Solutions to Homework Set No. 5 – Probability Theory (235A), Fall 2011
Solution to 1.
(a)
Z
∞
0
P
(
X
≥
x
)
dx
=
Z
∞
0
Z
∞
y
=
x
dF
(
y
)
dx
=
Z
∞
0
Z
y
0
dxdF
(
y
) =
Z
∞
0
ydF
(
y
) =
E
(
X
)
.
Here, “
dF
(
x
)” means
f
(
x
)
dx
if
X
has a density. Otherwise, the integral represents a
LebesgueStieltjes integral. Note that if the formula holds for all r.v.’s with density
(or discrete r.v.’s, for which it is also easy to verify) then by approximation it can
easily be seen to hold for general nonnegative r.v.’s.
(b) Let
P
(

X
n
 ≥
x
) =
P
(

X
 ≥
x
) =
g
(
x
) for all
n
. Since
g
(
x
) is decreasing,
∞
X
n
=1
P
(

X
n
 ≥
n
)
≤
Z
∞
0
P
(

X
n
 ≥
x
)
dx
=
E

X
n

=
E

X
1

,
and so if
E

X
1

<
∞
, by BorelCantelli lemma,
P
(

X
n
 ≥
n
i.o.) = 0. Similarly,
1 +
∞
X
n
=1
P
(

X
n
 ≥
n
)
≥
E

X
1

,
(1)
so if
E

X
1

=
∞
, then
P
(

X
n
 ≥
n
i.o.) = 1.
(c) Let
S
n
=
X
1
+
. . .
+
X
n
. Note that
S
n
n

S
n

1
n

1
=
X
n
n
+
S
n

1
n
(
n

1)
,
so on the event
{
(
X
n
)
+
≥
n
i.o.
} ∩ {
(
X
n
)

≥
n
i.o.
}
(which has probability 1 by the
previous section), if
S
n

1
n
(
n

1)
→
0 as
n
→ ∞
(which would have to happen in order
for lim
n
→∞
1
n
S
n
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 Fall '11
 DanRomik
 Probability, Probability theory, Order theory, dt, Monotonic function, Convex function

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