A similar argument shows that 4 is recurrent in

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Unformatted text preview: ith probability one, i.e., P0 (T0 = 1) = 1 and hence ⇢00 = 1. A similar argument shows that 4 is recurrent. In general if y is an absorbing state, i.e., if p(y, y ) = 1, then y is a very strongly recurrent state – the chain always stays there. To check the transience of the interior states, 1, 2, 3, we note that starting from 1, if the chain goes to 0, it will never return to 1, so the probability of never returning to 1, P1 (T1 = 1) p(1, 0) = 0.6 > 0 13 1.3. CLASSIFICATION OF STATES Similarly, starting from 2, the chain can go to 1 and then to 0, so P2 (T2 = 1) p(2, 1)p(1, 0) = 0.36 > 0 Finally, for starting from 3, we note that the chain can go immediately to 4 and never return with probability 0.4, so P3 (T3 = 1) p(3, 4) = 0.4 > 0 In some cases it is easy to identify recurrent states. Example 1.13. Social mobility. Recall that the transition probability is 1 2 3 1 .7 .3 .2 2 .2 .5 .4 3 .1 .2 .4 To begin we note that no matter where Xn is, there is a probability of at least 0.1...
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