How can I prove this result of the BC lemma?

Question

Corollory 4

If $\sum_{n = 1}^\infty P(A_n) = \infty$ and $$ \underset{n\rightarrow\infty}{\underline{\lim}} \frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} = 1, $$ then $P\Big(\underset{n\rightarrow\infty}{\overline{\lim}}A_n\Big) = 1$.

In this post, I am looking at Theorem 5 which states:

Theorem 5 If $\sum_{n=1}^\infty P(A_n) = \infty$ and if the $A_n$'s are negatively correlated in the sense that for ever $i,j \in \{1,2,...\}$ such that $i\neq j$,

$P(A_i \cap A_j) \leq P(A_i)P(A_j)$, then

$P\Big(\underset{n\rightarrow\infty}{\overline{\lim}}A_n\Big) = 1$.

Proof

By Corollary 4 it suffices to show that $$ \underset{n\rightarrow\infty}{\underline{\lim}} \frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} \leq 1. $$

The above statement is what I am having trouble with. Why is it that because of Corollary 4, all you need to show is that inequality? What does that inequality mean in words?

I'm even more confused because I am also looking at another proof which uses:

$$P(\bigcup_{k=1}^{n} A_k) \geq \frac{\bigg(\sum_{k=1}^n P(A_k)\bigg)^2}{\sum_{k,j=1}^n P(A_k\cap A_j)}$$

are these the same things?

This is the full proof of the second post I am looking at and I don't understand why they are both saying the same things:

can someone please show me how to adjust the proof for when C=1

Answer 1

1. Why is it that because of Corollary 4, all you need to show is that inequality?

Because $\liminf\frac{\sum P(A_i\cap A_j)}{(\sum P(A_i))^2}\ge1$ is rather obvious. To see this, note that $(\sum P(A_i))^2=(E\sum1_{A_i})^2\le E(\sum1_{A_i})^2=\sum P(A_i\cap A_j)$, where we used Jensen's inequality.

2. Can someone please show me how to adjust the proof for when $C=1$?

Our goal is to show that $P(A_n~i.o)\ge1$, i.e., $P(\cap_{n\ge1}\cup_{k\ge n}A_k)=1$.

Now if we write $B_n=\cup_{k\ge n}A_k$, then it will be enough to show $P(B_n)=1$ for all $n$.

Fix $n$. Next let us lower bound $P(B_n)$. Observe that for any $N>n$, we have that $$P(B_n)=P(\cup_{k=n}^\infty A_k)\ge P(\cup_{k=n}^N A_k)$$ and by Chung-Erdos Inequality we have $$P(\cup_{k=n}^N A_k)\ge\frac{(\sum_{n\le i\le N}P(A_i))^2}{\sum_{n\le i,j\le N}P(A_i\cap A_j)}.$$ As $P(A_i\cap A_j)\le P(A_i)P(A_j)$ for any $i\neq j$, it follows that $${\sum_{n\le i,j\le N}P(A_i\cap A_j)}\le\sum_{n\le i\le N}P(A_i)+\sum_{n\le i\neq j\le N}P(A_i)P(A_j)\le\sum_{n\le i\le N}P(A_i)+(\sum_{n\le i\le N}P(A_i))^2.$$ This implies that $$\frac{(\sum_{n\le i\le N}P(A_i))^2}{\sum_{n\le i,j\le N}P(A_i\cap A_j)}\ge\frac{{\sum_{n\le i\le N}P(A_i)}}{1+{\sum_{n\le i\le N}P(A_i)}}.$$ Putting everything together, we can conclude that for any $N>n$, it holds that $$P(B_n)\ge\frac{{\sum_{n\le i\le N}P(A_i)}}{1+{\sum_{n\le i\le N}P(A_i)}}.$$ From our assumption we know $\sum_{i=n}^\infty P(A_i)=\infty$. So when $N\rightarrow\infty$, $\frac{{\sum_{n\le i\le N}P(A_i)}}{1+{\sum_{n\le i\le N}P(A_i)}}\rightarrow 1$. Therefore $P(B_n)\ge1$ and thus $P(B_n)=1$.