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Jan3-09, 01:58 AM
P: 21

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,

[tex] f(s) = \left\{
\begin{array}{c l}
e^{-s} & if s \ge 0 \\
0 & otherwise

Let [tex]X_1, X_2, .... , X_n[/tex] be independently exponentially distributed random variables.

The PDF of the sum, [tex] X_1 + X_2 + ..... +X_n[/tex] is

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

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

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

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

= [tex]\int^b_a e^{-u} vol_{n-1} T_u du [/tex]

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

What would [tex] vol_{n-1}[/tex] be here?
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