0
$\begingroup$

Let $S_n$ denote the time of the $n$th event of the renewal process $\{N(t),t\geq0\}$ having interarrival distribution F.

I need to find $P(N(t)=n|S_n=y).$

Clearly, if $y$ is the time at which the $n$th event occurs, and if by time $t$, $n$ events have occurred, then $y$ must be less than or equal to $t$. Additionally, I know that $$P(N(t)=n|S_n=y)=\frac{P(N(t)=n,S_n=y)}{P(S_n=y)}$$

What I am not sure how to do is find each of these distributions. I know that $P(N(t)=n)=F_n(t)-F_{n+1}(t)$, which I think may be useful. But otherwise, I'm really not sure what to do.

$\endgroup$
0
$\begingroup$

It may help to define the iid inter-renewal times $\{X_i\}$, then possibly draw a picture.

Case 1: If $y>n$ then $P[N(t)=n|S_n=y] = [\mbox{fill in blank}]$.

Case 2: If $y\leq n$ then $P[N(t)=n|S_n=y] = P[X_{n+1} > \mbox{fill in blank}]$.

$\endgroup$
0
$\begingroup$

You basically gave the answer yourself; you know when the n$^{th}$ arrival happened, then t cannot be less than y, and the probability that no arrivals happen between t and y is just making sure that the n+1 arrival happens later than t; if you know the distribution of the arrival times, you can make calculate the probability that no arrivals happen between y and t.

$\endgroup$

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.