# Almost sure convergence of the series of independent random variables

Let $\{X_n:n\ge1\}$ be i.i.d. random variables with $\operatorname EX_1=0$ and $\operatorname E|X_1|^p<\infty$, where $1<p<2$. Let $\{b_n:n\ge1\}$ be a real sequence. Does the series $$\sum_{n=1}^\infty b_nX_n$$ converge almost surely if $b_n=O(n^{-1/p})$ as $n\to\infty$?

It seems that it should converge almost surely, but I'm not sure how to prove that.

I tried to use Kolmogorov's three-series theorem. I can show that the series $\sum_{n=1}^\infty\Pr\{|X_1|>|b_n|^{-1}\}$ and $\sum_{n=1}^\infty b_n\operatorname E[X_1I_{\{|X_1|\le|b_n|^{-1}\}}]$ converge, but I'm unable to show that the series $\sum_{n=1}^\infty b_n^2\operatorname{Var}[X_1I_{\{|X_1|\le|b_n|^{-1}\}}]$ converges. Since $b_n=O(n^{-1/p})$ as $n\to\infty$, we have that $$\sum_{n=N}^\infty b_n^2\operatorname{Var}\left[X_1I_{\{|X_1|\le|b_n|^{-1}\}}\right]\le M^2\sum_{n=N}^\infty n^{-2/p}\operatorname{Var}\left[X_1I_{\{|X_1|\le|b_n|^{-1}\}}\right],$$ where $M$ is a positive constant, but I'm not sure how to proceed.

Any help will be highly appreciated!

• How do you deduced convergence of the second series? – Davide Giraudo Apr 16 '15 at 17:55
• @DavideGiraudo Thank you for the question. I'm sorry I made a mistake. I was investigating a particular case and I didn't realize that I can't use the same idea in the general case. I changed my question (since there are no answers) to the particular case ($\operatorname E|X_1|^p<\infty$ with $1<p<2$ and $b_n=O(n^{-1/p})$ as $n\to\infty$). – Cm7F7Bb Apr 17 '15 at 7:06

For the last series, since the second is convergent it suffices to check that the series $\sum_{n\geqslant 1}b_n^2\mathbb E\left[X_1^2\mathbf 1\{|b_nX_1|\leqslant 1 \}\right]$ is convergent. To this aim, we start writing that $$b_n^2\mathbb E\left[X_1^2\mathbf 1\{|b_nX_1|\leqslant 1 \}\right]\leqslant \int_0^1 t\mathbb P\{|b_nX_1|\geqslant t\}\mathrm dt.$$ For simplicity, we assume $|b_n|\leqslant n^{-1/p}$, replacing if necessary $b_n$ by $b_n/M$. From the displayed equation, it follows that $$b_n^2\mathbb E\left[X_1^2\mathbf 1\{|b_nX_1|\leqslant 1 \}\right]\leqslant \int_0^1t\mathbb P\{|X_1|\geqslant tn^{1/p}\}\mathrm dt= \int_0^1t\mathbb P\left\{\frac{|X_1|^p}{t^p}\geqslant n\right\}\mathrm dt.$$ Summing over $n$ and using the fact that $\sum_n \mathbb P\{|Y|\geqslant n\}\leqslant\mathbb E|Y|$, we obtain $$b_n^2\mathbb E\left[X_1^2\mathbf 1\{|b_nX_1|\leqslant 1 \}\right]\leqslant \mathbb E|X_1|^p\int_0^1t^{1-p}\mathrm dt.$$
• Thank you very much for the answer! I would just like to clarify one thing. Does the first equality hold (I think it should be less than or equal and a constant is missing on the right side)? Correct me if I'm wrong, but we have that for any $r>0$, any $a>0$ and any random variable $X$, $$\operatorname E[|X|^rI_{\{|X|\le a\}}] =r\int_0^ax^{r-1}\Pr\{|X|>x\}\mathrm dx-a^r\Pr\{|X|>a\}.$$ – Cm7F7Bb Apr 17 '15 at 9:43