Solutions - Casella - Berger

Solutions - Casella - Berger

(Parte 1 de 3)

Solutions Manual for Statistical Inference, Second Edition

George Casella

University of Florida

Roger L. Berger North Carolina State University

Damaris Santana University of Florida

0-2 Solutions Manual for Statistical Inference

“When I hear you give your reasons,” I remarked, “the thing always appears to me to be so ridiculously simple that I could easily do it myself, though at each successive instance of your reasoning I am baffled until you explain your process.”

Dr. Watson to Sherlock Holmes A Scandal in Bohemia

0.1 Description

This solutions manual contains solutions for all odd numbered problems plus a large number of solutions for even numbered problems. Of the 624 exercises in Statistical Inference, Second Edition, this manual gives solutions for 484 (78%) of them. There is an obtuse pattern as to which solutions were included in this manual. We assembled all of the solutions that we had from the first edition, and filled in so that all odd-numbered problems were done. In the passage from the first to the second edition, problems were shuffled with no attention paid to numbering (hence no attention paid to minimize the new effort), but rather we tried to put the problems in logical order.

A major change from the first edition is the use of the computer, both symbolically through

Mathematicatm and numerically using R. Some solutions are given as code in either of these languages. Mathematicatm can be purchased from Wolfram Research, and R is a free download from Here is a detailed listing of the solutions included.

0.2 Acknowledgement

Many people contributed to the assembly of this solutions manual. We again thank all of those who contributed solutions to the first edition – many problems have carried over into the second edition. Moreover, throughout the years a number of people have been in constant touch with us, contributing to both the presentations and solutions. We apologize in advance for those we forget to mention, and we especially thank Jay Beder, Yong Sung Joo, Michael Perlman, Rob Strawderman, and Tom Wehrly. Thank you all for your help. And, as we said the first time around, although we have benefited greatly from the assistance and

ACKNOWLEDGEMENT 0-3 comments of others in the assembly of this manual, we are responsible for its ultimate correctness. To this end, we have tried our best but, as a wise man once said, “You pays your money and you takes your chances.”

George Casella Roger L. Berger Damaris Santana

December, 2001

Chapter 1 Probability Theory

“If any little problem comes your way, I shall be happy, if I can, to give you a hint or two as to its solution.”

Sherlock Holmes The Adventure of the Three Students

1.1 a. Each sample point describes the result of the toss (H or T) for each of the four tosses. So, for example THTT denotes T on 1st, H on 2nd, T on 3rd and T on 4th. There are 24 = 16 such sample points.

b. The number of damaged leaves is a nonnegative integer. So we might use S = {0,1,2,...}.

c. We might observe fractions of an hour. So we might use S = {t : t ≥ 0}, that is, the half infinite interval [0, ∞).

d. Suppose we weigh the rats in ounces. The weight must be greater than zero so we might use S = (0,∞). If we know no 10-day-old rat weighs more than 100 oz., we could use S = (0,100].

e. If n is the number of items in the shipment, then S = {0/n,1/n,...,1}.

1.2 For each of these equalities, you must show containment in both directions.

c. Similar to part a).

d. From part b).

1-2 Solutions Manual for Statistical Inference b. “A or B but not both” is (A ∩ Bc) ∪ (B ∩ Ac). Thus we have

c. “At least one of A or B” is A ∪ B. So we get the same answer as in a). d. “At most one of A or B” is (A ∩ B)c, and P((A ∩ B)c) = 1 − P(A ∩ B).

1.5 a. A ∩ B ∩ C = {a U.S. birth results in identical twins that are female}

These two equations imply u(1 − u) = 1/3, which has no solution in the real numbers. Thus, the probability assignment is not legitimate. 1.7 a.

P(scoring i points) = pir A

if i = 1,, 5.

P(scoring i points|board is hit) = P(scoring i points ∩ board is hit)

P(board is hit)

P(board is hit) = pir2

P(scoring i points ∩ board is hit) = pir2

i = 1,, 5.
i = 1,, 5

which is exactly the probability distribution of Example 1.2.7.

