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Math Help - Combined Poisson

  1. #1
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    Combined Poisson

    I am preparing for my test which is on Friday. There is this problem which is killing me and I don't have any idea how to solve it! It is in conditional probabilities chapter. The question is

    X ~ Poi (a)
    Y ~ Poi (b)

    Z = X^Y

    What is the Expected value of Z.

    I will be so thankful if anyone could help me.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Olivia787878 View Post
    I am preparing for my test which is on Friday. There is this problem which is killing me and I don't have any idea how to solve it! It is in conditional probabilities chapter. The question is

    X ~ Poi (a)
    Y ~ Poi (b)

    Z = X^Y

    What is the Expected value of Z.

    I will be so thankful if anyone could help me.
    Is it E[X^Y]?

    You know that E[s^Y]=e^{b(s-1)}, this is the moment generating function. Or maybe you know E[e^{tY}]=e^{b(e^t-1)}, which is the same with s=e^t. Anyway, you can compute this, even if you don't know it already.

    On the other hand, since (you should have mentioned it) X and Y are independent, the conditional distribution of Y given X=x is the same as the distribution of Y. Therefore, for all x\in\mathbb{N}, E[X^Y|X=x] = E[x^Y]=e^{b(x-1)}. Then E[X^Y]=E[E[X^Y|X]] = E[e^{b(X-1)}], and the last expectation can be computed using the moment generating function again.

    The last inequalities can also be written E[X^Y]=\sum_{x=0}^\infty E[X^Y|X=x] P(X=x) = \sum_{x=0}^\infty e^{b(x-1)} P(X=x) =E[e^{b(X-1)}].
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