# Joint distribution of consecutive renewal times

Consider a discrete analog to the Poisson process. Let the sequence $X_i$ be independent geometrically (with parameter $p$) distributed random variables that signify the inter arrival times of events. Let $S_k$ be the sequence of times at which the $k$th event occurs, ie. the renewal times. $S_k = X_1 + \ldots + X_k$ so that $S_k$ has a negative binomial distribution with parameters $k$ and $p$.

Now consider the joint distribution of two consecutive renewal times.

How do you show that: $P(S_k \leq n ; S_{k+1} = n+j) = P(S_{k+1} = n+1)(1-p)^{j-1}$ for $n \geq k$?

Let $Q_{k,n}$ be the probability that the first $k$ events occur by time $n$ (i.e., $S_k \le n$) and event $k+1$ does not occur by time $n$ (i.e., $S_{k+1} > n$). Then $$P(S_k\le n ; S_{k+1}=n+j) = p(1-p)^{j-1} Q_{k,n},$$ since the probability of $j-1$ consecutive non-events (at times $n+1,n+2,...,n+j-1$) followed by a single event is $p(1-p)^{j-1}$, while $$P(S_{k+1}=n+1) = p Q_{k,n}.$$ The desired equality follows.