Show that pairwise uncorrelated Random Variables $X_1,X_2,...$ with $P(X_m = m)=\frac{1}{2}= P(X_m=-m)$ don't obey the weak law of large numbers.

Question

Let $X_1,X_2,...$ be a sequence of pairwise uncorrelated Random Variables with $P(X_m = m)=\frac{1}{2}= P(X_m=-m)$. I want to show that this sequence doesn't obey the weak law of large numbers. Here's what I've got so far: Obviously, $E[X_m] =0, \hspace{2mm} \forall m \in \mathbb{N}$. So what we need to show is that $\exists \hspace{1mm} \epsilon$ such that $$\lim_{n \to \infty} P \left( \left|\frac{1}{n}\sum_{m=1}^{n}(X_m) \right| \geq \epsilon \right) > 0$$ if I'm not mistaken. Now intuitively what we need to show is that the sum of increasing (in absoulute value) integers with random signs doesn't zero out, i.e. the summands don't cancel in the long run. My guess is that I need to bound the expression $\left|\frac{1}{n}\sum_{m=1}^{n}(X_m) \right| $ from below somehow and that that doesn't converge (in probability) to zero either, but I'm very unsure, as I'm rather new to rigorous probability proofs.

Answer 1

The weak law of large numbers states that $\frac{S_n}{n} \xrightarrow{P} 0$ in our case, where $S_n = \sum_{i=1}^n X_i$. We want to contradict this.

Note that if this were true, then $P(\frac{S_n}{n} > \epsilon) \to 0$ as $n \to \infty$. Now, of course, note that $\frac{n-1}{n} \xrightarrow{P} 1$(think of them as constant random variables) , so we get: $$\frac{S_{n-1}}{n} = \frac{S_{n-1}}{n-1}\frac{n-1}{n} \xrightarrow{P}0 \times 1 = 0$$

(As an exercise, if $X_n \to X$ and $Y_n \to Y$ in probability, then $X_nY_n \to XY$ in probabilty).

Hence, $\frac{X_n}{n} = \frac{S_{n}-S_{n-1}}{n} \xrightarrow{P} 0$. However, this is not true in our case, since $\left|\frac{X_n}{n}\right| = 1$ for all $n$.

Hence, $X_n$ does not satisfy the weak law of large numbers.