统计推断(StatisticalInference)第二版练习题答案

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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 Sta tistical 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

Mathematica

and numerically using R. Some solutions are given as code in either of these lan-

guages. Mathematica

can be purchased from Wolfram Research, and R is a free download from

http://www.r-project.org/.

Here is a detailed listing of the solutions included.

Chapter Number of Exercises Number of Solutions Missing

1 55 51 26, 30, 36, 42

2 40 37 34, 38, 40

3 50 42 4, 6, 10, 20, 30, 32, 34, 36

4 65 52 8, 14, 22, 28, 36, 40

48, 50, 52, 56, 58, 60, 62

5 69 46 2, 4, 12, 14, 26, 28

all even problems from 36 − 68

6 43 35 8, 16, 26, 28, 34, 36, 38, 42

7 66 52 4, 14, 16, 28, 30, 32, 34,

36, 42, 54, 58, 60, 62, 64

8 58 51 36, 40, 46, 48, 52, 56, 58

9 58 41 2, 8, 10, 20, 22, 24, 26, 28, 30

32, 38, 40, 42, 44, 50, 54, 56

10 48 26 all even problems except 4 and 32

11 41 35 4, 20, 22, 24, 26, 40

12 31 16 all even problems

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 Pe rlman, 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

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 2

= 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.

a. x ∈ A\B ⇔ x ∈ A and x/∈ B ⇔ x ∈ A and x/∈ A ∩ B ⇔ x ∈ A\(A ∩ B). Also, x ∈ A and

x/∈ B ⇔ x ∈ A and x ∈ B

⇔ x ∈ A ∩ B

b. Suppose x ∈ B. Then either x ∈ A or x ∈ A

. If x ∈ A, then x ∈ B ∩ A, and, hence

x ∈ (B ∩A) ∪(B ∩A

). Thus B ⊂ (B ∩A) ∪(B ∩A

). Now suppose x ∈ (B ∩A) ∪(B ∩A

Then either x ∈ (B ∩ A) or x ∈ (B ∩ A

). If x ∈ (B ∩ A), then x ∈ B. If x ∈ (B ∩ A

then x ∈ B. Thus (B ∩ A) ∪ (B ∩ A

) ⊂ B. Since the containment goe s b oth ways, we have

B =(B ∩ A) ∪ (B ∩ A

). (Note, a more straightforward argument for this part simply uses

the Distributive Law to state that (B ∩A) ∪(B ∩ A

)=B ∩(A ∪A

)=B ∩S = B.)

c. Similar to part a).

d. From part b).

A ∪ B = A ∪ [(B ∩ A) ∪ (B ∩ A

)] = A ∪ (B ∩ A) ∪ A ∪ (B ∩ A

)=A ∪ [A ∪ (B ∩A

)] =

A ∪ (B ∩A

1.3 a. x ∈ A ∪ B ⇔ x ∈ A or x ∈ B ⇔ x ∈ B ∪ A

x ∈ A ∩ B ⇔ x ∈ A and x ∈ B ⇔ x ∈ B ∩ A.

b. x ∈ A ∪(B ∪ C) ⇔ x ∈ A or x ∈ B ∪C ⇔ x ∈ A ∪ B or x ∈ C ⇔ x ∈ (A ∪B) ∪ C.

(It can similarly b e shown that A ∪ (B ∪ C)=(A ∪ C) ∪ B.)

x ∈ A ∩ (B ∩ C) ⇔ x ∈ A and x ∈ B and x ∈ C ⇔ x ∈ (A ∩B) ∩ C.

c. x ∈ (A ∪B)

⇔ x/∈ A or x/∈ B ⇔ x ∈ A

and x ∈ B

⇔ x ∈ A

∩ B

x ∈ (A ∩ B)

⇔ x/∈ A ∩B ⇔ x/∈ A and x/∈ B ⇔ x ∈ A

or x ∈ B

⇔ x ∈ A

∪ B

1.4 a. “A or B or both” is A∪B. From Theorem 1.2.9b we have P (A∪B)=P (A)+P (B)−P (A∩B).

1-2 Solutions Manual for Statistical Inference

b. “A or B but not both” is (A ∩B

) ∪ (B ∩A

). Thus we have

P ((A ∩ B

) ∪ (B ∩A

)) = P (A ∩ B

)+P (B ∩ A

) (disjoint union)

=[P (A) − P (A ∩ B)] + [P (B) − P (A ∩ B)] (Theorem1.2.9a)

= P (A)+P (B) − 2P (A ∩ B).

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)

, and P ((A ∩ B)

)=1− P (A ∩ B).

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

b. P (A ∩ B ∩C)=

1.6

= (1 − u)(1 − w),p

= u(1 − w )+w(1 − u),p

= uw,

= p

⇒ u + w =1

= p

⇒ uw =1/3.

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) =



1 −

πr

if i =0

πr



(6−i)

−(5−i)



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) =

πr

P (scoring i points ∩ board is hit) =

πr



(6 − i)

− (5 − i)



i =1,...,5.

Therefore,

P (scoring i points|board is hit) =

(6 − i)

− (5 − i)

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) =

π(6 −i)

πr

−

π ((6−(i + 1)))

πr

(6 − i)

−(5 − i)

b. Expanding the squares in part a) we ﬁnd P (scoring exactly i points) =

11−2i

, which is

decreasing in i.

c. Let P (i)=

11−2i

. Since i ≤ 5, P (i) ≥ 0 for all i. P (S)=P (hitting the dartboard) = 1 by

deﬁnition. Lastly, P (i ∪ j) = area of i ring + area of j ring = P (i)+P (j).

1.9 a. Suppose x ∈ (∪

)

, by the deﬁnition of complement x �∈ ∪

, that is x �∈ A

for all

α ∈ Γ. Therefore x ∈ A

for all α ∈ Γ. Thus x ∈∩

and, by the deﬁnition of intersection

x ∈ A

for all α ∈ Γ. By the deﬁnition of complement x �∈ A

for all α ∈ Γ. Therefore

x �∈ ∪

. Thus x ∈ (∪

)

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