Contents · 24 2. ESTIMATOR QUALITY because we know that s2 is unbiased, so E [s2] 2. bias 2 ( 2) (n 1) 2 n 2 2 n The bias goes to 0 as n ! 1 , so the empirical variance is an asymptotically

Contents

I Probability Review 1

1 Probability Review 31.1 Functions and moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Probability distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Bernoulli distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2.2 Uniform distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2.3 Exponential distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Normal approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.5 Conditional probability and expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.6 Conditional variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

II Parameter Estimation 21

2 Estimator Quality 232.1 Bias . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2 Consistency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.3 E�ciency and Mean Square Error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25


3 Maximum Likelihood 413.1 Likelihood . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.2 Maximum Likelihood Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41


4 Variance of Maximum Likelihood Estimator 69Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

5 Su�cient Statistics 75Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

III Hypothesis Testing 83

6 Hypothesis Testing 856.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 856.2 Typical exam questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

CAS ST Study Manual 2nd edition 3rd printingCopyright ©2015 ASM

iii

iv CONTENTS

6.2.1 Calculate signi�cance or power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 886.2.2 Determine critical values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

7 Con�dence Intervals and Sample Size 1097.1 Con�dence intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1097.2 Sample size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110


8 Con�dence Intervals for Means 1178.1 χ2 distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1178.2 Student’s t distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1188.3 Testing the mean of a Bernoulli population . . . . . . . . . . . . . . . . . . . . . . . . . . . 1198.4 Testing the di�erence of means from two populations . . . . . . . . . . . . . . . . . . . . . 120

8.4.1 Two unpaired normal populations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1208.4.2 Two paired normal populations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1228.4.3 Two Bernoulli populations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

9 Chi Square Tests 1379.1 One-dimensional chi-square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1379.2 Two-dimensional chi-square . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140


10 Con�dence Intervals for Variances 15110.1 Testing variances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15110.2 Testing ratios of variances; the F distribution . . . . . . . . . . . . . . . . . . . . . . . . . . 152


11 Linear Regression 163Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

12 Linear Regression: Measures of Fit 17312.1 Standard error of the regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17312.2 R2: the coe�cient of determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17412.3 t statistic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17512.4 F statistic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17712.5 Multiple regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17812.6 Comparison of models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179


13 ANOVA 201Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213


CONTENTS v

14 Uniformly Most Powerful Critical Regions 219Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

15 Likelihood Ratio Tests 227Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

IV Bayesian Estimation 239

16 Bayesian Estimation 24116.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24116.2 Loss functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24316.3 Interval estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244


17 Beta-Bernoulli Conjugate Prior Pair 249Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

18 Normal-Normal Conjugate Prior Pair 255Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

19 Gamma-Poisson Conjugate Prior Pair 259Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

V Nonparametric Methods for Hypothesis Testing 271

20 Order Statistics 273Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280

21 Sign Test 287Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291

22 Wilcoxon Tests 29522.1 Signed rank test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29522.2 Rank sum test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298


23 The Runs Test 309Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315


vi CONTENTS

24 Rank Correlation Coe�cients 317Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323

VI Poisson Processes 327

25 The Poisson Process: Probabilities of Events 32925.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32925.2 Probabilities—Homogeneous Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33025.3 Probabilities—Non-Homogeneous Process . . . . . . . . . . . . . . . . . . . . . . . . . . . 332


26 The Poisson Process: Time To Next Event 345Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351

27 The Poisson Process: Thinning 35527.1 Constant Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35527.2 Non-Constant Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357


28 The Poisson Process: Sums and Mixtures 37128.1 Sums of Poisson Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37128.2 Mixtures of Poisson Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372


29 Compound Poisson Processes 38529.1 De�nition and Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38529.2 Sums of Compound Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387


VII Practice Exams 405

1 Practice Exam 1 407







CONTENTS vii

Appendices 455

A Solutions to the Practice Exams 457Solutions for Practice Exam 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457Solutions for Practice Exam 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463Solutions for Practice Exam 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470Solutions for Practice Exam 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478Solutions for Practice Exam 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485Solutions for Practice Exam 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 492

B Solutions to Statistics and Stochastic Process Questions on Old CAS 3 and 3L Exams 501B.1 Solutions to CAS Exam 3, Spring 2005 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 501B.2 Solutions to CAS Exam 3, Fall 2005 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503B.3 Solutions to CAS Exam 3, Spring 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505B.4 Solutions to CAS Exam 3, Fall 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509B.5 Solutions to CAS Exam 3, Spring 2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513B.6 Solutions to CAS Exam 3, Fall 2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516B.7 Solutions to CAS Exam 3L, Spring 2008 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 518B.8 Solutions to CAS Exam 3L, Fall 2008 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521B.9 Solutions to CAS Exam 3L, Spring 2009 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524B.10 Solutions to CAS Exam 3L, Fall 2009 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526B.11 Solutions to CAS Exam 3L, Spring 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529B.12 Solutions to CAS Exam 3L, Fall 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 532B.13 Solutions to CAS Exam 3L, Spring 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535B.14 Solutions to CAS Exam 3L, Fall 2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538B.15 Solutions to CAS Exam 3L, Spring 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 540B.16 Solutions to CAS Exam 3L, Fall 2012 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543B.17 Solutions to CAS Exam 3L, Spring 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545B.18 Solutions to CAS Exam 3L, Fall 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548B.19 Solutions to CAS Exam ST, Spring 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 551B.20 Solutions to CAS Exam ST, Fall 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558B.21 Solutions to CAS Exam ST, Spring 2015 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565

C Lessons Corresponding to Questions on Released and Practice Exams 571


viii CONTENTS


Lesson 2

Estimator Quality

We are about to learn how to estimate parameters of a distribution. In other words, we may believe thatthe random phenomenon we are analyzing follows a speci�c probability distribution, like exponentialor Pareto, but need to estimate the parameters of the distribution, θ for an exponential or θ and α for aPareto. Before we discuss estimation methods, let’s consider the following question: How do we measurethe quality of an estimator? There are several ways to measure the quality of an estimator.

In the following discussion, θ is a parameter to be estimated, θ is an estimator, and θn is an estimatorbased on n observations.

2.1 Bias

A desirable property of an estimator is that its expected value, based on the assumed underlying distri-bution, equals the parameter we’re estimating. In other words, E[θ] � θ. We de�ne bias, biasθ (θ) as

biasθ (θ) � E[θ] − θ (2.1)

If biasθ (θ) � 0, then we say that θ is an unbiased estimator of θ. If limn→∞ E[θn] � θ, then we say that θis asymptotically unbiased.

The sample mean is an unbiased estimator of the true mean. We can easily see this. The sample meanis de�ned by

x �

∑ni�1 xi

n

However, E[xi] � µ by de�nition of expected value. So

E[x] �∑n

i�1 E[xi]n

�nµn

� µ

proving that x is an unbiased estimator of µ.The sample variance, de�ned by

s2 �∑ (xi − x)2

n − 1is an unbiased estimator of the true variance, σ2. The empirical variance, de�ned by

σ2 �∑ (xi − x)2

n

is a biased estimator of the true variance, σ2. Its bias can be calculated as follows:

σ2 �n − 1

ns2

E[σ2] � n − 1n

E[s2] � (n − 1)σ2

n


23

24 2. ESTIMATOR QUALITY

because we know that s2 is unbiased, so E[s2] � σ2.

biasσ2 (σ2) �(n − 1)σ2

n− σ2 � −σ

2

nThe bias goes to 0 as n →∞, so the empirical variance is an asymptotically unbiased estimator of the truevariance.Example 2A You are given a sample x1 , x2 , . . . , xn . Which of the following estimators are unbiased?

