Let $\{X_n\}_{n\geq1}$ be a sequence of centered independent random variables such as $E(X_n^2)=2n$
and $$Y_n=\frac1{n^\alpha}\sum_{i=1}^{i=n}X_i\quad\quad\alpha\geq1$$
I am trying to prove that for $\alpha > \frac32$, $Y_n \rightarrow 0$ almost surely.
I started by calculating $Var(Y_n)$ which I found to be equal to $\frac{n(n+1)}{n^{2\alpha-2}}$ and since $E(Y_n)=0$ we get
$$\lim_{n\to\infty} E(|Y_n-0|^2)=0$$
which means that for $\alpha > 1$, $\{Y_n\}_{n\geq1}$ converges to $0$ in quadratic mean.
I am stuck going form there to the almost sure convergence.
Any help would be greatly appreciated!
Best Answer
Let $S_n=\sum_{k=1}^n X_k$ and note that $V(S_n)=\sum_{k=1}^n V(X_k)= \sum_{k=1}^n 2k = n(n+1)$.
For any $\epsilon >0$, by Markov's bound, $$P\left(\frac{|S_n|}{n^\alpha}\geq \epsilon \right) = P(|S_n|\geq n^\alpha\epsilon) \leq \frac{n(n+1)}{n^{2\alpha}\epsilon^2}\sim \frac{1}{n^{2\alpha-2}\epsilon^2}$$
When $\alpha >\frac 32$, we have $2\alpha-2>1$ and $\displaystyle \sum_n \frac{1}{n^{2\alpha-2}\epsilon^2}$ converges.
A standard criterion of almost sure convergence implies that $$\frac{|S_n|}{n^\alpha}\xrightarrow[]{a.s} 0$$