Suppose $Y_n$ are i.i.d. random variables with $E[Y_i]=0$, $\operatorname{Var}(Y_i)=C>0$, and denote $X_n=\frac{1}{n}Y_n$. Does $\sum_{i=1}^n X_i$ converge to $N(0,\sigma^2)$ ($\sigma^2=\frac{\pi}{6}\cdot \operatorname{Var}(Y_i)$) or is it possible for it to converge to something else? (From Kolmogorovs three-series theorem, the sum converges almost surely).

Is there a simple counter example?

  • $\begingroup$ I believe the variance of the sum should be $\sigma^2=\frac{\pi^2}6\cdot \operatorname{Var}(Y_i)$. $\endgroup$
    – robjohn
    Oct 24, 2018 at 9:31

1 Answer 1


It can converge to all kinds of distributions. If the common distribution is uniform on $(-1,1)$ then the characteristic function is $\phi (t)=\frac {sin (t)} t$. The characteristic function of $\sum X_n$ is $\prod \frac {sin (t/n)} {t/n}$ which has zeros. Since the characteristic function of an an infinitely divisible distribution never vanishes it follows that the the sum has a distribution that is not even infinitely divisible, so certainly not normal.


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