1. For an exponential distribution, x as an estimator for θ.

2. For a Pareto distribution with known θ, 1 + θx as an estimator for α.

3. For a uniform distribution on [0, θ], max xi as an estimator for θ.

Answer: 1. The sample mean is an unbiased estimator of the true mean, and θ is the mean of anexponential, so x is an unbiased estimator of θ.

2. The mean of a Pareto is µ �θα−1 . If θ is known, then we see from this that α � 1 + θ

µ . The expectedvalue of the proposed estimator is 1 + θ E[1/x]. In general, the expected value of a reciprocal is not thereciprocal of the expected value:

E[ 1

x

],

1E[x]

So this estimator of α is biased.3. We will discuss the distribution of the maximum of a sample in lesson 20, but let’s calculate it here.

Let Y be the maximum of the sample from a uniform distribution on [0, θ]. The probability that themaximum is less than x is the probability that all the observations are less than x, or

FY (x) �(xθ

)n

0 ≤ x ≤ θ

Di�erentiating,

fY (x) �nxn−1

θn 0 ≤ x ≤ θThe expected value of Y is

∫ θ

0x fY (x)dx �

∫ θ

0

nxndxθn

�

(nθn

) (1

n + 1

)xn+1

��θ

0

�nθ

n + 1So the bias of the maximum is nθ

n+1 − θ � − θn+1 .

However, as n →∞, the bias goes to 0. Therefore, the estimator is asymptotically unbiased. �

?Quiz 2-1 X is an observation from a uniform distribution on [0, θ]. 2X is an unbiased estimator of θ.

Calculate the bias of (2X)2 as an estimator for θ2.

While unbiasedness is desirable, it is not the onlymeasure of estimator quality. Many biased estimatorsare satisfactory as long as they’re asymptotically unbiased. Conversely, an unbiased estimator is good onthe average, but may be a poor estimator. It’s like the statistician with his head in the freezer and his feetin boiling water; he may be OK on the average, but quite uncomfortable. Good estimators are always closeto the correct value, and it’s not enough that they are correct on the average.


2.2. CONSISTENCY 25

2.2 Consistency

An estimator is (weakly) consistent if the probability that it is di�erent from the parameter by more thanε goes to 0 as n, the sample size, goes to in�nity. In other words

limn→∞Pr

(|θn − θ | > ε

)→ 0 for any ε > 0

A su�cient but not necessary condition for consistency is that the estimator is asymptotically unbiasedand its variance goes to 0 as n goes to in�nity. Thus, the sample mean is a consistent estimator of the truemean for an exponential or a gamma distribution, but may not be consistent for a Pareto distribution withα ≤ 2, since then the variance of the distribution and therefore of the sample mean is in�nite.

2.3 Efficiency and Mean Square Error

An estimator is e�cient if its variance is low. An estimator is more e�cient than another estimator if itsvariance is lower than the other estimator. For two estimators θ1 and θ2, the relative e�ciency of θ1 withrespect to θ2 is

Relative e�ciency of θ1 to θ2 �Var(θ2)Var(θ1)

(2.2)

The mean square error of an estimator is the expected value of the square di�erence between the esti-mator and the parameter:1

MSEθ (θ) � E[(θ − θ)2

](2.3)

The MSE is the sum of the bias squared and the variance:

MSEθ (θ) � biasθ (θ)2 + Var(θ) (2.4)

This is a convenient formula. It follows that if the estimator is unbiased, then the MSE is the variance.An estimator is called a uniformly minimum variance unbiased estimator (UMVUE) if it is unbiased and if

there is no other unbiased estimator with a smaller variance for any true value θ. It would make no senseto make a similar de�nition for biased estimators (i.e., a uniformly minimum MSE estimator), since theestimator equal to the constant happens to have an MSE of 0 if θ is that constant.Example 2B In an urn, there are four marbles numbered 5, 6, 7, and 8. You draw three marbles from theurn without replacement. Let θ be the maximum of the three marbles.

Calculate the bias and the mean square error of θ as an estimator for the maximummarble in the urn,θ.

Answer: There are four combinations of three marbles out of four. Three of the combinations have 8. Theremaining one is {5, 6, 7}, with a maximum of 7. Thus the expected value of θ is 3

4 (8) + 14 (7) � 7 3

4 , whereasthe true maximum is 8. The bias is 7 3

4 − 8 � − 14 .

The error is 1 one-fourth of the time, 0 otherwise, so the mean square error is 14 (12) � 1

4 .The variance of the estimator is

(0.25)(0.75)(12) �316 ,

and indeed(− 1

4

)2+ 3

16 �14—the bias squared plus the variance equals the mean square error. �

1Di�erent textbooks have di�erent conventions on the argument and subscript of MSE; some do the opposite of the formulahere and write MSEθ (θ) instead. The notation used here is the one you’ll encounter on Exam 4. Some old exam questions use theother notation.



Example 2C [4B-F96:21] (2 points) You are given the following:

• The expectation of a given estimator is 0.50.

• The variance of this estimator is 1.00.

• The bias of this estimator is 0.50.

Determine the mean square error of this estimator.

A. 0.75 B. 1.00 C. 1.25 D. 1.50 E. 1.75

Answer: MSEθ (θ) � 1.00 + 0.502 � 1.25 . (C) �

Example 2D For a uniform distribution on [0, θ], calculate the mean square error of Y � max xi as anestimator of θ.

Answer: We already calculated the bias in Example 2A. We showed that the density function of Y is

fY (x) �nxn−1

θn 0 ≤ x ≤ θ

Now let’s calculate the variance of the estimator. The second moment of Y is

E[Y2] �∫ θ

0

nxn+1dxθn

�

(nθn

) (1

n + 2

)xn+2

��θ

0�

nθ2

n + 2

The variance is

Var(Y) � E[X2] − E[X]2

�nθ2

n + 2 −(

nθn + 1

)2

�nθ2

(n + 2)(n + 1)2

So the MSE is

MSEθ (θ) � biasθ (θ)2 + Var(θ)

�θ2

(n + 1)2+ nθ2

(n + 2)(n + 1)2

�2θ2

(n + 1)(n + 2)�


2.3. EFFICIENCY ANDMEAN SQUARE ERROR 27

Table 2.1: Summary of Estimator Quality Concepts

• The bias of an estimator is the excess of its expected value over the true value:

biasθ (θ) � E[θ] − θ (2.1)

• An estimator is asymptotically unbiased, even if it isn’t unbiased, if the bias goes to 0 as the samplesize goes to in�nity.

• The sample mean is an unbiased estimator of the population mean. The sample variance (withdivision by n − 1) is an unbiased estimator of the population variance.

• An estimator is consistent if the probability that it di�ers from the true value by any amount goesto 0 as the sample size goes to in�nity, or

limn→∞Pr(|θ − θ | > ε) � 0 for ε > 0

• If an estimator is asymptotically unbiased and its variance goes to 0 as the sample size goes toin�nity, then it is consistent, but not conversely.

• The sample mean is a consistent estimator of the population mean if the population variance is�nite.

