# Questions on proving the lower bound for $P\left(\limsup\limits_{n\rightarrow \infty} A_n\right) \geq \frac{1}{C}$

This is Lyapunov's inequality for moments of a random variable (the paper can be accessed here):

Let $$\{F_{k}\}, k = 1, 2, ..., N$$ be an arbitrary sequence of events in $$(\Omega, F, P)$$.

We have, if $$P\left(\bigcup\limits_{k=1}^{n} F_k \right) > 0$$,

(1) $$2 \sum_{1\leq j < k \leq N} P(F_{j}F_k) \geq \Bigg[P\bigg(\bigcup_{k=1}^N F_k\bigg)\Bigg]^{-1}\Bigg(\sum_{k=1}^N P(F_k)\Bigg)^{2} - \sum_{k=1}^N P(F_k)$$

Proof: Define r.v. $$X_k(\omega)= \begin{cases} 0, & \text{if \omega \notin F_k} \\[2ex] 1, & \text{if \omega \in F_k} \end{cases}$$

The following identity is evident:

(2) $$2 \sum\limits_{1\leq j < k \leq N} P(F_j F_k) = E\left[\left(X_1+...+X_N\right)^{2}\right] - E\left(X_{1}^{2} +...+ X_{N}^{2}\right)$$

Now by the Schwarz inequality we have

(3) $$[E(X_1+...+X_N)]^2 \leq P(X_1+...+X_{N}>0)E[(X_{1}+...+X_N)]^2$$

Since $$E(X_k) = E(X_{k}^{2}) = P(F_k),$$

$$P(X_1+...+X_N>0) = P\left(\bigcup\limits_{k=1}^{N} F_k\right)$$ by definition, (1) follows from (2) and (3)

How can I strengthen Lyapunov's inequality for moments of a random variable so that this new inequality:

$$P\bigg(\bigcup_{k=n}^{N} A_k\bigg) \geq \frac{\big(\sum_{k=n}^{N} P(A_k)\big)^{2}}{\sum_{k,j=n}^{N} P(A_k A_j)}$$ holds?

• I don't quite understand. The inequality you displayed/proved is equivalent to the final inequality you hope to prove? Or are you talking about applying Chung-Erdos for $P(\limsup_n A_n)$? Mar 1 '19 at 15:36
• basically in Lyapunov's inequality, i have an extra term, $\sum_{k=1}^n F_k$ being subtracted and I don't know how i can strengthen that inequality so that term isn't there in the inequality I eventually want to show Mar 1 '19 at 15:54
• But $\sum_{j,k=1}^{N}P(F_j F_k) = 2\sum_{1 \le j < k \le N}^{N}P(F_j F_k) + \sum_{j=1}^{N}P(F_j)$. That extra term should be there, because it would actually be absorbed into the unexpanded sum $\sum_{j,k=1}^{N}P(F_j F_k)$ (with $F$'s replaced with $A$'s) in your final inequality. Mar 1 '19 at 16:19
• im really sorry, i don't know what you mean. can you please explain how it would be absorbed? i'm just confused about that term Mar 1 '19 at 16:33
• No problem, OP. I edited my answer below to reflect what I mean. Let me know if this is clear enough? Mar 1 '19 at 16:55

The inequality $$2 \sum_{1\leq j < k \leq N} P(F_{j}F_k) \geq \Bigg[P\bigg(\bigcup_{k=1}^N F_k\bigg)\Bigg]^{-1}\Bigg(\sum_{k=1}^N P(F_k)\Bigg)^{2} - \sum_{k=1}^N P(F_k)$$ can be rewritten as \begin{align*} P\bigg(\bigcup_{k=1}^N F_k\bigg) \ge \frac{\Bigg(\sum_{k=1}^N P(F_k)\Bigg)^{2}}{2 \sum_{1\leq j < k \leq N} P(F_{j}F_k) + \sum_{k=1}^N P(F_k)} \end{align*} Note that \begin{align*} \color{red}{2 \sum_{1\leq j < k \leq N} P(F_{j}F_k)} + \color{blue}{\sum_{k=1}^N P(F_k)} &= \color{red}{\sum_{j \neq k} P(F_{j}F_k)} + \color{blue}{\sum_{j = k}P(F_jF_k)} \\ &=\sum_{j,k} P(F_{j}F_k) \end{align*} Note that the red expressions equal because $$P(F_j F_k) = P(F_k F_j)$$, but since $$j < k$$ in the first sum, the $$2$$ doubles for $$j > k$$ as well. The blue expressions equal because $$P(F_k \cap F_k) = P(F_k)$$.

