Given a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, let $\{X_n: n\ge 1\}$ be sequence of square integrable random variables, i.e., $X_n \in L^2(\Omega, \mathcal{F}, \mathbb{P})$ for each $n\ge 1$. Further assume that $\mathbb{E}[X_i X_j] = 0$ whenever $i\neq j$ and $\sup_n \mathbb{E}[X_n^2] < \infty$. For each $n \ge 1 $ set $S_n := \sum_{j=1}^n X_j$.

I am trying to show that:

(i) for each $\alpha > \frac{1}{2}$, $\frac{S_n}{n^{\alpha}}\rightarrow 0$ in probability.

(ii) for each $\alpha > 1$, $\frac{S_n}{n^{\alpha}}\rightarrow 0$ a.s.

(iii) $\{\frac{S_n}{n}: n \in\mathbb{N}\}$ is uniformly integrable.

My questions:

In part (i) and (ii) is the assumption "$\mathbb{E}(X_n) = 0$ for each $n$" missing? Or these results hold without this extra assumption?

Note that the said assumption, along with $\mathbb{E}[X_i X_j] = 0$ imply that $X_j$'s are uncorrelated and hence one can use

  • an argument similar to the proof of WLLN (Chebyshev) to show $\mathbb{P}\left(\frac{|S_n|}{n^\alpha}\ge \epsilon\right) \le \frac{n \sup_n \mathbb{E}[X_n^2]}{n^{2\alpha}\epsilon^2}, \alpha > 1/2$ for part (i), and

  • an argument similar to the proof of SLLN (Rajchman) to show $\mathbb{P}\left(\frac{|S_n|}{n^\alpha}\ge \epsilon \text{ infinitely often}\right) \le \mathbb{P}\left(\frac{|S_n|}{n}\ge \epsilon \text{ infinitely often}\right) = 0, \alpha >1$.

If the mentioned assumption is not indeed necessary, how can one go about solving this problem? Also any hint for part (iii) is appreciated.

  • $\begingroup$ Why do you think that you need $\mathbb{E}(X_n)$ for the proof of (i) (using Chebyshev)? $\endgroup$ – saz Nov 1 '18 at 16:08
  • $\begingroup$ Since in Chebyshev LLN, where $\alpha =1$, what's proven is $\frac{S_n - \mathbb{E}[S_n]}{n} \rightarrow 0$, in prob. And WLOG, $\mathbb{E}[X_n] = 0$ is assumed. $\endgroup$ – math_enthuthiast Nov 1 '18 at 16:19
  • $\begingroup$ If you have a sequence $(Z_j)_j \subseteq L^2$ of uncorrelated identically distributed random variables, then $X_j := Z_j-\mathbb{E}(Z_j)$ satisfies the assumption of part (i), and you can apply it to recover the result which you mentioned in your previous comment. $\endgroup$ – saz Nov 1 '18 at 17:47

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