Sums of exponentially distributed rvs

[roadworx]

Hi,

Can anyone derive the sum of exponentially distributed random variables?

I have the derivation, but I'm confused about a number of steps in the derivation.

Here they are:

Random variable x has the PDF,

f(s) = \left\{<br /> \begin{array}{c l}<br /> e^{-s} &amp; if s \ge 0 \\<br /> 0 &amp; otherwise <br /> \end{array}<br /> \right. <br />

Let X_1, X_2, ... , X_n be independently exponentially distributed random variables.

The PDF of the sum, X_1 + X_2 + ... +X_n is

q(s) = e^{-(s_1+s_2+...+s_n)} where s s \ge 0

=> \int_{a \le s_1+s_2+...+s_n \le b} q(s) ds

= \int_{a \le s_1+s_2+...+s_n \le b} e^{-(s_1 + ... + s_n)} ds

Can anyone explain this stage? Going from the above integral to the following integral?

= \int^b_a e^{-u} vol_{n-1} T_u du

where T_u = [s_1+ ... + s_n = u]


What would vol_{n-1} be here?

 

[statdad]

Do you need to do the problem this way? I can suggest two procedures that would be easier than this approach:
1) Since the Xs are i.i.d exponential, you can calculate the moment generating function of the sum quite easily, and identify the distribution from that
2) Rather than using all n of the variables in a single integration, start with

<br /> S_2 = X_1 + X_2<br />

and find its distribution, then continue to the case you have by induction.