• An estimator is more e�cient than another estimator if its variance is lower.

• The relative e�ciency of θ1 with respect to θ2 is

Var(θ2)Var(θ1)

(2.2)

• The mean square error of an estimator is

MSEθ (θ) � E[(θ − θ)2] (2.3)

• A formula for the mean square error is

MSEθ (θ) � biasθ (θ)2 + Var(θ) (2.4)

• A uniformly minimum variance unbiased estimator is an unbiased estimator that has the lowest vari-ance of any unbiased estimator regardless of the true value of θ, the estimated parameter.



Exercises

2.1. [110-S83:15] Let T1 and T2 be estimators of a population parameter θ based upon the same randomsample. If Ti is distributed normally with mean θ and variance σ2i > 0, i � 1, 2, and if T � bT1 + (1− b)T2,then T is an unbiased estimator of θ.

Determine b to minimize the variance of T.A. σ2

σ1

B.σ22σ21

C.σ22

σ21 + σ22

D.σ22 − Cov(T1 , T2)

σ21 − 2Cov(T1 , T2) + σ22

E.σ22 − 1

2 Cov(T1 , T2)

σ21 − 2Cov(T1 , T2) + σ22

2.2. [110-S83:20] Let X be a random variable with mean 2. Let S and T be unbiased estimators of thesecond and third moments, respectively, of X about the origin.

Which of the following is an unbiased estimator of the third moment of X about its mean?A. T − 6S + 16B. T − 3S + 2C. (T − 2)3 − 3(S − 2)2

D. (T − 2)3

E. T − 8

2.3. [110-S88:36] Let X be a random variable with a binomial distribution with parameters m and q,and let q � X/m. Then q is an unbiased estimator of q.

Which of the following is an unbiased estimator of q(1 − q)?

A. q(1 − q) B.( 1

m−1)q(1 − q) C.

( 1m)q(1 − q) D.

(m−1m

)q(1 − q) E.

( mm−1

)q(1 − q)

2.4. [110-S83:33] Let X1, X2, . . . , Xn be a random sample of size n ≥ 2 from a Poisson distribution withmean λ. Consider the following three statistics as estimators of λ.

I. X �1n∑n

i�1 Xi

II. 1n−1

∑ni�1(Xi − X)2

III. 2X1 − X2

Which of these statistics are unbiased?

A. I only B. II only C. III only D. I, II, and IIIE. The correct answer is not given by A. , B. , C. , or D.


Exercises continue on the next page . . .

EXERCISES FOR LESSON 2 29

2.5. Which of the following statements are true?

I. An estimator that is asymptotically unbiased and whose variance approaches 0 as the sample sizegoes to in�nity is weakly consistent.

II. For an unbiased estimator, minimizing variance is equivalent to minimizing mean square error.III. The estimator S2 � 1

n∑n

j�1(X j − X)2 for the variance σ2 is asymptotically unbiased.

2.6. [4B-S96:12] (1 point) Which of the following must be true of a consistent estimator?

1. It is unbiased.

2. For a small quantity ε, the probability that the absolute value of the deviation of the estimator fromthe true parameter value is less than ε tends to 1 as the number of observations tends to in�nity.

3. It has minimal variance.

A. 1 B. 2 C. 3 D. 2,3 E. 1,2,3

2.7. Which of the following statements is false?A. If two estimators are unbiased, a weighted average of them is unbiased.B. The sample mean is an unbiased estimator of the population mean.C. The sample mean is a consistent estimator of the population mean.D. For a uniform distribution on [0, θ], the sample maximum is a consistent estimator of the popu-

lation maximum.E. The mean square error of an estimator cannot be less than the estimator’s variance.

2.8. θ is an estimator for θ. You are given:

• E[θ] � 3• E[θ2] � 13

If θ � 4, what is the mean square error of θ?

2.9. [4B-S92:2] (1 point) Which of the following are true?

1. The expected value of an unbiased estimator of a parameter is equal to the true value of the parameter.

2. If an estimator is e�cient, the probability that an estimate based on n observations di�ers from thetrue parameter by more than some �xed amount converges to zero as n grows large.

3. A consistent estimator is one with a minimal variance.

A. 1 only B. 3 only C. 1 and 2 only D. 1,2 and 3E. The correct answer is not given by A. , B. , C. , or D.




2.10. [4B-S91:28] (1 point) α is an estimator of α. Match each of these properties with the correct math-ematical description.

a. Consistent 1. E[α] � αb. Unbiased 2. Var[α] ≤ Var[α] where α is any other estimator of αc. E�cient 3. For any ε > 0, Pr(|α − α | < ε) → 1 as n → ∞, where n is the

sample size.

A. a � 1, b � 2, c � 3B. a � 2, b � 1, c � 3C. a � 1, b � 3, c � 2D. a � 3, b � 2, c � 1E. a � 3, b � 1, c � 2

2.11. [4-F04:40] Which of the following statements is true?A. A uniformly minimum variance unbiased estimator is an estimator such that no other estimator

has a smaller variance.B. An estimator is consistent whenever the variance of the estimator approaches zero as the sample

size increases to in�nity.C. A consistent estimator is also unbiased.D. For an unbiased estimator, the mean squared error is always equal to the variance.E. One computational advantage of using mean squared error is that it is not a function of the true

value of the parameter.

2.12. You are given a sample of 25 items from an exponential distribution. You consider the followingtwo estimators for the mean:

1. θ1 � x

2. θ2 � 0.9x

Calculate the relative e�ciency of θ2 with respect to θ1.

2.13. The mean of a uniform distribution on [0, θ] is estimated with the sample mean based on a samplewith 10 observations.

The bias of x2 as an estimator for the square of the mean of the distribution is cθ2.Determine c.

2.14. A population contains the values 1, 2, 4, 9. A sample of 3 without replacement is drawn from thisvariable. Let Y be the median of this sample.

Calculate the mean square error of Y as an estimator of the population mean.




2.15. [4B-F92:8] (1 point) You are given the following information:X is a random variable whose distribution function has parameter α � 2.00.Based on n random observations of X you have determined:

• E[α1] � 2.05, where α1 is an estimator of α having variance equal to 1.025.• E[α2] � 2.05, where α2 is an estimator of α having variance equal to 1.050.• As n increases to∞, Pr(|α1 − α | > ε) approaches 0 for any ε > 0.

Which of the following are true?

1. α1 is an unbiased estimator of α.2. α2 is an e�cient estimator of α.3. α1 is a consistent estimator of α.

A. 1 only B. 2 only C. 3 only D. 1,3 only E. 2,3 only

2.16. [4B-F93:13] (3 points) You are given the following:

• Two instruments are available for measuring a particular (non-zero) distance.• X is the random variable representing the measurement using the �rst instrument and Y is the ran-

dom variable representing the measurement using the second instrument.• X and Y are independent.• E[X] � 0.8m; E[Y] � m; Var(X) � m2; and Var(Y) � 1.5m2 where m is the true distance.

Consider the class of estimators of m which are of the form Z � αX + βY.Within this class of estimators of m, determine the value of α that makes Z an unbiased estimator with

minimum variance.A. Less than 0.45B. At least 0.45, but less than 0.50C. At least 0.50, but less than 0.55D. At least 0.55, but less than 0.60E. At least 0.60




2.17. [4B-S95:27] (2 points) Two di�erent estimators, ψ and φ, are available for estimating the parameter,β, of a given loss distribution.

