To Prove
In exercise 255 of Cacullos Theophilos, „Exercises in Probability“, I am asked to prove:
If the independent random variables $X_1, X_2, …$ satisfy the condition $V(X_i)≤C < ∞$ for all $i$, then the SLLN holds.
By which is meant that $1/n Σ_{i=1}^n (X_i - μᵢ) \to 0$ almost surely.
Help
Earlier in the chapter, Theophilios states Kolmogorov's variant of the SLLN, which i wanted to use:
Let $X_i$ be independent random variables such that $V(X_i)\to 0$. Then $1/nΣ_{i=1}^n (X_i-μ_i)→0$ almost surely.
In other words, if the averages $\overline{μ_n}$ of the $μ_i$ converge, then $\overline{X_n}\to \overline{μ}$.
My attempt
First centering the $X_i$ is a good idea, and dividing by $\sqrt{i}$ tightens up the variances: The random variables $Y_i≔(X_i-μ_i)/√i$ ($μ_i≔E[X_i])$ are still independent and satisfy \begin{align} E[Y_i]&=0\\ V[Y_i]&=\frac{V[X_i-μ_i]}{√i^2} = \frac{V[X_i]}{i}\leq \frac{C}{i}\to 0. \end{align} By Kolmogorov, we have \begin{align} 0\stackrel{a.s.}{←} \frac{1}{n} \sum_{i=1}^n (Y_i-E[Y_i]) = \frac{1}{n} \sum_{i=1}^n \frac{X_i-μ_i}{√i} \end{align}
I've noticed that I cannot use the converse of Cesaro-Stolz on the given assumption, because the sequence $n$ in the denominator satisfies $n/(n+1)\to 1$. I also cannot use simple estimates like $1/√i≥1/√n$ because the $X_i-μ_i$ are not guaranteed to be positive.
Is there a way to estimate out the $√i$ factor by a constant, or is my attempt at rescaling doomed? Or is there some other sequence $√c_i\to 0$ that I can use for rescaling for which this attempt works?