MATH 507b ASSIGNMENT 4 SOLUTIONS
SPRING 2008
Prof. Alexander
Chapter 5:
(2.1) Let
A
∈
σ
(
X
0
, .., X
n
) and
B
∈
σ
(
X
n
, X
n
+1
, ...
). Using Theorem 1.2 Chapter 4 (p. 224)
and the Markov property, since 1
A
∈ F
n
we get
P
μ
(
A
∩
B

X
n
) =
E
μ
(
E
μ
(1
A
1
B
 F
n
)

X
n
)
=
E
μ
(1
A
E
μ
(1
B
 F
n
)

X
n
)
=
E
μ
(1
A
E
μ
(1
B

X
n
)

X
n
)
=
E
μ
(1
B

X
n
)
E
μ
(1
A

X
n
)
.
Here the last equality uses the fact that
E
μ
(1
B

X
n
)
∈
σ
(
X
n
), so it can be pulled out of the
conditional expectation.
(2.2)(i) Let
a
be absorbing, let
δ >
0, let
D
=
{
X
n
=
a
for some
n
≥
1
}
, let
A
=
{
x
:
P
x
(
D
)
≥
δ
), and
h
(
x
) =
P
x
(
D
). We want to apply Theorem 2.3 with
B
m
=
{
a
}
for all
m
and
A
n
=
A
for all
n
. In fact we have
P
(
∪
∞
m
=
n
+1
{
X
m
=
a
} 
X
n
) =
P
X
n
(
D
)
≥
δ
on
{
X
n
∈
A
}
,
so that theorem says
P
(
{
X
n
∈
A
i.o.
}\{
X
n
=
a
i.o.
}
) = 0. Since
a
is absorbing, we have
{
X
n
=
a
i.o.
}
=
D
, so this is equivalent to
P
(
{
X
n
∈
A
i.o.
} ∩
D
c
) = 0. In other words, on
D
c
, except for a null set we have
X
n
∈
A
only finitely often, or equivalently,
h
(
X
n
)
< δ
for
all sufficiently large
n
. Since
δ
is arbitrary, this shows
h
(
X
n
)
→
0 on
D
c
, a.s.
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 Spring '08
 Alexander
 Math, Probability, Trigraph, Px, Xn, ty, eµ

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