4
$\begingroup$

(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.

$\endgroup$
5
  • $\begingroup$ Please give the series you arrived at (as a general principle, it is better to include what you know in the question). $\endgroup$
    – Did
    Nov 29, 2013 at 13:39
  • $\begingroup$ @Did ok, question edited. As I said before, I essentially did nothing here :( $\endgroup$
    – arax
    Nov 29, 2013 at 13:44
  • $\begingroup$ 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. $\endgroup$
    – Did
    Nov 29, 2013 at 13:57
  • $\begingroup$ @Did but there is the factor $e^{-n\lambda/\sqrt{\alpha}}$, how could I get rid of that? $\endgroup$
    – arax
    Nov 29, 2013 at 14:00
  • 1
    $\begingroup$ Do not get rid of it, it is just $b^n$ for $b=\mathrm e^{-\lambda/\sqrt{\alpha}}$ independent of $n$, right? Sooo... $\endgroup$
    – Did
    Nov 29, 2013 at 14:14

1 Answer 1

5
$\begingroup$

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 $\alpha$.

$\endgroup$
4
  • 1
    $\begingroup$ The moment generating function of a ST Normal is $e^{t^2 /2}$ $\endgroup$
    – JohnK
    Nov 29, 2013 at 14:31
  • $\begingroup$ @Ioannis yes, right! $\endgroup$
    – alexjo
    Nov 30, 2013 at 9:09
  • 1
    $\begingroup$ @alexjo There is a tiny typo in the second-to-last line: $\frac{t^3\alpha^{-3/2}}{6}$ should be $\frac{t^3\alpha^{-1/2}}{6}$. $\endgroup$
    – M.B.M.
    Jul 1, 2015 at 21:16
  • $\begingroup$ minor typo: should be $e^{-\alpha}$ in Poisson p.f. at 1st display eq. $\endgroup$
    – jdods
    Oct 29, 2019 at 11:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.