show that the regression of y
on x has the same R2 as the regression of x on y . (Both regressions are assumed to
include a constant term.) Easy, but you have to think!
Answer. The symmetry comes from the fact that, in this particular case, R2 is the squared
sample correlation coeﬃcient between x and y . Proof: y is an aﬃne transformation of x, and
ˆ
correlation coeﬃcients are invariant under aﬃne transformations (compare Problem 216). Problem 215. This Problem derives some relationships which are valid in simple
regression yt = α + βxt + εt but their generalization to multiple regression is not
obvious.
• a. 2 points Show that
ˆ2
R2 = β (14.3.17) (xt − x)2
¯
(yt − y )2
¯ Hint: show ﬁrst that yt − y = β (xt − x).
ˆ
¯ˆ
¯
ˆ
Answer. From yt = α + β xt and y = α + β x follows yt − y = β (xt − x). Therefore
ˆ
ˆˆ
¯ ˆ ˆ¯
ˆ
¯
¯
(14.3.18) R2 = (yt − y )2
ˆ
¯
(yt − y )2
¯ 2 ˆ
=β ( xt − x) 2
¯
(yt − y )2
¯ • b. 2 points Furthermore show that R2 is the sample correlation coeﬃcient
between y and x, i.e.,
2 (14.3.19) R2 = (xt − x)(yt − y )
¯
¯
(xt − x)2
¯ Hint: you are allowed to use (14.2.22). (yt − y )2
¯ . 170 14. MEANVARIANCE ANALYSIS IN THE LINEAR MODEL Answer.
2 (14.3.20) 2 ˆ
R =β
2 (xt − x)2
¯
(yt − y ) 2
¯ ( xt − x) 2
¯ (xt − x)(yt − y )
¯
¯
= 2 ( xt − x) 2
¯ (yt − y )2
¯ which simpliﬁes to (14.3.19). ˆˆ
• c. 1 point Finally show that R2 = βxy βyx , i.e., it is the product of the two
slope coeﬃcients one gets if one regresses y on x and x on y .
If the regression does not have a constant term, but a vector a exists with
ι = Xa, then the above mathematics remains valid. If a does not exist, then
the identity SST = SSR + SSE no longer holds, and (14.3.16) is no longer valid.
The fraction SST −SSE can assume negative values. Also the sample correlation
SST
coeﬃcient between y and y loses its motivation, since there will usually be other
ˆ
linear combinations of the columns of X that have higher sample correlation with y
than the ﬁtted values y .
ˆ
Equation (14.3.16) is still puzzling at this point: why do two quite diﬀerent simple
concepts, the sample correlation and the proportionate reduction of the SSE , give
the same numerical result? To explain this, we will take a short digression about
correlation coeﬃcients, in which it will be shown that correlation coeﬃcients always
denote proportionate reductions in the MSE. Since the SSE is (up to a constant
factor) the sample equivalent of the MSE of the prediction of y by y , this shows
ˆ
that (14.3.16) is simply the sample equivalent of a general fact about correlation
coeﬃcients.
But ﬁrst let us take a brief look at the Adjusted R2 . 14.4. The Adjusted RSquare
The coeﬃcient of determination R2 is often used as a criterion for the selection
of regressors. There are several drawbacks to this. [KA69, Chapter 8] shows that
the distribution function of R2 depends on both the unknown error variance and the
values taken by the explanatory variables; therefore the R2 belonging to diﬀerent
regressions cannot be compared.
A further drawback is that inclusion of more regressors always increases the
¯
R2 . The adjusted R2 is designed to remedy this. Starting from the formula R2 =
1 − SSE /SST , the “adjustment” consists in dividing both SSE and SST by their
degrees of freedom:
(14.4.1) SSE /(n − k )
n−1
¯
R2 = 1 −
= 1 − (1 − R2 )
.
SST /(n − 1)
n−k For given SST , i.e., when one looks at alternative regressions with the same depen¯
dent variable, R2 is therefore a declining function of s2 , the unbiased estimator of
2
¯
σ . Choosing the regression with the highest R2 amounts therefore to selecting that
2
regression which yields the lowest value for s .
¯
R2 has the following interesting property: (which we note here only for reference,
because we have not yet discussed the F test:) Assume one adds i more regressors:
¯
then R2 increases only if the F statistic for these additional regressors has a value
greater than one. One can also say: s2 decreases only if F > 1. To see this, write 14.4. THE ADJUSTED RSQUARE 171 this F statistic as
(14.4.2)
(14.4.3) F= (SSE k − SSE k+i )/i
n − k − i S SE k
=
−1
SSE k+i /(n − k − i)
i
SSE k+i
(n − k )s2
n−k−i
k
=
−1
i
(n − k − i)s2 +i
k (14.4.4) = (n − k )s2
n−k
k
−
+1
is2 +i
i
k (14.4.5) = (n − k ) s2
k
−1 +1
i
s2 +i
k From this the statement follows.
¯
Minimizing the adjusted R2 is equivalent to minimizing the unbiased variance
2
estimator s ; it still does not penalize the loss of degrees of freedom heavily enough,
i.e., it still admits too many variables into the model.
Alternatives minimize Amemiya’s prediction criterion or Akaike’s information
criterion, which minimize functions of the estimated variances and n and k . Akaike’s
information criterion minimizes an estimate of the KullbackLeibler discrepancy,
which was discussed on p. 126. CHAPTER 15 Digression about Correlation Coeﬃcients
15.1. A Uniﬁed Deﬁnition of Correlation Coeﬃcients
Correlation coeﬃcients measure linear association. The usual deﬁnition of the
simple correlation coeﬃcient between two variables ρxy (sometimes we also use the
notation corr[x, y ]) is their standardized covariance
(15.1.1) ρxy = cov[x, y ] . var[x] var[y ] Because of CauchySchwartz, its value lies between −1 and 1.
Problem 216. Given the constant scalars a = 0 and c = 0 and b and d arbitrary.
Show that corr[x, y ] = ± corr[ax + b, cy + d], with the + sign being valid if a and c
have the same sign, and the − sign otherwise.
Answer. Start with cov[ax + b, cy + d] = ac cov[x, y ] and go from there. Besides the simple correlation coeﬃcient ρxy between two scalar variables y and
x, one can also deﬁne the squared multiple correlation coeﬃcient ρ2 (x) between one
y
scalar variable y and a whole vector of variables x, and the partial correlation coefﬁcient ρ12.x between two scalar variables y 1 and y 2 , with a vector of other variables
x “partialled out.” The multiple correlation coeﬃcient measures the strength of
a linear association between y and all components of x together, and the partial
correlation coeﬃcient measures the strength of that part of the linear association
between y 1 and y 2 which cannot be attributed to their joint association with x. One
can also deﬁne partial multiple correlation coeﬃcients. If one wants to measure the
linear association between two vectors, then one number is no longer enough, but
one needs several numbers, the “canonical correlations.”
The multiple or partial correlation coeﬃcients are usually deﬁned as simple correlation coeﬃcients involving the best linear predictor or its residual. But all these
correlation coeﬃcients share the property that they indicate a proportionate reduction in the MSE. See e.g. [Rao73, pp. 268–70]. Problem 217 makes this point for
the simple correlation coeﬃcient:
Problem 217. 4 points Show that the proportionate reduction in the MSE of
the best predictor of y , if one goes from predictors of the form y ∗ = a to predictors
of the form y ∗ = a + bx, is equal to the squared correlation coeﬃcient between y and
x. You are allowed to use the results of Problems 191 and 202. To set notation, call
the minimum MSE in the ﬁrst prediction (Problem 191) MSE[constant term; y ], and
the minimum MSE in the second prediction (Problem 202) MSE[constant term and
x; y ]. Show that
(15.1.2)
MSE[constant term; y ] − MSE[constant term and x; y ]
(cov[y , x])2
2
=
= ρy x .
MSE[constant term; y ]
var[y ] var[x]
173 174 15. DIGRESSION ABOUT CORRELATION COEFFICIENTS Answer. The minimum MSE with only a constant is var[y ] and (14.2.32) says that MSE[constant
term and x; y ] = var[y ] − (cov[x, y ])2 / var[x]. Therefore the diﬀerence in MSE’s is (cov[x, y ])2 / var[x],
and if one divides by var[y ] to get the relative diﬀerence, one gets exactly the squared correlation
coeﬃcient. Multiple Correlation Coeﬃcients. Now assume x is a vector while y remains a
scalar. Their joint mean vector and dispersion matrix are
(15.1.3) Ω
x
µ
∼
, σ 2 xx
ω xy
y
ν ω xy
.
ωy y By theorem ??, the best linear predictor of y based on x has the formula
(15.1.4) y ∗ = ν + ω xy Ω −x (x − µ)
x y ∗ has the following additional extremal value property: no linear combination b x
has a higher squared correlation with y than y ∗ . This maximal value of the squared
correlation is called the squared multiple correlation coeﬃcient
(15.1.5) ρ2 ( x ) =
y ω xy Ω −xω xy
x
ωy y The multiple correlation coeﬃcient itself is the positive square root, i.e., it is always
nonnegative, while some other correlation coeﬃcients may take on negative values.
The squared multiple correlation coeﬃcient can also deﬁned in terms of proportionate reduction in MSE. It is equal to the proportionate reduction in the MSE of
the best predictor of y if one goes from predictors of the form y ∗ = a to predictors
of the form y ∗ = a + b x, i.e.,
MSE[constant term; y ] − MSE[constant term and x; y ]
2
(15.1.6)
ρy ( x ) =
MSE[constant term; y ]
There are therefore two natural deﬁnitions of the multiple correlation coeﬃcient.
These two deﬁnitions correspond to the two formulas for R2 in (14.3.6).
Partial Correlation Coeﬃcients. Now assume y = y 1 y 2
is a vector with
two elements and write Ω xx ω y 1 ω y 2
x
µ
y 1 ∼ ν1 , σ 2 ω y1 ω11 ω12 .
(15.1.7)
y2
ν2
ω y2 ω21 ω22
Let y ∗ be the best linear predictor of y based on x. The partial correlation coeﬃcient
ρ12.x is deﬁned to be the simple correlation between the residuals corr[(y 1 − y ∗ ), (y 2 −
1
y ∗ )]. This measures the correlation between y 1 and y 2 which is “local,” i.e., which
2
does not follow from their association with x. Assume for instance that both y 1 and
y 2 are highly correlated with x. Then they will also have a high correlation with
each other. Subtracting y ∗ from y i eliminates this dependency on x, therefore any
i
remaining correlation is “local.” Compare [Krz88, p. 475].
The partial correlation coeﬃcient can be deﬁned as the relative reduction in the
MSE if one adds y 2 to x as a predictor of y 1 :
(15.1.8)
MSE[constant term and x; y 2 ] − MSE[constant term, x, and y 1 ; y 2 ]
.
ρ2 . x =
12
MSE[constant term and x; y 2 ]
Problem 218. Using the deﬁnitions in terms of MSE’s, show that the following
relationship holds between the squares of multiple and partial correlation coeﬃcients:
(15.1.9) 1 − ρ2 x,1) = (1 − ρ2 .x )(1 − ρ2 x) )
21
2(
2( 15.1. A UNIFIED DEFINITION OF CORRELATION COEFFICIENTS 175 Answer. In terms of the MSE, (15.1.9) reads
(15.1.10)
MSE[constant term, x, and y 1 ; y 2 ]
MSE[constant term, x, and y 1 ; y 2 ] MSE[constant term and x; y 2 ]
=
.
MSE[constant term; y 2 ]
MSE[constant term and x; y 2 ]
MSE[constant term; y 2 ] From (15.1.9) follows the following weighted average formula:
ρ2 x,1) = ρ2 x) + (1 − ρ2 x) )ρ2 .x
21
2(
2(
2( (15.1.11) An alternative proof of (15.1.11) is given in [Gra76, pp. 116/17].
Mixed cases: One can also form multiple correlations coeﬃcients with some of
the variables partialled out. The dot notation used here is due to Yule, [Yul07]. The
notation, deﬁnition, and formula for the squared correlation coeﬃcient is
(15.1.12)
ρ2 (x).z =
y
(15.1.13) = MSE[constant term and z ; y ] − MSE[constant term, z , and x; y ]
MSE[constant term and z ; y ]
ω xy.z Ω −x.z ω xy.z
x
ω y y .z CHAPTER 16 Speciﬁc Datasets
16.1. Cobb Douglas Aggregate Production Function
Problem 219. 2 points The CobbDouglas production function postulates the
following relationship between annual output q t and the inputs of labor t and capital
kt :
(16.1.1) γ
q t = µ β kt exp(εt ).
t q t , t , and kt are observed, and µ, β , γ , and the εt are to be estimated. By the
variable transformation xt = log q t , yt = log t , zt = log kt , and α = log µ, one
obtains the linear regression
(16.1.2) xt = α + βyt + γzt + εt Sometimes the following alternative variable transformation is made: ut = log(q t / t ),
vt = log(kt / t ), and the regression
(16.1.3) ut = α + γvt + εt is estimated. How are the regressions (16.1.2) and (16.1.3) related to each other?
Answer. Write (16.1.3) as
(16.1.4) xt − yt = α + γ (zt − yt ) + εt and collect terms to get
(16.1.5) xt = α + (1 − γ )yt + γzt + εt From this follows that running the regression (16.1.3) is equivalent to running the regression (16.1.2)
with the constraint β + γ = 1 imposed. The assumption here is that output is the only random variable. The regression
model is based on the assumption that the dependent variables have more noise in
them than the independent variables. One can justify this by the argument that
any noise in the independent variables will be transferred to the dependent variable,
and also that variables which aﬀect other variables have more steadiness in them
than variables which depend on others. This justiﬁcation often has merit, but in the
speciﬁc case, there is much more measurement error in the labor and capital inputs
than in the outputs. Therefore the assumption that only the output has an error
term is clearly wrong, and problem 221 below will look for possible alternatives.
Problem 220. Table 1 shows the data used by Cobb and Douglas in their original
article [CD28] introducing the production function which would bear their name.
output is “Day’s index of the physical volume of production (1899 = 100)” described
in [DP20], capital is the capital stock in manufacturing in millions of 1880 dollars
[CD28, p. 145], labor is the “probable average number of wage earners employed in
manufacturing” [CD28, p. 148], and wage is an index of the real wage (1899–1908
= 100).
177 178 16. SPECIFIC DATASETS year 1899 1900 1901 1902 1903 1904 1905 1906 1907 1908 1909 output 100 101 112 122 124 122 143 152 151 126 155 159 capital 4449 4746 5061 5444 5806 6132 6626 7234 7832 8229 8820 9240 labor 4713 4968 5184 5554 5784 5468 5906 6251 6483 5714 6615 6807 wage 99 98 101 102 100 99 103 101 99 94 102 104 1911 1912 1913 1914 1915 1916 1917 1918 1919 1920 1921 1922 year 1910 output 153 177 184 169 189 225 227 223 218 231 179 240 capital 9624 10067 10520 10873 11840 13242 14915 16265 17234 18118 18542 19192 labor 6855 7167 7277 7026 7269 8601 9218 9446 9096 9110 6947 7602 wage 97 99 100 99 99 104 103 107 111 114 115 119 Table 1. Cobb Douglas Original Data • a. A text ﬁle with the data is available on the web at www.econ.utah.edu/
ehrbar/data/cobbdoug.txt, and a SDML ﬁle (XML for statistical data which can be
read by R, Matlab, and perhaps also SPSS) is available at www.econ.utah.edu/ehrbar/
data/cobbdoug.sdml. Load these data into your favorite statistics package.
Answer. In R, you can simply issue the command cobbdoug < read.table("http://www.
econ.utah.edu/ehrbar/data/cobbdoug.txt", header=TRUE). If you run R on unix, you can also
do the following: download cobbdoug.sdml from the www, and then ﬁrst issue the command
library(StatDataML) and then readSDML("cobbdoug.sdml"). When I tried this last, the XML package necessary for StatDataML was not available on windows, but chances are it will be when you
read this.
In SAS, you must issue the commands
data cobbdoug;
infile ’cobbdoug.txt’;
input year output capital labor;
run;
But for this to work you must delete the ﬁrst line in the ﬁle cobbdoug.txt which contains the
variable names. (Is it possible to tell SAS to skip the ﬁrst line?) And you may have to tell SAS
the full pathname of the text ﬁle with the data. If you want a permanent instead of a temporary
dataset, give it a twopart name, such as ecmet.cobbdoug.
Here are the instructions for SPSS: 1) Begin SPSS with a blank spreadsheet. 2) Open up a ﬁle
with the following commands and run:
SET
BLANKS=SYSMIS
UNDEFINED=WARN.
DATA LIST
FILE=’A:\Cbbunst.dat’ FIXED RECORDS=1 TABLE /1 year 14 output 59 capital
1016 labor 1722 wage 2327 .
EXECUTE.
This ﬁles assume the data ﬁle to be on the same directory, and again the ﬁrst line in the data ﬁle
with the variable names must be deleted. Once the data are entered into SPSS the procedures
(regression, etc.) are best run from the point and click environment. • b. The next step is to look at the data. On [CD28, p. 150], Cobb and Douglas
plot capital, labor, and output on a logarithmic scale against time, all 3 series
normalized such that they start in 1899 at the same level =100. Reproduce this graph
using a modern statistics package. 16.1. COBB DOUGLAS AGGREGATE PRODUCTION FUNCTION 179 • c. Run both regressions (16.1.2) and (16.1.3) on Cobb and Douglas’s original
dataset. Compute 95% conﬁdence intervals for the coeﬃcients of capital and labor
in the unconstrained and the cconstrained models.
Answer. SAS does not allow you to transform the data on the ﬂy, it insists that you ﬁrst
go through a data step creating the transformed data, before you can run a regression on them.
Therefore the next set of commands creates a temporary dataset cdtmp. The data step data cdtmp
includes all the data from cobbdoug into cdtemp and then creates some transformed data as well.
Then one can run the regressions. Here are the commands; they are in the ﬁle cbbrgrss.sas in
your data disk:
data cdtmp;
set cobbdoug;
logcap = log(capital);
loglab = log(labor);
logout = log(output);
logcl = logcaploglab;
logol = logoutloglab;
run;
proc reg data = cdtmp;
model logout = logcap loglab;
run;
proc reg data = cdtmp;
model logol = logcl;
run;
Careful! In R, the command lm(log(output)log(labor) ~ log(capital)log(labor), data=cobbdoug)
does not give the right results. It does not complain but the result is wrong nevertheless. The right
way to write this command is lm(I(log(output)log(labor)) ~ I(log(capital)log(labor)), data=cobbdoug). • d. The regression results are graphically represented in Figure 1. The big
ellipse is a joint 95% conﬁdence region for β and γ . This ellipse is a level line of the
SSE . The vertical and horizontal bands represent univariate 95% conﬁdence regions
for β and γ separately. The diagonal line is the set of all β and γ with β + γ = 1,
representing the constraint of constant returns to scale. The small ellipse is that level
line of the SSE which is tangent to the constraint. The point of tangency represents
the constrained estimator. Reproduce this graph (or as much of this graph as you
can) using your statistics package.
Remark: In order to make the hand computations easier, Cobb and Douglass
reduced the data for capital and labor to index numbers (1899=100) which were
rounded to integers, before running the regressions, and Figure 1 was constructed
using these rounded data. Since you are using the nonstandardized data, you may
get slightly diﬀerent results.
Answer. lines(ellipse.lm(cbbfit, which=c(2, 3))) Problem 221. In this problem we will treat the CobbDouglas data as a dataset
with errors in all three variables. See chapter ?? and problem ?? about that.
• a. Run the three elementary regressions for the whole period, then choose at
least two subperiods and run it for those. Plot all regression coeﬃcients as points
in a plane, using diﬀerent colors for the diﬀerent subperiods (you have to normalize
them in a special way that they all ﬁt on the same plot).
Answer. Here are the results in R:
> outputlm<lm(log(output)~log(capital)+log(labor),data=cobbdoug)
> capitallm<lm(log(capital)~log(labor)+log(output),data=cobbdoug)
> laborlm<lm(log(labor)~log(output)+log(capital),data=cobbdoug) 180 16. SPECIFIC DATASETS 1.0
0.9 d
d 0.8
0.7
0.6
0.5 d
d
d
d
d
d 0.4 d
d .
........................
...........................
..
......
......
...
......
...
......
......
...
....
......
......
....
.....
....
.....
....
.....
.....
....
....
.....
.....
....
....
.....
.....
....
.. .
.........
....
..........
.... ...
.....
.....
.....
..... ...
.....
..... ....
....
....
.....
........
.....
......
....
.....
....
.....
.
.....
....
....
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.....
....
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.....
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...
...
......
......
...
...
......
......
......
..
........
.
.........
.. ...
.................
.................
. 0.3 dq
dq
d 0.2
0.1 d
d
d 0.0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5
Figure 1. Coeﬃcients of capital (vertical) and labor (horizontal), dependent variable output, unconstrained and constrained,
1899–1922
> coefficients(outputlm)
(Intercept) log(capital)
log(labor)
0.1773097
0.2330535
0.8072782
> coefficients(capitallm)
(Intercept) log(labor) log(output)
2.72052726 0.08695944 1.67579357
> coefficients(laborlm)
(Intercept) log(output) log(capital)
1.27424214
0.73812541 0.01105754
#Here is the information for the confidence ellipse:
> summary(outputlm,correlation=T)
Call:
lm(formula = log(output) ~ log(capital) + log(labor), data = cobbdoug)
Residuals:
Min
1Q
Median
0.075282 0.035234 0.006439 3Q
0.038782 Max
0.142114 Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 0.17731 0.43429
0.408 0.68721
log(capital) 0.23305 0.06353
3.668 0.00143 **
log(labor)
0.80728 0.14508
5.565 1.6e05 ***
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘’ Residual standard error: 0.05814 on 21 degrees of freedom
Multiple RSquared: 0.9574,Adjusted Rsquared: 0.9534
Fstatistic: 236.1 on 2 and 21 degrees of freedom,pvalue: 3.997e15
Correlation of Coefficients:
(Intercept) log(capital)
log(capital)
0.7243
log(labor)
0.9451
0.9096 1 16.1. COBB DOUGLAS AGGREGATE PRODUCTION FUNCTION 181 #Quantile of the Fdistribution:
> qf(p=0.95, df1=2, df2=21)
[1] 3.4668 • b. The elementary regressions will give you three ﬁtted equations of the form
ˆ
ˆ
(16.1.6)
output = α1 + β12 labor + β13 capital + residual1
ˆ
(16.1.7)
(16.1.8) ˆ
ˆ
labor = α2 + β21 output + β23 capital + residual2
ˆ
ˆ
ˆ
capital = α3 + β31 output + β32 labor + residual3 .
ˆ In order to compare the slope parameters in these regressions, ﬁrst rearrange them
in the form
ˆ
ˆ
(16.1.9)
−output + β12 labor + β13 capital + α1 + residual1 = 0
ˆ
(16.1.10) ˆ
ˆ
β21 output − labor + β23 capital + α2 + residual2 = 0
ˆ (16.1.11) ˆ
ˆ
β31 output + β32 labor − capital + α3 + residual3 = 0
ˆ This gives the following table of coeﬃcients:
output
labor
capital
intercept
−1
0.8072782
0.2330535 −0.1773097
0.73812541
−1 −0.01105754
1.27424214
1.67579357 −0.08695944
−1 −2.72052726
Now divide the second and third rows by the negative of their ﬁrst coeﬃcient, so that
the coeﬃcient of output becomes −1:
out
labor
capital
intercept
−1
0.8072782
0.2330535
−0.1773097
−1
1/0.73812541 0.01105754/0.73812541 −1.27424214/0.73812541
−1 0.08695944/1.67579357
1/1.67579357
2.72052726/1.67579357
After performing the divisions the following numbers are obtained:
output
labor
capital
intercept
−1 0.8072782
0.2330535 −0.1773097
−1 1.3547833 0.014980570 −1.726322
−1 0.05189149 0.59673221
1.6234262
These results can also be rewritten in the form given by Table 2.
Intercept Slope of output Slope of output
wrt labor
wrt capital Regression of output
on labor and capital
Regression of labor on
output and capital
Regression of capital
on output and labor
Table 2. Comparison of coeﬃcients in elementary regressions
Fill in the values for the whole period and also for several sample subperiods.
Make a scatter plot of the contents of this table, i.e., represent each regression result
as a point in a plane, using diﬀerent colors for diﬀerent sample periods. 182 16. SPECIFIC DATASETS T
d
d
d
d
d
d
q d
d capital d
d
d
d
d
d dqoutput no error, crs
d c Cobb Douglas’s original result
output
dq
d
d
d
d
d qlabor E Figure 2. Coeﬃcients of capital (vertical) and labor (horizontal), dependent variable output, 1899–1922
1.0
0.9 d
d 0.8 d
d 0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0 q d ..... capital all errors
.....
.....
.....
.....
.....
.....
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.....
......
.....
.....
.....
.....
.....
......
.....
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.....
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...... output no error, crs
.....
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.....output all errors
.....
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.....
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..... labor
.....
.....
.....
....
.. d
d d d
dq
d dq
d d
d
d q all errors 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5
Figure 3. Coeﬃcient of capital (vertical) and labor (horizontal)
in the elementary regressions, dependent variable output, 1899–1922
Problem 222. Given a univariate problem with three variables all of which have
zero mean, and a linear constraint that the coeﬃcients of all variables sum to 0. (This
is the model apparently appropriate to the CobbDouglas data, with the assumption
of constant returns to scale, after taking out the means.) Call the observed variables
x, y , and z , with underlying systematic variables x∗ , y ∗ , and z ∗ , and errors u, v ,
and w.
• a. Write this model in the form (??). 16.1. COBB DOUGLAS AGGREGATE PRODUCTION FUNCTION 183 Answer.
x∗ y∗ −1
β
1−β z∗ (16.1.12)
x y x∗ = β y ∗ + (1 − β )z ∗
x = x∗ + u =0
or z = x∗ y∗ z∗ + u v w y = y∗ + v
z = z ∗ + w. • b. The moment matrix of the systematic variables can be written fully in terms
2
2
of σy∗ , σz∗ , σy∗ z∗ , and the unknown parameter β . Write out the moment matrix and
therefore the Frisch decomposition.
Answer. The moment matrix is the middle matrix in the following Frisch decomposition:
2
σx
σxy
σxz (16.1.13) (16.1.14)
2
2
β 2 σy∗ + 2β (1 − β )σy∗ z∗ + (1 − β )2 σz∗
2 + (1 − β )σ ∗ ∗
βσy∗
=
yz
2
βσy∗ z∗ + (1 − β )σz∗ σxy
2
σy
σy z σxz
σy z
2
σz = 2
βσy∗ + (1 − β )σy∗ z∗
2
σy ∗
2
σy ∗ 2
2
βσy∗ z∗ + (1 − β )σz∗
σu
σy ∗ z ∗
+0
2
0
σz ∗ • c. Show that the unknown parameters are not yet identiﬁed. However, if one
2
2
2
makes the additional assumption that one of the three error variances σu , σv , or σw
is zero, then the equations are identiﬁed. Since the quantity of output presumably
2
has less error than the other two variables, assume σu = 0. Under this assumption,
show that
σ 2 − σxz
(16.1.15)
β= x
σxy − σxz
and this can be estimated by replacing the variances and covariances by their sample
counterparts. In a similar way, derive estimates of all other parameters of the model.
Answer. Solving (16.1.14) one gets from the y z element of the covariance matrix
(16.1.16) σy ∗ z ∗ = σy z and from the xz element
(16.1.17) 2
σz ∗ = σxz − βσyz
1−β Similarly, one gets from the xy element:
(16.1.18) 2
σy ∗ = σxy − (1 − β )σyz
β Now plug (16.1.16), (16.1.17), and (16.1.18) into the equation for the xx element:
(16.1.19)
(16.1.20) 2
2
σx = β (σxy − (1 − β )σyz ) + 2β (1 − β )σyz + (1 − β )(σxz − βσyz ) + σu
2
= βσxy + (1 − β )σxz + σu 2
Since we are assuming σu = 0 this last equation can be solved for β : (16.1.21) β= 2
σx − σxz
σxy − σxz If we replace the variances and covariances by the sample variances and covariances, this gives an
estimate of β . • d. Evaluate these formulas numerically. In order to get the sample means and
the sample covariance matrix of the data, you may issue the SAS commands 0
2
σv
0 0
0.
2
σw 184 16. SPECIFIC DATASETS proc corr cov nocorr data=cdtmp;
var logout loglab logcap;
run;
These commands are in the ﬁle cbbcovma.sas on the disk.
Answer. Mean vector and covariance matrix are
(16.1.22) LOGOUT
LOGLAB
LOGCAP 5.07734
0.0724870714
4.96272 , 0.0522115563
5.35648
0.1169330807 ∼ 0.0522115563
0.0404318579
0.0839798588 0.1169330807
0.0839798588
0.2108441826 Therefore equation (16.1.15) gives
0.0724870714 − 0.1169330807
ˆ
β=
= 0.686726861149148
0.0522115563 − 0.1169330807
ˆ
ˆ
In Figure 3, the point (β, 1 − β ) is exactly the intersection of the long dotted line with the constraint. (16.1.23) • e. The fact that all 3 points lie almost on the same line indicates that there may
be 2 linear relations: log labor is a certain coeﬃcient times log output, and log capital
is a diﬀerent coeﬃcient times log output. I.e., y ∗ = δ1 + γ1 x∗ and z ∗ = δ2 + γ2 x∗ .
In other words, there is no substitution. What would be the two coeﬃcients γ1 and
γ2 if this were the case?
Answer. Now the Frisch decomposition is
(16.1.24) 2
σx
σxy
σxz σxy
2
σy
σy z σxz
σy z
2
σz 2
= σx∗ 1
γ1
γ2 γ1
2
γ1
γ1 γ2 γ2
γ1 γ2
2
γ2 + 2
σu
0
0 0
2
σv
0 0
0.
2
σw Solving this gives (obtain γ1 by dividing the 32element by the 31element, γ2 by dividing the
2
32element by the 12element, σx∗ by dividing the 21element by γ1 , etc.
(16.1.25)
0.0839798588
σy z
=
= 0.7181873452513939
γ1 =
σxy
0.1169330807
σy z
0.0839798588
γ2 =
=
= 1.608453467992104
σxz
0.0522115563
σyx σxz
0.0522115563 · 0.1169330807
2
= 0.0726990758
σx∗ =
=
σy z
0.0839798588 σyx σxz
= 0.0724870714 − 0.0726990758 = −0.000212
σy z
σxy σyz
2
2
σv = σy −
σxz
σxz σzy
2
2
σw = σz −
σxy
2
2
σu = σx − This model is just barely rejected by the data since it leads to a slightly negative variance for U . • f . The assumption that there are two linear relations is represented as the
lightblue line in Figure 3. What is the equation of this line?
Answer. If y = γ1 x and z = γ2 x then the equation x = β1 y + β2 z holds whenever β1 γ1 +
β2 γ2 = 1. This is a straight line in the β1 , β2 plane, going through the points and (0, 1/γ2 ) =
0.0522115563
(0, 0.0839798588 = 0.6217152189353289) and (1/γ1 , 0) = ( 0.1169330807 = 1.3923943475361023, 0).
0.0839798588
This line is in the ﬁgure, and it is just a tiny bit on the wrong side of the dotted line connecting
the two estimates. 16.2. Houthakker’s Data
For this example we will use Berndt’s textbook [Ber91], which discusses some
of the classic studies in the econometric literature.
One example described there is the estimation of a demand function for electricity [Hou51], which is the ﬁrst multiple regression with several variables run on a
computer. In this exercise you are asked to do all steps in exercise 1 and 3 in chapter
7 of Berndt, and use the additional facilities of R to perform other steps of data
analysis which Berndt did not ask for, such as, for instance, explore the best subset
of regressors using leaps and the best nonlinear transformation using avas, do some 16.2. HOUTHAKKER’S DATA 185 diagnostics, search for outliers or inﬂuential observations, and check the normality
of residuals by a probability plot.
Problem 223. 4 points The electricity demand date from [Hou51] are available on the web at www.econ.utah.edu/ehrbar/data/ukelec.txt. Import these
data into your favorite statistics package. For R you need the command ukelec <read.table("http://www.econ.utah.edu/ehrbar/data/ukelec.txt"). Make a
scatterplot matrix of these data using e.g. pairs(ukelec) and describe what you
see.
Answer. inc and cap are negatively correlated. cap is capacity of rented equipment and not
equipment owned. Apparently customers with higher income buy their equipment instead of renting
it.
gas6 and gas8 are very highly correlated. mc4, mc6, and mc8 are less hightly correlated, the
corrlation between mc6 and mc8 is higher than that between mc4 and mc6. It seem electicity prices
have been coming down.
kwh, inc, and exp are strongly positively correlated.
the stripes in all the plots which have mc4, mc6, or mc8 in them come from the fact that the
marginal cost of electricity is a round number.
electricity prices and kwh are negatively correlated.
There is no obvious positive correlation between kwh and cap or expen and cap.
Prices of electricity and gas are somewhat positively correlated, but not much.
When looking at the correlations of inc with the other variables, there are several outliers which
could have a strong “leverage” eﬀect.
in 1934, those with high income had lower electricity prices than those with low income. This
eﬀect dissipated by 1938.
No strong negative correlations anywhere.
cust negatively correlated with inc, because rich people live in smaller cities? If you simply type ukelec in R, it will print the data on the screen. The variables
have the following meanings:
cust Average number of consumers with twopart tariﬀs for electricity in 1937–
38, in thousands. Twopart tariﬀ means: they pay a ﬁxed monthly sum plus a certain
“running charge” times the number of kilowatt hours they use.
inc Average income of twopart consumers, in pounds per year. (Note that one
pound had 240 pence at that time.)
mc4 The running charge (marginal cost) on domestic twopart tariﬀs in 1933–34,
in pence per KWH. (The marginal costs are the costs that depend on the number of
kilowatt hours only, it is the cost of one additional kilowatt hour.
mc6 The running charge (marginal cost) on domestic twopart tariﬀs in 1935–36,
in pence per KWH
mc8 The running charge (marginal cost) on domestic twopart tariﬀs in 1937–38,
in pence per KWH
gas6 The marginal price of gas in 1935–36, in pence per therm
gas8 The marginal price of gas in 1937–38, in pence per therm
kwh Consumption on domestic twopart tariﬀs per consumer in 1937–38, in kilowatt hours
cap The average holdings (capacity) of heavy electric equipment bought on hire
purchase (leased) by domestic twopart consumers in 1937–38, in kilowatts
expen The average total expenditure on electricity by twopart consumers in
1937–38, in pounds
The function summary(ukelec) displays summary statistics about every variable. 186 16. SPECIFIC DATASETS Since every data frame in R is a list, it is possible to access the variables in ukelec
by typing ukelec$mc4 etc. Try this; if you type this and then a return, you will get
a listing of mc4. In order to have all variables available as separate objects and save
typing ukelec$ all the time, one has to “mount” the data frame by the command
attach(ukelec). After this, the individual data series can simply be printed on the
screen by typing the name of the variable, for instance mc4, and then the return key.
Problem 224. 2 points Make boxplots of mc4, mc6, and mc6 in the same graph
next to each other, and the same with gas6 and gas8.
Problem 225. 2 points How would you answer the question whether marginal
prices of gas vary more or less than those of electricity (say in the year 1936)?
Answer. Marginal gas prices vary a little more than electricity prices, although electricity
was the newer technology, and although gas prices are much more stable over time than the electricity prices. Compare sqrt(var(mc6))/mean(mc6) with sqrt(var(gas6))/mean(gas6). You get
0.176 versus 0.203. Another way would be to compute max(mc6)/min(mc6) and compare with
max(gas6)/min(gas6): you get 2.27 versus 2.62. In any case this is a lot of variation. Problem 226. 2 points Make a plot of the (empirical) density function of mc6
and gas6 and interpret the results.
Problem 227. 2 points Is electricity a big share of total income? Which command is better: mean(expen/inc) or mean(expen)/mean(inc)? What other options
are there? Actually, there is a command which is clearly better than at least one of
the above, can you ﬁgure out what it is?
Answer. The proportion is small, less than 1 percent. The two above commands give 0.89%
and 0.84%. The command sum(cust*expen) / sum(cust*inc) is better than mean(expen) / mean(inc),
because each component in expen and inc is the mean over many households, the number of households given by cust. mean(expen) is therefore an average over averages over diﬀerent population sizes, not a good idea. sum(cust*expen) is total expenditure in all households involved, and
sum(cust*inc) is total income in all households involved. sum(cust*expen) / sum(cust*inc) gives
the value 0.92%. Another option is median(expen/inc) which gives 0.91%. A good way to answer
this question is to plot it: plot(expen,inc). You get the line where expenditure is 1 percent of
income by abline(0,0.01). For higher incomes expenditure for electricity levels oﬀ and becomes a
lower share of income. Problem 228. Have your computer compute the sample correlation matrix of
the data. The Rcommand is cor(ukelec)
• a. 4 points Are there surprises if one looks at the correlation matrix?
Answer. Electricity consumption kwh is slightly negatively correlated with gas prices and
with the capacity. If one takes the correlation matrix of the logarithmic data, one gets the expected
positive signs.
marginal prices of gas and electricity are positively correlated in the order of 0.3 to 0.45.
higher correlation between mc6 and mc8 than between mc4 and mc6.
Correlation between expen and cap is negative and low in both matrices, while one should
expect positive correlation. But in the logarithmic matrix, mc6 has negative correlation with expen,
i.e., elasticity of electricity demand is less than 1.
In the logarithmic data, cust has higher correlations than in the nonlogarithmic data, and it
is also more nearly normally distributed.
inc has negative correlation with mc4 but positive correlation with mc6 and mc8. (If one looks
at the scatterplot matrix this seems just random variations in an essentially zero correlation).
mc6 and expen are positively correlated, and so are mc8 and expen. This is due to the one
outlier with high expen and high income and also high electricity prices.
The marginal prices of electricity are not strongly correlated with expen, and in 1934, they are
negatively correlated with income.
From the scatter plot of kwh versus cap it seems there are two datapoints whose removal
might turn the sign around. To ﬁnd out which they are do plot(kwh,cap) and then use the identify 16.2. HOUTHAKKER’S DATA 187 function: identify(kwh,cap,labels=row.names(ukelec)). The two outlying datapoints are Halifax
and Wallase. Wallase has the highest income of all towns, namely, 1422, while Halifax’s income of
352 is close to the minimum, which is 279. High income customers do not lease their equipment
but buy it. • b. 3 points The correlation matrix says that kwh is negatively related with cap,
but the correlation of the logarithm gives the expected positive sign. Can you explain
this behavior?
Answer. If one plots the date using plot(cap,kwh) one sees that the negative correlation
comes from the two outliers. In a logarithmic scale, these two are no longer so strong outliers. Problem 229. Berndt on p. 338 deﬁnes the intramarginal expenditure f <expenmc8*kwh/240. What is this, and what do you ﬁnd out looking at it?
After this preliminary look at the data, let us run the regressions.
Problem 230. 6 points Write up the main results from the regressions which in
R are run by the commands
houth.olsfit < lm(formula = kwh ~ inc+I(1/mc6)+gas6+cap)
houth.glsfit < lm(kwh ~ inc+I(1/mc6)+gas6+cap, weight=cust)
houth.olsloglogfit < lm(log(kwh) ~
log(inc)+log(mc6)+log(gas6)+log(cap))
Instead of 1/mc6 you had to type I(1/mc6) because the slash has a special meaning
in formulas, creating a nested design, therefore it had to be “protected” by applying
the function I() to it.
If you then type houth.olsfit, a short summary of the regression results will be
displayed on the screen. There is also the command summary(houth.olsfit), which
gives you a more detailed summary. If you type plot(houth.olsfit) you will get a
series of graphics relevant for this regression.
Answer. All the expected signs.
Gas prices do not play a great role in determining electricity consumption, despite the “cookers” Berndt talks about on p. 337. Especially the logarithmic regression makes gas prices highly
insigniﬁcant!
The weighted estimation has a higher R2 . Problem 231. 2 points The output of the OLS regression gives as standard
error of inc the value of 0.18, while in the GLS regression it is 0.20. For the other
variables, the standard error as given in the GLS regression is lower than that in the
OLS regression. Does this mean that one should use for inc the OLS estimate and
for the other variables the GLS estimates?
Problem 232. 5 points Show, using the leaps procedure om R or some other
selection of regressors, that the variables Houthakker used in his GLSregression are
the “best” among the following: inc, mc4, mc6, mc8, gas6, gas8, cap using either the Cp statistic or the adjusted R2 . (At this stage, do not transform the variables
but just enter them into the regression untransformed, but do use the weights, which
are theoretically well justiﬁed).
To download the leaps package, use install.packages("leaps", lib="C:/Documents
and Settings/420lab.420LAB/My Documents") and to call it up, use library(leaps,
lib.loc="C:/Documents and Settings/420lab.420LAB/My Documents"). If the
library ecmet is available, the command ecmet.script(houthsel) runs the following script: 188 16. SPECIFIC DATASETS library(leaps)
data(ukelec)
attach(ukelec)
houth.glsleaps<leaps(x=cbind(inc,mc4,mc6,mc8,gas6,gas8,cap),
y=kwh, wt=cust, method="Cp",
nbest=5, strictly.compatible=F)
ecmet.prompt("Plot Mallow’s Cp against number of regressors:")
plot(houth.glsleaps$size, houth.glsleaps$Cp)
ecmet.prompt("Throw out all regressions with a Cp > 50 (big gap)")
plot(houth.glsleaps$size[houth.glsleaps$Cp<50],
houth.glsleaps$Cp[houth.glsleaps$Cp<50])
ecmet.prompt("Cp should be roughly equal the number of regressors")
abline(0,1)
cat("Does this mean the best regression is overfitted?")
ecmet.prompt("Click at the points to identify them, left click to quit")
## First construct the labels
lngth < dim(houth.glsleaps$which)[1]
included < as.list(1:lngth)
for (ii in 1:lngth)
included[[ii]] < paste(
colnames(houth.glsleaps$which)[houth.glsleaps$which[ii,]],
collapse=",")
identify(x=houth.glsleaps$size, y=houth.glsleaps$Cp, labels=included)
ecmet.prompt("Now use regsubsets instead of leaps")
houth.glsrss< regsubsets.default(x=cbind(inc,mc4,mc6,mc8,gas6,gas8,cap),
y=kwh, weights=cust, method="exhaustive")
print(summary.regsubsets(houth.glsrss))
plot.regsubsets(houth.glsrss, scale="Cp")
ecmet.prompt("Now order the variables")
houth.glsrsord< regsubsets.default(x=cbind(inc,mc6,cap,gas6,gas8,mc8,mc4),
y=kwh, weights=cust, method="exhaustive")
print(summary.regsubsets(houth.glsrsord))
plot.regsubsets(houth.glsrsord, scale="Cp")
Problem 233. Use avas to determine the “best” nonlinear transformations of
the explanatory and the response variable. Since the weights are theoretically well
justiﬁed, one should do it for the weighted regression. Which functions do you think
one should use for the diﬀerent regressors?
Problem 234. 3 points Then, as a check whether the transformation interferred
with data selection, redo leaps, but now with the transformed variables. Show that
the GLSregression Houthakker actually ran is the “best” regression among the following variables: inc, 1/mc4, 1/mc6, 1/mc8, gas6, gas8, cap using either the
Cp statistic or the adjusted R2 .
Problem 235. Diagnostics, the identiﬁcation of outliers or inﬂuential observations is something which we can do easily with R, although Berndt did not ask for it.
The command houth.glsinf<lm.influence(houth.glsfit) gives you the building blocks for many of the regression disgnostics statistics. Its output is a list if three
ˆ
objects: A matrix whose rows are all the the least squares estimates β (i) when the
ith observation is dropped, a vector with all the s(i), and a vector with all the hii .
A more extensive function is influence.measures(houth.glsfit), it has Cook’s
distance and others. 16.3. LONG TERM DATA ABOUT US ECONOMY 189 In order to look at the residuals, use the command plot(resid(houth.glsfit),
type="h") or plot(rstandard(houth.glsfit), type="h") or plot(rstudent(houth.glsfit),
type="h"). To add the axis do abline(0,0). If you wanted to check the residuals
for normality, you would use qqnorm(rstandard(houth.glsfit)).
Problem 236. Which commands do you need to plot the predictive residuals?
Problem 237. 4 points Although there is good theoretical justiﬁcation for using
cust as weights, one might wonder if the data bear this out. How can you check this?
Answer. Do plot(cust, rstandard(houth.olsfit)) and plot(cust, rstandard(houth.glsfit)).
In the ﬁrst plot, smaller numbers of customers have larger residuals, in the second plot this is mitigated. Also the OLS plot has two terrible outliers, which are brought more into range with GLS. Problem 238. The variable cap does not measure the capacity of all electrical
equipment owned by the households, but only those appliances which were leased from
the Electric Utility company. A plot shows that people with higher income do not
lease as much but presumably purchase their appliances outright. Does this mean the
variable should not be in the regression?
16.3. Long Term Data about US Economy
The dataset uslt is described in [DL91]. Home page of the authors is www.cepremap.cnrs.fr/~evy/.
l
uslt has the variables kn, kg (net and gross capital stock in current $), kn2, kg2 (the
same in 1982$), hours (hours worked), wage (hourly wage in current dollars), gnp,
gnp2, nnp, inv2 (investment in 1982 dollars), r (proﬁt rate (nnp − wage × hours)/kn),
u (capacity utilization), kne, kge, kne2, kge2, inve2 (capital stock and investment
data for equipment), kns, kgs, kns2, kgs2, invs2 (the same for structures).
Capital stock data were estimated separately for structures and equipment and
then added up, i.e., kn2 = kne2 + kns2 etc. Capital stock since 1925 has been
constructed from annual investment data, and prior to 1925 the authors of the series
apparently went the other direction: they took someone’s published capital stock
estimates and constructed investment from it. In the 1800s, only a few observations
were available, which were then interpolated. The capacity utilization ratio is equal
to the ratio of gnp2 to its trend, i.e., it may be negative.
Here are some possible commands for your Rsession: data(uslt) makes the data
available; uslt.clean<na.omit(uslt) removes missing values; this dataset starts
in 1869 (instead of 1805). attach(uslt.clean) makes the variables in this dataset
available. Now you can plot various series, for instance plot((nnphours*wage)/nnp,
type="l") plots the proﬁt share, or plot(gnp/gnp2, kg/kg2, type="l") gives you
a scatter plot of the price level for capital goods versus that for gnp. The command
plot(r, kn2/hours, type="b") gives both points and dots; type = "o" will have
the dots overlaid the line. After the plot you may issue the command identify(r,
kn2/hours, label=1869:1989) and then click with the left mouse button on the
plot those data points for which you want to have the years printed.
If you want more than one timeseries on the same plot, you may do matplot(1869:1989,
cbind(kn2,kns2), type="l"). If you want the yaxis logarithmic, say matplot(1869:1989,
cbind(gnp/gnp2,kns/kns2,kne/kne2), type="l", log="y").
Problem 239. Computer assignment: Make a number of such plots on the
screen, and import the most interesting ones into your wordprocessor. Each class
participant should write a short paper which shows the three most insteresting plots,
together with a written explanation why these plots seem interesting. 190 16. SPECIFIC DATASETS To use pairs or xgobi, you should carefully select the variables you want to include, and then you need the following preparations: usltsplom < cbind(gnp2=gnp2,
kn2=kn2, inv2=inv2, hours=hours, year=1869:1989) dimnames(usltsplom)[[1]]
< paste(1869:1989) The dimnames function adds the row labels to the matrix, so
that you can see which year it is. pairs(usltsplom) or library(xgobi) and then
xgobi(usltsplom)
You can also run regressions with commands of the following sort: lm.fit <lm(formula = gnp2 ~ hours + kne2 + kns2). You can also ﬁt a “generalized additive model” with the formula gam.fit < gam(formula = gnp2 ~ s(hours) +
s(kne2) + s(kns2)). This is related to the avas command we talked about in
class. It is discussed in [CH93].
16.4. Dougherty Data
We have a new dataset, in both SAS and Splus, namely the data described in
[Dou92].
There are more data than in the tables at the end of the book; prelcosm for
instance is the relative price of cosmetics, it is 100*pcosm/ptpe, but apparently
truncated at 5 digits.
16.5. Wage Data
The two datasets used in [Ber91, pp. 191–209] are available in R as the data
frames cps78 and cps85. In R on unix, the data can be downloaded by cps78
< readSDML("http://www.econ.utah.edu/ehrbar/data/cps78.sdml"), and the
corresponding for cps85. The original data provided by Berndt contain many dummy
variables. The data frames in R have the same data coded as “factor” variables
instead of dummies. These “factor” variables automatically generate dummies when
included in the model statement.
cps78 consists of 550 randomly selected employed workers from the May 1978
current population survey, and cps85 consists of 534 randomly selected employed
workers from the May 1985 current population survey. These are surveys of 50,000
households conducted monthly by the U.S. Department of Commerce. They serve
as the basis for the national employment and unemployment statistics. Data are
collected on a number of individual characteristics as well as employment status.
The present extracts were performed by Leslie Sundt of the University of Arizona.
ed = years of education
ex = years of labor market experience (= age − ed − 6, or 0 if this is a negative
number).
lnwage = natural logarithm of average hourly earnings
age = age in years
ndep = number of dependent children under 18 in household (only in cps78).
region has levels North, South
race has levels Other, Nonwhite, Hispanic. Nonwhite is mainly the Blacks, and
Other is mainly the NonHispanic Whites.
gender has levels Male, Female
marr has levels Single, Married
union has levels Nonunion, Union
industry has levels Other, Manuf, and Constr
occupation has levels Other, Manag, Sales, Cler, Serv, and Prof
Here is a log of my commands for exercises 1 and 2 in [Ber91, pp. 194–197].
> cps78 < readSDML("http://www.econ.utah.edu/ehrbar/data/cps78.sdml") 16.5. WAGE DATA 191 > attach(cps78)
> ###Exercise 1a (2 points) in chapter V of Berndt, p. 194
> #Here is the arithmetic mean of hourly wages:
> mean(exp(lnwage))
[1] 6.062766
> #Here is the geometric mean of hourly wages:
> #(Berndt’s instructions are apparently misformulated):
> exp(mean(lnwage))
[1] 5.370935
> #Geometric mean is lower than arithmetic, due to Jensen’s inequality
> #if the year has 2000 hours, this gives an annual wage of
> 2000*exp(mean(lnwage))
[1] 10741.87
> #What are arithmetic mean and standard deviation of years of schooling
> #and years of potential experience?
> mean(ed)
[1] 12.53636
> sqrt(var(ed))
[1] 2.772087
> mean(ex)
[1] 18.71818
> sqrt(var(ex))
[1] 13.34653
> #experience has much higher standard deviation than education, not surprising.
> ##Exercise 1b (1 point) can be answered with the two commands
> table(race)
Hisp Nonwh Other
36
57
457
> table(race, gender)
gender
race
Female Male
Hisp
12
24
Nonwh
28
29
Other
167 290
> #Berndt also asked for the sample means of certain dummy variables;
> #This has no interest in its own right but was an intermediate
> #step in order to compute the numbers of cases as above.
> ##Exercise 1c (2 points) can be answered using tapply
> tapply(ed,gender,mean)
Female
Male
12.76329 12.39942
> #now the standard deviation:
> sqrt(tapply(ed,gender,var))
Female
Male
2.220165 3.052312
> #Women do not have less education than men; it is about equal,
> #but their standard deviation is smaller
> #Now the geometric mean of the wage rate:
> exp(tapply(lnwage,gender,mean))
Female
Male
4.316358 6.128320 192 16. SPECIFIC DATASETS > #Now do the same with race
> ##Exercise 1d (4 points)
> detach()
> ##This used to be my old command:
> cps85 < read.table("~/dpkg/ecmet/usr/share/ecmet/usr/lib/R/library/ecmet/data/cps85.txt", h
> #But this should work for everyone (perhaps only on linux):
> cps85 < readSDML("http://www.econ.utah.edu/ehrbar/data/cps85.sdml")
> attach(cps85)
> mean(exp(lnwage))
[1] 9.023947
> sqrt(var(lnwage))
[1] 0.5277335
> exp(mean(lnwage))
[1] 7.83955
> 2000*exp(mean(lnwage))
[1] 15679.1
> 2000*exp(mean(lnwage))/1.649
[1] 9508.248
> #real wage has fallen
> tapply(exp(lnwage), gender, mean)
Female
Male
7.878743 9.994794
> tapply(exp(lnwage), gender, mean)/1.649
Female
Male
4.777891 6.061125
> #Compare that with 4.791237 6.830132 in 1979:
> #Male real wages dropped much more than female wages
> ##Exercise 1e (3 points)
> #using cps85
> w < mean(lnwage); w
[1] 2.059181
> s < sqrt(var(lnwage)); s
[1] 0.5277335
> lnwagef < factor(cut(lnwage, breaks = w+s*c(4,2,1,0,1,2,4)))
> table(lnwagef)
lnwagef
(0.0518,1]
(1,1.53] (1.53,2.06] (2.06,2.59] (2.59,3.11] (3.11,4.17]
3
93
174
180
72
12
> ks.test(lnwage, "pnorm")
Onesample KolmogorovSmirnov test
data: lnwage
D = 0.8754, pvalue = < 2.2e16
alternative hypothesis: two.sided
> ks.test(lnwage, "pnorm", mean=w, sd =s)
Onesample KolmogorovSmirnov test
data: lnwage 16.5. WAGE DATA 193 D = 0.0426, pvalue = 0.2879
alternative hypothesis: two.sided
>
>
>
>
> #Normal distribution not rejected
#If we do the same thing with
wage < exp(lnwage)
ks.test(wage, "pnorm", mean=mean(wage), sd =sqrt(var(wage))) Onesample KolmogorovSmirnov test
data: wage
D = 0.1235, pvalue = 1.668e07
alternative hypothesis: two.sided
> #This is highly significant, therefore normality rejected
>
>
>
>
>
>
>
> #An alternative, simpler way to answer question 1e is by using qqnorm
qqnorm(lnwage)
qqnorm(exp(lnwage))
#Note that the SAS proc univariate rejects that wage is normally distributed
#but does not reject that lnwage is normally distributed.
###Exercise 2a (3 points), p. 196
summary(lm(lnwage ~ ed, data = cps78)) Call:
lm(formula = lnwage ~ ed, data = cps78)
Residuals:
Min
1Q
Median
2.123168 0.331368 0.007296 3Q
0.319713 Max
1.594445 Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 1.030445
0.092704 11.115 < 2e16 ***
ed
0.051894
0.007221
7.187 2.18e12 ***
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘’ Residual standard error: 0.469 on 548 degrees of freedom
Multiple RSquared: 0.08613,Adjusted Rsquared: 0.08447
Fstatistic: 51.65 on 1 and 548 degrees of freedom,pvalue: 2.181e12
> #One year of education increases wages by 5 percent, but low R^2.
> #Mincer (5.18) had 7 percent for 1959
> #Now we need a 95 percent confidence interval for this coefficient
> 0.051894 + 0.007221*qt(0.975, 548)
[1] 0.06607823
> 0.051894  0.007221*qt(0.975, 548)
[1] 0.03770977 1 194 16. SPECIFIC DATASETS > ##Exercise 2b (3 points): Include union participation
> summary(lm(lnwage ~ union + ed, data=cps78))
Call:
lm(formula = lnwage ~ union + ed, data = cps78)
Residuals:
Min
1Q
2.331754 0.294114 Median
0.001475 3Q
0.263843 Max
1.678532 Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 0.859166
0.091630
9.376 < 2e16 ***
unionUnion 0.305129
0.041800
7.300 1.02e12 ***
ed
0.058122
0.006952
8.361 4.44e16 ***
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ Residual standard error: 0.4481 on 547 degrees of freedom
Multiple RSquared: 0.1673,Adjusted Rsquared: 0.1642
Fstatistic: 54.93 on 2 and 547 degrees of freedom,pvalue: 0.1 ‘’ 0 > exp(0.058)
[1] 1.059715
> exp(0.305129)
[1] 1.3568
> # Union members have 36 percent higher wages
> # The test whether union and nonunion members have the same intercept
> # is the same as the test whether the union dummy is 0.
> # tvalue = 7.300 which is highly significant,
> # i.e., they are different.
> #The union variable is labeled unionUnion, because
> #it is labeled 1 for Union and 0 for Nonun. Check with the command
> contrasts(cps78$union)
Union
Nonun
0
Union
1
> #One sees it also if one runs
> model.matrix(lnwage ~ union + ed, data=cps78)
(Intercept) union ed
1
1
0 12
2
1
1 12
3
1
16
4
1
1 12
5
1
0 12
> #etc, rest of output flushed
> #and compares this with
> cps78$union[1:5]
[1] Nonun Union Union Union Nonun
Levels: Nonun Union 1 16.5. WAGE DATA >
>
>
>
>
>
>
>
>
> 195 #Consequently, the intercept for nonunion is 0.8592
#and the intercept for union is 0.8592+0.3051=1.1643.
#Can I have a different set of dummies constructed from this factor?
