Question-
$X_n$ can take only two values $n^a$ and $-n^a$ with equal probabilities. Show that we can apply weak law of large numbers to the sequence of independent random vatiables ${X_n}$ if $a<\frac{1}{2}$.
We have to show that$Var(\overline{X_n})$ $\to 0$ as $n\to\infty$. I can show that if $a>1/2$ then $Var(\overline{X_n})$ does not tend to $0$ but i can not prove that wlln can be applied if $a<1/2$.
Any help would be appreciated!
Best Answer
Since you don't provide us with any details on your calculations, it is hard to say where you went wrong.
Since the random variables $X_n$, $n \geq 1$, are independent and have mean $0$, it holds for $\bar{X}_n := n^{-1} \sum_{i=1}^n X_i$ that
$$\text{var}(\bar{X}_n) = \frac{1}{n^2} \sum_{i=1}^n \text{var}(X_i) = \frac{1}{n^2} \sum_{i=1}^n \mathbb{E}(X_i^2).$$
By assumption,
$$\mathbb{E}(X_i^2) = \frac{1}{2} (i^a)^2 +\frac{1}{2} (-i^a)^2 = i^{2a},$$
and so
$$\text{var}(\bar{X}_n)= \frac{1}{n^2} \sum_{i=1}^n i^{2a}.$$
Since $i^{2a} \leq n^{2a}$ for any $i \in \{1,\ldots,n\}$ this implies
$$\text{var}(\bar{X}_n)\leq \frac{n^{2a}}{n} \xrightarrow[2a<1]{n \to \infty} 0.$$
Applying Markov's inequality (/Tschebysheff inequality) it follows that the weak law of large numbers holds.