Central limit of independent indicator functions

Question

Suppose $\{A_n\}_{n=1}^\infty$ is a sequence of independent events, each with probability $\mathbb{P}(A_n) = p_n$ such that $\sum_{n=1}^\infty p_n = \infty$. The goal here is to prove a stronger version of the second Borel-Cantelli Lemma using Lindeberg-Feller Central Limit Theorem; here is my claim which I'm trying to prove $$ \frac{1}{\sigma_n}\left(\frac{\sum_{k=1}^n \mathbf{1}_{A_k}}{\sum_{k=1}^np_k}-1\right)\overset{\mathcal{D}}{\to}N(0,1)\quad\text{as}\quad n\to\infty. $$ where $\sigma_n^2:=\mathrm{Var}\left(\frac{\sum_{k=1}^n \mathbf{1}_{A_k}}{\sum_{k=1}^np_k}-1\right)$.

To prove this, I constructed a triangular array $\{X_{n,k}\}$ for $n\in\mathbb{N}$ and $k=1,\ldots,n$ with $$ X_{n,k}:=\frac{\mathbf{1}_{A_k}-p_k}{\sum_{k=1}^n p_k} $$ so that $\mathbb{E}[X_{n,k}]=0$ and $X_{n,1},\ldots,X_{n,n}$ are independent within each row $n$. The first hypothesis of Lindeberg-Feller CLT is satisfied: $$ \sum_{k=1}^n\mathbb{E}[(X_{n,k}/\sigma_n)^2] = 1 $$ for all $n\in\mathbb{N}$, so the limit as $n\to\infty$ of the equation above is $1>0$. To deduce the claim, I need to show that the other hypothesis is also satisfied: $$ \sum_{k=1}^n\mathbb{E}\left[(X_{n,k}^2/\sigma_n)^2\cdot\mathbf{1}_{|X_{n,k}|>\varepsilon}\right]\to 0\quad\text{as}\quad n\to\infty $$ for all $\varepsilon>0$.

Here are my questions:

1. How do I prove that the claim is true (if it is true at all)? More specifically, how do I show that the second hypothesis is satisfied?
2. The claim only shows convergence in distribution. How do I deduce the convergence almost surely to $0$ of the term in the bracket, i.e., $$ \frac{\sum_{k=1}^n \mathbf{1}_{A_k}}{\sum_{k=1}^np_k}-1\to 0\quad\text{a.s.?} $$

Answer 1

The conclusion you presented is not correct. Assuming $\lim_{n\to\infty}p_n = 1$ you will see that no central limiting behaviors happen. Specifically, consider the case $p_1 = p$, $p_2=p_3=\cdots = 1$. Then the quantitative Borel Cantelli Lemma still holds, but $\sum_k \mathbf{1}_{A_k}$ is a Bernoulli random variable.

Assuming $\lim_{n\to\infty}p_n \neq 1$ will be a necessity here. With this assumption, apply LF-CLT to $$X_{n,k} = \frac{\mathbf{1}_{A_k} - p_k}{\sqrt{\sum_{k=1}^n p_k(1-p_k)}}$$ you will see why we need $\lim_{n\to\infty}p_n \neq 1$.

Note that LF-CLT requires that $\sum_k\mathbb{E}X_{n,k}^2 \to \sigma^2>0$, your definition of $X_{n,k}$ does not satisfy this condition.