# Second Borel-Cantelli lemma via the moment method

Let $${E_1,E_2,\dots}$$ be a sequence of jointly independent events. If $${\sum_{n=1}^\infty {\bf P}(E_n) = \infty}$$, show that almost surely an infinite number of the $${E_n}$$ hold simultaneously. (Hint: compute the mean and variance of $${S_n= \sum_{i=1}^n 1_{E_i}}$$. One can also compute the fourth moment if desired, but it is not necessary to do so for this result.)

Question: From the hint, we want $${\bf P}(\lim_n S_n = \infty) = 1$$. i.e., $$S_n$$ diverges almost surely, yet I didn’t see immediately how the mean $${\bf E}(S_n) = \sum_{i=1}^n {\bf P}(E_i)$$ and the variance $${\bf Var}(S_n) = \sum_{i=1}^n {\bf P}(E_i)(1 - {\bf P}(E_i))$$ is applicable here, the Chebyshev’s inequality does not seem to convey too much information.

$$P\left(\vert S_n - E S_n \vert > \dfrac{1}{2}ES_n\right) \le \dfrac{4\text{Var}(S_n)}{(ES_n)^2} = \dfrac{4\sum_{k = 1}^n P(E_k)(1 - P(E_k))}{(\sum_{k = 1}^n P(E_k))^2} \le \dfrac{4}{\sum_{k = 1}^n P(E_k)} \rightarrow 0$$ Or, equivalently, $$P\left(\vert S_n - E S_n \vert \le \dfrac{1}{2}ES_n\right) \rightarrow 1$$ But, $$\{\vert S_n - ES_n \vert \le \dfrac{1}{2}ES_n\} \subseteq \{S_n \ge \dfrac{1}{2}ES_n\}$$ Therefore, $$P\left(S_n \ge \dfrac{1}{2}ES_n\right) \rightarrow 1$$ Now, for $$M > 0$$, we can choose $$n_0$$ such that $$\dfrac{1}{2} ES_n \ge M \ \forall n \ge n_0$$ Thus, $$P(S_n \ge M) \ge P\left(S_n \ge \dfrac{1}{2}ES_n\right) \forall n \ge n_0$$ Finally, since $$\bigcap_{M \in \mathbb{N}} \bigcup_{n \in \mathbb{N}} \{S_n \ge M\} \subseteq \{\lim_n S_n = \infty\}$$ we can conclude that $$\mathbb{P}(\lim_n S_n = \infty) = 1$$
• The independence assumption is used in the calculation of the variance: $$\text{Var}(S_n) = \sum_{k = 1}^n \text{Var}(1_{E_k}) = \sum_{k = 1}^n P(E_k)(1 - P(E_k))$$ Commented Feb 17 at 3:06
One can use the Paley-Zygmund inequality (the second moment method). As the sequence $$S_n$$ is non-decreasing, $$\lim_n S_n = \lim \sup_{n \to \infty} S_n$$, and it suffices to show that $${\bf P}(\lim \sup_{n \to \infty} S_n = \infty) = 1$$.
Fix some large $$M > 0$$. For any $$0 \leq \theta \leq 1$$ small, let $$n' = n(\theta)$$ be sufficiently large with $$\theta{\bf E}S_{n'} > M$$. For all $$n \geq n'$$, the Paley-Zygmund inequality gives $${\bf P}(S_n > M) \geq {\bf P}(S_n > \theta {\bf E}S_n) \geq (1 - \theta)^2\frac{({\bf E}S_n)^2}{({\bf E}|S_n|^2)} = (1 - \theta)^2$$ (by the independence hypothesis, $$({\bf E}S_n)^2 = {\bf E}|S_n|^2$$). In particular, $${\bf P}(S_n > M) \to 1$$.
We thus obtain $${\bf P}(\lim \sup_{n \to \infty} S_n > M) = \lim_{N \to \infty}{\bf P}(\bigvee_{n \geq N}(S_n > M)) \geq \lim_{N \to \infty} {\bf P}(S_N > M) = 1$$, since $$M$$ is arbitrary, the claim follows from continuity from above.