To test their performance, you have conducted 75 simulated trials of each estimator, using β � 2, withthe following results:

75∑

i�1ψi � 165,

75∑

i�1ψ2

i � 375,75∑

i�1φi � 147,

75∑

i�1φ2

i � 312.

Calculate MSEβ (ψ)/MSEβ (φ).A. Less than 0.50B. At least 0.50, but less than 0.65C. At least 0.65, but less than 0.80D. At least 0.80, but less than 0.95E. At least 0.95, but less than 1.00

2.18. [4B-S92:17] (2 points) You are given that the underlying size of loss distribution for disability claimsis a Pareto distribution with parameters α and θ � 6000.

You have determined the following for α, an estimator of α:E[α] � 2.20MSE(α) � 1.00

Determine the variance of α if α � 2.A. Less than 0.70B. At least 0.70, but less than 0.85C. At least 0.85, but less than 1.00D. At least 1.00, but less than 1.15E. At least 1.15

2.19. Losses follow a Pareto distribution with parameters α � 3, θ � 600. A sample of 100 is available.Determine the MSE of the sample mean as an estimator for the mean.

2.20. A sample of n elements, x1 , . . . , xn , is selected from a random variable having a uniform distri-bution on [0, θ]. Let Y � max(xi ). You wish to estimate the parameter θ with an estimator of the formkY.

You may use the following facts:

• E[Y] � nθn + 1.

• Var(Y) �nθ2

(n + 2)(n + 1)2.

Determine the k which minimizes the mean square error of the estimator.




2.21. [4-S00:18] You are given two independent estimates of an unknown quantity µ:

• Estimate A: E[µA] � 1000 and σ(µA) � 400.• Estimate B: E[µB] � 1200 and σ(µB) � 200.

Estimate C is a weighted average of the two estimates A and B, such that

µC � w · µA + (1 − w) · µB

Determine the value of w that minimizes σ(µC).

A. 0 B. 1/5 C. 1/4 D. 1/3 E. 1/2

2.22. [4-F02:31] You are given:

x 0 1 2 3Pr(X � x) 0.5 0.3 0.1 0.1

Using a sample of size n, the population mean is estimated by the sample mean X and the variance isestimated by S2

n �∑

(Xi − X)2/n.Calculate the bias of S2

n when n � 4.

A. −0.72 B. −0.49 C. −0.24 D. −0.08 E. 0.00

2.23. [4-S05:16] For the random variable X, you are given:

• E[X] � θ, θ > 0• Var(X) � θ2/25• θ �

(k/(k + 1)

)X, k > 0

• MSEθ (θ) � 2(biasθ (θ)

)2

Determine k.

A. 0.2 B. 0.5 C. 2 D. 5 E. 25

2.24. [CAS3-S05:21] An actuary obtains two independent, unbiased estimates, Y1 and Y2, for a certainparameter. The variance of Y1 is four times that of Y2.

A new unbiased estimator of the form k1Y1 + k2Y2 is to be constructed.What value of k1 minimizes the variance of the new estimate?A. Less than 0.18B. At least 0.18, but less than 0.23C. At least 0.23, but less than 0.28D. At least 0.28, but less than 0.33E. At least 0.33




2.25. [CAS3-F05:6] Claim sizes are uniformly distributed over the interval [0, θ]. A sample of 10 claims,denoted X1, X2, X3, . . . , X10, was observed and an estimate of θ was obtained as follows:

θ � Y � max(X1 ,X2 , . . . ,X10)

Recall that the probability density function for Y is:

fY (y) �10y9

θ10 for 0 ≤ y ≤ θ

Calculate the mean square error of θ for θ � 100.A. Less than 75B. At least 75, but less than 100C. At least 100, but less than 125D. At least 125, but less than 150E. At least 150

Additional old CAS Exam 3/3L questions: S06:3 (bias),4 (bias, consistent, su�cient), S08:2 (bias,consistent, MSE), F08:5 (MSE), S10:20 (bias), F11:17 (bias, MSE), F12:19 (MSE), S13:19 (bias), F13:17(consistent)Additional old CAS Exam ST questions: S14:5 (MSE), S15:4

Solutions

2.1. All you need for this exercise is the formula for the variance of a sum:

Var(aX + bY) � a2 Var(X) + 2ab Cov(X,Y) + b2 Var(Y)

So we’re minimizingVar(T) � b2σ21 + 2b(1 − b) Cov(T1 , T2) + (1 − b)2σ22

Di�erentiate with respect to b and set equal to zero.

2bσ21 + (2 − 4b) Cov(T1 , T2) − 2(1 − b)σ22 � 0b(2σ21 − 4Cov(T1 , T2) + 2σ22) + 2Cov(T1 , T2) − 2σ22 � 0

b �

σ22− Cov(T1 , T2)

σ21− 2Cov(T1 , T2) + σ2

2

(D)

2.2. The third central moment µ3 can be expressed in terms of the moments around the origin µ′n asfollows:

µ3 � µ′3 − 3µ′2µ + 2µ3

Since µ � 2, this reduces toµ3 � µ

′3 − 6µ′2 + 16

If the expected value of T is µ′3 and the expected value of S is µ′2, then the expected value of T − 6S + 16is µ3, so the answer is (A).


EXERCISE SOLUTIONS FOR LESSON 2 35

2.3. All �ve choices are multiples of q(1 − q), so let’s determine the expected value of that.

E[q] � q

E[q2] � E[X2]m2

E[X2] � E[X]2 + Var(X) � m2q2 + mq(1 − q) by the formula for moments of a binomial

E[q2] �m2q2 + mq(1 − q)

m2 �mq2 + q(1 − q)

mE[q(1 − q)] � E[q] − E[q2]

� q − mq2 + q(1 − q)m

�mq(1 − q) − q(1 − q)

m

�m − 1

mq(1 − q)

and therefore the estimator must be multiplied by mm−1 to make it unbiased. (E)

2.4. I is the sample mean, which is an unbiased estimator of the true mean λ.!II is the unbiased sample variance, which is an unbiased estimator of the true variance λ.!For III, E[2X1 − X2] � 2λ − λ � λ, making it an unbiased estimator.! (D)

2.5.I. As discussed in the lesson, true.!

II. MSEθ (θ) � Var(θ) +(biasθ (θ)

)2and biasθ (θ) � 0, so it is true.!

III. As discussed in the lesson, true.!

2.6. (B)2.7. These are all discussed in this lesson. (C) is false if the variance of the population isn’t �nite.2.8.

biasθ (θ) � 3 − 4 � −1Var(θ) � 13 − 32 � 4

MSEθ (θ) � 4 + (−1)2 � 5

2.9. Only 1 is true. The other two statements have interchangedde�nitions of consistency and e�ciency.(A)2.10. a � 3, b � 1, c � 2. (E)2.11. A correct version of (A) is “A uniformly minimum variance estimator is an estimator such than noother unbiased estimator has a smaller variance.”

An estimator which is a constant has no variance, but if it is not equal to the true parameter must beinconsistent, so (B) is false.

Consistency is an asymptotic property, so a biased estimator which is asymptotically unbiased couldbe consistent, making (C) false.

(D) is true.Mean square error is a function of the true value of the parameter; in fact, it is the expected value of

the square of the di�erence between the estimator and the true parameter, so (E) is false. Note however,that the variance of an estimator is not a function of the true parameter.



