# Thread: Sum of geometric number of exponentially distributed random variables

1. ## 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!

2. 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.

3. 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!

4. $\displaystyle M(t) =E(e^{Z_1 + Z_2 + ... + Z_K})=E(E(e^{Z_1 + Z_2 + ... + Z_K})|K)$

5. 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?

6. 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.