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.
Best Answer
$$ P\left(\vert S_n - E S_n \vert > \dfrac{1}{2}ES_n\right) \le \dfrac{4\text{Var}(S_n)}{(ES_n)^2} = \dfrac{4\sum_{k = 1}^n P(E_k)(1 - P(E_k))}{(\sum_{k = 1}^n P(E_k))^2} \le \dfrac{4}{\sum_{k = 1}^n P(E_k)} \rightarrow 0 $$ 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 $\mathbb{P}(\lim_n S_n = \infty) = 1$