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Summation of geometric number of iid exponentially distributed random variables

  1. Feb 24, 2010 #1
    Hello, I am having difficulty approaching this problem:

    Assume that K, Z_1, Z_2, ... are independent.
    Let K be geometrically distributed with parameter success = p, failure = q.
    P(K = k) = q^(k-1) * p , k >= 1

    Let Z_1, Z_2, ... be iid exponentially distributed random variables with parameter (lambda).
    f(z) =
    (lambda)*exp(-(lambda)x) , x >= 0
    0, otherwise

    Find the cdf of Z_1 + Z_2 + ... + Z_K

    I think there is some relation to the Gamma function here, but I'm not quite sure how...

    Any hints/suggestions?
  2. jcsd
  3. Feb 24, 2010 #2
    Well, I would try something like this. Let S be your "random number of random variables", ie. Z1+Z2+...+ZK

    P(S<X) = sum_n P(S<X | K=n) * P(K=n)

    Then analyze P(S<X | K=n) by finding the PDF or CDF for a random variable that is the sum of n exponential random variables. You could use the result that the resulting distribution function is the convolution of the n distribution functions.

    After you have found P(S<X | K=n), you can perform the sum of n.

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