MATH 507b ASSIGNMENT 5 SOLUTIONS
SPRING 2008
Prof. Alexander
Chapter 5:
(4.3) The stationary measure is unique up to constant multiples, which means the ratio of
the stationary measure at any two points
y
and
z
doesn’t depend on the base point, say
y
or some other
x
. That is,
μ
x
(
z
)
μ
x
(
y
)
=
μ
y
(
z
)
μ
y
(
y
)
=
μ
y
(
z
)
,
or equivalently,
μ
x
(
z
) =
μ
x
(
y
)
μ
y
(
z
). (Note that the chain being irreducible means that none
of these quantities is 0.)
(4.4)
μ
(
x
) = 1 for all
x
defines one stationary measure, and the measure
μ
0
corresponding
to the base point 0 is another. Uniqueness up to constant multiples means
1 =
μ
(
k
)
μ
(0)
=
μ
0
(
k
)
μ
0
(0)
=
μ
0
(
k
)
,
for all
k.
μ
0
(
k
) is exactly the expected number of visits to
k
between visits to 0.
(4.8) Irreducibility means there exits a path
γ
from
x
to
y
, say
x
→
x
1
→ · · · →
x
n

1
→
y
,
with all transitions having positive probability, and all
x
i
6
=
x
. Let
A
be the event that this
path is followed. Then
∞
> E
x
T
x
≥
E
x
(
T
x
1
A
) =
E
x
(
T
x

A
)
P
x
(
A
)
,
and
P
x
(
A
)
>
0, so
E
X
(
T
x

A
)
<
∞
. Hence by the Markov property,
∞
> E
x
(
T
x

A
) =
E
x
(
T
x

T
x
> n, X
n
=
y
) =
E
y
(
n
+
T
x
) =
n
+
E
y
T
x
.
Thus
E
y
T
x
<
∞
.
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 Spring '08
 Alexander
 Math, Probability, Trigraph, kN, stationary measure

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