Kolmogorov's Strong Law and Almost Sure Convergence

Question

I would like a help in the following problem

$(X)_{n \geq 1}$ iid, $E(X_i) = \xi$ and $Var(X_i) = \sigma_i$.

Show that $\dfrac{X_1+X_2+\cdots+X_n}{n} \rightarrow \xi$ with probability 1, when $\sigma_i = \Delta i^{\alpha}$ and $\sigma_i = i\Delta^i$ and show restrictions for $\Delta$ and $\alpha$.

I tried to solve using Borel-Cantelli Lemma, but won't work. Then, I tried the Kolmogorov Strong Law, but the same problem emerged, the series diverge. I would appreciate some hints. Thank you.

Answer 1

I think you meant independent, but not identically distributed.

I'm not sure about necessary conditions, but here's a theorem that will allow you to find sufficient conditions on $\Delta$ and $\alpha$ (source: Probability: A Graduate Course by Allen Gut):

Theorem: (The Kolmogorov sufficient condition)

Let $\{X_n\}_n$ be independent random variables with mean $\xi$ and finite variance $\sigma_n^2$, and set $S_n = X_1+...+X_n$. If,

$$ \sum\limits_{n=1}^\infty\frac{\sigma_n^2}{n^2} < \infty $$

then the SLLN holds, i.e., $\frac{1}{n}S_n \rightarrow \xi$ with probability $1$.

For example, in the first case, $\sigma_n^2 = \Delta^2 n^{2\alpha}$, we have,

$$ \sum\limits_{n=1}^\infty\frac{\sigma_n^2}{n^2} = \Delta^2\sum\limits_{n=1}^\infty \frac{1}{n^{2-2\alpha}} $$

The above series converges if $\alpha < 1/2$. Try to do something similar for the other case.