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Supposing that I have an interval $n$ units long with $m$ arrivals, and I model arrival using a poisson distribution with $\lambda = m/n$, it's pretty easy to show that the distance between successive arrivals has an exponential distribution $\lambda e^{-\lambda x}$.

However, I'm having a hard time generalizing this to a variation of the gamma distribution when I want to look more than $2$ arrivals (say I have $3$ successive or four successive arrivals). Can someone explain the logic there?

I believe the density should be $\frac{\lambda^{k-1}} {(k-1)!} x^{k-2}e^{-\lambda x}$ but I am not sure how!

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The sum of independent Gamma random variables with the same scale factor (equivalently, same rate factor) is a Gamma random variable with the same scale factor, and the order is the sum of the orders. See, for example, this question. So, apply this to your problem using the fact that exponential random variables are Gamma random variables of order $1$. Your density is not quite right, by the way: it should be $$\frac{\lambda(\lambda x)^{k-1}}{\Gamma(k)}\exp(-\lambda x) = \frac{\lambda^{k}x^{k-1}}{(k-1)!}\exp(-\lambda x)~~ \text{for}~ x>0.$$

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    $\begingroup$ Perfect, exactly what I needed. The paper here: ncbi.nlm.nih.gov/pmc/articles/PMC1950645 threw me a bit with formula 1 (since it showed my density) but after reading your's I realized that they are summing only k-1 terms. Thanks again. $\endgroup$ – user1357015 Mar 31 '13 at 3:50

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