#We will first do
##Exercise 2e (2 points)
contrasts(union)<matrix(c(1,0),nrow=2,ncol=1)
#This generates a new contrast matrix
#which covers up that in cps78
#Note that I do not say "data=cps78" in the next command:
summary(lm(lnwage ~ union + ed)) Call:
lm(formula = lnwage ~ union + ed)
Residuals:
Min
1Q
2.331754 0.294114 Median
0.001475 3Q
0.263843 Max
1.678532 Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 1.164295
0.090453 12.872 < 2e16
union1
0.305129
0.041800 7.300 1.02e12
ed
0.058122
0.006952
8.361 4.44e16
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ ***
***
***
0.05 ‘.’ Residual standard error: 0.4481 on 547 degrees of freedom
Multiple RSquared: 0.1673,Adjusted Rsquared: 0.1642
Fstatistic: 54.93 on 2 and 547 degrees of freedom,pvalue:
>
>
>
>
>
>
>
> 0.1 ‘’ 1 0 #Here the coefficients are different,
#but it is consistent with the above result.
##Ecercise 2c (2 points): If I want to have two contrasts from this one dummy, I have to do
contrasts(union,2)<matrix(c(1,0,0,1),nrow=2,ncol=2)
#The additional argument 2
#specifies different number of contrasts than it expects
#Now I have to supress the intercept in the regression
summary(lm(lnwage ~ union + ed  1)) Call:
lm(formula = lnwage ~ union + ed  1)
Residuals:
Min
1Q
2.331754 0.294114 Median
0.001475 3Q
0.263843 Max
1.678532 Coefficients:
Estimate Std. Error t value Pr(>t)
union1 0.859166
0.091630
9.376 < 2e16 ***
union2 1.164295
0.090453 12.872 < 2e16 ***
ed
0.058122
0.006952
8.361 4.44e16 *** 196 Signif. codes: 16. SPECIFIC DATASETS 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ Residual standard error: 0.4481 on 547 degrees of freedom
Multiple RSquared: 0.9349,Adjusted Rsquared: 0.9345
Fstatistic: 2617 on 3 and 547 degrees of freedom,pvalue:
>
>
>
> 0.1 ‘’ 1 0 #actually it was unnecessary to construct the contrast matrix.
#If we regress with a categorical variable without
#an intercept, R will automatically use dummies for all levels:
lm(lnwage ~ union + ed  1, data=cps85) Call:
lm(formula = lnwage ~ union + ed  1, data = cps85)
Coefficients:
unionNonunion
0.9926 unionUnion
1.2909 ed
0.0778 > ##Exercise 2d (1 point) Why is it not possible to include two dummies plus
> # an intercept? Because the two dummies add to 1,
> # you have perfect collinearity >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> ###Exercise 3a (2 points):
summary(lm(lnwage ~ ed + ex + I(ex^2), data=cps78))
#All coefficients are highly significant, but the R^2 is only 0.2402
#Returns to experience are positive and decline with increase in experience
##Exercise 3b (2 points):
summary(lm(lnwage ~ gender + ed + ex + I(ex^2), data=cps78))
contrasts(cps78$gender)
#We see here that gender is coded 0 for female and 1 for male;
#by default, the levels in a factor variable occur in alphabetical order.
#Intercept in our regression = 0.1909203 (this is for female),
#genderMale has coefficient = 0.3351771,
#i.e., the intercept for women is 0.5260974
#Gender is highly significant
##Exercise 3c (2 points):
summary(lm(lnwage ~ gender + marr + ed + ex + I(ex^2), data=cps78))
#Coefficient of marr in this is insignificant
##Exercise 3d (1 point) asks to construct a variable which we do
#not need when we use factor variables
##Exercise 3e (3 points): For interaction term do
summary(lm(lnwage ~ gender * marr + ed + ex + I(ex^2), data=cps78)) Call:
lm(formula = lnwage ~ gender * marr + ed + ex + I(ex^2), data = cps78)
Residuals:
Min 1Q Median 3Q Max 16.5. WAGE DATA 2.45524 0.24566 0.01969 0.23102 197 1.42437 Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept)
0.1893919 0.1042613
1.817 0.06984 .
genderMale
0.3908782 0.0467018
8.370 4.44e16 ***
marrSingle
0.0507811 0.0557198
0.911 0.36251
ed
0.0738640 0.0066154 11.165 < 2e16 ***
ex
0.0265297 0.0049741
5.334 1.42e07 ***
I(ex^2)
0.0003161 0.0001057 2.990 0.00291 **
genderMale:marrSingle 0.1586452 0.0750830 2.113 0.03506 *
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1
Residual standard error: 0.3959 on 543 degrees of freedom
Multiple RSquared: 0.3547,Adjusted Rsquared: 0.3476
Fstatistic: 49.75 on 6 and 543 degrees of freedom,pvalue: ‘’ 1 0 > #Being married raises the wage for men by 13% but lowers it for women by 3%
> ###Exercise 4a (5 points):
> summary(lm(lnwage ~ union + gender + race + ed + ex + I(ex^2), data=cps78))
Call:
lm(formula = lnwage ~ union + gender + race + ed + ex + I(ex^2),
data = cps78)
Residuals:
Min
1Q
2.41914 0.23674 Median
0.01682 3Q
0.21821 Max
1.31584 Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 0.1549723 0.1068589
1.450 0.14757
unionUnion
0.2071429 0.0368503
5.621 3.04e08 ***
genderMale
0.3060477 0.0344415
8.886 < 2e16 ***
raceNonwh
0.1301175 0.0830156 1.567 0.11761
raceOther
0.0271477 0.0688277
0.394 0.69342
ed
0.0746097 0.0066521 11.216 < 2e16 ***
ex
0.0261914 0.0047174
5.552 4.43e08 ***
I(ex^2)
0.0003082 0.0001015 3.035 0.00252 **
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ Residual standard error: 0.3845 on 542 degrees of freedom
Multiple RSquared: 0.3924,Adjusted Rsquared: 0.3846
Fstatistic: 50.01 on 7 and 542 degrees of freedom,pvalue: 0.1 ‘’ 0 > exp(0.1301175)
[1] 0.8779923
> #Being Hispanic lowers wages by 2.7%, byut being black lowers them 1 198 16. SPECIFIC DATASETS > #by 12.2 %
>
>
>
> #At what level of ex is lnwage maximized?
#exeffect = 0.0261914 * ex 0.0003082 * ex^2
#derivative = 0.0261914  2 * 0.0003082 * ex
#derivative = 0 for ex=0.0261914/(2*0.0003082) > 0.0261914/(2*0.0003082)
[1] 42.49091
>
>
>
>
>
>
> # age  ed  6 = 42.49091
# age = ed + 48.49091
# for 8, 12, and 16 years of schooling the max earnings
# are at ages 56.5, 60.5, and 64.5 years
##Exercise 4b (4 points) is a graph, not done here
##Exercise 4c (5 points)
summary(lm(lnwage ~ gender + union + race + ed + ex + I(ex^2) + I(ed*ex), data=cps78)) Call:
lm(formula = lnwage ~ gender + union + race + ed + ex + I(ex^2) +
I(ed * ex), data = cps78)
Residuals:
Min
1Q
2.41207 0.23922 Median
0.01463 3Q
0.21645 Max
1.32051 Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 0.0396495 0.1789073
0.222 0.824693
genderMale
0.3042639 0.0345241
8.813 < 2e16 ***
unionUnion
0.2074045 0.0368638
5.626 2.96e08 ***
raceNonwh
0.1323898 0.0830908 1.593 0.111673
raceOther
0.0319829 0.0691124
0.463 0.643718
ed
0.0824154 0.0117716
7.001 7.55e12 ***
ex
0.0328854 0.0095716
3.436 0.000636 ***
I(ex^2)
0.0003574 0.0001186 3.013 0.002704 **
I(ed * ex) 0.0003813 0.0004744 0.804 0.421835
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ Residual standard error: 0.3846 on 541 degrees of freedom
Multiple RSquared: 0.3932,Adjusted Rsquared: 0.3842
Fstatistic: 43.81 on 8 and 541 degrees of freedom,pvalue:
> #Maximum earnings ages must be computed as before
>
>
>
>
> 0.1 ‘’ 1 0 ##Exercise 4d (4 points) not done here
##Exercise 4e (6 points) not done here
###Exercise 5a (3 points):
#Naive approach to estimate impact of unionization on wages:
summary(lm(lnwage ~ gender + union + race + ed + ex + I(ex^2), data=cps78)) 16.5. WAGE DATA 199 Call:
lm(formula = lnwage ~ gender + union + race + ed + ex + I(ex^2),
data = cps78)
Residuals:
Min
1Q
2.41914 0.23674 Median
0.01682 3Q
0.21821 Max
1.31584 Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 0.1549723 0.1068589
1.450 0.14757
genderMale
0.3060477 0.0344415
8.886 < 2e16 ***
unionUnion
0.2071429 0.0368503
5.621 3.04e08 ***
raceNonwh
0.1301175 0.0830156 1.567 0.11761
raceOther
0.0271477 0.0688277
0.394 0.69342
ed
0.0746097 0.0066521 11.216 < 2e16 ***
ex
0.0261914 0.0047174
5.552 4.43e08 ***
I(ex^2)
0.0003082 0.0001015 3.035 0.00252 **
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ Residual standard error: 0.3845 on 542 degrees of freedom
Multiple RSquared: 0.3924,Adjusted Rsquared: 0.3846
Fstatistic: 50.01 on 7 and 542 degrees of freedom,pvalue:
> # What is wrong with the above? It assumes that unions
> # only affect the intercept, everything else is the same
> ##Exercise 5b (2 points)
> tapply(lnwage, union, mean)
Nonun
Union
1.600901 1.863137
> tapply(ed, union, mean)
Nonun
Union
12.76178 12.02381
> table(gender, union)
union
gender
Nonun Union
Female
159
48
Male
223
120
> table(race, union)
union
race
Nonun Union
Hisp
29
7
Nonwh
35
22
Other
318
139
> 7/(7+29)
[1] 0.1944444
> 22/(22+35)
[1] 0.3859649 0.1 0 ‘’ 1 200 16. SPECIFIC DATASETS > 139/(318+139)
[1] 0.3041575
> #19% of Hispanic, 39% of Nonwhite, and 30% of other (white) workers
> #in the sample are in unions
> ##Exercise 5c (3 points)
> summary(lm(lnwage ~ gender + race + ed + ex + I(ex^2), data=cps78, subset=union == "Union"))
Call:
lm(formula = lnwage ~ gender + race + ed + ex + I(ex^2), data = cps78,
subset = union == "Union")
Residuals:
Min
1Q
2.3307 0.1853 Median
0.0160 3Q
0.2199 Max
1.1992 Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 0.9261456 0.2321964
3.989 0.000101 ***
genderMale
0.2239370 0.0684894
3.270 0.001317 **
raceNonwh
0.3066717 0.1742287 1.760 0.080278 .
raceOther
0.0741660 0.1562131 0.475 0.635591
ed
0.0399500 0.0138311
2.888 0.004405 **
ex
0.0313820 0.0098938
3.172 0.001814 **
I(ex^2)
0.0004526 0.0002022 2.239 0.026535 *
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘’ 1 Residual standard error: 0.3928 on 161 degrees of freedom
Multiple RSquared: 0.2019,Adjusted Rsquared: 0.1721
Fstatistic: 6.787 on 6 and 161 degrees of freedom,pvalue: 1.975e06 > summary(lm(lnwage ~ gender + race + ed + ex + I(ex^2), data=cps78, subset=union == "Nonun"))
Call:
lm(formula = lnwage ~ gender + race + ed + ex + I(ex^2), data = cps78,
subset = union == "Nonun")
Residuals:
Min
1Q
1.39107 0.23775 Median
0.01040 3Q
0.23337 Max
1.29073 Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 0.0095668 0.1193399 0.080
0.9361
genderMale
0.3257661 0.0397961
8.186 4.22e15 ***
raceNonwh
0.0652018 0.0960570 0.679
0.4977
raceOther
0.0444133 0.0761628
0.583
0.5602
ed
0.0852212 0.0075554 11.279 < 2e16 ***
ex
0.0253813 0.0053710
4.726 3.25e06 ***
I(ex^2)
0.0002841 0.0001187 2.392
0.0172 *
 16.5. WAGE DATA Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 201 ‘*’ 0.05 ‘.’ Residual standard error: 0.3778 on 375 degrees of freedom
Multiple RSquared: 0.4229,Adjusted Rsquared: 0.4137
Fstatistic: 45.8 on 6 and 375 degrees of freedom,pvalue: 0.1 ‘’ 1 ‘’ 1 0 > #Are unionnonunion differences larger for females than males?
> #For this look at the intercepts for males and females in
> #the two regressions. Say for white males and females:
> 0.92614560.0741660+0.2239370
[1] 1.075917
> 0.92614560.0741660
[1] 0.8519796
> 0.0095668+0.0444133+0.3257661
[1] 0.3606126
> 0.0095668+0.0444133
[1] 0.0348465
> 1.0759170.3606126
[1] 0.7153044
> 0.85197960.0348465
[1] 0.8171331
>
>
>
>
>
>
>
> #White Males
White Females
#Union
1.075917
0.8519796
#Nonunion
0.3606126
0.0348465
#Difference
0.7153044
0.8171331
#Difference is greater for women
###Exercise 6a (5 points)
summary(lm(lnwage ~ gender + union + race + ed + ex + I(ex^2))) Call:
lm(formula = lnwage ~ gender + union + race + ed + ex + I(ex^2))
Residuals:
Min
1Q
2.41914 0.23674 Median
0.01682 3Q
0.21821 Max
1.31584 Coefficients:
Estimate Std. Error t value Pr(>t)
(Intercept) 0.1549723 0.1068589
1.450 0.14757
genderMale
0.3060477 0.0344415
8.886 < 2e16 ***
unionUnion
0.2071429 0.0368503
5.621 3.04e08 ***
raceNonwh
0.1301175 0.0830156 1.567 0.11761
raceOther
0.0271477 0.0688277
0.394 0.69342
ed
0.0746097 0.0066521 11.216 < 2e16 ***
ex
0.0261914 0.0047174
5.552 4.43e08 ***
I(ex^2)
0.0003082 0.0001015 3.035 0.00252 **
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 202 16. SPECIFIC DATASETS Residual standard error: 0.3845 on 542 degrees of freedom
Multiple RSquared: 0.3924,Adjusted Rsquared: 0.3846
Fstatistic: 50.01 on 7 and 542 degrees of freedom,pvalue: 0 > #To test whether Nonwh and Hisp have same intercept
> #one might generate a contrast matrix which collapses those
> #two and then run it and make an Ftest
> #or make a contrast matrix which has this difference as one of
> #the dummies and use the ttest for that dummy
> ##Exercise 6b (2 points)
> table(race)
race
Hisp Nonwh Other
36
57
457
> tapply(lnwage, race, mean)
Hisp
Nonwh
Other
1.529647 1.513404 1.713829
> tapply(lnwage, race, ed)
Error in get(x, envir, mode, inherits) : variable "ed" was not found
> tapply(ed, race, mean)
Hisp
Nonwh
Other
10.30556 11.71930 12.81400
> table(gender, race)
race
gender
Hisp Nonwh Other
Female
12
28
167
Male
24
29
290
> #Blacks, almost as many women than men, hispanic twice as many men,
> #Whites in between
>
>
>
>
>
>
>
>
>
>
>
>
>
> #Additional stuff:
#There are two outliers in cps78 with wages of less than $1 per hour,
#Both service workers, perhaps waitresses who did not report her tips?
#What are the commands for extracting certain observations
#by certain criteria and just print them? The split command.
#Interesting to do
loess(lnwage ~ ed + ex, data=cps78)
#loess is appropriate here because there are strong interation terms
#How can one do loess after taking out the effects of gender for instance?
#Try the following, but I did not try it out yet:
gam(lnwage ~ lo(ed,ex) + gender, data=cps78)
#I should put more plotting commands in! CHAPTER 17 The Mean Squared Error as an Initial Criterion of
Precision
The question how “close” two random variables are to each other is a central
concern in statistics. The goal of statistics is to ﬁnd observed random variables which
are “close” to the unobserved parameters or random outcomes of interest. These observed random variables are usually called “estimators” if the unobserved magnitude
is nonrandom, and “predictors” if it is random. For scalar random variables we will
ˆ
use the mean squared error as a criterion for closeness. Its deﬁnition is MSE[φ; φ]
ˆ as an estimator or predictor, whatever the case
(read it: mean squared error of φ
may be, of φ):
ˆ
ˆ
MSE[φ; φ] = E[(φ − φ)2 ] (17.0.1) ˆ
For our purposes, therefore, the estimator (or predictor) φ of the unknown parameter
˜
ˆ
(or unobserved random variable) φ is no worse than the alternative φ if MSE[φ; φ] ≤
˜ ; φ]. This is a criterion which can be applied before any observations are
MSE[φ
collected and actual estimations are made; it is an “initial” criterion regarding the
expected average performance in a series of future trials (even though, in economics,
usually only one trial is made).
17.1. Comparison of Two Vector Estimators
ˆ
˜
If one wants to compare two vector estimators, say φ and φ, it is often impossible
ˆ
to say which of two estimators is better. It may be the case that φ1 is better than
ˆ
˜
˜
φ1 (in terms of MSE or some other criterion), but φ2 is worse than φ2 . And even if
ˆ than by φ , certain linear combinations
˜
every component φi is estimated better by φi
i
˜
ˆ
t φ of the components of φ may be estimated better by t φ than by t φ.
ˆ
Problem 240. 2 points Construct an example of two vector estimators φ and
˜
ˆ
˜ of the same random vector φ = φ1 φ2 , so that MSE[φi ; φi ] < MSE[φi ; φi ] for
φ
ˆ
ˆ
˜
˜
i = 1, 2 but MSE[φ1 + φ2 ; φ1 + φ2 ] > MSE[φ1 + φ2 ; φ1 + φ2 ]. Hint: it is easiest to use
an example in which all random variables are constants. Another hint: the geometric
ˆ
˜
analog would be to ﬁnd two vectors in a plane φ and φ. In each component (i.e.,
ˆ is closer to the origin than φ. But in the projection on
˜
projection on the axes), φ
˜ is closer to the origin than φ.
ˆ
the diagonal, φ
ˆ
φ= Answer. In the simplest counterexample, all variables involved are constants: φ =
1 , and φ = −2 .
˜
1 0
0 , 2 ˆ
One can only then say unambiguously that the vector φ is a no worse estimator
˜ if its MSE is smaller or equal for every linear combination. Theorem 17.1.1
than φ
ˆ
will show that this is the case if and only if the MSE matrix of φ is smaller, by a
˜. If this is so, then theorem 17.1.1 says
nonnegative deﬁnite matrix, than that of φ
203 204 17. THE MEAN SQUARED ERROR AS AN INITIAL CRITERION OF PRECISION that not only the MSE of all linear transformations, but also all other nonnegative
deﬁnite quadratic loss functions involving these vectors (such as the trace of the
MSE matrix, which is an oftenused criterion) are minimized. In order to formulate
and prove this, we ﬁrst need a formal deﬁnition of the MSE matrix. We write MSE
ˆ
for the matrix and MSE for the scalar mean squared error. The MSE matrix of φ
as an estimator of φ is deﬁned as
ˆ
ˆ
ˆ
MSE [φ; φ] = E [(φ − φ)(φ − φ) ] . (17.1.1) ˆ
Problem 241. 2 points Let θ be a vector of possibly random parameters, and θ
an estimator of θ . Show that
(17.1.2) ˆ
ˆ
ˆ
ˆ
MSE [θ ; θ ] = V [θ − θ ] + (E [θ − θ ])(E [θ − θ ]) . Don’t assume the scalar result but make a proof that is good for vectors and scalars.
Answer. For any random vector x follows
E [xx ] = E (x − E [x] + E [x])(x − E [x] + E [x])
= E (x − E [x])(x − E [x]) − E (x − E [x]) E [x] − E E [x](x − E [x]) + E E [x] E [x] = V [x] − O − O + E [x] E [x] .
ˆ
Setting x = θ − θ the statement follows. ˆ
If θ is nonrandom, formula (17.1.2) simpliﬁes slightly, since in this case V [θ − θ ] =
ˆ]. In this case, the MSE matrix is the covariance matrix plus the squared bias
V [θ
ˆ
matrix. If θ is nonrandom and in addition θ is unbiased, then the MSE matrix
coincides with the covariance matrix.
ˆ
˜
Theorem 17.1.1. Assume φ and φ are two estimators of the parameter φ (which
is allowed to be random itself ). Then conditions (17.1.3), (17.1.4), and (17.1.5) are
equivalent:
(17.1.3) For every constant vector t, (17.1.4) ˜
ˆ
MSE [φ; φ] − MSE [φ; φ] (17.1.5) For every nnd Θ, ˆ
˜
MSE[t φ; t φ] ≤ MSE[t φ; t φ]
is a nonnegative deﬁnite matrix ˆ
ˆ
˜
˜
E (φ − φ) Θ(φ − φ) ≤ E (φ − φ) Θ(φ − φ) . ˜
ˆ
Proof. Call MSE [φ; φ] = σ 2 Ξ and MSE [φ; φ] = σ 2Ω . To show that (17.1.3)
ˆ
˜
implies (17.1.4), simply note that MSE[t φ; t φ] = σ 2 t Ω t and likewise MSE[t φ; t φ] =
2
Ω)t ≥ 0 for all t, which is the
σ t Ξt. Therefore (17.1.3) is equivalent to t (Ξ −
deﬁning property making Ξ − Ω nonnegative deﬁnite.
Here is the proof that (17.1.4) implies (17.1.5):
ˆ
ˆ
ˆ
ˆ
E[(φ − φ) Θ(φ − φ)] = E[tr (φ − φ) Θ(φ − φ) ] =
ˆ
ˆ
= E[tr Θ(φ − φ)(φ − φ) ˆ
ˆ = tr Θ E [(φ − φ)(φ − φ) ] = σ 2 tr ΘΩ and in the same way
˜
˜
E[(φ − φ) Θ(φ − φ)] = σ 2 tr ΘΞ .
The diﬀerence in the expected quadratic forms is therefore σ 2 tr Θ(Ξ − Ω ) . By
assumption, Ξ − Ω is nonnegative deﬁnite. Therefore, by theorem A.5.6 in the
Mathematical Appendix, or by Problem 242 below, this trace is nonnegative.
To complete the proof, (17.1.5) has (17.1.3) as a special case if one sets Θ =
tt . 17.1. COMPARISON OF TWO VECTOR ESTIMATORS 205 Problem 242. Show that if Θ and Σ are symmetric and nonnegative deﬁnite,
then tr(ΘΣ ) ≥ 0. You are allowed to use that tr(AB ) = tr(BA), that the trace of a
nonnegative deﬁnite matrix is ≥ 0, and Problem 118 (which is trivial).
Answer. Write Θ = RR ; then tr(ΘΣ ) = tr(RR Σ ) = tr(R Σ R) ≥ 0. Problem 243. Consider two very simpleminded estimators of the unknown
nonrandom parameter vector φ = φ1 . Neither of these estimators depends on any
φ2
ˆ
observations, they are constants. The ﬁrst estimator is φ = [ 11 ], and the second is
11
12 ].
˜
φ=[
8 • a. 2 points Compute the MSE matrices of these two estimators if the true
value of the parameter vector is φ = [ 10 ]. For which estimator is the trace of the
10
MSE matrix smaller?
ˆ
Answer. φ has smaller trace of the MSE matrix.
1
ˆ
φ−φ=
1
ˆ
ˆ
ˆ
MSE [φ; φ] = E [(φ − φ)(φ − φ) ]
= E[ 1
1 ˜
φ−φ= 4
−4 1 ] = E[ 1
1 1
1=
1
1 1
1 2
−2 ˜
MSE [φ; φ] = 1 −4
4 Note that both MSE matrices are singular, i.e., both estimators allow an errorfree look at certain
linear combinations of the parameter vector. ˆ
• b. 1 point Give two vectors g = [ g1 ] and h = h1 satisfying MSE[g φ; g φ] <
g2
h2
˜
ˆ
˜
MSE[g φ; g φ] and MSE[h φ; h φ] > MSE[h φ; h φ] (g and h are not unique;
there are many possibilities).
1
ˆ
˜
Answer. With g = −1 and h = 1 for instance we get g φ − g φ = 0, g φ −
1
ˆ; h φ = 2, h φ; h φ = 0, therefore MSE[g φ; g φ] = 0, MSE[g φ; g φ] = 16,
˜
ˆ
˜
g φ = 4, h φ
ˆ
˜
MSE[h φ; h φ] = 4, MSE[h φ; h φ] = 0. An alternative way to compute this is e.g. ˜
MSE [h φ; h φ] = 1 −1 4
−4 −4
4 1
= 16
−1 ˆ
˜
˜
• c. 1 point Show that neither MSE [φ; φ] − MSE [φ; φ] nor MSE [φ; φ] −
ˆ
MSE [φ; φ] is a nonnegative deﬁnite matrix. Hint: you are allowed to use the
mathematical fact that if a matrix is nonnegative deﬁnite, then its determinant is
nonnegative.
Answer.
(17.1.6) ˜
ˆ
MSE [φ; φ] − MSE [φ; φ] = 3
−5 −5
3 Its determinant is negative, and the determinant of its negative is also negative. CHAPTER 18 Sampling Properties of the Least Squares
Estimator
ˆ
The estimator β was derived from a geometric argument, and everything which
we showed so far are what [DM93, p. 3] calls its numerical as opposed to its statistical
ˆ
properties. But β has also nice statistical or sampling properties. We are assuming
right now the speciﬁcation given in (14.1.3), in which X is an arbitrary matrix of full
column rank, and we are not assuming that the errors must be Normally distributed.
The assumption that X is nonrandom means that repeated samples are taken with
the same X matrix. This is often true for experimental data, but not in econometrics.
The sampling properties which we are really interested in are those where also the X matrix is random; we will derive those later. For this later derivation, the properties
with ﬁxed X matrix, which we are going to discuss presently, will be needed as an
intermediate step. The assumption of ﬁxed X is therefore a preliminary technical
assumption, to be dropped later.
ˆ
In order to know how good the estimator β is, one needs the statistical properties
ˆ − β . This sampling error has the following formula:
of its “sampling error” β
ˆ
β − β = (X X )−1 X y − (X X )−1 X X β =
(18.0.7) = (X X )−1 X (y − Xβ ) = (X X )−1 X ε ˆ
From (18.0.7) follows immediately that β is unbiased, since E [(X X )−1 X ε ] = o.
Unbiasedness does not make an estimator better, but many good estimators are
unbiased, and it simpliﬁes the math.
We will use the MSE matrix as a criterion for how good an estimator of a vector
of unobserved parameters is. Chapter 17 gave some reasons why this is a sensible
criterion (compare [DM93, Chapter 5.5]).
18.1. The Gauss Markov Theorem
ˆ
Returning to the least squares estimator β , one obtains, using (18.0.7), that
ˆ
ˆ
ˆ
MSE [β ; β ] = E [(β − β )(β − β ) ] = (X X )−1 X E [εε ]X (X X )−1 =
(18.1.1) = σ 2 (X X )−1 . This is a very simple formula. Its most interesting aspect is that this MSE matrix
does not depend on the value of the true β . In particular this means that it is
bounded with respect to β , which is important for someone who wants to be assured
of a certain accuracy even in the worst possible situation.
Problem 244. 2 points Compute the MSE matrix MSE [ε; ε ] = E [(ε − ε )(ε −
ˆ
ˆ
ˆ
ε ) ] of the residuals as predictors of the disturbances.
Answer. Write ε − ε = M ε − ε = (M − I )ε = −X (X X )−1 X ε ; therefore MSE [ε; ε ] =
ˆ
ˆ
ˆ
E [X (X X )−1 X εε X (X X )−1 X = σ 2 X (X X )−1 X . Alternatively, start with ε − ε = y −
207 208 18. SAMPLING PROPERTIES OF THE LEAST SQUARES ESTIMATOR ˆ
ˆ
y −ε = Xβ −y = X (β −β ). This allows to use MSE [ε; ε ] = X MSE [β ; β ]X
ˆ
ˆ
ˆ = σ 2 X (X X )−1 X . Problem 245. 2 points Let v be a random vector that is a linear transformation
of y , i.e., v = T y for some constant matrix T . Furthermore v satisﬁes E [v ] = o.
Show that from this follows v = T ε. (In other words, no other transformation of y
ˆ
with zero expected value is more “comprehensive” than ε . However there are many
other transformation of y with zero expected value which are as “comprehensive” as
ε ).
Answer. E [v ] = T Xβ must be o whatever the value of β . Therefore T X = O , from which
follows T M = T . Since ε = M y , this gives immediately v = T ε. (This is the statistical implication
ˆ
ˆ
of the mathematical fact that M is a deﬁciency matrix of X .) ˆˆ
ˆ
Problem 246. 2 points Show that β and ε are uncorrelated, i.e., cov[β i , εj ] =
ˆ
ˆ, ε] as that matrix whose (i, j )
0 for all i, j . Deﬁning the covariance matrix C [β ˆ
ˆˆ
ˆˆ
element is cov[β i , εj ], this can also be written as C [β , ε] = O . Hint: The covariance
matrix satisﬁes the rules C [Ay , B z ] = A C [y , z ]B and C [y , y ] = V [y ]. (Other rules
for the covariance matrix, which will not be needed here, are C [z , y ] = (C [y , z ]) ,
C [x + y , z ] = C [x, z ] + C [y , z ], C [x, y + z ] = C [x, y ] + C [x, z ], and C [y , c] = O if c is
a vector of constants.)
Answer. A = (X X )−1 X
X (X X )−1 X ) = O . ˆˆ
and B = I −X (X X )−1 X , therefore C [β , ε] = σ 2 (X X )−1 X (I − Problem 247. 4 points Let y = Xβ + ε be a regression model with intercept, in
ˆ
which the ﬁrst column of X is the vector ι, and let β the least squares estimator of
ˆ
β . Show that the covariance matrix between y and β , which is deﬁned as the matrix
¯
(here consisting of one row only) that contains all the covariances
(18.1.2) yˆ
yˆ
C [¯, β ] ≡ cov[¯, β 1 ] cov[¯, β 2 ] · · ·
yˆ cov[¯, β k ]
yˆ 2
has the following form: C [¯, β ] = σ 1 0 · · · 0 where n is the number of obyˆ
n
servations. Hint: That the regression has an intercept term as ﬁrst column of the
X matrix means that Xe(1) = ι, where e(1) is the unit vector having 1 in the ﬁrst
place and zeros elsewhere, and ι is the vector which has ones everywhere. ˆ
Answer. Write both y and β in terms of y , i.e., y =
¯
¯ 1
ι
n ˆ
y and β = (X X )−1 X y . Therefore (18.1.3)
σ 2 (1)
1
σ2
σ 2 (1)
−1
ι X (X X )−1 =
e
e
yˆ
=
X X (X X )−1 =
.
C [¯, β ] = ι V [y ]X (X X )
n
n
n
n ˆ
Theorem 18.1.1. GaussMarkov Theorem: β is the BLUE (Best Linear Unbiased Estimator) of β in the following vector sense: for every nonrandom coeﬃcient
ˆ
vector t, t β is the scalar BLUE of t β , i.e., every other linear unbiased estimator
˜ = a y of φ = t β has a bigger MSE than t β .
ˆ
φ
˜
Proof. Write the alternative linear estimator φ = a y in the form
˜
(18.1.4)
φ = t (X X )−1 X + c y
then the sampling error is
˜
φ − φ = t (X X )−1 X + c
(18.1.5) −1 = t (X X ) X +c (X β + ε ) − t β
ε + c X β. 18.2. DIGRESSION ABOUT MINIMAX ESTIMATORS 209 By assumption, the alternative estimator is unbiased, i.e., the expected value of this
sampling error is zero regardless of the value of β . This is only possible if c X = o .
But then it follows
˜
˜
MSE[φ; φ] = E[(φ − φ)2 ] = E[ t (X X )−1 X + c
= σ 2 t (X X )−1 X + c X (X X )−1 t + c ] = εε X (X X )−1 t + c = σ 2 t (X X )−1 t + σ 2 c c, Here we needed again c X = o . Clearly, this is minimized if c = o, in which case
˜
ˆ
φ = t β.
˜
ˆ
Problem 248. 4 points Show: If β is a linear unbiased estimator of β and β is
˜; β ]−MSE [β ; β ]
ˆ
the OLS estimator, then the diﬀerence of the MSE matrices MSE [β
is nonnegative deﬁnite.
˜
Answer. (Compare [DM93, p. 159].) Any other linear estimator β of β can be written
˜ = (X X )−1 X + C y . Its expected value is E [β ] = (X X )−1 X X β + CXβ . For
˜
as β
˜
β to be unbiased, regardless of the value of β , C must satisfy CX = O . But then it follows
˜
˜
MSE [β ; β ] = V [β ] = σ 2 (X X )−1 X + C X (X X )−1 + C
= σ 2 (X X )−1 + σ 2 CC , i.e.,
ˆ
it exceeds the MSE matrix of β by a nonnegative deﬁnite matrix. 18.2. Digression about Minimax Estimators
Theorem 18.1.1 is a somewhat puzzling property of the least squares estimator,
since there is no reason in the world to restrict one’s search for good estimators
to unbiased estimators. An alternative and more enlightening characterization of
ˆ
β does not use the concept of unbiasedness but that of a minimax estimator with
respect to the MSE. For this I am proposing the following deﬁnition:
ˆ
Definition 18.2.1. φ is the linear minimax estimator of the scalar parameter φ
˜
with respect to the MSE if and only if for every other linear estimator φ there exists
a value of the parameter vector β 0 such that for all β 1
˜
ˆ
(18.2.1)
MSE[φ; φβ = β ] ≥ MSE[φ; φβ = β ]
0 1 ˜
In other words, the worst that can happen if one uses any other φ is worse than
ˆ . Using this concept one can prove the
the worst that can happen if one uses φ
following:
ˆ
Theorem 18.2.2. β is a linear minimax estimator of the parameter vector β
ˆ
in the following sense: for every nonrandom coeﬃcient vector t, t β is the linear
minimax estimator of the scalar φ = t β with respect to the MSE. I.e., for every
˜
˜
other linear estimator φ = a y of φ one can ﬁnd a value β = β 0 for which φ has a
ˆ.
larger MSE than the largest possible MSE of t β
Proof: as in the proof of Theorem 18.1.1, write the alternative linear estimator
˜
˜
φ in the form φ = t (X X )−1 X + c y , so that the sampling error is given by
(18.1.5). Then it follows
(18.2.2)
˜
˜
MSE[φ; φ] = E[(φ−φ)2 ] = E[ t (X X )−1 X +c ε +c X β ε X (X X )−1 t+c +β X c ]
(18.2.3) = σ 2 t (X X )−1 X + c X (X X )−1 t + c + c X ββ X c ˜
Now there are two cases: if c X = o , then MSE[φ; φ] = σ 2 t (X X )−1 t + σ 2 c c.
This does not depend on β and if c = o then this MSE is larger than that for c = o.
˜
If c X = o , then MSE[φ; φ] is unbounded, i.e., for any ﬁnite number ω one one 210 18. SAMPLING PROPERTIES OF THE LEAST SQUARES ESTIMATOR ˜
ˆ
can always ﬁnd a β 0 for which MSE[φ; φ] > ω . Since MSE[φ; φ] is bounded, a β 0
can be found that satisﬁes (18.2.1).
If we characterize the BLUE as a minimax estimator, we are using a consistent
and uniﬁed principle. It is based on the concept of the MSE alone, not on a mixture between the concepts of unbiasedness and the MSE. This explains why the
mathematical theory of the least squares estimator is so rich.
On the other hand, a minimax strategy is not a good estimation strategy. Nature
is not the adversary of the researcher; it does not maliciously choose β in such a way
that the researcher will be misled. This explains why the least squares principle,
despite the beauty of its mathematical theory, does not give terribly good estimators
(in fact, they are inadmissible, see the Section about the Stein rule below).
ˆ
β is therefore simultaneously the solution to two very diﬀerent minimization
problems. We will refer to it as the OLS estimate if we refer to its property of
minimizing the sum of squared errors, and as the BLUE estimator if we think of it
as the best linear unbiased estimator.
Note that even if σ 2 were known, one could not get a better linear unbiased
estimator of β .
18.3. Miscellaneous Properties of the BLUE
Problem 249.
• a. 1 point Instead of (14.2.22) one sometimes sees the formula
(xt − x)y t
¯
.
(xt − x)2
¯ ˆ
β= (18.3.1) for the slope parameter in the simple regression. Show that these formulas are mathematically equivalent.
y
¯ Answer. Equivalence of (18.3.1) and (14.2.22) follows from
(xt − x) = 0 and therefore also
¯
(xt − x) = 0. Alternative proof, using matrix notation and the matrix D deﬁned in Problem
¯ 161: (14.2.22) is
idempotent. xD
xD Dy
Dx x Dy
.
x D Dx and (18.3.1) is They are equal because D is symmetric and • b. 1 point Show that
σ2
(xi − x)2
¯ ˆ
var[β ] = (18.3.2)
Answer. Write (18.3.1) as
(18.3.3) ˆ
β= 1
( xt − x) 2
¯ (xt − x)y t
¯ ⇒ 1 ˆ
var[β ] = ( xt − x) 2
¯ 2 ( xt − x) 2 σ 2
¯ ˆ¯
• c. 2 points Show that cov[β , y ] = 0.
Answer. This is a special case of problem 247, but it can be easily shown here separately:
ˆ¯
cov[β , y ] = cov ( xs − x) y s 1
¯
,
( xt − x) 2 n
¯
t s yj =
j = n n 1
cov
( xt − x) 2
¯
t t 1
( xt − x) 2
¯ ( xs − x) y s ,
¯
s (xs − x)σ 2 = 0.
¯
s yj =
j 18.3. MISCELLANEOUS PROPERTIES OF THE BLUE 211 • d. 2 points Using (14.2.23) show that
x2
¯
(xi − x)2
¯ 1
+
n ˆ
var[α] = σ 2 (18.3.4) Problem 250. You have two data vectors xi and y i (i = 1, . . . , n), and the true
model is
y i = βxi + εi (18.3.5) where xi and εi satisfy the basic assumptions of the linear regression model. The
least squares estimator for this model is
xi y i
x2
i ˜
β = (x x)−1 x y = (18.3.6) ˜
• a. 1 point Is β an unbiased estimator of β ? (Proof is required.)
˜
Answer. First derive a nice expression for β − β :
xi y i ˜
β−β = x2
i x2
i x2
i
xi εi = since x2
i y i = βxi + εi xi εi ˜
E[β − β ] = E = x2 β
i xi ( y i − xi β ) = = − x2
i
E[xi εi ]
x2
i
xi E[εi ]
x2
i =0 since E εi = 0. ˜
• b. 2 points Derive the variance of β . (Show your work.)
Answer.
˜
˜
var β = E[β − β ]2 = = =
=
= 2 xi εi =E x2
i ( 1
E[
x2 )2
i ( 1
x2 )2
i (
( 1
x2 )2
i
1
σ2
x2 )2
i
σ2
.
x2
i E xi εi ]2 (xi εi )2 + 2 E (xi εi )(xj εj )
i<j E[xi εi ]2
x2
i since the εi ’s are uncorrelated, i.e., cov[εi , εj ] = 0 for i = j
since all εi have equal variance σ 2 212 18. SAMPLING PROPERTIES OF THE LEAST SQUARES ESTIMATOR Problem 251. We still assume (18.3.5) is the true model. Consider an alternative estimator:
¯
(xi − x)(y i − y )
¯
ˆ
(18.3.7)
β=
(xi − x)2
¯
i.e., the estimator which would be the best linear unbiased estimator if the true model
were (14.2.15).
ˆ
• a. 2 points Is β still an unbiased estimator of β if (18.3.5) is the true model?
(A short but rigorous argument may save you a lot of algebra here).
ˆ
Answer. One can argue it: β is unbiased for model (14.2.15) whatever the value of α or β ,
therefore also when α = 0, i.e., when the model is (18.3.5). But here is the pedestrian way:
ˆ
β=
= (xi − x)(y i − y )
¯
¯
( xi − x) 2
¯ (xi − x)y i
¯ = ( xi − x) 2
¯ (xi − x)(βxi + εi )
¯ =β
=β+ ( xi − x) xi
¯
( xi − x) 2
¯ y i = βxi + εi (xi − x)εi
¯ + ( xi − x) 2
¯ (xi − x)εi
¯ ( xi − x) xi =
¯ since ( xi − x) 2
¯ ˆ
Eβ = Eβ + E
=β+ since ( xi − x) 2
¯ (xi − x)¯ = 0
¯y since ( xi − x) 2
¯ (xi − x)εi
¯
( xi − x) 2
¯ (xi − x) E εi
¯ ˆ
since E εi = 0 for all i, i.e., β is unbiased. =β ( xi − x) 2
¯ ˆ
• b. 2 points Derive the variance of β if (18.3.5) is the true model.
ˆ
Answer. One can again argue it: since the formula for var β does not depend on what the
true value of α is, it is the same formula.
(18.3.8) ˆ
var β = var (18.3.9) = var (18.3.10) = (18.3.11) = β+ (xi − x)εi
¯
( xi − x) 2
¯ (xi − x)εi
¯
( xi − x) 2
¯
(xi − x)2 var εi
¯ ( (xi − x)2 )2
¯ since cov[εi εj ] = 0 σ2
.
( xi − x) 2
¯ ˆ
• c. 1 point Still assuming (18.3.5) is the true model, would you prefer β or the
˜
β from Problem 250 as an estimator of β ?
˜
ˆ
Answer. Since β and β are both unbiased estimators, if (18.3.5) is the true model, the pre˜
ˆ
ferred estimator is the one with the smaller variance. As I will show, var β ≤ var β and, therefore,
˜
ˆ
β is preferred to β . To show
(18.3.12) ˆ
var β = σ2
≥
( xi − x) 2
¯ σ2
˜
= var β
x2
i one must show
(18.3.13) ( xi − x) 2 ≤
¯ x2
i 18.3. MISCELLANEOUS PROPERTIES OF THE BLUE 213 ˆ
˜
which is a simple consequence of (9.1.1). Thus var β ≥ var β ; the variances are equal only if x = 0,
¯
˜
ˆ
i.e., if β = β . Problem 252. Suppose the true model is (14.2.15) and the basic assumptions
are satisﬁed.
xi y i ˜
• a. 2 points In this situation, β = is generally a biased estimator of β . x2
i Show that its bias is
nx
¯
x2
i ˜
E[β − β ] = α (18.3.14) Answer. In situations like this it is always worth while to get a nice simple expression for the
sampling error:
xi y i (18.3.15) ˜
β−β = (18.3.16) = (18.3.17) =α (18.3.18) =α x2
i xi (α + βxi + εi )
x2
i ˜
E[β − β ] = E α (18.3.19) −β (18.3.20) =α (18.3.21) =α xi
x2
i
xi
x2
i xi
xi
xi
x2
i + x2
i since y i = α + βxi + εi
xi εi
x2
i −β xi εi + x2
i
x2
i x2
i +β −β x2
i
xi εi +E + x2
i
xi E εi
x2
i +0=α nx
¯
x2
i This is = 0 unless x = 0 or α = 0.
¯ ˜
• b. 2 points Compute var[β ]. Is it greater or smaller than
σ2
(xi − x)2
¯ (18.3.22) which is the variance of the OLS estimator in this model?
Answer.
(18.3.23)
(18.3.24) xi y i ˜
var β = var
= x2
i
1
x2
i (18.3.25) = 1
x2
i (18.3.26) = = var[ xi y i ] 2 x2 var[y i ]
i 2 x2
i σ2
x2
i (18.3.27) 2 since all y i are uncorrelated and have equal variance σ 2 σ2
.
x2
i This variance is smaller or equal because x2 ≥
i ( xi − x) 2 .
¯ 214 18. SAMPLING PROPERTIES OF THE LEAST SQUARES ESTIMATOR ˜
• c. 5 points Show that the MSE of β is smaller than that of the OLS estimator
if and only if the unknown true parameters α and σ 2 satisfy the equation
α2 (18.3.28)
σ2 1
n + x2
¯
(xi −x)2
¯ <1 Answer. This implies some tedious algebra. Here it is important to set it up right.
αnx
¯
x2
i 2 αnx
¯
x2
i σ2
+
x2
i ˜
MSE[β ; β ] = 2 ≤ σ2
( xi − x) 2
¯ ≤ σ2
−
( xi − x) 2
¯ =
α2 n
x2
i = α2
1
n ( xi − x) 2 + x2
¯
¯
α2 σ2 1
n + x2
¯
(xi −x)2
¯ ≤ σ2
σ2
=
x2
i x2 −
i (xi − x)2
¯ ( xi − x) 2
¯ x2
i σ 2 nx2
¯
( xi − x) 2
¯ x2
i σ2
( xi − x) 2
¯ ≤1 Now look at this lefthand side; it is amazing and surprising that it is exactly the population
equivalent of the F test for testing α = 0 in the regression with intercept. It can be estimated by
replacing α2 with α2 and σ 2 with s2 (in the regression with intercept). Let’s look at this statistic.
ˆ
If α = 0 it has a F distribution with 1 and n − 2 degrees of freedom. If α = 0 it has what is called
a noncentral distribution, and the only thing we needed to know so far was that it was likely to
assume larger values than with α = 0. This is why a small value of that statistic supported the
hypothesis that α = 0. But in the present case we are not testing whether α = 0 but whether the
constrained MSE is better than the unconstrained. This is the case of the above inequality holds,
the limiting case being that it is an equality. If it is an equality, then the above statistic has a F
distribution with noncentrality parameter 1/2. (Here all we need to know that: if z ∼ N (µ, 1) then
z 2 ∼ χ2 with noncentrality parameter µ2 /2. A noncentral F has a noncentral χ2 in numerator and
1
a central one in denominator.) The testing principle is therefore: compare the observed value with
the upper α point of a F distribution with noncentrality parameter 1/2. This gives higher critical
values than testing for α = 0; i.e., one may reject that α = 0 but not reject that the MSE of the
contrained estimator is larger. This is as it should be. Compare [Gre97, 8.5.1 pp. 405–408] on
this. From the GaussMarkov theorem follows that for every nonrandom matrix R,
ˆ
ˆ
the BLUE of φ = Rβ is φ = Rβ . Furthermore, the best linear unbiased predictor
ˆ
ˆ
(BLUP) of ε = y − Xβ is the vector of residuals ε = y − X β .
˜
Problem 253. Let ε = Ay be a linear predictor of the disturbance vector ε in
the model y = Xβ + ε with ε ∼ (o, σ 2 I ).
˜
• a. 2 points Show that ε is unbiased, i.e., E[˜ − ε ] = o, regardless of the value
ε
of β , if and only if A satisﬁes AX = O .
Answer. E [Ay − ε ] = E [AXβ + Aε − ε ] = AXβ + o − o. This is = o for all β if and only if
AX = O ˜
• b. 2 points Which unbiased linear predictor ε = Ay of ε minimizes the MSE matrix E [(˜ − ε )(˜ − ε ) ]? Hint: Write A = I − X (X X )−1 X + C . What is the
ε
ε
minimum value of this MSE matrix?
Answer. Since AX = O , the prediction error Ay − ε = AXβ + Aε − ε = (A − I )ε ; therefore
one minimizes σ 2 (A − I )(A − I ) s. t. AX = O . Using the hint, C must also satisfy CX = O , and
(A − I )(A − I ) = (C − X (X X )−1 X )(C − X (X X )−1 X ) = X (X X )−1 X + CC ,
therefore one must set C = O . Minimum value is σ 2 X (X X )−1 X . 18.3. MISCELLANEOUS PROPERTIES OF THE BLUE 215 ˆ
• c. How does this best predictor relate to the OLS estimator β ?
ˆ
Answer. It is equal to the residual vector ε = y − X β .
ˆ ˆ
Problem 254. This is a vector generalization of problem 170. Let β the BLUE
˜ an arbitrary linear unbiased estimator of β .
of β and β
ˆ ˜ˆ
• a. 2 points Show that C [β − β , β ] = O .
˜
˜
˜
Answer. Say β = B y ; unbiasedness means BX = I . Therefore
−1
ˆ ˜ˆ
C [ β − β , β ] = C [ (X X ) X = (X X ) −1 X ˜
− B y , (X X )−1 X y ]
˜
− B V [y ]X (X X )−1 = σ 2 (X X )−1 X ˜
− B X ( X X ) −1 = σ 2 (X X )−1 − (X X )−1 = O . ˜
ˆ
˜ˆ
• b. 2 points Show that MSE [β ; β ] = MSE [β ; β ] + V [β − β ]
˜
ˆ
˜
ˆ
Answer. Due to unbiasedness, MSE = V , and the decomposition β = β + (β − β ) is an
˜
˜
ˆ˜ˆ
ˆ
ˆ˜ ˆ
uncorrelated sum. Here is more detail: MSE [β ; β ] = V [β ] = V [β + β − β ] = V [β ] + C [β , β − β ] +
˜ ˆˆ
˜ˆ
C [β − β , β ] + V [β − β ] but the two C terms are the null matrices. Problem 255. 3 points Given a simple regression y t = α + βxt + εt , where the
εt are independent and identically distributed with mean µ and variance σ 2 . Is it
possible to consistently estimate all four parameters α, β , σ 2 , and µ? If yes, explain
how you would estimate them, and if no, what is the best you can do?
Answer. Call ˜t = εt − µ, then the equation reads y t = α + µ + βxt + ˜t , with well behaved
ε
ε
disturbances. Therefore one can estimate α + µ, β , and σ 2 . This is also the best one can do; if
α + µ are equal, the y t have the same joint distribution. Problem 256. 3 points The model is y = Xβ + ε but all rows of the X matrix
are exactly equal. What can you do? Can you estimate β ? If not, are there any linear
combinations of the components of β which you can estimate? Can you estimate σ 2 ?
Answer. If all rows are equal, then each column is a multiple of ι. Therefore, if there are more
than one column, none of the individual components of β can be estimated. But you can estimate
x β (if x is one of the row vectors of X ) and you can estimate σ 2 . Problem 257. This is [JHG+ 88, 5.3.32]: Consider the loglinear statistical
model
(18.3.29) y t = αxβ exp εt = zt exp εt
t with “wellbehaved” disturbances εt . Here zt = αxβ is the systematic portion of y t ,
t
which depends on xt . (This functional form is often used in models of demand and
production.)
• a. 1 point Can this be estimated with the regression formalism?
Answer. Yes, simply take logs:
(18.3.30) log y t = log α + β log xt + εt • b. 1 point Show that the elasticity of the functional relationship between xt and
zt
(18.3.31) η= ∂zt /zt
∂xt /xt 216 18. SAMPLING PROPERTIES OF THE LEAST SQUARES ESTIMATOR does not depend on t, i.e., it is the same for all observations. Many authors talk
about the elasticity of y t with respect to xt , but one should really only talk about the
elasticity of zt with respect to xt , where zt is the systematic part of yt which can be
estimated by yt .
ˆ
Answer. The systematic functional relationship is log zt = log α + β log xt ; therefore
∂ log zt
1
=
∂zt
zt (18.3.32)
which can be rewritten as ∂zt
= ∂ log zt ;
zt (18.3.33)
The same can be done with xt ; therefore ∂zt /zt
∂ log zt
=
=β
∂xt /xt
∂ log xt (18.3.34) What we just did was a tricky way to take a derivative. A less tricky way is:
∂zt
= αβxβ −1 = βzt /xt
t
∂xt (18.3.35)
Therefore ∂zt xt
=β
∂xt zt (18.3.36) Problem 258.
• a. 2 points What is the elasticity in the simple regression y t = α + βxt + εt ?
Answer.
(18.3.37) ηt = ∂ z t /z t
∂ z t xt
βxt
βxt
=
=
=
∂xt /xt
∂xt z t
zt
α + βxt This depends on the observation, and if one wants one number, a good way is to evaluate it at
x.
¯ • b. Show that an estimate of this elasticity evaluated at x is h =
¯ ˆ¯
βx
y.
¯ Answer. This comes from the fact that the ﬁtted regression line goes through the point x, y .
¯¯
If one uses the other deﬁnition of elasticity, which Greene uses on p. 227 but no longer on p. 280,
and which I think does not make much sense, one gets the same formula:
(18.3.38) ηt = ∂ y t xt
βxt
∂ y t /y t
=
=
∂xt /xt
∂xt y t
yt This is diﬀerent than (18.3.37), but if one evaluates it at the sample mean, both formulas give the
same result ˆ¯
βx
.
y
¯ • c. Show by the delta method that the estimator
(18.3.39) h= ˆ¯
βx
y
¯ of the elasticity in the simple regression model has the estimated asymptotic variance (18.3.40) s2 −h
y
¯ x(1−h)
¯
y
¯ 1x
¯
x x2
¯¯ −1 −h
y
¯
x(1−h)
¯
y
¯ 18.3. MISCELLANEOUS PROPERTIES OF THE BLUE 217 • d. Compare [Gre97, example 6.20 on p. 280]. Assume
(18.3.41) 1
1x
¯
1q
(X X ) =
→Q=
qr
x x2
¯¯
n where we assume for the sake of the argument that q is known. The true elasticity
of the underlying functional relationship, evaluated at lim x, is
¯
qβ
(18.3.42)
η=
α + qβ
Then
ˆ
qβ
(18.3.43)
h=
ˆ
α + qβ
ˆ
is a consistent estimate for η .
A generalization of the loglinear model is the translog model, which is a secondorder approximation to an unknown functional form, and which allows to model
secondorder eﬀects such as elasticities of substitution etc. Used to model production,
cost, and utility functions. Start with any function v = f (u1 , . . . , un ) and make a
secondorder Taylor development around u = o:
(18.3.44) v = f (o) + ui ∂f
∂ui u=o + 1
2 ui uj
i,j ∂2f
∂ui ∂uj u=o Now say v = log(y ) and ui = log(xi ), and the values of f and its derivatives at o are
the coeﬃcients to be estimated:
1
βi log xi +
(18.3.45)
log(y ) = α +
γij log xi log xj + ε
2 i,j
Note that by Young’s theorem it must be true that γkl = γlk .
The semilog model is often used to model growth rates:
(18.3.46) log y t = xt β + εt Here usually one of the columns of X is the time subscript t itself; [Gre97, p. 227]
writes it as
(18.3.47) log y t = xt β + tδ + εt where δ is the autonomous growth rate. The logistic functional form is appropriate
for adoption rates 0 ≤ y t ≤ 1: the rate of adoption is slow at ﬁrst, then rapid as the
innovation gains popularity, then slow again as the market becomes saturated:
exp(xt β + tδ + εt )
1 + exp(xt β + tδ + εt )
This can be linearized by the logit transformation:
yt
= xt β + tδ + εt
(18.3.49)
logit(y t ) = log
1 − yt
(18.3.48) yt = Problem 259. 3 points Given a simple regression y t = αt + βxt which deviates
from an ordinary regression in two ways: (1) There is no disturbance term. (2) The
“constant term” αt is random, i.e., in each time period t, the value of αt is obtained
by an independent drawing from a population with unknown mean µ and unknown
variance σ 2 . Is it possible to estimate all three parameters β , σ 2 , and µ, and to
“predict” each αt ? (Here I am using the term “prediction” for the estimation of a
random parameter.) If yes, explain how you would estimate it, and if not, what is
the best you can do? 218 18. SAMPLING PROPERTIES OF THE LEAST SQUARES ESTIMATOR Answer. Call εt = αt − µ, then the equation reads y t = µ + βxt + εt , with well behaved
disturbances. Therefore one can estimate all the unknown parameters, and predict αt by µ + εt .
ˆ 18.4. Estimation of the Variance
The formulas in this section use ginverses (compare (A.3.1)) and are valid even
if not all columns of X are linearly independent. q is the rank if X . The proofs are
not any more complicated than in the case that X has full rank, if one keeps in mind
identity (A.3.3) and some other simple properties of ginverses which are tacitly used
at various places. Those readers who are only interested in the fullrank case should
simply substitute (X X )−1 for (X X )− and k for q (k is the number of columns
of X ).
SSE , the attained minimum value of the Least Squares objective function, is a
random variable too and we will now compute its expected value. It turns out that
E[SSE ] = σ 2 (n − q ) (18.4.1) ˆ
Proof. SSE = ε ε, where ε = y − X β = y − X (X X )− X y = M y ,
ˆˆ
ˆ
−
with M = I − X (X X ) X . From M X = O follows ε = M (Xβ + ε ) =
ˆ
M ε . Since M is idempotent and symmetric, it follows ε ε = ε M ε , therefore
ˆˆ
E[ε ε] = E[tr ε M ε ] = E[tr M εε ] = σ 2 tr M = σ 2 tr(I − X (X X )− X ) =
ˆˆ
σ 2 (n − tr(X X )− X X ) = σ 2 (n − q ).
Problem 260.
• a. 2 points Show that
(18.4.2) SSE = ε M ε where M = I − X (X X )− X ˆ
Answer. SSE = ε ε, where ε = y − X β = y − X (X X )− X y = M y where M =
ˆˆ
ˆ
I − X (X X )− X . From M X = O follows ε = M (Xβ + ε ) = M ε . Since M is idempotent and
ˆ
symmetric, it follows ε ε = ε M ε .
ˆˆ • b. 1 point Is SSE observed? Is ε observed? Is M observed?
