3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ @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, 2016 at 16:02

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.