2.12. Let V be the variance of the sample mean. Then Var(θ1) � V , Var(θ2) � 0.92V � 0.81V . Therelative e�ciency of θ2 to θ1 is V/(0.81V) � 1.2346 . But note that θ1 is unbiased while θ2 is biased.2.13. The square of the mean is θ2/4. Now let’s calculate the expected value of x2.

E[x2] � E(∑10

i�1 Xi

10

)2�

1100

*,E[ 10∑

i�1X2

i

]+ E

[ ∑

1≤i≤101≤ j≤10

i, j

XiX j

]+-For a uniform on [0, θ], E[X2] � θ2/3. Also, E[XiX j] � E[Xi]E[X j] because observations in a randomsample are independent, so E[XiX j] � θ2/4, the square of E[Xi]. Therefore,

E[x2] � 1100

*.,10(θ2

3

)+ 90

(θ2

4

)+/-� θ2

(3101200

)

The bias is 31120θ

2 − 14θ

2 � θ2

120 , and c � 1/120 .2.14. Half the time the samplemedian is 2 and the other half the time it is 4. Themean is (1+2+4+9)/4 � 4.So the MSE is 1

2 (2 − 4)2 � 2 .2.15. Only 3 is true. α2 has higher variance than α1 and the same bias, so it is less e�cient. (C)2.16.

E[αX + βY] � m0.8α + β � 1

Minimize g(α) � Var(αX + βY) � α2 Var(X) + β2 Var(Y)

� α2m2 + (1 − 0.8α)2(1.5m2)

org(α)m2 � α2 + 1.5 − 2.4α + 0.96α2

� 1.96α2 − 2.4α + 1.5

g(α) is minimized at α � 2.4/3.92 � 0.6122 . (E)2.17. We must estimate the variance of each estimator. The question is vague on whether to use theempirical variance (divide by 75) or the sample variance (divide by 74). The original exam question saidto work out the answer according to a speci�c textbook that used the empirical variance. We then get:

MSEβ (ψ) �(16575 − 2

)2+ *.,

37575 −

(16575

)2+/- � 0.04 + 0.16 � 0.2

MSEβ (φ) �(14775 − 2

)2+ *.,

31275 −

(14775

)2+/- � 0.0016 + 0.3184 � 0.32


EXERCISE SOLUTIONS FOR LESSON 2 37

0.20.32 � 0.625 (B)

If the sample variance were used, we would multiply 0.16 and 0.3184 by 75/74 to get 0.1622 and 0.3227.The resulting quotient, (0.04 + 0.1622)/(0.0016 + 0.3227) � 0.6234, which still leads to answer B.2.18.

biasα (α) � 2.20 − 2 � 0.2

Var(α) � MSE(α) −(biasα (α)

)2� 1 − 0.22 � 0.96 (C)

2.19. The estimator is unbiased because the sample mean is an unbiased estimator of the populationmean, so the mean square error equals the variance. The variance of the estimator is:

Var(X) �Var(X)100 �

2(600)22·1 − (600

2)2

100 � 2700 .

2.20. The bias of kY isk

nθn + 1 − θ � θ

(n(k − 1) − 1

n + 1

).

The variance of kY isk2nθ2

(n + 2)(n + 1)2.

The MSE is thenk2nθ2

(n + 2)(n + 1)2+ θ2 *..,

(n(k − 1) − 1

)2

(n + 1)2+//-.

We shall minimize this by di�erentiating with respect to k. To simplify matters, divide the entire expres-sion by θ2 and multiply it by (n + 1)2; this has no e�ect on the minimizing k:

f (k) �k2n

n + 2 +(n(k − 1) − 1

)2

f ′(k) �2knn + 2 + 2n

(n(k − 1) − 1

)� 0

kn + 2 + n(k − 1) − 1 � 0

k( 1

n + 2 + n)� n + 1

k(n(n + 2) + 1

)� (n + 1)(n + 2)

k(n + 1)2 � (n + 1)(n + 2)

k �n + 2n + 1



2.21. The variance of the weighted average is

σ2C � w2σ2 + (1 − w)2σ2B� 160,000w2 + 40,000(1 − w)2

Di�erentiating,

2(160,000)w − 2(40,000)(1 − w) � 0200,000w � 40,000

w � 1/5 (B)

2.22. We know that S2 �∑

(Xi − X)2/(n − 1) is an unbiased estimator; in other words, E[S2] � σ2. Butthen

E[S2n] �

n − 1n

E[S2] � n − 1n

σ2

and the bias isE[S2

n] − σ2 �( n − 1

n− 1

)σ2 � −σ

2

nIn this case, the true mean µ � 0.5(0) + 0.3(1) + 0.1(2) + 0.1(3) � 0.8 and the true variance is

σ2 � 0.5(0 − 0.8)2 + 0.3(1 − 0.8)2 + 0.1(2 − 0.8)2 + 0.1(3 − 0.8)2 � 0.96

So the bias is −0.96/4 � −0.24 . (C)2.23. Since

MSEθ (θ) �(biasθ (θ)

)2+ Var(θ)

by (iv) (biasθ (θ)

)2� Var(θ)

so we calculate biasθ (θ) and Var(θ).

biasθ (θ) � E[θ] − θ

� E[

kk + 1X

]− θ

�

(kθ

k + 1

)− θ � − θ

k + 1

Var(θ) � Var(

kk + 1X

)

�

(k

k + 1

)2 (θ2

25

)

biasθ (θ) � E[θ] − θ(θ

k + 1

)2�

(k

k + 1

)2 (θ2

25

)

k2

25 � 1

k � 5 (D)

Since k > 0, we reject k � −5.


QUIZ SOLUTIONS FOR LESSON 2 39

2.24. Without loss of generality, assume the variance of Y2 is 1. This is anyway a positive multiplicativeconstant, and such constants don’t a�ect the minimum.

Let the estimated parameter by θ. Since the new estimator for θ, which we’ll call Y, and the oldestimators are unbiased,

θ � E[Y] � k1 E[Y1] + k2 E[Y2] � (k1 + k2)θ

so k2 � 1 − k1. The variance of Y isVar(Y) � k21 (4) + (1 − k1)2

Di�erentiate and set equal to 0.

8k1 − 2(1 − k1) � 010k1 − 2 � 0

k1 � 0.2 (B)

2.25. See Example 2D, which solves this in general and derives the formula

MSEθ (θ) �2θ2

(n + 1)(n + 2)

Here, this is2(1002)(11)(12)

�20,000132 � 151.5152 (E)

Quiz Solutions

2-1. For a uniform distribution on [0, θ],

E[X2] �∫ θ

0

x2dxθ

�θ2

3

Therefore,

E[(2X)2] � 4E[X2] � 4θ2

3

The bias is 43θ

2 − θ2 � θ2

3 .




Practice Exam 1

1. Cars arrive at a toll booth in a Poisson process at the rate of 6 per minute.Determine the probability that the third car will arrive between 30 and 40 seconds from now.A. Less than 0.18B. At least 0.18, but less than 0.21C. At least 0.21, but less than 0.24D. At least 0.24, but less than 0.27E. At least 0.27

2. A business receives 50 pieces of mail every day in a Poisson process. One tenth of the mail containschecks. The logarithm of the amount of each check has a normal distribution with parameters µ � 3,σ2 � 9.

Determine the average number of checks for amounts greater than 10,000 that the business receives ina seven day week.

