- #1
roadworx
- 21
- 0
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,
[tex] f(s) = \left\{
\begin{array}{c l}
e^{-s} & if s \ge 0 \\
0 & otherwise
\end{array}
\right.
[/tex]
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?
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
\end{array}
\right.
[/tex]
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?