show that the random geometric sum of independent and identical exponential random variables is again an exponential random variable
Answers
I too prefer to call the random variables X and Y. You can think of X and Y as waiting times for two independent things (say A and B respectively) to happen. Suppose we wait until the first of these happens. If it is A, then (by the lack-of-memory property of the exponential distribution) the further waiting time until B happens still has the same exponential distribution as Y; if it is B, the further waiting time until A happens still has the same exponential distribution as X. That says that the conditional distribution of X−Y given X>Y is the distribution of X, and the conditional distribution of X−Y given X<Y is the distribution of −Y. Since P(X>Y)=
λ
μ+λ
, that says the PDF for X−Y is
f(x)=
λμ
λ+μ
{ e−μx if x>0 eλx if x<0
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