• c. 3 points Under the usual assumption that X has full column rank, show that
E[SSE ] = σ 2 (n − k ) (18.4.3) Answer. E[ε ε] = E[tr ε M ε ] = E[tr M εε ] = σ 2 tr M = σ 2 tr(I − X (X X )− X ) =
ˆˆ
σ 2 (n − tr(X X )− X X ) = σ 2 (n − k). Problem 261. As an alternative proof of (18.4.3) show that SSE = y M y
and use theorem ??.
From (18.4.3) follows that SSE /(n − q ) is an unbiased estimate of σ 2 . Although
it is commonly suggested that s2 = SSE /(n − q ) is an optimal estimator of σ 2 , this
is a fallacy. The question which estimator of σ 2 is best depends on the kurtosis of
the distribution of the error terms. For instance, if the kurtosis is zero, which is the
case when the error terms are normal, then a diﬀerent scalar multiple of the SSE ,
namely, the TheilSchweitzer estimator from [TS61]
(18.4.4) σT S =
ˆ2 1
1
y My =
n−q+2
n−q+2
2 n ε2 ,
ˆi
i=1 is biased but has lower MSE than s . Compare problem 163. The only thing one
can say about s2 is that it is a fairly good estimator which one can use when one
does not know the kurtosis (but even in this case it is not the best one can do). 18.5. MALLOW’S CPSTATISTIC 219 18.5. Mallow’s CpStatistic as Estimator of the Mean Squared Error
Problem 262. We will compute here the MSE matrix of y as an estimator of
ˆ
E [y ] in a regression which does not use the correct X matrix. For this we assume
that y = η +ε with ε ∼ (o, σ 2 I ). η = E [y ] is an arbitrary vector of constants, and we
do not assume that η = Xβ for some β , i.e., we do not assume that X contains all
the necessary explanatory variables. Regression of y on X gives the OLS estimator
ˆ
β = (X X )− X y .
ˆ
• a. 2 points Show that the MSE matrix of y = X β as estimator of η is
ˆ
ˆ
(18.5.1)
MSE [X β ; η ] = σ 2 X (X X )− X + M ηη M
where M = I − X (X X )− X .
• b. 1 point Formula (18.5.1) for the MSE matrix depends on the unknown σ 2
and η and is therefore useless for estimation. If one cannot get an estimate of the
whole MSE matrix, an oftenused second best choice is its trace. Show that
ˆ
(18.5.2)
tr MSE [X β ; η ] = σ 2 q + η M η .
where q is the rank of X .
• c. 3 points If an unbiased estimator of the true σ 2 is available (call it s2 ), then
an unbiased estimator of the righthand side of (18.5.2) can be constructed using this
s2 and the SSE of the regression SSE = y M y . Show that
(18.5.3) E[SSE − (n − 2q )s2 ] = σ 2 q + η M η . Hint: use equation (??). If one does not have an unbiased estimator s2 of σ 2 , one
usually gets such an estimator by regressing y on an X matrix which is so large that
one can assume that it contains the true regressors.
The statistic
SSE
+ 2q − n
s2
ˆ
is called Mallow’s Cp statistic. It is a consistent estimator of tr MSE [X β ; η ]/σ 2 . If
X contains all necessary variables, i.e., η = Xβ for some β , then (18.5.2) becomes
ˆ
tr MSE [X β ; η ] = σ 2 q , i.e., in this case Cp should be close to q . Therefore the
selection rule for regressions should be here to pick that regression for which the
Cp value is closest to q . (This is an explanation; nothing to prove here.)
(18.5.4) Cp = If one therefore has several regressions and tries to decide which is the right one,
it is recommended to plot Cp versus q for all regressions, and choose one for which
this value is small and lies close to the diagonal. An example of this is given in
problem 232. CHAPTER 19 Nonspherical Positive Deﬁnite Covariance Matrix
The socalled “Generalized Least Squares” model speciﬁes y = Xβ + ε with
ε ∼ (o, σ 2 Ψ) where σ 2 is an unknown positive scalar, and Ψ is a known positive
deﬁnite matrix.
This is simply the OLS model in disguise. To see this, we need a few more
facts about positive deﬁnite matrices. Ψ is nonnegative deﬁnite if and only if a Q
exists with Ψ = QQ . If Ψ is positive deﬁnite, this Q can be chosen square and
nonsingular. Then P = Q−1 satisﬁes P P Ψ = P P QQ = I , i.e., P P = Ψ−1 ,
and also P ΨP = P QQ P = I . Premultiplying the GLS model by P gives
therefore a model whose disturbances have a spherical covariance matrix:
P y = P Xβ + P ε (19.0.5) P ε ∼ (o, σ 2 I ) The OLS estimate of β in this transformed model is
(19.0.6) ˆ
β = (X P P X )−1 X P P y = (X Ψ−1 X )−1 X Ψ−1 y . ˆ
This β is the BLUE of β in model (19.0.5), and since estimators which are linear
ˆ
in P y are also linear in y and vice versa, β is also the BLUE in the original GLS
model.
Problem 263. 2 points Show that
ˆ
β − β = (X Ψ−1 X )−1 X Ψ−1ε (19.0.7) ˆ
ˆ
and derive from this that β is unbiased and that MSE [β ; β ] = σ 2 (X Ψ−1 X )−1 .
Answer. Proof of (19.0.7) is very similar to proof of (18.0.7). The objective function of the associated least squares problem is
(19.0.8) ˆ
β=β (y − Xβ ) Ψ−1 (y − Xβ ). minimizes The normal equations are
(19.0.9) ˆ
X Ψ−1 X β = X Ψ−1 y ˆ
If X has full rank, then X Ψ−1 X is nonsingular, and the unique β minimizing
(19.0.8) is
(19.0.10) ˆ
β = (X Ψ−1 X )−1 X Ψ−1 y Problem 264. [Seb77, p. 386, 5] Show that if Ψ is positive deﬁnite and X has
full rank, then also X Ψ−1 X is positive deﬁnite. You are allowed to use, without
proof, that the inverse of a positive deﬁnite matrix is also positive deﬁnite.
Answer. From X Ψ−1 Xa = o follows a X Ψ−1 Xa = 0, and since Ψ−1 is positive deﬁnite, it follows Xa = o, and since X has full column rank, this implies a = o.
221 222 19. NONSPHERICAL COVARIANCE MATRIX ˆ
The least squares objective function of the transformed model, which β = β
minimizes, can be written
(19.0.11) (P y − P Xβ ) (P y − P Xβ ) = (y − Xβ ) Ψ−1 (y − Xβ ), and whether one writes it in one form or the other, 1/(n − k ) times the minimum
value of that GLS objective function is still an unbiased estimate of σ 2 .
Problem 265. Show that the minimum value of the GLS objective function can
be written in the form y M y where M = Ψ−1 − Ψ−1 X (X Ψ−1 X )−1 X Ψ−1 .
Does M X = O still hold? Does M 2 = M or a similar simple identity still hold?
Show that M is nonnegative deﬁnite. Show that E[y M y ] = (n − k )σ 2 .
ˆ
ˆ
ˆ
Answer. In (y − X β ) Ψ−1 (y − X β ) plug in β = (X Ψ−1 X )−1 X Ψ−1 y and multiply out
to get y M y . Yes, M X = O holds. M is no longer idempotent, but it satisﬁes M ΨM = M .
One way to show that it is nnd would be to use the ﬁrst part of the question: for all z , z M z =
ˆ
ˆ
(z − X β ) (z − X β ), and another way would be to use the second part of the question: M nnd
because M ΨM = M . To show expected value, show ﬁrst that y M y = εM ε, and then use those
tricks with the trace again. The simplest example of Generalized Least Squares is that where Ψ is diagonal
(heteroskedastic data). In this case, the GLS objective function (y − Xβ ) Ψ−1 (y −
Xβ ) is simply a weighted least squares, with the weights being the inverses of the
diagonal elements of Ψ. This vector of inverse diagonal elements can be speciﬁed
with the optional weights argument in R, see the helpﬁle for lm. Heteroskedastic
data arise for instance when each data point is an average over a diﬀerent number
of individuals.
If one runs OLS on the original instead of the transformed model, one gets an
ˆ
estimator, we will calle it here β OLS , which is still unbiased. The estimator is usually
also consistent, but no longer BLUE. This not only makes it less eﬃcient than the
GLS, but one also gets the wrong results if one relies on the standard computer
printouts for signiﬁcance tests etc. The estimate of σ 2 generated by this regression
is now usually biased. How biased it is depends on the X matrix, but most often
it seems biased upwards. The estimated standard errors in the regression printouts
not only use the wrong s, but they also insert this wrong s into the wrong formula
ˆ
σ 2 (X X )−1 instead of σ 2 (X Ψ−1 X )−1 for V [β ].
Problem 266. In the generalized least squares model y = Xβ + ε with ε ∼
(o, σ 2 Ψ), the BLUE is
(19.0.12) ˆ
β = (X Ψ−1 X )−1 X Ψ−1 y . ˆ
We will write β OLS for the ordinary least squares estimator
(19.0.13) ˆ
β OLS = (X X )−1 X y which has diﬀerent properties now since we do not assume ε ∼ (o, σ 2 I ) but ε ∼
(o, σ 2 Ψ).
ˆ
• a. 1 point Is β OLS unbiased?
ˆ
• b. 2 points Show that, still under the assumption ε ∼ (o, σ 2 Ψ), V [β OLS ] −
ˆ
ˆ
ˆ
V [β ] = V [β OLS − β ]. (Write down the formulas for the left hand side and the right
hand side and then show by matrix algebra that they are equal.) (This is what one
should expect after Problem 170.) Since due to unbiasedness the covariance matrices
ˆ
ˆ
are the MSE matrices, this shows that MSE [β OLS ; β ] − MSE [β ; β ] is nonnegative
deﬁnite. 19. NONSPHERICAL COVARIANCE MATRIX 223 Answer. Verify equality of the following two expressions for the diﬀerences in MSE matrices:
2
−1
−1
−1
−1
ˆ
ˆ
=
V [β OLS ] − V [β ] = σ (X X ) X ΨX (X X ) − (X Ψ X ) = σ 2 (X X )−1 X − (X Ψ−1 X )−1 X Ψ−1 Ψ X (X X )−1 − Ψ−1 X (X Ψ−1 X )−1 Examples of GLS models are discussed in chapters ?? and ??. CHAPTER 20 Best Linear Prediction
Best Linear Prediction is the second basic building block for the linear model,
in addition to the OLS model. Instead of estimating a nonrandom parameter β
about which no prior information is available, in the present situation one predicts
a random variable z whose mean and covariance matrix are known. Most models to
be discussed below are somewhere between these two extremes.
Christensen’s [Chr87] is one of the few textbooks which treat best linear prediction on the basis of known ﬁrst and second moments in parallel with the regression
model. The two models have indeed so much in common that they should be treated
together.
20.1. Minimum Mean Squared Error, Unbiasedness Not Required
Assume the expected values of the random vectors y and z are known, and their
joint covariance matrix is known up to an unknown scalar factor σ 2 > 0. We will
write this as
y
Ω yz
µ
Ω
(20.1.1)
∼
, σ 2 yy
,
σ 2 > 0.
z
Ω zy Ω zz
ν
y is observed but z is not, and the goal is to predict z on the basis of the observation
of y .
There is a unique predictor of the form z ∗ = B ∗ y + b∗ (i.e., it is linear with a constant term, the technical term for this is “aﬃne”) with the following two properties:
it is unbiased, and the prediction error is uncorrelated with y , i.e.,
(20.1.2) ∗
C [z − z , y ] = O . The formulas for B ∗ and b∗ are easily derived. Unbiasedness means ν = B ∗ µ + b∗ ,
the predictor has therefore the form
(20.1.3) z ∗ = ν + B ∗ (y − µ). Since
(20.1.4) z ∗ − z = B ∗ (y − µ) − (z − ν ) = B ∗ −I y−µ
,
z−ν the zero correlation condition (20.1.2) translates into
(20.1.5) B ∗Ω y y = Ω z y , which, due to equation (A.5.13) holds for B ∗ = Ω zy Ω −y . Therefore the predictor
y
(20.1.6) z ∗ = ν + Ω zy Ω −y (y − µ)
y satisﬁes the two requirements.
Unbiasedness and condition (20.1.2) are sometimes interpreted to mean that z ∗
is an optimal predictor. Unbiasedness is often naively (but erroneously) considered
to be a necessary condition for good estimators. And if the prediction error were
correlated with the observed variable, the argument goes, then it would be possible to
225 226 20. BEST LINEAR PREDICTION improve the prediction. Theorem 20.1.1 shows that despite the ﬂaws in the argument,
the result which it purports to show is indeed valid: z ∗ has the minimum MSE of
all aﬃne predictors, whether biased or not, of z on the basis of y .
Theorem 20.1.1. In situation (20.1.1), the predictor (20.1.6) has, among all
predictors of z which are aﬃne functions of y , the smallest MSE matrix. Its MSE
matrix is
Ω
(20.1.7) MSE [z ∗ ; z ] = E [(z ∗ − z )(z ∗ − z ) ] = σ 2 (Ω zz − Ω zy Ω −y Ω yz ) = σ 2Ω zz.y .
y
˜
˜
˜
˜
Proof. Look at any predictor of the form z = B y + b. Its bias is d = E [˜ − z ] =
z
˜
˜
Bµ + b − ν , and by (17.1.2) one can write
˜˜
z
z
z
E [(˜ − z )(˜ − z ) ] = V [(˜ − z )] + dd (20.1.8)
(20.1.9) ˜
=V B −I y
z (20.1.10) ˜
= σ2 B −I Ω yy
Ω zy ˜˜
+ dd
Ω yz
Ω zz ˜
B
˜˜
+ dd .
−I This MSE matrix is minimized if and only if d∗ = o and B ∗ satisﬁes (20.1.5). To see
˜
˜
this, take any solution B ∗ of (20.1.5), and write B = B ∗ + D . Since, due to theorem
−
−
∗
A.5.11, Ω zy = Ω zy Ω yy Ω yy , it follows Ω zy B
= Ω zy Ω yy Ω yy B ∗ = Ω zy Ω −y Ω yz .
y
Therefore
Ω yz
Ω zz ˜
B∗ + D
−I ˜
MSE [˜ ; z ] = σ 2 B ∗ + D
z −I Ω yy
Ω zy (20.1.11) ˜
= σ2 B ∗ + D −I ˜
Ω yy D
˜
−Ω zz.y + Ω zy D (20.1.12) ˜˜
˜
˜
Ω
= σ 2 (Ω zz.y + DΩ yy D ) + dd . ˜˜
+ dd ˜˜
+ dd The MSE matrix is therefore minimized (with minimum value σ 2Ω zz.y ) if and only
˜
˜
˜
if d = o and DΩ yy = O which means that B , along with B ∗ , satisﬁes (20.1.5).
Problem 267. Show that the solution of this minimum MSE problem is unique
in the following sense: if B ∗ and B ∗ are two diﬀerent solutions of (20.1.5) and y
1
2
is any feasible observed value y , plugged into equations (20.1.3) they will lead to the
same predicted value z ∗ .
Answer. Comes from the fact that every feasible observed value of y can be written in the
form y = µ + Ω yy q for some q , therefore B ∗ y = B ∗Ω yy q = Ω zy q .
i
i The matrix B ∗ is also called the regression matrix of z on y , and the unscaled
covariance matrix has the form
Ω yy Ω yz
Ω yy
Ω yy X
(20.1.13)
Ω=
=
Ω zy Ω zz
X Ω y y X Ω y y X + Ω z z .y
Where we wrote here B ∗ = X in order to make the analogy with regression clearer.
A ginverse is
(20.1.14) Ω− = −
Ω −y + X Ω zz.y X
y
−X Ω −z.y
z −X Ω −z.y
z
−
Ω z z .y and every ginverse of the covariance matrix has a ginverse of Ω zz.y as its z z partition. (Proof in Problem 392.) 20.1. MINIMUM MEAN SQUARED ERROR, UNBIASEDNESS NOT REQUIRED 227 Ω yy Ω yz
Ω
is nonsingular, 20.1.5 is also solved by B ∗ = −(Ω zz )−Ω zy
Ω zy Ω zz
zz
zy
−1
where Ω and Ω are the corresponding partitions of the inverse Ω . See Problem
392 for a proof. Therefore instead of 20.1.6 the predictor can also be written
If Ω = (20.1.15) z ∗ = ν − Ω zz −1 Ω zy (y − µ) (note the minus sign) or
(20.1.16) z ∗ = ν − Ω zz.y Ω zy (y − µ). Problem 268. This problem utilizes the concept of a bounded risk estimator,
which is not yet explained very well in these notes. Assume y , z , µ, and ν are
jointly distributed random vectors. First assume ν and µ are observed, but y and z
are not. Assume we know that in this case, the best linear bounded MSE predictor
of y and z is µ and ν , with prediction errors distributed as follows:
(20.1.17) y−µ
o
Ω
∼
, σ 2 yy
z−ν
Ω zy
o Ω yz
.
Ω zz This is the initial information. Here it is unnecessary to specify the unconditional
distributions of µ and ν , i.e., E [µ] and E [ν ] as well as the joint covariance matrix
of µ and ν are not needed, even if they are known.
Then in a second step assume that an observation of y becomes available, i.e.,
now y , ν , and µ are observed, but z still isn’t. Then the predictor
(20.1.18) z ∗ = ν + Ω zy Ω −y (y − µ)
y is the best linear bounded MSE predictor of z based on y , µ, and ν .
• a. Give special cases of this speciﬁcation in which µ and ν are constant and y
and z random, and one in which µ and ν and y are random and z is constant, and
one in which µ and ν are random and y and z are constant.
Answer. If µ and ν are constant, they are written µ and ν . From this follows µ = E [y ] and
Ω yy Ω yz
y
= V[ and every linear predictor has bounded MSE . Then the
ν = E [z ] and σ 2
Ω zy Ω zz
rx
proof is as given earlier in this chapter. But an example in which µ and ν are not known constants
but are observed random variables, and y is also a random variable but z is constant, is (21.0.26).
Another example, in which y and z both are constants and µ and ν random, is constrained least
squares (22.4.3). • b. Prove equation 20.1.18.
Answer. In this proof we allow all four µ and ν and y and z to be random. A linear
˜
˜
predictor based on y , µ, and ν can be written as z = B y + C µ + D ν + d, therefore z − z =
B (y − µ) + (C + B )µ + (D − I )ν − (z − ν ) + d. E [˜ − z ] = o + (C + B ) E [µ] + (D − I ) E [ν ] − o + d.
z
Assuming that E [µ] and E [ν ] can be anything, the requirement of bounded MSE (or simply the
requirement of unbiasedness, but this is not as elegant) gives C = −B and D = I , therefore
˜
˜
z = ν + B (y − µ) + d, and the estimation error is z − z = B (y − µ) − (z − ν ) + d. Now continue
as in the proof of theorem 20.1.1. I must still carry out this proof much more carefully! Problem 269. 4 points According to (20.1.2), the prediction error z ∗ − z is
uncorrelated with y . If the distribution is such that the prediction error is even
independent of y (as is the case if y and z are jointly normal), then z ∗ as deﬁned
in (20.1.6) is the conditional mean z ∗ = E [z y ], and its MSE matrix as deﬁned in
(20.1.7) is the conditional variance V [z y ]. 228 20. BEST LINEAR PREDICTION Answer. From independence follows E [z ∗ − z y ] = E [z ∗ − z ], and by the law of iterated
expectations E [z ∗ − z ] = o. Rewrite this as E [z y ] = E [z ∗ y ]. But since z ∗ is a function of y ,
E [z ∗ y ] = z ∗ . Now the proof that the conditional dispersion matrix is the MSE matrix:
∗
∗
V [z y ] = E [(z − E [z y ])(z − E [z y ]) y ] = E [(z − z )(z − z ) y ] (20.1.19) = E [(z − z ∗ )(z − z ∗ ) ] = MSE [z ∗ ; z ]. Problem 270. Assume the expected values of x, y and z are known, and their
joint covariance matrix is known up to an unknown scalar factor σ 2 > 0. Ω xx
x
λ
y ∼ µ , σ 2 Ω xy
z
ν
Ω xz (20.1.20) Ω xy
Ω yy
Ω yz Ω xz
Ω yz .
Ω zz x is the original information, y is additional information which becomes available,
and z is the variable which we want to predict on the basis of this information.
• a. 2 points Show that y ∗ = µ + Ω xy Ω −x (x − λ) is the best linear predictor
x
−
of y and z ∗ = ν + Ω xz Ω xx (x − λ) the best linear predictor of z on the basis of the
observation of x, and that their joint MSE matrix is
y∗ − y
z∗ − z (y ∗ − y ) = σ2 (z ∗ − z ) Ω yy − Ω xy Ω −xΩ xy
x
Ω yz − Ω xz Ω −xΩ xy
x = σ2 E Ω y y .x
Ω y z .x Ω yz − Ω xy Ω −xΩ xz
x
Ω zz − Ω xz Ω −xΩ xz
x which can also be written
Ω y z .x
.
Ω z z .x Answer. This part of the question is a simple application of the formulas derived earlier. For
the MSE matrix you ﬁrst get
σ2 Ω yy
Ω yz Ω yz
Ω xy
−
Ω −x Ω xy
x
Ω zz
Ω xz Ω xz • b. 5 points Show that the best linear predictor of z on the basis of the observations of x and y has the form
z ∗∗ = z ∗ + Ω yz.xΩ −y.x (y − y ∗ )
y (20.1.21) This is an important formula. All you need to compute z ∗∗ is the best estimate
z ∗ before the new information y became available, the best estimate y ∗ of that new
information itself, and the joint MSE matrix of the two. The original data x and
the covariance matrix (20.1.20) do not enter this formula.
Answer. Follows from
z ∗∗ = ν + Ωxz Ωyz Ω xx
Ω xy Ω xy
Ω yy − x−λ
=
y−µ 20.1. MINIMUM MEAN SQUARED ERROR, UNBIASEDNESS NOT REQUIRED 229 Now apply (A.8.2):
= ν + Ω xz Ω yz −
Ω −x + Ω −xΩ xy Ω −y.xΩ xy Ω xx
x
x
y
−
−
−Ω yy.xΩ xy Ω xx −
−Ω −xΩ xy Ω yy.x
x
−
Ω yy.x = ν + Ω xz Ω yz Ω −x (x − λ) + Ω −xΩ xy Ω −y.x (y ∗ − µ) − Ω −xΩ xy Ω −y.x (y − µ)
x
x
y
x
y
=
−
−Ω −y.x (y ∗ − µ) + Ω yy.x (y − µ)
y = ν + Ω xz Ω yz Ω −x (x − λ) − Ω −xΩ xy Ω −y.x (y − y ∗ )
x
x
y
=
Ω−
+Ω yy.x (y − y ∗ ) x−λ
=
y−µ = ν + Ω xz Ω −x (x − λ) − Ω xz Ω −xΩ xy Ω −y.x (y − y ∗ ) + Ω yz Ω −y.x (y − y ∗ ) =
x
x
y
y
= z ∗ + Ω yz − Ω xz Ω −xΩ xy Ω −y.x (y − y ∗ ) = z ∗ + Ω yz.xΩ −y.x (y − y ∗ )
x
y
y Problem 271. Assume x, y , and z have a joint probability distribution, and
the conditional expectation E [z x, y ] = α∗ + A∗ x + B ∗ y is linear in x and y .
• a. 1 point Show that E [z x] = α∗ + A∗ x + B ∗ E [y x]. Hint: you may use the
law of iterated expectations in the following form: E [z x] = E E [z x, y ] x .
Answer. With this hint it is trivial: E [z x] = E α∗ + A∗ x + B ∗ y x = α∗ + A∗ x + B ∗ E [y x]. • b. 1 point The next three examples are from [CW99, pp. 264/5]: Assume
E[z x, y ] = 1 + 2x + 3y , x and y are independent, and E[y ] = 2. Compute E[z x].
Answer. According to the formula, E[z x] = 1 + 2x + 3E[y x], but since x and y are independent, E[y x] = E[y ] = 2; therefore E[z x] = 7 + 2x. I.e., the slope is the same, but the intercept
changes. • c. 1 point Assume again E[z x, y ] = 1 + 2x + 3y , but this time x and y are not
independent but E[y x] = 2 − x. Compute E[z x].
Answer. E[z x] = 1 + 2x + 3(2 − x) = 7 − x. In this situation, both slope and intercept change,
but it is still a linear relationship. • d. 1 point Again E[z x, y ] = 1 + 2x + 3y , and this time the relationship between
x and y is nonlinear: E[y x] = 2 − ex . Compute E[z x].
Answer. E[z x] = 1 + 2x + 3(2 − ex ) = 7 + 2x − 3ex . This time the marginal relationship
between x and y is no longer linear. This is so despite the fact that, if all the variables are included,
i.e., if both x and y are included, then the relationship is linear. • e. 1 point Assume E[f (z )x, y ] = 1 + 2x + 3y , where f is a nonlinear function,
and E[y x] = 2 − x. Compute E[f (z )x].
Answer. E[f (z )x] = 1 + 2x + 3(2 − x) = 7 − x. If one plots z against x and z , then the plots
should be similar, though not identical, since the same transformation f will straighten them out.
This is why the plots in the top row or right column of [CW99, p. 435] are so similar. Connection between prediction and inverse prediction: If y is observed and z
is to be predicted, the BLUP is z ∗ − ν = B ∗ (y − µ) where B ∗ = Ω zy Ω −y . If z
y
is observed and y is to be predicted, then the BLUP is y ∗ − µ = C ∗ (z − ν ) with
C ∗ = Ω yz Ω −z . B ∗ and C ∗ are connected by the formula
z
(20.1.22) Ω yy B ∗ = C ∗Ω z z . This relationship can be used for graphical regression methods [Coo98, pp. 187/8]:
If z is a scalar, it is much easier to determine the elements of C ∗ than those of
B ∗ . C ∗ consists of the regression slopes in the scatter plot of each of the observed
variables against z . They can be read oﬀ easily from a scatterplot matrix. This 230 20. BEST LINEAR PREDICTION works not only if the distribution is Normal, but also with arbitrary distributions as
long as all conditional expectations between the explanatory variables are linear.
Problem 272. In order to make relationship (20.1.22) more intuitive, assume x
and ε are Normally distributed and independent of each other, and E[ε] = 0. Deﬁne
y = α + β x + ε.
• a. Show that α + β x is the best linear predictor of y based on the observation
of x.
Answer. Follows from the fact that the predictor is unbiased and the prediction error is
uncorrelated with x. • b. Express β in terms of the variances and covariances of x and y .
Answer. cov[x, y ] = β var[x], therefore β = cov[x,y ]
var[x] • c. Since x and y are jointly normal, they can also be written x = γ + δ y + ω
where ω is independent of y . Express δ in terms of the variances and covariances of
x and y , and show that var[y ]β = γ var[x].
Answer. δ = cov[x,y ]
.
var[y ] • d. Now let us extend the model a little: assume x1 , x2 , and ε are Normally
distributed and independent of each other, and E[ε] = 0. Deﬁne y = α + β1 x1 +
β2 x2 + ε. Again express β1 and β2 in terms of variances and covariances of x1 , x2 ,
and y .
Answer. Since x1 and x2 are independent, one gets the same formulas as in the univariate
cov[x1 ,y ]
case: from cov[x1 , y ] = β1 var[x1 ] and cov[x2 , y ] = β2 var[x2 ] follows β1 = var[x ] and β2 =
1 cov[x2 ,y ]
.
var[x2 ] • e. Since x1 and y are jointly normal, they can also be written x1 = γ1 +δ1 y +ω 1 ,
where ω 1 is independent of y . Likewise, x2 = γ2 + δ2 y + ω 2 , where ω 2 is independent
of y . Express δ1 and δ2 in terms of the variances and covariances of x1 , x2 , and y ,
and show that
0
δ1
var[x1 ]
var[y ] =
0
var[x2 ]
δ2 (20.1.23) β1
β2 This is (20.1.22) in the present situation.
Answer. δ1 = cov[x1 ,y ]
var[y ] and δ2 = cov[x2 ,y ]
.
var[y ] 20.2. The Associated Least Squares Problem
For every estimation problem there is an associated “least squares” problem. In
the present situation, z ∗ is that value which, together with the given observation y ,
“blends best” into the population deﬁned by µ, ν and the dispersion matrix Ω , in
−
the following sense: Given the observed value y , the vector z ∗ = ν + Ω zy Ω yy (y − µ)
y
is that value z for which
has smallest Mahalanobis distance from the population
z
µ
Ω
Ω yz
deﬁned by the mean vector
and the covariance matrix σ 2 yy
.
ν
Ω zy Ω zz
In the case of singular Ω zz , it is only necessary to minimize among those z
which have ﬁnite distance from the population, i.e., which can be written in the form 20.3. PREDICTION OF FUTURE OBSERVATIONS IN THE REGRESSION MODEL z = ν + Ω zz q for some q . We will also write r = rank
solves the following “least squares problem:”
(20.2.1)
z = z ∗ min. 1
rσ 2 y−µ
z−ν Ω yy
Ω zy Ω yz
Ω zz − Ωyy Ωyz
Ω zy Ω zz 231 . Therefore, z ∗ y−µ
s. t. z = ν + Ω zz q for some q .
z−ν To prove this, use (A.8.2) to invert the dispersion matrix:
Ω yy
Ω zy (20.2.2) Ω yz
Ω zz − = Ω −y + Ω −y Ω yz Ω −z.y Ω zy Ω −y
y
y
z
y
−
−Ω −z.y Ω zy Ω yy
z −Ω −y Ω yz Ω −z.y
y
z
.
−
Ω z z .y If one plugs z = z ∗ into this objective function, one obtains a very simple expression:
(20.2.3)
(y −µ) I Ω −y Ω yz
y Ω −y + Ω −y Ω yz Ω −z.y Ω zy Ω −y
y
y
z
y
−Ω −z.y Ω zy Ω −y
z
y −
−Ω yy Ω yz Ω −z.y
z
Ω −z.y
z I
(y −µ) =
−
Ω zyΩ yy −
= (y − µ) Ω yy (y − µ). (20.2.4) Now take any z of the form z = ν + Ωzz q for some q and write it in the form
z = z ∗ + Ωzz d, i.e.,
y−µ
y−µ
o
=∗
+
.
z−ν
z −ν
Ω zz d
Then the cross product terms in the objective function disappear:
(20.2.5)
o d Ω zz Ω −y + Ω −y Ω yz Ω −z.y Ω zy Ω −y
y
y
z
y
−
−Ω zz.y Ω zy Ω −y
y −Ω −y Ω yz Ω −z.y
y
z
Ω −z.y
z =o d Ω zz I
(y −µ) =
Ω zy Ω −y
y Ω −y
y
(y − µ) = 0
O Therefore this gives a larger value of the objective function.
Problem 273. Use problem 379 for an alternative proof of this.
From (20.2.1) follows that z ∗ is the mode of the normal density function, and
since the mode is the mean, this is an alternative proof, in the case of nonsingular
covariance matrix, when the density exists, that z ∗ is the normal conditional mean.
20.3. Prediction of Future Observations in the Regression Model
For a moment let us go back to the model y = Xβ +ε with spherically distributed
disturbances ε ∼ (o, σ 2 I ). This time, our goal is not to estimate β , but the situation
is the following: For a new set of observations of the explanatory variables X 0 the
values of the dependent variable y 0 = X 0 β + ε 0 have not yet been observed and we
ˆ
want to predict them. The obvious predictor is y ∗ = X 0 β = X 0 (X X )−1 X y .
0
Since
(20.3.1) y ∗ − y 0 = X 0 (X X )−1 X y − y 0 =
0 = X 0 (X X )−1 X X β +X 0 (X X )−1 X ε −X 0 β −ε 0 = X 0 (X X )−1 X ε −ε 0 232 20. BEST LINEAR PREDICTION one sees that E[y ∗ − y 0 ] = o, i.e., it is an unbiased predictor. And since ε and ε 0
0
are uncorrelated, one obtains
(20.3.2) MSE [y ∗ ; y 0 ] = V [y ∗ − y 0 ] = V [X 0 (X X )−1 X ε ] + V [ε 0 ]
0
0
= σ 2 (X 0 (X X )−1 X 0 + I ). (20.3.3) Problem 274 shows that this is the Best Linear Unbiased Predictor (BLUP) of y 0 on
the basis of y .
Problem 274. The prediction problem in the Ordinary Least Squares model can
be formulated as follows:
(20.3.4) y
X
ε
=
β+
y0
X0
ε0 ε
o
E[ ε ] = o
0 ε
2I
V[ ε ] = σ O
0 O
.
I X and X 0 are known, y is observed, y 0 is not observed.
ˆ
• a. 4 points Show that y ∗ = X 0 β is the Best Linear Unbiased Predictor (BLUP)
0
ˆ
of y 0 on the basis of y , where β is the OLS estimate in the model y = Xβ + ε .
˜
˜
˜
Answer. Take any other predictor y 0 = B y and write B = X 0 (X X )−1 X + D . Unbiasedy
ness means E [˜ 0 − y 0 ] = X 0 (X X )−1 X X β + DXβ − X 0 β = o, from which follows DX = O .
y
y
Because of unbiasedness we know MSE [˜ 0 ; y 0 ] = V [˜ 0 − y 0 ]. Since the prediction error can be
˜
written y 0 − y = X 0 (X X )−1 X y
V [˜ 0 − y 0 ] = X 0 (X X )−1 X y
, one obtains
y0 −I +D +D −I V [ y
X (X X )−1 X 0 + D
y0
−I
X (X X )1 X 0 + D
−I = σ 2 X 0 (X X )−1 X +D −I = σ 2 X 0 ( X X ) −1 X +D X 0 (X X )−1 X = σ 2 X 0 (X X )−1 X 0 + DD +D + σ2 I +I . This is smallest for D = O . • b. 2 points From our formulation of the GaussMarkov theorem in Theorem
ˆ
18.1.1 it is obvious that the same y ∗ = X 0 β is also the Best Linear Unbiased Es0
timator of X 0 β , which is the expected value of y 0 . You are not required to reˆ
prove this here, but you are asked to compute MSE [X 0 β ; X 0 β ] and compare it with
MSE [y ∗ ; y 0 ]. Can you explain the diﬀerence?
0
Answer. Estimation error and MSE are
ˆ
ˆ
X 0 β − X 0 β = X 0 (β − β ) = X 0 (X X )−1 X ε due to (??) ˆ
ˆ
MSE [X 0 β ; X 0 β ] = V [X 0 β − X 0 β ] = V [X 0 (X X )−1 X ε ] = σ 2 X 0 (X X )−1 X 0 .
It diﬀers from the prediction MSE matrix by σ 2 I , which is the uncertainty about the value of the
new disturbance ε 0 about which the data have no information. [Gre97, p. 369] has an enlightening formula showing how the prediction intervals
increase if one goes away from the center of the data.
Now let us look at the prediction problem in the Generalized Least Squares
model
Ψ
C
y
X
ε
ε
o
ε
2
(20.3.5)
=
β+
.
Eε =o
V ε =σ
y0
X0
ε0
C
Ψ0
0
0
X and X 0 are known, y is observed, y 0 is not observed, and we assume Ψ is positive
ˆ
ˆ
deﬁnite. If C = O , the BLUP of y 0 is X 0 β , where β is the BLUE in the model 20.3. PREDICTION OF FUTURE OBSERVATIONS IN THE REGRESSION MODEL 233 y = Xβ + ε . In other words, all new disturbances are simply predicted by zero. If
past and future disturbances are correlated, this predictor is no longer optimal.
In [JHG+ 88, pp. 343–346] it is proved that the best linear unbiased predictor
of y 0 is
ˆ
ˆ
(20.3.6)
y ∗ = X 0 β + C Ψ−1 (y − X β ).
0
ˆ
where β is the generalized least squares estimator of β , and that its MSE matrix
MSE [y ∗ ; y 0 ] is
0
(20.3.7) σ 2 Ψ0 − C Ψ−1 C +(X 0 − C Ψ−1 X )(X Ψ−1 X )−1 (X 0 − X Ψ−1 C ) .
Problem 275. Derive the formula for the MSE matrix from the formula of
the predictor, and compute the joint MSE matrix for the predicted values and the
parameter vector.
Answer. The prediction error is, using (19.0.7),
ˆ
ˆ
y ∗ − y 0 = X 0 β − X 0 β + X 0 β − y 0 + C Ψ−1 (y − Xβ + Xβ − X β )
0 (20.3.8)
(20.3.9) ˆ
ˆ
= X 0 (β − β ) − ε 0 + C Ψ−1 (ε − X (β − β )) (20.3.10) ˆ
= C Ψ−1ε + (X 0 − C Ψ−1 X )(β − β ) − ε 0 (20.3.11) = C Ψ−1 + (X 0 − C Ψ−1 X )(X Ψ−1 X )−1 X Ψ−1 −I ε
ε0 The MSE matrix is therefore E [(y ∗ − y 0 )(y ∗ − y 0 ) ] =
0
0
= σ 2 C Ψ−1 + (X 0 − C Ψ−1 X )(X Ψ−1 X )−1 X Ψ−1 (20.3.12) Ψ
C C
Ψ0 −1 Ψ −1 C+Ψ −1 X (X Ψ −I X )−1 (X 0 − X Ψ−1 C ) −I ˆ
and the joint MSE matrix with the sampling error of the parameter vector β − β is
σ2 (20.3.13)
Ψ
C C
Ψ0 (20.3.14) −1 CΨ C Ψ−1 + (X 0 − C Ψ−1 X )(X Ψ−1 X )−1 X Ψ−1
(X Ψ−1 X )−1 X Ψ−1
Ψ−1 C + Ψ−1 X (X Ψ−1 X )−1 (X 0 − X Ψ−1 C )
−I = σ2 −I
O
Ψ−1 X (X Ψ−1 X )−1
O C Ψ−1 + (X 0 − C Ψ−1 X )(X Ψ−1 X )−1 X Ψ−1
(X Ψ−1 X )−1 X Ψ−1 X (X Ψ−1 X )−1 (X 0 − X Ψ−1 C )
C + C Ψ−1 X (X Ψ−1 X )−1 (X 0 − X Ψ−1 C ) − Ψ0 = −I
O X (X Ψ−1 X )−1
C Ψ−1 X (X Ψ−1 X )−1 If one multiplies this out, one gets
(20.3.15)
Ψ0 − C Ψ−1 C + (X 0 − C Ψ−1 X )(X Ψ−1 X )−1 (X 0 − X Ψ−1 C )
(X Ψ−1 X )−1 (X 0 − X Ψ−1 C ) (X 0 − C Ψ−1 X )(X Ψ−1 X )−1
(X Ψ−1 X )−1 The upper left diagonal element is as claimed in (20.3.7). The strategy of the proof given in ITPE is similar to the strategy used to obtain
the GLS results, namely, to transform the data in such a way that the disturbances
are well behaved. Both data vectors y and y 0 will be transformed, but this transformation must have the following additional property: the transformed y must be
a function of y alone, not of y 0 . Once such a transformation is found, it is easy to
predict the transformed y 0 on the basis of the transformed y , and from this one also
obtains a prediction of y 0 on the basis of y . 234 20. BEST LINEAR PREDICTION Here is some heuristics in order to understand formula (20.3.6). Assume for a
moment that β was known. Then you can apply theorem ?? to the model
Ψ
Xβ
y
∼
, σ2
X 0β
y0
C (20.3.16) C
Ψ0 to get y ∗ = X 0 β + C Ψ−1 (y − Xβ ) as best linear predictor of y 0 on the basis of
0
y . According to theorem ??, its MSE matrix is σ 2 (Ψ0 − C Ψ−1 C ). Since β is
ˆ
ˆ
not known, replace it by β , which gives exactly (20.3.6). This adds MSE [X 0 β +
−1
ˆ); X 0 β +C Ψ−1 (y −Xβ )] to the MSE matrix, which gives (20.3.7).
C Ψ (y −X β
Problem 276. Show that
(20.3.17) ˆ
ˆ
MSE [X 0 β + C Ψ−1 (y − X β ); X 0 β + C Ψ−1 (y − Xβ )] =
= σ 2 (X 0 − C Ψ−1 X )(X Ψ−1 X )−1 (X 0 − X Ψ−1 C ). Answer. What is predicted is a random variable, therefore the MSE matrix is the covariance
ˆ
matrix of the prediction error. The prediction error is (X 0 − C Ψ−1 )(β − β ), its covariance matrix
is therefore σ 2 (X 0 − C Ψ−1 X )(X Ψ−1 X )−1 (X 0 − X Ψ−1 C ). Problem 277. In the following we work with partitioned matrices. Given the
model
(20.3.18) y
X
ε
=
β+
y0
X0
ε0 E[ Ψ
ε
2
V[ ε ] = σ
C
0 ε
o=
ε0
o C
.
Ψ0 X has full rank. y is observed, y 0 is not observed. C is not the null matrix.
ˆ
ˆ
• a. Someone predicts y 0 by y ∗ = X 0 β , where β = (X Ψ−1 X )−1 X Ψ−1 y is
0
the BLUE of β . Is this predictor unbiased?
ˆ
Answer. Yes, since E [y 0 ] = X 0 β , and E [β ] = β . ˆ
• b. Compute the MSE matrix MSE [X 0 β ; y 0 ] of this predictor. Hint: For any
y
matrix B , the diﬀerence B y − y 0 can be written in the form B −I
. Hint:
y0
For an unbiased predictor (or estimator), the MSE matrix is the covariance matrix
of the prediction (or estimation) error.
Answer.
(20.3.19) E[(B y − y 0 )(B y − y 0 ) ] = V [B y − y 0 ] (20.3.20) =V B −I y
y0 (20.3.21) = σ2 B −I Ψ
C (20.3.22) = σ 2 B ΨB −C B C
Ψ0 B
−I − CB + Ψ0 . Now one must use B = X 0 (X Ψ−1 X )−1 X Ψ−1 . One ends up with
(20.3.23)
ˆ
MSE [X 0 β ; y 0 ] = σ 2 X 0 (X Ψ−1 X )−1 X 0 −C Ψ−1 X (X Ψ−1 X )−1 X 0 −X 0 (X Ψ−1 X )−1 X Ψ−1 C +Ψ0 . • c. Compare its MSE matrix with formula (20.3.7). Is the diﬀerence nonnegative deﬁnite? 20.3. PREDICTION OF FUTURE OBSERVATIONS IN THE REGRESSION MODEL 235 Answer. To compare it with the minimum MSE matrix, it can also be written as
(20.3.24)
ˆ
MSE [X 0 β ; y 0 ] = σ 2 Ψ0 +(X 0 −C Ψ−1 X )(X Ψ−1 X )−1 (X 0 −X Ψ−1 C )−C Ψ−1 X (X Ψ−1 X )−1 X Ψ−1 C .
i.e., it exceeds the minimum MSE matrix by C (Ψ−1 − Ψ−1 X (X Ψ−1 X )−1 X Ψ−1 )C . This
is nnd because the matrix in parentheses is M = M ΨM , refer here to Problem 265. CHAPTER 21 Updating of Estimates When More Observations
become Available
The theory of the linear model often deals with pairs of models which are nested
in each other, one model either having more data or more stringent parameter restrictions than the other. We will discuss such nested models in three forms: in
the remainder of the present chapter 21 we will see how estimates must be updated
when more observations become available, in chapter 22 how the imposition of a
linear constraint aﬀects the parameter estimates, and in chapter 23 what happens if
one adds more regressors.
ˆ
Assume you have already computed the BLUE β on the basis of the observations
y = Xβ + ε , and afterwards additional data y 0 = X 0 β + ε 0 become available. Then
ˆ
β can be updated using the following principles:
Before the new observations became available, the information given in the origˆ
inal dataset not only allowed to estimate β by β , but also yielded a prediction
∗
ˆ of the additional data. The estimation error β − β and the prediction
ˆ
y0 = X 0 β
∗
error y 0 − y 0 are unobserved, but we know their expected values (the zero vectors),
and we also know their joint covariance matrix up to the unknown factor σ 2 . After
the additional data have become available, we can compute the actual value of the
prediction error y ∗ − y 0 . This allows us to also get a better idea of the actual value of
0
the estimation error, and therefore we can get a better estimator of β . The following
steps are involved:
(1) Make the best prediction y ∗ of the new data y 0 based on y .
0
(2) Compute the joint covariance matrix of the prediction error y ∗ − y 0 of the
0
new data by the old (which is observed) and the sampling error in the old regression
ˆ
β − β (which is unobserved).
ˆ
(3) Use the formula for best linear prediction (??) to get a predictor z ∗ of β − β .
ˆ
ˆ
ˆ
(4) Then β = β − z ∗ is the BLUE of β based on the joint observations y and
y0 .
(5) The sum of squared errors of the updated model minus that of the basic
model is the standardized prediction error SSE ∗ − SSE = (y ∗ − y 0 ) Ω −1 (y ∗ − y 0 )
0
0
ˆ
ˆ
ˆ
ˆ
where SSE ∗ = (y − X β ) (y − X β ) V [y ∗ − y 0 ] = σ 2Ω .
0
In the case of one additional observation and spherical covariance matrix, this
procedure yields the following formulas:
ˆ
Problem 278. Assume β is the BLUE on the basis of the observation y =
Xβ + ε , and a new observation y 0 = x0 β + ε0 becomes available. Show that the
updated estimator has the form
(21.0.25) ˆˆ
ˆ
β = β + (X X )−1 x0 ˆ
y 0 − x0 β
.
1 + x0 (X X )−1 x0 237 238 21. ADDITIONAL OBSERVATIONS Answer. Set it up as follows:
ˆ
y 0 − x0 β
0
x0 (X X )−1 x0 + 1
∼
, σ2
ˆ
o
(X X )−1 x0
β−β (21.0.26) x0 (X X )−1
(X X )−1 and use (20.1.18). By the way, if the covariance matrix is not spherical but is Ψ
c c
ψ0 we get from (20.3.6)
ˆ
ˆ
y ∗ = x0 β + c Ψ−1 (y − X β )
0 (21.0.27)
and from (20.3.15)
y0 − y∗
0
ˆ
β−β (21.0.28) ∼ 0
, σ2
o ψ0 − c Ψ−1 c + (x0 − c Ψ−1 X )(X Ψ−1 X )−1 (x0 − X Ψ−1 c)
(X Ψ−1 X )−1 (x0 − X Ψ−1 c) (x0 − c Ψ−1 X )(X Ψ−1 X )−1
(X Ψ−1 X )−1 ˆ
• a. Show that the residual ε0 from the full regression is the following nonrandom
ˆ
ˆ
multiple of the “predictive” residual y 0 − x0 β :
1
ˆ
ˆ
ˆ
ˆ
ε0 = y 0 − x0 β =
ˆ
(21.0.29)
(y 0 − x0 β )
1 + x0 (X X )−1 x0
Interestingly, this is the predictive residual divided by its relative variance (to standardize it one would have to divide it by its relative standard deviation). Compare
this with (24.2.9).
Answer. (21.0.29) can either be derived from (21.0.25), or from the following alternative
application of the updating principle: All the information which the old observations have for the
ˆ
estimate of x0 β is contained in y 0 = x0 β . The information which the updated regression, which
ˆ
includes the additional observation, has about x0 β can therefore be represented by the following
two “observations”:
y0
ˆ
y0 (21.0.30) = 1
δ
x β+ 1
10
δ2 δ1
δ2 ∼ 0
x0 (X X )−1 x0
, σ2
0
0 0
1 This is a regression model with two observations and one unknown parameter, x0 β , which has a
nonspherical error covariance matrix. The formula for the BLUE of x0 β in model (21.0.30) is
(21.0.31)
ˆ
y0 =
ˆ
(21.0.32) = (21.0.33) = 1 x0 (X X )−1 x0
0 1
1 1+ 0
1 −1 1
1 y0
ˆ 1
x0 (X X )−1 x0 1
1 + x0 (X X )−1 x0 x0 (X X )−1 x0 −1 1 1 x0 (X X )−1 x0
0 + y0 (y 0 + x0 (X X )−1 x0 y 0 ).
ˆ Now subtract (21.0.33) from y 0 to get (21.0.29). Using (21.0.29), one can write (21.0.25) as
ˆˆ
ˆ
ˆ
β = β + (X X )−1 x0 ε0
ˆ (21.0.34) Later, in (25.4.1), one will see that it can also be written in the form
ˆˆ
ˆ
ˆ
β = β + (Z Z )−1 x0 (y 0 − x0 β ) (21.0.35)
where Z = X
.
x0 0
1 −1 y0
ˆ
y0 21. ADDITIONAL OBSERVATIONS 239 Problem 279. Show the following fact which is point (5) in the above updating
principle in this special case: If one takes the squares of the standardized predictive
residuals, one gets the diﬀerence of the SSE for the regression with and without the
additional observation y 0
ˆ
(y 0 − x0 β )2
SSE ∗ − SSE =
(21.0.36)
1 + x0 (X X )−1 x0
ˆ
ˆ
Answer. The sum of squared errors in the old regression is SSE = (y − X β ) (y − X β );
ˆ
ˆ
∗ = (y − X β ) (y − X β ) + ε 2 . From
ˆ0
ˆ
ˆ
the sum of squared errors in the updated regression is SSE
ˆ
(21.0.34) follows
(21.0.37) ˆ
ˆ
ˆ
ˆ
y − X β = y − X β − X (X X )−1 x0 ε0 .
ˆ ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
If one squares this, the cross product terms fall away: (y − X β ) (y − X β ) = (y − X β ) (y − X β ) +
ˆ0 x (X X )−1 x0 ε0 . Adding ε0 2 to both sides gives SSE ∗ = SSE + ε0 2 (1 + x (X X )−1 x0 ).
ˆ
ˆ
ˆ
ε0
ˆ
ˆ
ˆ
ˆ
0
Now use (21.0.29) to get (21.0.36). CHAPTER 22 Constrained Least Squares
One of the assumptions for the linear model was that nothing is known about
the true value of β . Any k vector γ is a possible candidate for the value of β . We
˜
used this assumption e.g. when we concluded that an unbiased estimator B y of β
˜
must satisfy BX = I . Now we will modify this assumption and assume we know
that the true value β satisﬁes the linear constraint Rβ = u. To ﬁx notation, assume
y be a n × 1 vector, u a i × 1 vector, X a n × k matrix, and R a i × k matrix.
In addition to our usual assumption that all columns of X are linearly independent
(i.e., X has full column rank) we will also make the assumption that all rows of R
are linearly independent (which is called: R has full row rank). In other words, the
matrix of constraints R does not include “redundant” constraints which are linear
combinations of the other constraints.
22.1. Building the Constraint into the Model
Problem 280. Given a regression with a constant term and two explanatory
variables which we will call x and z , i.e.,
(22.1.1) y t = α + βxt + γzt + εt • a. 1 point How will you estimate β and γ if it is known that β = γ ?
Answer. Write
(22.1.2) y t = α + β (xt + zt ) + εt • b. 1 point How will you estimate β and γ if it is known that β + γ = 1?
Answer. Setting γ = 1 − β gives the regression
(22.1.3) y t − zt = α + β (xt − zt ) + εt • c. 3 points Go back to a. If you add the original z as an additional regressor
into the modiﬁed regression incorporating the constraint β = γ , then the coeﬃcient
of z is no longer an estimate of the original γ , but of a new parameter δ which is a
linear combination of α, β , and γ . Compute this linear combination, i.e., express δ
in terms of α, β , and γ . Remark (no proof required): this regression is equivalent to
(22.1.1), and it allows you to test the constraint.
Answer. It you add z as additional regressor into (22.1.2), you get y t = α + β (xt + zt )+ δzt + εt .
Now substitute the right hand side from (22.1.1) for y to get α + βxt + γzt + εt = α + β (xt + zt ) +
δzt + εt . Cancelling out gives γzt = βzt + δzt , in other words, γ = β + δ . In this regression,
therefore, the coeﬃcient of z is split into the sum of two terms, the ﬁrst term is the value it should
be if the constraint were satisﬁed, and the other term is the diﬀerence from that. • d. 2 points Now do the same thing with the modiﬁed regression from part b
which incorporates the constraint β + γ = 1: include the original z as an additional
regressor and determine the meaning of the coeﬃcient of z .
241 242 22. CONSTRAINED LEAST SQUARES What Problem 280 suggests is true in general: every constrained Least Squares
problem can be reduced to an equivalent unconstrained Least Squares problem with
fewer explanatory variables. Indeed, one can consider every least squares problem to
be “constrained” because the assumption E [y ] = Xβ for some β is equivalent to a
linear constraint on E [y ]. The decision not to include certain explanatory variables
in the regression can be considered the decision to set certain elements of β zero,
which is the imposition of a constraint. If one writes a certain regression model as
a constrained version of some other regression model, this simply means that one is
interested in the relationship between two nested regressions.
Problem 219 is another example here.
22.2. Conversion of an Arbitrary Constraint into a Zero Constraint
This section, which is nothing but the matrix version of Problem 280, follows
[DM93, pp. 16–19]. By reordering the elements of β one can write the constraint
Rβ = u in the form
(22.2.1) R1 R2 β1
≡ R1 β 1 + R2 β 2 = u
β2 where R1 is a nonsingular i × i matrix. Why can that be done? The rank of R is i,
i.e., all the rows are linearly independent. Since row rank is equal to column rank,
there are also i linearly independent columns. Use those for R1 . Using this same
partition, the original regression can be written
(22.2.2) y = X 1 β1 + X 2 β2 + ε Now one can solve (22.2.1) for β 1 to get
(22.2.3) β 1 = R−1 u − R−1 R2 β 2
1
1 Plug (22.2.3) into (22.2.2) and rearrange to get a regression which is equivalent to
the constrained regression:
(22.2.4) y − X 1 R−1 u = (X 2 − X 1 R−1 R2 )β 2 + ε
1
1 or
(22.2.5) y∗ = Z 2 β2 + ε One more thing is noteworthy here: if we add X 1 as additional regressors into
(22.2.5), we get a regression that is equivalent to (22.2.2). To see this, deﬁne the
diﬀerence between the left hand side and right hand side of (22.2.3) as γ 1 = β 1 −
R−1 u + R−1 R2 β 2 ; then the constraint (22.2.1) is equivalent to the “zero constraint”
1
1
γ 1 = o, and the regression
(22.2.6) y − X 1 R−1 u = (X 2 − X 1 R−1 R2 )β 2 + X 1 (β 1 − R−1 u + R−1 R2 β 2 ) + ε
1
1
1
1
is equivalent to the original regression (22.2.2). (22.2.6) can also be written as
(22.2.7) y∗ = Z 2 β2 + X 1 γ 1 + ε The coeﬃcient of X 1 , if it is added back into (22.2.5), is therefore γ 1 .
Problem 281. [DM93] assert on p. 17, middle, that
(22.2.8) R[X 1 , Z 2 ] = R[X 1 , X 2 ]. where Z 2 = X 2 − X 1 R−1 R2 . Give a proof.
1 22.3. LAGRANGE APPROACH TO CONSTRAINED LEAST SQUARES 243 Answer. We have to show
(22.2.9) {z : z = X 1 γ + X 2 δ } = {z : z = X 1 α + Z 2 β } First ⊂: given γ and δ we need a α and β with
(22.2.10) X 1 γ + X 2 δ = X 1 α + ( X 2 − X 1 R −1 R 2 ) β
1 This can be accomplished with β = δ and α = γ + R−1 R2 δ . The other side is even more trivial:
1
given α and β , multiplying out the right side of (22.2.10) gives X 1 α + X 2 β − X 1 R−1 R2 β , i.e.,
1
δ = β and γ = α − R−1 R2 β .
1 22.3. Lagrange Approach to Constrained Least Squares
ˆ
ˆ
The constrained least squares estimator is that k × 1 vector β = β which minimizes SSE = (y − Xβ ) (y − Xβ ) subject to the linear constraint Rβ = u.
Again, we assume that X has full column and R full row rank.
The Lagrange approach to constrained least squares, which we follow here, is
given in [Gre97, Section 7.3 on pp. 341/2], also [DM93, pp. 90/1]:
The Constrained Least Squares problem can be solved with the help of the
“Lagrange function,” which is a function of the k × 1 vector β and an additional i × 1
vector λ of “Lagrange multipliers”:
(22.3.1) L(β , λ) = (y − Xβ ) (y − Xβ ) + (Rβ − u) λ λ can be considered a vector of “penalties” for violating the constraint. For every
˜
possible value of λ one computes that β = β which minimizes L for that λ (This is
an unconstrained minimization problem.) It will turn out that for one of the values
ˆ
ˆ
ˆ
ˆ
λ = λ∗ , the corresponding β = β satisﬁes the constraint. This β is the solution of
the constrained minimization problem we are looking for.
ˆ
ˆ
Problem 282. 4 points Show the following: If β = β is the unconstrained
minimum argument of the Lagrange function
L(β , λ∗ ) = (y − Xβ ) (y − Xβ ) + (Rβ − u) λ∗
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
for some ﬁxed value λ∗ , and if at the same time β satisﬁes Rβ = u, then β = β
minimizes (y − Xβ ) (y − Xβ ) subject to the constraint Rβ = u.
