# Sum of geometric number of exponentially distributed random variables

• February 23rd 2010, 06:25 PM
redflame34
Sum of geometric number of exponentially distributed random variables
Hello, I am having difficulty approaching this problem:

Assume that K, Z_1, Z_2, ... are independent.
Let K be geometrically distributed with parameter success = p, failure = q.

Let Z_1, Z_2, ... be iid exponentially distributed random variables with parameter (lambda).

Find the cdf of Z_1 + Z_2 + ... + Z_K

I think there is some relation to the Gamma function here, but I'm not quite sure how...

Any help would be greatly appreciated, thanks in advance!
• February 23rd 2010, 07:07 PM
matheagle
Use the MGF of your sum and obtain a double expectation, where you condition on K.
But I need to know how you write your exponentials and also the geometric.
• February 23rd 2010, 07:19 PM
redflame34
Hmm, I guess I am still kind of confused about how the MGF and double expectation are used.

Here is more detail:

P(K = k) = q^(k-1) * p , k >= 1

f(z) =
(lambda)*exp(-(lambda)x) , x >= 0
0, otherwise

Thanks!
• February 23rd 2010, 09:30 PM
matheagle
$M(t) =E(e^{Z_1 + Z_2 + ... + Z_K})=E(E(e^{Z_1 + Z_2 + ... + Z_K})|K)$
• February 23rd 2010, 10:11 PM
redflame34
I'm sorry, I don't quite understand how that allows me to get the cumulative distribution function...

A stupid question: how would I evaluate that double expectation?
• February 24th 2010, 06:42 AM
matheagle
So, treat K as a constant.
You must know that the sum of iid exponetials is a Gamma.
So plug in that MGF and then get the expectation wrt K.