A. Less than 0.66B. At least 0.66, but less than 0.69C. At least 0.69, but less than 0.75D. At least 0.75, but less than 0.75E. At least 0.75

3. ATM withdrawals occur in a Poisson process at varying rates throughout the day, as follows:11PM–6AM 3 per hour6AM–8AM Linearly increasing from 3 per hour to 30 per hour8AM–5PM 30 per hour5PM–11PM Linearly decreasing from 30 per hour to 3 per hour

Withdrawal amounts are uniformly distributed on (100, 500), and are independent of each other andthe number of withdrawals.

Using the normal approximation, estimate the amount of money needed to be adequate for all with-drawals for a day 95% of the time.

A. Less than 137,500B. At least 137,500, but less than 138,000C. At least 138,000, but less than 138,500D. At least 138,500, but less than 139,000E. At least 139,000


407 Exercises continue on the next page . . .

408 PART VII. PRACTICE EXAMS

4. An estimator θ for θ has the following properties:

E[θ] � 4Var(θ) � 20

If θ � 6, calculate the bias of θ2 as an estimator for θ2.A. Less than −3B. At least −3, but less than −1C. At least −1, but less than 1D. At least 1, but less than 3E. At least 3

5. For 2 estimators of θ, θ and θ, you are given:

•θ θ

Expected value 4 5Variance 2 3

• θ � 5• Cov(θ, θ) � −1Determine the mean square error of 1

2 (θ + θ) as an estimator of θ.A. Less than 1.25B. At least 1.25, but less than 1.75C. At least 1.75, but less than 2.25D. At least 2.25, but less than 2.75E. At least 2.75

6. For a set of 3 biased coins, the probability of head is p. The 3 coins are tossed 10 times, with thefollowing results:

Number of heads Number of times0 41 32 23 1

Determine the maximum likelihood estimate of p.

A. 1/5 B. 1/4 C. 1/3 D. 2/5 E. 1/2



PRACTICE EXAM 1 409

7. A sample of 6 observed claim sizes is

10 25 30 52 70 90

These observations are �tted to a Lognormal distribution with µ � 2 using maximum likelihood.Determine the variance of the �tted distribution.A. Less than 21,000B. At least 21,000, but less than 23,000C. At least 23,000, but less than 25,000D. At least 25,000, but less than 27,000E. At least 27,000

8. For two baseball teams A and B:

• Team A wins 7 out of 10 games.• Team B wins x out of 14 games.• The null hypothesis is that the two teams are equally likely to win games.• The alternative hypothesis is that the two teams are not equally likely to win games.

Determine the highest value of x for which the null hypothesis is accepted at 5% signi�cance.

A. 10 B. 11 C. 12 D. 13 E. 14

9. For a Normally distributed variable X with σ2 � 2500, you test H0: µ � 100 against H1: µ < 100using the sample mean of 30 observations. The test is constructed to have 1% signi�cance.

Determine the power of the test at 70.A. Less than 0.72B. At least 0.72, but less than 0.76C. At least 0.76, but less than 0.80D. At least 0.80, but less than 0.84E. At least 0.84

10. A sample of 20 items from a normal distribution yields the following summary statistics:∑

Xi � 120∑

X2i � 1100

Construct a 99% con�dence interval of the form (0, a) for the variance.Determine a.

A. 10.0 B. 10.1 C. 10.5 D. 48.5 E. 49.8




11. X is a random variable having probability density function

f (x) � αxα−1 0 < x < 1

You test H0: α � 1 against H1: α > 1 using 2 observations, x1 and x2.Determine the form of the uniformly most powerful critical region for this test.A. x1 + x2 < kB. x1 + x2 > kC. x1x2 < kD. x1x2 > kE. 1

x1+ 1

x2 < k

12. The amount of time your trip to work takes is a Normally distributed random variable with meanx minutes and variance 25. You would like to test the hypothesis H0: x � 30 against the alternative H1:x > 30. The test should have 5% signi�cance and 90% power at 35.

Determine the minimum number of trips you will need in order to perform this test.

A. 9 B. 10 C. 11 D. 12 E. 13

13. A Normal random variable is known to have mean 5. For a sample of �ve observations from thevariable,

∑5i�1(xi − 5)2 � 175.

Construct a 95% con�dence interval of the form (a ,∞) for the variance.Determine a.A. Less than 12B. At least 12, but less than 14C. At least 14, but less than 16D. At least 16, but less than 18E. At least 18

14. You are given a sample of size 4 from a distribution with probability density function

f (x) � 2x 0 ≤ x ≤ 1

Y1, . . . , Y4 are the order statistics.Determine Pr(Y2 > 0.5).A. Less than 0.5B. At least 0.5, but less than 0.6C. At least 0.6, but less than 0.7D. At least 0.7, but less than 0.8E. At least 0.8



PRACTICE EXAM 1 411

15. You are given the following information from a group of students regarding time spent studyingfor an exam and the score on the exam:

Time (minutes) 372 405 428 457 500Score 85 78 82 100 92

Calculate Spearman’s ρ relating study time and exam score.A. Less than 0.3B. At least 0.3, but less than 0.4C. At least 0.4, but less than 0.5D. At least 0.5, but less than 0.6E. At least 0.6

16. For a random variable X, the null hypothesis is

H0: the median is 820

and the alternative hypothesis is

H1: the median is not 820.

For a sample of size 48, the 18th order statistic is 815 and the 19st order statistic is 822.Which of the following statements is true?A. Reject H0 at 1% signi�cance.B. Accept H0 at 1% signi�cance but not at 2.5% signi�cance.C. Accept H0 at 2.5% signi�cance but not at 5% signi�cance.D. Accept H0 at 5% signi�cance but not at 10% signi�cance.E. Accept H0 at 10% signi�cance.

17. In a certain town, the natural logarithm of annual wages is hypothesized to be symmetrically dis-tributed with mean 10.5. The wages of six people are

20,000 30,000 50,000 80,000 110,000 200,000

Using the Wilcoxon signed rank test, calculate the p-value of the hypothesis.A. Less than 0.05B. At least 0.05, but less than 0.10C. At least 0.10, but less than 0.15D. At least 0.15, but less than 0.20E. At least 0.20




18. It is hypothesized that the price of a restaurant meal (P) has a linear relationship to its star rating(S):

P � α + βS + ε

You are given the following data from six restaurants:

Price 15 15 19 20 22 35Stars 1 2 3 3 4 5

P � 21 S � 36∑

i�1(P − P)2 � 274

6∑

i�1(S − S)2 � 10

6∑

i�1(P − P)(S − S) � 47

Calculate the t statistic to test the signi�cance of the star rating as a factor in the restaurant price.A. Less than 2.0B. At least 2.0, but less than 3.0C. At least 3.0, but less than 4.0D. At least 4.0, but less than 5.0E. At least 5.0

19. In a regression model of the form

Y � α + βX + ε

you are given

• There are 8 observations.•

∑Xi � 85

•∑

X2i � 1547

•∑

Yi � 199•

∑XiYi � 3616

• The standard error of the regression is 19.36059.

