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Sums of exponentially distributed rvs

  1. Jan 3, 2009 #1

    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?
  2. jcsd
  3. Jan 3, 2009 #2


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    Homework Helper

    Do you need to do the problem this way? I can suggest two procedures that would be easier than this approach:
    1) Since the [tex] X [/tex]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 [tex] n [/tex] 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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