# Sums of exponentially distributed rvs

1. Jan 3, 2009

### 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,

$$f(s) = \left\{ \begin{array}{c l} e^{-s} & if s \ge 0 \\ 0 & otherwise \end{array} \right.$$

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?

2. Jan 3, 2009

### 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 $$X$$s 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

$$S_2 = X_1 + X_2$$

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

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