Math 280A, Fall 2011
Homework 6
1.
[Problem 2 (page 153)] (a) Choose
r
= 2
/
3, and apply H¨
older’s inequality with
p
= 3
/
2 (and
conjugate exponent
q
= 3) to obtain
1 =
E
[
X
2
] =
E
[

X

2
/
3

X

4
/
3
]
≤
(
E

X

)
2
/
3
(
E
[
X
4
]
)
1
/
3
.
Taking the 3
/
2 power of the extreme terms in this string of inequalities, we have
1
≤
E

X
 ·
radicalbig
E
[
X
4
]
,
which immediately implies the desired inequality.
(b) More generally, if
E
[
X
2
] = 1 and
E
[
X
2
m
]
<
∞
for some
m
≥
1, then
E

X
 ≥
1
(
E
[
X
2
m
])
1
/
(2
m

2)
.
To see this do the same as in part (a) but now with
r
= (2
m
−
2)
/
(2
m
−
1) and
p
= 1
/r
= (2
m
−
1)
/
(2
m
−
2).
2.
[Problem 5 (page 154)] Splitting
E
[
X
]:
E
[
X
] =
E
[
X
;
X > λ
E
[
X
]] +
E
[
X
;
X
≤
λ
E
[
X
]]
≤
E
[
X
;
X > λ
E
[
X
]] +
λ
E
[
X
]
Moving the extreme righthand term to the other side of the inequality, we obtain
(1
−
λ
)
E
[
X
]
≤
E
[
X
;
X > λ
E
[
X
]]
.
Let us now apply the CauchySchwarz inequality (aka, H¨
older’s inequality with
p
=
q
= 2) to the right
side of this inequality to obtain
(1
−
λ
)
E
[
X
]
≤
(
E
[
X
2
]
)
1
/
2
·
(
P
[
X > λ
E
[
X
]])
1
/
2
.
Squaring this yields
(1
−
λ
)
2
(
E
[
X
])
2
≤
E
[
X
2
]
·
P
[
X > λ
E
[
X
]]
,
which yields the stated inequality upon dividing by
E
[
X
2
].
3.
In this problem let
X
and
Y
be independent random variables with cumulative distribution functions
F
and
G
; thus, for example,
F
(
x
) =
P
[
X
≤
x
]
for
x
∈
R
.
(a) Use Tonelli’s theorem to show that
E
[
F
(
Y
)] =
P
[
X
≤
Y
]
and
E
[
G
(
X
)] =
P
[
Y
≤
X
]
.
(b) Deduce from (a) that
E
[
F
(
Y
)] +
E
[
G
(
X
)] = 1 +
P
[
X
=
Y
]
.
(c) Use Tonelli’s theorem again to show that
P
[
X
=
Y
] =
∑
b
Δ
F
(
b
)
·
Δ
G
(
b
)
, where
Δ
F
(
b
) :=
F
(
b
)
−
F
(
b
−
)
,
G
(
b
) :=
G
(
b
)
−
G
(
b
−
)
, and the sum extends over the (at most countable) set of real
numbers
b
with
Δ
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 Fall '08
 Driver,B
 Math, Probability, Convergence, dt, Xn, Dominated convergence theorem, Tonelli

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