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Math 136  Stochastic Processes
Homework Set 5, Fall 2008, Due: October 29
See Victor Hu on questions on grading
1. Exercise 3.2.21.
Consider the random variables
b
S
k
of Example 1.4.13.
(a) Applying Proposition 3.2.6 verify that the corresponding characteristic functions are
Φ
b
S
k
(
θ
) = [cos(
θ/
√
k
)]
k
.
ANS:
Let
X
i
for
i
= 1
...k
be i.i.d. RVs with
P
(
X
i
=

1) =
P
(
X
i
= 1) = 1
/
2. Then using
Proposition 3.2.6 for the ﬁrst equality we have
Φ
b
S
k
(
θ
) =
k
Y
i
=1
Φ
X
i
/
√
k
(
θ
) =
{
Φ
X
1
/
√
k
(
θ
)
}
k
=
{
E
(
e
iθX
1
/
√
k
)
}
k
=
{
(
e

iθ/
√
k
+
e
iθ/
√
k
)
/
2
}
k
=
{
cos(
θ/
√
k
)
}
k
(b) Recalling that
δ

2
log(cos
δ
)
→ 
0
.
5 as
δ
→
0, ﬁnd the limit of Φ
b
S
k
(
θ
) as
k
→ ∞
while
θ
∈
IR is
ﬁxed.
ANS:
Note that Φ
b
S
k
(
θ
) = exp
{
k
log[cos(
θ/
√
k
)]
}
. Taking
δ
=
θ/
√
k
and exploiting the continuity
of the exponential function we get Φ
b
S
k
(
θ
)
→
e

θ
2
/
2
.
(c) Suppose random vectors
X
(
k
)
and
X
in IR
n
are such that Φ
X
(
k
)
(
θ
)
→
Φ
X
(
θ
) as
k
→ ∞
, for any
ﬁxed
θ
. It can be shown that then the laws of
X
(
k
)
, as probability measures on IR
n
, must converge
weakly in the sense of Deﬁnition 1.4.20 to the law of
X
. Explain how this fact allows you to verify
the C.L.T. statement
b
S
n
L
→
G
of Example 1.4.13.
ANS:
From the previous part we see that Φ
b
S
k
(
θ
)
→
Φ
G
(
θ
) for all
θ
, where
G
is a standard normal
random variable. Then what has been stated above implies that
b
S
k
L
→
G
.
2. Exercise 3.2.22.
Consider the random vectors
X
(
k
)
= (
1
√
k
S
k/
2
,
1
√
k
S
k
) in IR
2
, where
k
= 2
,
4
,
6
,...
is
even, and
S
k
is the
simple random walk
of Deﬁnition 3.1.2, with
P
(
ξ
1
=

1) =
P
(
ξ
1
= 1) = 0
.
5.
(a) Verify that Φ
X
(
k
)
(
θ
) = [cos((
θ
1
+
θ
2
)
/
√
k
)]
k/
2
[cos(
θ
2
/
√
k
)]
k/
2
,
where
θ
= (
θ
1
,θ
2
).
ANS:
Here Φ
X
(
k
)
(
θ
) =
E
exp(
θ
1
S
k/
2
/
√
k
+
θ
2
S
k
/
√
k
) and since
S
k
=
S
k/
2
+
˜
S
k/
2
where
˜
S
k/
2
is
independent, identically distributed copy of
S
k/
2
, we have
E
exp(
θ
1
S
k/
2
/
√
k
+
θ
2
S
k
/
√
k
) =
E
exp[(
θ
1
+
θ
2
)
S
k/
2
/
√
k
]
E
exp[
θ
2
S
k/
2
/
√
k
]
The required result now follows by noting that
S
k
/
√
k
has the same distribution as
b
S
k
from Exercise
3.2.21, so their characteristic functions are equal.
1
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View Full Document(b) Find the mean vector
μ
and the covariance matrix Σ of a Gaussian random vector
X
for which
Φ
X
(
k
)
(
θ
) converges to Φ
X
(
θ
) as
k
→ ∞
.
ANS:
Same approach as in part (b) of the Exercise 3.2.21 gives Φ
X
(
k
)
(
θ
)
→
e

(
θ
1
+
θ
2
)
2
/
4
e

θ
2
2
/
4
. We
now need
μ
and Σ such that
exp[

(
θ
,
Σ
θ
)
/
2 +
i
(
θ
,μ
)] = exp[(

θ
2
1
/
2

θ
1
θ
2

θ
2
2
)
/
2]
which gives
μ
= (0
,
0), Σ
11
= Σ
12
= Σ
21
= 1
/
2 and Σ
22
= 1.
(c) Upon appropriately generalizing what you did in part (b), I claim that the
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 Fall '08
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