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;


Further for the variance;


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.

  • 1
    $\begingroup$ For $\alpha \le 3$ the variance of $S_n$ is diverging, so be careful. Maybe check the Lindeberg conditions. $\endgroup$ Commented Dec 10, 2018 at 7:59
  • $\begingroup$ 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 $S_n$? $\endgroup$
    – whuber
    Commented Dec 10, 2018 at 15:42

1 Answer 1


You're on the right track, but one modification is needed: the variance of $S_n$ is a finite sum, because $S_n$ is a sum of a finite number of independent variables (whence their variances add):

$$\operatorname{Var}(S_n) = \operatorname{Var}\left(\sum_{i=1}^n X_i\right) = \sum_{i=1}^n \operatorname{Var}(X_i) = \sum_{i=1}^n \frac{i^{2-\alpha}}{2}.$$

The variances of the $X_i$ were computed from the definition (suitable for any discrete variable)

$$\begin{aligned} \operatorname{Var}(X_i) &= \sum_{x\in\mathbb{R}}\Pr(X_i=x) (x-E[X_i])^2\\ &= \frac{i^{-\alpha}}{4}(-i-0)^2+\frac{i^{-\alpha}}{4}(i-0)^2 + \left(1-\frac{i^{-\alpha}}{2}\right)(0-0)^2 \\ &= \frac{i^{2-\alpha}}{2}. \end{aligned}$$

The variance of $S_n$ can be closely approximated because it is an upper and lower Riemann sum for two simple integrals,

$$\frac{n^{3-\alpha}-1}{3-\alpha}=\int_1^n x^{2-\alpha}\,\mathrm{d}x \le \sum_{i=1}^n i^{2-\alpha} \le \int_0^n x^{2-\alpha}\,\mathrm{d}x=\frac{n^{3-\alpha}}{3-\alpha}.$$

Although this does grow without bound, for every $n$ it is finite and its square root can be used to normalize $S_n$ approximately, indicating you should evaluate the limiting behavior of

$$\frac{S_n - 0}{\sqrt{n^{3-\alpha}/(3-\alpha) + O(1)}} = n^{(\alpha-3)/2}S_n + O(n^{(\alpha-3)/2})$$

You should conclude that $b_n(\alpha)$ ought to be asymptotically close to $n^{(\alpha-3)/2}$ (and you can take it to equal that expression).


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