# Variants of weak and strong LLN

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.

• Why do you think that you need $\mathbb{E}(X_n)$ for the proof of (i) (using Chebyshev)? – saz Nov 1 '18 at 16:08
• 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. – math_enthusiast Nov 1 '18 at 16:19
• 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. – saz Nov 1 '18 at 17:47