What if the random variable are not IID, and we sum them?

Question

Let $a_n > 0$ be decreasing with $\sum a_n^2=1$. Starting at the origin we do drunken random walk on the real line, with step size $a_1,a_2...$, so the total deviation is $\sum a_n^2=1$. If our position is $s_n$ at step $n$, is $s_n$ approaching to the normal distribution? That would be expected, and cool.

Answer 1

No. Intuition comes from the limiting case $a_1=1, a_2=a_3=...=0$, in which case the process only takes one step and its distribution does not converge to a Gaussian. A positive counter-example comes whenever $a_1=1-\epsilon$ for suitably small $\epsilon>0$ (shown below).

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Fix $a_1 \in (1/2,1)$. Let $\{a_n\}_{n=2}^{\infty}$ be any positive numbers that satisfy $\sum_{n=2}^{\infty} a_n^2 = 1-a_1^2$. Let $\{X_i\}_{i=1}^{\infty}$ be iid with $P[X_i=-1]=P[X_i=1]=1/2$ and define $S_n = \sum_{i=1}^n a_i X_i$. The variance of $S_n$ indeed converges to $1$.

However, for $n \in \{2, 3, 4, ...\}$:
$$ \left\{S_n \in [-1/4 , 1/4] \right\} \subseteq \left\{\left|\sum_{i=2}^n a_iX_i\right|\geq 1/4\right\}$$ Thus: \begin{align} P[S_n \in [-1/4, 1/4]] &\leq P\left[\left|\sum_{i=2}^n a_iX_i\right|> 1/4\right]\\ &\leq \frac{E\left[(\sum_{i=2}^n a_iX_i)^2\right]}{(1/4)^2}\\ &= 16\sum_{i=2}^n a_i^2 \\ &\leq 16(1-a_1^2) \end{align} This can be made arbitrarily small by choosing $a_1$ very close to 1. So then when $a_1$ is sufficiently close to $1$, $P\left[S_n \in [-1/4,1/4]\right]$ cannot converge to $\int_{-1/4}^{1/4}\frac{1}{\sqrt{2\pi}} e^{-x^2}dx$.

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A deterministic example is when $a_i = \frac{2^{-i}}{2}$ for $i \in \{2, 3, 4, ...\}$ and $a_1^2 = 1-\sum_{i=2}^{\infty} a_i^2$. Then $\sum_{i=2}^{\infty} a_i = 1/4$, $\sum_{i=2}^{\infty} a_i^2 < 1/4$, and $a_1>\sqrt{3/4}>1/2$. So then the process moves a distance larger than 1/2 on the first jump, but a sum total distance at most 1/4 on all other jumps. So $P[S_n \in [-1/4, 1/4]]=0$ for all $n$.