Solutions Manual for
Statistical Inference, Second Edition
George Casella University of Florida
Roger L. Berger
North Carolina State University Damaris Santana
University of Florida
Second Edition 3-13
c. (i) h(x) = 1 I
(x), c(α) = α α > 0, w
(α) = α, w
(α) = α,
α
x {0 {1, -2, 2.5}].
2λ
3.35 a. In Exercise 3.34(a) w1(λ) = 1
and for a n(eθ, eθ ), w1(θ) = 1 .
2eθ
b. EX = μ = αβ, then β = μ . Therefore h(x) = 1 I
(x),
α
α x {0 0, w1
(α) = α, w2
(α) = α , t1(x) = log(x), t2(x) = ?x.
c. From (b) then (α1, . . . , αn, β1, . . . , βn) = (α1, . . . , αn, α1 , . . . , αn )
μ μ
3.37 The pdf ( 1 )f ( (x?μ) ) is symmetric about μ because, for any ? > 0,
o σ
1 f .(μ+?)?μ . =
σ σ
1 f . ? . =
σ σ
1 f . ? . =
?
σ σ
1 f .(μ??)?μ . . σ σ
Thus, by Exercise 2.26b, μ is the median.
3.38 P (X > xα) = P (σZ + μ > σzα + μ) = P (Z > zα) by Theorem 3.5.6.
3.39 First take μ = 0 and σ = 1.
a. The pdf is symmetric about 0, so 0 must be the median. Verifying this, write
? ∞ 1 1
1 1 .∞
1 . π . 1
.
P (Z ≥ 0) =
0
π 1+z2 dz =
tan?
π
(z). =
.0 π
2 ?0 = 2 .
b. P (Z ≥ 1) = 1 tan?1(z).∞ = 1
π ? π = 1 . By symmetry this is also equal to P (Z ≤ ?1).
π .1
π . 2 4 . 4
2
4
Writing z = (x ? μ)/σ establishes P (X ≥ μ) = 1
and P (X ≥ μ + σ) = 1 .
3.40
σ
Let X ～ f (x) have mean μ and variance σ2. Let Z = X ?μ . Then
EZ =
. 1 .
σ
E(X ? μ) = 0
and
VarZ = Var
. X ? μ .
σ
. 1 .
= σ2
Var(X ? μ) =
. 1 .
σ2
σ2
VarX =
σ2
= 1.
Then compute the pdf of Z, fZ (z) = fx(σz + μ)· σ = σfx(σz + μ) and use fZ (z) as the standard
pdf.
3.41 a. This is a special case of Exercise 3.42a.
b. This is a special case of Exercise 3.42b.
3.42 a. Let θ1 > θ2. Let X1 ～ f (x ? θ1) and X2 ～ f (x ? θ2). Let F (z) be the cdf corresponding to
f (z) and let Z ～ f (z).Then
F (x | θ1) = P (X1 ≤ x) = P (Z + θ1 ≤ x) = P (Z ≤ x ? θ1) = F (x ? θ1)
≤ F (x ? θ2) = P (Z ≤ x ? θ2) = P (Z + θ2 ≤ x) = P (X2 ≤ x)
= F (x | θ2).
3-14 Solutions Manual for Statistical Inference
The inequality is because x ? θ2 > x ? θ1, and F is nondecreasing. To get strict inequality for some x, let (a, b] be an interval of length θ1 ? θ2 with P (a < Z ≤ b) = F (b) ? F (a) > 0.
Let x = a + θ1. Then
F (x | θ1) = F (x ? θ1) = F (a + θ1 ? θ1) = F (a)
< F (b) = F (a + θ1 ? θ2) = F (x ? θ2) = F (x | θ2).
b. Let σ1 > σ2. Let X1 ～ f (x/σ1) and X2 ～ f (x/σ2). Let F (z) be the cdf corresponding to
f (z) and let Z ～ f (z). Then, for x > 0,
F (x | σ1) = P (X1 ≤ x) = P (σ1Z ≤ x) = P (Z ≤ x/σ1) = F (x/σ1)
≤ F (x/σ2) = P (Z ≤ x/σ2) = P (σ2Z ≤ x) = P (X2 ≤ x)
= F (x | σ2).
The inequality is because x/σ2 > x/σ1 (because x > 0 and σ1 > σ2 > 0), and F is nondecreasing. For x ≤ 0, F (x | σ1) = P (X1 ≤ x) = 0 = P (X2 ≤ x) = F (x | σ2). To
get strict inequality for some x, let (a, b] be an interval such that a > 0, b/a = σ1/σ2 and
P (a < Z ≤ b) = F (b) ? F (a) > 0. Let x = aσ1. Then
F (x | σ1) = F (x/σ1) = F (aσ1/σ1) = F (a)
< F (b) = F (aσ1/σ2) = F (x/σ2)
= F (x | σ2).
y
3.43 a. FY (y|θ) = 1 ? FX ( 1 |θ) y > 0, by Theorem 2.1.3. For θ1 > θ2,
FY (y|θ1) = 1 ? FX
. 1 . .
.
. θ1
y
.
≤ 1 ? FX
. 1 . .
.
. θ2
y
.
= FY (y|θ2)
for all y, since FX (x|θ) is stochastically increasing and if θ1 > θ2, FX (x|θ2) ≤ FX (x|θ1) for all x. Similarly, FY (y|θ1) = 1 ? FX ( 1 |θ1) < 1 ? FX ( 1 |θ2) = FY (y|θ2) for some y, since if
y y
θ1 > θ2, FX (x|θ2) < FX (x|θ1) for some x. Thus FY (y|θ) is stochastically decreasing in θ.
b.
θ2
FX (x|θ) is stochastically increasing in θ. If θ1 > θ2 and θ1, θ2 > 0 then 1
> 1 . Therefore
θ
1
FX (x| 1 ) ≤ FX (x| 1 ) for all x and FX (x| 1 ) < FX (x| 1 ) for some x. Thus FX (x| 1 ) is
θ1 θ2
θ1 θ2 θ
stochastically decreasing in θ.
3.44 The function g(x) = |x| is a nonnegative function. So by Chebychev’s Inequality,
P (|X| ≥ b) ≤ E|X|/b.
Also, P (|X| ≥ b) = P (X2 ≥ b2). Since g(x) = x2 is also nonnegative, again by Chebychev’s Inequality we have
P (|X| ≥ b) = P (X2 ≥ b2) ≤ EX2/b2.
For X ～ exponential(1), E|X| = EX = 1 and EX2 = VarX + (EX)2 = 2 . For b = 3,
E|X|/b = 1/3 > 2/9 = EX2/b2.
Thus EX2/b2 is a better bound. But for b = √2,
√
E|X|/b = 1/
Thus E|X|/b is a better bound.
2 < 1 = EX2/b2.
Second Edition 3-15
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