1.8 a. P(scoring exactly i points) = P(inside circle i) − P(inside circle i + 1). Circle i has radius (6 − i)r/5, so

P(sscoring exactly i points) = pi(6 − i)2r2

b. Expanding the squares in part a) we find P(scoring exactly i points) = 1−2i25 , which is decreasing in i.

1.9 a. Suppose x ∈ (∪αAα)c, by the definition of complement x 6∈ ∪αAα, that is x 6∈ Aα for all α ∈ Γ. Therefore x ∈ Acα for all α ∈ Γ. Thus x ∈ ∩αAcα and, by the definition of intersection x ∈ Acα for all α ∈ Γ. By the definition of complement x 6∈ Aα for all α ∈ Γ. Therefore x 6∈ ∪αAα. Thus x ∈ (∪αAα)c.

Second Edition 1-3 b. Suppose x ∈ (∩αAα)c, by the definition of complement x 6∈ (∩αAα). Therefore x 6∈ Aα for some α ∈ Γ. Therefore x ∈ Acα for some α ∈ Γ. Thus x ∈ ∪αAcα and, by the definition of union, x ∈ Acα for some α ∈ Γ. Therefore x 6∈ Aα for some α ∈ Γ. Therefore x 6∈ ∩αAα. Thus x ∈ (∩αAα)c.

Ac i

Proof of (i): If x ∈ (∪Ai)c, then x /∈ ∪Ai. That implies x /∈ Ai for any i, so x ∈ Aci for every i and x ∈ ∩Ai.

Proof of (i): If x ∈ (∩Ai)c, then x /∈ ∩Ai. That implies x ∈ Aci for some i, so x ∈ ∪Aci. 1.1 We must verify each of the three properties in Definition 1.2.1.

b. (1) The empty set ∅ is a subset of any set, in particular, ∅ ⊂ S. Thus ∅ ∈ B. (2) If A ∈ B, then A ⊂ S. By the definition of complementation, Ac is also a subset of S, and, hence,

Ac ∈ B. (3) If A1,A2,∈ B, then, for each i,Ai ⊂ S. By the definition of union, ∪Ai ⊂ S.
A1,A2,∈ B1 ∩ B2, then A1,A2,... ∈ B1 and A1,A2,... ∈ B2. Therefore, since B1 and B2

are both sigma algebra, ∪∞

Ai) (Ais are disjoint)

Ai) (finite additivity)

Now define Bk = ⋃∞ i=k Ai. Note that Bk+1 ⊂ Bk and Bk → φ as k → ∞. (Otherwise the sum of the probabilities would be infinite.) ThusP (∞⋃ so A and B cannot be disjoint.

1.14 If S = {s1,...,sn}, then any subset of S can be constructed by either including or excluding si, for each i. Thus there are 2n possible choices.

1.15 Proof by induction. The proof for k = 2 is given after Theorem 1.2.14. Assume true for k, that is, the entire job can be done in n1 × n2 × · × nk ways. For k + 1, the k + 1th task can be done in nk+1 ways, and for each one of these ways we can complete the job by performing

1-4 Solutions Manual for Statistical Inference the remaining k tasks. Thus for each of the nk+1 we have n1 × n2 × · × nk ways of completing the job by the induction hypothesis. Thus, the number of ways we can do the job is

1.17 There are ( n

2) = n(n − 1)/2 pieces on which the two numbers do not match. (Choose 2 out of n numbers without replacement.) There are n pieces on which the two numbers match. So the total number of different pieces is n + n(n − 1)/2 = n(n + 1)/2.

n = (n−1)(n−1)! 2n . There are many ways to obtain this. Here is one. The denominator is n because this is the number of ways to place n balls in n cells. The numerator is the number of ways of placing the balls such that exactly one cell is empty. There are n ways to specify the empty cell. There are n − 1 ways of choosing the cell with two balls. There are( n

2) ways of picking the 2 balls to go into this cell. And there are (n − 2)! ways of placing the remaining n − 2 balls into the n − 2 cells, one ball in each cell. The product of these is the

b. Think of the n variables as n bins. Differentiating with respect to one of the variables is equivalent to putting a ball in the bin. Thus there are r unlabeled balls to be placed in n unlabeled bins, and there are ( n+r−1 r ) ways to do this.

