# CLT and convergence of Variance

I am looking at a problem where the sum of the individual $$X_i$$ is $$S_n=X_1+\dotsm+X_n$$. The probability is given as, $$P(X_i=i)=P(X_i=-i)=\frac{i^{-\alpha}}{4}$$ and $$P(X_i=0)=1-\frac{i^{-\alpha}}{2}$$.

The task is to find two functions of $$\alpha$$ such that $$(S_{n}- a_n(\alpha))/b_n(\alpha) \implies N(0,1)$$ where $$\alpha \in (0,1)$$.

By CLT $$\frac{S_n-n\mu}{\sigma\sqrt{n}} \implies N(0,1)$$

So this implies that $$a_n(\alpha)=n\mu$$ and $$b_n(\alpha)=\sigma\sqrt{n}$$.

I startet calculating the expected value, which is found as;

$$EX_i=0$$

Further for the variance;

$$Var(X_i)=\sum_{i=1}^{\infty}\frac{i^{2-\alpha}}{2}$$

But when I looked at the converging sum it is clear since $$\alpha \in (0,1)$$ the sum is divergence.

Therefore am I wondering if I have done the wrong approach, and if so how it should be done.

• For $\alpha \le 3$ the variance of $S_n$ is diverging, so be careful. Maybe check the Lindeberg conditions. – kjetil b halvorsen Dec 10 '18 at 7:59
• Hint: although the variance of $S_n$ may be expressed as a finite (not infinite!) sum with no simple closed form, it can be closely approximated by an integral whose value asymptotically is $n^{3-\alpha}/(3-\alpha).$ What happens when you use this variance to standardize $S_n$? – whuber Dec 10 '18 at 15:42