INSTRUCTOR’S
SOLUTIONS MANUAL
I NTRODUCTION TO
M ATHEMATICAL S TATISTICS
Robert Hogg
University of Iowa
Joseph McKean
Western Michigan University
Allen Craig
Boston Columbus Indianapolis New York San Francisco Upper Saddle River
Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto
Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
,Contents
1 Probability and Distributions 1
2 Multivariate Distributions 11
3 Some Special Distributions 19
4 Some Elementary Statistical Inferences 31
5 Consistency and Limiting Distributions 49
6 Maximum Likelihood Methods 53
7 Sufficiency 65
8 Optimal Tests of Hypotheses 77
9 Inferences about Normal Models 83
10 Nonparametric and Robust Statistics 93
11 Bayesian Statistics 103
iii
Copyright ©
, Chapter 1
Probability and Distributions
1.2.1 Part (c): C1 ∩ C2 = {(x, y) : 1 < x < 2, 1 < y < 2}.
1.2.3 C1 ∩ C2 = {mary,mray}.
1.2.6 Ck = {x : 1/k ≤ x ≤ 1 − (1/k)}.
1.2.7 Ck = {(x, y) : 0 ≤ x ≤ 1/k, 0 ≤ y ≤ 1/k}.
1.2.8 limk→∞ Ck = {x : 0 < x < 3}. Note: neither the number 0 nor the number 3
is in any of the sets Ck , k = 1, 2, 3, . . .
1.2.9 Part (b): limk→∞ Ck = φ, because no point is in all the sets Ck , k = 1, 2, 3, . . .
1.2.11 Because f (x) = 0 when 1 ≤ x < 10,
Z 10 Z 1
Q(C3 ) = f (x) dx = 6x(1 − x) dx = 1.
0 0
1.2.13 Part (c): Draw the region C carefully, noting that x < 2/3 because 3x/2 < 1.
Thus
Z 2/3 "Z 3x/2 # Z 2/3
Q(C) = dy dx = x dx = 2/9.
0 x/2 0
1.2.16 Note that
25 = Q(C) = Q(C1 ) + Q(C2 ) − Q(C1 ∩ C2 ) = 19 + 16 − Q(C1 ∩ C2 ).
Hence, Q(C1 ∩ C2 ) = 10.
1.2.17 By studying a Venn diagram with 3 intersecting sets, it should be true that
11 ≥ 8 + 6 + 5 − 3 − 2 − 1 = 13.
It is not, and the accuracy of the report should be questioned.
1
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, 2 Probability and Distributions
1.3.3
1 1 1 1/2
P (C) = + + + ··· = = 1.
2 4 8 1 − (1/2)
1.3.6 Z Z Z
∞ 0 ∞
P (C) = e−|x| dx = ex dx + e−x dx = 2 6= 1.
−∞ −∞ 0
We must multiply by 1/2.
1.3.8
P (C1c ∪ C2c ) = P [(C1 ∩ C2 )c ] = P (C) = 1,
because C1 ∩ C2 = φ and φc = C.
1.3.11 The probability that he does not win a prize is
990 1000
/ .
5 5
1.3.13 Part (a): We must have 3 even or one even, 2 odd to have an even sum.
Hence, the answer is
10 10 10 10
3
20
0 + 1
20
2 .
3 3
1.3.14 There are 5 mutual exclusive ways this can happen: two “ones”, two “twos”,
two “threes”, two “reds”, two “blues.” The sum of the corresponding proba-
bilities is
2 6 2 6 2 6 5 3 3 5
2 0 + 2 0 + 2 0 + 2 0 + 2 0
8
.
2
1.3.15
48 2
5
(a) 1− 50
0
5
48 2
n 1
(b) 1− 50
0 ≥ , Solve for n.
n
2
1 1
1.3.20 Choose an integer n0 > max{a−1 , (1−a)−1 }. Then {a} = ∩∞
n=n0 a − n , a + n .
Hence by (1.3.10),
1 1 2
P ({a}) = lim P a− ,a+ = = 0.
n→∞ n n n
1.4.2
P [(C1 ∩ C2 ∩ C3 ) ∩ C4 ] = P [C4 |C1 ∩ C2 ∩ C3 ]P (C1 ∩ C2 ∩ C3 ),
and so forth. That is, write the last factor as
P [(C1 ∩ C2 ) ∩ C3 ] = P [C3 |C1 ∩ C2 ]P (C1 ∩ C2 ).
