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Let $X$ have the Gamma$(s,1)$ and given $X=x$, let $Y$ have the Possion distribution with parameter $x$. Show that $$\frac{Y-E(Y)}{\sqrt{\operatorname{var}(Y)}}\longrightarrow W$$ where $\longrightarrow$ means converges in distribution as $s$ goes to infinity. And $W$ needs to be identified.

I have worked out the moment generating function of $Y$, $$ M_Y(t)=\left(\frac{1}{2-e^{t}}\right)^s$$ Then I work out the mgf of $\frac{Y-E(Y)}{\sqrt{\operatorname{var}(Y)}}$, $$ M(t)=e^{-\frac{s}{\sqrt{2s}}t}\left(\frac{1}{2-e^{\frac{t}{\sqrt{2s}}}}\right)^s$$ But I don't know what does it converges to.

Anything wrong with my above calculation?


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take a look at that –  Ilya Dec 9 '11 at 13:41
You said "with parameter $x$". Did you actually mean "with parameter $X$"? That would at least make the question make sesne. –  Michael Hardy Dec 9 '11 at 17:37
@MichaelHardy I think they are the same, since my question is given X=x and Y hase Poisson with parameter x. –  John Dec 9 '11 at 18:09
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1 Answer

I haven't worked out the whole thing, but here's my scratchwork.

The moment-generating function of $X$ is $$ M_X(t) = E(e^{tX}) = \left(\frac{1}{1-t}\right)^s. $$

For every positive $x$, the moment-generating function of a Poisson-distributed random variable $Y_x$ with expectation $x$ is $$ E(e^{tY_x}) = e^{x(e^t-1)}, $$ hence $$ E(e^{tY}\mid X) = e^{X(e^t-1)}, $$ so that is the conditional moment-generating function of $Y$ given $X$.

So the "unconditional" moment-generating function of $Y$ is $$ M_Y(t) = E(e^{tY}) = E(E(e^{tY} \mid X)) = E\left( e^{X(e^t-1)} \right) = M_X(u), $$ where $u=e^t-1$, that is, $$ M_Y(t) = \left(\frac{1}{1-u}\right)^s = \left(1-(e^t - 1)\right)^{-s}. $$

Let's see what the variance and expectation of $Y$ are: $$ E(Y) = E(E(Y\mid X)) = E(X) = s. $$ $$ \operatorname{var}(Y) = \operatorname{var}(E(Y \mid X)) + E(\operatorname{var}(Y \mid X)) = \operatorname{var}(X) + E(X) = s+s. $$

Maybe I'll add more here later.

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For the mgf of Y do you mean $(\frac{1}{2-e^t})^s$ –  John Dec 9 '11 at 18:43
Corrected a formula giving (wrongly) $E(e^{tY})$ as a function of $X$, and a sign error in the exponent $s$ at the end. –  Did Apr 8 '12 at 8:29
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