And so \begin{align*} P\bigg(\bigcup_{k=1}^N F_k\bigg) \ge \frac{\Bigg(\sum_{k=1}^N P(F_k)\Bigg)^{2}}{\sum_{j,k=1}^{N}P(F_j F_k) } \end{align*} We can reindex the lower index of $$1$$ to $$n$$, and change variables $$F$$ to $$A$$, and we have the desired final inequality.

Proving $$P(\limsup_n A_n) \ge 1/C$$ with this bound.

Here is the result I will prove:

If $$\sum_{n=1}^{\infty}P(A_n) = \infty$$, then
\begin{align*} P\left(\limsup_n A_n\right) \ge \limsup_{n\rightarrow\infty}\frac{(\sum_{k=1}^{n}P(A_k))^2}{\sum_{i,j=1}^{n}P(A_iA_j)} \end{align*}
In particular, it implies a stronger version of the second Borel-Cantelli: If $$P(A_i A_j) \le P(A_i)P(A_j)$$ for all $$i \neq j$$, then $$P\left(\limsup_n A_n\right) = 1$$.

Proof. Let $$a_n = (\sum_{k=1}^{n}P(A_k))^2$$ and $$b_n = \sum_{i,j=1}^n P(A_i A_j)$$. By assumption, $$a_n \rightarrow \infty$$, and by Chung-Erdos, so does $$b_n \rightarrow \infty$$. Note that $$(\sum_{k=m+1}^{n}P(A_k))^2 = (\sqrt{a_n}-\sqrt{a_m})^2$$ and \begin{align*} \sum_{i,j = m+1}^{n}P(A_iA_k) = b_n - b_m - \sum_{i=1}^{n}\sum_{j=m+1}^{n}P(A_iA_j) - \sum_{i=1}^{n}\sum_{j=m+1}^{n}P(A_iA_j) \le b_n - b_m \end{align*} And so \begin{align*} P\left(\bigcup_{k=m+1}^{\infty}A_k\right) = \lim_{n\rightarrow \infty}P\left(\bigcup_{k=m+1}^{n}A_k\right) \ge \limsup_{n\rightarrow \infty}\frac{(\sqrt{a}_n-\sqrt{a}_m)^2}{b_n - b_m} = \limsup_{n\rightarrow \infty}\frac{a_n}{b_n} \end{align*} And so $$P\left(\limsup_n A_n\right) = \lim_{m\rightarrow \infty}P\left(\bigcup_{k=m+1}^{\infty}A_k\right) \ge \limsup_{n\rightarrow \infty}\frac{a_n}{b_n}.$$

• thank you for being so patient with me. can you just explain how you know $2 \sum_{1\leq j < k \leq N} P(F_{j}F_k) + \sum_{k=1}^N P(F_k) = \sum_{j,k=1}^{N}P(F_k)$ this is true Mar 1 '19 at 17:42
• No worries. I edited my answer with more details on this step! And realized a typo; should be $\sum_{j,k=1}^{N} P(F_j F_k)$, not $\sum_{j,k=1}^{N} P(F_k)$ Mar 1 '19 at 19:14
• i think theres a typo again because you said $\sum P(F_k F_k) = \sum P(F_k)$. also are you assuming these events are independent? bc they aren't supposed to be independent so how do you know that $P(F_k F_j)$ is just $P(F_k)$? Mar 2 '19 at 5:29
• is it true because they are indicator random variables? Mar 2 '19 at 5:32
• The blue portion is $\sum_{j=k} P(F_j F_k)$, not $\sum_{j,k=1}^{N} P(F_j F_k)$; that is, we sum over when $j = k$. So $\sum_{j=k} P(F_j F_k) = \sum_{k=1}^N P(F_k F_k) = \sum_{k=1}^{N} P(F_k)$. Mar 2 '19 at 13:46