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Math Help - Conditional distributions

  1. #1
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    Apr 2009
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    Conditional distributions

    I've got a homework question that I can't for the life of me solve.

    Let X1, X2, X3 ... be an independently identically distributed sequence of binomial random variables with parameters n = 100 and p = e/100. Let N be a Poisson random variable with mean parameter 5 that is independent of all the Xi's. Define the random variable Y by

    Y = X1 * X2 * X3 * ... * XN (X sub N, the Poisson random variable)

    Determine P(Y = 1 | N = n)
    Determine P(Y = 1)
    Find E(Y | N)
    Find E(Y)

    For the first two, I keep finding different formulas and examples that do the problems differently, and the vast amount of random variables are making my head spin. I know that P(Y = 1 | N = n) = (P(N = n | Y = 1) * P(Y = 1)) / P(N = n) but I don't know if I can use that formula, seeing as how I don't know most of those variables.

    For the second two, I know that E(Y) = E(E(Y | N)) but how would I find E(Y | N) in the first place?
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  2. #2
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    Apr 2009
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    my work

    I worked on the problem last night, and these are the answers I came up with. Could anyone verify if they're right or wrong?

    a) \binom{100}{1} * \frac{e}{100} * (1 - \frac{e}{100})^{99} = .178

    So P(Y = 1 | N = n) = .178^{n}

    b) \sum_{n=1}^{100} (.178)^{n} * e^{-5} * \frac{-5^n}{n!} = .96

    c) My instinct is telling me that it's e^n but I can't articulate why. Is that right?

    d) Going with c...

    \sum_{n=1}^{100} (e)^{n} * e^{-5} * \frac{-5^n}{n!} = 5389.14
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