If $X_1,X_2, ....$ independent but not identical distributed such that $E[X_i] = \mu_i$ and $\sum_n \mu_i = \infty$

Question

If $X_1,X_2, ....$ independent but not identical distributed such that $E[X_i] = \mu_i$ and $\sum_n \mu_i = \infty$ Let $S_n = \sum_{k = 1}^n X_i $

Is it true in general that $\frac{S_n}{E[S_n]} \to 1$ a.e.?

I know it is true in the case where $X_i \sim$ Poisson$(\lambda_i)$.

I wanted to prove it in a similar manner to the Poisson case by in which I use Chebychev's inequality to get convergence in probability then convergence a.e. using the squeeze theorem.

However, it seems I can't use that method in general, or at least is not clear that I can.

Answer 1

Suppose $X_i\sim \mathcal N(\frac 1i, i)$ and let $Y_n:=\frac{S_n}{E(S_n)}$ so that $$Y_n\sim \mathcal N\left(1,\frac{\sum_{i=1}^n i}{\left(\sum_{i=1}^n \frac 1i\right)^2}\right)$$

The variance $V(Y_n)=\displaystyle \frac{\sum_{i=1}^n i}{\left(\sum_{i=1}^n \frac 1i\right)^2}$ is asymptotically equivalent to $\displaystyle \frac{n^2}{2\log^2n}$ which goes to $\infty$.

Suppose for the sake of contradiction that $Y_n$ converges a.s. to $1$. This implies convergence in distribution to $1$, hence for all $t\in \mathbb R$, $\lim_ne^{it-V(Y_n)t^2/2} = e^{it}$, hence $\lim_n e^{-V(Y_n)t^2/2}=1$. But $\lim_n V(Y_n)=\infty$, a contradiction.