Show that a series converges almost surely

convergence-divergencemartingalesstochastic-processes

Let $\{X,X_1,X_2,…\}$ be i.i.d random variables with mean zero. Assume that $E[X\log(1+|X|)] < \infty$, prove that the series $\sum_{n=1}^\infty (X_n/n)$ converges almost surely.

My attempt:

$S_n = \sum_{i=1}^n(X_n/i)$ is a martingale. $E[S_{n+1}|\mathscr{F}_n] = E[S_n + (X_{n+1}/(n+1)) | \mathscr{F}_n] = S_n + E[X_{n+1}/(n+1)] = S_n$.
Therefore, if $\langle S \rangle_{\infty} < \infty$, then we have that $\lim_{n \to \infty}S_n$ exists almost surely.

However, $\langle S \rangle_n = \sum_{i=1}^n (E[X_i^2]/i^2)$ and we are not given that this sequence have finite second moment. I'm guessing here is where I need to use $E[X\log(1+|X|)] < \infty$ but I'm not completely sure how that applies.

I thought that since $\log(1+|X|) \sim |X|$ we could say $E[X\log(1+|X|)] < \infty \Rightarrow E[X^2] < \infty$ but I don't think that's enough to assume $\sum_{i=1}^\infty (E[X_i^2])/i^2 < \infty$

Best Answer

Here is a proof using Kolmogorov's three-series theorem: Let $Y_n = \frac{X_n}{n} \mathbf{1}_{\{|X_n|\leq n\}}$. Then by the theorem, it suffices to show that all of the following three series converge:

$$ \sum_{n=1}^{\infty} \mathsf{P}(|X_n| > n), \qquad \sum_{n=1}^{\infty} \mathsf{E}[Y_n], \qquad \sum_{n=1}^{\infty} \mathsf{Var}[Y_n] < \infty. $$

  1. First, we have $$ \sum_{n=1}^{\infty} \mathsf{P}(|X_n| > n) = \sum_{n=1}^{\infty} \mathsf{P}(|X| > n) \leq \mathsf{E} [|X|] < \infty $$

  2. Next, using $\mathsf{E}[X] = 0$, we have $\mathsf{E}[Y_n]=-\mathsf{E}[\frac{X}{n}\mathsf{1}_{\{|X|>n\}}]$ and hence \begin{align*} \sum_{n=1}^{\infty} \left| \mathsf{E}[Y_n] \right| &\leq \sum_{n=1}^{\infty} \mathsf{E}\biggl[\frac{|X|}{n}\mathsf{1}_{\{|X|>n\}} \biggr] = \mathsf{E}\biggl[|X|\biggl( \sum_{n=1}^{\infty} \frac{1}{n}\mathsf{1}_{\{|X|>n\}} \biggr) \biggr] \\ &\leq \mathsf{E}[|X| (1 + \log(1+|X|))] < \infty. \end{align*} Here, we utilized the fact that $ \sum_{n=1}^{\infty} \frac{1}{n}\mathbf{1}_{\{ x > n\}} \leq 1 + \log(1+x) $ holds for all $x > 0$.

  3. Finally, \begin{align*} \sum_{n=1}^{\infty} \mathsf{Var}(Y_n) &\leq \sum_{n=1}^{\infty} \mathsf{E}\biggl[\frac{|X|^2}{n^2}\mathsf{1}_{\{|X|\leq n\}} \biggr] = \mathsf{E}\biggl[|X|^2\biggl( \sum_{n=1}^{\infty} \frac{1}{n^2}\mathsf{1}_{\{|X|\leq n\}} \biggr) \biggr] \\ &\leq \mathsf{E}[1+|X|] < \infty. \end{align*} Here, we utilized the fact that $ \sum_{n=1}^{\infty} \frac{1}{n^2}\mathbf{1}_{\{ x \leq n\}} \leq \frac{1}{x^2} + \frac{1}{x} $ holds for all $x > 0$.

Therefore the desired conclusion follows.

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