5
$\begingroup$

$X_1,X_2,\ldots$ is a sequence of random variables that are complete convergent to $X$ if $$\sum_{n=1}^{\infty} P(\mid X_n-X\mid >\epsilon)<\infty \space\forall\epsilon > 0$$. Show if $X_n$ are independent, then complete convergence is equivalent to convergence a.s.

I showed that complete convergence implies convergence a.s. using the Borel-Cantelli lemma, but I'm not sure how to use show the converse using independence. This is what I have so far:

$$X_n\rightarrow_{a.s.} X \implies X_n\rightarrow_{p} 0$$ WLOG, $X=0$.

$$\forall\epsilon >0, P(\mid X_n\mid > \epsilon)\rightarrow 0$$ $$P(\limsup_{n\rightarrow\infty} \mid X_n\mid > \epsilon)=0$$

Since they are independent, then $\{\mid X\mid > \epsilon\}$ are independent, and can I use Borel-Cantelli (ii) to say $\sum P(\mid X_n\mid >\epsilon) < \infty$?

$\endgroup$
  • $\begingroup$ en.m.wikipedia.org/wiki/Borel–Cantelli_lemma has a converse result using independence, but that you might have to prove it anyway. $\endgroup$ – Evan Sep 28 '13 at 18:54
  • 1
    $\begingroup$ Your proof looks fine. $\endgroup$ – Etienne Sep 28 '13 at 19:23
5
$\begingroup$

Your idea is fine and there is small step to finish it. I write the proof anyway.


Suppose that you do have the a.s. convergence but not the complete convergence. Therefore you have: $$ \sum_{n=1}^{\infty} P(\mid X_n-X\mid >\epsilon)=\infty \space\exists\epsilon > 0 $$ For such $\epsilon$, you can use second Borel-Cantelli to prove that: $$ P(\limsup_{n\rightarrow\infty} \mid X_n-X\mid > \epsilon)=1 $$ But from a.s. convergence you have $\displaystyle P(\lim_{n\rightarrow\infty} X_n=X)=1$ which means that for all $\epsilon>0$, $\displaystyle P(\limsup_{n\rightarrow\infty} \mid X_n-X\mid > \epsilon)=0$ which is a contradiction. Therefore $X_n$ should have complete convergence.

$\endgroup$
  • $\begingroup$ how does $X_n\rightarrow_{a.s.}X$ imply $\sum P(|X_n-X|>\epsilon)=\infty$? $\endgroup$ – lightfish Sep 28 '13 at 22:12
  • $\begingroup$ It does not imply. If the complete convergence does not hold then you will get this. $\endgroup$ – Arash Sep 28 '13 at 22:14
  • 1
    $\begingroup$ Oh I see now, it is a proof by contradiction. Thank you! $\endgroup$ – lightfish Sep 28 '13 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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