Calculate the t statistic to test the hypothesis α � 0.A. Less than 0.01B. At least 0.01, but less than 0.02C. At least 0.02, but less than 0.03D. At least 0.03, but less than 0.04E. At least 0.04



PRACTICE EXAM 1 413

20. Four di�erent treatments are tried on two �elds apiece, with the following results:

Treatment 1 Treatment 2 Treatment 3 Treatment 480 60 62 7270 30 95 41

Calculate the F ratio to test whether the mean results of the treatments are equal.A. Less than 0.5B. At least 0.5, but less than 1.0C. At least 1.0, but less than 1.5D. At least 1.5, but less than 2.0E. At least 2.0

21. You are given:

• Claim counts follow a binomial distribution with m � 10 and Q.• Q varies by policyholder.• Q follows a beta distribution with a � 0.1, b � 0.9, θ � 1.

A policyholder submits 2 claims in 1 year.Calculate the expected number of claims from this policyholder in the next year.A. Less than 1.2B. At least 1.2, but less than 1.4C. At least 1.4, but less than 1.6D. At least 1.6, but less than 1.8E. At least 1.8

22. You are given:

• Claim counts follow a Poisson distribution. The probability of 0 claims is θ.• The distribution of θ over the entire population has density function

f (θ) � 3θ2 0 < θ < 1

A policyholder submits no claims for 2 years.Calculate the posterior probability that this policyholder submits no claims in the third year.

A. 0.50 B. 0.67 C. 0.75 D. 0.80 E. 0.83

23. A small department store chain has 5 stores. At each store, daily sales are normally distributedwith variance 5,000,000. Mean daily sales at each store are µ, where µ is normally distributed with mean50,000 and variance 10,000,000.

At one of the stores, total sales over a 7 day week are 420,000.Determine the posterior mean daily sales for this store.A. Less than 52,000B. At least 52,000, but less than 54,000C. At least 54,000, but less than 56,000D. At least 56,000, but less than 58,000E. At least 58,000




24. Claim counts are Poisson with mean λ. The distribution of λ is exponential with mean 0.1.For a randomly selected policyholder, 6 claims are observed in 2 years.Determine the posterior expected number of claims from this policyholder.A. Less than 0.4B. At least 0.4, but less than 0.5C. At least 0.5, but less than 0.6D. At least 0.6, but less than 0.7E. At least 0.7

25. The monthly number of losses on an insurance coverage follows a Poisson distribution with meanλ. The prior distribution of λ is gamma with parameters α and θ.

A randomly selected insured is observed for n months and submits no claims.Determine the smallest n such that the expected number of claims for this policyholder is half of the

expected number of claims for the general population.

A. θ B. 1/θ C. αθ D. α/θ E. θ/α

Solutions to the above questions begin on page 457.


Appendix A. Solutions to the Practice Exams

Answer Key for Practice Exam 11 B 6 C 11 D 16 D 21 E2 B 7 D 12 A 17 E 22 E3 B 8 D 13 C 18 D 23 E4 C 9 D 14 D 19 A 24 C5 A 10 E 15 E 20 C 25 B

Practice Exam 1

1. [Lesson 25] The probability that the third car will arrive in the interval (30, 40) is the probabilityof at least 3 cars in 40 seconds minus the probability of at least 3 cars in 30 seconds. For 40 seconds, thePoisson parameter is 4 and the probability is

1 − e−4(1 + 4 + 42

2

)� 1 − 0.238103

For 30 seconds, the Poisson parameter is 3 and the probability is

1 − e−3(1 + 3 + 32

2

)� 1 − 0.423190

The di�erence is 0.423190 − 0.238103 � 0.185087 . (B)

2. [Lesson 27] The probability of a check greater than 10,000 is

1 −Φ(ln 10,000 − 3

3

)� 1 −Φ(2.07) � 1 − 0.9808 � 0.0192

The Poisson distribution of just the checks over 10,000 in one week has parameter 7(50)(0.1)(0.0192) �

0.672 . (B)

3. [Lesson 29] The Poisson parameter per day is computed by adding up the rates over the 4 periods.For 11PM–6AM, we have 7 hours times 3 per hour, or 21. For 8AM–5PM we have 9 hours times 30 perhour, or 270. For the other two periods, because of the linear increase or decrease, the average per hour isthe midpoint, or (30 + 3)/2 � 16.5, and there are 8 hours with varying rates, for a total of 8 × 16.5 � 132.The total number of withdrawals per day is 21 + 270 + 132 � 423. The mean aggregate withdrawals is(423)(300) � 126,900.

The second moment of the uniform distribution on (100, 500) is the variance plus the mean squared.The variance of a uniform distribution is the range squared divided by 12, or 4002/12. Therefore, thesecond moment of the uniform distribution is 4002/12 + 3002 � 103,333 1

3 . The variance of aggregatewithdrawals, by the compound variance formula (29.2), is λ E[X2] � (423)(103,333 1

3 ) � 43,710,000.


457

458 PRACTICE EXAM 1, SOLUTIONS TO QUESTIONS 4–7

The amount of money needed to be adequate 95% of the time is

126,900 + 1.645√43,710,000 � 137,775.68 (B)

4. [Lesson 2] The bias is the expected value of the estimator minus the true value of the parameter.

E[θ2] � Var(θ) + E[θ]2 � 20 + 42 � 36

and θ2 � 62 � 36, sobiasθ2 (θ2) � 36 − 36 � 0 (C)

5. [Lesson 2] E[12 (θ + θ)

]�

12 (4+5) � 4.5, so the bias is 4.5−5 � −0.5. The variance of the estimator

is (12

)2 (Var(θ + θ)

)�

(14

) (Var(θ) + Var(θ) + 2Cov(θ, θ)

)�

(14

) (2 + 3 + 2(−1)

)� 0.75

Therefore, the mean square error is 0.52 + 0.75 � 1 . (A)

6. [Lesson 3] For a binomial with �xed m � 3, maximum likelihood estimates q the same way as themethod of moments. For 30 tosses (10 tosses of 3 coins) we have (1)(3) + (2)(2) + (3)(1) � 10 heads, soq �

1030 �

13 . (C)

7. [Lesson 3] The likelihood function in terms of the 6 observations xi , dropping multiplicativeconstants such as 1

xi√2π

, is

L(σ) �1σ6

e−∑6

i�1 (ln xi−2)2

2σ2

6∑

i�1(ln xi − 2)2 � 0.091558 + 1.485658 + 1.963354 + 3.807352 + 5.055731 + 6.249048 � 18.652701

l(σ) � −6 ln σ − 18.6527012σ2

dldσ � − 6

σ+ 18.652701

σ3� 0

− 6σ2 + 18.652701 � 0

σ2 �18.652701

6 � 3.108784

The moments of the �tted distribution are

E[X] � e2+3.108784/2 � 34.9666E[X2] � e4+2(3.108784)

� 27,380

Var(X) � 27,380 − 34.96662 � 26,157 (D)


PRACTICE EXAM 1, SOLUTIONS TO QUESTIONS 8–12 459

8. [Subsection 8.4.3] The number of games won is binomial. The pooled mean games won is (7 +x)/24. For a two-sided test with 5% signi�cance, we need the Z statistic to be no higher than 1.96, the97.5th percentile of a standard normal distribution. The Z statistic is

Z �

x14 − 7

10√(7+x24

) (17−x24

) (110 +

114

)

We set this equal to 1.96 and solve for x.

x14 − 0.7 �

1.9624

√0.171429(7 + x)(17 − x)

2.112446x − 20.701967 �√

(7 + x)(17 − x)

4.462426x2 − 87.463556x + 428.5714 � −x2 + 10x + 1195.462426x2 − 97.463556x + 309.5714 � 0

x � 13.71, 4.13

Thus we accept the null hypothesis when x is between 4 and 13 . (D)It may be easier to solve this question by plugging in the answer choices for x in the original equation

setting Z equal to 1.96.

