# 9 which looks the same as the result in discrete time

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Unformatted text preview: . .} Vm = 0 otherwise Vm = 1 if a renewal occurs at time m, i.e., if Tn visits m. Let An = min{n m : m n, Vm = 1} be the age and let Zn = min{m n : m n, Vm = 1} be the residual life. An example should help clarify the deﬁnitions: n Vn An Zn 0 1 0 0 1 0 1 3 2 0 2 2 3 0 3 1 4 1 0 0 5 0 1 2 6 0 2 1 7 1 0 0 8 1 0 0 9 0 1 4 10 0 2 3 11 0 3 2 12 0 4 1 13 1 0 0 As we can see from the concrete example, values taken in an excursion away from 0 are j, j1 , . . . 1 in the residual life chain and 1, 2, . . . j in the age chain so we will have n n 1X 1X P (Am = i) = lim P (Zm = i) n!1 n n!1 n m=1 m=1 lim From this we see that it is enough to study one of the two chains. We choose with Zn since it is somewhat simpler. It is clear that if Zn = i &gt; 0 then Zn+1 = i 1. When Zn = 0, a renewal has just occurred. If the time to the next renewal is k then Zn+1 = k 1. To check this note that Z4 = 0 and the time to the next renewal is 3 (it occurs at time 7) so Z5 = 2. Thus Zn is a Markov chain with state space S = {0, 1, 2, . . .} and transition probabili...
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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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