Sum of Poisson Random variables converges a.s.

Question

I'm stuck on this question. My thoughts are to try and first make each of these random variables have mean 0 by subtracting out $\lambda_i$. Then I think you can use some kind of borel cantelli argument by trying to show $P(|S_n-\sum_{i=1}^n\lambda_i|>\epsilon \text{ i.o })=0$ Is this the right track? Any hints or solutions would be appreciated

Source: Old Qual problem

https://dornsife.usc.edu/assets/sites/990/docs/Spring_2012/20121_507a.pdf

Answer 1

Yes, you want to do a Borel-Cantelli type argument. A hint for how to check the decay property that you need:

$$\sum_{n=N}^\infty \frac{\lambda^n}{n!} \leq \frac{e^\lambda \lambda^N}{N!}$$

which follows from the Lagrange remainder formula.

Answer 2

Here is a quick argument which I learned from other posting.

Write $S_n = X_1 + \cdots + X_n$ and notice that $S_n$ is a Poisson random variable of rate $\sum_{i=1}^{n} \lambda_i$. Also notice that,

$$ \forall t > 0 \ : \quad \mathbb{E}\left[ e^{-t S_n} \right] = \exp\left\{ -(1-e^{-t})\sum_{i=1}^{n} \lambda_i \right\}. $$

From this, writing $S_{\infty} = \lim_{n\to\infty} S_n$ and $\mu = \sum_{i=1}^{\infty} \lambda_i$, bounded convergence theorem tells

$$ \mathbb{E}\left[ e^{-t S_{\infty}} \right] = \lim_{n\to\infty} \mathbb{E}\left[ e^{-t S_n} \right] = e^{-(1-e^{-t})\mu}. $$

(Here, we are using the convention $e^{-\infty} = 0$.) So

• If $\mu = \infty$, then $e^{-t S_{\infty}} = 0$ a.s. and hence $S_{\infty} = \infty$ a.s.

• If $\mu < \infty$, then letting $t \downarrow 0$ together with bounded convergence theorem tells that

  $$ \mathbb{P}(S_{\infty} < \infty) = \mathbb{E}\left[ \mathbf{1}_{\{S_{\infty} < \infty\}} \right] = \mathbb{E}\left[ \lim_{t\downarrow 0} e^{-t S_{\infty}} \right] = \lim_{t\downarrow 0} e^{-(1-e^{-t})\mu} = 1. $$

  Alternatively, $\mathbb{E}[ e^{-t S_{\infty}}] > 0$ implies that $\mathbb{P}(S_{\infty} < \infty) > 0$. By the Kolmogorov 0-1 law, this implies that $\mathbb{P}(S_{\infty} < \infty) = 1$.

  Yet another solution is that, if a basic theory of Laplace transform is available, then you can deduce that $S_{\infty}$ has the Poisson distribution of rate $\mu$, which also proves that $S_{\infty}$ is finite a.s.