(22.3.2) ˆ
ˆ
Answer. Since β minimizes the Lagrange function, we know that
ˆ
ˆ
ˆ
˜
˜
˜
ˆ
ˆ
ˆ
(y − X β ) (y − X β ) + (Rβ − u) λ∗ ≥ (y − X β ) (y − X β ) + (Rβ − u) λ∗
(22.3.3)
ˆ
˜
ˆ
for all β . Since by assumption, β also satisﬁes the constraint, this simpliﬁes to:
(22.3.4) ˆ
ˆ
˜
˜
˜
ˆ
ˆ
(y − X β ) (y − X β ) + (Rβ − u) λ∗ ≥ (y − X β ) (y − X β ). ˜
˜
This is still true for all β . If we only look at those β which satisfy the constraint, we get
(22.3.5) ˆ
ˆ
˜
˜
ˆ
ˆ
(y − X β ) ( y − X β ) ≥ (y − X β ) (y − X β ). ˆ
ˆ
This means, β is the constrained minimum argument. Instead of imposing the constraint itself, one imposes a penalty function which
has such a form that the agents will “voluntarily” heed the constraint. This is
a familiar principle in neoclassical economics: instead of restricting pollution to a
certain level, tax the polluters so much that they will voluntarily stay within the
desired level.
ˆ
ˆ
The proof which follows now not only derives the formula for β but also shows
ˆ satisﬁes Rβ = u.
ˆ
∗
ˆ
ˆ
that there is always a λ for which β 244 22. CONSTRAINED LEAST SQUARES Problem 283. 2 points Use the simple matrix diﬀerentiation rules ∂ (w β )/∂ β =
w and ∂ (β M β )/∂ β = 2β M to compute ∂ L/∂ β where
(22.3.6) L(β ) = (y − Xβ ) (y − Xβ ) + (Rβ − u) λ Answer. Write the objective function as y y − 2y X β + β X X β + λ Rβ − λ u to get
(22.3.7). ˆ
ˆ
ˆ
ˆ
Our goal is to ﬁnd a β and a λ∗ so that (a) β = β minimizes L(β , λ∗ ) and (b)
ˆ = u. In other words, β and λ∗ together satisfy the following two conditions: (a)
ˆ
ˆ
ˆ
Rβ
they must satisfy the ﬁrst order condition for the unconstrained minimization of L
ˆ
ˆ
with respect to β , i.e., β must annul
(22.3.7) ∂ L/∂ β = −2y X + 2β X X + λ∗ R, ˆ
ˆ
and (b) β must satisfy the constraint (22.3.9).
(22.3.7) and (22.3.9) are two linear matrix equations which can indeed be solved
ˆ
ˆ
for β and λ∗ . I wrote (22.3.7) as a row vector, because the Jacobian of a scalar
function is a row vector, but it is usually written as a column vector. Since this
conventional notation is arithmetically a little simpler here, we will replace (22.3.7)
with its transpose (22.3.8). Our starting point is therefore
(22.3.8)
(22.3.9) ˆ
ˆ
2X X β = 2X y − R λ∗
ˆ
ˆ
Rβ − u = o Some textbook treatments have an extra factor 2 in front of λ∗ , which makes the
math slightly smoother, but which has the disadvantage that the Lagrange multiplier
can no longer be interpreted as the “shadow price” for violating the constraint.
ˆ
ˆ
ˆ
ˆ
Solve (22.3.8) for β to get that β which minimizes L for any given λ∗ :
(22.3.10) 1
ˆ
ˆ
ˆ1
β = (X X )−1 X y − (X X )−1 R λ∗ = β − (X X )−1 R λ∗
2
2 ˆ
Here β on the right hand side is the unconstrained OLS estimate. Plug this formula
ˆ into (22.3.9) in order to determine that value of λ∗ for which the corresponding
ˆ
for β
ˆ
ˆ
β satisﬁes the constraint:
(22.3.11) ˆ1
Rβ − R(X X )−1 R λ∗ − u = o.
2 Since R has full row rank and X full column rank, R(X X )−1 R
(Problem 284). Therefore one can solve for λ∗ :
(22.3.12) λ∗ = 2 R(X X )−1 R −1 has an inverse ˆ
(Rβ − u) If one substitutes this λ∗ back into (22.3.10), one gets the formula for the constrained
least squares estimator:
(22.3.13) ˆˆ
ˆ
β = β − (X X )−1 R R(X X )−1 R −1 ˆ
(Rβ − u). Problem 284. If R has full row rank and X full column rank, show that
R(X X )−1 R has an inverse.
Answer. Since it is nonnegative deﬁnite we have to show that it is positive deﬁnite. b R(X X )−1 R b =
0 implies b R = o b ecause (X X )−1 is positive deﬁnite, and this implies b = o because R has
full row rank. 22.4. CONSTRAINED LEAST SQUARES AS THE NESTING OF TWO SIMPLER MODELS 245 Problem 285. Assume ε ∼ (o, σ 2 Ψ) with a nonsingular Ψ and show: If one
minimizes SSE = (y − Xβ ) Ψ−1 (y − Xβ ) subject to the linear constraint Rβ = u,
ˆ
ˆ
the formula for the minimum argument β is the following modiﬁcation of (22.3.13):
(22.3.14) ˆˆ
ˆ
β = β − (X Ψ−1 X )−1 R R(X Ψ−1 X )−1 R −1 ˆ
(Rβ − u) ˆ
where β = (X Ψ−1 X )−1 X Ψ−1 y . This formula is given in [JHG+ 88, (11.2.38)
on p. 457]. Remark, which you are not asked to prove: this is the best linear unbiased
estimator if ε ∼ (o, σ 2 Ψ) among all linear estimators which are unbiased whenever
the true β satisﬁes the constraint Rβ = u.)
Answer. Lagrange function is
L(β , λ) = (y − Xβ ) Ψ−1 (y − Xβ ) + (Rβ − u) λ
= y y − 2y Ψ−1 Xβ + β X Ψ−1 Xβ + λ Rβ − λ u
Jacobian is
∂ L/∂ β = −2y Ψ−1 X + 2β X Ψ−1 X + λ R, Transposing and setting it zero gives
ˆ
ˆ
2X Ψ−1 X β = 2X Ψ−1 y − R λ∗ (22.3.15)
ˆ
ˆ
Solve (22.3.15) for β : 1
ˆ
ˆ1
ˆ
(22.3.16) β = (X Ψ−1 X )−1 X Ψ−1 y − (X Ψ−1 X )−1 R λ∗ = β − (X Ψ−1 X )−1 R λ∗
2
2
ˆ
ˆ is the unconstrained GLS estimate. Plug β into the constraint (22.3.9):
ˆ
Here β
ˆ
Rβ − (22.3.17) 1
R(X Ψ−1 X )−1 R λ∗ − u = o.
2 Since R has full row rank and X full column rank and Ψ is nonsingular, R(X Ψ−1 X )−1 R
has an inverse. Therefore
λ∗ = 2 R(X Ψ−1 X )−1 R (22.3.18) −1 still ˆ
(Rβ − u) ∗ Now substitute this λ back into (22.3.16):
(22.3.19) ˆˆ
ˆ
β = β − (X Ψ−1 X )−1 R R(X Ψ−1 X )−1 R −1 ˆ
( R β − u) . 22.4. Constrained Least Squares as the Nesting of Two Simpler Models
The imposition of a constraint can also be considered the addition of new information: a certain linear transformation of β , namely, Rβ , is observed without
error.
ˆ
Problem 286. Assume the random β ∼ (β , σ 2 (X X )−1 ) is unobserved, but
one observes Rβ = u.
• a. 2 points Compute the best linear predictor of β on the basis of the observation
u. Hint: First write down the joint means and covariance matrix of u and β .
Answer.
(22.4.1) u
∼
β −1 R
ˆ
Rβ
2 R (X X )
ˆ ,σ
( X X ) −1 R
β R(X X )−1
( X X ) −1 . Therefore application of formula (??) gives
(22.4.2) ˆ
β ∗ = β + (X X )−1 R R(X X )−1 R −1 ˆ
( u − R β ). 246 22. CONSTRAINED LEAST SQUARES • b. 1 point Look at the formula for the predictor you just derived. Have you
seen this formula before? Describe the situation in which this formula is valid as a
BLUEformula, and compare the situation with the situation here.
Answer. Of course, constrained least squares. But in contrained least squares, β is nonrandom
ˆ
and β is random, while here it is the other way round. In the unconstrained OLS model, i.e., before the “observation” of u = Rβ , the
ˆ
ˆ
best bounded MSE estimators of u and β are Rβ and β , with the sampling errors
having the following means and variances:
(22.4.3) ˆ
u − Rβ
ˆ∼
β−β o
R(X X )−1 R
, σ2
o
(X X )−1 R R(X X )−1
(X X )−1 After the observation of u we can therefore apply (20.1.18) to get exactly equation
ˆ
ˆ
(22.3.13) for β . This is probably the easiest way to derive this equation, but it derives
constrained least squares by the minimization of the MSE matrix, not by the least
squares problem.
22.5. Solution by Quadratic Decomposition
An alternative purely algebraic solution method for this constrained minimization problem rewrites the OLS objective function in such a way that one sees immediately what the constrained minimum value is.
Start with the decomposition (14.2.12) which can be used to show optimality of
the OLS estimate:
ˆ
ˆ
ˆ
ˆ
(y − Xβ ) (y − Xβ ) = (y − X β ) (y − X β ) + (β − β ) X X (β − β ).
ˆˆ
ˆ
Split the second term again, using β − β = (X X )−1 R
u):
ˆ
ˆ
ˆ
ˆ
ˆ
ˆˆ
(β − β ) X X (β − β ) = β − β − (β − β ) R(X X )−1 R −1 ˆ
(Rβ − ˆ
ˆ
ˆ
ˆˆ
X X β − β − (β − β ) ˆ
ˆ
ˆ
ˆ
= (β − β ) X X (β − β )
ˆ
ˆ
− 2(β − β ) X X (X X )−1 R R(X X )−1 R −1 ˆ
(Rβ − u) ˆˆ
ˆ
ˆˆ
ˆ
+ (β − β ) X X (β − β ).
−1 ˆ
The cross product terms can be simpliﬁed to −2(Rβ −u) R(X X )−1 R
(Rβ −
ˆ − u) R(X X )−1 R −1 (Rβ − u). Therefore the
ˆ
u), and the last term is (Rβ
objective function for an arbitrary β can be written as
ˆ
ˆ
(y − Xβ ) (y − Xβ ) = (y − X β ) (y − X β )
ˆ
ˆ
ˆ
ˆ
+ (β − β ) X X (β − β )
− 2(Rβ − u)
ˆ
+ (Rβ − u) R(X X )−1 R
R(X X )−1 R −1
−1 ˆ
(Rβ − u) ˆ
(Rβ − u) The ﬁrst and last terms do not depend on β at all; the third term is zero whenever
ˆ
ˆ
β satisﬁes Rβ = u; and the second term is minimized if and only if β = β , in which
case it also takes the value zero. 22.6. SAMPLING PROPERTIES OF CONSTRAINED LEAST SQUARES 247 22.6. Sampling Properties of Constrained Least Squares
Again, this variant of the least squares principle leads to estimators with desirable
ˆ
ˆ
ˆ
ˆ
sampling properties. Note that β is an aﬃne function of y . We will compute E [β − β ]
ˆ; β ] not only in the case that the true β satisﬁes Rβ = u, but also in
ˆ
and MSE [β
the case that it does not. For this, let us ﬁrst get a suitable representation of the
sampling error:
ˆ
ˆˆ
ˆ
ˆ
ˆ
β − β = (β − β ) + (β − β ) =
(22.6.1) ˆ
= (β − β ) − (X X )−1 R
−(X X )−1 R −1 R(X X )−1 R
R(X X )−1 R −1 ˆ
R(β − β ) (Rβ − u). The last term is zero if β satisﬁes the constraint. Now use (18.0.7) twice to get
(22.6.2) ˆ
ˆ
β − β = W X ε −(X X )−1 R R(X X )−1 R −1 (Rβ − u) where
(22.6.3) W = (X X )−1 − (X X )−1 R −1 R(X X )−1 R R(X X )−1 . ˆ
ˆ
If β satisﬁes the constraint, (22.6.2) simpliﬁes to β − β = W X ε . In this case,
ˆ is unbiased and MSE [β ; β ] = σ 2 W (Problem 287). Since (X X )−1 −
ˆ
ˆ
ˆ
therefore, β
ˆ
ˆ; β ] is smaller than MSE [β ; β ] by a nonnegative
ˆ
W is nonnegative deﬁnite, MSE [β
ˆ
ˆ
ˆ
deﬁnite matrix. This should be expected, since β uses more information than β .
Problem 287.
• a. Show that W X X W = W (i.e., X X is a ginverse of W ).
Answer. This is a tedious matrix multiplication. ˆ
ˆ
• b. Use this to show that MSE [β ; β ] = σ 2 W .
(Without proof:) The GaussMarkov theorem can be extended here as follows:
the constrained least squares estimator is the best linear unbiased estimator among
all linear (or, more precisely, aﬃne) estimators which are unbiased whenever the true
β satisﬁes the constraint Rβ = u. Note that there are more estimators which are
unbiased whenever the true β satisﬁes the constraint than there are estimators which
are unbiased for all β .
ˆ
ˆ
If Rβ = u, then β is biased. Its bias is
(22.6.4) ˆ
−1
ˆ
E [β − β ] = −(X X ) R R(X X )−1 R −1 (Rβ − u). Due to the decomposition (17.1.2) of the MSE matrix into dispersion matrix plus
squared bias, it follows
(22.6.5) ˆ
ˆ
MSE [β ; β ] = σ 2 W +
+ (X X )−1 R R(X X )−1 R
· (Rβ − u) −1 (Rβ − u) · R(X X )−1 R −1 R(X X )−1 . Even if the true parameter does not satisfy the constraint, it is still possible
that the constrained least squares estimator has a better MSE matrix than the 248 22. CONSTRAINED LEAST SQUARES unconstrained one. This is the case if and only if the true parameter values β and
σ 2 satisfy
(22.6.6) (Rβ − u) R(X X )−1 R )−1 (Rβ − u) ≤ σ 2 . This equation, which is the same as [Gre97, (827) on p. 406], is an interesting
result, because the obvious estimate of the lefthand side in (22.6.6) is i times the
value of the F test statistic for the hypothesis Rβ = u. To test for this, one has to
use the noncentral F test with parameters i, n − k , and 1/2.
ˆ
ˆ
Problem 288. 2 points This Problem motivates Equation (22.6.6). If β is a
ˆ = u is also a better estimator of Rβ than
ˆ
ˆ
better estimator of β than β , then Rβ
ˆ
Rβ . Show that this latter condition is not only necessary but already suﬃcient,
ˆ
i.e., if MSE [Rβ ; Rβ ] − MSE [u; Rβ ] is nonnegative deﬁnite then β and σ 2 satisfy
(22.6.6). You are allowed to use, without proof, theorem A.5.9 in the mathematical
Appendix.
Answer. We have to show
σ 2 R(X X )−1 R (22.6.7) is nonnegative deﬁnite. Since Ω =
leads to (22.6.6). σ 2 R (X − (Rβ − u)(Rβ − u) X )−1 R has an inverse, theorem A.5.9 immediately 22.7. Estimation of the Variance in Constrained OLS
Next we will compute the expected value of the minimum value of the constrained
ˆ
ˆˆ
ˆ
ˆ
OLS objective funtion, i.e., E[ε ε] where ε = y − X β , again without necessarily
ˆˆ
ˆ
making the assumption that Rβ = u:
−1
ˆˆ
ˆ
ˆ
ˆ
(22.7.1)
ε = y − X β = ε + X (X X )−1 R R(X X )−1 R
ˆ
(Rβ − u).
Since X ε = o, it follows
ˆ
(22.7.2) ˆˆ ˆˆ
ˆ
ε ε = ε ε + (Rβ − u)
ˆˆ R(X X )−1 R −1 ˆ
(Rβ − u). ˆ
ˆ
Now note that E [Rβ − u] = Rβ − u and V [Rβ − u] = σ 2 R(X X )−1 R . Therefore
−1
−1
R(X X )−1 R
use (??) in theorem ?? and tr R(X X ) R
= i to get
(22.7.3) ˆ
E[(Rβ − u) R(X X )−1 R −1 ˆ
(Rβ − u)] = = σ 2 i+(Rβ − u) R(X X )−1 R −1 (Rβ − u) Since E[ε ε] = σ 2 (n − k ), it follows
ˆˆ
(22.7.4) ˆˆ
E[ε ε] = σ 2 (n + i − k )+(Rβ − u)
ˆˆ R(X X )−1 R −1 (Rβ − u). ˆˆ
In other words, ε ε/(n + i − k ) is an unbiased estimator of σ 2 if the constraint
ˆˆ
holds, and it is biased upwards if the constraint does not hold. The adjustment of
the degrees of freedom is what one should expect: a regression with k explanatory
variables and i constraints can always be rewritten as a regression with k − i diﬀerent
explanatory variables (see Section 22.2), and the distribution of the SSE does not
depend on the values taken by the explanatory variables at all, only on how many
there are. The unbiased estimate of σ 2 is therefore
ˆ
ˆˆ
(22.7.5)
σ 2 = ε ε/(n + i − k )
ˆ
ˆˆ
ˆˆ
Here is some geometric intuition: y = X β + ε is an orthogonal decomposition, since ε is orthogonal to all columns of X . From orthogonality follows y y =
ˆ 22.7. ESTIMATION OF THE VARIANCE IN CONSTRAINED OLS 249 ˆ
ˆ
ˆ ˆˆ
ˆˆ
β X X β + ε ε. If one splits up y = X β + ε, one should expect this to be orthogˆ
onal as well. But this is only the case if u = o. If u = o, one ﬁrst has to shift the
origin of the coordinate system to a point which can be written in the form Xβ 0
where β 0 satisﬁes the constraint:
ˆ
ˆ
Problem 289. 3 points Assume β is the constrained least squares estimate, and
β 0 is any vector satisfying Rβ 0 = u. Show that in the decomposition
ˆ
ˆ
ˆ
y − Xβ 0 = X (β − β 0 ) + ε
ˆ (22.7.6) the two vectors on the righthand side are orthogonal.
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆˆ
ˆ
Answer. We have to show (β − β 0 ) X ε = 0. Since ε = y − X β = y − X β + X (β − β ) =
ˆ
ˆ
ˆ
ˆ − β ), and we already know that X ε = o, it is necessary and suﬃcient to show that
ˆ
ε + X (β
ˆ
ˆ
ˆ
ˆ
ˆˆ
ˆ
(β − β 0 ) X X (β − β ) = 0. By (22.3.13),
ˆ
ˆ
ˆ
ˆˆ
ˆ
ˆ
(β − β 0 ) X X (β − β ) = (β − β 0 ) X X (X X )−1 R
= (u − u) −1 R(X X )−1 R R(X X )−1 R −1 ˆ
( R β − u) ˆ
(Rβ − u) = 0. ˆˆ
If u = o, then one has two orthogonal decompositions: y = y + ε, and y = y + ε.
ˆˆ
ˆˆ
And if one connects the footpoints of these two orthogonal decompositions, one
obtains an orthogonal decomposition into three parts:
ˆ
ˆ
Problem 290. Assume β is the constrained least squares estimator subject to
ˆ is the unconstrained least squares estimator.
the constraint Rβ = o, and β
ˆ
ˆ
ˆ
ˆ
• a. 1 point With the usual notation y = X β and y = X β , show that
ˆ
ˆ
ˆ
y = y + (y − y ) + ε
ˆ
ˆˆ
ˆ
ˆ (22.7.7) Point out these vectors in the reggeom simulation.
ˆ
ˆ
Answer. In the reggeomsimulation, y is the purple line; X β is the red line starting at the
ˆ) = y − y is the light blue line, and ε is the green line which
ˆ
ˆˆ
origin, one could also call it y ; X (β − β
ˆ
ˆˆ
ˆ
ˆ
does not start at the origin. In other words: if one projects y on a plane, and also on a line in that
plane, and then connects the footpoints of these two projections, one obtains a zigzag line with
two right angles. ˆ
• b. 4 points Show that in (22.7.7) the three vectors y , y − y , and ε are orthogˆˆ ˆ
ˆ
ˆ
onal. You are allowed to use, without proof, formula (22.3.13):
Answer. One has to verify that the scalar products of the three vectors on the right hand side
ˆ
ˆˆ ˆ
ˆˆ
ˆ
of (22.7.7) are zero. y ε = β X ε = 0 and (y − y ) ε = (β − β ) X ε = 0 follow from X ε = o;
ˆ
ˆ
ˆˆˆ
ˆ
ˆ
ˆ
ˆ
geometrically on can simply say that y and y are in the space spanned by the columns of X , and
ˆ
ˆ
ˆˆ
ˆ
ε is orthogonal to that space. Finally, using (22.3.13) for β − β ,
ˆ
ˆ
ˆ
ˆ
ˆˆ
ˆ
y (y − y ) = β X X (β − β ) =
ˆˆˆ
ˆ
ˆ
ˆ
= β X X (X X )−1 R
ˆ
ˆ
=β R R(X X )−1 R ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
because β satisﬁes the constraint Rβ = o, hence β R Problem 291. R(X X )−1 R
−1 ˆ
Rβ = 0 =o . −1 ˆ
Rβ = 250 22. CONSTRAINED LEAST SQUARES • a. 3 points In the model y = β + ε , where y is a n × 1 vector, and ε ∼ (o, σ 2 I ),
ˆˆ
ˆ2
ˆˆ
ˆ
subject to the constraint ι β = 0, compute β , ε, and the unbiased estimate σ . Give
general formulas and the numerical results for the case y = −1 0 1 2 . All
you need to do is evaluate the appropriate formulas and correctly count the number
of degrees of freedom.
ˆ
Answer. The unconstrained least squares estimate of β is β = y , and since X = I , R = ι ,
ˆ = y − ι(ι ι)−1 (ι y ) = y − ιy by (22.3.13). If
ˆ
and u = 0, the constrained LSE has the form β
¯
ˆ
ˆ
y = [−1, 0, 1, 2] this gives β = [−1.5, −0.5, 0.5, 1.5]. The residuals in the constrained model are
ˆ
ˆ
therefore ε = ιy , i.e., ε = [0.5, 0.5, 0.5, 0.5]. Since one has n observations, n parameters and 1
ˆ
¯
ˆ
ˆ2
ˆˆ
constraint, the number of degrees of freedom is 1. Therefore σ = ε ε/1 = ny 2 which is = 1 in our
ˆ
ˆˆ
¯
case. • b. 1 point Can you think of a practical situation in which this model might be
appropriate?
Answer. This can occur if one measures data which theoretically add to zero, and the measurement errors are independent and have equal standard deviations. • c. 2 points Check your results against a SAS printout (or do it in any other
statistical package) with the data vector y = [ −1 0 1 2 ]. Here are the sas commands:
data zeromean;
input y x1 x2 x3 x4;
cards;
1 1 0 0 0
00100
10010
20001
;
proc reg;
model y= x1 x2 x3 x4 /
noint;
restrict x1+x2+x3+x4=0;
output out=zerout
residual=ehat;
run;
proc print data=zerout;
run;
Problem 292. Least squares estimates of the coeﬃcients of a linear regression
model often have signs that are regarded by the researcher to be ‘wrong’. In an effort to obtain the ‘right’ signs, the researcher may be tempted to drop statistically
insigniﬁcant variables from the equation. [Lea75] showed that such attempts necessarily fail: there can be no change in sign of any coeﬃcient which is more signiﬁcant
than the coeﬃcient of the omitted variable. The present exercise shows this, using
a diﬀerent proof than Leamer’s. You will need the formula for the constrained least
squares estimator subject to one linear constraint r β = u, which is
(22.7.8) ˆˆ
ˆ
β =β−Vr r Vr −1 ˆ
(r β − u). where V = (X X )−1 .
• a. In order to assess the sensitivity of the estimate of any linear combination
of the elements of β , φ = t β , due to imposition of the constraint, it makes sense 22.9. APPLICATION: BIASED ESTIMATORS AND PRETEST ESTIMATORS 251 ˆ
ˆ
ˆ
ˆ
to divide the change t β − t β by the standard deviation of t β , i.e., to look at
ˆˆ
ˆ
t (β − β )
(22.7.9)
.
σ t (X X )−1 t
Such a standardization allows you to compare the sensitivity of diﬀerent linear comˆ
binations. Show that that linear combination of the elements of β which is aﬀected
most if one imposes the constraint r β = u is the constraint t = r itself. If this
ˆ
value is small, then no other linear combination of the elements of β will be aﬀected
much by the imposition of the constraint either.
Answer. Using (22.7.8) and equation (25.4.1) one obtains
max
t ˆ
( r β − u)
= r (X −1
X )−1 r
(r σ2 ˆ
β − u)
= ˆ
ˆˆ
(t (β − β ))2
σ 2 t (X X )−1 t = ˆˆ
ˆ
ˆˆ
ˆ
(β − β ) X X ( β − β )
=
2
σ ˆ
( r β − u)2
σ2 r (X X )−1 r 22.8. Inequality Restrictions
With linear inequality restrictions, it makes sense to have R of deﬁcient rank,
these are like two diﬀerent half planes in the same plane, and the restrictions deﬁne
a quarter plane, or a triangle, etc.
One obvious approach would be: compute the unrestricted estimator, see what
restrictions it violates, and apply these restrictions with equality. But this equality
restricted estimator may then suddenly violate other restrictions.
One brute force approach would be: impose all combinations of restrictions and
see if the so partially restricted parameter satisﬁes the other restrictions too; and
among those that do, choose the one with the lowest SSE.
[Gre97, 8.5.3 on pp. 411/12] has good discussion. The inequality restricted
estimator is biased, unless the true parameter value satisﬁes all inequality restrictions
ˆ
with equality. It is always a mixture between the unbiased β and some restricted
estimator which is biased if this condition does not hold.
ˆ
Its variance is always smaller than that of β but, incredibly, its MSE will someˆ
times be larger than that of β . Don’t understand how this comes about.
22.9. Application: Biased Estimators and PreTest Estimators
The formulas about Constrained Least Squares which were just derived suggest
that it is sometimes advantageous (in terms of MSE) to impose constraints even if
they do not really hold. In other words, one should not put all explanatory variables into a regression which have an inﬂuence, but only the main ones. A logical
extension of this idea is the common practice of ﬁrst testing whether some variables
have signiﬁcant inﬂuence and dropping the variables if they do not. These socalled
pretest estimators are very common. [DM93, Chapter 3.7, pp. 94–98] says something about them. Pretest estimation this seems a good procedure, but the graph
regarding MSE shows it is not: the pretest estimator never has lowest MSE, and it
has highest MSE exactly in the area where it is most likely to be applied. CHAPTER 23 Additional Regressors
A good detailed explanation of the topics covered in this chapter is [DM93, pp.
19–24]. [DM93] use the addition of variables as their main paradigm for going from
a more restrictive to a less restrictive model.
In this chapter, the usual regression model is given in the form
(23.0.1) y = X 1 β 1 + X 2 β 2 + ε = X 1 X2 β1
+ ε = Xβ + ε ,
β2 ε ∼ (o, σ 2 I ) β1
.
β2
We take a sequential approach to this regression. First we regress y on X 1
ˆ
ˆ
alone, which gives the regression coeﬃcient β 1 . This by itself is an inconsistent
estimator of β 1 , but we will use it as a stepping stone towards the full regression.
We make use of the information gained by the regression on X 1 in our computation
of the full regression. Such a sequential approach may be appropriate in the following
situations:
• If regression on X 1 is much simpler than the combined regression, for instance if X 1 contains dummy or trend variables, and the dataset is large.
Example: model (??).
• If we want to ﬁt the regressors in X 2 by graphical methods and those in
X 1 by analytical methods (added variable plots).
• If we need an estimate of β 2 but are not interested in an estimate of β 1 .
• If we want to test the joint signiﬁcance of the regressors in X 2 , while X 1
consists of regressors not being tested.
ˆˆ
ˆ
ˆˆ
ˆ
If one regresses y on X 1 , one gets y = X 1 β 1 + ε. Of course, β 1 is an inconsistent
ˆ
ˆ
estimator of β 1 , since some explanatory variables are left out. And ε is orthogonal
to X 1 but not to X 2 .
ˆ
The iterative “backﬁtting” method proceeds from here as follows: it regresses ε
ˆ
on X 2 , which gives another residual, which is again orthogonal on X 2 but no longer
orthogonal on X 1 . Then this new residual is regressed on X 1 again, etc.
where X = X 1 X 2 has full column rank, and the coeﬃcient vector is β = Problem 293. The purpose of this Problem is to get a graphical intuition of the
issues in sequential regression. Make sure the standalone program xgobi is installed
on your computer (in Debian GNULinux do aptget install xgobi), and the Rinterface xgobi is installed (the Rcommand is simply install.packages("xgobi"),
or, on a Debian system the preferred argument is install.packages("xgobi", lib
= "/usr/lib/R/library")). You have to give the commands library(xgobi) and
then reggeom(). This produces a graph in the XGobi window which looks like [DM93,
Figure 3b on p. 22]. If you switch from the XYPlot view to the Rotation view, you
will see the same lines rotating 3dimensionally, and you can interact with this graph.
You will see that this graph shows the dependent variable y , the regression of y on
x1 , and the regression of y on x1 and x2 .
253 254 23. ADDITIONAL REGRESSORS • a. 1 point In order to show that you have correctly identiﬁed which line is y ,
please answer the following two questions: Which color is y : red, yellow, light blue,
dark blue, green, purple, or white? If it is yellow, also answer the question: Is it that
yellow line which is in part covered by a red line, or is it the other one? If it is red,
green, or dark blue, also answer the question: Does it start at the origin or not?
• b. 1 point Now answer the same two questions about x1 .
• c. 1 point Now answer the same two questions about x2 .
ˆ
• d. 1 point Now answer the same two questions about ε, the residual in the
ˆ
regression of y on x1 .
• e. Now assume x1 is the vector of ones. The R2 of this regression is a ratio
of the squared lengths of two of the lines in the regression. Which lines?
ˆ
ˆ
• f . 2 points If one regresses ε on x2 , one gets a decomposition ε = h + k,
ˆ
ˆ
where h is a multiple of x2 and k orthogonal to x2 . This is the next step in the
backﬁtting algorithm. Draw this decomposition into the diagram. The points are
already invisibly present. Therefore you should use the line editor to connect the
points. You may want to increase the magniﬁcation scale of the ﬁgure for this. (In
my version of XGobi, I often lose lines if I try to add more lines. This seems to be
a bug which will probably be ﬁxed eventually.) Which label does the corner point of
the decomposition have? Make a geometric argument that the new residual k is no
longer orthogonal to x2 .
• g. 1 point The next step in the backﬁtting procedure is to regress k on x1 .
The corner point for this decomposition is again invisibly in the animation. Identify the two endpoints of the residual in this regression. Hint: the Rcommand
example(reggeom) produces a modiﬁed version of the animation in which the backﬁtting prodecure is highlighted. The successive residuals which are used as regressors
are drawn in dark blue, and the quickly improving approximations to the ﬁtted value
are connected by a red zigzag line.
• h. 1 point The diagram contains the points for two more backﬁtting steps.
Identify the endpoints of both residuals.
• i. 2 points Of the ﬁve cornerpoints obtained by simple regressions, c, p, q , r,
and s, three lie on one straight line, and the other two on a diﬀerent straight line,
with the intersection of these straight lines being the corner point in the multiple
regression of y on x1 and x2 . Which three points are on the same line, and how can
these two lines be characterized?
• j. 1 point Of the lines cp, pq , qr, and rs, two are parallel to x1 , and two
parallel to x2 . Which two are parallel to x1 ?
• k. 1 point Draw in the regression of y on x2 .
• l. 3 points Which two variables are plotted against each other in an addedvariable plot for x2 ?
Here are the coordinates of some of the points in this animation:
x1 x2 y y y
ˆˆ
ˆ
5 1 3 3 3
0
4330
0
0400 23. ADDITIONAL REGRESSORS 255 In the dataset which R submits to XGobi, all coordinates are multiplied by 1156,
which has the eﬀect that all the points included in the animation have integer coordinates.
Problem 294. 2 points How do you know that the decomposition
0
3
4 ˆˆ
ˆˆ
is y = y + ε in the regression of y = 3
3
4 on x1 = 5
0
0 3
3
4 = 3
0
0 + ? ˆ
ˆ
ˆ
Answer. Besides the equation y = y + ε we have to check two things: (1) y is a linear
ˆ
ˆ
ˆ
ˆ
combination of all the explanatory variables (here: is a multiple of x1 ), and (2) ε is orthogonal to
ˆ
all explanatory variables. Compare Problem ??. Problem 295. 3 points In the same way, check that the decomposition
3
3
0 + 0
0
4 ˆ
is y = y + ε in the regression of y = 3
3
4 on x1 = 5
0
0 and x2 = 3
3
4
−1
4
0 =
. ˆ
ˆ
ˆ
Answer. Besides the equation y = y + ε we have to check two things: (1) y is a linear
ˆ
ˆ
ˆ
combination of all the explanatory variables. Since both x1 and x2 have zero as third coordinate,
and they are linearly independent, they span the whole plane, therefore y , which also has the
ˆ
ˆ
third coordinate zero, is their linear combination. (2) ε is orthogonal to both explanatory variables
ˆ
because its only nonzero coordinate is the third. ˆ
ˆ
ˆ
ˆ
The residuals ε in the regression on x1 are y − y = 3
3
4 3
0
0 − = 0
3
4 . This −1
ˆ
vector is clearly orthogonal to x1 =
. Now let us regress ε =
on x2 = 4 .
ˆ
0
Say h is the vector of ﬁtted values and k the residual vector in this regression. We
saw in problem 293 that this is the next step in backﬁtting, but k is not the same
as the residual vector ε in the multiple regression, because k is not orthogonal to
ˆ
x1 . In order to get the correct residual in the joint regression and also the correct
ˆ
coeﬃcient of x2 , one must regress ε only on that part of x2 which is orthogonal to
ˆ
x1 . This regressor is the dark blue line starting at the origin.
ˆ
ˆ
In formulas: One gets the correct ε and β 2 by regressingx ε = M 1 y not on X 2
ˆ
ˆ
but on M 1 X 2 , where M 1 = I − X 1 (X 1 X 1 )−1 X 1 is the matrix which forms the
residuals under the regression on X 1 . In other words, one has to remove the inﬂuence
of X 1 not only from the dependent but also the independent variables. Instead of
ˆ
regressing the residuals ε = M 1 y on X 2 , one has to regress them on what is new
ˆ
about X 2 after we know X 1 , i.e., on what remains of X 2 after taking out the eﬀect
ˆ
of X 1 , which is M 1 X 2 . The regression which gets the correct β 2 is therefore
3
0
0 0
3
4 ˆ
M 1 y = M 1 X 2 β2 + ε
ˆ (23.0.2)
ˆ
In formulas, the correct β 2 is
(23.0.3) ˆ
β 2 = (X 2 M 1 X 2 )−1 X 2 M 1 y . This regression also yields the correct covariance matrix. (The only thing which
is not right is the number of degrees of freedom). The regression is therefore fully
ˆ
representative of the additional eﬀect of x2 , and the plot of ε against M 1 X 2 with
ˆ
ˆ
the ﬁtted line drawn (which has the correct slope β 2 ) is called the “added variable
plot” for X 2 . [CW99, pp. 244–246] has a good discussion of added variable plots.
ˆ
Problem 296. 2 points Show that in the model (23.0.1), the estimator β 2 =
−1
ˆ
(X 2 M 1 X 2 ) X 2 M 1 y is unbiased. Compute MSE [β 2 ; β 2 ].
ˆ
Answer. β 2 −β 2 = (X 2 M 1 X 2 )−1 X 2 M 1 (X 1 β 1 +X 2 β 2 +ε )−β 2 = (X 2 M 1 X 2 )−1 X 2 M 1ε ;
ˆ
therefore MSE [β 2 ; β 2 ] = σ 2 (X 2 M 1 X 2 )−1 X 2 M 1 M 1 X 2 (X 2 M 1 X 2 )−1 = σ 2 (X 2 M 1 X 2 )−1 . 256 23. ADDITIONAL REGRESSORS ˆ
In order to get an estimate of β 1 , one can again do what seems intuitive, namely,
ˆ2 on X 1 . This gives
regress y − X 2 β
ˆ
ˆ
(23.0.4)
β 1 = (X 1 X 1 )−1 X 1 (y − X 2 β 2 ).
This regression also gives the right residuals, but not the right estimates of the
covariance matrix.
Problem 297. The three Figures in [DM93, p. 22] can be seen in XGobi if
you use the instructions in Problem 293. The purple line represents the dependent
variable y , and the two yellow lines the explanatory variables x1 and x2 . (x1 is the
one which is in part red.) The two green lines represent the unconstrained regression
ˆˆ
y = y + ε, and the two red lines the constrained regression y = y + ε where y is
ˆˆ
ˆˆ
only regressed on x1 . The two dark blue lines, barely visible against the dark blue
background, represent the regression of x2 on x1 .
• a. The ﬁrst diagram which XGobi shows on startup is [DM93, diagram (b)
on p. 22]. Go into the Rotation view and rotate the diagram in such a way that the
view is [DM93, Figure (a)]. You may want to delete the two white lines, since they
are not shown in Figure (a).
• b. Make a geometric argument that the light blue line, which represents y − y =
ˆˆ
ˆ
ˆ), is orthogonal on the green line ε (this is the green line which ends at the
ˆ−β
ˆ
ˆ
X (β
point y , i.e., not the green line which starts at the origin).
Answer. The light blue line lies in the plane spanned by x1 and x2 , and ε is orthogonal to
ˆ
this plane. • c. Make a geometric argument that the light blue line is also orthogonal to the
ˆ
red line y emanating from the origin.
ˆ
ˆ
Answer. This is a little trickier. The red line ε is orthogonal to x1 , and the green line ε is
ˆ
ˆ
ˆ
also orthogonal to x1 . Together, ε and ε span therefore the plane orthogonal to x1 . Since the light
ˆ
ˆ
ˆ
blue line lies in the plane spanned by ε and ε, it is orthogonal to x1 .
ˆ
ˆ ˆ
Question 297 shows that the decomposition y = y + (y − y ) + ε is orthogonal,
ˆ
ˆˆ
ˆ
ˆ
ˆ
ˆ , y − y , and ε are orthogonal to each other. This is (22.7.6) in the
ˆ
i.e., all 3 vectors y ˆ ˆ
ˆ
special case that u = o and therefore β 0 = o.
One can use this same animation also to show the following: If you ﬁrst project
the purple line on the plane spanned by the yellow lines, you get the green line in the
plane. If you then project that green line on x1 , which is a subspace of the plane,
then you get the red section of the yellow line. This is the same result as if you
had projected the purple line directly on x1 . A matrixalgebraic proof of this fact is
given in (A.6.3).
The same animation allows us to verify the following:
ˆ
ˆ
ˆ
• In the regression of y on x1 , the coeﬃcient is β 1 , and the residual is ε.
ˆ
ˆ , β , and the
ˆ
• In the regression of y on x1 and x2 , the coeﬃcients are β 1 2
residual is ε.
ˆ
ˆˆ
ˆ
• In the regression of y on x1 and M 1 x2 , the coeﬃcients are β 1 , β 2 , and the
residual is ε. The residual is ε because the space spanned by the regressors
ˆ
ˆ
ˆ
is the same as in the regression on x1 and x2 , and ε only depends on that
space.
ˆ
• In the regression of y on M 1 x2 , the coeﬃcient is β 2 , because the regressor I am leaving out is orthogonal to M 1 x2 . The residual contains the
contribution of the leftout variable, i.e., it is ε + β 1 x1 .
ˆˆ 23. ADDITIONAL REGRESSORS 257 ˆ
ˆ
• But in the regression of ε = M 1 y on M 1 x2 , the coeﬃcient is β 2 and the
ˆ
residual ε.
ˆ
This last statement is (23.0.3).
Now let us turn to proving all this mathematically. The “brute force” proof, i.e.,
the proof which is conceptually simplest but has to plow through some tedious mathematics, uses (14.2.4) with partitioned matrix inverses. For this we need (23.0.5).
Problem 298. 4 points This is a simpliﬁed version of question 393. Show the
following, by multiplying X X with its alleged inverse: If X = X 1 X 2 has full
column rank, then (X X )−1 is the following partitioned matrix:
(23.0.5)
−1
X1 X1 X1 X2
(X 1 X 1 )−1 + K 1 X 2 (X 2 M 1 X 2 )−1 X 2 K 1 −K 1 X 2 (X 2 M 1 X 2 )−1
=
X2 X1 X2 X2
−(X 2 M 1 X 2 )−1 X 2 K 1
(X 2 M 1 X 2 )−1
where M 1 = I − X 1 (X 1 X 1 )−1 X 1 and K 1 = X 1 (X 1 X 1 )−1 .
From (23.0.5) one sees that the covariance matrix in regression (23.0.3) is the
lower left partition of the covariance matrix in the full regression (23.0.1).
ˆ
Problem 299. 6 points Use the usual formula β = (X X )−1 X y together
with (23.0.5) to prove (23.0.3) and (23.0.4).
Answer. (14.2.4) reads here
(23.0.6)
ˆ
(X 1 X 1 )−1 + K 1 X 2 (X 2 M 1 X 2 )−1 X 2 K 1
β1
ˆ=
−(X M 1 X 2 )−1 X K 1
β2
2 2 −K 1 X 2 (X 2 M 1 X 2 )−1
(X 2 M 1 X 2 )−1 X1 y
X2 y Since M 1 = I − K 1 X 1 , one can simplify
(23.0.7)
ˆ
β 2 = −(X 2 M 1 X 2 )−1 X 2 K 1 X 1 y + (X 2 M 1 X 2 )−1 X 2 y
(23.0.8)
= (X 2 M 1 X 2 )−1 X 2 M y
(23.0.9)
ˆ
β 1 = (X 1 X 1 )−1 X 1 y + K 1 X 2 (X 2 M 1 X 2 )−1 X 2 K 1 X 1 y − K 1 X 2 (X 2 M 1 X 2 )−1 X 2 y
(23.0.10)
= K 1 y − K 1 X 2 (X 2 M 1 X 2 )−1 X 2 (I − K 1 X 1 )y
(23.0.11)
= K 1 y − K 1 X 2 (X 2 M 1 X 2 )−1 X 2 M 1 y
(23.0.12)
ˆ
= K 1 (y − X 2 β 2 ) [Gre97, pp. 245–7] follow a diﬀerent proof strategy: he solves the partitioned
normal equations
(23.0.13) X1 X1
X2 X1 X1 X2
X2 X2 ˆ
β1
ˆ
β2 X1 y
X2 y directly, without going through the inverse. A third proof strategy, used by [Seb77,
pp. 65–72], is followed in Problems 301 and 302. 258 23. ADDITIONAL REGRESSORS Problem 300. 5 points [Gre97, problem 18 on p. 326]. The following matrix
gives the slope in the simple regression of the column variable on the row variable:
y
1
0.4
1.2 (23.0.14) x1
0.03
1
0.075 x2
0.36
0.3
1 y
x1
x2 For example, if y is regressed on x1 , the slope is 0.4, but if x1 is regressed on y , the
slope is 0.03. All variables have zero means, so the constant terms in all regressions
are zero. What are the two slope coeﬃcients in the multiple regression of y on x1
and x2 ? Hint: Use the partitioned normal equation as given in [Gre97, p. 245] in
the special case when each of the partitions of X has only one colum.
Answer.
x1 x1
x2 x1 (23.0.15) x1 x2
x2 x2 ˆ
β1
ˆ
β2 = x1 y
x2 y The ﬁrst row reads
ˆ
ˆ
β1 + (x1 x1 )−1 x1 x2 β2 = (x1 x1 )−1 x1 y (23.0.16) ˆ
ˆ
which is the upper line of [Gre97, (6.24) on p, 245], and in our numbers this is β1 = 0.4 − 0.3β2 .
The second row reads
ˆ
ˆ
(23.0.17)
(x2 x2 )−1 x2 x1 β1 + β2 = (x2 x2 )−1 x2 y
ˆ
ˆ
ˆ
or in our numbers 0.075β2 + β2 = 1.2. Plugging in the formula for β1 gives 0.075 · 0.4 − 0.075 ·
ˆ
ˆ
ˆ
ˆ
0.3β2 + β2 = 1.2. This gives β2 = 1.17/0.9775 = 1.196931 = 1.2 roughly, and β1 = 0.4 − 0.36 =
0.0409207 = 0.041 roughly. Problem 301. Derive (23.0.3) and (23.0.4) from the ﬁrst order conditions for
minimizing
(23.0.18) (y − X 1 β 1 − X 2 β 2 ) (y − X 1 β 1 − X 2 β 2 ). Answer. Start by writing down the OLS objective function for the full model. Perhaps we
can use the more sophisticated matrix diﬀerentiation rules?
(23.0.19)
(y −X 1 β 1 −X 2 β 2 ) (y −X 1 β 1 −X 2 β 2 ) = y y +β 1 X 1 X 1 β 1 +β 2 X 2 X 2 β 2 −2y X 1 β 1 −2y X 2 β 2 +2β 2 X 2 X 1 β 1 .
Taking partial derivatives with respect to β 1 and β 2 gives
(23.0.20)
2β 1 X 1 X 1 − 2y X 1 + 2β 2 X 2 X 1 or, transposed 2X 1 X 1 β 1 − 2X 1 y + 2X 1 X 2 β 2 (23.0.21)
2β 2 X 2 X 2 − 2y X 2 + 2β 1 X 1 X 2 or, transposed 2X 2 X 2 β 2 − 2X 2 y + 2X 2 X 1 β 1 ˆ
ˆ
Setting them zero and replacing β 1 by β 1 and β 2 by β 2 gives
(23.0.22) ˆ
ˆ
X 1 X 1 β 1 = X 1 (y − X 2 β 2 ) (23.0.23) ˆ
ˆ
X 2 X 2 β 2 = X 2 (y − X 1 β 1 ). Premultiply (23.0.22) by X 1 (X 1 X 1 )−1 :
(23.0.24) ˆ
ˆ
X 1 β 1 = X 1 (X 1 X 1 )−1 X 1 (y − X 2 β 2 ). Plug this into (23.0.23):
(23.0.25)
(23.0.26) ˆ
X 2 X 2 β2 = X 2 ˆ
y − X 1 (X 1 X 1 )−1 X 1 y + X 1 (X 1 X 1 )−1 X 1 X 2 β 2 ˆ
X 2 M 1 X 2 β2 = X 2 M 1 y. (23.0.26) is the normal equation of the regression of M 1 y on M 1 X 2 ; it immediately implies (23.0.3).
ˆ
ˆ
Once β 2 is known, (23.0.22) is the normal equation of the regression of y − X 2 β 2 on X 1 , which
gives (23.0.4). 23. ADDITIONAL REGRESSORS 259 Problem 302. Using (23.0.3) and (23.0.4) show that the residuals in regression
(23.0.1) are identical to those in the regression of M 1 y on M 1 X 2 .
Answer.
(23.0.27) ˆ
ˆ
ε = y − X 1 β1 − X 2 β2
ˆ (23.0.28) ˆ
ˆ
= y − X 1 (X 1 X 1 )−1 X 1 (y − X 2 β 2 ) − X 2 β 2 (23.0.29) ˆ
= M 1 y − M 1 X 2 β2 . Problem 303. The following problem derives one of the main formulas for
adding regressors, following [DM93, pp. 19–24]. We are working in model (23.0.1).
• a. 1 point Show that, if X has full column rank, then X X , X 1 X 1 , and
X 2 X 2 are nonsingular. Hint: A matrix X has full column rank if Xa = o implies
a = o.
Answer. From X X a = o follows a X X a = 0 which can also be written X a = 0.
Therefore Xa = o, and since the columns are linearly independent, it follows a = o. X 1 X 1 and
X 2 X 2 are nonsingular because, along with X , also X 1 and X 2 have full column rank. • b. 1 point Deﬁne M = I − X (X X )−1 X and M 1 = I − X 1 (X 1 X 1 )−1 X 1 .
Show that both M and M 1 are projection matrices. (Give the deﬁnition of a projection matrix.) Which spaces do they project on? Which space is bigger?
Answer. A projection matrix is symmetric and idempotent. That M M = M is easily veriﬁed.
M projects on the orthogonal complement of the column space of X , and M 1 on that of X 1 . I.e.,
M 1 projects on the larger space. • c. 2 points Prove that M 1 M = M and that M X 1 = O as well as M X 2 = O .
You will need each these equationse below. What is their geometric meaning?
Answer. X 1 = X 1 X2 I
= XA, say. Therefore M 1 M = (I −XA(A X X A)−1 A X )M =
O M because X M = O . Geometrically this means that the space on which M projects is a subspace
of the space on which M 1 projects. To show that M X 2 = O note that X 2 can be written in the
O
form X 2 = XB , too; this time, B =
. M X 2 = O means geometrically that M projects on a
I
space that is orthogonal to all columns of X 2 . • d. 2 points Show that M 1 X 2 has full column rank.
Answer. If M 1 X 2 b = o, then X 2 b = X 1 a for some a. We showed this in Problem 196.
−a
o
−a
Therefore X 1 X 2
=
, and since X 1 X 2 has full column rank, it follows
=
b
o
b
o
, in particular b = o.
o • e. 1 point Here is some more notation: the regression of y on X 1 and X 2 can
also be represented by the equation
(23.0.30) ˆ
ˆ
y = X 1 β1 + X 2 β2 + ε
ˆ The diﬀerence between (23.0.1) and (23.0.30) is that (23.0.30) contains the parameter
estimates, not their true values, and the residuals, not the true disturbances. Explain
the diﬀerence between residuals and disturbances, and between the ﬁtted regression
line and the true regression line. 260 23. ADDITIONAL REGRESSORS • f . 1 point Verify that premultiplication of (23.0.30) by M 1 gives
(23.0.31) ˆ
ˆ
M 1 y = M 1 X 2 β2 + ε Answer. We need M 1 X 1 = O and M 1 ε = M 1 M y = M y = ε (or this can also besseen
ˆ
ˆ
because X 1 ε = o).
ˆ • g. 2 points Prove that (23.0.31) is the ﬁt which one gets if one regresses M 1 y
on M 1 X 2 . In other words, if one runs OLS with dependent variable M 1 y and
ˆ
explanatory variables M 1 X 2 , one gets the same β 2 and ε as in (23.0.31), which are
ˆ
ˆ2 and ε as in the complete regression (23.0.30).
the same β
ˆ
Answer. According to Problem ?? we have to check X 2 M 1 ε = X 2 M 1 M y = X 2 M y =
ˆ
O y = o. ˆ
• h. 1 point Show that V [β 2 ] = (X 2 M 1 X 2 )−1 . Are the variance estimates
and conﬁdence intervals valid, which the computer automatically prints out if one
regresses M 1 y on M 1 X 2 ?
Answer. Yes except for the number of degrees of freedom. • i. 4 points If one premultiplies (23.0.1) by M 1 , one obtains
(23.0.32) M 1 y = M 1 X 2 β 2 + M 1ε , M 1ε ∼ (o, σ 2 M 1 ) Although the covariance matrix of the disturbance M 1ε in (23.0.32) is no longer
ˆ
spherical, show that nevertheless the β 2 obtained by running OLS on (23.0.32) is the
BLUE of β 2 based on the information given in (23.0.32) (i.e., assuming that M 1 y
and M 1 X 2 are known, but not necessarily M 1 , y , and X 2 separately). Hint: this
proof is almost identical to the proof that for spherically distributed disturbances the
OLS is BLUE (e.g. given in [DM93, p. 159]), but you have to add some M 1 ’s to
your formulas.
˜
˜
Answer. Any other linear estimator γ of β 2 can be written as γ = (X 2 M 1 X 2 )−1 X 2 +
˜
γ
C M 1 y . Its expected value is E[˜ ] = (X 2 M 1 X 2 )−1 X 2 M 1 X 2 β 2 + CM 1 X 2 β 2 . For γ to be unbiased, regardless of the value of β 2 , C must satisfy CM 1 X 2 = O . From this follows MSE [˜ ; β 2 ] =
γ
γ
= σ 2 (X 2 M 1 X 2 )−1 +σ 2 CM 1 C ,
V [˜ ] = σ 2 (X 2 M 1 X 2 )−1 X 2 +C M 1 X 2 (X 2 M 1 X 2 )−1 +C
ˆ
i.e., it exceeds the MSE matrix of β by a nonnegative deﬁnite matrix. Is it unique? The formula
for the BLUE is not unique, since one can add any C with CM 1 C = O or equivalently CM 1 = O
or C = AX for some A. However such a C applied to a dependent variable of the form M 1 y will
give the null vector, therefore the values of the BLUE for those values of y which are possible are
indeed unique. ˆ
• j. 1 point Once β 2 is known, one can move it to the left hand side in (23.0.30)
to get
ˆ
ˆ
(23.0.33)
y − X 2 β2 = X 1 β1 + ε
ˆ
ˆ
ˆ
Prove that one gets the right values of β 1 and of ε if one regresses y − X 2 β 2 on X 1 .
ˆ
Answer. The simplest answer just observes that X 1 ε = o. Or: The normal equation for this
ˆ
ˆ
ˆ
pseudoregression is X 1 y − X 1 X 2 β 2 = X 1 X 1 β 1 , which holds due to the normal equation for the
full model. ˆ
• k. 1 point Does (23.0.33) also give the right covariance matrix for β 1 ?
ˆ
Answer. No, since y − X 2 β 2 has a diﬀerent covariance matrix than σ 2 I . This following Problems gives some applications of the results in Problem 303.
You are allowed to use the results of Problem 303 without proof. 23. ADDITIONAL REGRESSORS 261 Problem 304. Assume your regression involves an intercept, i.e., the matrix of
regressors is ι X , where X is the matrix of the “true” explanatory variables with
no vector of ones built in, and ι the vector of ones. The regression can therefore be
written
y = ια + Xβ + ε . (23.0.34) • a. 1 point Show that the OLS estimate of the slope parameters β can be obtained
by regressing y on X without intercept, where y and X are the variables with their
1
means taken out, i.e., y = D y and X = DX , with D = I − n ιι .
Answer. This is called the “sweeping out of means.” It follows immediately from (23.0.3).
This is the usual procedure to do regression with a constant term: in simple regression y i =
α + βxi + εi , (23.0.3) is equation (14.2.22):
ˆ
β= (23.0.35) (xi − x)(y i − y )
¯
¯
(xi − x)2
¯ . ¯
• b. Show that the OLS estimate of the intercept is α = y − x β where x
ˆ
¯ ¯ˆ
1
¯
the row vector of column means of X , i.e., x = n ι X . is Answer. This is exactly (23.0.4). Here is a more speciﬁc argument: The intercept α is obˆ
ˆ
ˆ
tained by regressing y − X β on ι. The normal equation for this second regression is ι y − ι X β =
¯
ˆ
¯
ι ια. If y is the mean of y , and x the row vector consisting of means of the colums of X , then this
gives y = x β + α. In the case of simple regression, this was derived earlier as formula (14.2.23).
¯ ¯ˆˆ ˆ
ˆ
• c. 2 points Show that MSE [β ; β ] = σ 2 (X X )−1 . (Use the formula for β .)
Answer. Since
ι
X (23.0.36) ι X= n
¯
xn ¯
nx
,
XX it follows by Problem 393
(23.0.37) ( ι
X ι X )−1 = ¯
¯
1/n + x (X X )−1 x
¯
−(X X )−1 x ¯
−x (X X )−1
(X X )−1 In other words, one simply does as if the actual regressors had been the data with their means
removed, and then takes the inverse of that design matrix. The only place where on has to be
careful is the number of degrees of freedom. See also Seber [Seb77, section 11.7] about centering
and scaling the data. ˆ
• d. 3 points Show that y − ιy = X β .
ˆ
¯
1
¯
¯
Answer. First note that X = X + n ιι X = X + ιx where x is the row vector of means
ˆ = ια + X β + ιx β = ι( α + x β ) + X β = ιy + X β .
ˆ
ˆ
ˆ
¯ˆ
of X . By deﬁnition, y = ια + X β
ˆ
ˆ
ˆ
ˆ ¯ˆ
¯ • e. 2 points Show that R2 = y X (X X )−1 X y
yy Answer.
(23.0.38) R2 = ˆ
ˆ
( y − y ι) ( y − y ι)
ˆ¯
ˆ¯
β X Xβ
=
yy
yy ˆ
and now plugging in the formula for β the result follows. 262 23. ADDITIONAL REGRESSORS • f . 3 points Now, split once more X = X 1 x2 where the second partition
x2 consists of one column only, and X is, as above, the X matrix with the column
ˆ
β
ˆ
means taken out. Conformably, β = ˆ1 . Show that
β2
(23.0.39) ˆ
var[β 2 ] = σ2
1
2
x x (1 − R2· ) 2
where R2· is the R2 in the regression of x2 on all other variables in X . This is in
ˆ
[Gre97, (9.3) on p. 421]. Hint: you should ﬁrst show that var[β 2 ] = σ 2 /x2 M 1 x2
−1
where M 1 = I −X 1 (X 1 X 1 ) X 1 . Here is an interpretation of (23.0.39) which you
don’t have to prove: σ 2 /x x is the variance in a simple regression with a constant
2
term and x2 as the only explanatory variable, and 1/(1 − R2· ) is called the variance
inﬂation factor. Answer. Note that we are not talking about the variance of the constant term but that of all
the other terms.
x X (X 1 X 1 )−1 X 1 x2
(23.0.40) x2 M 1 x2 = x2 x2 + x2 X 1 (X 1 X 1 )−1 X 1 x2 = x2 x2 1 + 2 1
x2 x2
2
and since the fraction is R2· , i.e., it is the R2 in the regression of x2 on all other variables in X , we
get the result. CHAPTER 24 Residuals: Standardized, Predictive, “Studentized”
24.1. Three Decisions about Plotting Residuals
After running a regression it is always advisable to look at the residuals. Here
one has to make three decisions.
