Consider $\{ X_{i} \}$ are independent random variables with Poisson distribution. We want to know about convergence in distribution of $\frac{S_{n}-\mathbb{E}(S_{n})}{\sqrt{\operatorname{Var}(S_{n})}}$.

There are two cases :

Case 1 : $\sum_{i}\lambda_{i}$ converges. Then we can say variance and mean value of $S_{n}$ converges to some fixed value. Also we know that sum of Poisson random variables is also the Poisson random variable with parameter equals sum of previous ones. So in my opinion $\frac{S_{n}-\mathbb{E}(S_{n})}{\sqrt{\operatorname{Var}(S_{n})}}$ converges to $\frac{\mathrm{Pois}(c) - c}{\sqrt{c}}$ , where $c = \sum_{i} \lambda_{i}$

Case 2 : we have that $\sum_{i} \lambda_{i}$ diverges. Then I guess we can try to use Central limit theorem and satisfy that $\frac{S_{n}-\mathbb{E}(S_{n})}{\sqrt{\operatorname{Var}(S_{n})}}$ converges to $N(0,1)$

Am I right ? Or where have I problems to fix ?

  • 1
    $\begingroup$ How do you expand "i.i.d."? In standard terminology if $X_i$'s are i.i.d Poisson then the parameters $\lambda_i$ are all the same, so CLT applies. $\endgroup$ – Kavi Rama Murthy May 11 '18 at 8:04
  • $\begingroup$ @KaviRamaMurthy my bad! $\endgroup$ – openspace May 11 '18 at 8:04
  • $\begingroup$ @KaviRamaMurthy changed $\endgroup$ – openspace May 11 '18 at 8:15

The answer is easy if you compute the characteristic function explicitly. Let $u_n=\lambda_1+\lambda_2++...+\lambda_n$. Then $Ee^{it(S_n-ES_n)/\sqrt (Var(S_n)}=e^{-it\sqrt (u_n)}\prod_{j=1}^{n}(Ee^{itX_j /\sqrt(u_n)}$. The Poisson characteristic function (paramater $\lambda $) is $e^{-\lambda (1- e^{it})}$. We get $e^{-it\sqrt (u_n)}e^{-u_n (1-e^{it/\sqrt (u_n)})}$. Using the Taylor expansion of $e^{it/\sqrt (u_n)}$ up to the term in $t^{2}$ you will see that that characteristic function indeed converges to $e^{-t^{2}/2}$ if $\sum \lambda_j =\infty$. The characteristic function also converges when $\sum \lambda_j <\infty$ and you can write down the limiting characteristic function explictly.


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