Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am working through Achim Klenke's text entitled "Probability Theory", and I came across the following interesting exercise:

Let $(X_i)_{i\in\mathbb{N}}$ be independent, square-integrable random variables with $\mathbb{E}(X_i)=0$ for all $i$. Suppose that $\sum_{i=1}^\infty \mathbb{E}(X_i^2)<\infty$. Conclude that there exists a real random variable $X$ with $\sum_{i=1}^n X_i \xrightarrow{n\to\infty} X$ almost surely.

I attempted to prove this via Borel-Cantelli, namely, I tried to show that $\mathbb{P}(\{\omega:\sum_{i=n}^\infty X_i(\omega)\xrightarrow{n\to\infty}0 \})=1$, since the sequence will be summable if and only if the remainders are going to zero. In the details of B-C, though, for a fixed $\epsilon>0$ this requires showing that $\mathbb{P}(|\sum_{i=n}^\infty X_i| > \epsilon \;\;\;i.o.) =0$. An application of Chebyshev's inequality and using independence then gives

$$\mathbb{P}\left(\left|\sum_{i=n}^\infty X_i\right|>\epsilon\right) \leq \frac{1}{\epsilon^2}\sum_{i=n}^\infty \mathbb{E}(X_i^2)<\infty.$$ But now we certainly need not have that this is summable over all $n$ (take $X_i$ to be Bernoulli with possible values $\pm 1/i$).

I imagine my choice of Chebyshev's wasn't strong enough, or the entire approach is off. Suggestions?

share|cite|improve this question
Another neat proof of this fact, though maybe not the one you want right now, is that $M_n = \sum_{i=1}^n X_i$ is a martingale which is $L^2$-bounded and hence uniformly integrable. – Nate Eldredge Oct 22 '12 at 13:01
The three series theorem of Kolmogorov gives necessary and sufficient conditions for almost sure convergence of random series, so I imagine it'd be useful here:'s_three-series_theorem – dsaxton Jun 28 '15 at 20:03
up vote 0 down vote accepted

We can use this answer to see that we just have to check convergence in probability, what it's done in the OP.

share|cite|improve this answer
I still wonder whether there is a Borel-Cantelli approach to this problem. – Matt Spencerman Oct 23 '12 at 12:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.