The Solutions Manual is also available.

*The book is copyright (c) by Michael J. Evans and Jeffrey S.
Rosenthal. It may be copied and distributed without restriction,
provided it is not altered, appropriate attribution is given
and no money is charged. If you are using the book in a course you are teaching please let us know.
*

- Front page
- Table of Contents
- Preface
- Chapter 1 - Probability Models
- Chapter 2 - Random Variables and Distributions
- Chapter 3 - Expectation
- Chapter 4 - Sampling Distributions and Limits
- Chapter 5 - Statistical Inference
- Chapter 6 - Likelihood Inference
- Chapter 7 - Bayesian Inference
- Chapter 8 - Optimal Inferences
- Chapter 9 - Model Checking
- Chapter 10 - Relationsships Among Variables
- Chapter 11 - Stochastic Processes
- Appendix A - Mathematical Background
- Appendix B - Computations
- Appendix C - Common Distributions
- Appendix D - Tables
- Appendix E - Answers to Odd-Numbered Exercises
- Index

If you have errata or comments please send these to the authors.

- bottom of page 7, $A\B^c$ should be $A \cap B^c$.
- Challenges 1.3.10 solution should have $+P(A \cap B \cap C)$ not $-P(A \cap B \cap C)$
- Problem 1.4.15 solution: for the value 11, another possibility is three dice each show 2 and the remaining five dice each show 1.
- Problem 1.4.16 solution: \sum_{i=1}^6 \binom{6}{i} / 2^6 doesn't equal 1, it equals 63/64.
- Problem 1.4.19 solution: There are 16 cards of value 10, not 20 of them, so "20/52" should be "16/52" (twice). Then, the answer 79/1024 is not correct and should be removed. The numerical final answer is 0.07707332 (as opposed to 79/1024 = 0.07714844) but that still rounds to 0.0771 so actually the "0.0771" can be left as is.
- Example 1.5.2 P(B)=P(A)P(B|A)+P(A^c)P(B|A^c)= (1/2)(7/11)+(1/2)(2/5).
- Challenge 1.5.16 solution: In the denominator \sum_{j=7}^{12}q_{j} should be \sum_{j=6}^{12}q_{j} otherwise solution is correct.
- Exercise 2.1.4 solution: the last 3 need to be adjusted so that we have 4^3, 5^3 and 6^3 instead of 4^4, 5^5, and 6^6.
- Exercise 2.1.8(a) solution: "1-1+0 = 0" should be "1-0+0 = 1".
- Exervise 2.3.15(b) solution: .65 should be raised to the power 10 not 9.
- Exercise 2.3.15(b,c) statements are unclear, and should instead be written as: (b) What is the probability the player scores their first basket on their 10th attempt? (c) What is the probability the player scores their second basket on their 10th attempt?
- Exercise 2.3.16(c): The question's word "until" is unclear, and would be clearer to ask the probability of the fifth black ball being on the 15th draw. Then in the solution, here r=5 and k=10 so the answer should be {14 \choose 4} (4/9)^5 (5/9)^{10}.
- Exercise 2.3.19 solution: "Binomial(10,1/1000)" should be "Binomial(100,1/1000)".
- Problem 2.3.27 statement: $X$ should be $X_n$ (twice), since they're different random variables for each $n$.
- Exercise 2.4.1 solution: part (e) is missing; the answer should be 2/3.
- Exercise 2.4.3(d) solution: final answer should be $e^{-4 (26)^{1/4})$; the first 4 is missing.
- Exercise 2.4.4(c) solution: the final answer is written awkwardly, and could instead be written as $c = (3/2) 2^{-3/2} = 3/4\sqrt(2)$.
- Exercise 2.5.8(a) solution: The end should be $F_Y(3/4) - F_Y(1/3) = 1-(1/4)^3 - (1/3)^3 = 1637/1728$.
- Exercise 2.5.13(a) solution: the graph isn't correct for the function F it is supposed to demonstrate.
- Exercise 2.5.14(b) solution: It should say $1-G(4) = 1 - (1-e^{-4^2}) = e^{-16}$, and $G(2)-G(-1) = (1-e^{-2^2}) - 0 = 1 - e^{-4}$.
- Exercise 2.6.9(b) solution: $0 < x <\sqrt{2}$ should be $0 < z\sqrt{2}$.
- Exercise 2.7.1 solution: ``1/3'' should be ``2/3''.
- Exercise 2.7.2 solution: The function definition misses the case where $0 \le x < 1$ and $-7 \le y < 0$, where it should equal 0. (Equivalently, the case $x<1$ and $y< -7$ could be extended to be $x<1$ and $y<0$.) Also, the ``1/4'' and ``3/4'' should be swapped.
- Exercise 2.9.2(a) solution: $e^{-x}$ should be $e^{-3x}$.
- Exercise 2.9.3 (c) solution: Jacobian of the transformation should be (2x)(2y)+(2x)(2y)=8xy.
- Exercise 3.3.13 solution: The end should read "Cov(Z,W) = E(ZW) - E(Z) E(W) = 1/6 - (5/6)(1/6) = 1/36, and Corr(Z,W) = Cov(Z,W) / \sqrt{Var(Z) Var(W)} = (1/36)/\sqrt{(17/36)(17/36)} = 1/17."
- Exercise 3.3.14 solution: The final sentence should say that Cov(Z,W) = E(ZW) - E(Z) E(W) = -1/2 - (1/2)(1/2) = - 3/4, and Corr(Z,W) = - 3/4 / \sqrt{(5/4)(5/4)} = - 3/5.
- Problem 3.4.19 solution: The third integral should have the lower limit equal to - infinity and the dx should be dz.
- Exercise 3.6.12(d): Should ask for Cov(X,W) instead of Cov(W,Z), to fit with the solution and also with the comparison to part (a).
- Exercise 4.3.1 solution: ``7-U'' should be ``U-7'' (twice).
- Exercise 4.3.6(c) solution: A_n should be M_n, and m should be mu.
- Exercise 4.4.22: replace "N(mu,sigma^2)" by "N(0,1)".
- Exercise 4.5.2 solution: The first "Bernoulli(2/3)" should be "Binomial(m,2/3)".