Math 136  Stochastic Processes
Homework Set 4, Winter 2008, Due: February 6
1. Exercise 2.3.16.
Let
Z
= (
X, Y
) be a uniformly chosen point on (0
,
1)
2
. That is,
X
and
Y
are independent
random variables, each having the
U
(0
,
1) measure of Example 1.1.11. Set
T
=
I
A
(
Z
) + 5
I
B
(
Z
) where
A
=
{
0
< x <
1
/
4
,
3
/
4
< y <
1
}
and
B
=
{
3
/
4
< x <
1
,
0
< y <
1
/
2
}
.
(a) Find an explicit formula for the conditional expectation
W
=
E
(
T

X
) and use it to determine the
conditional expectation
U
=
E
(
TX

X
).
ANS:
Note
A
=
A
1
×
A
2
for
A
1
=
{
x
∈
(0
,
1
/
4)
}
, A
2
=
{
y
∈
(3
/
4
,
1)
}
hence
I
A
(
x, y
) =
I
A
1
(
x
)
I
A
2
(
y
).
Similarly
I
B
(
x, y
) =
I
B
1
(
x
)
I
B
2
(
y
) for
B
1
=
{
x
∈
(3
/
4
,
1)
}
, B
2
=
{
y
∈
(0
,
1
/
2)
}
.
Consequently,
T
=
I
A
1
(
X
)
I
A
2
(
Y
)+5
I
B
1
(
X
)
I
B
2
(
Y
). Thus, by the linearity of the C.E. and “taking
out what is known” (Proposition 2.3.15) we have that
W
=
E
(
T

X
) =
I
A
1
(
X
)
E
(
I
A
2
(
Y
)

X
) + 5
I
B
1
(
X
)
E
(
I
B
2
(
Y
)

X
)
.
Further, since
X
and
Y
are independent,
I
A
2
(
Y
) and
I
B
2
(
Y
) are independent of
X
. Thus, we have
that
E
(
I
A
2
(
Y
)

X
) =
E
I
A
2
(
Y
) =
P
(
Y
∈
A
2
) =
1
4
,
with the rightmost identity due to
Y
being uniformly chosen on (0
,
1) with
A
2
an interval of length
1
/
4. Similarly,
E
(
I
B
2
(
Y
)

X
) = 1
/
2, so we have that
W
=
1
4
I
A
1
(
X
) +
5
2
I
B
1
(
X
)
.
Since
X
is bounded, we know that
U
=
E
(
TX

X
) =
XW
by Proposition 2.3.15.
(b) Find the value of
E
((
T
−
W
) sin (
e
X
)).
ANS:
Since sin (
e
X
)
∈
L
2
(Ω
, σ
(
X
)
,
P
) =
H
X
and
W
=
E
(
T

X
) for
T
squareintegrable, this is
zero by Proposition 2.1.2.
(c) Without any computation decide whether
E
W
2
−
E
T
2
is negative, zero, or positive. Explain your
answer.
ANS:
Recall Proposition 2.1.2 that
E
((
T
−
W
)
W
) = 0 (since
W
∈ H
X
).
Hence, with
T
=
W
+ (
T
−
W
) we have that
E
T
2
=
E
W
2
+ 2
E
(
T
−
W
)
W
+
E
(
T
−
W
)
2
=
E
W
2
+
E
(
T
−
W
)
2
.
By part (a) we already know that
P
(
T
negationslash
=
W
)
>
0, hence
E
(
T
−
W
)
2
>
0, implying that
E
W
2
−
E
T
2
is negative.
We end by remarking that a perhaps shorter and more geometric argument can be
1
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made by recalling that the conditional expectation
W
of a square integrable
T
given
G
=
σ
(
X
)
is just an orthogonal projection in a Hilbert space;
T
not being measurable with respect to
G
is
equivalent this projection being strictly normreducing.
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 Fall '08
 2
 Normal Distribution, Variance, Probability theory, Multivariate normal distribution, X

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