assignment9

32 we will show that the chain is reversible by

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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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