Unformatted text preview: from the recurrence σ + Y {σ 2 Y 2 = σY n−1 n −1
, for n = 1.
Y −1 n−1 Y n−2 (n − 1 − 1)/(Y − 1)} n (Y − 1)/(Y − 1). Which completes the inductive argument. For Y = 1, we have V ar(Xn ) = σ 2 +
V ar(Xn−1 ) = 2σ 2 + V ar(Xn−2 ) = nσ 2 + V ar(X0 ) = nσ 2 .
Exercise 5.7:
Using theorem 5.3.2, we will show that the chain is reversible by demonstrating a set
{πi } of steady state probabilities for which πi Pij = πj Pji for all i, j . Thus, we want to ﬁnd
{πi } satisfying
(1) dij
dij
= πj �
πi �
k dik
k djk Where we have used dij = dji in the upper right corner of the above equation. This
�
equation will be satisﬁed if we choose πi to be proportional to k dik . Normalizing {πi }
�
to satisfy i πi = 1, we see that eq. (1) is satisﬁed by
�
k dik
πi = � �
j
k djk 2 Exercise 5.8:
Note that if�i is summed over i, the numerator term becomes the same as the denomi
π
�
nator, so that i π i = 1. Thus, using theorem 5.3.2., it suﬃces to show that π i Pij = π j Pj�i .
We have
πi...
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 Spring '11
 RobertGallager
 Probability theory, Markov chain, Andrey Markov, Random walk, πi Pij, steady state probabilities

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