Borel-Cantelli lemma variation Consider $(X_i)$ independent events such that $\sum_{i = 1}^\infty P(X_i) = \infty$
Let $S_n = \sum_{i = 1}^n \mathbb{1}_{X_i}$, then $\frac{S_n}{\mathbb{E}S_n} \to 1$ almost surely.
Do you have any ideas how to prove this? I've been thinkint about applying law of large numbers or usual Borel-Cantelli lemma, but I could not come up with anything.
 A: Note that $1_{X_i} \sim \text{Bernoulli}(p_i)$ where $p_i = P(X_i)$.
If we consider a sequence of independent Bernoulli random variables $B_i$ where $P(B_i = 1)=p_i \in (0,1)$ then:
$$E[S_n] = \sum_{i=1}^n p_i,\;\; Var[S_n]= \sum_{i=1}^n p_i(1-p_i) $$
For independent but not necessarily identically distributed random variables, the following version of the strong law of large numbers holds:
$$\sum_{i=1}^{\infty} \frac{1}{k^2}Var[B_i] < \infty \implies \bar{B}_n-E[\bar{B}_n] \xrightarrow{a.s.} 0$$
Where $\bar{B}_n := \frac1n \sum_{i=1}^n B_i$ (sample average).
Since $0<Var[B_i] < 1$ we know that:
$$0<\sum_{i=1}^{\infty} \frac{1}{k^2}Var[B_i] < \sum_{i=1}^{\infty} \frac{1}{k^2} < \infty$$
Therefore,
$$\frac1n \sum_{i=1}^n B_i - \frac1n \sum_{i=1}^n p_i \xrightarrow{a.s.} 0$$
If $\frac{\sum_{i=1}^n p_i}{n} \to p$ (Cesaro convergence) and $\sum_{i=1}^n p_i > 0\;\forall n$ then :
$$\frac{\sum_{i=1}^n B_i}{\sum_{i=1}^n p_i} - 1 \xrightarrow{a.s.} 0 \implies \frac{\sum_{i=1}^n B_i}{\sum_{i=1}^n p_i} \xrightarrow{a.s.} 1$$
Stated more suggestively,
$$\frac{S_n}{E[S_n]} \xrightarrow{a.s.} 1\;\;\square$$