The ﬁrst decision is whether to look at the ordinary residuals
(24.1.1) ˆ
εi = y i − xi β
ˆ (xi is the ith row of X ), or the “predictive” residuals, which are the residuals
computed using the OLS estimate of β gained from all the other data except the
ˆ
data point where the residual is taken. If one writes β (i) for the OLS estimate
without the ith observation, the deﬁning equation for the ith predictive residual,
which we call εi (i), is
ˆ
(24.1.2) ˆ
εi (i) = y i − xi β (i).
ˆ The second decision is whether to standardize the residuals or not, i.e., whether
to divide them by their estimated standard deviations or not. Since ε = M y , the
ˆ
variance of the ith ordinary residual is
(24.1.3) var[εi ] = σ 2 mii = σ 2 (1 − hii ),
ˆ and regarding the predictive residuals it will be shown below, see (24.2.9), that
(24.1.4) var[εi (i)] =
ˆ σ2
σ2
=
.
mii
1 − hii Here
(24.1.5) hii = xi (X X )−1 xi . (Note that xi is the ith row of X written as a column vector.) hii is the ith diagonal
element of the “hat matrix” H = X (X X )−1 X , the projector on the column
ˆ
space of X . This projector is called “hat matrix” because y = H y , i.e., H puts the
“hat” on y .
Problem 305. 2 points Show that the ith diagonal element of the “hat matrix”
H = X (X X )−1 X is xi (X X )−1 xi where xi is the ith row of X written as a
column vector.
Answer. In terms of ei , the nvector with 1 on the ith place and 0 everywhere else, xi =
X ei , and the ith diagonal element of the hat matrix is ei Hei = ei X i (X X )−1 X ei =
xi (X X )−1 xi . Problem 306. 2 points The variance of the ith disturbance is σ 2 . Is the variance
of the ith residual bigger than σ 2 , smaller than σ 2 , or equal to σ 2 ? (Before doing the
math, ﬁrst argue in words what you would expect it to be.) What about the variance
of the predictive residual? Prove your answers mathematically. You are allowed to
use (24.2.9) without proof.
263 264 24. RESIDUALS Answer. Here is only the math part of the answer: ε = M y . Since M = I − H is idempotent
ˆ
and symmetric, we get V [M y ] = σ 2 M , in particular this means var[εi ] = σ 2 mii where mii is the
ˆ
ith diagonal elements of M . Then mii = 1 − hii . Since all diagonal elements of projection matrices
are between 0 and 1, the answer is: the variances of the ordinary residuals cannot be bigger than
σ 2 . Regarding predictive residuals, if we plug mii = 1 − hii into (24.2.9) it becomes
(24.1.6) εi (i) =
ˆ 1
εi
ˆ
mii therefore var[εi (i)] =
ˆ 12
σ2
σ mii =
2
mii
mii which is bigger than σ 2 . Problem 307. Decide in the following situations whether you want predictive
residuals or ordinary residuals, and whether you want them standardized or not.
• a. 1 point You are looking at the residuals in order to check whether the associated data points are outliers and do perhaps not belong into the model.
Answer. Here one should use the predictive residuals. If the ith observation is an outlier
which should not be in the regression, then one should not use it when running the regression. Its
inclusion may have a strong inﬂuence on the regression result, and therefore the residual may not
be as conspicuous. One should standardize them. • b. 1 point You are looking at the residuals in order to assess whether there is
heteroskedasticity.
Answer. Here you want them standardized, but there is no reason to use the predictive
residuals. Ordinary residuals are a little more precise than predictive residuals because they are
based on more observations. • c. 1 point You are looking at the residuals in order to assess whether the
disturbances are autocorrelated.
Answer. Same answer as for b. • d. 1 point You are looking at the residuals in order to assess whether the
disturbances are normally distributed.
Answer. In my view, one should make a normal QQplot of standardized residuals, but one
should not use the predictive residuals. To see why,√ us ﬁrst look at the distribution of the
let
ˆ
standardized residuals before division by s. Each εi / 1 − hii is normally distributed with mean
zero and standard deviation σ . (But diﬀerent such residuals are not independent.) If one takes a
QQplot of those residuals against the normal distribution, one will get in the limit a straight line
with slope σ . If one divides every residual by s, the slope will be close to 1, but one will again get
something approximating a straight line. The fact that s is random does not aﬀect the relation
of the residuals to each other, and this relation is what determines whether or not the QQplot
approximates a straight line.
But Belsley, Kuh, and Welsch on [BKW80, p. 43] draw a normal probability plot of the
studentized, not the standardized, residuals. They give no justiﬁcation for their choice. I think it
is the wrong choice. • e. 1 point Is there any situation in which you do not want to standardize the
residuals?
Answer. Standardization is a mathematical procedure which is justiﬁed when certain conditions hold. But there is no guarantee that these conditions acutally hold, and in order to get
a more immediate impression of the ﬁt of the curve one may want to look at the unstandardized
residuals. The third decision is how to plot the residuals. Never do it against y . Either
do it against the predicted y , or make several plots against all the columns of the
ˆ
X matrix.
In time series, also a plot of the residuals against time is called for. 24.2. RELATIONSHIP BETWEEN ORDINARY AND PREDICTIVE RESIDUALS 265 Another option are the partial residual plots, see about this also (23.0.2). Say
ˆ[h] is the estimated parameter vector, which is estimated with the full model, but
β
after estimation we drop the hth parameter, and X [h] is the X matrix without
the hth column, and xh is the hth column of the X matrix. Then by (23.0.4), the
estimate of the hth slope parameter is the same as that in the simple regression of
ˆ
ˆ
y − X [h]β [h] on xh . The plot of y − X [h]β [h] against xh is called the hth partial
residual plot.
To understand this better, start out with a regression y i = α + βxi + γzi + εi ;
which gives you the ﬁtted values y i = α + β xi + γ zi + εi . Now if you regress y i − α − β xi
ˆˆ
ˆ
ˆ
ˆˆ
on xi and zi then the intercept will be zero and the estimated coeﬃcient of xi will
ˆ
ˆ
be zero, and the estimated coeﬃcient of zi will be γ , and the residuals will be εi .
The plot of y i − α − β xi versus zi is the partial residuals plot for z .
ˆˆ
24.2. Relationship between Ordinary and Predictive Residuals
ˆ
In equation (24.1.2), the ith predictive residuals was deﬁned in terms of β (i),
the parameter estimate from the regression of y on X with the ith observation left
out. We will show now that there is a very simple mathematical relationship between
the ith predictive residual and the ith ordinary residual, namely, equation (24.2.9).
(It is therefore not necessary to run n diﬀerent regressions to get the n predictive
residuals.)
We will write y (i) for the y vector with the ith element deleted, and X (i) is the
matrix X with the ith row deleted.
Problem 308. 2 points Show that
(24.2.1) X (i) X (i) = X X − xi xi (24.2.2) X (i) y (i) = X y − xi y i . Answer. Write (24.2.2) as X y = X (i) y (i) + xi y i , and observe that with our deﬁnition of
xi as column vectors representing the rows of X , X = x1 · · · xn . Therefore y1 (24.2.3) X y = x1 ... xn . = x1 y1 + · · · + xn yn .
.
.
yn An important stepping stone towards the proof of (24.2.9) is equation (24.2.8),
which gives a relationship between hii and
(24.2.4) hii (i) = xi (X (i) X (i))−1 xi . ˆ
y i (i) = xi β (i) has variance σ 2 hii (i). The following problems give the steps necesˆ
sary to prove (24.2.8). We begin with a simpliﬁed version of theorem A.8.2 in the
Mathematical Appendix:
Theorem 24.2.1. Let A be a nonsingular k × k matrix, δ = 0 a scalar, and b a
k × 1 vector with b A−1 b + δ = 0. Then
(24.2.5) A+ bb
δ −1 = A−1 − A−1 bb A−1
.
δ + b A−1 b Problem 309. Prove (24.2.5) by showing that the product of the matrix with its
alleged inverse is the unit matrix. 266 24. RESIDUALS Problem 310. As an application of (24.2.5) show that
(24.2.6)
(X X )−1 xi xi (X X )−1
(X X )−1 +
is the inverse of
1 − hii X (i) X (i). Answer. This is (24.2.5), or (A.8.20), with A = X X , b = xi , and δ = −1. Problem 311. Using (24.2.6) show that
1
(X X )−1 xi ,
1 − hii
and using (24.2.7) show that hii (i) is related to hii by the equation
1
(24.2.8)
1 + hii (i) =
1 − hii
[Gre97, (937) on p. 445] was apparently not aware of this relationship.
(24.2.7) (X (i) X (i))−1 xi = Problem 312. Prove the following mathematical relationship between predictive
residuals and ordinary residuals:
1
(24.2.9)
εi (i) =
ˆ
εi
ˆ
1 − hii
which is the same as (21.0.29), only in a diﬀerent notation.
Answer. For this we have to apply the above mathematical tools. With the help of (24.2.7)
(transpose it!) and (24.2.2), (24.1.2) becomes
εi (i) = y i − xi (X (i) X (i))−1 X (i) y (i)
ˆ
1
x (X X )−1 (X y − xi y i )
1 − hii i
1
1
ˆ
= yi −
x β+
x (X X )−1 xi y i
1 − hii i
1 − hii i
1
hii
ˆ
= yi 1 +
−
xβ
1 − hii
1 − hii i
1
ˆ
(y − xi β )
=
1 − hii i
This is a little tedious but simpliﬁes extremely nicely at the end.
= yi − The relationship (24.2.9) is so simple because the estimation of ηi = xi β can be
done in two steps. First collect the information which the n − 1 observations other
than the ith contribute to the estimation of ηi = xi β is contained in y i (i). The
ˆ
information from all observations except the ith can be written as
(24.2.10) y i (i) = η i + δ i
ˆ δ i ∼ (0, σ 2 hii (i)) Here δ i is the “sampling error” or “estimation error” y i (i) − η i from the regression of
ˆ
y (i) on X (i). If we combine this compound “observation” with the ith observation
y i , we get
(24.2.11) y i (i)
ˆ
1
δ
=
η+ i
yi
1i
εi δi
∼
εi 0
h (i) 0
, σ 2 ii
0
0
1 This is a regression model similar to model (14.1.1), but this time with a nonspherical
covariance matrix.
Problem 313. Show that the BLUE of η i in model (24.2.11) is
(24.2.12) y i = (1 − hii )y i (i) + hii y i = y i (i) + hii εi (i)
ˆ
ˆ
ˆ
ˆ Hint: apply (24.2.8). Use this to prove (24.2.9). 24.3. STANDARDIZATION 267 Answer. As shown in problem 178, the BLUE in this situation is the weighted average of the
observations with the weights proportional to the inverses of the variances. I.e., the ﬁrst observation
has weight
1/hii (i)
1
=
= 1 − hii .
1/hii (i) + 1
1 + hii (i) (24.2.13) Since the sum of the weights must be 1, the weight of the second observation is hii .
Here is an alternative solution, using formula (19.0.6) for the BLUE, which reads here
yi =
ˆ 1 = hii 1 hii
1−hii 1 0
1 0
1 1−hii
hii 0 −1 0
1 1
1 −1 1 1 hii
1−hii 0 0
1 −1 y i (i)
ˆ
=
yi y i (i)
ˆ
= (1 − hii )y i (i) + hii y i .
ˆ
yi Now subtract this last formula from y i to get y i − y i = (1 − hii )(y i − y i (i)), which is (24.2.9).
ˆ
ˆ 24.3. Standardization
In this section we will show that the standardized predictive residual is what is
sometimes called the “studentized” residual. It is recommended not to use the term
“studentized residual” but say “standardized predictive residual” instead.
The standardization of the ordinary residuals has two steps: every εi is divided
ˆ
√
by its “relative” standard deviation 1 − hii , and then by s, an estimate of σ , the
standard deviation of the true disturbances. In formulas,
εi
ˆ
(24.3.1)
the ith standardized ordinary residual = √
.
s 1 − hii
Standardization of the ith predictive residual has the same two steps: ﬁrst divide
the predictive residual (24.2.9) by the relative standard deviation, and then divide by
s(i). But a look at formula (24.2.9) shows that the ordinary and the predictive residual diﬀer only by a nonrandom factor. Therefore the ﬁrst step of the standardization
yields exactly the same result whether one starts with an ordinary or a predictive
residual. Standardized predictive residuals diﬀer therefore from standardized ordinary residuals only in the second step:
εi
ˆ
(24.3.2)
the ith standardized predictive residual =
.
√
s(i) 1 − hii
Note that equation (24.3.2) writes the standardized predictive residual as a function
of the ordinary residual, not the predictive residual. The standardized predictive
residual is sometimes called the “studentized” residual.
Problem 314. 3 points The ith predictive residual has the formula
1
(24.3.3)
εi (i) =
ˆ
εi
ˆ
1 − hii
You do not have to prove this formula, but you are asked to derive the standard
deviation of εi (i), and to derive from it a formula for the standardized ith predictive
ˆ
residual.
This similarity between these two formulas has lead to widespread confusion.
Even [BKW80] seem to have been unaware of the signiﬁcance of “studentization”;
they do not work with the concept of predictive residuals at all.
The standardized predictive residuals have a tdistribution, because they are
a normally distributed variable divided by an independent χ2 over its degrees of
freedom. (But note that the joint distribution of all standardized predictive residuals
is not a multivariate t.) Therefore one can use the quantiles of the tdistribution to 268 24. RESIDUALS judge, from the size of these residuals, whether one has an extreme observation or
not.
Problem 315. Following [DM93, p. 34], we will use (23.0.3) and the other
formulas regarding additional regressors to prove the following: If you add a dummy
variable which has the value 1 for the ith observation and the value 0 for all other
observations to your regression, then the coeﬃcient estimate of this dummy is the ith
predictive residual, and the coeﬃcient estimate of the other parameters after inclusion
ˆ
of this dummy is equal to β (i). To ﬁx notation (and without loss of generality),
assume the ith observation is the last observation, i.e., i = n, and put the dummy
variable ﬁrst in the regression:
o
y (n)
=
yn
1 (24.3.4) X (n)
xn ε(i)
ˆ
α
+
εn
ˆ
β • a. 2 points With the deﬁnition X 1 = en = or y = en X α
+ε
β o
, write M 1 = I −X 1 (X 1 X 1 )−1 X 1
1 as a 2 × 2 partitioned matrix.
Answer.
(24.3.5) M1 = I
o o
o
−
1
1 o 1= I
o o
;
0 I
o o
0 z (i)
z (i)
=
zi
0 i.e., M 1 simply annulls the last element. • b. 2 points Either show mathematically, perhaps by evaluating (X 2 M 1 X 2 )−1 X 2 M 1 y ,
or give a good heuristic argument (as [DM93] do), that regressing M 1 y on M 1 X
gives the same parameter estimate as regressing y on X with the nth observation
dropped.
Answer. (23.0.2) reads here
X (n) ˆ
y (n)
ε(i)
ˆ
=
β (i) +
o
0
0 (24.3.6) ˆ
in other words, the estimate of β is indeed β (i), and the ﬁrst n − 1 elements of the residual are
indeed the residuals one gets in the regression without the ith observation. This is so ugly because
the singularity shows here in the zeros of the last row, usually it does not show so much. But this
way one also sees that it gives zero as the last residual, and this is what one needs to know!
To have a mathematical proof that the last row with zeros does not aﬀect the estimate, evaluate
(23.0.3)
ˆ
β 2 = (X 2 M 1 X 2 )−1 X 2 M 1 y
= X (n) xn I
o o
0 X (n)
xn −1 X (n) xn I
o o
0 y (n)
yn ˆ
= (X (n) X (n))−1 X (n) y (n) = β (n) • c. 2 points Use the fact that the residuals in the regression of M 1 y on M 1 X
are the same as the residuals in the full regression (24.3.4) to show that α is the nth
ˆ
predictive residual.
ˆ
Answer. α is obtained from that last row, which reads y n = α+xn β (i), i.e., α is the predictive
ˆ
ˆ
ˆ
residual. • d. 2 points Use (23.0.3) with X 1 and X 2 interchanged to get a formula for α.
ˆ
Answer. α = (X 1 M X 1 )−1 X 1 M y =
ˆ 1
ε
ˆ
mnn n = 1
ε,
ˆ
1−hnn n here M = I − X (X X )−1 X . 24.3. STANDARDIZATION 269 ˆ
ˆ
• e. 2 points From (23.0.4) follows that also β 2 = (X 2 X 2 )−1 X 2 (y − X 1 β 1 ).
Use this to prove
1
ˆˆ
(24.3.7)
β − β (i) = (X X )−1 xi εi
ˆ
1 − hii
which is [DM93, equation (1.40) on p. 33].
ˆ
Answer. For this we also need to show that one gets the right β (i) if one regresses y − en α,
ˆ
ˆ
or, in other words y − en εn (n), on X . In other words, β (n) = (X X )−1 X (y − en εn (n)), which
ˆ
ˆ
is exactly (25.4.1). CHAPTER 25 Regression Diagnostics
“Regression Diagnostics” can either concentrate on observations or on variables.
Regarding observations, it looks for outliers or inﬂuential data in the dataset. Regarding variables, it checks whether there are highly collinear variables, or it keeps
track of how much each variable contributes to the MSE of the regression. Collinearity is discussed in [DM93, 6.3] and [Gre97, 9.2]. Regression diagnostics needs ﬁve
to ten times more computer resources than the regression itself, and often relies on
graphics, therefore it has only recently become part of the standard procedures.
Problem 316. 1 point Deﬁne multicollinearity.
• a. 2 points What are the symptoms of multicollinearity?
• b. 2 points How can one detect multicollinearity?
• c. 2 points How can one remedy multicollinearity?
25.1. Missing Observations
First case: data on y are missing. If you use a least squares predictor then this
will not give any change in the estimates and although the computer will think it is
more eﬃcient it isn’t.
What other schemes are there? Filling in the missing y by the arithmetic mean
of the observed y does not give an unbiased estimator.
General conclusion: in a singleequation context, ﬁlling in missing y not a good
idea.
Now missing values in the X matrix.
If there is only one regressor and a constant term, then the zero order ﬁlling in
of x “results in no changes and is equivalent with dropping the incomplete data.”
¯
The alternative: ﬁlling it with zeros and adding a dummy for the data with
missing observation amounts to exactly the same thing.
The only case where ﬁlling in missing data makes sense is: if you have multiple
regression and you can predict the missing data in the X matrix from the other data
in the X matrix.
25.2. Grouped Data
If single observations are replaced by arithmetic means of groups of observations,
then the error variances vary with the size of the group. If one takes this into
consideration, GLS still has good properties, although having the original data is of
course more eﬃcient.
25.3. Inﬂuential Observations and Outliers
The following discussion focuses on diagnostics regarding observations. To be
more precise, we will investigate how each single observation aﬀects the ﬁt established
271 272 25. REGRESSION DIAGNOSTICS by the other data. (One may also ask how the addition of any two observations aﬀects
the ﬁt, etc.)
25.3.1. The “Leverage”. The ith diagonal element hii of the “hat matrix”
is called the “leverage” of the ith observation. The leverage satisﬁes the following
identity
y i = (1 − hii )y i (i) + hii y i
ˆ
ˆ (25.3.1) ˆ
hii is therefore is the weight which y i has in the least squares estimate y i of ηi = xi β ,
ˆ
ˆ
compared with all other observations, which contribute to y i through y i (i). The
larger this weight, the more strongly this one observation will inﬂuence the estimate
of ηi (and if the estimate of ηi is aﬀected, then other parameter estimates may be
aﬀected too).
Problem 317. 3 points Explain the meanings of all the terms in equation (25.3.1)
and use that equation to explain why hii is called the “leverage” of the ith observation. Is every observation with high leverage also “inﬂuential” (in the sense that its
removal would greatly change the regression estimates)?
Answer. y i is the ﬁtted value for the ith observation, i.e., it is the BLUE of ηi , of the expected
ˆ
value of the ith observation. It is a weighted average of two quantities: the actual observation y i
(which has ηi as expected value), and y i (i), which is the BLUE of ηi based on all the other
ˆ
observations except the ith. The weight of the ith observation in this weighted average is called the
“leverage” of the ith observation. The sum of all leverages is always k, the number of parameters
in the regression. If the leverage of one individual point is much greater than k/n, then this point
has much more inﬂuence on its own ﬁtted value than one should expect just based on the number
of observations,
Leverage is not the same as inﬂuence; if an observation has high leverage, but by accident
ˆ
the observed value y i is very close to y i (i), then removal of this observation will not change the
regression results much. Leverage is potential inﬂuence. Leverage does not depend on any of the
observations, one only needs the X matrix to compute it. Those observations whose xvalues are away from the other observations have
“leverage” and can therefore potentially inﬂuence the regression results more than the
others. hii serves as a measure of this distance. Note that hii only depends on the X matrix, not on y , i.e., points may have a high leverage but not be inﬂuential, because
the associated y i blends well into the ﬁt established by the other data. However,
regardless of the observed value of y , observations with high leverage always aﬀect
ˆ
the covariance matrix of β .
(25.3.2) hii = det(X X ) − det(X (i) X (i))
,
det(X X ) where X (i) is the X matrix without the ith observation.
Problem 318. Prove equation (25.3.2).
Answer. Since X (i)X (i) = X X − xi xi , use theorem A.7.3 with W = X X , α = −1,
and d = xi . Problem 319. Prove the following facts about the diagonal elements of the socalled “hat matrix” H = X (X X )−1 X , which has its name because H y = y ,
ˆ
i.e., it puts the hat on y .
• a. 1 point H is a projection matrix, i.e., it is symmetric and idempotent.
Answer. Symmetry follows from the laws for the transposes of products: H = (ABC ) =
C B A = H where A = X , B = (X X )−1 which is symmetric, and C = X . Idempotency
X (X X )−1 X X (X X )−1 X = X (X X )−1 X . 25.4. SENSITIVITY OF ESTIMATES TO OMISSION OF ONE OBSERVATION 273 • b. 1 point Prove that a symmetric idempotent matrix is nonnegative deﬁnite.
Hg Answer. If H is symmetric and idempotent, then for arbitrary g , g H g = g H H g =
2 ≥ 0. But g H g ≥ 0 for all g is the criterion which makes H nonnegative deﬁnite. • c. 2 points Show that
0 ≤ hii ≤ 1 (25.3.3) Answer. If ei is the vector with a 1 on the ith place and zeros everywhere else, then ei Hei =
hii . From H nonnegative deﬁnite follows therefore that hii ≥ 0. hii ≤ 1 follows because I − H is
symmetric and idempotent (and therefore nonnegative deﬁnite) as well: it is the projection on the
orthogonal complement. • d. 2 points Show: the average value of the hii is
hii /n = k/n, where k is
the number of columns of X . (Hint: for this you must compute the trace tr H .)
Answer. The average can be written as
1
n tr(H ) = 1
n tr(X (X X )−1 X ) = 1
n tr(X X (X X )−1 ) = 1
n tr(I k ) = k
.
n Here we used tr BC = tr CB (Theorem A.1.2). • e. 1 point Show that
ones. 1
n ιι is a projection matrix. Here ι is the nvector of 1
• f . 2 points Show: If the regression has a constant term, then H − n ιι
projection matrix. is a Answer. If ι, the vector of ones, is one of the columns of X (or a linear combination
of these columns), this means there is a vector a with ι = Xa. From this follows Hιι =
1
X (X X )−1 X X aι = Xaι = ιι . One can use this to show that H − n ιι is idempotent:
1
1
1
1
1
1
1
1
1
(H − n ιι )(H − n ιι ) = HH − H n ιι − n ιι H + n ιι n ιι = H − n ιι − n ιι + n ιι =
1
H − n ιι . • g. 1 point Show: If the regression has a constant term, then one can sharpen
inequality (25.3.3) to 1/n ≤ hii ≤ 1.
Answer. H − ιι /n is a projection matrix, therefore nonnegative deﬁnite, therefore its diagonal elements hii − 1/n are nonnegative. • h. 3 points Why is hii called the “leverage” of the ith observation? To get full
points, you must give a really good verbal explanation.
ˆ
Answer. Use equation (24.2.12). Eﬀect on any other linear combination of β is less than the
eﬀect on y i . Distinguish from inﬂuence. Leverage depends only on X matrix, not on y .
ˆ hii is closely related to the test statistic testing whether the xi comes from the
same multivariate normal distribution as the other rows of the X matrix. Belsley,
Kuh, and Welsch [BKW80, p. 17] say those observations i with hii > 2k/n, i.e.,
more than twice the average, should be considered as “leverage points” which might
deserve some attention.
25.4. Sensitivity of Estimates to Omission of One Observation
The most straightforward approach to sensitivity analysis is to see how the estimates of the parameters of interest are aﬀected if one leaves out the ith observation.
In the case of linear regression, it is not necessary for this to run n diﬀerent regressions, but one can derive simple formulas for the changes in the parameters of
interest. Interestingly, the various sensitivity measures to be discussed below only
depend on the two quantities hii and εi .
ˆ 274 25. REGRESSION DIAGNOSTICS ˆ
25.4.1. Changes in the Least Squares Estimate. Deﬁne β (i) to be the
OLS estimate computed without the ith observation, and εi (i) = 1−1 ii εi the ith
ˆ
hˆ
predictive residual. Then
(25.4.1) ˆˆ
β − β (i) = (X X )−1 xi εi (i)
ˆ Problem 320. Show (25.4.1) by methods very similar to the proof of (24.2.9)
Answer. Here is this bruteforce proof, I think from [BKW80]: Let y (i) be the y vector with
the ith observation deleted. As shown in Problem 308, X (i)y (i) = X y − xi y i . Therefore by
(24.2.6)
ˆ
β (i) = (X (i)X (i))−1 X (i)y (i) = (X X )−1 + (X X )−1 xi xi (X X )−1 X y − xi y i
1 − hii
hii
1
ˆ
ˆ
= β − (X X )−1 xi y i +
(X X )−1 xi xi β −
(X X )−1 xi y i
1 − hii
1 − hii
1
1
1
ˆ
ˆˆ
(X X )−1 xi y i +
(X X )−1 xi xi β = β −
(X X )−1 xi εi
ˆ
=β−
1 − hii
1 − hii
1 − hii = To understand (25.4.1), note the following fact which is interesting in its own
ˆ
right: β (i), which is deﬁned as the OLS estimator if one drops the ith observation,
can also be obtained as the OLS estimator if one replaces the ith observation by the
prediction of the ith observation on the basis of all other observations, i.e., by y i (i).
ˆ
Writing y ((i)) for the vector y whose ith observation has been replaced in this way,
one obtains
(25.4.2) ˆ
β = (X X )−1 X y ; ˆ
β (i) = (X X )−1 X y ((i)). Since y − y ((i)) = ei εi (i) and xi = X ei (25.4.1) follows.
ˆ
ˆ
ˆ
The quantities hii , β (i)−β , and s2 (i) are computed by the Rfunction lm.influence.
Compare [CH93, pp. 129–131].
25.4.2. Scaled Measures of Sensitivity. In order to assess the sensitivity of
the estimate of any linear combination of the elements of β , φ = t β , it makes sense
ˆ
to divide the change in t β due to omission of the ith observation by the standard
ˆ
deviation of t β , i.e., to look at
ˆˆ
t (β − β (i)) (25.4.3)
σ . t (X X )−1 t Such a standardization makes it possible to compare the sensitivity of diﬀerent
ˆ
linear combinations, and to ask: Which linear combination of the elements of β is
aﬀected most if one drops the ith observation? Interestingly and, in hindsight, perhaps not surprisingly, the linear combination which is most sensitive to the addition
of the ith observation, is t = xi .
For a mathematical proof we need the following inequality, which is nothing but
the CauchySchwartz inequality in disguise:
Theorem 25.4.1. If Ω is positive deﬁnite symmetric, then
(25.4.4) max
g (g x)2
g Ωg = x Ω −1 x. If the denominator in the fraction on the lefthand side is zero, then g = o and
therefore the numerator is necessarily zero as well. In this case, the fraction itself
should be considered zero. 25.4. SENSITIVITY OF ESTIMATES TO OMISSION OF ONE OBSERVATION 275 Proof: As in the derivation of the BLUE with nonsperical covariance matrix, pick
a nonsingular Q with Ω = QQ , and deﬁne P = Q−1 . Then it follows P ΩP = I .
Deﬁne y = P x and h = Q g . Then h y = g x, h h = g Ωg , and y y =
x Ω −1 x. Therefore (25.4.4) follows from the CauchySchwartz inequality (h y )2 ≤
(h h)(y y ).
Using Theorem 25.4.1 and equation (25.4.1) one obtains
(25.4.5) max
t ˆˆ
(t (β − β (i)))2
σ2 t (X X )−1 t = 1ˆ ˆ
ˆˆ
(β − β (i)) X X (β − β (i)) =
σ2 1
hii
x (X X )−1 X X (X X )−1 xi ε2 (i) = 2 ε2 (i)
ˆi
ˆ
σ2 i
σi
Now we will show that the linear combination which attains this maximum, i.e.,
which is most sensitive to the addition of the ith observation, is t = xi . If one
premultiplies (25.4.1) by xi one obtains
= hii
εi = hii εi (i)
ˆ
ˆ
1 − hii
If one divides (25.4.6) by the standard deviation of y i , i.e., if one applies the conˆ
struction (25.4.3), one obtains
√
√
y i − y i (i)
ˆ
ˆ
hii
hii
√
(25.4.7)
εi (i) =
ˆ
εi
ˆ
=
σ
σ (1 − hii )
σ hii
ˆ
ˆ
If y i changes only little (compared with the standard deviation of y i ) if the ith
ˆ
observation is removed, then no other linear combination of the elements of β will
be aﬀected much by the omission of this observation either.
The righthand side of (25.4.7), with σ estimated by s(i), is called by [BKW80]
and many others DFFITS (which stands for DiFference in FIT, Standardized). If
one takes its square, divides it by k , and estimates σ 2 by s2 (which is more consistent
ˆ
than using s2 (i), since one standardizes by the standard deviation of t β and not
ˆ(i)), one obtains Cook’s distance [Coo77]. (25.4.5) gives an equation
by that of t β
ˆˆ
for Cook’s distance in terms of β − β (i):
(25.4.8)
ˆˆ
ˆˆ
(β − β (i)) X X (β − β (i))
hii
hii
Cook’s distance =
ˆi
ε2
ˆ
= 2 ε2 (i) = 2
2
ks
ks
k s (1 − hii )2 i
(25.4.6) ˆ
ˆ
y i − y i (i) = xi β − xi β (i) =
ˆ
ˆ Problem 321. Can you think of a situation in which an observation has a small
residual but a large “inﬂuence” as measured by Cook’s distance?
Answer. Assume “all observations are clustered near each other while the solitary odd observation lies a way out” as Kmenta wrote in [Kme86, p. 426]. If the observation happens to lie
on the regression line, then it can be discovered by its inﬂuence on the variancecovariance matrix
(25.3.2), i.e., in this case only the hii count. Problem 322. The following is the example given in [Coo77]. In R, the command data(longley) makes the data frame longley available, which has the famous
Longleydata, a standard example for a highly multicollinear dataset. These data
are also available on the web at www.econ.utah.edu/ehrbar/data/longley.txt.
attach(longley) makes the individual variables available as Robjects.
• a. 3 points Look at the data in a scatterplot matrix and explain what you
see. Later we will see that one of the observations is in the regression much more
inﬂuential than the rest. Can you see from the scatterplot matrix which observation
that might be? 276 25. REGRESSION DIAGNOSTICS Answer. In linux, you ﬁrst have to give the command x11() in order to make the graphics window available. In windows, this is not necessary. It is important to display the data in a reasonable
order, therefore instead of pairs(longley) you should do something like attach(longley) and then
pairs(cbind(Year, Population, Employed, Unemployed, Armed.Forces, GNP, GNP.deflator)). Put
Year ﬁrst, so that all variables are plotted against Year on the horizontal axis.
Population vs. year is a very smooth line.
Population vs GNP also quite smooth.
You see the huge increase in the armed forced in 1951 due to the Korean War, which led to a
(temporary) drop in unemployment and a (not so temporary) jump in the GNP deﬂator.
Otherwise the unemployed show the stopandgo scenario of the ﬁfties.
unemployed is not correlated with anything.
One should expect a strong negative correlation between employed and unemployed, but this
is not the case. • b. 4 points Run a regression of the model Employed ~ GNP.deflator + GNP
+ Unemployed + Armed.Forces + Population + Year and discuss the result.
Answer. To ﬁt a regression run longley.fit < lm(Employed ~ GNP + Unemployed + Armed.Forces
+ Population + Year). You can see the regression results by typing summary(longley.fit).
Armed forces and unemployed are signiﬁcant and have negative sign, as expected.
GNP and Population are insigniﬁcant and have negative sign too, this is not expected. GNP,
Population and Year are highly collinear. • c. 3 points Make plots of the ordinary residuals and the standardized residuals
against time. How do they diﬀer? In R, the commands are plot(Year, residuals(longley.fit),
type="h", ylab="Ordinary Residuals in Longley Regression"). In order to
get the next plot in a diﬀerent graphics window, so that you can compare them,
do now either x11() in linux or windows() in windows, and then plot(Year,
rstandard(longley.fit), type="h", ylab="Standardized Residuals in Longley
Regression").
Answer. You see that the standardized residuals at the edge of the dataset are bigger than
the ordinary residuals. The datapoints at the edge are better able to attract the regression plane
than those in the middle, therefore the ordinary residuals are “too small.” Standardization corrects
for this. • d. 4 points Make plots of the predictive residuals. Apparently there is no special
command in R to do this, therefore you should use formula (24.2.9). Also plot the
standardized predictive residuals, and compare them.
Answer. The predictive residuals are plot(Year, residuals(longley.fit)/(1hatvalues(longley.fit)),
type="h", ylab="Predictive Residuals in Longley Regression"). The standardized predictive
residuals are often called studentized residuals, plot(Year, rstudent(longley.fit), type="h",
ylab="Standardized predictive Residuals in Longley Regression").
A comparison shows an opposite eﬀect as with the ordinary residuals: the predictive residuals
at the edge of the dataset are too large, and standardization corrects this.
Speciﬁc results: standardized predictive residual in 1950 smaller than that in 1962, but predictive residual in 1950 is very close to 1962.
standardized predictive residual in 1951 smaller than that in 1956, but predictive residual in
1951 is larger than in 1956.
Largest predictive residual is 1951, but largest standardized predictive residual is 1956. • e. 3 points Make a plot of the leverage, i.e., the hii values, using plot(Year,
hatvalues(longley.fit), type="h", ylab="Leverage in Longley Regression"),
and explain what leverage means.
• f . 3 points One observation is much more inﬂuential than the others; which
is it? First look at the plots for the residuals, then look also at the plot for leverage, 25.4. SENSITIVITY OF ESTIMATES TO OMISSION OF ONE OBSERVATION 277 and try to guess which is the most inﬂuential observation. Then do it the right way.
Can you give reasons based on your prior knowledge about the time period involved
why an observation in that year might be inﬂuential?
Answer. The “right” way is to use Cook’s distance: plot(Year, cooks.distance(longley.fit),
type="h", ylab="Cook’s Distance in Longley Regression")
One sees that 1951 towers above all others. It does not have highest leverage, but it has
secondhighest, and a bigger residual than the point with the highest leverage.
1951 has the largest distance of .61. The second largest is the last observation in the dataset,
1962, with a distance of .47, and the others have .24 or less. Cook says: removal of 1951 point will
ˆ
move the least squares estimate to the edge of a 35% conﬁdence region around β . This point is
probably so inﬂuential because 1951 was the ﬁrst full year of the Korean war. One would not be
able to detect this point from the ordinary residuals, standardized or not! The predictive residuals
are a little better; their maximum is at 1951, but several other residuals are almost as large. 1951
is so inﬂuential because it has an extremely high hatvalue, and one of the highest values for the
ordinary residuals! At the end don’t forget to detach(longley) if you have attached it before.
25.4.3. Changes in the Sum of Squared Errors. For the computation of
s2 (i) from the regression results one can take advantage of the following simple
relationship between the SSE for the regression with and without the ith observation: SSE − SSE (i) = (25.4.9) ε2
ˆi
1 − hii Problem 323. Use (25.4.9) to derive the following formula for s2 (i):
s2 (i) = (25.4.10) ε2
ˆi
1
(n − k )s2 −
n−k−1
1 − hii Answer. This merely involves rewriting SSE and SSE (i) in terms of s2 and s2 (i).
s2 (i) = (25.4.11) ε2
ˆi
1
SSE (i)
=
S SE −
n−1−k
n−k−1
1 − hii Proof of equation (25.4.9):
ˆ
(y j − xj β (i))2 = SSE (i) =
j : j =i =
j : j =i εj +
ˆ
j ε2 +
ˆj =
j 2 j : j =i εj +
ˆ = ˆ
ˆ
ˆ
y j − xj β − xj (β (i) − β ) hji
εi
ˆ
1 − hii hji
εi
ˆ
1 − hii 2ε i
ˆ
1 − hii 2 2 − 1
εi
ˆ
1 − hii hij εj +
ˆ
j 2 εi
ˆ
1 − hii 2 h2 −
ji
j εi
ˆ
1 − hii 2 In the last line the ﬁrst term is SSE . The second term is zero because H ε = o.
ˆ
Furthermore, hii = j h2 because H is symmetric and idempotent, therefore the
ji
sum of the last two items is −ε2 /(1 − hii ).
ˆi
Note that every single relationship we have derived so far is a function of εi and
ˆ
hii . 278 25. REGRESSION DIAGNOSTICS Problem 324. 3 points What are the main concepts used in modern “Regression
Diagnostics”? Can it be characterized to be a careful look at the residuals, or does it
have elements which cannot be inferred from the residuals alone?
Answer. Leverage (sometimes it is called “potential”) is something which cannot be inferred
from the residuals, it does not depend on y at all. Problem 325. An observation in a linear regression model is “inﬂuential” if
its omission causes large changes to the regression results. Discuss how you would
ascertain in practice whether a given observation is inﬂuential or not.
• a. What is meant by leverage? Does high leverage necessarily imply that an
observation is inﬂuential?
Answer. Leverage is potential inﬂuence. It only depends of X , not on y . It is the distance
of the observation from the center of gravity of all observations. Whether this is actual inﬂuence
depends on the y values. • b. How are the concepts of leverage and inﬂuence aﬀected by sample size?
• c. What steps would you take when alerted to the presence of an inﬂuential
observation?
Answer. Make sure you know whether the results you rely on are aﬀected if that inﬂuential
observation is dropped. Try to ﬁnd out why this observation is inﬂuential (e.g. in the Longley data
the observations in the year when the Korean War started are inﬂuential). • d. What is a “predictive residual” and how does it diﬀer from an ordinary
residual?
• e. Discuss situations in which one would want to deal with the “predictive”
residuals rather than the ordinary residuals, and situations in which one would want
residuals standardized versus situations in which it would be preferable to have the
unstandardized residuals.
Problem 326. 6 points Describe what you would do to ascertain that a regression
you ran is correctly speciﬁed?
Answer. Economic theory behind that regression, size and sign of coeﬃcients, plot residuals
versus predicted values, time, and every independent variable, run all tests: F test, ttests, R2 ,
DW, portmanteau test, forecasting, multicollinearity, inﬂuence statistics, overﬁtting to see if other
variables are signiﬁcant, try to defeat the result by using alternative variables, divide time period into
subperiods in order to see if parameters are constant over time, pretest speciﬁcation assumptions. CHAPTER 26 Asymptotic Properties of the OLS Estimator
A much more detailed treatment of the contents of this chapter can be found in
[DM93, Chapters 4 and 5].
Here we are concerned with the consistency of the OLS estimator for large samples. In other words, we assume that our regression model can be extended to
encompass an arbitrary number of observations. First we assume that the regressors
are nonstochastic, and we will make the following assumption:
1
(26.0.12)
Q = lim X X exists and is nonsingular.
n→∞ n
Two examples where this is not the case. Look at the model y t = α + βt + εt . Here 11
1 2 1 + 1 + 1 + ··· + 1 1 + 2 + 3 + ··· + n =
X = 1 3 . Therefore X X =
1 + 2 + 3 + · · · + n 1 + 4 + 9 + · · · + n2
. . . . ..
1n
n
n(n + 1)/2
1∞
1
, and n X X →
. Here the assumption
n(n + 1)/2 n(n + 1)(2n + 1)/6
∞∞
(26.0.12) does not hold, but one can still prove consistency and asymtotic normality,
the estimators converge even faster than in the usual case.
The other example is the model y t = α + βλt + εt with a known λ with −1 <
λ < 1. Here
1 + 1 + ··· + 1
λ + λ 2 + · · · + λn
X X=
=
2
n
λ + λ + ··· + λ
λ2 + λ 4 + · · · + λ 2n
= (λ − λ n
)/(1 − λ) n+1 (λ − λn+1 )/(1 − λ)
.
(λ2 − λ2n+2 )/(1 − λ2 ) 10
, which is singular. In this case, a consistent estimate of
00
λ does not exist: future observations depend on λ so little that even with inﬁnitely
many observations there is not enough information to get the precise value of λ.
ˆ
We will show that under assumption (26.0.12), β and s2 are consistent. However
this assumption is really too strong for consistency. A weaker set of assumptions
is the Grenander conditions, see [Gre97, p. 275]. To write down the Grenander
conditions, remember that presently X depends on n (in that we only look at the
ﬁrst n elements of y and ﬁrst n rows of X ), therefore also the column vectors xj also
depend of n (although we are not indicating this here). Therefore xj xj depends
on n as well, and we will make this dependency explicit by writing xj xj = d2 .
nj
Then the ﬁrst Grenander condition is limn→∞ d2 = +∞ for all j . Second: for all i
nj
and k , limn→∞ maxi=1···n xij /d2 = 0 (here is a typo in Greene, he leaves the max
nj
out). Third: Sample correlation matrix of the columns of X minus the constant
term converges to a nonsingular matrix.
Therefore 1
nX X→ 279 280 26. ASYMPTOTIC PROPERTIES OF OLS Consistency means that the probability limit of the estimates converges towards
ˆ
ˆ
the true value. For β this can be written as plimn→∞ β n = β . This means by
ˆn − β  ≤ ε] = 1.
deﬁnition that for all ε > 0 follows limn→∞ Pr[β
The probability limit is one of several concepts of limits used in probability
theory. We will need the following properties of the plim here:
(1) For nonrandom magnitudes, the probability limit is equal to the ordinary
limit.
(2) It satisﬁes the Slutsky theorem, that for a continuous function g ,
(26.0.13) plim g (z ) = g (plim(z )). (3) If the MSE matrix of an estimator converges towards the null matrix, then
the estimator is consistent.
(4) Kinchine’s theorem: the sample mean of an i.i.d. distribution is a consistent
estimate of the population mean, even if the distribution does not have a population
variance.
26.1. Consistency of the OLS estimator
ˆ
For the proof of consistency of the OLS estimators β and of s2 we need the
following result:
1
X ε = o.
n
I.e., the true ε is asymptotically orthogonal to all columns of X . This follows immediately from MSE [o; X ε /n] = E [X εε X /n2 ] = σ 2 X X /n2 , which converges
towards O .
ˆ
ˆ
In order to prove consistency of β and s2 , transform the formulas for β and s2
in such a way that they are written as continuous functions of terms each of which
ˆ
converges for n → ∞, and then apply Slutsky’s theorem. Write β as
(26.1.1) (26.1.2)
(26.1.3)
(26.1.4) plim XX
ˆ
β = β + (X X )−1 X ε = β +
n
X X −1
Xε
ˆ
plim β = β + lim
plim
n
n
−1
= β + Q o = β. −1 X ε n Let’s look at the geometry of this when there is only one explanatory variable.
The speciﬁcation is therefore y = xβ + ε . The assumption is that ε is asymptotically
orthogonal to x. In small samples, it only happens by sheer accident with probability
0 that ε is orthogonal to x. Only ε is. But now let’s assume the sample grows
ˆ
larger, i.e., the vectors y and x become very highdimensional observation vectors,
i.e. we are drawing here a twodimensional subspace out of a very highdimensional
space. As more and more data are added, the observation vectors also become
√
lengths of these
longer and longer. But if we divide each vector by n, then the√
normalized lenghts stabilize. The squared length of the vector ε / n has the plim
1
of σ 2 . Furthermore, assumption (26.0.12) means in our case that plimn→∞ n x x
1
exists and is nonsingular. This is the squared length of √n x. I.e., if we normalize the
√
vectors by dividing them by n, then they do not get longer but converge towards
1
a ﬁnite length. And the result (26.1.1) plim n x ε = 0 means now that with this
√
√
normalization, ε / n becomes more and more orthogonal to x/ n. I.e., if n is large
enough, asymptotically, not only ε but also the true ε is orthogonal to x, and this
ˆ
ˆ
means that asymptotically β converges towards the true β . 26.2. ASYMPTOTIC NORMALITY OF THE LEAST SQUARES ESTIMATOR 281 For the proof of consistency of s2 we need, among others, that plim ε nε = σ 2 ,
which is a consequence of Kinchine’s theorem. Since ε ε = ε M ε it follows
ˆˆ
εε
ˆˆ
I
n
X X X −1 X
ε
ε=
=
−
n−k
n−k
n
n
n
n
n
ε ε ε X X X −1 X ε
=
−
→ 1 · σ 2 − o Q−1 o .
n−k n
n
n
n
26.2. Asymptotic Normality of the Least Squares Estimator
√ To show asymptotic normality of an estimator, multiply the sampling error by
n, so that the variance is stabilized.
1
1
We have seen plim n X ε = o. Now look at √n X ε n . Its mean is o and its covariance matrix σ 2 X n X . Shape of distribution, due to a variant of the Central Limit
1
Theorem, is asymptotically normal: √n X ε n → N (o, σ 2 Q). (Here the convergence
is convergence in distribution.)
−1
√ˆ
1
We can write n(β n − β ) = X n X
( √n X ε n ). Therefore its limiting covari√ˆ
ance matrix is Q−1 σ 2 QQ−1 = σ 2 Q−1 , Therefore n(β n − β ) → N (o, σ 2 Q−1 ) in disˆ
tribution. One can also say: the asymptotic distribution of β is N (β , σ 2 (X X )−1 ).
√
ˆn − Rβ ) → N (o, σ 2 RQ−1 R ), and therefore
From this follows n(Rβ
(26.2.1) ˆ
n(Rβ n − Rβ ) RQ−1 R −1 ˆ
(Rβ n − Rβ ) → σ 2 χ2 .
i Divide by s2 and replace in the limiting case Q by X X /n and s2 by σ 2 to get
−1
ˆ
ˆ
(Rβ n − Rβ ) R(X X )−1 R
(Rβ n − Rβ )
→ χ2
i
2
s
in distribution. All this is not a proof; the point is that in the denominator, the
distribution is divided by the increasingly bigger number n − k , while in the numerator, it is divided by the constant i; therefore asymptotically the denominator can
be considered 1.
The central limit theorems only say that for n → ∞ these converge towards the
χ2 , which is asymptotically equal to the F distribution. It is easily possible that
before one gets to the limit, the F distribution is better. (26.2.2) ˆ
Problem 327. Are the residuals y − X β asymptotically normally distributed?
√ Answer. Only if the disturbances are normal, otherwise of course not! We can show that
√
ˆ
n(ε − ε) = nX (β − β ) ∼ N (o, σ 2 XQX ).
ˆ Now these results also go through if one has stochastic regressors. [Gre97, 6.7.7]
shows that the above condition (26.0.12) with the lim replaced by plim holds if xi
and ε i are an i.i.d. sequence of random variables.
Problem 328. 2 points In the regression model with random regressors y =
1
1
X β +ε , you only know that plim n X X = Q is a nonsingular matrix, and plim n X ε =
o. Using these two conditions, show that the OLS estimate is consistent.
ˆ
Answer. β = (X X )−1 X y = β + (X X )−1 X ε due to (18.0.7), and
plim(X X )−1 X ε = plim( X X −1 X ε
)
= Qo = o.
n
n CHAPTER 27 Least Squares as the Normal Maximum Likelihood
Estimate
Now assume ε is multivariate normal. We will show that in this case the OLS
ˆ
estimator β is at the same time the Maximum Likelihood Estimator. For this we
need to write down the density function of y . First look at one y t which is y t ∼ x1
.
2
N (xt β , σ ), where X = . , i.e., xt is the tth row of X . It is written as a
.
xn
column vector, since we follow the “column vector convention.” The (marginal)
density function for this one observation is
(27.0.3) fyt (yt ) = √ 1
2πσ 2 e−(yt −xt β )2 /2σ 2 . Since the y i are stochastically independent, their joint density function is the product,
which can be written as
1
(27.0.4)
fy (y ) = (2πσ 2 )−n/2 exp − 2 (y − Xβ ) (y − Xβ ) .
2σ
To compute the maximum likelihood estimator, it is advantageous to start with
the log likelihood function:
(27.0.5) log fy (y ; β , σ 2 ) = − n
n
1
log 2π − log σ 2 − 2 (y − Xβ ) (y − Xβ ).
2
2
2σ Assume for a moment that σ 2 is known. Then the MLE of β is clearly equal to
ˆ
ˆ
the OLS β . Since β does not depend on σ 2 , it is also the maximum likelihood
2
ˆ
estimate when σ is unknown. β is a linear function of y . Linear transformations
of normal variables are normal. Normal distributions are characterized by their
mean vector and covariance matrix. The distribution of the MLE of β is therefore
ˆ
β ∼ N (β , σ 2 (X X )−1 ).
ˆ
If we replace β in the log likelihood function (27.0.5) by β , we get what is called
the log likelihood function with β “concentrated out.”
ˆ
(27.0.6) log fy (y ; β = β , σ 2 ) = − n
n
1
ˆ
ˆ
log 2π − log σ 2 − 2 (y − X β ) (y − X β ).
2
2
2σ One gets the maximum likelihood estimate of σ 2 by maximizing this “concentrated”
log likelihoodfunction. Taking the derivative with respect to σ 2 (consider σ 2 the
name of a variable, not the square of another variable), one gets
∂
n1
1
ˆ
ˆ
ˆ
log fy (y ; β ) = −
+ 4 (y − X β ) (y − X β )
∂σ 2
2 σ2
2σ
Setting this zero gives
(27.0.7) (27.0.8) σ2 =
˜ ˆ
ˆ
(y − X β ) (y − X β )
εε
ˆˆ
=
.
n
n
283 284 27. LEAST SQUARES AS THE NORMAL MAXIMUM LIKELIHOOD ESTIMATE This is a scalar multiple of the unbiased estimate s2 = ε ε/(n − k ) which we
ˆˆ
had earlier.
Let’s look at the distribution of s2 (from which that of its scalar multiples follows
easily). It is a quadratic form in a normal variable. Such quadratic forms very often
have χ2 distributions.
Now recall equation 7.4.9 characterizing all the quadratic forms of multivariate
normal variables that are χ2 ’s. Here it is again: Assume y is a multivariate normal
vector random variable with mean vector µ and covariance matrix σ 2 Ψ, and Ω is a
symmetric nonnegative deﬁnite matrix. Then (y − µ) Ω (y − µ) ∼ σ 2 χ2 iﬀ
k
ΨΩ ΨΩ Ψ = ΨΩ Ψ, (27.0.9) and k is the rank of ΨΩ .
This condition is satisﬁed in particular if Ψ = I (the identity matrix) and
Ω2 = Ω, and this is exactly our situation.
(27.0.10) σ2 =
ˆ ˆ
ˆ
(y − X β ) (y − X β )
ε (I − X (X X )−1 X )ε
ε Mε
=
=
n−k
n−k
n−k where M 2 = M and rank M = n − k . (This last identity because for idempotent
matrices, rank = tr, and we computed its tr above.) Therefore s2 ∼ σ 2 χ2 −k /(n − k ),
n
from which one obtains again unbiasedness, but also that var[s2 ] = 2σ 4 /(n − k ), a
result that one cannot get from mean and variance alone.
ˆ
Problem 329. 4 points Show that, if y is normally distributed, s2 and β are
independent.
ˆ
Answer. We showed in question 246 that β and ε are uncorrelated, therefore in the normal
ˆ
ˆ
case independent, therefore β is also independent of any function of ε, such as σ 2 .
ˆ
ˆ Problem 330. Computer assignment: You run a regression with 3 explanatory
variables, no constant term, the sample size is 20, the errors are normally distributed
and you know that σ 2 = 2. Plot the density function of s2 . Hint: The command
dchisq(x,df=25) returns the density of a Chisquare distribution with 25 degrees of
freedom evaluated at x. But the number 25 was only taken as an example, this is not
the number of degrees of freedom you need here.
• a. In the same plot, plot the density function of the TheilSchweitzer estimate.
Can one see from the comparison of these density functions why the TheilSchweitzer
estimator has a better MSE?
Answer. Start with the TheilSchweitzer plot, because it is higher. > x < seq(from = 0, to
= 6, by = 0.01) > Density < (19/2)*dchisq((19/2)*x, df=17) > plot(x, Density, type="l",
lty=2) > lines(x,(17/2)*dchisq((17/2)*x, df=17)) > title(main = "Unbiased versus TheilSchweitzer
Variance Estimate, 17 d.f.") Now let us derive the maximum likelihood estimator in the case of nonspherical but positive deﬁnite covariance matrix. I.e., the model is y = Xβ + ε , ε ∼
N (o, σ 2 Ψ). The density function is
−1/2 (27.0.11) fy (y ) = (2πσ 2 )−n/2 det Ψ exp − 1
(y − Xβ ) Ψ−1 (y − Xβ ) .
2σ 2 Problem 331. Derive (27.0.11) as follows: Take a matrix P with the property
that P ε has covariance matrix σ 2 I . Write down the joint density function of P ε .
Since y is a linear transformation of ε , one can apply the rule for the density function
of a transformed random variable. 27. LEAST SQUARES AS THE NORMAL MAXIMUM LIKELIHOOD ESTIMATE 285 Answer. Write Ψ = QQ with Q nonsingular and deﬁne P = Q−1 and v = P ε . Then
V [v ] = σ 2 P QQ P = σ 2 I , therefore
(27.0.12) fv (v ) = (2πσ 2 )−n/2 exp − 1
vv.
2σ 2 For the transformation rule, write v , whose density function you know, as a function of y , whose
density function you want to know. v = P (y − Xβ ); therefore the Jacobian matrix is ∂ v /∂ y =
∂ (P y − P Xβ )/∂ y = P , or one can see it also element by element ∂ v1 ∂ y1 (27.0.13) .
.
.
∂ vn
∂ y1 ··· ∂ v1
∂ yn .. .
.
. .
··· ∂ vn
∂ yn = P, therefore one has to do two things: ﬁrst, substitute P (y − Xβ ) for v in formula (27.0.12), and
secondly multiply by the absolute value of the determinant of the Jacobian. Here is how to express the determinant of the Jacobian in terms of Ψ: From Ψ−1 = (QQ )−1 = (Q )−1 Q−1 =
√
(Q−1 ) Q−1 = P P follows (det P )2 = (det Ψ)−1 , hence det P  = det Ψ. From (27.0.11) one obtains the following log likelihood function:
(27.0.14)
n
n
1
1
log fy (y ) = − ln 2π − ln σ 2 − ln det[Ψ] − 2 (y − Xβ ) Ψ−1 (y − Xβ ).
2
2
2
2σ
Here, usually not only the elements of β are unknown, but also Ψ depends on
unknown parameters. Instead of concentrating out β , we will ﬁrst concentrate out
σ 2 , i.e., we will compute the maximum of this likelihood function over σ 2 for any
given set of values for the data and the other parameters:
(27.0.15)
(27.0.16) ∂
(y − Xβ ) Ψ−1 (y − Xβ )
n1
log fy (y ) = −
+
2
2
∂σ
2σ
2σ 4
−1
(y − Xβ ) Ψ (y − Xβ )
σ2 =
˜
.
n Whatever the value of β or the values of the unknown parameters in Ψ, σ 2 is the
˜
value of σ 2 which, together with the given β and Ψ, gives the highest value of the
likelihood function. If one plugs this σ 2 into the likelihood function, one obtains the
˜
socalled “concentrated likelihood function” which then only has to be maximized
over β and Ψ:
(27.0.17)
n
n
1
log fy (y ; σ 2 ) = − (1 + ln 2π − ln n) − ln(y − Xβ ) Ψ−1 (y − Xβ ) − ln det[Ψ]
˜
2
2
2
This objective function has to be maximized with respect to β and the parameters
ˆ
entering Ψ. If Ψ is known, then this is clearly maximized by the β minimizing
(19.0.11), therefore the GLS estimator is also the maximum likelihood estimator.
