Suppose that $X_n$ are independent random variables where $\sup_n E|X_n| < \infty$.

Then, $\frac{\sum_{i=1}^{n}X_i}{n^a}$ converges to 0 almost surely for all $a>1$.

I think that $\frac{1}{n^{a-1}}$ converges to 0 and $\frac{\sum_{i=1}^{n}X_i}{n}$ converges because the sup of expectation is bounded, but since $X_n$ are not identical, the SLLN cannot be applied directly. And it makes me hard to follow the process of proving SLLN. Could you please help me?


1 Answer 1


Let $\Delta\equiv \sup_n\mathsf{E}|X_n|$. Then $$ \sum_{k=1}^\infty\mathsf{P}\left(\left|\frac{X_k}{k^\alpha}\right|\ge 1\right)\le \sum_{k=1}^\infty \frac{\Delta}{k^\alpha}<\infty, $$ and for $Y_n=X_n/n^{\alpha}1\{|X_n|\le n^{\alpha}\}$, $$ \sum_{k=1}^\infty\mathsf{E}Y_k \quad\text{and}\quad \sum_{k=1}^\infty \operatorname{Var}(Y_k) $$ converge as well$^{(*)}$. Hence, by Kolmogorov's three-series theorem, $\sum_{k=1}^n X_k/k^{\alpha}$ converges a.s. in $\mathbb{R}$. Convergence of $\sum_{k=1}^n X_k/n^{\alpha}$ follows from Kroneker's lemma.

$^{(*)}$ \begin{align} \sum_{k=1}^\infty \operatorname{Var}(Y_k)&\le \sum_{k=1}^\infty \frac{1}{k^{2\alpha}}\mathsf{E}[X_k^21\{|X_k|\le k^{\alpha}\}] \\ &\le \sum_{k=1}^\infty \frac{1}{k^{\alpha}}\mathsf{E}[|X_k|1\{|X_k|\le k^{\alpha}\}]\le \sum_{k=1}^\infty\frac{\Delta}{k^{\alpha}}<\infty. \end{align}


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