Statistical Inference, 2nd Edition (George Casella) – Complete Solutions Manual | Step-by-Step Worked Solutions

SOLUTION MANUAL
All Chapters Included




SECOND EDITION

, Solutions Manual for Statistical Inference, Second Edition


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.”
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.

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 ∈ Bc ⇔ x ∈ A ∩ Bc.
b. Suppose x ∈ B. Then either x ∈ A or x ∈ Ac. If x ∈ A, then x ∈ B ∩ A,
and, hence x ∈ (B ∩ A) ∪ (B ∩ Ac). Thus B ⊂ (B ∩ A) ∪ (B ∩ Ac). Now suppose
x ∈ (B ∩ A) ∪ (B ∩ Ac). Then either x ∈ (B ∩ A) or x ∈ (B ∩ Ac). If x ∈ (B
∩ A), then x ∈ B. If x ∈ (B ∩ Ac), then x ∈ B. Thus (B ∩ A) ∪ (B ∩ Ac) ⊂ B.
Since the containment goes both ways, we have B = (B ∩ A) ∪ (B ∩ Ac). (Note, a
more straightforward argument for this part simply uses the Distributive Law to
state that (B ∩ A) ∪ (B ∩ Ac) = B ∩ (A ∪ Ac) = B ∩ S = B.)
c. Similar to part a).
d. From part b).
A ∪ B = A ∪ [(B ∩ A) ∪ (B ∩ Ac)] = A ∪ (B ∩ A) ∪ A ∪ (B ∩ Ac) = A ∪ [A ∪
(B ∩ Ac)] =
A ∪ (B ∩ Ac).

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 be 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)c ⇔ x ∈/ A or x ∈
/ B ⇔ x ∈ A c and x ∈ Bc ⇔ x ∈ Ac ∩ Bc
c
x ∈ (A ∩ B) ⇔ x ∈/ A∩B ⇔ x∈ / A and x ∈
/ B ⇔ x ∈ Ac or x ∈ Bc ⇔ x ∈ A c
∪B.c


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 ∩ Bc) ∪ (B ∩ Ac). Thus we have

P ((A ∩ Bc) ∪ (B ∩ Ac)) = P (A ∩ Bc) + P (B ∩ Ac) (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)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}
b. P (A ∩ B ∩ C) = 1 × 1 × 1
90 3 2
1.6
p0 = (1 − u)(1 − w), p1 = u(1 − w) + w(1 − u), p2 = uw,

p0 = p2 ⇒ u+w =1
p1 = p2 ⇒ 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.
(
1 −h πr2 if i = 0
P (scoring i points) A
2 2i
= 2

πr (6−i)
−(5−i) if i = 1, . . . , 5.
A 52

b.
P (scoring i points ∩ board is hit)
P(scoring i points|board is hit) =
P (board is hit)
πr2
P (board is hit)
= A
πr2 (6 − i)2 − (5 − i)2
P (scoring i points ∩ board is hit) = i = 1, . . . , 5.
A 52
Therefore,
(6 — i )2 − (5 − i )2
P (scoring i points|board is hit) i = 1, . . . , 5
52
=
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
π(6 — i)2r2 π ((6−(i + 1)))2r2 (6 − i)2−(5 − i)2
P (sscoring exactly i points) −= .
= 52πr2 52πr2 52

b. Expanding the squares in part a) we find P (scoring exactly i points) =25 11−2i , which is
decreasing in i.
c. Let P (i) = 11−2i
25
. 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 ∈ (∪αAα)c, by the definition of complement x /∈ ∪αAα, that is x /∈ Aα
for all
α ∈ Γ. Therefore x ∈αAc for all α ∈ Γ. Thus x ∈α∩αAc and, by the definition of
intersection
x ∈ Aαc for all α ∈ Γ. By the definition of complement x /∈ Aα for all α ∈ Γ.
Therefore
x /∈ ∪αAα. Thus x ∈ (∪αAα)c.

, Second Edition 1-3

b. Suppose x ∈ (∩αAα)c, by the definition of complement x /∈ (∩αAα). Therefore x /∈ Aα
for
some α ∈ Γ. Therefore x ∈α Ac for some α ∈ Γ. Thus x ∈α ∪αAc and, by the
definition of
union, x ∈ Aαc for some α ∈ Γ. Therefore x /∈ Aα for some α ∈ Γ. Therefore x /∈ ∩αAα.
Thus
x ∈ (∩αAα)c.
1.10 For A1, . . . , An
!c !c
[
n n
\ c \
n n
[ c
(i) Ai = Ai (ii) Ai = Ai
i=1 i=1 i=1 i=1

Proof of (i): If x ∈ (∪Ai)c, then x ∈/ ∪Ai. That implies x ∈/ Ai for any i, iso x ∈ Ac for
every i
and x ∈ ∩Ai.
Proof of (ii): If x ∈ (∩Ai)c, then x ∈/ ∩Ai. That implies x ∈ Ac for some i, so x ∈
∪Ac.
i i
1.11 We must verify each of the three properties in Definition 1.2.1.
a. (1) The empty set ∅ ∈ {∅ , S}. Thus ∅ ∈ B. (2) ∅ c = S ∈ B and Sc = ∅ ∈ B. (3)
∅ ∪S = S ∈ B.
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. Hence, ∪Ai ∈ B.
c. Let B1 and B2 be the two sigma algebras. (1) ∅ ∈ B1 and ∅ ∈ B2 since B1 and
B2 are sigma algebras. Thus ∅ ∈ B1 ∩ B2. (2) If A ∈ B1 ∩ B2, then A ∈ B1
and A ∈ B2. Since B1 and B2 are both sigma algebra Ac ∈ B1 and Ac ∈ B2.
Therefore Ac ∈ B1 ∩ B2. (3) If A1, A2, . . . ∈ B1 ∩ B2, then A1, A2, . . . ∈ B1 and A1,
A2, . . . ∈ B2. Therefore, since B1 and B2
are both sigma algebra,i=1 ∪∞ Ai ∈ B1 and ∞
i=1 ∪ Ai ∈ B2. Thus
∞
i=1 ∪ A i ∈ B1 ∩ B2.

1.12 First write
! [ !
[
∞ [
n

Ai = P Ai Ai
P i= n
i= !
i= 1 ∪ ∞ [
1 ∞
n
[ ! +1


= P Ai + P Ai (Ais are disjoint)
i=1 i=n+1
!
nΣ [
∞
= P (Ai) + P Ai (finite additivity)
i=1 i=n+1
S∞
Now define Bk = Ai. Note that Bk+1 ⊂ Bk and Bk → φ as k → ∞. (Otherwise the

sum
i=k
of the probabilities would be infinite.) Thus
! ! " n P (Ai) + P ∞
[∞
[∞ Σ # Σ
P Ai = lim Ai = = P (Ai).
P lim i=
i= n→∞ i= n→ i= (Bn+1)
1 1 ∞ 1 1


1.13 If A and B are disjoint, P (A ∪ B) = P (A) + P (B)3 = 14 + 3 12
= 13 , which is impossible.
More
generally, if A and B are disjoint, then ⊂A Bc and P (A)
≤ P (Bc). But here P (A) > P
(Bc), 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.