Im having trouble finding any good resources or examples for finding the conditional distribution of two variables. I've tried using double expectations but cant get it to work out. Thanks

I found the pmf of X not conditioned on N below

$ P(x=x) = \frac{e^{-\lambda p}(\lambda p)^x}{x!} $

Then used $ E(e^{tx}) $ to find the mgf

$ M_x(t) = e^{\lambda p(e^t-1) } $

  • 2
    $\begingroup$ Can you share what you've done so far? $\endgroup$
    – gunes
    Nov 13, 2020 at 14:33

1 Answer 1

  1. You want the moment generating function $M_X(t)=\mathbb{E}[e^{tX}]$.
  2. Read this to understand why $\mathbb{E}[e^{tX}]=\mathbb{E}[\mathbb{E}[e^{tX}\mid N]]$.
  3. The information given in the problem statement gives you $$\mathbb{E}[e^{tX}\mid N]=(1-p+e^tp)^N. \qquad\qquad \text{(why?)}$$
  4. Defining $e^u:=1-p+e^tp$, and using the expression of the moment generating function of a random variable with Poisson distribution given in the problem statement, you're done; just express the result as a function of $t$.

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