$X_n$ 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!

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    $\begingroup$ $Var(\bar{X_n})$ does not tend to 1/2 for $a> 1/2$. $\endgroup$ – zoidberg Dec 8 '18 at 7:08
  • $\begingroup$ Sorry i meant it is greater than $1/2$ $\endgroup$ – user587126 Dec 8 '18 at 7:09
  • $\begingroup$ OK. I have a hard time seeing how an argument for $a>1/2$ doesn't immediately lead to the corresponding argument for $a<1/2$. Can you sketch out your argument for $a>1/2$? $\endgroup$ – zoidberg Dec 8 '18 at 7:13
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    $\begingroup$ I think you want $2a$ in those exponents instead of $a$, but basically yes. I guess you weren't sure how to sum the numerator in general. saz below provides a solution by just bounding each term by $n^{2a}$ which certainly works. Do you know about the integral test? It isn't hard to show that $1+2^{2a}+...+n^{2a}$ grows asymptotically as $n^{2a+1}/(2a+1)$. Then plugging in $a<1/2$ shows you that the expression is $o(n^2)$. $\endgroup$ – zoidberg Dec 8 '18 at 7:22
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    $\begingroup$ You compare $1+2^{2a}+...+n^{2a}$ to $\int_1^n x^{2a}dx$ $\endgroup$ – zoidberg Dec 8 '18 at 8:19

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.

  • $\begingroup$ I don't think that last step is quite right. The bound isn't $1/n$. For instance, if $a=1/2$, the variance tends to 1/2. $\endgroup$ – zoidberg Dec 8 '18 at 7:12
  • $\begingroup$ @norfair Sorry, was a bit too rough; should be fine now. $\endgroup$ – saz Dec 8 '18 at 7:15

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