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February 4th, 2018, 11:56 AM   #1
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Exclamation Joint Moment Generating Function from Conditional and Marginal Distribution

Suppose that that random variable $N$ follows a Poisson distribution with mean $\lambda=6$. Suppose that the conditional distribution of the random variable $X$, given that $N=n$, follows that of a $Binomial(n,0.6)$.

Find the Joint Moment Generating Function of $(N, X)$.

Initially I just tried to use the definition. I found the joint PMF using the definition of the conditional distribution, but then I have to sum over both of them in order to find the joint MGF, and this was the step I was stuck at because trying to do a double sum over the product of the binomial and Poisson PMFs doesn't exactly go over very nicely.

Since this problem does NOT assume independence, I can't exactly attempt to use that to my advantage either... so now I'm stuck. How can I calculate this joint MGF?
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February 4th, 2018, 12:08 PM   #2
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So in terms of the probability mass functions, what is the formula for $p(x|n)$?
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February 4th, 2018, 12:15 PM   #3
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It's just the standard binomial PMF. $n \choose{x}$$ 0.6^x0.4^{n-x}$
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February 4th, 2018, 12:24 PM   #4
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What is the theoretical definition of p(x|n) ?
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February 4th, 2018, 12:26 PM   #5
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I'm just asking things you've already done according to your OP, but if you're not going to share any detailed calculations, there's nothing more I can do...
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February 4th, 2018, 12:32 PM   #6
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$f(X|N)=\frac{f(X,N)}{f(N)}$

That's why this implies that, in order to get the joint PDF, I have to multiply the binomial and poisson PMFs. Then I have to take the expected value of that, $E(e^{t_1X+t_2Y})$ which gives me that nasty double sum I mentioned earlier. I don't know how to work with that. This is why I was hoping there was a better way.
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