# Deducing that the sum of exonentials has a gamma distribution

If I have got $\ X_1,...X_n$ all independent and identically distributed with the exponential distribution with parameter $\ 1/\theta$. Then how can I deduce that $\ S=X_1+...+X_n$ has the gamma distribution with paramters $\ (n,\theta)$? The previous question asks me to calculate the moment generating function of S ( I don't know if this is meant to help me do this part) and it says I can use the fact that $\int x^{n-1}e^{-\beta x}dx ~~ = ~~(n-1)!/\beta^n$

This is not a homework question but is a past paper question. Thanks

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The mgf uniquely determines he istribution, so find the mgf of S and show it is the same as the Gamma mgf. The mgf of S is easy o get because it is a sum of independent components. –  guy Jun 12 '12 at 15:57
You can calculate the convolution integral to do this by induction. –  Michael Chernick Jun 12 '12 at 17:18