MATH 587
Solutions to Assignment 4
November 30, 2004
(1)
E
(
Y
−
X
)
2
=
E
[(
Y
−
E
(
X
G
)) + (
E
(
X
G
)
−
X
)]
2
=
E
[
Y
−
E
(
X
G
)]
2
+
E
[(
Y
−
E
(
X
G
))(
E
(
X
G
)
−
X
)] +
E
[
E
(
X
G
)
−
X
)]
2
. But
E
[(
Y
−
E
(
X
G
))(
E
(
X
G
)
−
X
)] =
E
[
E
(
Y
−
X
G
)(
E
(
X
G
)
−
X
)] =
E
{
E
[
E
(
Y
−
X
G
)(
E
(
X
G
)
−
X
)
G
]
}
=
E
{
E
(
Y
−
X
G
)
E
[
E
(
X
G
)
−
X
G
]
}
= 0. Hence
E
(
Y
−
X
)
2
=
E
[
Y
−
E
(
X
G
)]
2
+
E
[
E
(
X
G
)
−
X
)]
2
. The LHS will be a minimun when
Y
=
E
(
X
G
).
(2)
(a) By Fubini,
E
∑
∞
i
=1
X
i
=
∑
∞
i
=1
EX
i
<
∞
, so
∑
∞
i
=1
Y
i
<
∞
a.s. and therefore
Y
n
→
0 a.s.
(b) Let
W
i
=
X
i
−
µ
. Then
E
(
S
n
−
nµ
)
4
=
E
(
W
1
+
· · ·
+
W
n
)
4
. By independence and since
EW
= 0,
the only terms in the expansion with nonzero expectation are of the form
EW
4
and
E
(
W
2
i
)
E
(
W
2
j
)
where
i
=
j
. It follows that
E
(
S
n
−
nµ
)
4
=
nρ
+ 3
n
(
n
−
1)
σ
4
, where
ρ
=
E
(
X
i
−
µ
)
4
<
∞
. Then
∑
∞
n
=1
EY
n
<
∞
, so
Y
n
→
0 a.s., so
S
n
n
→
µ
a.s.
(c) It is equivalent to show that
1
n
∑
n
i
=1
log
X
i
→ −
1 a.s. This follows from part (b) once we observe
that
E
log
X
=
1
0
(log
x
)
4
dx
=
∞
0
w
4
e
−
w
dw
= Γ(5) = 24 (where
w
=
−
log
x
), and
E
log
X
=
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 '07
 RomanVershynin
 Probability, Trigraph, Sn, g, Cantor set, Lebesgue measure, Sn − nµ

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