# Convergence of Distribution

Let $X_1,X_2,\dots$ be independent random variables such that $P(X_j=+a_j)=P(X_j=-a_j)=1/2$. Is it the case that for all sequences {$a_j: j\geq 1$}, where $a_j>0$ and $\sum_{j=1}^{\infty } a_j^{2}=\infty$, $\frac{X_1+\cdots+X_n}{\sqrt{a_1^2+\cdots+a_n^2}}$ converges to a standard normal distribution?

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No.    – Did Nov 14 '11 at 21:06
What do you know, what did you try ? – Did Nov 14 '11 at 21:07
You are definitely gonna need to put additional conditions on $\{a_i\}$ because if $a_i=2^i$ then $X_1+...+X_n$ will never be zero, so the limit must be zero for the value $0$, which means it cannot be normal. – Thomas Andrews Nov 14 '11 at 21:33

The characteristic function for $T_n:=\frac{X_1+\cdots+X_n}{\sqrt{a_1^2+\cdots+a_n^2}}$ is $$\phi_n(t)=\prod_{j=1}^n \cos\left(t a_j/ \sqrt{a_1^2+\cdots+a_n^2}\right).$$ If $T_n$ converges in distribution to a standard normal, we have $\phi_n(t)\to \exp(-t^2/2)$ for all $t\in\mathbb{R}$.

Here's a way to make this fail. Set $a_1=1$, and inductively define $a_n$ to be the positive square root of $$(\pi/2)^2(a_1^2+\cdots+a_{n-1}^2)/(4-(\pi/2)^2).$$ This guarantees that $\phi_n(2)=0$ for all $n>1$, so $\phi_n(t)\not\to\exp(-t^2/2)$ at $t=2$.

Update: As Didier Piau points out in a comment, an easier solution is to take $a_n=2^n$. Then the sum of random variables is smaller than $2\cdot2^n$ and the square root is greater than $2^n$, hence the ratio is at most $2$, almost surely.

Note that the normalized sum of bounded, mean-zero random variables does converge to a normal limit under mild conditions. The OP is missing some condition, like Lindeberg's, so that the contribution from every summand is comparable. That is, for each finite sequence $a_1, a_2,\dots, a_n$, no one $a_j$ should dominate.

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Byron: nice. Maybe simpler: take $a_n=2^n$. Then the sum of random variables is smaller than $2\cdot2^n$ and the square root is greater than $2^n$, hence the ratio is at most $2$, almost surely. – Did Nov 27 '11 at 12:01
Didier You're right, that is a better solution. If you write it up, I'll vote for it and we can put this question to bed. – Byron Schmuland Nov 27 '11 at 15:11
Byron: No need to, please add it to your post. – Did Nov 27 '11 at 16:09