From: "Charles H. Giffen" <chg4k@virginia.edu>
Subject: Re: Q: random sum of indep. exponentials?
Date: Thu, 20 May 1999 12:39:50 -0400
Newsgroups: sci.math
To: David Grabiner <grabiner@math.lsa.umich.edu>
Keywords: sum of random (number of) exponentials

David Grabiner wrote:
> 
> research@grimm.cogsci.uiuc.edu (research) writes:
> 
> 
> > Hello!
> 
> > I would like to know how to get
> > the random sum of the indep. exponentials.
> > If someone can give me the idea or the source,
> > I really appreciate that.
> > Thank you.
> 
> The sum of n independent, identically distributed exponentials is a
> gamma distribution, with density
> 
> x^(n-1)e^(-x)/(n-1)!
> 
> if the exponentials have parameter 1.
> 
> --
> David Grabiner, grabiner@math.lsa.umich.edu
> http://www.math.lsa.umich.edu/~grabiner
> Shop at the Mobius Strip Mall: Always on the same side of the street!
> Klein Glassworks, Torus Coffee and Donuts, Projective Airlines, etc.

It seems as if the original question might really be interpreted
as what is the distribution of the sum of a random number
of independent (identically distributed?) exponentials.

So, let us assume that  N ~ Geom(p)  is geometrically distributed,
and that  (Y|N=n) = X_1 + ... + X_n  is a sum of independent
exponentials  X_1,...,X_n  each with mean  b, and thus the
(conditional) density of  (Y|N=n)  is that of a gamma random
variable  

	f(y|N=n) = b^{-n} y^{n-1} \exp(-y/b) / (n-1)! .

Then the unconditional density of  Y  is

	f(y) = \sum_{n=1}^\infty f(y|N=n) p(1-p)^{n-1}

   = pb^{-1} \exp(-y/b) \sum_{n=1}^\infty ((1-p)y/b)^{n-1} / (n-1)!

   = pb^{-1} \exp(-y/b) \exp((1-p)y/b)

   = (b/p)^{-1} \exp(-y/(b/p))

and so, unconditionally,  Y  is geometric with mean  b/p .

This is a pretty standard problem from my post-calculus
probability course.

--Chuck Giffen