If Ψ depends on unknown parameters, it is interesting to compare the maximum likelihood estimator with the nonlinear least squares estimator. The objective
function minimized by nonlinear least squares is (y − Xβ ) Ψ−1 (y − Xβ ), which
is the sum of squares of the innovation parts of the residuals. These two objective
1
functions therefore diﬀer by the factor (det[Ψ]) n , which only matters if there are
unknown parameters in Ψ. Asymptotically, the objective functions are identical.
Using the factorization theorem for suﬃcient statistics, one also sees easily that
ˆ
σ 2 and β together form suﬃcient statistics for σ 2 and β . For this use the identity
ˆ
ˆ
ˆ
ˆ
ˆ
(y − Xβ ) (y − Xβ ) = (y − X β ) (y − X β ) + (β − β ) X X (β − β )
(27.0.18) ˆ
ˆ
= (n − k )s2 + (β − β ) X X (β − β ). 286 27. LEAST SQUARES AS THE NORMAL MAXIMUM LIKELIHOOD ESTIMATE Therefore the observation y enters the likelihood function only through the two
ˆ
statistics β and s2 . The factorization of the likelihood function is therefore the trivial
factorization in which that part which does not depend on the unknown parameters
but only on the data is unity.
Problem 332. 12 points The log likelihood function in the linear model is given
by (27.0.5). Show that the inverse of the information matrix is
σ 2 (X X )−1
o (27.0.19) o
2σ 4 /n The information matrix can be obtained in two diﬀerent ways. Its typical element
has the following two forms:
∂ ln
∂ θi
or written as matrix derivatives
∂ ln
(27.0.21)
E[
∂θ
(27.0.20) E[ ∂ ln
∂ 2 ln = − E[,
∂ θk
∂ θi ∂θk
∂ ln
∂ 2 ln = − E[.
∂θ
∂ θ∂ θ β
. The expectation is taken under the assumption that the
σ2
parameter values are the true values. Compute it both ways. In our case θ = Answer. The log likelihood function can be written as
n
n
1
(27.0.22)
ln = − ln 2π − ln σ 2 −
(y y − 2 y X β + β X X β ) .
2
2
2σ 2
The ﬁrst derivatives were already computed for the maximum likelihood estimators:
∂
1
1
1
ln = − 2 (2y X + 2β X X ) = 2 (y − Xβ ) X = 2 ε X
2σ
σ
σ
∂β
n
1
1
∂
n
ln = − 2 +
(y − Xβ ) (y − Xβ ) = − 2 +
(27.0.24)
εε
∂σ 2
2σ
2σ 4
2σ
2σ 4
By the way, one sees that each of these has expected value zero, which is a fact that is needed to
prove consistency of the maximum likelihood estimator.
The formula with only one partial derivative will be given ﬁrst, although it is more tedious:
(27.0.23) By doing ∂
∂β ∂
∂β we get a symmetric 2 × 2 partitioned matrix with the diagonal elements (27.0.25) E[ 1
1
X εε X ] = 2 X X
σ4
σ and
1
n
n
1
1
+
2nσ 4 =
ε ε ] = var[ 4 ε ε ] =
2σ 2
2σ 4
2σ
4σ 8
2σ 4
n
1
One of the oﬀdiagonal elements is ( 2σ4 + 2σ6 ε ε )ε X . Its expected value is zero: E [ε ] = o,
and also E [εε ε ] = o since its ith component is E[εi
ε2 ] =
E[εi ε2 ]. If i = j , then εi is
j
jj
j (27.0.26) E[ − n
1
+
εε
2σ 2
2σ 4 2 = var[− independent of ε2 , therefore E[εi ε2 ] = 0 · σ 2 = 0. If i = j , we get E[ε3 ] = 0 since εi has a symmetric
j
j
i
distribution.
It is easier if we diﬀerentiate once more:
∂2
1
(27.0.27)
ln = − 2 X X
σ
∂ β∂ β
(27.0.28)
(27.0.29) ∂2
1
1
ln = − 4 X (y − Xβ ) = − 4 X ε
∂ β ∂σ 2
σ
σ
∂2
n
1
n
1
ln =
− 6 (y − Xβ ) (y − Xβ ) =
− 6ε ε
(∂σ 2 )2
2σ 4
σ
2σ 4
σ This gives the top matrix in [JHG+ 88, (6.1.24b)]:
(27.0.30) 1
− σ2 X X
1
− σ 4 (X y − X X β ) n
2σ 4 1
− σ 4 (X y − X X β )
1
− σ6 (y − Xβ ) (y − Xβ ) 27. LEAST SQUARES AS THE NORMAL MAXIMUM LIKELIHOOD ESTIMATE 287 Now assume that β and σ 2 are the true values, take expected values, and reverse the sign. This
gives the information matrix
(27.0.31) σ −2 X X
o o
n/(2σ 4 ) For the lower righthand side corner we need that E[(y − Xβ ) (y − Xβ )] = E[ε ε ] = nσ 2 .
Taking inverses gives (27.0.19), which is a lower bound for the covariance matrix; we see that
s2 with var[s2 ] = 2σ 4 /(n − k) does not attain the bound. However one can show with other means
that it is nevertheless eﬃcient. CHAPTER 28 Random Regressors
Until now we always assumed that X was nonrandom, i.e., the hypothetical
repetitions of the experiment used the same X matrix. In the nonexperimental
sciences, such as economics, this assumption is clearly inappropriate. It is only
justiﬁed because most results valid for nonrandom regressors can be generalized to
the case of random regressors. To indicate that the regressors are random, we will
write them as X . 28.1. Strongest Assumption: Error Term Well Behaved Conditionally
on Explanatory Variables
The assumption which we will discuss ﬁrst is that X is random, but the classical
assumptions hold conditionally on X , i.e., the conditional expectation E [ε X ] = o,
and the conditional variancecovariance matrix V [εX ] = σ 2 I . In this situation,
the least squares estimator has all the classical properties conditionally on X , for
ˆ
ˆ
instance E [β X ] = β , V [β X ] = σ 2 (X X )−1 , E[s2 X ] = σ 2 , etc.
Moreover, certain properties of the Least Squares estimator remain valid unconditionally. An application of the law of iterated expectations shows that the least
ˆ
squares estimator β is still unbiased. Start with (18.0.7):
(28.1.1)
(28.1.2)
(28.1.3) ˆ
β − β = (X X )−1 X ε
−1
−1
ˆ
E [β − β X ] = E [(X X ) X ε X ] = (X X ) X E [ε X ] = o.
ˆ
ˆ
E [β − β ] = E E [β − β X ] = o. Problem 333. 1 point In the model with random explanatory variables X you
˜
˜
are considering an estimator β of β . Which statement is stronger: E [β ] = β , or
˜
[β X ] = β . Justify your answer.
E
Answer. The second statement is stronger. The ﬁrst statement follows from the second by
the law of iterated expectations. Problem 334. 2 points Assume the regressors X are random, and the classical
assumptions hold conditionally on X , i.e., E [ε X ] = o and V [ε X ] = σ 2 I . Show
that s2 is an unbiased estimate of σ 2 .
Answer. From the theory with nonrandom explanatory variables follows E[s2 X ] = σ 2 .
Therefore E[s2 ] = E E[s2 X ] = E[σ 2 ] = σ 2 . In words: if the expectation conditional on X
does not depend on X , then it is also the unconditional expectation.
289 290 28. RANDOM REGRESSORS The law of iterated expectations can also be used to compute the unconditional
ˆ
MSE matrix of β :
ˆ
ˆ
ˆ
(28.1.4)
MSE [β ; β ] = E [(β − β )(β − β ) ]
(28.1.5) ˆ
ˆ
= E E [(β − β )(β − β ) X ] (28.1.6) = E [σ 2 (X X )−1 ] (28.1.7) = σ 2 E [(X X )−1 ]. ˆ
Problem 335. 2 points Show that s2 (X X )−1 is unbiased estimator of MSE [β ; β ].
Answer.
(28.1.8) −1
−1
2
2
E [s ( X X ) ] = E E [s ( X X ) X ] (28.1.9) = E [σ 2 (X X )−1 ] (28.1.10) = σ 2 E [(X X )−1 ] (28.1.11) ˆ
= MSE [β ; β ] by (28.1.7). ˜
The GaussMarkov theorem generalizes in the following way: Say β is an estima˜X ] = β (which is stronger
tor, linear in y , but not necessarily in X , satisfying E [β
˜
ˆ
than unbiasedness); then MSE [β ; β ] ≥ MSE [β ; β ]. Proof is immediate: we know by
˜
ˆ
the usual GaussMarkov theorem that MSE [β ; β X ] ≥ MSE [β ; β X ], and taking
˜
ˆ
expected values will preserve this inequality: E MSE [β ; β X ] ≥ E MSE [β ; β X ] ,
but this expected value is exactly the unconditional MSE .
The assumption E [ε X ] = o can also be written E [y X ] = X β , and V [ε X ] =
σ 2 I can also be written as V [y X ] = σ 2 I . Both of these are assumptions about the
conditional distribution y X = X for all X . This suggests the following broadening
of the regression paradigm: y and X are jointly distributed random variables, and
one is interested how y X = X depends on X . If the expected value of this distribution depends linearly, and the variance of this distribution is constant, then this
is the linear regression model discussed above. But the expected value might also
depend on X in a nonlinear fashion (nonlinear least squares), and the variance may
not be constant—in which case the intuition that y is some function of X plus some
error term may no longer be appropriate; y may for instance be the outcome of a binary choice, the probability of which depends on X (see chapter ??; the generalized
linear model).
28.2. Contemporaneously Uncorrelated Disturbances
In many situations with random regressors, the condition E [ε X ] = o is not
satisﬁed. Instead, the columns of X are contemporaneously uncorrelated with ε , but
they may be correlated with past values of ε . The main example here is regression
with a lagged dependent variable. In this case, OLS is no longer unbiased, but
asymptotically it still has all the good properties, it is asymptotically normal with the
covariance matrix which one would expect. Asymptotically, the computer printout
is still valid. This is a very important result, which is often used in econometrics,
but most econometrics textbooks do not even start to prove it. There is a proof in
[Kme86, pp. 749–757], and one in [Mal80, pp. 535–539].
Problem 336. Since least squares with random regressors is appropriate whenever the disturbances are contemporaneously uncorrelated with the explanatory variables, a friend of yours proposes to test for random explanatory variables by checking 28.3. DISTURBANCES CORRELATED WITH REGRESSORS IN SAME OBSERVATION 291 whether the sample correlation coeﬃcients between the residuals and the explanatory
variables is signiﬁcantly diﬀerent from zero or not. Is this an appropriate statistic?
Answer. No. The sample correlation coeﬃcients are always zero! 28.3. Disturbances Correlated with Regressors in Same Observation
But if ε is contemporaneously correlated with X , then OLS is inconsistent. This
can be the case in some dynamic processes (lagged dependent variable as regressor,
and autocorrelated errors, see question ??), when there are, in addition to the relation which one wants to test with the regression, other relations making the righthand
side variables dependent on the lefthand side variable, or when the righthand side
variables are measured with errors. This is usually the case in economics, and econometrics has developed the technique of simultaneous equations estimation to deal
with it.
Problem 337. 3 points What does one have to watch out for if some of the
regressors are random? CHAPTER 29 The Mahalanobis Distance
Everything in this chapter is unpublished work, presently still in draft form. The
aim is to give a motivation for the least squares objective function in terms of an
initial measure of precision. The case of prediction is mathematically simpler than
that of estimation, therefore this chapter will only discuss prediction. We assume
that the joint distribution of y and z has the form
X
Ω
y
∼
β , σ 2 yy
Ωzy
W
z Ω yz
,
Ωzz (29.0.1) σ 2 > 0, otherwise unknown
β unknown as well. y is observed but z is not and has to be predicted. But assume we are not interested
in the MSE since we do the experiment only once. We want to predict z in such a
way that, whatever the true value of β , the predicted value z ∗ “blends in” best with
the given data y .
There is an important conceptual diﬀerence between this criterion and the one
based on the MSE . The present criterion cannot be applied until after the data are
known, therefore it is called a “ﬁnal” criterion as opposed to the “initial” criterion
of the MSE . See Barnett [Bar82, pp. 157–159] for a good discussion of these issues.
How do we measure the degree to which a given data set “blend in,” i.e., are not
outliers for a given distribution? Hypothesis testing uses this criterion. The most
oftenused testing principle is: reject the null hypothesis if the observed value of a
certain statistic is too much an outlier for the distribution which this statistic would
have under the null hypothesis. If the statistic is a scalar, and if under the null
hypothesis this statistic has expected value µ and standard deviation σ , then one
often uses an estimate of x − µ /σ , the number of standard deviations the observed
value is away from the mean, to measure the “distance” of the observed value x from
the distribution (µ, σ 2 ). The Mahalanobis distance generalizes this concept to the
case that the test statistic is a vector random variable.
29.1. Deﬁnition of the Mahalanobis Distance
Since it is mathematically more convenient to work with the squared distance
than with the distance itself, we will make the following thought experiment to
motivate the Mahalanobis distance. How could one generalize the squared scalar
distance (y − µ)2 /σ 2 for the distance of a vector value y from the distribution of
the vector random variable y ∼ (µ, σ 2Ω )? If all y i have same variance σ 2 , i.e., if
Ω = I , one might measure the squared distance of y from the distribution (µ, σ 2Ω )
1
by σ2 maxi (yi − µi )2 , but since the maximum from two trials is bigger than the value
from one trial only, one should divide this perhaps by the expected value of such
2
a maximum. If the variances are diﬀerent, say σi , one might want to look a the
number of standard deviations which the “worst” component of y is away from what
would be its mean if y were an observation of y , i.e., the squared distance of the
µ2
obsrved vector from the distribution would be maxi (yi −2 i ) , again normalized by its
σi
expected value.
293 294 29. THE MAHALANOBIS DISTANCE The principle actually used by the Mahalanobis distance goes only a small step
further than the examples just cited. It is coordinatefree, i.e., any linear combinations of the elements of y are considered on equal footing with these elements
themselves. In other words, it does not distinguish between variates and variables.
The distance of a given vector value from a certain multivariate distribution is deﬁned
to be the distance of the “worst” linear combination of the elements of this vector
from the univariate distribution of this linear combination, normalized in such a way
that the expected value of this distance is 1.
Definition 29.1.1. Given a random nvector y which has expected value and a
nonsingular covariance matrix. The squared “Mahalanobis distance” or “statistical
distance” of the observed value y from the distribution of y is deﬁned to be
g y − E[g y ]
1
MHD[y ; y ] = max
g
n
var[g y ] (29.1.1) 2 . If the denominator var[g y ] is zero, then g = o, therefore the numerator is zero as
well. In this case the fraction is deﬁned to be zero.
Theorem 29.1.2. Let y be a vector random variable with E [y ] = µ and V [y ] =
σ 2Ω , σ 2 > 0 and Ω positive deﬁnite. The squared Mahalanobis distance of the value
y from the distribution of y is equal to
1
(29.1.2)
MHD[y ; y ] =
(y − µ) Ω −1 (y − µ)
nσ 2
Proof. (29.1.2) is a simple consequence of (25.4.4). It is also somewhat intuitive
since the righthand side of (29.1.2) can be considered a division of the square of y − µ
by the covariance matrix of y .
The Mahalanobis distance is an asymmetric measure; a large value indicates a
bad ﬁt of the hypothetical population to the observation, while a value of, say, 0.1
does not necessarily indicate a better ﬁt than a value of 1.
Problem 338. Let y be a random nvector with expected value µ and nonsingular
covariance matrix σ 2Ω . Show that the expected value of the Mahalobis distance of
the observations of y from the distribution of y is 1, i.e.,
(29.1.3) E MHD[y ; y ] = 1 Answer.
(29.1.4)
1
1
1
1
E[ 2 (y − µ) Ω −1 (y − µ)] = E[tr
Ω −1 (y − µ)(y − µ) ] tr( 2 Ω −1 σ 2Ω ) = tr(I ) = 1.
nσ
nσ 2
nσ
n (29.1.2) is, up to a constant factor, the quadratic form in the exponent of the
normal density function of y . For a normally distributed y , therefore, all observations
located on the same density contour have equal distance from the distribution.
The Mahalanobis distance is also deﬁned if the covariance matrix of y is singular.
In this case, certain nonzero linear combinations of the elements of y are known with
certainty. Certain vectors can therefore not possibly be realizations of y , i.e., the set
of realizations of y does not ﬁll the whole Rn .
Problem 339. 2 points The random vector y =
covariance matrix 1
3 2 −1 −1
−1 2 −1
−1 −1 2 y1
y2
y3 has mean 1
2
−3 and . Is this covariance matrix singular? If so, give a 29.1. DEFINITION OF THE MAHALANOBIS DISTANCE 295 linear combination of the elements of y which is known with certainty. And give a
value which can never be a realization of y . Prove everything you state.
Answer. Yes, it is singular;
(29.1.5) 2 −1 −1
−1 2 −1
−1 −1 2 1
1
1 = 0
0
0 I.e., y 1 + y 2 + y 3 = 0 because its variance is 0 and its mean is zero as well since [ 1 1 1 ] 1
2
−3 = 0. Definition 29.1.3. Given a vector random variable y which has a mean and
a covariance matrix. A value y has inﬁnite statistical distance from this random
variable, i.e., it cannot possibly be a realization of this random variable, if a vector
of coeﬃcients g exists such that var[g y ] = 0 but g y = g E [y ]. If such a g does
not exist, then the squared Mahalanobis distance of y from y is deﬁned as in (29.1.1),
Ω
with n replaced by rank[Ω ]. If the denominator in (29.1.1) is zero, then it no longer
necessarily follows that g = o but it nevertheless follows that the numerator is zero,
and the fraction should in this case again be considered zero.
If Ω is singular, then the inverse Ω −1 in formula (29.1.2) must be replaced by a
“ginverse.” A ginverse of a matrix A is any matrix A− which satisﬁes AA− A = A.
Ginverses always exist, but they are usually not unique.
Problem 340. a is a scalar. What is its ginverse a− ?
Theorem 29.1.4. Let y be a random variable with E [y ] = µ and V [y ] = σ 2Ω ,
σ > 0. If it is not possible to express the vector y in the form y = µ + Ω a for some
a, then the squared Mahalanobis distance of y from the distribution of y is inﬁnite,
i.e., MHD[y ; y ] = ∞; otherwise
1
(29.1.6)
MHD[y ; y ] = 2
(y − µ) Ω − (y − µ)
Ω
σ rank[Ω ]
2 Now we will dicuss how a given observation vector can be extended by additional
observations in such a way that the Mahalanobis distance of the whole vector from
its distribution is minimized. CHAPTER 30 Interval Estimation
We will ﬁrst show how the least squares principle can be used to construct
conﬁdence regions, and then we will derive the properties of these conﬁdence regions.
30.1. A Basic Construction Principle for Conﬁdence Regions
The least squares objective function, whose minimum argument gave us the
BLUE, naturally allows us to generate conﬁdence intervals or higherdimensional
conﬁdence regions. A conﬁdence region for β based on y = Xβ +ε can be constructed
as follows:
ˆ
• Draw the OLS estimate β into k dimensional space; it is the vector which
ˆ
ˆ
minimizes SSE = (y − X β ) (y − X β ).
˜ one can deﬁne the sum of squared errors associ• For every other vector β
˜
˜
ated with that vector as SSE β = (y − X β ) (y − X β ). Draw the level
˜
hypersurfaces (if k = 2: level lines) of this function. These are ellipsoids
ˆ
centered on β .
• Each of these ellipsoids is a conﬁdence region for β . Diﬀerent conﬁdence
regions diﬀer by their coverage probabilities.
• If one is only interested in certain coordinates of β and not in the others, or
in some other linear transformation β , then the corresponding conﬁdence
regions are the corresponding transformations of this ellipse. Geometrically
this can best be seen if this transformation is an orthogonal projection; then
the conﬁdence ellipse of the transformed vector Rβ is also a projection or
shadow” of the conﬁdence region for the whole vector. Projections of the
same conﬁdence region have the same conﬁdence level, independent of the
direction in which this projection goes.
The conﬁdence regions for β with coverage probability π will be written here
as B β;π or, if we want to make its dependence on the observation vector y explicit,
Bβ;π (y ). These conﬁdence regions are level lines of the SSE , and mathematically,
it is advantageous to deﬁne these level lines by their level relative to the minimum
˜
level, i.e., as as the set of all β for which the quotient of the attained SSE β =
˜
ˆ
ˆ
˜) (y − X β ) divided by the smallest possible SSE = (y − X β ) (y − X β )
˜
(y − X β
is smaller or equal a given number. In formulas,
˜
˜
(y − X β ) (y − X β )
˜
(30.1.1)
β ∈ Bβ;π (y ) ⇐⇒
≤ cπ;n−k,k
ˆ
ˆ
(y − X β ) (y − X β )
It will be shown below, in the discussion following (30.2.1), that cπ;n−k,k only depends
on π (the conﬁdence level), n − k (the degrees of freedom in the regression), and k
(the dimension of the conﬁdence region).
To get a geometric intuition of this principle, look at the case k = 2, in which
˜
the parameter vector β has only two components. For each possible value β of the
˜) (y −
parameter vector, the associated sum of squared errors is SSE β = (y − X β
˜
297 298 30. INTERVAL ESTIMATION ˜
˜
X β ). This a quadratic function of β , whose level lines form concentric ellipses as
shown in Figure 1. The center of these ellipses is the unconstrained least squares
estimate. Each of the ellipses is a conﬁdence region for β for a diﬀerent conﬁdence
level.
If one needs a conﬁdence region not for the whole vector β but, say, for i linearly
independent linear combinations Rβ (here R is a i × k matrix with full row rank),
˜
then the above principle applies in the following way: the vector u lies in the conﬁdence region for Rβ generated by y for conﬁdence level π , notation B Rβ;π , if and
˜
only if there is a β in the conﬁdence region (30.1.1) (with the parameters adjusted
˜˜
˜
to reﬂect the dimensionality of u) which satisﬁes Rβ = u:
(30.1.2)
˜
˜
(y − X β ) (y − X β )
˜
˜
˜
˜
u ∈ BRβ;π (y ) ⇐⇒ exist β with u = Rβ and
≤ cπ;n−k,i
ˆ) (y − X β )
ˆ
(y − X β
Problem 341. Why does one have to change the value of c when one goes over
to the projections of the conﬁdence regions?
Answer. Because the projection is a manytoone mapping, and vectors which are not in the
original ellipsoid may still end up in the projection. Again let us illustrate this with the 2dimensional case in which the conﬁdence
region for β is an ellipse, as drawn in Figure 1, called Bβ;π (y ). Starting with this
ellipse, the above criterion deﬁnes individual conﬁdence intervals for linear combina˜
˜˜
tions u = r β by the rule: u ∈ Br β;π (y ) iﬀ a β ∈ Bβ (y ) exists with r β = u. For
˜
1 ], this interval is simply the projection of the ellipse on the horizontal axis,
r = [0
and for r = [ 0 ] it is the projection on the vertical axis.
1
The same argument applies for all vectors r with r r = 1. The inner product
of two vectors is the length of the ﬁrst vector times the length of the projection
˜
of the second vector on the ﬁrst. If r r = 1, therefore, r β is simply the length
˜ on the line generated by the vector r . Therefore
of the orthogonal projection of β
the conﬁdence interval for r β is simply the projection of the ellipse on the line
generated by r . (This projection is sometimes called the “shadow” of the ellipse.)
˜
The conﬁdence region for Rβ can also be deﬁned as follows: u lies in this
ˆ
ˆ
ˆ which satisﬁes Rβ = u lies in the
ˆ
˜
conﬁdence region if and only if the “best” β
ˆ
ˆ
conﬁdence region (30.1.1), this best β being, of course, the constrained least squares
˜
estimate subject to the constraint Rβ = u, whose formula is given by (22.3.13).
˜
The conﬁdence region for Rβ consists therefore of all u for which the constrained
−1
ˆˆ
−1
ˆ˜
ˆ
least squares estimate β = β − (X X ) R R(X X )−1 R
(Rβ − u) satisﬁes
condition (30.1.1):
(30.1.3) ˜
u ∈ BRβ (y ) ⇐⇒ ˆ
ˆ
ˆ
ˆ
(y − X β ) (y − X β )
≤ cπ;n−k,i
ˆ
ˆ
(y − X β ) (y − X β ) One can also write it as
(30.1.4) ˜
u ∈ BRβ (y ) ⇐⇒ SSE constrained
≤ cπ;n−k,i
SSE unconstrained ˜
i.e., those u are in the conﬁdence region which, if imposed as a constraint on the
regression, will not make the SSE too much bigger. 30.1. CONSTRUCTION OF CONFIDENCE REGIONS −2
−3 −1 0 1 2
−3 ......
..........................................
............ ................................
.....
...........
..........
...
.. ...
..
.
.
.
.
....................................... ............
.......................................... ............
.
.
..
......... ...........
...
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..
......
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.
...... ......
..
.
..
...... ....
..
......
..
..
...
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..
...
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...
...
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.
.
.... ....
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..
..
..
..
..
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.
..
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..
.
.
...... ........
..
.
........
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..
.
......
.
......
.
...... ................
.
...... .....................
.
.......
.
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.
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........................... −4 299 −4 −5 −5
−2 −1 2 1 0 Figure 1. Conﬁdence Ellipse with “Shadows”
In order to transform (30.1.3) into a mathematically more convenient form, write
it as
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
(y − X β ) (y − X β ) − (y − X β ) (y − X β )
˜
u ∈ BRβ;π (y ) ⇐⇒
≤ cπ;n−k,i − 1
ˆ) (y − X β )
ˆ
(y − X β
and then use (22.7.2) to get
(30.1.5)
ˆ˜
(Rβ − u)
˜
u ∈ BRβ;π (y ) ⇐⇒ −1
ˆ˜
R(X X )−1 R
(Rβ − u)
≤ cπ;n−k,i − 1
ˆ
ˆ
(y − X β ) (y − X β ) ˆ
ˆ
This formula has the great advantage that β no longer appears in it. The condition
ˆ
˜
whether u belongs to the conﬁdence region is here formulated in terms of β alone.
Problem 342. Using (14.2.12), show that (30.1.1) can be rewritten as
(30.1.6) ˆ˜
ˆ˜
(β − β ) X X (β − β )
˜
≤ cπ;n−k,k − 1
β ∈ Bβ;π (y ) ⇐⇒
ˆ
ˆ
(y − X β ) (y − X β ) Verify that this is the same as (30.1.5) in the special case R = I .
Problem 343. You have run a regression with intercept, but you are not interested in the intercept per se but need a joint conﬁdence region for all slope parameters.
Using the notation of Problem 304, show that this conﬁdence region has the form
(30.1.7) ˆ˜
ˆ˜
(β − β ) X X (β − β )
˜
≤ cπ;n−k,k−1 − 1
β ∈ Bβ;π (y ) ⇐⇒
ˆ
ˆ
(y − X β ) (y − X β ) I.e., we are sweeping the means out of both regressors and dependent variables, and
then we act as if the regression never had an intercept and use the formula for the
full parameter vector (30.1.6) for these transformed data (except that the number of
degrees of freedom n − k still reﬂects the intercept as one of the explanatory variables).
Answer. Write the full parameter vector as α
β and R = o I . Use (30.1.5) but instead ˜
˜
of u write β . The only tricky part is the following which uses (23.0.37):
(30.1.8)
¯
¯
¯
1/n + x (X X )−1 x −x (X X )−1
R(X X )−1 R = o I
¯
−(X X )−1 x
(X X )−1 o
I = (X X )−1 300 30. INTERVAL ESTIMATION Figure 2. Conﬁdence Band for Regression Line
ˆ
ˆ
The denominator is (y − ια − X β ) (y − ια − X β ), but since α = y x β , see problem 204, this
ˆ
ˆ
ˆ ¯ ¯ˆ
ˆ
ˆ
denominator can be rewritten as (y − X β ) (y − X β ). Problem 344. 3 points We are in the simple regression y t = α + βxt + εt . If
one draws, for every value of x, a 95% conﬁdence interval for α + βx, one gets a
“conﬁdence band” around the ﬁtted line, as shown in Figure 2. Is the probability
that this conﬁdence band covers the true regression line over its whole length equal
to 95%, greater than 95%, or smaller than 95%? Give a good verbal reasoning for
your answer. You should make sure that your explanation is consistent with the fact
that the conﬁdence interval is random and the true regression line is ﬁxed.
30.2. Coverage Probability of the Conﬁdence Regions
˜
The probability that any given known value u lies in the conﬁdence region
(30.1.3) depends on the unknown β . But we will show now that the “coverage probability” of the region, i.e., the probability with which the conﬁdence region contains
the unknown true value u = Rβ , does not depend on any unknown parameters.
˜
To get the coverage probability, we must substitute u = Rβ (where β is the true
parameter value) in (30.1.5). This gives
(30.2.1)
−1
ˆ
ˆ
(Rβ − Rβ ) R(X X )−1 R
(Rβ − Rβ )
Rβ ∈ BRβ;π (y ) ⇐⇒
≤ cπ;n−k,i − 1
ˆ
ˆ
(y − X β ) (y − X β )
Let us look at numerator and denominator separately. Under the Normality assumpˆ
tion, Rβ ∼ N (Rβ , σ 2 R(X X )−1 R ). Therefore, by (7.4.9), the distribution of the
numerator of (30.2.1) is
(30.2.2) ˆ
(Rβ − Rβ ) R(X X )−1 R −1 ˆ
(Rβ − Rβ ) ∼ σ 2 χ2 .
i This probability distribution only depends on one unknown parameter, namely, σ 2 .
ˆ
ˆ
Regarding the denominator, remember that, by (18.4.2), (y − X β ) (y − X β ) =
ε M ε , and if we apply (7.4.9) to this we can see that
(30.2.3) ˆ
ˆ
(y − X β ) (y − X β ) ∼ σ 2 χ2 −k
n Furthermore, numerator and denominator are independent. To see this, look ﬁrst
ˆ
at β and ε. By Problem 246 they are uncorrelated, and since they are also jointly
ˆ
ˆ
Normal, it follows that they are independent. If β and ε are independent, any
ˆ
ˆ are independent of any functions of ε. The numerator in the test
functions of β
ˆ 30.4. INTERPRETATION IN TERMS OF STUDENTIZED MAHALANOBIS DISTANCE 301 ˆ
statistic (30.2.1) is a function of β and the denominator is a function of ε; therefore
ˆ
they are independent, as claimed. Lastly, if we divide numerator by denominator,
the unknown “nuisance parameter” σ 2 in their probability distributions cancels out,
i.e., the distribution of the quotient is fully known.
˜
˜
To sum up: if u is the true value u = Rβ , then the test statistic in (30.2.1)
can no longer be observed, but its distribution is is known; it is a χ2 divided by an
i
independent χ2 −k . Therefore, for every value c, the probability that the conﬁdence
n
region (30.1.5) contains the true Rβ can be computed, and conversely, for any desired
coverage probability, the appropriate critical value c can be computed. As claimed,
this critical value only depends on the conﬁdence level π and n − k and i.
30.3. Conventional Formulas for the Test Statistics
In order to get this test statistic into the form in which it is conventionally
tabulated, we must divide both numerator and denominator of (30.1.5) by their
degrees of freedom, to get a χ2 /i divided by an independent χ2 −k /(n − k ). This
i
n
quotient is called a F distribution with i and n − k degrees of freedom.
χ2 /i
i
The F distribution is deﬁned as F i,j = χ2 /j instead of the seemingly simpler
j formula χ2
i
,
χ2
j because the division by the degrees of freedom makes all F distributions and the associated critical values similar; an observed value below 4 is insigniﬁcant,
but greater values may be signﬁcant depending on the number of parameters.
˜
Therefore, instead of , the condition deciding whether a given vector u lies in the
conﬁdence region for Rβ with conﬁdence level π = 1 − α is formulated as follows:
(30.3.1)
(SSE constrained − SSE unconstrained )/number of constraints
≤ F(i,n−k;α)
SSE unconstr. /(numb. of obs. − numb. of coeﬀ. in unconstr. model)
Here the constrained SSE is the SSE in the model estimated with the constraint
˜
Rβ = u imposed, and F(i,n−k;α) is the upper α quantile of the F distribution
with i and n − k degrees of freedom, i.e., it is that scalar c for which a random
variable F which has a F distribution with i and n − k degrees of freedom satisﬁes
Pr[F ≥ c] = α.
30.4. Interpretation in terms of Studentized Mahalanobis Distance
The division of numerator and denominator by their degrees of freedom also gives
us a second intuitive interpretation of the test statistic in terms of the Mahalanobis
distance, see chapter 29. If one divides the denominator by its degrees of freedom,
one gets an unbiased estimate of σ 2
(30.4.1) s2 = 1
ˆ
ˆ
(y − X β ) (y − X β ).
n−k Therefore from (30.1.5) one gets the following alternative formula for the joint conﬁdence region B (y ) for the vector parameter u = Rβ for conﬁdence level π = 1 − α:
(30.4.2)
˜
u ∈ BRβ;1−α (y ) ⇐⇒ 1
ˆ˜
(Rβ − u)
s2 R(X X )−1 R −1 ˆ˜
(Rβ − u) ≤ iF(i,n−k;α) ˆ
ˆ
ˆ
Here β is the least squares estimator of β , and s2 = (y − X β ) (y − X β )/(n − k ) the
2
ˆ = s2 (X X )−1 is the estimated covariance
unbiased estimator of σ . Therefore Σ
ˆ
matrix as available in the regression printout. Therefore V = s2 R(X X )−1 R 302 30. INTERVAL ESTIMATION ˆ
is the estimate of the covariance matrix of Rβ . Another way to write (30.4.2) is
therefore
(30.4.3) ˆ ˜ ˆ −1 ˆ ˜
˜
B (y ) = {u ∈ Ri : (Rβ − u) V (Rβ − u) ≤ iF(i,n−k;α) }. ˜
This formula allows a suggestive interpretation. whether u lies in the conﬁdence
ˆ
region or not depends on the Mahalanobis distance of the actual value of Rβ would
ˆ would have if the true parameter vector were
have from the distribution which Rβ
˜
to satisfy the constraint Rβ = u. It is not the Mahalanobis distance itself but only
an estimate of it because σ 2 is replaced by its unbiased estimate s2 .
These formulas are also useful for drawing the conﬁdence ellipses. The r which
you need in equation (7.3.22) in order to draw the conﬁdence ellipse is r = iF(i,n−k;α) .
This is the same as the local variable mult in the following Sfunction to draw this
ellipse: its arguments are the center point (a 2vector d), the estimated covariance
matrix (a 2 × 2 matrix C), the degrees of freedom in the denominator of the F distribution (the scalar df), and the conﬁdence level (the scalar level between 0
and 1 which defaults to 0.95 if not speciﬁed).
confelli <function(b, C, df, level = 0.95, xlab = "", ylab = "", add=T, prec=51)
#
#
#
#
#
#
#
#
#
#
# Plot an ellipse with "covariance matrix" C, center b, and Pcontent
level according the F(2,df) distribution.
Sent to SNEWS on May 19, 1999 by Roger Koenker
Department of Economics
University of Illinois
Champaign, IL 61820
url: http://www.econ.uiuc.edu
email roger@ysidro.econ.uiuc.edu
vox: 2173334558
fax: 2172446678.
Included in the ecmet package with his permission. {
d < sqrt(diag(C))
dfvec < c(2, df)
phase < acos(C[1, 2]/(d[1] * d[2]))
angles < seq(  (PI), PI, len = prec)
mult < sqrt(dfvec[1] * qf(level, dfvec[1], dfvec[2]))
xpts < b[1] + d[1] * mult * cos(angles)
ypts < b[2] + d[2] * mult * cos(angles + phase)
if(add) lines(xpts, ypts)
else plot(xpts, ypts, type = "l", xlab = xlab, ylab = ylab)
}
The mathematics why this works is in Problem 146.
Problem 345. 3 points In the regression model y = Xβ + ε you observe y and
the (nonstochastic) X and you construct the following conﬁdence region B (y ) for
Rβ , where R is a i × k matrix with full row rank:
(30.4.4)
ˆ
ˆ
B (y ) = {u ∈ Ri : (Rβ − u) (R(X X )−1 R )−1 (Rβ − u) ≤ is2 F(i,n−k;α) }.
Compute the probability that B contains the true Rβ . 30.4. INTERPRETATION IN TERMS OF STUDENTIZED MAHALANOBIS DISTANCE 303 Answer.
(30.4.5) Pr[B (y )
(30.4.6) = Pr[ ˆ
ˆ
Rβ ] = Pr[(Rβ − Rβ ) (R(X X )−1 R )−1 (Rβ − Rβ ) ≤ iF(i,n−k;α) s2 ] =
ˆ
ˆ
(Rβ − Rβ ) (R(X X )−1 R )−1 (Rβ − Rβ )/i
≤ F(i,n−k;α) ] = 1 − α
2
s This interpretation with the Mahalanobis distance is commonly used for the
construction of tIntervals. A tinterval is a special case of the above conﬁdence
region for the case i = 1. The conﬁdence interval with conﬁdence level 1 − α for the
scalar parameter u = r β , where r = o is a vector of constant coeﬃcients, can be
written as
ˆ
(30.4.7)
B (y ) = {u ∈ R : u − r β  ≤ t(n−k;α/2) s ˆ}.
rβ ˆ
What do those symbols mean? β is the least squares estimator of β . t(n−k;α/2) is
the upper α/2quantile of the t distribution with n − k degrees of freedom, i.e., it
is that scalar c for which a random variable t which has a t distribution with n − k
degrees of freedom satisﬁes Pr[t ≥ c] = α/2. Since by symmetry Pr[t ≤ −c] = α/2
as well, one obtains the inequality relevant for a twosided test:
Pr[t ≥ t(n−k;α/2) ] = α. (30.4.8) ˆ
Finally, sr β is the estimated standard deviation of r β .
ˆ
It is computed by the following three steps: First write down the variance of
ˆ
r β:
ˆ
(30.4.9)
var[r β ] = σ 2 r (X X )−1 r .
ˆ
ˆ
Secondly, replace σ 2 by its unbiased estimator s2 = (y − X β ) (y − X β )/(n − k ),
and thirdly take the square root. This gives sr ˆ
β = s r (X X )−1 r . Problem 346. Which element(s) on the right hand side of (30.4.7) depend(s)
on y ?
ˆ
Answer. β depends on y , and also sr ˆ
β depends on y through s2 . Let us verify that the coverage probability, i.e., the probability that the conﬁdence interval constructed using formula (30.4.7) contains the true value r β , is, as
claimed, 1 − α:
(30.4.10)
Pr[B (y ) ˆ
r β ] = Pr[ r β − r β ≤ t(n−k;α/2) sr ˆ
β] (30.4.11) = Pr r (X X )−1 X ε ≤ t(n−k;α/2) s r (X X )−1 r (30.4.12) = Pr[ r (X X )−1 X ε ≤ t(n−k;α/2) ] s r (X X )−1 r
(30.4.13) = Pr[ r (X X )−1 X ε
σ r (X X )−1 r s
≤ t(n−k;α/2) ] = 1 − α,
σ This last equality holds because the expression left of the big slash is a standard
normal, and the expression on the right of the big slash is the square root of an 304 30. INTERVAL ESTIMATION independent χ2 −k divided by n − k . The random variable between the absolute signs
n
has therefore a tdistribution, and (30.4.13) follows from (30.4.8).
In R, one obtains t(n−k;α/2) by giving the command qt(1alpha/2,np). Here
qt stands for tquantile [BCW96, p. 48]. One needs 1alpha/2 instead of alpha/2
because it is the usual convention for quantiles (or cumulative distribution functions)
to be deﬁned as lower quantiles, i.e., as the probabilities of a random variable being
≤ a given number, while test statistics are usually designed in such a way that the
signiﬁcant values are the high values, i.e., for testing one needs the upper quantiles.
There is a basic duality between conﬁdence intervals and hypothesis tests. Chapter 31 is therefore a discussion of the same subject under a slightly diﬀerent angle: CHAPTER 31 Three Principles for Testing a Linear Constraint
We work in the model y = Xβ + ε with normally distributed errors ε ∼
N (o, σ 2 I ). There are three basic approaches to test the null hypothesis Rβ = u. In
the linear model, these three approaches are mathematically equivalent, but if one
goes over to nonlinear least squares or maximum likelihood estimators, they lead to
diﬀerent (although asymptotically equivalent) tests.
ˆ
(1) (“Wald Criterion”) Compute the vector of OLS estimates β , and reject the
ˆ
null hypothesis if Rβ is “too far away” from u. For this criterion one only needs the
unconstrained estimator, not the constrained one.
(2) (“Likelihood Ratio Criterion”) Estimate the model twice: once with the
constraint Rβ = u, and once without the constraint. Reject the null hypothesis if
the model with the constraint imposed has a much worse ﬁt than the model without
the constraint.
(3) (“Lagrange Multiplier Criterion”) This third criterion is based on the constrained estimator only. It has two variants. In its “score test” variant, one rejects
the null hypothesis if the vector of derivatives of the unconstrained least squares
ˆ
ˆ
objective function, evaluated at the constrained estimate β , is too far away from o.
In the variant which has given this Criterion its name, one rejects if the vector of
Lagrange multipliers needed for imposing the constraint is too far away from o.
Many textbooks inadvertently and implicitly distinguish between (1) and (2)
as follows: they introduce the ttest for one parameter by principle (1), and the
F test for several parameters by principle (2). Later, the student is surprised to
ﬁnd out that the ttest and the F test in one dimension are equivalent, i.e., that
the diﬀerence between ttest and F test has nothing to do with the dimension of
the parameter vector to be tested. Some textbooks make the distinction between
(1) and (2) explicit. For instance [Chr87, p. 29ﬀ] distinguishes between “testing
linear parametric functions” and “testing models.” However the distinction between
all 3 principles has been introduced into the linear model only after the discovery
that these three principles give diﬀerent but asymptotically equivalent tests in the
Maximum Likelihood estimation. Compare [DM93, Chapter 3.6] about this. 31.1. Mathematical Detail of the Three Approaches
ˆ
(1) For the “Wald criterion” we must specify what it means that Rβ is “too
far away” from u. The Mahalanobis distance gives such a criterion: If the true β
ˆ
satisﬁes Rβ = u, then Rβ ∼ (u, σ 2 R(X X )−1 R ), and the Mahalanobis distance
ˆ
of the observed value of Rβ from this distribution is a logical candidate for the Wald
criterion. The only problem is that σ 2 is not known, therefore we have to use the
“studentized” Mahalanobis distance in which σ 2 is replaced by s2 . Conventionally,
in the conterxt of linear regression, the Mahalanobis distance is also divided by the
number of degrees of freedom; this normalizes its expected value to 1. Replacing σ 2
305 306 31. THREE PRINCIPLES FOR TESTING A LINEAR CONSTRAINT by s2 and dividing by i gives the test statistic
(31.1.1) ˆ
1 (Rβ − u)
i R(X X )−1 R
s2 −1 ˆ
(Rβ − u) . (2) Here are the details for the second approach, the “goodnessofﬁt criterion.”
In order to compare the ﬁt of the models, we look at the attained SSE ’s. Of course,
the constrained SSE r is always larger than the unconstrained SSE u , even if the
true parameter vector satisﬁes the constraint. But if we divide SSE r by its degrees
of freedom n + i − k , it is an unbiased estimator of σ 2 if the constraint holds and it
is biased upwards if the constraint does not hold. The unconstrained SSE u , divided
by its degrees of freedom, on the other hand, is always an unbiased estimator of σ 2 .
If the constraint holds, the SSE ’s divided by their respective degrees of freedom
should give roughly equal numbers. According to this, a feasible test statistic would
be
(31.1.2) SSE r /(n + i − k )
SSE u /(n − k ) and one would reject if this is too much > 1. The following variation of this is more
convenient, since its distribution does not depend on n, k and i separately, but only
through n − k and i.
(31.1.3) (SSE r − SSE u )/i
SSE u /(n − k ) It still has the property that the numerator is an unbiased estimator of σ 2 if the constraint holds and biased upwards if the constraint does not hold, and the denominator
is always an unbiased estimator. Furthermore, in this variation, the numerator and
denominator are independent random variables. If this test statistic is much larger
than 1, then the constraints are incompatible with the data and the null hypothesis
must be rejected. The statistic (31.1.3) can also be written as
(31.1.4)
(SSE constrained − SSE unconstrained )/number of constraints
SSE unconstrained /(numb. of observations − numb. of coeﬃcients in unconstr. model)
The equivalence of formulas (31.1.1) and (31.1.4) is a simple consequence of (22.7.2).
(3) And here are the details about the score test variant of the Lagrange multiplier criterion: The Jacobian of the least squares objective function is
(31.1.5) ∂
(y − Xβ ) (y − Xβ ) = −2(y − Xβ ) X .
∂β This is a row vector consisting of all the partial derivatives. Taking its transpose,
in order to get a column vector, and plugging the constrained least squares estimate
ˆ
ˆ
ˆ
ˆ
β into it gives −2X (y − X β ). Again we need the Mahalanobis distance of this
observed value from the distribution which the random variable
(31.1.6) ˆ
ˆ
−2X (y − X β ) ˆ
ˆ
has if the true β satisﬁes Rβ = u. If this constraint is satisﬁed, β is unbiased, therefore (31.1.6) has expected value zero. Furthermore, if one premulti−1
ˆ
ˆ
ˆ
plies (22.7.1) by X one gets X (y − X β ) = R R(X X )−1 R
(Rβ − u),
−1
ˆ
ˆ
therefore V [X (y − X β )] = R R(X X )−1 R
R; and now one can see that 31.1. MATHEMATICAL DETAIL OF THE THREE APPROACHES 307 X )−1 is a ginverse of this covariance matrix. Therefore the Malahalanobis
distance of the observed value from the distribution is
1
ˆ
ˆ
ˆ
ˆ
(31.1.7)
(y − X β ) X (X X )−1 X (y − X β )
σ2
1
σ 2 (X The Lagrange multiplier statistic is based on the restricted estimator alone. If one
wanted to take this principle seriously one would have to to replace σ 2 by the unbiased
estimate from the restricted model to get the “score form” of the Lagrange Multiplier
Test statistic. But in the linear model this leads to it that the denominator in the
test statistic is no longer independent of the numerator, and since the test statistic as
a function of the ratio of the constrained and unconstrained estimates of σ 2 anyway,
one will only get yet another monotonic transformation of the same test statistic.
If one were to use the unbiased estimate from the unrestricted model, one would
exactly get the Wald statistic back, as one can verify using (22.3.13).
This same statistic can also be motivated in terms of the Lagrange multipliers,
and this is where this testing principle has its name from, although the applications
usually use the score form. According to (22.3.12), the Lagrange multiplier is λ =
−1
ˆ
2 R(X X )−1 R
(Rβ − u). If the constraint holds, then E [λ] = o, and V [λ] =
4σ 2 R(X X )−1 R
distribution is −1 (31.1.8) λ (V [λ])−1 λ = . The Mahalanobis distance of the observed value from this
1
λ R(X X )−1 R λ
4σ 2 Using (22.7.1) one can verify that this is the same as (31.1.7).
Problem 347. Show that (31.1.7) is equal to the righthand side of (31.1.8).
ˆˆ ˆˆ
Problem 348. 10 points Prove that ε ε − ε ε can be written alternatively in
ˆˆ
the following ﬁve ways:
(31.1.9) ˆˆ ˆˆ
ˆˆ
ˆ
ˆˆ
ˆ
ε ε − ε ε = (β − β ) X X (β − β )
ˆˆ (31.1.12) ˆ
ˆ
= (Rβ − u) (R(X X )−1 R )−1 (Rβ − u)
1
= λ R(X X )−1 R λ
4
ˆ
ˆ
= ε X (X X )−1 X ε
ˆ
ˆ (31.1.13) ˆˆˆˆ
= (ε − ε) (ε − ε)
ˆ
ˆ (31.1.10)
(31.1.11) Furthermore show that
(31.1.14) XX is σ 2 times a ginverse of (31.1.15) (R(X X )−1 R )−1
1
R(X X )−1 R
4 is σ 2 times the inverse of ˆˆ
ˆ
V [β − β ]
ˆ
V [Rβ − u] is σ 2 times the inverse of V [λ] (31.1.17) (X X )−1 is σ 2 times a ginverse of (31.1.18) I is σ 2 times a ginverse of ˆ
ˆ
V [X (y − X β )]
ˆˆ
ˆ
V [ε − ε] (31.1.16) ˆ
ˆ
and show that −2X (y − X β ) is the gradient of the SSE objective function evaluated
ˆ
ˆ
at β . By the way, one should be a little careful in interpreting (31.1.12) because
ˆ
X (X X )−1 X is not σ 2 times the ginverse of V [ε].
ˆ 308 31. THREE PRINCIPLES FOR TESTING A LINEAR CONSTRAINT Answer.
(31.1.19) ˆ
ˆ
ˆ
ˆ
ˆˆ
ˆ
ˆˆ
ˆ
ˆ
ˆ
ε = y − X β = X β + ε − X β = X (β − β ) + ε , ˆ
and since X ε = o, the righthand decomposition is an orthogonal decomposition. This gives
(31.1.9) above:
(31.1.20) ˆˆ
ˆˆ
ˆ
ˆˆ
ˆ
ˆˆ
ˆˆ
ε = (β − β ) X X (β − β ) + ε ε , −1
ˆ
ˆˆ
Using (22.3.13) one obtains V [β − β ] = σ 2 (X X )−1 R R(X X )−1 R
R(X X )−1 . This is
1
a singular matrix, and one veriﬁes immediately that σ2 X X is a ginverse of it.
ˆ
To obtain (31.1.10), which is (22.7.2), one has to plug (22.3.13) into (31.1.20). Clearly, V [Rβ − u] = σ 2 R(X X )−1 R .
For (31.1.11) one needs the formula for the Lagrange multiplier (22.3.12). The test statistic deﬁned alternatively either by (31.1.1) or (31.1.4) or (31.1.7)
or (31.1.8) has the following nice properties:
• E(SSE u ) = E(ε ε) = σ 2 (n − k ), which holds whether or not the constraint
ˆˆ
is true. Furthermore it was shown earlier that
(31.1.21) E(SSE r − SSE u ) = σ 2 i + (Rβ − u) (R(X X )−1 R )−1 (Rβ − u), i.e., this expected value is equal to σ 2 i if the constraint is true, and larger
otherwise. If one divides SSE u and SSE r − SSE u by their respective
degrees of freedom, as is done in (31.1.4), one obtains therefore: the denominator is always an unbiased estimator of σ 2 , regardless of whether the
null hypothesis is true or not. The numerator is an unbiased estimator of
σ 2 when the null hypothesis is correct, and has a positive bias otherwise.
• If the distribution of ε is normal, then numerator and denominator are
ˆ
independent. The numerator is a function of β and the denominator one
ˆ and ε are independent.
ˆ
ˆ
of ε, and β
• Again under assumption of normality, numerator and denominator are distributed as σ 2 χ2 with i and n − k degrees of freedom, divided by their
respective degrees of freedom. If one divides them, the common factor σ 2
cancels out, and the ratio has a F distribution. Since both numerator and
denominator have the same expected value σ 2 , the value of this F distribution should be in the order of magnitude of 1. If it is much larger than
that, the null hypothesis is to be rejected. (Precise values in the F tables).
31.2. Examples of Tests of Linear Hypotheses
Some tests can be read oﬀ directly from the computer printouts. One example is
the ttests for an individual component of β . The situation is y = Xβ +ε , where β =
β1 · · · βk , and we want to test βj = u. Here R = ej = [ 0 ··· 0 1 0 ··· 0 ], with the
1 on the j th place, and u is the 1vector u, and i = 1. Therefore R(X X )−1 R =
djj , the j th diagonal element of (X X )−1 , and (31.1.1) becomes
(31.2.1) ˆ
(β j − u)2
∼ F1,n−k
s2 djj when H is true. This is the square of a random variable which has a tdistribution:
ˆ
βj − u
∼ tn−k when H is true.
(31.2.2)
s djj
ˆ
This latter test statistic is simply β j − u divided by the estimated standard deviation
ˆ.
of β j 31.2. EXAMPLES OF TESTS OF LINEAR HYPOTHESES 309 If one wants to test that a certain linear combination of the parameter values is
equal to (or bigger than or smaller than) a given value, say r β = u, one can use a
ˆ
ttest as well. The test statistic is, again, simply r β − u divided by the estimated
ˆ:
standard deviation of r β
ˆ
r β−u
(31.2.3)
∼ tn−k when H is true.
s r (X X )−1 r
By this one can for instance also test whether the sum of certain regression coeﬃcients
is equal to 1, or whether two regression coeﬃcients are equal to each other (but not
the hypothesis that three coeﬃcients are equal to each other).
Many textbooks use the Wald criterion to derive the ttest, and the LikelihoodRatio criterion to derive the F test. Our approach showed that the Wald criterion
can be used for simultaneous testing of several hypotheses as well. The ttest is
equivalent to an F test if only one hypothesis is tested, i.e., if R is a row vector.
The only diﬀerence is that with the ttest one can test onesided hypotheses, with
the F test one cannot.
Next let us discuss the test for the existence of a relationship, “the” F test
which every statistics package performs automatically whenever the regression has
a constant term: it is the test whether all the slope parameters are zero, such that
only the intercept may take a nonzero value.
Problem 349. 4 points In the model y = Xβ + ε with intercept, show that the
test statistic for testing whether all the slope parameters are zero is
ˆ
(y X β − ny 2 )/(k − 1)
¯
(31.2.4)
ˆ
(y y − y X β )/(n − k )
This is [Seb77, equation (4.26) on p. 110]. What is the distribution of this test
statistic if the null hypothesis is true (i.e., if all the slope parameters are zero)?
Answer. The distribution is ∼ F k−1,n−k . (31.2.4) is most conveniently derived from (31.1.4).
In the constrained model, which has only a constant term and no other explanatory variables, i.e.,
y = ιµ + ε , the BLUE is µ = y . Therefore the constrained residual sum of squares SSE const. is
ˆ
¯
what is commonly called SST (“total” or, more precisely, “corrected total” sum of squares):
(31.2.5)
SSE const. = SST = (y − ιy ) (y − ιy ) = y (y − ιy ) = y y − ny 2
¯
¯
¯
¯
while the unconstrained residual sum of squares is what is usually called SSE :
(31.2.6)
ˆ
ˆ
ˆ
ˆ
SSE unconst. = SSE = (y − X β ) (y − X β ) = y (y − X β ) = y y − y X β .
ˆ
This last equation because X (y − X β ) = X ε = o. A more elegant way is perhaps
ˆ
(31.2.7)
ˆ
SSE unconst. = SSE = ε ε = y M M y = y M y = y y − y X (X X )−1 X y = y y − y X β
ˆˆ
According to (14.3.12) we can write SSR = SST − SSE , therefore the F statistic is
(31.2.8) ˆ
SSR/(k − 1)
(y X β − ny 2 )/(k − 1)
¯
=
∼ F k−1,n−k
ˆ
SSE /(n − k)
( y y − y X β )/ (n − k ) if H0 is true. Problem 350. 2 points Can one compute the value of the F statistic testing for
the existence of a relationship if one only knows the coeﬃcient of determination R2 =
SSR/SST , the number of observations n, and the number of regressors (counting the
constant term as one of the regressors) k ? 310 31. THREE PRINCIPLES FOR TESTING A LINEAR CONSTRAINT Answer.
(31.2.9) F= SSR/(k − 1)
n−k
SSR
n − k R2
=
.
=
SSE /(n − k)
k − 1 SST − SSR
k − 1 1 − R2 Other, similar F tests are: the F test that all among a number of additional
variables have the coeﬃcient zero, the F test that three or more coeﬃcients are
equal. One can use the ttest for testing whether two coeﬃcients are equal, but not
for three. It may be possible that the ttest for β1 = β2 does not reject and the ttest
for β2 = β3 does not reject either, but the ttest for β1 = β3 does reject!
Problem 351. 4 points [Seb77, exercise 4b.5 on p. 109/10] In the model y =
β + ε with ε ∼ N (o, σ 2 I ) and subject to the constraint ι β = 0, which we had in
Problem 291, compute the test statistic for the hypothesis β1 = β3 .
Answer. In this problem, the “unconstrained” model for the purposes of testing is already
constrained, it is subject to the constraint ι β 0. The “constrained” model has the additional
=
β1
.
constraint Rβ = 1 0 −1 0 · · · 0 . = 0. In Problem 291 we computed the “uncon.
βk
ˆ
strained” estimates β = y − ιy and s2 = ny 2 = (y 1 + · · · + y n )2 /n. You are allowed to use this
¯
¯
ˆ
without proving it again. Therefore Rβ = y 1 − y 3 ; its variance is 2σ 2 , and the F test statistic
n(y −y )2 1
is 2(y +···+3 )2 ∼ F1,1 . The “unconstrained” model had 4 parameters subject to one constraint,
yn
1
therefore it had 3 free parameters, i.e.,k = 3, n = 4, and j = 1. Another important F test is the “Chow test” named by its popularizer Chow
[Cho60]: it tests whether two regressions have equal coeﬃcients (assuming that
the disturbance variances are equal). For this one has to run three regressions. If
the ﬁrst regression has n1 observations and sum of squared error SSE 1 , and the
second regression n2 observations and SSE 2 , and the combined regression (i.e., the
restricted model) has SSE r , then the test statistic is
(31.2.10) (SSE r − SSE 1 − SSE 2 )/k
.
