Convergence of Distribution for a Modified Sequence of IID Random Variables

Question

I have a sequence of IID random variables $S_i$ where $i = 1, 2, \dots, n$. Each $S_i$ can take values (-1, 1) with probability $\left(\frac{1}{2}, \frac{1}{2}\right)$. As we know, the distribution of the random variable $$ S = \frac{S_1 + S_2 + \dots + S_n}{n} $$ when $n$ approaches infinity, $S \sqrt n $ will approximate the normal distribution because of central limit theorem. Now, consider another set of random variables $$ T_i = \frac{S_i}{i} $$ where $i = 1, 2, \dots, n$. Define T: $$ T = \frac{T_1 + T_2 + \dots + T_n}{n} $$ what will be the distribution of $ T \sqrt n $ when $n$ approaches infinity?

Answer 1

The variance of $T_i$ is $\frac 1{i^2}$ and so $\text{Var}(\sum_{i = 1}^n T_i) = \sum_{i=1}^n \frac 1{i^2}\to \frac {\pi^2} 6$.

This means that as $n$ goes to $\infty$, the variance of $\sqrt n T$ goes to $0$, so $\sqrt n T$ converges to $0$ in $L^2$. This implies that $\sqrt n T$ goes to $0$ in distribution. As $T = \frac{\sqrt n T}{\sqrt n}$, $T$ also goes to $0$ (in $L^2$/ distribution).

Generally you want your limiting distribution to have non-zero, but finite variance and $\sum_{i=1}^\infty T_i$ should have that, but I am not sure whether that will lead to a normal distribution: The characteristic function of $T_i$ is $\cos (\frac t i)$, so the characteristic function of the sum is a product $$ \prod_{i = 1}^n \cos \left( \frac t i\right) $$ and you'll have to check whether this converges pointwise to a continuous function of $t$ (continuous at $0$ suffices).

Here is some simulation code (in R) that shows what happens with $T$ and $\sqrt{n} T$ as $n \to \infty$:

Si <- matrix(rbinom(1000 * 100, 1, .5) * 2 - 1, 1000)
Ti <- Si*1/seq(nrow(Si))

T <- apply(Ti, 2, cumsum)  / seq(nrow(Ti))
sqrtn_T <- sqrt(seq(nrow(T))) * T

par(mfrow=c(2,1))
matplot(T, type='l')
matplot(sqrtn_T, type='l', ylab = latex2exp::TeX("$\\sqrt{n} T$"))

^{Created on 2023-10-22 with reprex v2.0.2}