1
$\begingroup$

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!

enter image description here

$\endgroup$
3
$\begingroup$

$\Pr(N(t)=n\text{ and }T_{N(t)+1} > t+s)$ 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 $$ \frac{(\lambda t)^n e^{-\lambda t}}{n!}\cdot e^{-\lambda s} = \frac{(\lambda t)^n e^{-\lambda (t+s)}}{n!}. $$

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.