# Problem on Weak Law of Large Numbers

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!

• $Var(\bar{X_n})$ does not tend to 1/2 for $a> 1/2$. – zoidberg Dec 8 '18 at 7:08
• Sorry i meant it is greater than $1/2$ – user587126 Dec 8 '18 at 7:09
• 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 $a>1/2$? – zoidberg Dec 8 '18 at 7:13
• 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 $o(n^2)$. – zoidberg Dec 8 '18 at 7:22
• You compare $1+2^{2a}+...+n^{2a}$ to $\int_1^n x^{2a}dx$ – 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.

• 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. – zoidberg Dec 8 '18 at 7:12
• @norfair Sorry, was a bit too rough; should be fine now. – saz Dec 8 '18 at 7:15