9. [Lesson 6] To achieve 1% signi�cance, the critical value for a normal random variable must be2.326 times the standard deviation below the mean, or 100 − 2.326

( 50√30)� 78.76. The power of the test at

70 is the probability of rejecting the null hypothesis if µ � 70, or

Pr(X < 70) � Φ(78.76 − 7050/√30

)� Φ(0.960) � 0.831 (D)

10. [Lesson 10] The sample variance is

S2�

2019

*,110020 −

(12020

)2+- � 20

σ2 �19S2

W , where W is chi-square with 19 degrees of freedom. To make σ2 large, make W small: pick its1st percentile, 7.633. Then σ2 � 19(20)

7.633 � 49.8 is the upper bound of the interval. (E)

11. [Lesson 14] The likelihood ratio is (α0 � 1)

(x1x2)α0−1

α2(x1x2)α−1�

(1α2

)(x1x2)1−α

This should be less than a constant k. The �rst factor is a positive constant and can be incorporated in k.Since 1 − α < 0, we will have this expression less than a constant if x1x2 > k. (D)

12. [Lesson 7] If the critical value is x and n is the number of trips, we need x � 30 + 1.645( 5√

n

)for the

signi�cance condition, and we need x ≤ 35 − 1.282( 5√

n

)for the power condition. Thus we have

(1.645 + 1.282)(5)√n

� 5



√n � 2.927n � 8.567

Rounding up to the next integer, 9 trips are needed. (A)

13. [Lesson 10] Let Xi be an observation of the normal random variable and σ2 the variance of Xi .Let W �

∑5i�1(Xi −5)2/σ2. Then by the de�nition of the chi-square distribution, W is a chi-square random

variable with 5 degrees of freedom. The observed value of W is 175/si gma2, so

σ2 ∼ 175W

To �nd the lower bound a of a 95% con�dence interval, we use the 95th percentile of W , or 11.070:

a �175

11.070 � 15.808 (C)

14. [Lesson 20] The probability that one item X is greater than 0.5 is

Pr(X > 0.5) �∫ 1

0.52x dx � 1 − 0.52 � 0.75

The probability that Y2 is greater than 0.5 is the probability that three or four items are above 0.5, or

Pr(Y2 ≥ 0.5) �(43

)(0.753)(0.25) +

(44

)(0.754)

� 0.4218785 + 0.31640625 � 0.7383 (D)

15. [Lesson 24] The ranks of time are in order: 1, 2, 3, 4, 5. The ranks of scores are 3, 1, 2, 5, 4. Usingformula (24.5),

ρ � 1 −6((1 − 3)2 + (2 − 1)2 + (3 − 2)2 + (4 − 5)2 + (5 − 4)2

)

5(24)� 0.6 (E)

16. [Lesson 21] There are k � 30 numbers higher than the hypothesizedmedian. The sign test statisticis

Z �k − n/2√

n/2�

30 − 24√48/2

� 1.732

In the standard normal table, this is greater than the 95th percentile, which is 1.645, but less than the 97.5thpercentile, which is 1.96. For a two-sided test, this means accepting H0 at 5% signi�cance but not at 10%signi�cance. (D)

17. [Section 22.1] The logarithms of the six numbers are 9.9, 10.3, 10.8, 11.3, 11.6, 12.2. After subtract-ing 10.5, we have −0.6, −0.2, 0.3, 0.8, 1.1, 1.7. The ranks are 3, 1, 2, 4, 5, 6. The sum of the ranks of thepositive numbers is 2 + 4 + 5 + 6 � 17. Since the maximum statistic is n(n + 1)/2 � (6)(7)/2 � 21, theprobability Pr(T ≥ 17) is the same as Pr(T ≤ 4). To get a statistic of 4 or less as a sum of unequal numbers,you’d need no numbers, or 1, 2, 3, 4, or 1 + 2, 1 + 3. That’s a total of 7 ways out of 26, or 7/64 � 0.109375.Since we are performing a two-sided test, the p-value is 2(0.109375) � 0.219 . (E)


PRACTICE EXAM 1, SOLUTIONS TO QUESTIONS 18–21 461

18. [Section 12.3] The estimated value of β is

β �

∑x y∑x2 �

4710 � 4.7

The standard error of β is

SSE � 274 − 4.72(10) � 53.1

s2 �53.14 � 13.275

sβ �√13.275/10 � 1.1522

The t statistic is 4.7/1.1522 � 4.0793 . (D)

19. [Section 12.3] First let’s calculate α.∑

(Xi − X)2 � 1547 − 8528 � 643.875

∑(Xi − X)(Yi − Y) � 3616 − (85)(199)

8 � 1501.625

β �1501.625643.875 � 2.3322

α � Y − βx �1998 − 2.3322

(858

)� 0.09571

The variance of α is (19.360592643.875

) (15478

)� 112.57

The t statistic to test α � 0 is 0.09571/√112.57 � 0.0090 . (A)

20. [Lesson 13] The total is 80 + 70 + 60 + 30 + 62 + 95 + 72 + 41 � 510. The total sum of squares is

802 + 702 + · · · + 412 − 51028 � 3021.5

The sums of each treatment are 150, 90, 157, 113. The treatment sum of squares is

1502 + 902 + 1572 + 11322 − 5102

8 � 1496.5

It has 3 degrees of freedom.The error sum of squares is 3021.5 − 1496.5 � 1525. It has 4 degrees of freedom.The F ratio is

F3,4 �1496.5/31525/4 � 1.308 (C)

21. [Lesson 17] There are 10 possible claims in a year; 2 materialized, 8 didn’t.

a → 0.1 + 2 � 2.1b → 0.9 + 8 � 8.9

The expected number of claims in the next year is 10( 2.12.1+8.9

)� 1.9091 . (E)



22. [Lesson 17] Since there are only two possibilities (either 0 claims are submitted or not), the modelis Bernoulli. The prior is beta with a � 3, b � 1, which in the posterior go to a′ � 3 + 2 � 5, b′ � 1 + 0 � 1,and the posterior expected value of θ is then a/(a + b) � 5/6 � 0.8333 , which is the posterior probabilityof no claims. (E)

23. [Lesson 18] We are given 7 days of experience, so n � 7. We are given that v � 5,000,000, a �

10,000,000, and nx � 420,000. Using formula (18.1),

µ∗ �5,000,000(50,000) + 420,000(10,000,000)

5,000,000 + 7(10,000,000)� 59,333 (E)

24. [Lesson 19] An exponential is a gamma with α � 1. Let γ � 1/θ.

α � 1→ 1 + 6 � 7γ � 10→ 10 + 2 � 12

Posterior expected claims is 712 . (C)

25. [Lesson 19] Let γ � 1/θ. Then after n months, α → α and γ → γ + n. We want

αγ + n

�12αγ

This means n � γ, or n �1θ . (B)


Documents

Contents · 24 2. ESTIMATOR QUALITY because we know that s2 is unbiased, so E [s2] 2. bias 2 ( 2) (n 1) 2 n 2 2 n The bias goes to 0 as n ! 1 , so the empirical variance is an asymptotically