View Single Post
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?
Phys.Org News Partner Science news on
Climate change increases risk of crop slowdown in next 20 years
Researcher part of team studying ways to better predict intensity of hurricanes
New molecule puts scientists a step closer to understanding hydrogen storage