Homework 1 Solutions
January 24, 2009
Problem 1
First we check that
X
(
t
)
is stationary.
Let (
a
1
, . . . , a
n
) be any
n
tuple with
a
i
=
±
1. We need
to verify that
P
(
X
(
t
1
) =
a
1
, X
(
t
2
) =
a
2
, . . . , X
(
t
n
) =
a
n
) =
P
(
X
(
t
1
+
h
) =
a
1
, X
(
t
2
+
h
) =
a
2
, . . . , X
(
t
n
+
h
) =
a
n
)
The left hand side has the form
P
[
X
(
t
1
) =
a
1
, X
(
t
2
) =
a
2
, . . . , X
(
t
n
) =
a
n
] =
=
P
[
ξ
(

1)
N
t
1
=
a
1
, ξ
(

1)
N
t
2
=
a
2
, . . . , ξ
(

1)
N
tn
=
a
n
] =
=
P
[
ξ
(

1)
N
t
1
=
a
1
, ξ
(

1)
N
t
1
+
N
t
2

N
t
1
=
a
2
, . . . , ξ
(

1)
N
t
n

1
+
N
tn

N
t
n

1
=
a
n
] =
=
P
[
ξ
(

1)
N
t
1
=
a
1
,
(

1)
N
t
2

N
t
1
=
a
2
a
1
, . . . ,
(

1)
N
tn

N
t
n

1
=
a
n
a
n

1
] =
=
P
[
ξ
(

1)
N
t
1
=
a
1
]
P
[(

1)
N
t
2

N
t
1
=
a
2
a
1
]
. . . P
[(

1)
N
tn

N
t
n

1
=
a
n
a
n

1
]
,
where in the last step we used that the Poisson process
N
t
has independent increments and
ξ
is independent
of it. In the same way the right hand side has the form
P
[
ξ
(

1)
N
t
1
=
a
1
]
P
[(

1)
N
t
2
+
h

N
t
1
+
h
=
a
2
a
1
]
. . . P
[(

1)
N
tn
+
h

N
t
n

1
+
h
=
a
n
a
n

1
]
Now noting that the random variables
N
t
i
+
h

N
t
i

1
+
h
and
N
t
i

N
t
i

1
have the same (Poisson, parameter
t
i

t
i
+1
) distribution we see that the above expressions are identical except for the first terms. But since
ξ
and
N
t
are independent
P
[
ξ
(

1)
N
s
= 1] =
P
[
ξ
= 1]
P
[(

1)
N
s
= 1] +
P
[
ξ
=

1]
P
[(

1)
N
s
=

1] =
= 1
/
2
(
P
[(

1)
N
s
= 1] +
P
[(

1)
N
s
=

1]
)
= 1
/
2
and similarly
P
[
ξ
(

1)
N
s
=

1] = 1
/
2, so they do not depend on
s
. This means that the first terms in the
products are also the same. We have thus verified directly that
X
(
t
) is stationary.
Now we compute the mean and covariance (autocorrelation).
The mean is easy, since
E
[
ξ
] = 0 implies
E
[
X
(
t
)] =
E
[
ξ
(

1)
N
t
] =
E
[
ξ
]
E
[(

1)
N
t
] = 0. To find the covariance, suppose that
s
≤
t
, so that
R
(
X
t
, X
s
) =
E
[
X
t
X
s
] =
E
[

ξ
2
(

1)
N
t
+
N
s

] =
E
[(

1)
N
t

N
s
+2
N
s
] =
E
[(

1)
N
t

N
s
] =
= (

1)
P
(
N
t

N
s
= odd) + (1)
P
(
N
t

N
s
= even) = 1

2
P
(
N
t

N
s
= odd)
To compute the latter observe that
N
t

N
s
is Poisson with parameter
t

s
and so
P
(
N
t

N
s
= odd) =
∞
summationdisplay
k
=0
e

λ
λ
2
k
+1
(2
k
+ 1)!
=
e

λ
e
λ

e

λ
2
=
1

e

2
λ
2
.
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 Winter '08
 Papanicolaou
 Differential Equations, Equations, Normal Distribution, Probability theory, 1 k, 2k, 0 k, 22k

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