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The pair of random variables $(X, Y )$ is distributed as follows. $Y$ has probability mass function $\text{Poisson}(1).$ Given $Y , X$ has probability mass function $\text{Poisson}(Y ).$ Show that the probability generation function of $ X + Y $ equals: $$G_{X+Y}(s)=\exp[se^{s-1}-1]$$

I tried to do the following: $$G_{X+Y}(s)=\mathbb{E}(s^{X+Y})=\mathbb{E}(\mathbb{E}(s^{X+Y}|Y))=\mathbb{E}(s^Y\mathbb{E}(s^X|Y))$$

I am stuck with the $\mathbb{E}(s^X|Y)$ part. By definition we know : $\mathbb{E}(X|Y))=\sum_{1}^{\infty}x\mathbb{P}(X=x|Y)$, but how do we apply and determine it for $s^X$?

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  • $\begingroup$ I didn't downvote. But isn't it given that $X\sim\textrm{Poi}(Y) $? $\endgroup$ Commented Jun 4, 2023 at 12:50
  • $\begingroup$ @User1865345 Yes it is given $\endgroup$ Commented Jun 4, 2023 at 12:53
  • $\begingroup$ Do you know the definition of $\mathbb{E}(f(X)|Y)$? $\endgroup$ Commented Jun 4, 2023 at 14:37
  • $\begingroup$ @jbowman The definition is : $\mathbb{E}(f(X)|Y)=\sum_{x\in Im X}f(x)\mathbb{P}(f(X)=f(x)|Y)=\sum_{x\in Im X}f(x)p_{X|Y}(X=x|Y)$. I think I solved it can you check my answer below $\endgroup$ Commented Jun 4, 2023 at 15:01

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The pgf of a Poisson random variable with parameter $\lambda$ is $G(s)=e^{\lambda(s-1)}$. Hence \begin{align} G_{X+Y}(s)&=E(s^{X+Y}) &&\text{(by definition of the pgf)} \\&=E(E(s^{X+Y}|Y)) &&\text{(law of total expectation)} \\&=E(s^YE(s^X|Y)) \\&=E(s^YG_{X|Y}(s)) &&\text{(by definition of the pgf)} \\&=E(s^Ye^{Y(s-1)}) &&\text{(since $X|Y\sim\operatorname{Poisson}(Y)$)} \\&=E((se^{s-1})^Y) \\&=G_Y(se^{s-1}) &&\text{(by definition of the pgf)} \\&=e^{se^{s-1}-1}. &&\text{(since $Y\sim\operatorname{Poisson}(1)$)} \end{align}

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  • $\begingroup$ For a moment, I thought you switched X and Y. But now it is correct. +1. The info was given and OP couldn't drive through (that's why those comments above). $\endgroup$ Commented Jun 4, 2023 at 15:05

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