1.20 A sample point specifies on which day (1 through 7) each of the 12 calls happens. Thus there are 712 equally likely sample points. There are several different ways that the calls might be assigned so that there is at least one call each day. There might be 6 calls one day and 1 call each of the other days. Denote this by 6111111. The number of sample points with this pattern

6 ) 6!. There are 7 ways to specify the day with 6 calls. There are (

6 ) to specify which of the 12 calls are on this day. And there are 6! ways of assigning the remaining 6 calls to the remaining 6 days. We will now count another pattern. There might be 4 calls on one day, 2 calls on each of two days, and 1 call on each of the remaining four days. Denote this by 4221111.

The number of sample points with this pattern is 7( 12

4!. (7 ways to pick day with 4calls, ( 12

4 ) to pick the calls for that day, (

2) to pick two days with two calls, (

2) ways to pick two calls for lowered numbered day, ( 6

2) ways to pick the two calls for higher numbered day,

4! ways to order remaining 4 calls.) Here is a list of all the possibilities and the counts of the sample points for each one.

pattern number of sample points

2r) ways of choosing 2r shoes from a total of 2n shoes.

Thus there are ( 2n

2r) equally likely sample points. The numerator is the number of sample points for which there will be no matching pair. There are ( n

2r) ways of choosing 2r different shoes

Second Edition 1-5 styles. There are two ways of choosing within a given shoe style (left shoe or right shoe), which gives 22r ways of arranging each one of the ( n

2r) arrays. The product of this is the numerator(

P( same number of heads ) = n∑

1.24 a.

i=1 P(A wins on ith toss) c. d dp

An ambiguity may arise if order is not acknowledged, the space is S′ = {(B,B),(B,G),(G,G)}, with each outcome equally likely.

1.27 a. For n odd the proof is straightforward. There are an even number of terms in the sum n−k) , which are equal, have opposite signs. Thus, all pairs cancel and the sum is zero. If n is even, use the following identity, which is the basis of Pascal’s

. Then, for n even

b. Use the fact that for k > 0, k( n

= 0 from part a).

1.28 The average of the two integrals is

Let dn = logn!−[(n + 1/2)logn − n], and we want to show that limn→∞ mdn = c, a constant. This would complete the problem, since the desired limit is the exponential of this one. This is accomplished in an indirect way, by working with differences, which avoids dealing with the factorial. Note that

log(1 + x), as the remaining adjacent pairs are negative. Hence

It therefore follows, by the comparison test, that the series ∑∞ 1 dn−dn+1 converges. Moreover, the partial sums must approach a limit. Hence, since the sum telescopes,

Thus limn→∞ dn = d1 − c, a constant.

1.29 a.

Unordered Ordered

Unordered Ordered

b. Same as (a).

d. If the k objects were distinguishable then there would be k! possible ordered arrangements.

Since we have k1,...,km different groups of indistinguishable objects, once the positions of the objects are fixed in the ordered arrangement permutations within objects of the same group won’t change the ordered arrangement. There are k1!k2!···km! of such permutations for each ordered component. Thus there would be k! k !k !···k ! different ordered components.

e. Think of the m distinct numbers as m bins. Selecting a sample of size k, with replacement, is the same as putting k balls in the m bins. This is ( k+m−1 k ) , which is the number of distinct bootstrap samples. Note that, to create all of the bootstrap samples, we do not need to know what the original sample was. We only need to know the sample size and the distinct values.