(SSE 1 + SSE 2 )/(n1 + n2 − 2k ) If n2 < k , the second regression cannot be run by itself. In this case, the unconstrained model has “too many” parameters: they give an exact ﬁt for the second
group of observations SSE 2 = 0, and in addition not all parameters are identiﬁed.
In eﬀect this second regression has only n2 parameters. These parameters can be
considered dummy variables for every observation, i.e., this test can be interpreted
to be a test whether the n2 additional observations come from the same population
as the n1 ﬁrst ones. The test statistic becomes
(31.2.11) (SSE r − SSE 1 )/n2
.
SSE 1 /(n1 − k ) This latter is called the “predictive Chow test,” because in its Wald version it looks
at the prediction errors involving observations in the second regression.
The following is a special case of the Chow test, in which one can give a simple
formula for the test statistic.
Problem 352. Assume you have n1 observations uj ∼ N (µ1 , σ 2 ) and n2 observations v j ∼ N (µ2 , σ 2 ), all independent of each other, and you want to test whether
µ1 = µ2 . (Note that the variances are known to be equal).
• a. 2 points Write the model in the form y = Xβ + ε. 31.2. EXAMPLES OF TESTS OF LINEAR HYPOTHESES 311 Answer.
ι µ + ε1
u
= 11
v
ι2 µ 2 + ε 2 (31.2.12) ι1
o = µ1
ε
+ 1.
µ2
ε2 o
ι2 here ι1 and ι2 are vectors of ones of appropriate lengths. • b. 2 points Compute (X X )−1 in this case.
Answer.
(31.2.13) X X= (31.2.14) (X X )−1 = ι1
o o
ι2 1
n1 ι1
o o
ι2 n1
0 0
n2 0 = 1
n2 0 ˆ
• c. 2 points Compute β = (X X )−1 X y in this case.
Answer.
(31.2.15) X y= (31.2.16) ˆ
β = (X X )−1 X y = ι1
o o
ι2 1
n1 0 n1
u
i=1 i
n2
v
j =1 j 1
n2 0 n1
u
i=1 i
n2
v
j =1 j u
=
v u
¯
v
¯ = ˆ
ˆ
• d. 3 points Compute SSE = (y − X β ) (y − X β ) and s2 , the unbiased esti2
mator of σ , in this case.
Answer.
ˆ
y − Xβ = (31.2.17) u
ι
−1
v
o u
u − ι1 u
¯
¯
=
v
¯
v − ι2 u
¯ o
ι2 n1 (31.2.18) SSE = s2 = (31.2.19) n2 ¯
( ui − u) 2 +
i=1
n1
(u
i=1 i ¯
(v j − v )2
j =1 − u) 2 +
¯ n2
(v
j =1 j − v )2
¯ n1 + n2 − 2 • e. 1 point Next, the hypothesis µ1 = µ2 must be written in the form Rβ = u.
Since in the present case R has just has one row, it should be written as a rowvector
R = r , and since the vector u has only one component, it should be written as a
scalar u, i.e., the hypothesis should be written in the form r β = u. What are r and
u in our case?
Answer. Since β = (31.2.20) 1 −1 µ1
, the constraint can be written as
µ2
µ1
µ2 =0 i.e., r= 1
−1 and u=0 ˆ
• f . 2 points Compute the standard deviation of r β .
Answer. First compute the variance and then take the square root.
(31.2.21) ˆ
var[r β ] = σ 2 r (X X )−1 r = σ 2 1 −1 1
n1 0 0
1
n2 1
1
1
= σ2
+
−1
n1
n2 1
1
One can also see this without matrix algebra. var[¯ = σ 2 n , var[¯ = σ 2 n , and since u and v are
u
v
¯
¯
1
2
independent, the variance of the diﬀerence is the sum of the variances. 312 31. THREE PRINCIPLES FOR TESTING A LINEAR CONSTRAINT • g. 2 points Use (31.2.3) to derive the formula for the ttest.
Answer. The test statistic is u − v divided by its estimated standard deviation, i.e.,
¯¯
u−v
¯¯ (31.2.22)
s 1
n1 + ∼ tn1 +n2 −2 when H is true. 1
n2 Problem 353. [Seb77, exercise 4d3] Given n + 1 observations yj from a
N (µ, σ 2 ). After the ﬁrst n observations, it is suspected that a sudden change in the
mean of the distribution occurred, i.e., that y n+1 ∼ N (ν, σ 2 ) with ν = µ. We will
use here three diﬀerent approaches to derive the same test statistic for testing the
hypothesis that the n + 1st observation has the same population mean as the previous
observations, i.e., that ν = µ, against the twosided alternative. The formulas for
this statistic should be given in terms of the observations yi . It is recommended to
n
n+1
1
¯
use the notation y = n i=1 yi and y = n+1 j =1 yj .
¯1
• a. 3 points First you should derive this statistic by testing whether ν − µ = 0
(the “Wald principle”). For this you must compute the BLUE of ν − µ and its
standard deviation and construct the t statistic from this.
n 1
Answer. BLUE of µ is y = n
¯
y , and that of ν is y n+1 . BLUE of ν − µ is y − y n+1 .
¯
i=1 i
y
y
Because of independence var[¯ − y n+1 ] = var[¯]+var[y n+1 ] = σ 2 ((1/n)+1) = σ 2 (n +1)/n. Standard deviation is σ (n + 1)/n.
For the denominator in the tstatistic you need the s2 from the unconstrained regression, which
is
(31.2.23) s2 = 1
n−1 n (y j − y )2
¯
j =1 What happened to the (n + 1)st observation here? It always has a zero residual. And the factor
1/(n − 1) should really be written 1/(n + 1 − 2): there are n + 1 observations and 2 parameters.
Divide y − y n+1 by its standard deviation and replace σ by s (the square root of s2 ) to get the
¯
t statistic
y − y n+1
¯
(31.2.24)
1
s 1+ n • b. 2 points One can interpret this same formula also diﬀerently (and this is
why this test is sometimes called the “predictive” Chow test). Compute the Best
Linear Unbiased Predictor of y n+1 on the basis of the ﬁrst n observations, call it
ˆ
ˆ
y (n + 1)n+1 . Show that the predictive residual y n+1 − y (n + 1)n+1 , divided by the
ˆ
ˆ
ˆ(n + 1)n+1 ; y
square root of MSE[y
ˆ
n+1 ], with σ replaced by s (based on the ﬁrst n
observations only), is equal to the above t statistic.
Answer. BLUP of y n+1 based on ﬁrst n observations is y again. Since it is unbiased,
¯
MSE[¯; y n+1 ] = var[¯ − y n+1 ] = σ 2 (n + 1)/n. From now on everything is as in part a.
y
y • c. 6 points Next you should show that the above two formulas are identical
to the statistic based on comparing the SSE s of the constrained and unconstrained
models (the likelihood ratio principle). Give a formula for the constrained SSE r , the
unconstrained SSE u , and the F statistic.
Answer. According to the Likelihood Ratio principle, one has to compare the residual sums of
squares in the regressions under the assumption that the mean did not change with that under the
¯
assumption that the mean changed. If the mean did not change (constrained model), then y is the 31.2. EXAMPLES OF TESTS OF LINEAR HYPOTHESES 313 OLS of µ. In order to make it easier to derive the diﬀerence between constrained and unconstrained
SSE , we will write the constrained SSE as follows:
n+1 n+1 n+1
2
¯
yj − (n + 1)y 2 = ¯
( yj − y ) 2 = SSE r =
j =1 2
yj − 1
(ny + yn+1 )2
¯
n+1 j =1 j =1 If one allows the mean to change (unconstrained model), then y is the BLUE of µ, and yn+1 is the
¯
BLUE of ν .
n n
2
¯
yj − ny 2 . (yj − y )2 + (yn+1 − yn+1 )2 =
¯ SSE u =
j =1 j =1 Now subtract:
1
(ny + yn+1 )2
¯
n+1
1
2
2
= yn+1 + ny 2 −
¯
(n2 y 2 + 2ny yn+1 + yn+1 )
¯
¯
n+1
n2
n
1
= (1 −
)y 2
+ (n −
)¯2 −
y
2¯yn+1
y
n + 1 n+1
n+1
n+1
n
=
(yn+1 − y )2 .
¯
n+1 2
SSE r − SSE u = yn+1 + ny 2 −
¯ Interestingly, this depends on the ﬁrst n observations only through y .
¯
Since the unconstrained model has n + 1 observations and 2 parameters, the test statistic is
(31.2.25) SSE r − SSE u
=
SSE u /(n + 1 − 2) n
(y
− y )2
¯
n+1 n+1
n
(yj − y ) 2 / (n −
¯
1 1) = (yn+1 − y )2 n(n − 1)
¯
n
(y
1j − y )2 (n + 1)
¯ ∼ F1,n−1 This is the square of the t statistic (31.2.24). 31.2.1. Goodness of Fit Test.
Problem 354. [Seb77, pp. 117–119] Given a regression model with k independent variables. There are n observations of the vector of independent variables, and
for each of these n values there is not one but r > 1 diﬀerent replicated observations
of the dependent variable. This model can be written
k (31.2.26) y mq = xmj βj + εmq or y mq = xm β + εmq , j =1 where m = 1, . . . , n, j = 1, . . . , k , q = 1, . . . , r, and xm is the mth row of the X matrix. For simplicity we assume that r does not depend on m, each observation of
the independent variables has the same number of repetitions. We also assume that
the n × k matrix X has full column rank.
• a. 2 points In this model it is possible to test whether the regression line is in
fact a straight line. If it is not a straight line, then each observation of the dependent
variables xm has a diﬀerent coeﬃcient vector β m associated with it, i.e., the model
is
k (31.2.27) y mq = xmj βmj + εmq or y mq = xm β m + εmq . j =1 This unconstrained model does not have enough information to estimate any of the
individual coeﬃcients βmj . Explain how it is nevertheless still possible to compute
SSE u . 314 31. THREE PRINCIPLES FOR TESTING A LINEAR CONSTRAINT Answer. Even though the individual coeﬃcients βmj are not identiﬁed, their linear combinak tion ηm = xm β m =
xβ
is identiﬁed; one unbiased estimator, although by far not the
j =1 mj mj
best one, is any individual observation y mq . This linear combination is all one needs to compute
SSE u , the sum of squared errors in the unconstrained model. • b. 2 points Writing your estimate of ηm = xm β m as η m , give the formula of the
˜
sum of squared errors of this estimate, and by taking the ﬁrst order conditions, show
that the unconstrained least squares estimate of ηm is η m = y m· for m = 1, . . . , n,
ˆ
¯
r
where y m· = 1 q=1 y mq (i.e., the dot in the subscript indicates taking the mean).
¯
r
Answer. If we know the η m the sum of squared errors no longer depents on the independent
˜
observations xm but is simply
(31.2.28) (y mq − η m )2
˜ SSE u =
m,q First order conditions are
(31.2.29) ∂
∂ ηh
˜ (y mq − η m )2 =
˜
m,q ∂
∂ ηh
˜ (y hq − η h )2 = −2
˜
q (y hq − η h ) = 0
˜
q • c. 1 point The sum of squared errors associated with this least squares estimate
is the unconstrained sum of squared errors SSE u . How would you set up a regression
with dummy variables which would give you this SSE u ?
Answer. The unconstrained model should be regressed in the form y mq = ηm + εmq . I.e.,
string out the matrix Y as a vector and for each column of Y introduce a dummy variable which
is = 1 if the given observation was originally in this colum. • d. 2 points Next turn to the constrained model (31.2.26). If X has full column
˜
rank, then it is fully identiﬁed. Writing β j for your estimates of βj , give a formula
for the sum of squared errors of this estimate. By taking the ﬁrst order conditions,
ˆ
show that the estimate β is the same as the estimate in the model without replicated
observations
k (31.2.30) zm = xmj βj + εm ,
j =1 where z m = y m· as deﬁned above.
¯
• e. 2 points If SSE c is the SSE in the constrained model (31.2.26) and SSE b
the SSE in (31.2.30), show that SSE c = r · SSE b + SSE u .
ˆ
(y mq − xm β )2 =
ˆ
)2 + r
(y − xm β )2 ; Answer. For every m we have
therefore SSE c = m,q (y mq − y m·
¯ q m q ˆ
(y mq − y m· )2 + r (y m· − xm β )2 ;
¯ m· • f . 3 points Write down the formula of the F test in terms of SSE u and SSE c
with a correct accounting of the degrees of freedom, and give this formula also in
terms of SSE u and SSE b .
Answer. Unconstrained model has n parameters, and constrained model has k parameters;
the number of additional “constraints” is therefore n − k. This gives the F statistic
(31.2.31) (SSE c − SSE u )/(n − k)
r SSE b /(n − k)
=
SSE u /n(r − 1)
SSE u /n(r − 1) 31.4. TESTS OF NONLINEAR HYPOTHESES 315 31.3. The FTest Statistic is a Function of the Likelihood Ratio
Problem 355. The critical region of the generalized likelihood ratio test can be
written as
C = {y1 , . . . , yn : (31.3.1) supθ∈Ω (y1 , . . . , yn ; θ1 , . . . , θk )
≥ k },
supθ∈ω (y1 , . . . , yn ; θ1 , . . . , θk ) where ω refers to the null and Ω to the alternative hypothesis (it is assumed that the
hypotheses are nested, i.e., ω ⊂ Ω). In other words, one rejects the hypothesis if the
maximal achievable likelihood level with the restriction imposed is much lower than
ˆ
ˆ
θ
that without the restriction. If θ is the unrestricted and ˆ the restricted maximum
likelihood estimator, then the test statistic is
ˆ
ˆ
θ
LR = 2(log (y , θ ) − log (y , ˆ )) → χ2
i (31.3.2) where i is the number of restrictions. In this exercise we are proving that the F test
in the linear model is equivalent to the generalized likelihood ratio test. (You should
assume here that both β and σ 2 are unknown.) All this is in [Gre97, p. 304].
• a. 1 point Since we only have constraints on β and not on σ 2 , it makes sense
to ﬁrst compute the concentrated likelihood function with σ 2 concentrated out. Derive
the formula for this concentrated likelihood function which is given in [Gre97, just
above (6.88)].
Answer.
(31.3.3) Concentrated log (y ; β ) = − n
1
1 + log 2π + log (y − Xβ ) (y − Xβ )
2
n • b. 2 points In the case of a linear restriction, show that LR is connected with
the F statistic F as follows:
(31.3.4) LR = n log 1 + Answer. LR = −n log 1
ε
ˆ
n ε − log
ˆ 1ˆ
ε
ˆ
n i
F
n−k ˆ
ε = n log
ˆ ˆˆ
εε
ˆˆ
εε
ˆˆ In order to connect this with the F statistic note that
(31.3.5) F= n−k
i ˆˆ
εε
ˆˆ
−1
εε
ˆˆ 31.4. Tests of Nonlinear Hypotheses
Make linear approximation, need Jacobian for this. Here is an example where a
nonlinear hypothesis arises naturally:
Problem 356. [Gre97, Example 7.14 on p. 361]: The model
(31.4.1) Ct = α + βYt + γCt−1 + εt has diﬀerent long run and short run propensities to consume. Give formulas for both.
Answer. Shortrun is β ; to compute the long run propensity, which would prevail in the
stationary state when Ct = Ct−1 , write C∞ = α + βY∞ + γC∞ + ε∞ or C∞ (1 − γ ) = α + βY∞ + ε∞
or C∞ = α/(1 − γ ) + β/(1 − γ )Y∞ + εt /(1 − γ ). Therefore long run propensity is δ = β/(1 − γ ). 316 31. THREE PRINCIPLES FOR TESTING A LINEAR CONSTRAINT 31.5. Choosing Between Nonnested Models
Throwing all regressors into the same regression is a straightforward way out but
not very good. Jtest (the J comes from “joint”) is better: throw the predicted values
of one of the two models as a regressor into the other model and test whether this
predicted value has a nonzero coeﬃcient. Here is more detail: if the null hypothesis
is that model 1 is right, then throw the predicted value of model 2 into model 1
and test the null hypothesis that the coeﬃcient of this predicted value is zero. If
Model 1 is right, then this additional regressor leaves all other estimators unbiased,
and the true coeﬃcient of the additional regressor is 0. If Model 2 is right, then
asymptotically, this additional regressor should be the only regressor in the combined
model with a nonzero coeﬃcient (its coeﬃcient is = 1 asymptotically, and all the
other regressors should have coeﬃcient zero.) Whenever nonnested hypotheses are
tested, is is possible that both hypotheses are rejected, or that neither hypothesis is
rejected by this criterion. CHAPTER 32 Instrumental Variables
Compare here [DM93, chapter 7] and [Gre97, Section 6.7.8]. Greene ﬁrst introduces the simple instrumental variables estimator and then shows that the generalized one picks out the best linear combinations for forming simple instruments. I will
follow [DM93] and ﬁrst introduce the generalized instrumental variables estimator,
and then go down to the simple one.
In this chapter, we will discuss a sequence of models y n = X n β + ε n , where
ε n ∼ (on , σ 2 I n ), and X n are n × k matrices of random regressors, and the number
1
of observations n → ∞. We do not make the assumption plim n X n ε n = o which
would ensure consistency of the OLS estimator (compare Problem 328). Instead, a
sequence of n × m matrices of (random or nonrandom) “instrumental variables” W n
is available which satisﬁes the following three conditions:
(32.0.1)
(32.0.2)
(32.0.3) plim 1
W εn = o
nn 1
W W n = Q exists, is nonrandom and nonsingular
nn
1
plim W n X n = D exists, is nonrandom and has full column rank
n plim Full column rank in (32.0.3) is only possible if m ≥ k .
In this situation, regression of y on X is inconsistent. But if one regresses y
on the projection of X on R[W ], the column space of W , one obtains a consistent
estimator. This is called the instrumental variables estimator.
If xi is the ith column vector of X , then W (W W )−1 W xi is the projection of xi on the space spanned by the columns of W . Therefore the matrix
W (W W )−1 W X consists of the columns of X projected on R[W ]. This is what
we meant by the projection of X on R[W ]. With these projections as regressors,
the vector of regression coeﬃcients becomes the “generalized instrumental variables
estimator”
(32.0.4) ˜
β = X W (W W )−1 W X −1 X W (W W )−1 W y Problem 357. 3 points We are in the model y = X β + ε and we have a
matrix W of “instrumental variables” which satisﬁes the following three conditions:
1
1
plim n W ε = o, plim n W W = Q exists, is nonrandom and positive deﬁnite, and
1
plim n W X = D exists, is nonrandom and has full column rank. Show that the
instrumental variables estimator
(32.0.5) ˜
β = X W (W W )−1 W X ˜
is consistent. Hint: Write β n − β = B n ·
matrices B n has a plim.
317 −1 1
nW X W (W W )−1 W y ε and show that the sequence of 318 32. INSTRUMENTAL VARIABLES Answer. Write it as
˜
βn = X W (W W )−1 W X −1 X W (W W )−1 W (X β + ε ) = β + X W (W W )−1 W X −1 X W (W W )−1 W ε 1
1
1
= β + ( X W )( W W )−1 ( W X )
n
n
n −1 1
1
1
( X W )( W W )−1 W ε ,
n
n
n i.e., the B n and B of the hint are as follows:
1
1
1
( X W )( W W )−1 ( W X )
n
n
n
B = plim B n = (D Q−1 D )−1 D Q−1 Bn = −1 1
1
( X W )( W W )−1
n
n 1
1
Problem 358. Assume plim n X X exists, and plim n X ε exists. (We only
need the existence, not that the ﬁrst is nonsingular and the second zero). Show that
1
˜
˜
σ 2 can be estimated consistently by s2 = n (y − X β ) (y − X β ). ˜
˜
˜
Answer. y − X β = X β + ε − X β = ε − X (β − β ). Therefore
1
1
2
˜
˜
˜
˜
( y − X β ) ( y − X β ) = ε ε − ε X ( β − β ) + (β − β )
n
n
n 1
˜
X X (β − β ).
n All summands have plims, the plim of the ﬁrst is σ 2 and those of the other two are zero. Problem 359. In the situation of Problem 357, add the stronger assumption
√˜
ε → N (o, σ 2 Q), and show that n(β n − β ) → N (o, σ 2 (D Q−1 D )−1 ) 1
√W
n √˜
1
˜
Answer. β n − β = B n n W n ε n , therefore n(β n − β ) = B n n−1/2 W n ε n → B N (o, σ 2 Q) =
N (o, σ 2 BQB ). Since B = (D Q−1 D )−1 D Q−1 , the result follows. ˜
From Problem 359 follows that for ﬁnite samples approximately β n − β ∼
2
1
˜
N o, σ (D Q−1 D )−1 . Since n (D Q−1 D )−1 = (nD (nQ)−1 nD )−1 , MSE [β ; β ]
n
−1 can be estimated by s2 X W (W W )−1 W X
The estimator (32.0.4) is sometimes called the two stages least squares estimate,
because the projection of X on the column space of W can be considered the predicted values if one regresses every column of X on W . I.e., instead of regressing y
on X one regresses y on those linear combinations of the columns of W which best
approximate the columns of X . Here is more detail: the matrix of estimated coeﬃˆ
cients in the ﬁrst regression is Π = (W W )−1 W X , and the predicted values in
ˆ = W Π = W (W W )−1 W X . The second regression, which
ˆ
this regression are X
ˆ
regresses y on X , gives the coeﬃcient vector
(32.0.6) ˆˆ
ˆ
˜
β = (X X )−1 X y . If you plug this in you see this is exactly (32.0.4) again.
Now let’s look at the geometry of instrumental variable regression of one variable
y on one other variable x with w as an instrument. The speciﬁcation is y = xβ + ε .
On p. 280 we visualized the asymptotic results if ε is asymptotically orthogonal to x.
Now let us assume ε is asymptotically not orthogonal to x. One can visualize this as
√
three vectors, again normalized by dividing by n, but now even in the asymptotic
case the ε vector is not orthogonal to x. (Draw ε vertically, and make x long enough
that β < 1.) We assume n is large enough so that the asymptotic results hold for
the sample already (or, perhaps better, that the diﬀerence between the sample and
its plim is only inﬁnitesimal). Therefore the OLS regression, with estimates β by 32. INSTRUMENTAL VARIABLES 319 x y /x x, is inconsistent. Let O be the origin, A the point on the xvector where
ε branches oﬀ (i.e., the end of xβ ), furthermore let B be the point on the xvector
where the orthogonal projection of y comes down, and C the end of the xvector.
¯¯
¯2
¯
¯
Then x y = OC OB and x x = OC , therefore x y /x x = OB/OC , which would
be the β if the errors were orthogonal. Now introduce a new variable w which is
orthogonal to the errors. (Since ε is vertical, w is on the horizontal axis.) Call D the
projection of y on w, which is the prolongation of the vector ε , and call E the end of
¯¯
the wvector, and call F the projection of x on w. Then w y = OE OD, and w x =
¯¯
¯¯
¯
¯
¯
¯
¯ OF . Therefore w y /w x = (OE OD)(OE OF ) = OD/OF = OA/OC = β .
¯
OE
Or geometrically it is obvious that the regression of y on the projection of x on w
ˆ
will give the right β . One also sees here why the s2 based on this second regression
is inconsistent.
If I allow two instruments, the two instruments must be in the horizontal plane
perpendicular to the vector ε which is assumed still vertical. Here we project x on
this horizontal plane and then regress the y , which stays where it is, on this x. In
this way the residuals have the right direction!
What if there is one instrument, but it does not not lie in the same plane as
x and y ? This is the most general case as long as there is only one regressor and
one instrument. This instrument w must lie somewhere in the horizontal plane. We
have to project x on it, and then regress y on this projection. Look at it this way:
take the plane orthogonal to w which goes through point C . The projection of x
on w is the intersection of the ray generated by w with this plane. Now move this
plane parallel until it intersects point A. Then the intersection with the wray is the
projection of y on w. But this latter plane contains ε , since ε is orthogonal to w.
This makes sure that the regression gives the right results.
Problem 360. 4 points The asymptotic MSE matrix of the instrumental variables estimator with W as matrix of instruments is σ 2 plim X W (W W )−1 W X
Show that if one adds more instruments, then this asymptotic MSE matrix can only
decrease. It is suﬃcient to show that the inequality holds before going over to the
plim, i.e., if W = U V , then
(32.0.7) X U (U U )−1 U X is nonnegative deﬁnite.
not required). (2) Note
partitioned matrix form?
W G(G W W G)−1 G −1 − X W (W W )−1 W X −1 Hints: (1) Use theorem A.5.5 in the Appendix (proof is
that U = W G for some G. Can you write this G in
(3) Show that, whatever W and G, W (W W )−1 W −
W is idempotent. Answer.
(32.0.8) U= U V I
= WG
O where G= I
.
O Problem 361. 2 points Show: if a matrix D has full column rank and is square,
then it has an inverse.
Answer. Here you need that column rank is row rank: if D has full column rank it also
has full row rank. And to make the proof complete you need: if A has a left inverse L and a
right inverse R, then L is the only left inverse and R the only right inverse and L = R. Proof:
L = L(AR) = (LA)R = R. −1 320 32. INSTRUMENTAL VARIABLES Problem 362. 2 points If W X is square and has full column rank, then it is
nonsingular. Show that in this case (32.0.4) simpliﬁes to the “simple” instrumental
variables estimator:
˜
(32.0.9)
β = (W X )−1 W y
Answer. In this case the big inverse can be split into three:
(32.0.10)
(32.0.11) ˜
β= X W ( W W ) −1 W X −1 X W (W W )−1 W y = = (W X )−1 W W (X W )−1 X W (W W )−1 W y Problem 363. We only have one regressor with intercept, i.e., X = ι x , and
we have one instrument w for x (while the constant term is its own instrument),
i.e., W = ι w . Show that the instrumental variables estimators for slope and
intercept are
(wt − w)(y t − y )
¯
¯
˜
(32.0.12)
β=
¯
¯
(wt − w)(xt − x)
˜x
(32.0.13)
α = y − β¯
˜¯
Hint: the math is identical to that in question 200.
Problem 364. 2 points Show that, if there are as many instruments as there are
observations, then the instrumental variables estimator (32.0.4) becomes identical to
OLS.
Answer. In this case W has an inverse, therefore the projection on R[W ] is the identity.
Staying in the algebraic paradigm, (W W )−1 = W −1 (W )−1 . An implication of Problem 364 is that one must be careful not to include too
many instruments if one has a small sample. Asymptotically it is better to have more
instruments, but for n = m, the instrumental variables estimator is equal to OLS, i.e.,
the sequence of instrumental variables estimators starts at the (inconsistent) OLS.
If one uses fewer instruments, then the asymptotic MSE matrix is not so good, but
one may get a sequence of estimators which moves away from the inconsistent OLS
more quickly. APPENDIX A Matrix Formulas
In this Appendix, eﬀorts are made to give some of the familiar matrix lemmas in
their most general form. The reader should be warned: the concept of a deﬁciency
matrix and the notation which uses a thick fraction line multiplication with a scalar
ginverse are my own.
A.1. A Fundamental Matrix Decomposition
Theorem A.1.1. Every matrix B which is not the null matrix can be written
as a product of two matrices B = CD , where C has a left inverse L and D a right
inverse R, i.e., LC = DR = I . This identity matrix is r × r, where r is the rank
of B .
A proof is in [Rao73, p. 19]. This is the fundamental theorem of algebra, that
every homomorphism can be written as a product of epimorphism and monomorphism, together with the fact that all epimorphisms and monomorphisms split, i.e.,
have onesided inverses.
One such factorization is given by the singular value theorem: If B = P ΛQ
is the svd as in Theorem A.9.2, then one might set e.g. C = P Λ and D = Q,
consequently L = Λ−1 P and R = Q . In this decomposition, the ﬁrst row/column
carries the largest weight and gives the best approximation in a least squares sense,
etc.
The trace of a square matrix is deﬁned as the sum of its diagonal elements. The
rank of a matrix is deﬁned as the number of its linearly independent rows, which is
equal to the number of its linearly independent columns (row rank = column rank).
Theorem A.1.2. tr BC = tr CB .
Problem 365. Prove theorem A.1.2.
Problem 366. Use theorem A.1.1 to prove that if BB = B , then rank B =
tr B .
Answer. Premultiply the equation CD = CDCD by L and postmultiply it by R to get
DC = I r . This is useful for the trace: tr B = tr CD = tr DC = tr I r = r . I have this proof from
[Rao73, p. 28]. Theorem A.1.3. B = O if and only if B B = O .
A.2. The Spectral Norm of a Matrix
The spectral norm of a matrix extends the Euclidean norm z from vectors
to matrices. Its deﬁnition is A = max z =1 Az . This spectral norm is the
maximum singular value µmax , and if A is square, then A−1 = 1/µmin . It is a
true norm, i.e., A = 0 if and only if A = O , furthermore λA = λ· A , and the
triangle inequality A + B ≤ A + B . In addition, it obeys AB ≤ A · B .
Problem 367. Show that the spectral norm is the maximum singular value.
321 322 A. MATRIX FORMULAS Answer. Use the deﬁnition
(A.2.1) A 2 = max z A Az
zz . Write A = P ΛQ as in (A.9.1), Then z A Az = z Q Λ2 Qz . Therefore we can ﬁrst show:
there is a z in the form z = Q x which attains this maximum. Proof: for every z which has a
nonzero value in the numerator of (A.2.1), set x = Qz . Then x = o, and Q x attains the same
value as z in the numerator of (A.2.1), and a smaller or equal value in the denominator. Therefore
one can restrict the search for the maximum argument to vectors of the form Q x. But for them
2
the objective function becomes x x Λx x , which is maximized by x = i1 , the ﬁrst unit vector (or
column vector of the unit matrix). Therefore the squared spectral norm is λ2 , and therefore the
ii
spectral norm itself is λii . A.3. Inverses and gInverses of Matrices
A ginverse of a matrix A is any matrix A− satisfying
(A.3.1) A = AA− A. It always exists but is not always unique. If A is square and nonsingular, then A1
is its only ginverse.
Problem 368. Show that a symmetric matrix Ω has a ginverse which is also
symmetric.
Answer. Use Ω −ΩΩ − . The deﬁnition of a ginverse is apparently due to [Rao62]. It is sometimes called
the “conditional inverse” [Gra83, p. 129]. This ginverse, and not the MoorePenrose
generalized inverse or pseudoinverse A+ , is needed for the linear model, The MoorePenrose generalized inverse is a ginverse that in addition satisﬁes A+ AA+ = A+ ,
and AA+ as well as A+ A symmetric. It always exists and is also unique, but the
additional requirements are burdensome ballast. [Gre97, pp. 445] also advocates
the MoorePenrose inverse, but he does not really use it. If he were to try to use it,
he would probably soon discover that it is not appropriate. The book [Alb72] does
the linear model with the MoorePenrose inverse. It is a good demonstration of how
complicated everything gets if one uses an inappropriate mathematical tool.
Problem 369. Use theorem A.1.1 to prove that every matrix has a ginverse.
Answer. Simple: a null matrix has its transpose as ginverse, and if A = O then RL is such
a ginverse. The ginverse of a number is its inverse if the number is nonzero, and is arbitrary
otherwise. Scalar expressions written as fractions are in many cases the multiplication
by a ginverse. We will use a fraction with a thick horizontal rule to indicate where
this is the case. In other words, by deﬁnition,
a
a
= b− a.
Compare that with the ordinary fraction
.
(A.3.2)
b
b
This idiosyncratic notation allows to write certain theorems in a more concise form,
but it requires more work in the proofs, because one has to consider the additional
case that the denominator is zero. Theorems A.5.8 and A.8.2 are examples.
Theorem A.3.1. If B = AA− B holds for one ginverse A− of A, then it holds
for all ginverses. If A is symmetric and B = AA− B , then also B = B A− A.
If B = BA− A and C = AA− C then BA− C is independent of the choice of ginverses. A.4. DEFICIENCY MATRICES 323 Proof. Assume the identity B = AA+ B holds for some ﬁxed ginverse A+
(which may be, as the notation suggests, the Moore Penrose ginverse, but this is
not necessary), and let A− be an diﬀerent ginverse. Then AA− B = AA− AA+ B =
AA+ B = B . For the second statement one merely has to take transposes and note
that a matrix is a ginverse of a symmetric A if and only if its transpose is. For the
third statement: BA+ C = BA− AA+ AA− C = BA− AA− C = BA− C . Here +
signiﬁes a diﬀerent ginverse; again, it is not necessarily the MoorePenrose one.
Problem 370. Show that x satisﬁes x = Ba for some a if and only if x =
BB − x.
Theorem A.3.2. Both A (AA )− and (A A)− A are ginverses of A.
Proof. We have to show
(A.3.3) A = AA (AA )− A which is [Rao73, (1b.5.5) on p. 26]. Deﬁne D = A − AA (AA )− A and show, by
multiplying out, that DD = O .
A.4. Deﬁciency Matrices
Here is again some idiosyncratic terminology and notation. It gives an explicit
algebraic formulation for something that is often done implicitly or in a geometric
paradigm. A matrix G will be called a “left deﬁciency matrix” of S , in symbols,
G ⊥ S , if GS = O , and for all Q with QS = O there is an X with Q = XG. This
factorization property is an algebraic formulation of the geometric concept of a null
space. It is symmetric in the sense that G ⊥ S is also equivalent with: GS = O ,
and for all R with GR = O there is a Y with R = SY . In other words, G ⊥ S and
S ⊥ G are equivalent.
This symmetry follows from the following characterization of a deﬁciency matrix
which is symmetric:
Theorem A.4.1. T ⊥ U iﬀ T U = O and T T + U U nonsingular. Proof. This proof here seems terribly complicated. There must be a simpler
way. Proof of “⇒”: Assume T ⊥ U . Take any γ with γ T T γ + γ U U γ =
0, i.e., T γ = o and γ U = o . From this one can show that γ = o: since
T γ = o, there is a ξ with γ = U ξ , therefore γ γ = γ U ξ = 0. To prove
“⇐” assume T U = O and T T + U U is nonsingular. To show that T ⊥ U
take any B with BU = O . Then B = B (T T + U U )(T T + U U )−1 =
BT T (T T + U U )−1 . In the same way one gets T = T T T (T T + U U )−1 .
Premultiply this last equation by T T (T T T T )− T and use theorem A.3.2 to get
T T (T T T T )− T T = T T (T T + U U )−1 . Inserting this into the equation
for B gives B = BT T (T T T T )− T T , i.e., B factors over T .
The R/Splusfunction Null gives the transpose of a deﬁciency matrix.
Theorem A.4.2. If for all Y , BY = O implies AY = O , then a X exists with
A = XB .
Problem 371. Prove theorem A.4.2.
Answer. Let B ⊥ C . Choosing Y = B follows AB = O , hence X exists. Problem 372. Show that I − SS − ⊥ S .
Answer. Clearly, (I − SS − )S = O . Now if QS = O , then Q = Q(I − SS − ), i.e., the X
whose existence is postulated in the deﬁnition of a deﬁciency matrix is Q itself. 324 A. MATRIX FORMULAS Problem 373. Show that S ⊥ U if and only if S is a matrix with maximal rank
which satisﬁes SU = O . In other words, one cannot add linearly independent rows
to S in such a way that the new matrix still satisﬁes T U = O .
Answer. First assume S ⊥ U and take any additional row t
exists a Q
S
such that
r
t = Q
S , i.e., SQ = S , and t
r so that S
t U= O
. Then
o = r S . But this last equation means that t is a linear combination of the rows of S with the ri as coeﬃcients. Now conversely, assume
S
O
S is such that one cannot add a linearly independent row t such that
U=
, and let
t
o
P U = O . Then all rows of P must be linear combinations of rows of S (otherwise one could add
such a row to S and get the result which was just ruled out), therefore P = SS where A is the
matrix of coeﬃcients of these linear combinations. The deﬁciency matrix is not unique, but we will use the concept of a deﬁciency
matrix in a formula only then when this formula remains correct for every deﬁciency
matrix. One can make deﬁciency matrices unique if one requires them to be projection matrices.
Problem 374. Given X and a symmetric nonnegative deﬁnite Ω such that X =
Ω W for some W . Show that X ⊥ U if and only if X Ω − X ⊥ U .
Answer. One has to show that XY = O is equivalent to X Ω − XY = O . ⇒ clear; for
⇐ note that X Ω − X = W Ω W , therefore XY = Ω W Y = Ω W (W Ω W )− W Ω W Y =
Ω W (W Ω W )− X Ω − XY = O . A matrix is said to have full column rank if all its columns are linearly independent, and full row rank if its rows are linearly independent. The deﬁciency matrix
provides a “holistic” deﬁnition for which it is not necessary to look at single rows
and columns. X has full column rank if and only if X ⊥ O , and full row rank if and
only if O ⊥ X .
Problem 375. Show that the following three statements are equivalent: (1) X
has full column rank, (2) X X is nonsingular, and (3) X has a left inverse.
Answer. Here use X ⊥ O as the deﬁnition of “full column rank.” Then (1) ⇔ (2) is theorem
A.4.1. Now (1) ⇒ (3): Since IO = O , a P exists with I = P X . And (3) ⇒ (1): if a P exists with
I = P X , then any Q with QO = O can be factored over X , simply say Q = QP X . Note that the usual solution of linear matrix equations with ginverses involves
a deﬁciency matrix:
Theorem A.4.3. The solution of the consistent matrix equation T X = A is
(A.4.1) X = T −A + U W where T ⊥ U and W is arbitrary.
Proof. Given consistency, i.e., the existence of at least one Z with T Z = A,
(A.4.1) deﬁnes indeed a solution, since T X = T T − T Z . Conversely, if Y satisﬁes
T Y = A, then T (Y − T − A) = O , therefore Y − T − A = U W for some W .
Theorem A.4.4. Let L ⊥ T ⊥ U and J ⊥ HU ⊥ R; then
L
−J HT − O
T
⊥
⊥ U R.
H
J A.4. DEFICIENCY MATRICES 325 Proof. First deﬁciency relation: Since I −T T − = U W for some W , −J HT − T +
T
J H = O , therefore the matrix product is zero. Now assume A B
= O.
H
Then BHU = O , i.e., B = DJ for some D . Then AT = −DJ H , which
has as general solution A = −DJ HT − + CL for some C . This together gives
L
O
AB=CD
. Now the second deﬁciency relation: clearly,
−J HT − J
the product of the matrices is zero. If M satisﬁes T M = O , then M = U N
for some N . If M furthermore satisﬁes HM = O , then HU N = O , therefore
N = RP for some P , therefore M = U RP .
Theorem A.4.5. Assume Ω is nonnegative deﬁnite symmetric and K is such
that KΩ is deﬁned. Then the matrix
(A.4.2) Ξ = Ω − Ω K (KΩ K )− KΩ has the
(1)
(2)
(3) following properties:
Ξ does not depend on the choice of ginverse of KΩ K used in (A.4.2).
Any ginverse of Ω is also a ginverse of Ξ, i.e. ΞΩ − Ξ = Ξ.
Ξ is nonnegative deﬁnite and symmetric.
K
(4) For every P ⊥ Ω follows
⊥Ξ
P
K
K
(5) If T is any other right deﬁciency matrix of
, i.e., if
⊥ T , then
P
P (A.4.3) Ξ = T (T Ω − T )− T . Hint: show that any D satisfying Ξ = T DT is a ginverse of T Ω− T .
In order to apply (A.4.3) show that the matrix T = SK where K ⊥ S and
K
P S ⊥ K is a right deﬁciency matrix of
.
P
Proof of theorem A.4.5: Independence of choice of ginverse follows from theorem
A.5.10. That Ω − is a ginverse is also an immediate consequence of theorem A.5.10.
From the factorization Ξ = ΞΩ − Ξ follows also that Ξ is nnd symmetric (since every
nnd symmetric Ω also has a symmetric nnd ginverse). (4) Deﬁciency property:
K
From
Q = O follows KQ = O and P Q = O . From this second equation
P
and P ⊥ Ω follows Q = Ω R for some R. Since KΩ R = KQ = O , it follows
Ω
Q = Ω R = (Ω − Ω K (KΩ K )− KΩ )R.
K
Proof of (5): Since
Ξ = O it follows Ξ = T A for some A, and therefore
P
Ξ = ΞΩ − Ξ = T AΩ − A T = T DT where D = AΩ − A .
Ω
Before going on we need a lemma. Since (I − ΩΩ − )Ω = O , there exists a N
with I − ΩΩ − = N P , therefore T − ΩΩ − T = N P T = O or
(A.4.4) T = ΩΩ − T Using (A.4.4) one can show the hint: that any D satisfying Ξ = T DT
ginverse of T Ω − T : is a Ω
Ω
(A.4.5) T Ω − T DT Ω − T ≡ T Ω − (Ω − Ω K (KΩ K )− KΩ )Ω − T = T Ω − T .
To complete the proof of (5) we have to show that the expression T (T Ω − T )− T
does not depend on the choice of the ginverse of T Ω − T . This follows from
T (T Ω − T )− T = ΩΩ − T (T Ω − T )− T Ω −Ω and theorem A.5.10. 326 A. MATRIX FORMULAS Theorem A.4.6. Given two matrices T and U . Then T ⊥ U if and only if for
any D the following two statements are equivalent:
(A.4.6) TD = O and
(A.4.7) For all C which satisfy CU = O follows CD = O .
A.5. Nonnegative Deﬁnite Symmetric Matrices By deﬁnition, a symmetric matrix Ω is nonnegative deﬁnite if a Ω a ≥ 0 for all
vectors a. It is positive deﬁnite if a Ω a > 0 for all vectors a = o.
Theorem A.5.1. Ω nonnegative deﬁnite symmetric if and only if it can be written in the form Ω = A A for some A.
Theorem A.5.2. If Ω is nonnegative deﬁnite, and a Ω a = 0, then already
Ω a = o.
Theorem A.5.3. A is positive deﬁnite if and only it is nonnegative deﬁnite and
nonsingular.
Theorem A.5.4. If the symmetric matrix A has a nnd ginverse then A itself
is also nnd.
Theorem A.5.5. If Ω and Σ are positive deﬁnite, then Ω − Σ is positive (nonnegative) deﬁnite if and only if Σ −1 − Ω −1 is.
Ω
Theorem A.5.6. If Ω and Σ are nonnegative deﬁnite, then tr(ΩΣ ) ≥ 0.
Problem 376. Prove theorem A.5.6.
Ω
Answer. Find any factorization Σ = P P . Then tr(ΩΣ ) = tr(P Ω P ) ≥ 0. Theorem A.5.7. If Ω is nonnegative deﬁnite symmetric, then
(A.5.1) (g Ω a)2 ≤ g Ω g a Ω a, for arbitrary vectors a and g . Equality holds if and only if Ω g and Ω a are linearly
dependent, i.e., α and β exist, not both zero, such that Ω g α + Ω aβ = o.
Proof: First we will show that the condition for equality is suﬃcient. Therefore
assume Ω g α + Ω aβ = 0 for a certain α and β , which are not both zero. Without
loss of generality we can assume α = 0. Then we can solve a Ω g α + a Ω aβ = 0 to
get a Ω g = −(β/α)a Ω a, therefore the lefthand side of (A.5.1) is (β/α)2 (a Ω a)2 .
Furthermore we can solve g Ω g α + g Ω aβ = 0 to get g Ω g = −(β/α)g Ω a =
(β/α)2 a Ω a, therefore the righthand side of (A.5.1) is (β/α)2 (a Ω a)2 as well—i.e.,
(A.5.1) holds with equality.
Secondly we will show that (A.5.1) holds in the general case and that, if it holds
with equality, Ω g and Ω a are linearly dependent. We will split this second half of
the proof into two substeps. First verify that (A.5.1) holds if g Ω g = 0. If this is
the case, then already Ω g = o, therefore the Ω g and Ω a are linearly dependent and,
by the ﬁrst part of the proof, (A.5.1) holds with equality.
The second substep is the main part of the proof. Assume g Ω g = 0. Since Ω
is nonnegative deﬁnite, it follows
(A.5.2)
(g Ω a)2 (g Ω a)2
(g Ω a)2
g Ωa
g Ωa
0 ≤ a−g
Ω a−g
= a Ω a−2
+
= a Ω a−
.
g Ωg
g Ωg
g Ωg
g Ωg
g Ωg A.5. NONNEGATIVE DEFINITE SYMMETRIC MATRICES From this follows (A.5.1). If (A.5.2) is an equality, then already Ω a − g g
g
o, which means that Ω g and Ω a are linearly dependent. 327
Ωa
Ωg = Theorem A.5.8. In the situation of theorem A.5.7, one can take ginverses as
follows without disturbing the inequality
(g Ω a)2 ≤ a Ω a.
g Ωg
Equality holds if and only if a γ = 0 exists with Ω g = Ω aγ . (A.5.3) Problem 377. Show that if Ω is nonnegative deﬁnite, then its elements satisfy
2
ωij ≤ ωii ωjj (A.5.4) Answer. Let a and b be the ith and j th unit vector. Then
(A.5.5) (b Ω a)2 ≤ max (g Ω a)2 g b Ωb g Ωg = a Ω a. Problem 378. Assume Ω nonnegative deﬁnite symmetric. If x satisﬁes x = Ω a
for some a, show that
(g x)2 = x Ω − x.
Ωg
g
Furthermore show that equality holds if and only if Ω g = xγ for some γ = 0.
(A.5.6) max
g Answer. From x = Ω a follows g x = g Ω a and x Ω − x = a Ω a; therefore it follows from
theorem A.5.8. Problem 379. Assume Ω nonnegative deﬁnite symmetric, x satisﬁes x = Ω a
for some a, and R is such that Rx is deﬁned. Show that
(A.5.7) x R (RΩ R )− Rx ≤ x Ω − x Answer. Follows from
(A.5.8) max (h Rx)2 ≤ max (g x)2 g Ωg
h RΩ R h
because on the term on the lhs maximization is done over the smaller set of g which have the
form Rh. An alternative proof would be to show that Ω − Ω r (RΩ R )− RΩ is nnd (it has Ω − as
ginverse).
h g Problem 380. Assume Ω nonnegative deﬁnite symmetric. Show that
(A.5.9) max
g:
Ω
g =Ω a
for some a (g x)2
g Ω−g = x Ω x. Answer. Since g = Ω a for some a, maximize over a instead of g . This reduces it to theorem
A.5.8:
( g x) 2
(a Ω x)2
(A.5.10)
max
= max
= x Ωx
Ω
g : g =Ω a for some a g Ω − g
a
a Ωa Theorem A.5.9. Let Ω be symmetric and nonnegative deﬁnite, and x an arbitrary vector. Then Ω − xx is nonnegative deﬁnite if and only if the following two
conditions hold: x can be written in the form x = Ω a for some a, and x Ω − x ≤ 1
for one (and therefore for all) ginverses Ω − of Ω . 328 A. MATRIX FORMULAS Problem 381. Prove theorem A.5.9.
Answer. Assume x = Ω a and x Ω − x = a Ω a ≤ 1; then for any g , g (Ω − xx )g =
g Ω g − g Ω aa Ω g ≥ a Ω ag Ω g − g Ω aa Ω g ≥ 0 by theorem A.5.7.
Conversely, assume x cannot be written in the form x = Ω a for some a; then a g exists with
g Ω = o but g x = o. Then g (Ω − xx )g < 0, therefore not nnd.
Finally assume x Ω − x = a Ω a > 1; then a (Ω − xx )a = a Ω a − (a Ω a)2 < 0, therefore
again not nnd. Theorem A.5.10. If Ω and Σ are nonnegative deﬁnite symmetric, and K a
matrix so that Σ KΩ is deﬁned, then
KΩ = (KΩ K + Σ )(KΩ K + Σ )− KΩ . (A.5.11) Furthermore, Ω K (KΩ K + Σ )− KΩ is independent of the choice of ginverses.
Problem 382. Prove theorem A.5.10.
Answer. To see that (A.5.11) is a special case of (A.3.3), take any Q with Ω = QQ and P
with Σ = P P and deﬁne A = K Q P . The independence of the choice of ginverses follows
from theorem A.3.1 together with (A.5.11). The following was apparently ﬁrst shown in [Alb69] for the special case of the
MoorePenrose pseudoinverse:
Theorem A.5.11. The symmetric partitioned matrix Ω = Ω yy
Ω yz Ω yz
is nonΩ zz negative deﬁnite if and only if the following conditions hold:
(A.5.12)
Ω yy and Ω zz.y := Ω zz − Ω yz Ω −y Ω yz
y are both nonnegative deﬁnite, and Ω yz = Ω yy Ω −y Ω yz
y (A.5.13) Reminder: It follows from theorem A.3.1 that (A.5.13) holds for some ginverse
if and only if it holds for all, and that, if it holds, Ω zz.y is independent of the choice
of the ginverse.
Proof of theorem A.5.11: First we prove the necessity of the three conditions
in the theorem. If the symmetric partitioned matrix Ω is nonnegative deﬁnite,
Ω yy Ω yz
there exists a R with Ω = R R. Write R = Ry Rz to get
=
Ω yz Ω zz
Ry Ry Ry Rz
. Ω yy is nonnegative deﬁnite because it is equal to Ry Ry ,
Rz Ry Rz Rz
and (A.5.13) follows from (A.5.11): Ω yy Ω −y Ω yz = Ry Ry (Ry Ry )− Ry Rz =
y
Ry Rz = Ω yz . To show that Ω zz.y is nonnegative deﬁnite, deﬁne S = (I −
Ry (Ry Ry )− Ry )Rz . Then S S = Rz I − Ry (Ry Ry )− Ry Rz = Ω zz.y .
To show suﬃciency of the three conditions of theorem A.5.11, assume the symΩyy Ωyz
metric
satisﬁes them. Pick two matrices Q and S so that Ω yy = Q Q
Ω yz Ω zz
and Ω zz.y = S S . Then
Ω yy
Ω yz Q
Ω yz
=
Ω zz
Ω yz Ω −y Q
y O
S Q
O QΩ −y Ω yz
y
,
S therefore nonnegative deﬁnite.
Problem 383. [SM86, A 3.2/11] Given a positive deﬁnite matrix Q and a
positive deﬁnite Q with Q∗ = Q − Q nonnegative deﬁnite. A.6. PROJECTION MATRICES 329 • a. Show that Q − QQ−1 Q is nonnegative deﬁnite.
−1 Answer. We know that Q −1 − Q∗−1 is nnd, therefore QQ Q − QQ∗−1 Q nnd. • b. This part is more diﬃcult: Show that also Q∗ − Q∗ Q−1 Q∗ is nonnegative
deﬁnite.
Answer. We will write it in a symmetric form from which it is obvious that it is nonnegative
deﬁnite:
(A.5.14) Q∗ − Q∗ Q−1 Q∗ = Q∗ − Q∗ (Q + Q∗ )−1 Q∗ (A.5.15) = Q∗ (Q + Q∗ )−1 (Q + Q∗ − Q∗ ) = Q∗ (Q + Q∗ )−1 Q (A.5.16) = Q(Q + Q∗ )−1 (Q + Q∗ )Q (A.5.17) = QQ−1 (Q∗ + Q∗ Q −1 −1 Q∗ (Q + Q∗ )−1 Q Q∗ )Q−1 Q. Problem 384. Given the vector h = o. For which values of the scalar γ is
the matrix I − hh singular, nonsingular, nonnegative deﬁnite, a projection matrix,
γ
orthogonal?
Answer. It is nnd iﬀ γ ≥ h h, because of theorem A.5.9. One easily veriﬁes that it is
orthogonal iﬀ γ = h h/2, and it is a projection matrix iﬀ γ = h h. Now let us prove that it is
singular iﬀ γ = h h: if this condition holds, then the matrix annuls h; now assume the condition
does not hold, i.e., γ = h h, and take any x with (I − hh )x = o. It follows x = hα where
γ
α = h x/γ , therefore (I − hh )x = hα(1 − h h/γ ). Since h = o and 1 − h h/γ = 0 this can
γ
only be the null vector if α = 0. A.6. Projection Matrices
Problem 385. Show that X (X X )− X is the projection matrix on the range
space R[X ] of X , i.e., on the space spanned by the columns of X . This is true
whether or not X has full column rank.
Answer. Idempotence requires theorem A.3.2, and symmetry the invariance under choice of
ginverse. Furthermore one has to show X (X X )− Xa = a holds if and only if a = Xb for some
b. ⇒ is clear, and ⇐ follows from theorem A.3.2. Theorem A.6.1. Let P and Q be projection matrices, i.e., both are symmetric
and idempotent. Then the following ﬁve conditions are equivalent, each meaning that
the space on which P projects is a subspace of the space on which Q projects:
(A.6.1)
(A.6.2)
(A.6.3) R[P ] ⊂ R[Q]
QP = P
PQ = P (A.6.4) Q−P projection matrix (A.6.5) Q−P nonnegative deﬁnite. (A.6.2) is geometrically trivial. It means: if one ﬁrst projects on a certain space,
and then on a larger space which contains the ﬁrst space as a subspace, then nothing
happens under this second projection because one is already in the larger space.
(A.6.3) is geometrically not trivial and worth remembering: if one ﬁrst projects on a
certain space, and then on a smaller space which is a subspace of the ﬁrst space, then
the result is the same as if one had projected directly on the smaller space. (A.6.4)
means: the diﬀerence Q − P is the projection on the orthogonal complement of R[P ] 330 A. MATRIX FORMULAS in R[Q]. And (A.6.5) means: the projection of a vector on the smaller space cannot
be longer than that on the larger space.
Problem 386. Prove theorem A.6.1.
Answer. Instead of going in a circle it is more natural to show (A.6.1) ⇐⇒ (A.6.2) and
(A.6.3) ⇐⇒ (A.6.2) and then go in a circle for the remaining conditions: (A.6.2), (A.6.3) ⇒
(A.6.4) ⇒ (A.6.3) ⇒ (A.6.5).
(A.6.1) ⇒ (A.6.2): R[P ] ⊂ R[Q] means that for every c exists a d with P c = Qd. Therefore
far all c follows QP c = QQd = Qd = P c, i.e., QP = P .
(A.6.2) ⇒ (A.6.1): if P c = QP c for all c, then clearly R[P ] ⊂ R[Q].
(A.6.2) ⇒ (A.6.3) by symmetry of P and Q: If QP = P then P Q = P Q = (QP ) =
P = P.
(A.6.3) ⇒ (A.6.2) follows in exactly the same way: If P Q = P then QP = Q P = (P Q) =
P = P.
(A.6.2), (A.6.3) ⇒ (A.6.4): Symmetry of Q − P clear, and (Q − P )(Q − P ) = Q − P − P + P =
Q − P.
(A.6.4) ⇒ (A.6.5): c (Q − P )c = c (Q − P ) (Q − P )c ≥ 0.
(A.6.5) ⇒ (A.6.3): First show that, if Q − P nnd, then Qc = o implies P c = o. Proof: from
Q − P nnd and Qc = o follows 0 ≤ c (Q − P )c = −c P c ≤ 0, therefore equality throughout, i.e.,
0 = c P c = c P P c = P c 2 and therefore P c = o. Secondly: this is also true for matrices:
QC = O implies P C = O , since it is valid for every column of C . Thirdly: Since Q(I − Q) = O ,
it follows P (I − Q) = O , which is (A.6.3). Problem 387. If Y = XA for some A, show that Y (Y Y )− Y X (X X )− X
Y (Y Y )− Y . = Answer. Y = XA means that every column of Y is a linear combination of columns of A:
(A.6.6) y1 ··· y m = X a1 ··· am = X a1 ··· Xam . Therefore geometrically the statement follows from the fact shown in Problem 385 that the
above matrices are projection matrices on the columnn spaces. But it can also be shown algebraically: Y (Y Y )− Y X (X X )− X = Y (Y Y )− A X X (X X )− X = Y (Y Y )− Y . Problem 388. (Not eligible for inclass exams) Let Q be a projection matrix
(i.e., a symmetric and idempotent matrix) with the property that Q = XAX for
˜
some A. Deﬁne X = (I − Q)X . Then
(A.6.7) X (X X )− X ˜
˜˜ ˜
= X (X X )− X + Q. Hint: this can be done through a geometric argument. If you want to do it
algebraically, you might want to use the fact that (X X )− is also a ginverse of
˜˜
X X. Answer. Geometric argument: Q is a projector on a subspace of the range space of X . The
˜
columns of X are projections of the columns of X on the orthogonal complement of the space on
which Q projects. The equation which we have to prove shows therefore that the projection on the
column space of X is the sum of the projections on the space Q projects on plus the projection on
the orthogonal complement of that space in X .
