Suppose $X_1$, $X_2$,... is a sequence of non negative independent random variables where $E[X_n]=\dfrac{1}{n^2}$. Show that if $Y_n=\sum_{i=1}^{n} X_i$ and $Y=\sum_{i=1}^{\infty} X_{i}$, $Y_n$ converges to $Y$ almost surely. I tried to use Markov inequality and then Borel Cantelli Lemma to prove that but I couldn't.

$P(\sum_{i=n+1}^{\infty} X_i> \epsilon)=\dfrac{\sum_{i=n+1}^{\infty}E(X_i)}{\epsilon}=\dfrac{1}{\epsilon}\sum_{i=n+1}^{\infty}{\dfrac{1}{i^2}}$

Any help would be appreciated. Thank you.

  • $\begingroup$ @Xi'an :can we say this sum is 0 and then $\sum_{n=1}^{\infty}p(\sum_{i=n+1}^{\infty}> \epsilon)=0$ therefore$ Y_n\ converges\ to\ Y\ almost\ surely$? Thank you in advance $\endgroup$ – SAm Dec 15 '16 at 16:02

As expressed in the question, $$\mathbb{P}\left(\sum_{i=n+1}^{\infty} X_i> \epsilon\right)=\dfrac{1}{\epsilon}\sum_{i=n+1}^{\infty}\mathbb{E}(X_i)=\dfrac{1}{\epsilon}\sum_{i=n+1}^{\infty}{\dfrac{1}{i^2}}$$which is incorrect: a consequence of Markov's inequality is that $$\mathbb{P}\left(\sum_{i=n+1}^{\infty} X_i> \epsilon\right)\le\dfrac{1}{\epsilon}\sum_{i=n+1}^{\infty}\mathbb{E}(X_i)=\dfrac{1}{\epsilon}\sum_{i=n+1}^{\infty}{\dfrac{1}{i^2}}$$ Now, since the series$$\sum_{i=1}^{\infty}i^{-2}=\dfrac{\pi^2}{6}$$converges, the remainder$$\sum_{i=n+1}^{\infty}i^{-2}$$converges to zero as $n$ grows to infinity. This implies that$$\mathbb{P}\left(\vert Y_n-Y\vert> \epsilon\right)=\mathbb{P}\left(\sum_{i=n+1}^{\infty} X_i> \epsilon\right)$$converges to zero.


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