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(I've read the related questions here but found no satisfying answer, as I would prefer a rigorous proof for this because this is a homework problem)

Prove: If $X_\alpha$ follows the Poisson distribution $\pi(\alpha)$, then $$\lim_{\alpha\rightarrow\infty}P\{\frac{X_\alpha-\alpha}{\sqrt{\alpha}} \leq u \} = \Phi(u)$$

where $\Phi(u)$ is the cdf of normal distribution $N(0,1)$

Hint: use the Laplace transform $E(e^{-\lambda(X_\alpha-\alpha)/\sqrt{\alpha}})$, show that as $\alpha\rightarrow\infty$ it converges to $e^{\lambda^2/2}$

I did the transform but failed to sum the series(which is essentially doing nothing)

Here's what I got:

$$g(\lambda)=\sum_{n=0}^{\infty} \frac{e^{-\alpha}}{n!}\alpha^n e^{-\frac{\lambda(n-\alpha)}{\sqrt{\alpha}}}$$

and $\lim_{\alpha\rightarrow\infty} g(\lambda)=e^{-\lambda^2}$ is what I'm trying to arrive at. I tried L'Hospital only to find that the result is identical to the original ratio.

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Please give the series you arrived at (as a general principle, it is better to include what you know in the question). – Did Nov 29 '13 at 13:39
@Did ok, question edited. As I said before, I essentially did nothing here :( – arax Nov 29 '13 at 13:44
Factoring everything independent of $n$ in the series defining $g(\lambda)$ leaves one with a constant times the sum of a well known series. You might want to go through these steps and see what happens. – Did Nov 29 '13 at 13:57
@Did but there is the factor $e^{-n\lambda/\sqrt{\alpha}}$, how could I get rid of that? – arax Nov 29 '13 at 14:00
Do not get rid of it, it is just $b^n$ for $b=\mathrm e^{-\lambda/\sqrt{\alpha}}$ independent of $n$, right? Sooo... – Did Nov 29 '13 at 14:14
up vote 3 down vote accepted

Let $X_{\alpha}$ Poisson $\pi(\alpha)$, for $\alpha = 1, 2, \ldots$ The probability mass function of $X_{\alpha}$ is $$f_{X_{\alpha}}(x)=\frac{{\alpha}^x\operatorname{e}^{-x}}{x!} \qquad \alpha = 1, 2, \ldots$$ The moment generating function of $X_{\alpha}$ is $$ M_{X_{\alpha}}=\Bbb{E}\left(\operatorname{e}^{tX_{\alpha}}\right)=\operatorname{e}^{\alpha(\operatorname{e}^{t}-1)}\qquad t\in(-1,1) $$ Now consider a “standardized” Poisson random variable $Z_{\alpha}=\frac{X_{\alpha}-{\alpha}}{\sqrt{\alpha}}$ which has limiting moment generating function $$ \begin{align} \lim_{\alpha\to\infty}M_{Z_{\alpha}}&= \lim_{\alpha\to\infty}\Bbb{E}\left(\exp{(tZ_{\alpha})}\right)\\ &=\lim_{\alpha\to\infty}\Bbb{E}\left(\exp{\left(t\frac{X_{\alpha}-{\alpha}}{\sqrt{\alpha}}\right)}\right)\\ &=\lim_{\alpha\to\infty}\exp(-t\sqrt\alpha)\Bbb{E}\left(\exp{\left(\frac{tX_{\alpha}}{\sqrt{\alpha}}\right)}\right)\\ &=\lim_{\alpha\to\infty}\exp(-t\sqrt\alpha)\exp\left(\alpha(\operatorname{e}^{t/\sqrt\alpha}-1)\right)\\ &=\lim_{\alpha\to\infty}\exp\left(-t\sqrt\alpha +\alpha\left[t\alpha^{-1/2}+\frac{t^2\alpha^{-1}}{2}+\frac{t^3\alpha^{-3/2}}{6}+\cdots\right]\right)\\ &=\lim_{\alpha\to\infty}\exp\left(\frac{t^2}{2}+\frac{t^3\alpha^{-3/2}}{6}+\cdots\right)\\ &=\exp\left(\frac{t^2}{2}\right) \end{align} $$ by using the moment generating function of a Poisson random variable and expanding the exponential function as a Taylor series. This can be recognized as the moment generating function of a standard normal random variable. This implies that the associated unstandardized random variable $X_{\alpha}$ has a limiting distribution that is normal with mean $\alpha$ and variance $\alpha$.

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The moment generating function of a ST Normal is $e^{t^2 /2}$ – JohnK Nov 29 '13 at 14:31
@Ioannis yes, right! – alexjo Nov 30 '13 at 9:09
@alexjo There is a tiny typo in the second-to-last line: $\frac{t^3\alpha^{-3/2}}{6}$ should be $\frac{t^3\alpha^{-1/2}}{6}$. – M.B.M. Jul 1 '15 at 21:16

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