I am having difficulty with the following problem. I tried conditioning on T_n but I am unsure how to proceed with that conditional expectation. Thanks for the help!

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$\Pr(N(t)=n\text{ and }T_{N(t)+1} > t+s)$ is the probability that the number of renewals before time $t$ is exactly $n$ and the time until the next renewal after that is more than $s$. This is $\Pr(N(t)=n) \cdot \Pr(T_{N(t)+1} > t+s)$, because the two events are independent because they are about non-overlapping time intervals in a Poisson process. The waiting time until the next renewal is exponentially distributed with mean $1/\lambda$. So the probability that it exceeds $s$ is $e^{-\lambda s}$. Hence the probability you're looking for is $$ \frac{(\lambda t)^n e^{-\lambda t}}{n!}\cdot e^{-\lambda s} = \frac{(\lambda t)^n e^{-\lambda (t+s)}}{n!}. $$

  • $\begingroup$ Thank you! Would there also be an alternative way of doing this by conditioning on T_n? Much appreciated. $\endgroup$ – icobes Mar 6 '12 at 4:19
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    $\begingroup$ Another way to look at it is that you're looking for the probability that exactly $n$ renewals happen before time $t$ and none happen between then and time $t+s$. So you have two independent Poisson random variables, one with expectation $\lambda t$ and the other $\lambda(s-t)$. The probability that the first equals $n$ is $(\lambda t)^n e^{-\lambda t}/n!$. The probability that the second equals $0$ is $e^{-\lambda s}$. $\endgroup$ – Michael Hardy Mar 6 '12 at 13:50

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