1.31 a. The number of ordered samples drawn with replacement from the set {x1,...,xn} is n. The number of ordered samples that make up the unordered sample {x1,...,xn} is n!. Therefore

Second Edition 1-7 has probability n! n . Any other unordered outcome from {x1,...,xn}, distinct from the un- where k1 + k2 + · + km = n with at least one of the ki’s satisfying 2 ≤ ki ≤ n. The probability of obtaining the corresponding average of such outcome is

Therefore the outcome with average x +x +·+xn is the most likely. b. Stirling’s approximation is that, as n → ∞, n! ≈ √

c. Since we are drawing with replacement from the set {x1,...,xn}, the probability of choosing any xi is 1n. Therefore the probability of obtaining an ordered sample of size n without xi is (1 − 1n)n. To prove that limn→∞(1 − 1n)n = e−1, calculate the limit of the log. That is

L’Hopital’s rule shows that the limit is −1, establishing the result. See also Lemma 2.3.14.

1.32 This is most easily seen by doing each possibility. Let P(i) = probability that the candidate hired on the ith trial is best. Then

,, P(i) =
,,P(N) = 1.

N −i+1 1.3 Using Bayes rule

1.34 a. P(Brown Hair) b. Use Bayes Theorem

1.35 Clearly P(·|B) ≥ 0, and P(S|B) = 1. If A1,A2,are disjoint, thenP

1-8 Solutions Manual for Statistical Inference 1.37 a. Using the same events A, B, C and W as in Example 1.3.4, we have

b. By Exercise 1.35, P(·|W) is a probability function. A, B and C are a partition. So P(A|W)+P(B|W)+P(C|W) = 1.

And also,

1.39 a. Suppose A and B are mutually exclusive. Then A ∩ B = ∅ and P(A ∩ B) = 0. If A and B are independent, then 0 = P(A ∩ B) = P(A)P(B). But this cannot be since P(A) > 0 and P(B) > 0. Thus A and B cannot be independent.

b. If A and B are independent and both have positive probability, then

This implies A ∩ B 6= ∅, that is, A and B are not mutually exclusive.

P( dash sent | dash rec)

= P( dash rec | dash sent)P( dash sent)

P( dash rec | dash sent)P( dash sent) + P( dash rec | dot sent)P( dot sent)

b. By a similar calculation as the one in (a) P(dot sent|dot rec) = 27/434. Then we have

P( dash sent|dot rec) = 1643. Given that dot-dot was received, the distribution of the four possibilities of what was sent are

Event Probability

1.43 a. For Boole’s Inequality,

since Pi ≥ Pj if i ≤ j and therefore the terms −P2k + P2k+1 ≤ 0 for k = 1,..., n−12 when n is odd. When n is even the last term to consider is −Pn ≤ 0. For Bonferroni’s Inequality apply the inclusion-exclusion identity to the Aci, and use the argument leading to (1.2.10).

b. We illustrate the proof that the Pi are increasing by showing that P2 ≥ P3. The other arguments are similar. Write

Now to get to P3 we drop terms from this last expression. That is n−1∑ i=1 n∑ j=i+1 [ n∑

1-10 Solutions Manual for Statistical Inference c. If all of the Ai are equal, all of the probabilities in the inclusion-exclusion identity are the same. Thus

P(A),,Pj =

and the sequence of upper bounds on P(∪iAi) = P(A) becomes


)] which eventually sum to one, so the last bound is exact. For the lower bounds we get


which start out negative, then become positive, with the last one equaling P(A) (see Schwager 1984 for details).

1.45 X is finite. Therefore B is the set of all subsets of X. We must verify each of the three properties

A1,A2,∈ B and pairwise disjoint then

where the second inequality follows from the fact the P is a probability function.

1.46 This is similar to Exercise 1.20. There are 7 equally likely sample points. The possible values of

X3 are 0, 1 and 2. Only the pattern 331 (3 balls in one cell, 3 balls in another cell and 1 ball in a third cell) yields X3 = 2. The number of sample points with this pattern is ( 7

So P(X3 = 2) = 14,700/7 ≈ .0178. There are 4 patterns that yield X3 = 1. The number of sample points that give each of these patterns is given below.

pattern number of sample points

1.47 All of the functions are continuous, hence right-continuous. Thus we only need to check the limit, and that they are nondecreasing

( pi

d dx b. See Example 1.5.5.

Second Edition 1-1


(Parte 1 de 3)