Math 280A, Fall 2011
Homework 5 Solutions
1.
Problem 16 (page 116). If
X
and
Y
are independent random variables, then
X
and 1
B
(
Y
) are
also independent random variables, for each Borel set
B
. Also
X
is integrable by hypothesis, and
so is 1
B
(
Y
) (being bounded). The product
X
·
1
B
(
Y
) is therefore integrable. Finally,
E
[
X
1
{
Y
∈
B
}
] =
E
[
X
·
1
B
(
Y
)] =
E
[
X
]
·
E
[1
B
(
Y
)] =
E
[
X
]
·
P
[
Y
∈
B
]
.
2.
Problem 21 (page 117). I shall let
X
n
denote the
n
th
binary digit of
X
, so that
X
1
, X
2
, . . .
is
an iid sequence of Bernoulli random variables with
P
[
X
n
= 1] =
P
[
X
n
= 0] = 1
/
2 for all
n
.
(a) As noted in class, for each
k
= 0
,
1
,
2
, . . .
,
{
l
n
=
k
}
=
{
X
n
= 0
, X
n
+1
= 0
, . . . , X
n
+
k

1
= 0
, X
n
+
k
= 1
}
so that
P
[
l
n
=
k
] = 2

k

1
,
k
= 0
,
1
,
2
, . . .
by independence.
(b) Let us fix
k
≥
0 and define
A
n
:=
{
l
n
=
k
}
for
n
= 1
,
2
,
. . . .
Observe that
A
1
depends
only on
X
1
, X
2
, . . . , X
k
+1
;
A
k
+2
depends only on
X
k
+2
,
X
k
+3
, . . . , X
2
k
+2
; etc. The point is that
the sequence of events
A
(
n

1)(
k
+1)+1
,
n
= 1
,
2
,
3
, . . . ,
is an iid sequence with
P
[
A
(
n

1)(
k
+1)+1
] = 1
/
2
k
+1
for all
n
. Since
∑
n
2

k

1
= +
∞
, the second
BorelCantelli lemma tells us that
P
[
A
(
n

1)(
k
+1)+1
i.o.(
n
)] = 1
.