3
$\begingroup$

$\sum_{n=1}^{\infty} E|X_n - X| < \infty$ imples $X_n$ converges to $X$ almost surely

I'm sure this is an obvious one line proof and I'm being stupid here, but I cannot see how to show the above. I can see that $\sum_{n=1}^{\infty} P(|X_n - X|>\epsilon) < \infty$ would give the result via the Borel Cantelli lemma, am not sure how I can use that fact. Could I maybe approximate $|X_n - X|$ by simple functions? Then turn the expected value operator into a probability.

$\endgroup$

1 Answer 1

2
$\begingroup$

That's quite strange that you came up with the BC but cannot attain a proof :) (It's very close!)

By the BC, if we let $A_n(\epsilon)=\left \{|X_n-X|>\epsilon\right\}$, $A(\epsilon)=\limsup A_n(\epsilon)$, then $P(A(\epsilon))=0$. So the set $B=\cup A(\frac{1}{n})$ has probability 0 and $X_n(\omega) \rightarrow X(\omega)$ for $\omega \in B^c$.


Ah, I used the Markov inequality in the very beginning.

$\endgroup$
3
  • $\begingroup$ I see where you are going, but have you not just shown ∑P(|Xn−X|>ϵ)<∞ implies convergence. What I am wondering is how you go from ∑E|Xn−X|<∞ to the convergence. If I am being stupid and it is obvious then I apologise $\endgroup$
    – James
    Nov 18, 2012 at 2:51
  • $\begingroup$ Do you know the Markov ineq? $\endgroup$
    – cjackal
    Nov 18, 2012 at 3:33
  • $\begingroup$ Ah, now it makes perfect sense. I didn't know it, but I probably should have been able to derive it, my brains not really working that well this evening. Thanks $\endgroup$
    – James
    Nov 18, 2012 at 3:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .