Dr. Hackney STA Solutions pg 78 - 5-12 Solutions Manual for...

5-12 Solutions Manual for Statistical Inference
c. S 2
n
→ σ 2 in probability. By a), S
n
= S 2
n
→
√
σ 2 = σ in probability. By b), σ/S
n
→ 1 in
probability.
5.33 For all > 0 there exist N such that if n > N , then P ( X
n
+ Y
n
> c ) > 1 - . Choose N
1
such
that P ( X
n
> - m ) > 1 - / 2 and N
2
such that P ( Y
n
> c + m ) > 1 - / 2. Then
P ( X
n
+ Y
n
> c ) ≥ P ( X
n
> - m, + Y
n
> c + m ) ≥ P ( X
n
> - m ) + P ( Y
n
> c + m ) - 1 = 1 - .
5.34 Using E
¯
X
n
= μ and Var
¯
X
n
= σ 2 /n , we obtain
E
√
n (
¯
X
n
- μ )
σ
=
√
n
σ
E(
¯
X
n
- μ ) =
√
n
σ
( μ - μ ) = 0 .
Var
√
n (
¯
X
n
- μ )
σ
=
n
σ 2
Var(
¯
X
n
- μ ) =
n
σ 2
Var
¯
X =
n
σ 2
σ 2
n
= 1 .
5.35 a. X
i
∼ exponential(1). μ
X
= 1, Var X = 1. From the CLT,
¯
X
n
is approximately n(1 , 1 /n ). So
¯
X
n
- 1
1 /
√
n
→ Z ∼ n(0 , 1) and P
¯
X
n
- 1
1 /
√
n
≤ x → P ( Z ≤ x ) .
b.
d
dx
P ( Z ≤ x ) =
d
dx
F
Z
( x ) = f
Z
( x ) =
1
√
2 π
e
- x
2
/ 2
.
d
dx
P
¯
X
n
- 1
1 /
√
n
≤ x
=
d
dx
n
i =1
X
i
≤ x
√
n + n W =
n
i =1
X
i
∼ gamma( n, 1)
=
d
dx
F
W
( x
√
n + n ) = f
W
( x
√
n + n ) ·
√
n =
1
Γ( n )
( x
√
n + n )
n - 1
e
- ( x
√
n + n )
√
n.
Therefore, (1 / Γ( n ))( x
√
n + n )
n - 1
e
- ( x
√
n + n )
√
n ≈
1
√
2 π
e
- x
2
/ 2
as n → ∞ . Substituting x = 0
yields n ! ≈ n
n +1 / 2
e
- n
√