I'm trying to solve the question below and after an hour no success yet! Here is the question:

$X_1, X_2, ...$ are independent poisson random variables with $EX_j = \lambda_j$. Assume that $ 0 < \lambda_j < 1$. Define $S_n = \sum_{i = 1}^{n} X_i$. We need to show that:

$$ \text{if } \sum_{j = 1}^{\infty} \lambda_j = \infty \text{ then} \frac{S_n}{ES_n} \rightarrow 1 \text{ almost surely}$$.

I appreciate if you could give me some hints on how I should approach this question.

There is a hint on this question that Borel-Cantelli lemma should be used. As you know Borel-Cantelli (lemma 2) says:

Let $A_1, A_2, ...$ be events in the probability space. Then:

if $A_1, A_2, ...$ are independent and $\sum_{k = 1}^{\infty} p(A_k) = \infty$ then $\rightarrow p(A_k, \text{ infinitely often}) = 1$

Thank you for your help.

  • $\begingroup$ How are $j$ and $k$ related? $\endgroup$ – joriki Nov 27 '12 at 8:56
  • $\begingroup$ Why does the title say "Application of Borel Cantelli"? The body doesn't mention the Borel–Cantelli lemma. $\endgroup$ – joriki Nov 27 '12 at 8:59
  • $\begingroup$ Hi @joriki, I edited the question...The hint on this question is to use Borel-Cantelli and that's why I use the title. $\endgroup$ – Sam Nov 27 '12 at 13:28
  • $\begingroup$ That missing information should be added to the question itself; people shouldn't have to read the comments to understand the question. It's currently not clear from the title whether you would like to apply the Borel–Cantelli lemma or whether this is a hint that's part of the question. $\endgroup$ – joriki Nov 27 '12 at 13:31
  • $\begingroup$ You are absolutely right. Lemme add that too... $\endgroup$ – Sam Nov 27 '12 at 13:32

First we remark that $S_n \sim \text{Poi} \left(\mu_n \right)$ where $\mu_n := \sum_{j=1}^n \lambda_j$. Moreover, $\mathbb{E}S_n = var S_n=\mu_n$.

Let $\varepsilon>0$. By applying Tschebyscheff-inequality

$$\mathbb{P}\left[ \left| \frac{S_n}{\mathbb{E}S_n} -1 \right|>\varepsilon \right] \leq \frac{1}{(\varepsilon \cdot \mathbb{E}S_n)^2} \cdot \text{var} S_n = \frac{1}{\varepsilon^2} \cdot \frac{1}{\mu_n} \to 0 \quad (n \to \infty)$$

since $\mu_n= \sum_{j=1}^n \lambda_j \to \infty$. Hence $\frac{S_n}{\mathbb{E}S_n} \to 1$ in probability. Now let $n_k := \inf\{n; \mathbb{E}S_n \geq k^2\}$. Then the same calculation shows

$$\mathbb{P}\left[ \left| \frac{S_{n_k}}{\mathbb{E}S_{n_k}} -1 \right|>\varepsilon \right] \leq \frac{1}{\varepsilon^2} \cdot \frac{1}{k^2} \\ \Rightarrow \sum_k \mathbb{P}\left[ \left| \frac{S_{n_k}}{\mathbb{E}S_{n_k}} -1 \right|>\varepsilon \right] \leq \sum_k \frac{1}{\varepsilon^2} \cdot \frac{1}{k^2} < \infty$$

By applying the Lemma of Borel-Cantelli we obtain $\frac{S_{n_k}}{\mathbb{E}S_{n_k}} \to 1$ almost surely as $k \to \infty$. Since

$$\frac{S_{n_k}(w)}{\mathbb{E}S_{n_{k+1}}} \leq \frac{S_n(w)}{\mathbb{E}S_n} \leq \frac{S_{n_{k+1}}(w)}{\mathbb{E}S_{n_k}}$$

for all $n_k<n \leq n_{k+1}$ and $\frac{\mathbb{E}S_{n_{k+1}}}{\mathbb{E}S_{n_k}} \to 1$ as $k \to \infty$ we conclude $\frac{S_n}{\mathbb{E}S_n} \to 1$ almost surely.

  • $\begingroup$ What a nice answer...I've read it several times and I enjoy the way you use different tools to solve this question...thank a lot. $\endgroup$ – Sam Nov 28 '12 at 0:23

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