Copyright ©
SOLUTIONS MANUAL
I NTRODUCTION TO
M ATHEMATICAL S TATISTICS
Robert Hogg
University of Iowa
Joseph McKean
Western Michigan University
Allen Craig
Boston Columbus Indianapolis New York San Francisco Upper Saddle River
Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto
Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo
,Contents
1 Probability and Distributions 1
2 Multivariate Distributions 11
3 Some Special Distributions 19
4 Some Elementary Statistical Inferences 31
5 Consistency and Limiting Distributions 49
6 Maximum Likelihood Methods 53
7 Sufficiency 65
8 Optimal Tests of Hypotheses 77
9 Inferences about Normal Models 83
10 Nonparametric and Robust Statistics 93
11 Bayesian Statistics 103
iii
Copyright ©
, Chapter 1
Probability and Distributions
1.2.1 Part (c): C1 ∩ C2 = {(x, y) : 1 < x < 2, 1 < y < 2}.
1.2.3 C1 ∩ C2 = {mary,mray}.
1.2.6 Ck = {x : 1/k ≤ x ≤ 1 − (1/k)}.
1.2.7 Ck = {(x, y) : 0 ≤ x ≤ 1/k, 0 ≤ y ≤ 1/k}.
1.2.8 limk→∞ Ck = {x : 0 < x < 3}. Note: neither the number 0 nor the number 3
is in any of the sets Ck , k = 1, 2, 3, . . .
1.2.9 Part (b): limk→∞ Ck = φ, because no point is in all the sets Ck , k = 1, 2, 3, . . .
1.2.11 Because f (x) = 0 when 1 ≤ x < 10,
Z 10 Z 1
Q(C3 ) = f (x) dx = 6x(1 − x) dx = 1.
0 0
1.2.13 Part (c): Draw the region C carefully, noting that x < 2/3 because 3x/2 < 1.
Thus
Z 2/3 "Z 3x/2 # Z 2/3
Q(C) = dy dx = x dx = 2/9.
0 x/2 0
1.2.16 Note that
25 = Q(C) = Q(C1 ) + Q(C2 ) − Q(C1 ∩ C2 ) = 19 + 16 − Q(C1 ∩ C2 ).
Hence, Q(C1 ∩ C2 ) = 10.
1.2.17 By studying a Venn diagram with 3 intersecting sets, it should be true that
11 ≥ 8 + 6 + 5 − 3 − 2 − 1 = 13.
It is not, and the accuracy of the report should be questioned.
1
Copyright ©
, 2 Probability and Distributions
1.3.3
1 1 1 1/2
P (C) = + + + ··· = = 1.
2 4 8 1 − (1/2)
1.3.6 Z Z Z
∞ 0 ∞
P (C) = e−|x| dx = ex dx + e−x dx = 2 6= 1.
−∞ −∞ 0
We must multiply by 1/2.
1.3.8
P (C1c ∪ C2c ) = P [(C1 ∩ C2 )c ] = P (C) = 1,
because C1 ∩ C2 = φ and φc = C.
1.3.11 The probability that he does not win a prize is
990 1000
/ .
5 5
1.3.13 Part (a): We must have 3 even or one even, 2 odd to have an even sum.
Hence, the answer is
10 10 10 10
3
20
0 + 1
20
2 .
3 3
1.3.14 There are 5 mutual exclusive ways this can happen: two “ones”, two “twos”,
two “threes”, two “reds”, two “blues.” The sum of the corresponding proba-
bilities is
2 6 2 6 2 6 5 3 3 5
2 0 + 2 0 + 2 0 + 2 0 + 2 0
8
.
2
1.3.15
48 2
5
(a) 1− 50
0
5
48 2
n 1
(b) 1− 50
0 ≥ , Solve for n.
n
2
1 1
1.3.20 Choose an integer n0 > max{a−1 , (1−a)−1 }. Then {a} = ∩∞
n=n0 a − n , a + n .
Hence by (1.3.10),
1 1 2
P ({a}) = lim P a− ,a+ = = 0.
n→∞ n n n
1.4.2
P [(C1 ∩ C2 ∩ C3 ) ∩ C4 ] = P [C4 |C1 ∩ C2 ∩ C3 ]P (C1 ∩ C2 ∩ C3 ),
and so forth. That is, write the last factor as
P [(C1 ∩ C2 ) ∩ C3 ] = P [C3 |C1 ∩ C2 ]P (C1 ∩ C2 ).
Copyright ©
