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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Oct 22 '12 at 13:01
  • $\begingroup$ 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: en.wikipedia.org/wiki/Kolmogorov's_three-series_theorem $\endgroup$ – dsaxton Jun 28 '15 at 20:03

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

  • $\begingroup$ I still wonder whether there is a Borel-Cantelli approach to this problem. $\endgroup$ – Matt Spencerman Oct 23 '12 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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