# For states i 0 we let pi i 1 1 in words the chain

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Unformatted text preview: ber is even, then it will be odd on the next one step. This alternation between even and odd means that it is impossible to be back where we started after an odd number of steps. In symbols, if n is odd then pn (x, x) = 0 for all x. To see that the problem in the last example can occur for multiples of any number N consider: Example 1.22. Renewal chain. We will explain the name in Section 3.3. For the moment we will use it to illustrate “pathologies.” Let fk be a distribution on the positive integers and let p(0, k 1) = fk . For states i &gt; 0 we let p(i, i 1) = 1. In words the chain jumps from 0 to k 1 with probability fk and then walks back to 0 one step at a time. If X0 = 0 and the jump is to k 1 then it returns to 0 at time k . If say f5 = f15 = 1/2 then pn (0, 0) = 0 unless n is a multiple of 5. The period of a state is the largest number that will divide all the n 1 for which pn (x, x) &gt; 0. That is, it is the greatest common divisor of Ix = {n 1 : pn (x, x) &gt; 0}. To check that this deﬁnition works correctly, we note that in Example 1.21, {n 1 : pn (x, x) &gt; 0}...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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