2
$\begingroup$

Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of independent random variables with $E[X_n]=0$ for all $n\in\mathbb{N}$ and $\sum_{n=1}^\infty E[X_n^2] < \infty$. Then $S_n = \sum_{j=1}^n X_j$ converges almost surely for $n\rightarrow\infty$.

One can prove this in various ways, e.g. with

  • Martingale convergence theorem
  • Kolmogorov's three series theorem
  • Using Chebyshev's inequality to prove convergence in probability and then using Lévy's equivalence theorem

I want to do it the third way (done with the others):

Because the random variables are independent, $$\mathbb{E}[|S_n-S_m|^2] = \mathbb{E}\bigg[\bigg|\sum_{j=m+1}^n X_j\bigg|^2\bigg] = \sum_{j=m+1}^n \mathbb{E}[|X_j|^2] \rightarrow 0 \quad\text{for m, n}\rightarrow\infty$$ where the convergence holds since $\sum_{n=1}^\infty E[X_n^2] < \infty$, i.e. $b_n = \sum_{j=1}^n E[X_j^2]$ is a Cauchy sequence. Then, by Chebyshev's inequality $$P(|S_n-S_m|\geq \epsilon) \leq \frac{\mathbb{E}[|S_n-S_m|^2]}{\epsilon^2}$$ so $S_n$ is a Cauchy sequence in probability, i.e. it converges in probability, which by Lévy's equivalence theorem is equivalent to almost sure converges.

$\endgroup$
  • 1
    $\begingroup$ Because the series of variances converges, the partial sums $b_n:=\sum_{i=1}^n E[X_i^2]$, $n\ge 1$, form a Cauchy sequence. $\endgroup$ – John Dawkins Nov 24 '15 at 20:34
  • $\begingroup$ I've edited my post. Is what I wrote precise? I'm a bit unsure as I'm not very used to using Cauchy sequences. $\endgroup$ – jarchitect Nov 25 '15 at 1:58

Your Answer

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

Browse other questions tagged or ask your own question.