Adding waiting times, from exponential to Gamma

Question

Suppose the gaps between consecutive trades are independent exponential waits, each with rate $\lambda$ . You want to know the distribution of the total time until the $k$ -th trade.

Identify the distribution of the sum of $k$ independent exponentials, give its mean and variance, and connect it to the Poisson process.

Accepted Answer

The sum is Gamma (Erlang). Adding k independent Exponential(\lambda) waits gives a Gamma(k, \lambda) distribution, when the shape k is a positive integer this is also called the Erlang distribution. Its density is f(t) = \frac{\lambda^k\, t^{k-1} e^{-\lambda t}}{(k-1)!}, \qquad t 0. One clean way to see this: each exponential has MGF \lambda/(\lambda - s), so the sum of k independent ones has MGF \big(\lambda/(\lambda-s)\big)^k, which is exactly the Gamma(k,\lambda) MGF. Mean and variance. Expectation and variance add across the independent pieces. A single Exponential(\lambda) has mean 1/\la…

Adding waiting times, from exponential to Gamma

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