# Formula for the nth convolution of a laplace random variable

Let x_1, x_2, ... be iid draws from a laplace distribution with scale parameter b. Is there a relatively nice closed form for x_1+x_2+...x_n? I've seen a derivation floating around for when b=1, but I couldn't figure out a generalisation.

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Did any of you work out the exact expression in terms of the scale parameter b? I'm having trouble taking the inverse Fourier transform :( Thank you! –  user3150 Jan 11 '10 at 7:30

The distribution of the n-th convolution of the Laplace distribution can be computed from the characteristic function: exp(i mu t)/(1+b^2 t^2). (See the Wikipedia article). The characteristic function of the n-th convolution becomes exp(i n mu t)/(1+b^2 t^2)n = exp(i n mu t)/((1 - i b t)^n ((1 + i b t)^n) . The inverse Fourier transform can be computed using the residue theorem. The integration contour is closed from the upper or lower half plane according to the sign of (x-n mu).

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One way to generate a Laplace random variable is to generate two IID (independent and identically distributed) exponential random variables and then subtract them: x_i = y_i - z_i with y_i and z_i ~ exponential(parameter=b), and of course everything independent. Then the sum of the x_i is simply (sum y_i) - (sum z_i); each of those two sums have Gamma distributions. To be more specific, since we are summing an integer number of terms, they have Erlang distributions. The difference of two Gammas is called "bilateral gamma", and there are a few papers out there on it. A quick search just found:

Bilateral gamma distributions and processes in financial mathematics Uwe Küchlera, Stefan Tappe

On the shapes of bilateral Gamma densities Uwe Küchlera, Stefan Tappe

It would be nice if someone would write a Wikipedia article about bilateral Gammas, I guess.

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