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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!

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  • $\begingroup$ How do you deduced convergence of the second series? $\endgroup$ – Davide Giraudo Apr 16 '15 at 17:55
  • $\begingroup$ @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 $n\to\infty$). $\endgroup$ – Cm7F7Bb Apr 17 '15 at 7:06
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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 $$b_n^2\mathbb E\left[X_1^2\mathbf 1\{|b_nX_1|\leqslant 1 \}\right]\leqslant \mathbb E|X_1|^p\int_0^1t^{1-p}\mathrm dt.$$

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  • $\begingroup$ 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 $X$, $$ \operatorname E[|X|^rI_{\{|X|\le a\}}] =r\int_0^ax^{r-1}\Pr\{|X|>x\}\mathrm dx-a^r\Pr\{|X|>a\}. $$ $\endgroup$ – Cm7F7Bb Apr 17 '15 at 9:43
  • $\begingroup$ How did you get the last term? $\endgroup$ – Davide Giraudo Apr 17 '15 at 9:49
  • $\begingroup$ \begin{align*} \operatorname E[|X|^rI_{\{|X|\le a\}}] &=r\int_0^ax^{r-1}\Pr\{|X|I_{\{|X|\le a\}}>x\}\mathrm dx\\ &=r\int_0^ax^{r-1}\Pr\{|X|>x\}\mathrm dx-r\Pr\{|X|>a\}\int_0^ax^{r-1}\mathrm dx\\ &=r\int_0^ax^{r-1}\Pr\{|X|>x\}\mathrm dx-a^r\Pr\{|X|>a\}. \end{align*} $\endgroup$ – Cm7F7Bb Apr 17 '15 at 9:51

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