assignment9

# 7 the steady state probabilities for the sampled time

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Unformatted text preview: = πi−1 = 3 i≥1 i≥0 Thus, π0 = πi = 1 6 �� 5 2i , for i ≥ 1. 12 3 b) The transition rates are qij = Pij vi = Pij 2i . Therefore, q01 = 1, qi,i+1 = (2/5)2i and qi,i−1 = (3/5)2i for i = 1, 2, ....The steady state probabilities pi for the Markov process are proportional to πi /vi = (5/12)(1/3)i for i ≥ 1 and π0 /v0 = 1/6. Normalizing so that pi ’s sum to one,we get: 6 p0 = pi = 4 9 �� 10 1 i , for i ≥ 1 93 The pi ’s decay faster than πi ’s. This is because the transition rate vi increases with i. So, the mean time until a transition from the current state decreases with i. c) Since vi is growing unboundedly with increasing i and Pi,i+1 and Pi,i−1 is constant for all i > 0, qij = Pij vi is also growing unboundedly with i for j = i + 1 and j = i − 1. Thus for any δ > 0, and for large enough n, the transit...
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## This document was uploaded on 03/19/2014 for the course EECS 6.262 at MIT.

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