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?
Best Answer
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 $\lambda$.