# show partial sum converges almost surely

I am trying to show giving a sequence $(X_n) \subseteq L^2$ of independent random variables with zero mean and $\sum_{n\in N}E[X_n^2]< \infty$ , $\sum_{n\in N}X_n$ converges a.s.

For this I wanna get $\sum_{n\in N} \mathbb{P}(|\sum_{k=1}^{n}X_k|>\epsilon)<\infty$. As if I can show it , then by Borel-Cantelli lemma, $\sum_{n\in N}X_n$ converges to zero almost surely.

But I dont know how to get $\sum_{n\in N} \mathbb{P}(|\sum_{k=1}^{n}X_k|>\epsilon)<\infty$,

I tried this:

By Chebychev's inequality, $\mathbb{P}(|\sum_{k=1}^{n}X_k|>\epsilon)\leq \mathbb{E}[X_n^2]/\epsilon^2$, follows that $\sum_{n\in N}\mathbb{P}(|\sum_{k=1}^{n}X_k|>\epsilon)\leq\sum_{n\in N}\mathbb{E}[X_n^2]/\epsilon^2$. However, letting $\epsilon\longrightarrow 0$ doesn't make the right side bounded. Seems doesnt work...

I am wondering if anyone can show the correct way to do it. Thanks in advance.

• I don't understand: if you can show what you said after the "I can get", then almost surely, $\sum_nX_n=0$. But it doesn't need to be the case. Aug 13, 2012 at 15:32
• I need some help to prove that $\sum_{n\in N} \mathbb{P}(|\sum_{k=1}^{n}X_k|>\epsilon)<\infty$ Aug 13, 2012 at 16:31
• @lindamac This inequality is false. That's why you can't prove it.
– user940
Aug 13, 2012 at 16:40

1. $\sum_{j=1}^{+\infty}P(|X_j|>1)$ is convergent. It's the case, as $P(|X_j|>1)\leq EX_j^2$.
2. $\sum_{j=1}^{+\infty}E[X_j\chi_{\{X_j\leq 1\}}]$ is convergent. We have, as $EX_j=0$, $$E[X_j\chi_{\{X_j\leq 1\}}]=-E[X_j\chi_{\{|X_j|>1\}}]\leq \sqrt{EX_j^2}\sqrt{P(|X_j|\geq 1)}\leq EX_j^2.$$
3. $\sum_{j=1}^{+\infty}\operatorname{Var}(X_j\chi_{\{|X_j|\leq 1\}})$ is convergent. But since the expectation of $X_j$ is $0$, $$\operatorname{Var}(X_j\chi_{\{|X_j|\leq 1\}})\leq \operatorname{Var}(X_j)=EX_j^2,$$ and we can conclude.