˜˜
Now an algebraic proof: First let us show that (X X )− is a ginverse of X X , i.e., let us
evaluate
(A.6.8)
X (I −Q)X (X X )− X (I −Q)X = X X (X X )− X X −X X (X X )− X QX −X QX (X X )− X X +X QX (X X )− X (A.6.9) = X X − X QX − X QX + X QX = X (I − Q)X . Only for the fourth term did we need the condition Q = XAX :
(A.6.10) X X AX X (X X )− X X AX X = X X AX X AX X = X QQX = X X . A.7. DETERMINANTS 331 Using this ginverse we have
(A.6.11) X (X X )− X ˜
˜
− X (X X )− X = X (X X )− X − (I − Q)X (X X )− X (I − Q) = (A.6.12)
= X (X X )− X −X (X X )− X +X (X X )− X Q+QX (X X )− X −QX (X X )− X Q = X (X X )− X −X (X X )− X + Problem 389. Given any projection matrix P . Show that its ith diagonal element can be written
(A.6.13) p2 .
ij pii =
j Answer. From idempotence P = P P follows pii =
(A.6.13). j pij pji , now use symmetry to get A.7. Determinants
Theorem A.7.1. The determinant of a blocktriangular matrix is the product of
the determinants of the blocks in the diagonal. In other words,
A
O (A.7.1) B
= A D 
D For the proof recall the deﬁnition of a determinant. A mapping π : {1, . . . , n} →
{1, . . . , n} is called a permutation if and only if it is onetoone if and only if it is
onto. Permutations can be classiﬁed as even or odd according to whether they can
be written as the product of an even or odd number of transpositions. Then the
determinant is deﬁned as
(A.7.2) a1π(1) · · · anπ(n) − det(A) =
π : π even a1π(1) · · · anπ(n)
π : π odd Now assume A is m × m, 1 ≤ m < n. If a j ≤ m exists with π (j ) > m then
not all i ≤ m can be images of other points j ≤ m, i.e., there must be at least one
j > m with π (j ) ≤ m. Therefore, in a block triangular matrix in which all aij = 0
for i ≤ m, j > m, only those permutations give a nonzero product which remain in
the two submatrices straddling the diagonal.
Theorem A.7.2. If B = AA− B , then the following identity is valid between
determinants:
(A.7.3) A
C B
= A E 
D where E = D − CA− B . Proof: Postmultiply by a matrix whose determinant, by lemma A.7.1, is one,
and then apply lemma A.7.1 once more:
(A.7.4)
A
O
AB
A B I −A− B
=
=
= A D − CA− B .
CD
CDO
I
C D − CA− B
Problem 390. Show the following counterpart of theorem A.7.2: If C = DD − C ,
then the following identity is valid between determinants:
(A.7.5) A
C B
= A − BD − C D  .
D 332 A. MATRIX FORMULAS Answer.
A
C (A.7.6) A
B
=
C
D I
−D − C B
D A − BD − C
O
=
O
I B
= A − BD − D D  .
D Problem 391. Show that whenever BC and CB are deﬁned, it follows I − BC  =
I − CB 
Answer. Set A = I and D = I in (A.7.3) and (A.7.5). Theorem A.7.3. Assume that d = W W − d. Then
det(W + α · dd ) = det(W )(1 + αd W − d). (A.7.7) Proof: If α = 0, then there is nothing to prove. Otherwise look at the determinant of the matrix
W
d
(A.7.8)
H=
d
−1/α
Equations (A.7.3) and (A.7.5) give two expressions for it:
1
(A.7.9)
det(H ) = det(W )(−1/α − d W − d) = − det(W + αdd ).
α
A.8. More About Inverses
AB
which satisﬁes B = AA− B
CD
and C = CA− A. (These conditions hold for instance, due to theorem A.5.11, if
AB
is nonnegative deﬁnite symmetric, but it also holds in the nonsymmetric
CD
case if A is nonsingular, which by theorem A.7.2 is the case if the whole partioned
matrix is nonsingular.) Deﬁne E = D − CA− B , F = A− B , and G = CA− .
Problem 392. Given a partitioned matrix • a. Prove that in terms of A, E , F , and G, the original matrix can be written
as
A
C (A.8.1) B
A
=
D
GA AF
E + GAF (this is trivial), and that (this is the nontrivial part)
A− + F E − G −F E −
−E − G
E− (A.8.2) is a ginverse of A
C B
.
D Answer. This here is not the shortest proof because I was still wondering if it could be
formulated in a more general way. Multiply out but do not yet use the conditions B = AA− B and
C = CA− A:
(A.8.3) A
C B
D A− + F E − G
−E − G −F E −
E− = AA− − (I − AA− )BE − G
(I − EE − )G (I − AA− )BE −
EE − and
(A.8.4) AA− − (I − AA− )BE − G
(I − EE − )G
= (I − AA− )BE −
EE − A
C A + (I − AA− )BE − C (I − A− A)
C − (I − EE − )C (I − A− A) B
=
D
B − (I − AA− )B (I − E − E )
D One sees that not only the conditions B = AA− B and C = CA− A, but also the conditions B =
AA− B and C = EE − C , or alternatively the conditions B = BE − E and C = CA− A imply the
statement. I think one can also work with the conditions AA− B = BD − D and DD − C = CA− A.
Note that the lower right partition is D no matter what. A.8. MORE ABOUT INVERSES U
W • b. If V
X is a ginverse of A
GA 333 AF
, show that X is a gE + GAF inverse of E .
Answer. The ginverse condition means
A
GA (A.8.5) AF
E + GAF U
W V
X A
GA AF
E + GAF = A
GA AF
E + GAF The ﬁrst matrix product evaluated is
(A.8.6) A
GA AF
E + GAF U
W V
X = AU + AF W
GAU + EW + GAF W The ginverse condition means therefore
(A.8.7)
AU + AF W
AV + AF X
GAU + EW + GAF W GAV + EX + GAF X A
GA AV + AF X
.
GAV + EX + GAF X AF
E + GAF = A
GA AF
E + GAF For the upper left partition this means AU A + AF W A + AV GA + AF XGA = A, and for the upper
right partition it means AU AF + AF W AF + AV E + AV GAF + AF XE + AF XGAF = AF .
Postmultiply the upper left equation by F and subtract from the upper right to get AV E +
AF XE = O . For the lower left we get GAU A + EW A + GAF W A + GAV GA + EXGA +
GAF XGA = GA. Premultiplication of the upper left equation by G and subtraction gives EW A +
EXGA = O . For the lower right corner we get GAU AF + EW AF + GAF W AF + GAV E +
EXE + GAF XE + GAV GAF + EXGAF + GAF XGAF = E + GAF . Since AV E + AF XE = O
and EW A + EXGA = O , this simpliﬁes to GAU AF + GAF W AF + EXE + GAV GAF +
GAF XGAF = E + GAF . And if one premultiplies the upper right corner by G and postmultiplies
it by F and subtracts it from this one gets EXE = E . Problem 393. Show that a ginverse of the matrix
X1 X1
X2 X1 (A.8.8) X1 X2
X2 X2 has the form
(A.8.9) (X 1 X 1 )− + D 1 X 2 (X 2 M 1 X 2 )− X 2 D 1
−(X 2 M 1 X 2 )− X 2 D 1 −D 1 X 2 (X 2 M 1 X 2 )−
(X 2 M 1 X 2 )− where M 1 = I − X 1 (X 1 X 1 )− X 1 and D 1 = X 1 (X 1 X 1 )− .
Answer. Either show it by multiplying it out, or apply Problem 392. Problem 394. Show that the following are ginverses:
(A.8.10)
−
−
I
X
IO
(X X )−
XX X
=
=
OO
X
XX
X
I
O O
I − X (X X )− X Answer. Either do it by multiplying it out, or apply problem 392. Problem 395. Assume again B = AA− B and C = CAA− , but assume this
AB
time that
nonsingular. Then A is nonsingular,
CD
(A.8.11)
−1
A
PQ
AB
AB
and if
=
, then the determinant
=
.
RS
CD
CD
S 
Answer. The determinant is, by (A.7.3), A E  where E = D − CA− B . By assumption,
this determinant is nonzero, therefore also A and E  are nonzero, i.e., A and E are nonsingular.
Therefore (A.8.2) reads
(A.8.12) P
R Q
A−1 + F E −1 G
=
S
−E −1 G −F E −1
,
E −1 334 A. MATRIX FORMULAS i.e., S = E −1 = (D − CA− B )−1 . hence A E  = A / S . Theorem A.8.1. Given a m × n matrix A, a m × h matrix B , a k × n matrix C ,
and a k × h matrix D satisfying AA− B = BD − D and DD − C = CA− A. Then
the following are ginverses:
(A.8.13) (A.8.14) D + CA− B − A + BD − C − = A− − A− B (D + CA− B )− CA− = D − − D − C (A + BD − C )− BD − . Problem 396. Prove theorem A.8.1.
Answer. Proof: Deﬁne E = D + CA− B . Then it follows from the assumptions that
(A.8.15)
(A + BD − C )(A− − A− BE − CA− ) = AA− − BD − DE − CA− + BD − CA− − BD − CA− BE − CA− =
= AA− + BD − (I − EE − )CA− (A.8.16) Since AA− (A + BD − C ) = A + BD − C , we have to show that the second term on the rhs. annulls
(A + BD − C ). Indeed,
BD − (I − EE − )CA− (A + BD − C ) = (A.8.17) (A.8.18) = BD − CA− A + BD − CA− BD − C − BD − EE − CA− A − BD − EE − CA− BD − C =
(A.8.19)
= BD − (D + CA− B − EE − D − EE − CA− B )D − C = BD − (E − EE − E )D − C = O . Theorem A.8.2. (ShermanMorrisonWoodbury theorem) Given a m × n matrix
A, a m × 1 vector b satisfying AA− b = b, a n × 1 vector c satisfying c AA− = c ,
and a scalar δ . If A− is a ginverse of A, then
(A.8.20) A− − A− bc A− is a ginverse of − c A b+δ A+ bc
δ Problem 397. Prove theorem A.8.2.
Answer. It is a special case of theorem A.8.1. Theorem A.8.3. For any symmetric nonnegative deﬁnite r × r matrix A,
(A.8.21) (det A) e−(tr A) ≤ e−r , with equality holding if and only if A = I .
Problem 398. Prove Theorem A.8.3. Hint: Let λ1 , . . . , λr be the eigenvalues of
A. Then det A = i λi , and tr A = i λi .
Answer. Therefore the inequality reads
r λi e−λi ≤ e−r (A.8.22)
i=1 For this it is suﬃcient to show for each value of λ
(A.8.23) λe−λ ≤ e−1 , which follows immediately by taking the derivatives: e−λ − λe−λ = 0 gives λ = 1. The matrix with
all eigenvalues being equal to 1 is the identity matrix. A.9. EIGENVALUES AND SINGULAR VALUE DECOMPOSITION 335 A.9. Eigenvalues and Singular Value Decomposition
Every symmetric matrix B has real eigenvalues and a system of orthogonal
eigenvectors which span the whole space. If one normalizes these eigenvectors and
combines them as row vectors into a matrix T , then orthonormality means T T = I ,
and since T is square, T T = I also implies T T = I , i.e., T is an orthogonal
matrix. The existence of a complete set of real eigenvectors is therefore equivalent
to the following matrix algebraic result: For every symmetric matrix B there is an
orthogonal transformation T so that BT = T Λ where Λ is a diagonal matrix.
Equivalently one could write B = T ΛT . And if B has rank r, then r of the
diagonal elements are nonzero and the others zero. If one removes those eigenvectors
from T which belong to the eigenvalue zero, and calls the remaining matrix P , one
gets the following:
Theorem A.9.1. If B is a symmetric n × n matrix of rank r, then a r × n
matrix P exists with P P = I (any P satisfying this condition which is not a
square matrix is called incomplete orthogonal), and B = P ΛP , where Λ is a r × r
diagonal matrix with all diagonal elements nonzero.
Proof. Let T be an orthogonal matrix whose rows are eigenvectors of B , and
P
partition it T =
where P consists of all eigenvectors with nonzero eigenvalue
Q
ΛO
(there are r of them). The eigenvalue property reads B P
;
Q =P
Q
OO
ΛOP
therefore by orthogonality T T = I follows B = P
=
Q
OOQ
IO
P
=
,
P ΛP . Orthogonality also means T T = I , i.e.,
P
Q
OI
Q
therefore P P = I .
Problem 399. If B is a n × n symmetric matrix of rank r and B 2 = B , i.e.,
B is a projection, then a r × n matrix P exists with B = P P and P P = I .
Answer. Let t be an eigenvector of the projection matrix B with eigenvalue λ. Then B 2 t =
Bt, i.e., λ2 t = λt, and since t = o, λ2 = λ. This is a quadratic equation with solutions λ = 0 or
λ = 1. The matrix Λ from theorem A.9.1, whose diagonal elements are the nonzero eigenvalues, is
therefore an identity matrix. A theorem similar to A.9.1 holds for arbitrary matrices. It is called the “singular
value decomposition”:
Theorem A.9.2. Let B be a m × n matrix of rank r. Then B can be expressed
as
(A.9.1) B = P ΛQ where Λ is a r × r diagonal matrix with positive diagonal elements, and P P = I
as well as QQ = I . The diagonal elements of Λ are called the singular values of
B.
Proof. If P ΛQ is the svd of B then P ΛQQ ΛP = P Λ2 Q is the eigenvalue decomposition of BB . We will use this fact to construct P and Q, and then
verify condition (A.9.1). P and Q have r rows each, write them p1
q1
.
.
(A.9.2)
P = . and Q = . .
.
.
pr qr 336 A. MATRIX FORMULAS Then the pi are orthonormal eigenvectors of BB corresponding to the nonzero
eigenvalues λ2 , and q i = B pi λ−1 . The proof that this deﬁnition is symmetric is
i
i
left as exercise problem 400 below.
Now ﬁnd pr+1 , . . . , pm such that p1 , . . . , pm is a complete set of orthonormal
vectors, i.e., p1 p1 + · · · + pm pm = I . Then
(A.9.3) B = (p1 p1 + · · · + pm pm )B (A.9.4) = (p1 p1 + · · · + pr pr )B (A.9.5) = (p1 q 1 λ1 + · · · + pr q r λr ) = P ΛQ. because pi B = o for i > r Problem 400. Show that the q i are orthonormal eigenvectors of B B corresponding to the same eigenvalues λ2 .
i
Answer.
(A.9.6)
(A.9.7) q i q j = λ−1 pi BB pj λ−1 = λ−1 pi pj λ2 λ−1 = δij
jj
i
j
i
B B qi = B B B pi λ−1
i = B pi λi = Kronecker symbol q i λ2
i Problem 401. Show that Bq i = λi pi and B pi = λi q i .
Answer. The second condition comes from the deﬁnition q i = B pi λ−1 , and premultiply
i
this deﬁnition by B to get Bq i = BB pi λ−1 = λ2 pi λ1 = λpi .
i
i P
Q
and
are orthogonal. Then the singular
P0
Q0
value decomposition can also be written in the full form, in which the matrix in the
middle is m × n:
ΛO Q
(A.9.8)
B= P
P0
O O Q0
Let P 0 and Q0 be such that Problem 402. Let λ1 be the biggest diagonal element of Λ, and let c and d be
two vectors with the properties that c B d is deﬁned and c c = 1 as well as d d = 1.
Show that c B d ≤ λ1 . The other singular values maximize among those who are
orthogonal to the prior maximizers.
Answer. c B d = c P ΛQd = h Λk where we call P c = h and Qd = k. By CauchySchwartz (A.5.1), (h Λk)2 ≤ (h Λh)(k Λk). Now (h Λk) =
λii h2 ≤
λ11 h2 = λ11 h h.
i
i
Now we only have to show that h h ≤ 1: 1 − h h = c c − c P P c = c (I − P P )c =
c (I − P P )(I − P P )c ≥ 0, here we used that P P = I , therefore P P idempotent, therefore
also I − P P idempotent. APPENDIX B Arrays of Higher Rank
This chapter was presented at the Array Programming Languages Conference in
Berlin, on July 24, 2000.
Besides scalars, vectors, and matrices, also higher arrays are necessary in statistics; for instance, the “covariance matrix” of a random matrix is really an array of
rank 4, etc. Usually, such higher arrays are avoided in the applied sciences because
of the diﬃculties to write them on a twodimensional sheet of paper. The following
symbolic notation makes the structure of arrays explicit without writing them down
element by element. It is hoped that this makes arrays easier to understand, and that
this notation leads to simple highlevel user interfaces for programming languages
manipulating arrays.
B.1. Informal Survey of the Notation
Each array is symbolized by a rectangular tile with arms sticking out, similar
to a molecule. Tiles with one arm are vectors, those with two arms matrices, those
with more arms are arrays of higher rank (or “valence” as in [SS35], [Mor73], and
[MS86, p. 12]), and those without arms are scalars. The arrays considered here are
rectangular, not “ragged,” therefore in addition to their rank we only need to know
the dimension of each arm; it can be thought of as the number of ﬁngers associated
with this arm. Arrays can only hold hands (i.e., “contract” along two arms) if the
hands have the same number of ﬁngers.
Sometimes it is convenient to write the dimension of each arm at the end of the
A
n . Matrix products
arm, i.e., a m × n matrix A can be represented as m
are represented by joining the obvious arms: if B is n × q , then the matrix product
AB is m
A
n
B
q or, in short,
A
B
. The notation allows
the reader to always tell which arm is which, even if the arms are not marked. If
C
m
r is m × r, then the product C A is (B.1.1) C A= r C
m A n=r C m A n. In the second representation, the tile representing C is turned by 180 degrees. Since
the white part of the frame of C is at the bottom, not on the top, one knows that
the West arm of C , not its East arm, is concatenated with the West arm of A. The
C
r is r
C
m , i.e., it is not a diﬀerent entity but
transpose of m
the same entity in a diﬀerent position. The order in which the elements are arranged
on the page (or in computer memory) is not a part of the deﬁnition of the array
itself. Likewise, there is no distinction between row vectors and column vectors.
Vectors are usually, but not necessarily, written in such a way that their arm
points West (column vector convention). If
a and
b are vectors, their
337 338 B. ARRAYS OF HIGHER RANK scalar product a b is the concatenation a b which has no free arms, i.e., it is a scalar, and their outer product ab is
a
b
, which is a matrix.
Juxtaposition of tiles represents the outer product, i.e., the array consisting of all
the products of elements of the arrays represented by the tiles placed side by side.
Q
is the concatenation
, which
Q
The trace of a square matrix
is a scalar since no arms are sticking out. In general, concatenation of two arms of
the same tile represents contraction, i.e., summation over equal values of the indices
associated with these two arms. This notation makes it obvious that tr XY =
X
Y
tr Y X , because by deﬁnition there is no diﬀerence between
and
Y X . Also X Y or X Y etc. represent the same array (here array of rank zero, i.e., scalar). Each of these tiles can be evaluated in essentially
two diﬀerent ways. One way is
(1) Juxtapose the tiles for X and Y , i.e., form their outer product, which is
an array of rank 4 with typical element xmp yqn .
(2) Connect the East arm of X with the West arm of Y . This is a contraction, resulting in an array of rank 2, the matrix product XY , with typical
element p xmp ypn .
(3) Now connect the West arm of X with the East arm of Y . The result of
this second contraction is a scalar, the trace tr XY = p,m xmp ypm .
An alternative sequence of operations evaluating this same graph would be
(1) Juxtapose the tiles for X and Y .
(2) Connect the West arm of X with the East arm of Y to get the matrix
product Y X .
(3) Now connect the East arm of X with the West arm of Y to get tr Y X .
The result is the same, the notation does not specify which of these alternative evaluation paths is meant, and a computer receiving commands based on this notation
can choose the most eﬃcient evaluation path. Probably the most eﬃcient evaluation
path is given by (B.2.8) below: take the elementbyelement product of X with the
transpose of Y , and add all the elements of the resulting matrix.
If the user speciﬁes tr(XY ), the computer is locked into one evaluation path: it
ﬁrst has to compute the matrix product XY , even if X is a column vector and Y a
row vector and it would be much more eﬃcient to compute it as tr(Y X ), and then
form the trace, i.e., throw away all oﬀdiagonal elements. If the trace is speciﬁed
as X Y , the computer can choose the most eﬃcient of a number of diﬀerent evaluation paths transparently to the user. This advantage of the graphical notation
is of course even more important if the graphs are more complex.
There is also the “diagonal” array, which in the case of rank 3 can be written
n ∆ (B.1.2) n n
∆ or
n n n or similar conﬁgurations. It has 1’s down the main diagonal and 0’s elsewhere. It
can be used to construct the diagonal matrix diag(x) of a vector (the square matrix B.2. AXIOMATIC DEVELOPMENT OF ARRAY OPERATIONS 339 with the vector in the diagonal and zeros elsewhere) as
n
(B.1.3) ∆ n diag(x) = ,
x the diagonal vector of a square matrix (i.e., the vector containing its diagonal elements) as
∆ (B.1.4) A, and the “Hadamard product” (elementbyelement product) of two vectors x ∗ y as
x
(B.1.5) x∗y= ∆ .
y All these are natural operations involving vectors and matrices, but the usual matrix
notation cannot represent them and therefore adhoc notation must be invented for
it. In our graphical representation, however, they all can be built up from a small
number of atomic operations, which will be enumerated in Section B.2.
Each such graph can be evaluated in a number of diﬀerent ways, and all these
evaluations give the same result. In principle, each graph can be evaluated as follows:
form the outer product of all arrays involved, and then contract along all those pairs
of arms which are connected. For practical implementations it is more eﬃcient to
develop functions which connect two arrays along one or several of their arms without
ﬁrst forming outer products, and to perform the array concatenations recursively in
such a way that contractions are done as early as possible. A computer might be
programmed to decide on the most eﬃcient construction path for any given array.
B.2. Axiomatic Development of Array Operations
The following sketch shows how this axiom system might be built up. Since I
am an economist I do not plan to develop the material presented here any further.
Others are invited to take over. If you are interested in working on this, I would be
happy to hear from you; email me at ehrbar@econ.utah.edu
There are two kinds of special arrays: unit vectors and diagonal arrays.
For every natural number m ≥ 1, m unit vectors m
i (i = 1, . . . , m) exist.
Despite the fact that the unit vectors are denoted here by numbers, there is no
intrinsic ordering among them; they might as well have the names “red, green, blue,
. . . ” (From (B.2.4) and other axioms below it will follow that each unit vector can
be represented as a mvector with 1 as one of the components and 0 elsewhere.)
For every rank ≥ 1 and dimension n ≥ 1 there is a unique diagonal array denoted
by ∆. Their main properties are (B.2.1) and (B.2.2). (This and the other axioms
must be formulated in such a way that it will be possible to show that the diagonal
arrays of rank 1 are the “vectors of ones” ι which have 1 in every component; diagonal
arrays of rank 2 are the identity matrices; and for higher ranks, all arms of a diagonal
array have the same dimension, and their ijk · · · element is 1 if i = j = k = · · ·
and 0 otherwise.) Perhaps it makes sense to deﬁne the diagonal array of rank 0
and dimension n to be the scalar n, and to declare all arrays which are everywhere
0dimensional to be diagonal. 340 B. ARRAYS OF HIGHER RANK There are only three operations of arrays: their outer product, represented by
writing them side by side, contraction, represented by the joining of arms, and the
direct sum, which will be deﬁned now:
The direct sum is the operation by which a vector can be built up from scalars,
a matrix from its row or column vectors, an array of rank 3 from its layers, etc. The
direct sum of a set of r similar arrays (i.e., arrays which have the same number of
arms, and corresponding arms have the same dimensions) is an array which has one
additional arm, called the reference arm of the direct sum. If one “saturates” the
reference arm with the ith unit vector, one gets the ith original array back, and this
property deﬁnes the direct sum uniquely:
m m m m r Ai n=r S ⇒ n i r S n= Ai n. i=1 q q q q It is impossible to tell which is the ﬁrst summand and which the second, direct sum
is an operation deﬁned on ﬁnite sets of arrays (where diﬀerent elements of a set may
be equal to each other in every respect but still have diﬀerent identities).
There is a broad rule of associativity: the order in which outer products and
contractions are performed does not matter, as long as the at the end, the right arms
are connected with each other. And there are distributive rules involving (contracted)
outer products and direct sums.
Additional rules apply for the special arrays. If two diﬀerent diagonal arrays
join arms, the result is again a diagonal array. For instance, the following three
concatenations of diagonal threeway arrays are identical, and they all evaluate to
the (for a given dimension) unique diagonal array or rank 4:
∆
(B.2.1) = ∆ ∆ ∆ = ∆ = ∆ ∆
The diagonal array of rank 2 is neutral under concatenation, i.e., it can be written
as
(B.2.2) n ∆ n = . because attaching it to any array will not change this array. (B.2.1) and (B.2.2) make
it possible to represent diagonal arrays simply as the branching points of several arms.
This will make the array notation even simpler. However in the present introductory
article, all diagonal arrays will be shown explicitly, and the vector of ones will be
denoted m
ι instead of m
∆ or perhaps m
δ.
Unit vectors concatenate as follows:
(B.2.3) i m j = 1
0 if i = j
otherwise. and the direct sum of all unit vectors is the diagonal array of rank 2:
n (B.2.4) i
i=1 n = n ∆ n = . B.2. AXIOMATIC DEVELOPMENT OF ARRAY OPERATIONS 341 I am sure there will be modiﬁcations if one works it all out in detail, but if done
right, the number of axioms should be fairly small. Elementbyelement addition of
arrays is not an axiom because it can be derived: if one saturates the reference arm
of a direct sum with the vector of ones, one gets the elementbyelement sum of the
arrays in this direct sum. Multiplication of an array by a scalar is also contained in
the above system of axioms: it is simply the outer product with an array of rank
zero.
Problem 403. Show that the saturation of an arm of a diagonal array with the
vector of ones is the same as dropping this arm.
Answer. Since the vector of ones is the diagonal array of rank 1, this is a special case of the
general concantenation rule for diagonal arrays. Problem 404. Show that the diagonal matrix of the vector of ones is the identity
matrix, i.e.,
n ∆ n (B.2.5) = . ι
Answer. In view of (B.2.2), this is a special case of Problem 403. Problem 405. A trivial array operation is the addition of an arm of dimension
1; for instance, this is how a nvector can be turned into a n × 1 matrix. Is this
operation contained in the above system of axioms?
Answer. It is a special case of the direct sum: the direct sum of one array only, the only eﬀect
of which is the addition of the reference arm. From (B.2.4) and (B.2.2) follows that every array of rank k can be represented
as a direct sum of arrays of rank k − 1, and recursively, as iterated direct sums of
those scalars which one gets by saturating all arms with unit vectors. Hence the
following “extensionality property”: if the arrays A and B are such that for all
possible conformable choices of unit vectors κ1 · · · κ8 follows
κ3
κ2 A κ7 = κ4 κ5 κ2 B κ6 κ1 κ6 κ8 κ3 κ5 κ1 (B.2.6) κ4 κ8 κ7 then A = B . This is why the saturation of an array with unit vectors can be
considered one of its “elements,” i.e.,
κ3 κ5 κ2 A κ6 κ1 (B.2.7) κ4 κ8 κ7 = aκ1 κ2 κ3 κ4 κ5 κ6 κ7 κ8 . From (B.2.3) and (B.2.4) follows that the concatenation of two arrays by joining
one or more pairs of arms consists in forming all possible products and summing over
those subscripts (arms) which are joined to each other. For instance, if
m A n B r =m C r, 342 B. ARRAYS OF HIGHER RANK
n then cµρ = ν =1 aµν bνρ . This is one of the most basic facts if one thinks of arrays
as collections of elements. From this point of view, the proposed notation is simply
a graphical elaboration of Einstein’s summation convention. But in the holistic
approach taken by the proposed system of axioms, which is informed by category
theory, it is an implication; it comes at the end, not the beginning.
Instead of considering arrays as bags ﬁlled with elements, with the associated
false problem of specifying the order in which the elements are packed into the
bag, this notation and system of axioms consider each array as an abstract entity,
associated with a certain ﬁnite graph. These entities can be operated on as speciﬁed
in the axioms, but the only time they lose their abstract character is when they are
fully saturated, i.e., concatenated with each other in such a way that no free arms are
left: in this case they become scalars. An array of rank 1 is not the same as a vector,
although it can be represented as a vector—after an ordering of its elements has
been speciﬁed. This ordering is not part of the deﬁnition of the array itself. (Some
vectors, such as time series, have an intrinsic ordering, but I am speaking here of
the simplest case where they do not.) Also the ordering of the arms is not speciﬁed,
and the order in which a set of arrays is packed into its direct sum is not speciﬁed
either. These axioms therefore make a strict distinction between the abstract entities
themselves (which the user is interested in) and their various representations (which
the computer worries about).
Maybe the following examples may clarify these points. If you specify a set of
colors as {red, green, blue}, then this representation has an ordering built in: red
comes ﬁrst, then green, then blue. However this ordering is not part of the deﬁnition
of the set; {green, red, blue} is the same set. The two notations are two diﬀerent representations of the same set. Another example: mathematicians usually distinguish
between the outer products A ⊗ B and B ⊗ A; there is a “natural isomorphism”
between them but they are two diﬀerent objects. In the system of axioms proposed
here these two notations are two diﬀerent representations of the same object, as in
the set example. This object is represented by a graph which has A and B as nodes,
but it is not apparent from this graph which node comes ﬁrst. Interesting conceptual
issues are involved here. The proposed axioms are quite diﬀerent than e.g. [Mor73].
Problem 406. The trace of the product of two matrices can be written as
tr(XY ) = ι (X ∗ Y )ι. (B.2.8) I.e., one forms the elementbyelement product of X and Y and takes the sum of
all the elements of the resulting matrix. Use tile notation to show that this gives
indeed tr(XY ).
Answer. In analogy with (B.1.5), the Hadamard product of the two matrices X and Z , i.e.,
their element by element multiplication, is
X
X∗Z= ∆ ∆
Z If Z = Y , one gets
X
X∗Y = ∆ ∆
Y . B.4. EQUALITY OF ARRAYS AND EXTENDED SUBSTITUTION 343 Therefore one gets, using (B.2.5):
X
ι (X ∗ Y )ι = ι X ∆ ∆ ι = = tr(XY )
Y Y B.3. An Additional Notational Detail
Besides turning a tile by 90, 180, or 270 degrees, the notation proposed here also
allows to ﬂip the tile over. The tile
(here drawn without its arms) is simply the
tile
laid on its face; i.e., those parts of the frame, which are black on the side
visible to the reader, are white on the opposite side and vice versa. If one ﬂips a
tile, the arms appear in a mirrorsymmetric manner. For a matrix, ﬂipping over is
equivalent to turning by 180 degrees, i.e., there is no diﬀerence between the matrix
A
and the matrix
A
. Since sometimes one and sometimes the other
notation seems more natural, both will be used. For higher arrays, ﬂipping over
arranges the arms in a diﬀerent fashion, which is sometimes convenient in order to
keep the graphs uncluttered. It will be especially useful for diﬀerentiation. If one
allows turning in 90 degree increments and ﬂipping, each array can be represented
in eight diﬀerent positions, as shown here with a hypothetical array of rank 3:
m
k L n k m L n n k L m n L
k
k m L m k m m n L k n n L L k
m n The blackandwhite pattern at the edge of the tile indicates whether and how much
the tile has been turned and/or ﬂipped over, so that one can keep track which arm
is which. In the above example, the arm with dimension k will always be called the
West arm, whatever position the tile is in.
B.4. Equality of Arrays and Extended Substitution
Given the ﬂexibility of representing the same array in various positions for concatenation, speciﬁc conventions are necessary to determine when two such arrays in
generalized positions are equal to each other. Expressions like
A =B or K = K are not allowed. The arms on both sides of the equal sign must be parallel, in order
to make it clear which arm corresponds to which. A permissible way to write the 344 B. ARRAYS OF HIGHER RANK above expressions would therefore be
A =B and K = K One additional beneﬁt of this tile notation is the ability to substitute arrays with
diﬀerent numbers of arms into an equation. This is also a necessity since the number
of possible arms is unbounded. This multiplicity can only be coped with because
each arm in an identity written in this notation can be replaced by a bundle of many
arms.
Extended substitution also makes it possible to extend deﬁnitions familiar from
matrices to higher arrays. For instance we want to be able to say that the array
Ω
is symmetric if and only if
Ω
=
Ω
. This notion of symmetry is not limited to arrays of rank 2. The arms of this array may symbolize not just
a single arm, but whole bundles of arms; for instance an array of the form
satisfying Σ = Σ Σ is symmetric according to this deﬁnition, and so is every scalar. Also the notion of a nonnegative deﬁnite matrix, or of a matrix inverse
or generalized inverse, or of a projection matrix, can be extended to arrays in this
way.
B.5. Vectorization and Kronecker Product
One conventional generally accepted method to deal with arrays of rank > 2 is
the Kronecker product. If A and B are both matrices, then the outer product in tile
notation is
A
(B.5.1)
B
Since this is an array of rank 4, there is no natural way to write its elements down
on a sheet of paper. This is where the Kronecker product steps in. The Kronecker
product of two matrices is their outer product written again as a matrix. Its deﬁnition
includes a protocol how to arrange the elements of an array of rank 4 as a matrix.
Alongside the Kronecker product, also the vectorization operator is useful, which is
a protocol how to arrange the elements of a matrix as a vector, and also the socalled
“commutation matrices” may become necessary. Here are the relevant deﬁnitions:
B.5.1. Vectorization of a Matrix. If A is a matrix, then vec(A) is the vector
obtained by stacking the column vectors on top of each other, i.e., a1
.
(B.5.2)
if A = a1 · · · an
then vec(A) = . .
.
an
The vectorization of a matrix is merely a diﬀerent arrangement of the elements of
the matrix on paper, just as the transpose of a matrix.
Problem 407. Show that tr(B C ) = (vec B ) vec C . B.5. VECTORIZATION AND KRONECKER PRODUCT 345 Answer. Both sides are
bji cji . (B.5.28) is a proof in tile notation which does not have to
look at the matrices involved element by element. By the way, a better protocol for vectorizing would have been to assemble all
rows into one long row vector and then converting it into a column vector. In other
words b1
.
if B = . . then vec(B ) should have been deﬁned as b1
. . .
.
bm bm The usual protocol of stacking the columns is inconsistent with the lexicograpical
ordering used in the Kronecker product. Using the alternative deﬁnition, equation
(B.5.19) which will be discussed below would be a little more intelligible; it would
read
vec(ABC ) = (A ⊗ C ) vec B with the alternative deﬁnition of vec and also the deﬁnition of vectorization in tile notation would be a little less awkward;
instead of (B.5.24) one would have m
mn vec A = mn Π A
n But this is merely a side remark; we will use the conventional deﬁnition (B.5.2)
throughout. B.5.2. Kronecker Product of Matrices. Let A and B be two matrices, say
A is m × n and B is r × q . Their Kronecker product A ⊗ B is the mr × nq matrix
which in partitioned form can be written (B.5.3) a11 B
.
A⊗B = .
.
am1 B ···
..
.
··· a1n B
.
.
.
amn B This convention of how to write the elements of an array of rank 4 as a matrix is not
symmetric, so that usually A ⊗ C = C ⊗ A. Both Kronecker products represent the
same abstract array, but they arrange it diﬀerently on the page. However, in many
other respects, the Kronecker product maintains the properties of outer products. 346 B. ARRAYS OF HIGHER RANK Problem 408. [The71, pp. 303–306] Prove the following simple properties of
the Kronecker product:
(A ⊗ B ) = A ⊗ B (B.5.4) (A ⊗ B ) ⊗ C = A ⊗ (B ⊗ C ) (B.5.5) I ⊗I =I (B.5.6)
(B.5.7) (A ⊗ B )(C ⊗ D ) = AC ⊗ BD (B.5.8) (A ⊗ B )−1 = A−1 ⊗ B −1 (B.5.9) (A ⊗ B )− = A− ⊗ B − (B.5.10) A ⊗ (B + C ) = A ⊗ B + A ⊗ C (B.5.11) (A + B ) ⊗ C = A ⊗ C + B ⊗ C
(cA) ⊗ B = A ⊗ (cB ) = c(A ⊗ B ) (B.5.12) A12
A11 ⊗ B
⊗B =
A22
A21 ⊗ B A11
A21 (B.5.13) A12 ⊗ B
A22 ⊗ B rank(A ⊗ B ) = (rank A)(rank B ) (B.5.14) tr(A ⊗ B ) = (tr A)(tr B ) (B.5.15)
If a is a 1 × 1 matrix, then a ⊗ B = B ⊗ a = aB (B.5.16) det(A ⊗ B ) = (det(A))n (det(B ))k (B.5.17) where A is k × k and B is n × n.
Answer. For the determinant use the following facts: if a is an eigenvector of A with eigenvalue
α and b is an eigenvector of B with eigenvalue β , then a ⊗ b is an eigenvector of A ⊗ B with eigenvalue
αβ . The determinant is the product of all eigenvalues (multiple eigenvalues being counted several
times). Count how many there are.
An alternative approach would be to write A ⊗ B = (A ⊗ I )(I ⊗ B ) and then to argue that
det(A ⊗ I ) = (det(A))n and det(I ⊗ B ) = (det(B ))k .
The formula for the rank can be shown using rank(A) = tr(AA− ). compare Problem 366. Problem 409. 2 points [JHG+ 88, pp. 962–4] Write down the Kronecker product
of
(B.5.18) A= 1
2 3
0 and B= 2
1 2
0 0
.
3 Show that A ⊗ B = B ⊗ A. Which other facts about the outer product do not carry
over to the Kronecker product?
Answer. 2
1
A⊗B =
4
2 2
0
4
0 0
3
0
6 6
3
0
0 6
0
0
0 0
9
0
0 2
4
B⊗A=
1
2 Partitioning of the matrix on the right does not carry over. Problem 410. [JHG+ 88, p. 965] Show that
(B.5.19) vec(ABC ) = (C ⊗ A) vec(B ). 6
0
3
0 2
4
0
0 6
0
0
0 0
0
3
6 0
0
9
0 B.5. VECTORIZATION AND KRONECKER PRODUCT 347 a1
.
Answer. Assume A is k × m, B is m × n, and C is n × p. Write A = . and B =
.
ak
b1 ··· ⊗ A) vec B = bn . Then (C c11 A c12 A
= .
.
.
c 1p A c21 A
c22 A
.
.
.
c 2p A c11 a1 b1 + c21 a1 b2 + · · · + cn1 a1 bn
c11 a2 b1 + c21 a2 b2 + · · · + cn1 a2 bn . .
. c11 a b1 + c21 a b2 + · · · + cn1 a bn k
k
k c12 a1 b1 + c22 a1 b2 + · · · + cn2 a1 bn cn1 A c12 a2 b1 + c22 a2 b2 + · · · + cn2 a2 bn b1 cn2 A
. .
.
.
.
. . = . . c12 a b1 + c22 a b2 + · · · + cn2 a bn .
k
k
k bn
cnp A .
. . c1p a1 b1 + c2p a1 b2 + · · · + cnp a1 bn c a b + c a b + ··· + c a b np 2 n 1p 2 1 2p 2 2 . .
.
c1p ak b1 + c2p ak b2 + · · · + cnp ak bn ···
···
..
.
··· One obtains the same result by vectorizing the matrix a1 b1
a2 b1
ABC = .
.
.
ak b1 a1 b2
a2 b2
.
.
.
ak b2 ···
···
..
.
··· a1 bn
c11
a2 bn c21
. .
. .
.
.
ak bn
cn1 a1 b1 c11 + a1 b2 c21 + · · · + a1 bn cn1 a2 b1 c11 + a2 b2 c21 + · · · + a2 bn cn1
=
. .
.
ak b1 c11 + ak b2 c21 + · · · + ak bn cn1 c12
c22
.
.
.
cn2 ···
···
..
.
··· c 1p
c 2p . =
.
.
cnp a1 b1 c12 + a1 b2 c22 + · · · + a1 bn cn2
a2 b1 c12 + a2 b2 c22 + · · · + a2 bn cn2
.
.
.
ak b1 c12 + ak b2 c22 + · · · + ak bn cn2
···
···
..
.
··· ···
···
..
.
··· a1 b1 c1p + a1 b2 c2p + · · · + a1 bn cnp
a2 b1 c1p + a2 b2 c2p + · · · + a2 bn cnp .
. .
.
ak b1 c1p + ak b2 c2p + · · · + ak bn cnp The main challenge in this automatic proof is to ﬁt the many matrix rows, columns, and single
elements involved on the same sheet of paper. Among the shuﬄing of matrix entries, it is easy to
lose track of how the result comes about. Later, in equation (B.5.29), a compact and intelligible
proof will be given in tile notation. The dispersion of a random matrix Y is often given as the matrix V [vec Y ], where
the vectorization is usually not made explicit, i.e., this matrix is denoted V [Y ].
Problem 411. If V [vec Y ] = Σ ⊗ Ω and P and Q are matrices of constants,
show that V [vec P Y Q] = (Q Σ Q) ⊗ (P Ω P ).
Answer. Apply (B.5.19):
Now apply (B.5.7). Σ
V [vec P Y Q] = V [(Q ⊗ P ) vec Y ] = (Q ⊗ P )(Σ ⊗ Ω )(Q ⊗ P ). Problem 412. 2 points If α and γ are vectors, then show that vec(αγ ) =
γ ⊗ α.
Answer. One sees this by writing down the matrices, or one can use (B.5.19) with A = α,
B = 1, the 1 × 1 matrix, and C = γ . Problem 413. 2 points If α is a nonrandom vector and δ a random vector,
show that V [δ ⊗ α] = V [δ ] ⊗ (αα ). 348 B. ARRAYS OF HIGHER RANK Answer. α var[δ 1 ]α α cov[δ 2 , δ 1 ]α
V [δ ⊗ α] = . .
.
α cov[δ n , δ 1 ]α αδ 1
.
δ ⊗α = . .
αδ n var[δ 1 ]αα cov[δ 2 , δ 1 ]αα
=
. .
.
cov[δ n , δ 1 ]αα α cov[δ 1 , δ 2 ]α
α var[δ 2 ]α
.
.
.
α cov[δ n , δ 2 ]α
···
···
..
.
··· cov[δ 1 , δ 2 ]αα
var[δ 2 ]αα
.
.
.
cov[δ n , δ 2 ]αα ···
···
..
.
··· cov[δ 1 , δ n ]αα
cov[δ 2 , δ n ]αα
.
.
.
cov[δ n , δ n ]αα α cov[δ 1 , δ n ]α
α cov[δ 2 , δ n ]α
.
.
.
α cov[δ n , δ n ]α = = V [δ ] ⊗ αα B.5.3. The Commutation Matrix. Besides the Kronecker product and the
vectorization operator, also the “commutation matrix” [MN88, pp. 46/7], [Mag88,
p. 35] is needed for certain operations involving arrays of higher rank. Assume A
is m × n. Then the commutation matrix K (m,n) is the mn × mn matrix which
transforms vec A into vec(A ):
K (m,n) vec A = vec(A ) (B.5.20) The main property of the commutation matrix is that it allows to commute the
Kronecker product. For any m × n matrix A and r × q matrix B follows
K (r,m) (A ⊗ B )K (n,q) = B ⊗ A (B.5.21) Problem 414. Use (B.5.20) to compute K (2,3) .
Answer. K (2,3) (B.5.22) 1
0
0
=
0
0
0 0
0
0
1
0
0 0
1
0
0
0
0 0
0
0
0
1
0 0
0
1
0
0
0 0
0
0 0 0
1 B.5.4. Kronecker Product and Vectorization in Tile Notation. The
Kronecker product of m
A
n and r
B
q is the following concatenation of A and B with members of a certain family of threeway arrays Π(i,j ) :
m
mr A⊗B nq = mr A n r (B.5.23) B q Π Π nq Strictly speaking we should have written Π(m,r) and Π(n,q) for the two Πarrays in
(B.5.23), but the superscripts can be inferred from the context: the ﬁrst superscript
is the dimension of the Northeast arm, and the second that of the Southeast arm.
Vectorization uses a member of the same family Π(m,n) to convert the matrix
A
n
m into the vector
m
(B.5.24) mn vec A = mn A Π
n B.5. VECTORIZATION AND KRONECKER PRODUCT 349 This equation is a little awkward because the A is here a n × m matrix, while elsewhere it is a m × n matrix. It would have been more consistent with the lexicographical ordering used in the Kronecker product to deﬁne vectorization as the stacking
of the row vectors; then some of the formulas would have looked more natural.
m
exists for every m ≥ 1 and n ≥ 1. The
Π
The array Π(m,n) = mn
n
dimension of the West arm is always the product of the dimensions of the two East
arms. The elements of Π(m,n) will be given in (B.5.30) below; but ﬁrst I will list
three important properties of these arrays and give examples of their application.
First of all, each Π(m,n) satisﬁes
m
Π (B.5.25) m m
mn Π m = n . n n n Let us discuss the meaning of (B.5.25) in detail. The lefthand side of (B.5.25) shows
the concatenation of two copies of the threeway array Π(m,n) in a certain way that
yields a 4way array. Now look at the righthand side. The arm m
m by itself
(which was bent only in order to remove any doubt about which arm to the left of
the equal sign corresponds to which arm to the right) represents the neutral element
under concatenation (i.e., the m × m identity matrix). Writing two arrays next to
each other without joining any arms represents their outer product, i.e., the array
whose rank is the sum of the ranks of the arrays involved, and whose elements are
all possible products of elements of the ﬁrst array with elements of the second array.
The second identity satisﬁed by Π(m,n) is
m
(B.5.26) mn Π Π mn mn . mn = n
Finally, there is also associativity: (B.5.27) mnp m
n Π m
= Π Π
mnp Π p n
p Here is the answer to Problem 407 in tile notation:
tr B C = B
(B.5.28) C=B
= vec B Π Π
vec C = C=
(vec B ) vec C 350 B. ARRAYS OF HIGHER RANK Equation (B.5.25) was central for obtaining the result. The answer to Problem 410
also relies on equation (B.5.25):
C
C ⊗A Π vec B = Π Π B A
C
B Π = A
(B.5.29) vec ABC = B.5.5. Looking Inside the Kronecker Arrays. It is necessary to open up
the arrays from the Π family and look at them “element by element,” in order to
verify (B.5.23), (B.5.24), (B.5.25), (B.5.26), and (B.5.27). The elements of Π(m,n) ,
which can be written in tile notation by saturating the array with unit vectors, are
m
(B.5.30) (m,n)
πθµν =θ mn µ Π =
n ν 1
0 if θ = (µ − 1)n + ν
otherwise.
(m,n) Note that for every θ there is exactly one µ and one ν such that πθµν
other values of µ and ν ,
Writing ν (m,n)
πθµν A = 1; for all = 0. µ = aνµ and θ (B.5.31) vec A = cθ , (B.5.24) reads (m,n) cθ = πθµν aνµ ,
µ,ν which coincides with deﬁnition (B.5.2) of vec A.
One also checks that (B.5.23) is (B.5.3). Calling A ⊗ B = C , it follows from
(B.5.23) that
(B.5.32) (m,r ) cφθ = (n,q ) πφµρ aµν bρκ πθνκ .
µ,ν,ρ,κ
(m,r ) For 1 ≤ φ ≤ r one gets a nonzero πφµρ only for µ = 1 and ρ = φ, and for 1 ≤ θ ≤ q
(n,q ) one gets a nonzero πθνκ only for ν = 1 and κ = θ. Therefore cφθ = a11 bφθ for all
elements of matrix C with φ ≤ r and θ ≤ q . Etc.
The proof of (B.5.25) uses the fact that for every θ there is exactly one µ and
(m,n)
one ν such that πθµν = 0:
θ =mn
(m,n) (m,n) (B.5.33) πθµν πθωσ
θ =1 = 1
0 if µ = ω and ν = σ
otherwise Similarly, (B.5.26) and (B.5.27) can be shown by elementary but tedious proofs.
The best veriﬁcation of these rules is their implementation in a computer language,
see Section ?? below. B.5. VECTORIZATION AND KRONECKER PRODUCT 351 B.5.6. The Commutation Matrix in Tile Notation. The simplest way to
represent the commutation matrix K (m,n) in a tile is
m
K (m,n) = mn (B.5.34) Π Π mn . n
This should not be confused with the lefthand side of (B.5.26): K (m,n) is composed
of Π(m,n) on its West and Π(n,m) on its East side, while (B.5.26) contains Π(m,n)
twice. We will therefore use the following representation, mathematically equivalent
to (B.5.34), which makes it easier to see the eﬀects of K (m,n) :
m
K (m,n) = mn (B.5.35) Π Π mn . n
Problem 415. Using the deﬁnition (B.5.35) show that K (m,n) K (n,m) = I mn ,
the mn × mn identity matrix.
Answer. You will need (B.5.25) and (B.5.26). Problem 416. Prove (B.5.21) in tile notation.
Answer. Start with a tile representation of K (r,m) (A ⊗ B )K (n,q) :
r
rm Π m
rm A n r Π B q Π Π m = nq
n
nq Π Π nq q
Now use (B.5.25) twice to get
r
= rm A n m B q Π Π nq = r
= rm B q m A n Π Π nq . APPENDIX C Matrix Diﬀerentiation
C.1. First Derivatives
Let us ﬁrst consider the scalar case and then generalize from there. The derivative
of a function f is often written
dy
(C.1.1)
= f (x)
dx
Multiply through by dx to get dy = f (x) dx. In order to see the meaning of this
equation, we must know the deﬁnition dy = f (x + dx) − f (x). Therefore one obtains
f (x + dx) = f (x) + f (x) dx. If one holds x constant and only varies dx this formula
shows that in an inﬁnitesimal neighborhood of x, the function f is an aﬃne function
of dx, i.e., a linear function of dx with a constant term: f (x) is the intercept, i.e.,
the value for dx = 0, and f (x) is the slope parameter.
Now let us transfer this argument to vector functions y = f (x). Here y is a
nvector and x a mvector, i.e., f is a ntuple of functions of m variables each y1
f1 (x1 , . . . , xm )
. .
.
(C.1.2) . = .
.
yn fn (x1 , . . . , xm ) One may also say, f is a nvector, each element of which depends on x. Again, under
certain diﬀerentiability conditions, it is possible to write this function inﬁnitesimally
as an aﬃne function, i.e., one can write
(C.1.3) f (x + dx) = f (x) + Adx. Here the coeﬃcient of dx is no longer a scalar but necessarily a matrix A (whose
elements again depend on x). A is called the Jacobian matrix of f . The Jacobian
matrix generalizes the concept of a derivative to vectors. Instead of a prime denoting
the derivative, as in f (x), one writes A = Df .
Problem 417. 2 points If f is a scalar function of a vector argument x, is its
Jacobian matrix A a row vector or a column vector? Explain why this must be so.
The Jacobian A deﬁned in this way turns out to have a very simple functional
form: its elements are the partial derivatives of all components of f with respect to
all components of x:
∂fi
.
(C.1.4)
aij =
∂xj
Since in this matrix f acts as column and x as a row vector, this matrix can be
written, using matrix diﬀerentiation notation, as A(x) = ∂ f (x)/∂ x .
Strictly speaking, matrix notation can be used for matrix diﬀerentiation only if
we diﬀerentiate a column vector (or scalar) with respect to a row vector (or scalar),
or if we diﬀerentiate a scalar with respect to a matrix or a matrix with respect to a
scalar. If we want to diﬀerentiate matrices with respect to vectors or vectors with
353 354 C. MATRIX DIFFERENTIATION respect to matrices or matrices with respect to each other, we need the tile notation
for arrays. A diﬀerent, much less enlightening approach is to ﬁrst “vectorize” the
matrices involved. Both of those methods will be discussed later.
If the dependence of y on x can be expressed in terms of matrix operations
or more general array concatenations, then some useful matrix diﬀerentiation rules
exist.
The simplest matrix diﬀerentiation rule, for f (x) = w x with x1
w1
.
.
and x = . (C.1.5)
w= . .
.
xn wn
is
(C.1.6) ∂ w x/∂ x = w Here is the proof of (C.1.6):
∂w x
∂
= ∂x1 (w1 x1 + · · · + wn xn ) · · ·
∂x
= w1 · · · wn = w ∂
∂xn (w1 x1 + · · · + wn xn ) The second rule, for f (x) = x M x and M symmetric, is:
(C.1.7) ∂ x M x/∂ x = 2x M . To show (C.1.7), write
x Mx =
+ x1 m11 x1
x2 m21 x1 + .
.
.
xn mn1 x1 +
+ x1 m12 x2
x2 m22 x2 + .
.
.
xn mn2 x2 +
+ + ···
··· +
+ x1 m1n xn
x2 m2n xn ··· .
.
.
+ xn mnn xn +
+ and take the partial derivative of this sum with respect to each of the xi . For instance,
diﬀerentiation with respect to x1 gives
∂ x M x/∂x1 =
+ + 2m11 x1
x2 m21 +
+ m12 x2 + ··· + m1n xn + .
.
.
xn mn1 Now split the upper diagonal element, writing it as m11 x1 + x1 m11 , to get
=
+
+ + m11 x1
x1 m11
x2 m21 +
+
+ m12 x2 + ··· + m1n xn + .
.
.
xn mn1 The sum of the elements in the ﬁrst row is the ﬁrst element of the column vector
M x, and the sum of the elements in the column underneath is the ﬁrst element of C.1. FIRST DERIVATIVES 355 the row vector x M . Overall this has to be arranged as a row vector, since we
diﬀerentiate with respect to ∂ x , therefore we get
(C.1.8) ∂ x M x/∂ x = x (M + M ). This is true for arbitrary M , and for symmetric M , it simpliﬁes to (C.1.7). The
formula for symmetric M is all we need, since a quadratic form with an unsymmetric
M is identical to that with the symmetric (M + M )/2.
Here is the tile notation for matrix diﬀerentiation: If n
y depends on
x , then m
(C.1.9) A m =∂ A n y dx = dy , dx = ∂x dy is that array which satisﬁes
(C.1.10)
i.e.,
(C.1.11) ∂ y ∂x Extended substitutability applies here: n
y and m
x are not necessarily
vectors; the arms with dimension m and n can represent diﬀerent bundles of several
arms.
In tiles, (C.1.6) is
(C.1.12) ∂w x ∂x = w and (C.1.8) is
x
(C.1.13) ∂M x
∂x +M =M .
x x In (C.1.6) and (C.1.7), we took the derivatives of scalars with respect to vectors.
The simplest example of a derivative of a vector with respect to a vector is a linear
function. This gives the most basic matrix diﬀerentiation rule: If y = Ax is a linear
vector function, then its derivative is that same linear vector function:
(C.1.14) ∂ Ax/∂ x = A, or in tiles
(C.1.15) ∂ A x = ∂x A Problem 418. Show that
∂ tr AX
= A.
∂X (C.1.16)
In tiles it reads
m
(C.1.17) X ∂A ∂X = A. n
Answer. tr(AX ) = i,j aij xji i.e., the coeﬃcient of xji is aij . 356 C. MATRIX DIFFERENTIATION Here is a diﬀerentiation rule for a matrix with respect to a matrix, ﬁrst written
element by element, and then in tiles: If Y = AXB , i.e., yim = j,k aij xjk bkm ,
then ∂yim = aij akm , because for every ﬁxed i and m this sum contains only one term
∂xjk
which has xjk in it, namely, aij xjk bkm . In tiles:
A
A
(C.1.18) ∂X ∂X =
B B
Equations (C.1.17) and (C.1.18) can be obtained from (C.1.12) and (C.1.15) by
extended substitution, since a bundle of several arms can always be considered as
one arm. For instance, (C.1.17) can be written
∂A X ∂X = A and this is a special case of (C.1.12), since the two parallel arms can be treated as
one arm. With a better development of the logic underlying this notation, it will not
be necessary to formulate them as separate theorems; all matrix diﬀerentiation rules
given so far are trivial applications of (C.1.15).
Problem 419. As a special case of (C.1.18) show that ∂x Ay
∂A = yx . Answer.
x
x
(C.1.19) ∂A ∂A =
y y Here is a basic diﬀerentiation rule for bilinear array concatenations: if
x
y= (C.1.20) A
x then one gets the following simple generalization of (C.1.13):
x x
(C.1.21) ∂ ∂x A
x = A + A
x Proof. yi = j,k aijk xj xk . For a given i, this has x2 in the term aipp x2 , and
p
p
it has xp in the terms aipk xp xk where p = k , and in aijp xj xp where j = p. The
derivatives of these terms are 2aipp xp + k=p aipk xk + j =p aijp xj , which simpliﬁes
to k aipk xk + j aijp xj . This is the i, pelement of the matrix on the rhs of (C.1.21). C.1. FIRST DERIVATIVES 357 But there are also other ways to have the array X occur twice in a concatenation
Y . If Y = X X then yik = j xji xjk and therefore ∂yik /∂xlm = 0 if m = i and
m = k . Now assume m = i = k : ∂yik /∂xli = ∂xli xlk /∂xli = xlk . Now assume
m = k = i: ∂yik /∂xlk = ∂xli xlk /∂xlk = xli . And if m = k = i then one gets the
sum of the two above: ∂yii /∂xli = ∂x2 /∂xli = 2xli . In tiles this is
li
i
l
(C.1.22) ∂X X
∂X X
= X
∂X ∂ = X
+ . X
m
k This rule is helpful for diﬀerentiating the multivariate Normal likelihood function.
A computer implementation of this tile notation should contain algorithms to
automatically take the derivatives of these array concatenations.
Here are some more matrix diﬀerentiation rules:
Chain rule: If g = g (η ) and η = η (β ) are two vector functions, then
(C.1.23) ∂ g /∂ β = ∂ g /∂ η · ∂ η /∂ β For instance, the linear least squares objective function is SSE = (y − Xβ ) (y −
Xβ ) = ε ε where ε = y − Xβ . Application of the chain rule gives ∂ SSE /∂ β =
ˆˆ
ˆ
ˆ
ˆ
ˆ
∂ SSE /∂ ε · ∂ ε/∂ β = 2ε (−X ) which is the same result as in (14.2.2).
If A is nonsingular then
∂ log det A
= A−1
(C.1.24)
∂A
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