Central Limit Theorem with Bounded Sum of Variances?

Question

I have a sequence of bounded independent random variables $X_1,...,X_n,...$ satisfying $\sum_{i=1}^{\infty} \mathbb{E}[X_i] < \infty$, $\sum_{i=1}^{\infty} Var[X_i] < \infty$. Most versions of the central limit theorem I've found online require $\sum_{i=1}^{\infty} Var[X_i] = +\infty$ in order to be applied. Is there some limit theorem which can give me the limiting distribution of $$Pr[\frac{\sum_{i=1}^{N}{X_i}-\mathbb{E}[X_i]}{\sqrt{\sum_{i=1}^{N} Var[X_i]}}\leq y].$$

Is there a reason why the central limit theorem would not apply?

Answer 1

Heuristically, if the sum of the variances isn't infinite, there is some residual shape information in the sum about some of the individual variables. For example, if $$\mathrm{var}[X_1]=\epsilon\sum_i \mathrm{var}[X_i]$$ then $X_1$ makes up $\epsilon>0$ of the limiting random variable and the shape of $X_i$ (tails, moments, etc) has $\epsilon$ effect on the shape of the limiting distribution.

One way to think about why there's a problem is the Levy-Cramér theorem, which says that if $Y_1$ and $Y_2$ are independent and not constant, and $Y_1+Y_2$ has a Normal distribution, then both $Y_1$ and $Y_2$ have Normal distributions.

Now take $Y_1$ to be $X_1$ and $Y_2$ to be the sum of the rest of the sequence. If $\sum_i \mathrm{var}[X_i]=S^2<\infty$, then $Y_1$ and $Y_2$ are non-constant independent random variables. Unless they are both Normal, their sum isn't Normal -- and so the sum of the series isn't Normal. You can see this argument breaks down if $S^2$ is infinite, as then $Y_1/S$ is constant.

In special cases you can think about this in terms of moments. For example, suppose $Y_2$ is Normal but $Y_1$ has non-zero skewness. Then $Y_1+Y_2$ will have non-zero skewness. The Levy-Cramér argument does the same sort of thing, only a lot more general.

[If only finitely many $\mathrm{var}[X_i]$ are non-zero, the Levy-Cramér argument is much more direct, since the ones with non-zero variance immediately have to be Normal, but that's a special case]