Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n \geq k$?

share|improve this question
add comment

2 Answers

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.

share|improve this answer
    
+1. Nice decomposition. –  Did Jun 19 '12 at 7:43
add comment

\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*}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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