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Math Help - Sum of geometric number of exponentially distributed random variables

  1. #1
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    Sum of geometric number of exponentially distributed random variables

    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.

    Let Z_1, Z_2, ... be iid exponentially distributed random variables with parameter (lambda).

    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 help would be greatly appreciated, thanks in advance!
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  2. #2
    MHF Contributor matheagle's Avatar
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    Use the MGF of your sum and obtain a double expectation, where you condition on K.
    But I need to know how you write your exponentials and also the geometric.
    So, show your work.
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  3. #3
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    Hmm, I guess I am still kind of confused about how the MGF and double expectation are used.

    Here is more detail:

    P(K = k) = q^(k-1) * p , k >= 1

    f(z) =
    (lambda)*exp(-(lambda)x) , x >= 0
    0, otherwise

    Thanks!
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  4. #4
    MHF Contributor matheagle's Avatar
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    M(t) =E(e^{Z_1 + Z_2 + ... + Z_K})=E(E(e^{Z_1 + Z_2 + ... + Z_K})|K)
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  5. #5
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    I'm sorry, I don't quite understand how that allows me to get the cumulative distribution function...

    A stupid question: how would I evaluate that double expectation?
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  6. #6
    MHF Contributor matheagle's Avatar
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    So, treat K as a constant.
    You must know that the sum of iid exponetials is a Gamma.
    So plug in that MGF and then get the expectation wrt K.
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