7
$\begingroup$

Let $X_1,X_2\dots$ all be independent, Poisson distributed with parameter $l_i$ each. Then it is known that for each n $S_n:=\sum_{i=1}^n X_i\sim \text{po}(\lambda_n)$ where $\lambda_n:=\sum_{i=1}^n l_i$.

Now assume $\sum_{i=1}^\infty l_i = \lambda < \infty$ is it then true that $S:=\sum_{i=1}^\infty X_i \sim \text{po}(\lambda)$?

I'm thinking for each $x\in \mathbb{N}_0$ by dominated convergence and continuity

$$P(S=x)=E[1_{\{S=x\}}]=\lim_{n\to \infty} E[1_{\{S_n=x\}}]=\lim_{n\to \infty}P(S_n=x)=\lim_{n\to \infty} \frac{\lambda_n^x}{x!}e^{-\lambda_n}=\frac{\lambda^x}{x!}e^{-\lambda}$$

But I can't find a result like this anywhere. Is my argument correct? Is the statement?

$\endgroup$

2 Answers 2

7
$\begingroup$

The random variable $S$ is almost surely finite because $E[S]$ is finite. Hence $S_n\to S$ almost surely. Almost sure convergence implies convergence in probability, which implies convergence in distribution. Hence $S_n\to S$ in distribution. For discrete random variables, convergence in distribution means convergence of the weight at each atom. In the present case the weight at each $k$ of the distribution of $S_n$ converges to the weight at $k$ of the Poisson distribution with parameter $\lambda$. Hence $S$ is Poisson with parameter $\lambda$.

$\endgroup$
2
  • $\begingroup$ Thats a really nice argument :) proving it using the probability function directly is right too? For "layman" doing it the direct way can sometimes feel slightly more comfortable. $\endgroup$
    – htd
    Commented Apr 6, 2013 at 11:03
  • $\begingroup$ The argument you suggest is correct, provided your "layman" justifies carefully that $\mathbf 1_{S_n=x}\to\mathbf 1_{S=x}$ pointwise. $\endgroup$
    – Did
    Commented Apr 6, 2013 at 11:55
1
$\begingroup$

Your argument seems correct to me. I think the correct way to define $S$ is as the limit in probability of $S_n$. The limit exists because the sequence $S_n$ is increasing. By Borel Cantelli, you know that $P(S< \infty) = 1$, and the distribution of $S_n$ is by construction $$ P(S = x ) = \lim_{n \to \infty} P(S_n = x) $$ which is what you calculated.

$\endgroup$
2
  • 1
    $\begingroup$ It is definitely nice to note that $S_n$ is increasing in n. By monotonce convergence then $E[S]=\sum _{i=1}^\infty E[X_i]=\sum _{i=1}^\infty l_i<\infty$ which is maybe simpler than B.C. Why define it as limit in probability? We know the limit exists (from the increasing fact) and is finite almost surely? $\endgroup$
    – htd
    Commented Apr 6, 2013 at 9:49
  • 1
    $\begingroup$ Yes you're right it is simpler to define it as the limit almost surely, my bad. $\endgroup$
    – roger
    Commented Apr 6, 2013 at 13:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .