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
√
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- Spring '12
- Dr.Hackney
- Statistics, Probability, Xn
-
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