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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 $n \geq k$?

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2 Answers 2

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.

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+1. Nice decomposition. –  Did Jun 19 '12 at 7:43

\begin{align*} P(S_k \le n, S_{k+1} = n+j) &= \sum_{\nu = k}^n P(S_k = \nu, S_{k+1} = k+j)\\\ &= \sum_{\nu = k}^n P(S_k = \nu, X_{k+1} = n+j-\nu)\\\ &= \sum_{\nu = k}^n P(S_k = \nu)P(X_{k+1} = n+j-\nu)\\\ &= \sum_{\nu = k}^n P(S_k = \nu)(1-p)^{n+j-\nu -1}p\\\ &= (1-p)^{j-1}\sum_{\nu = k}^n P(S_k = \nu)(1-p)^{n-\nu}p\\\ &= (1-p)^{j-1}\sum_{\nu = k}^n P(S_k = \nu)P(X_{k+1} = n+1-\nu)\\\ &= (1-p)^{j-1}\sum_{\nu = k}^n P(S_k = \nu, X_{k+1} = n+1-\nu)\\\ &= (1-p)^{j-1} P(S_k \le n, S_{k+1} = n+1)\\\ &= (1-p)^{j-1} P(S_{k+1} = n+1). \end{align*}

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