If I can show for some event $A_n$, $$ \sum_n P(A_n) < \infty. $$ Then by First Borel-Cantelli Lemma, I get $P(A_n \; i.o.) = 0$; But I still confused about how this infinitely often connects to almost sure convergence.

Is it true that the complement $\{A_n i.o. \}^c$ happens almost surely?

For example, suppose $P( \{ X_n = C \; i.o.\}) =0$ for some constant $C$, where $X_n$ is random variable. Can I conclude $X_n \neq C$ almost surely for free?

Thank you

  • 3
    $\begingroup$ In your second example, you probably mean $\{X_n = C\}$ or something like that, because the "infinitely often" part is useless otherwise. In that case, you can conclude that almost surely, $X_n = C$ only for finitely many $n$. $\endgroup$ – PhoemueX Oct 21 '14 at 4:50
  • $\begingroup$ Thanks! @PhoemueX , I revised the question. So there is no connection to say the complement event $X_n \neq C$ almost surely? $\endgroup$ – Fianra Oct 21 '14 at 7:17

You are using almost surely here in two senses. First, in the $i.o.$ case, you are referring to a sequence of events, whereas in your second question, you are asking a pointwise question.

In short: you cannot conclude $X_n \neq C, a.s.$ as this would imply $P(X_n=C)=0\;\;\forall n$.

However, i think you mean not the above event $\left(\{X_n \neq C\}\right)$but the event $\{X_n\neq C\;\; i.o\}$, correct? If so, then we know that that tail $\sigma-$algebra does not contain $\{X=C\}$, thus, the tail $\sigma-$algebra will be $\sigma\left(\{X\neq C\}\right)$ which implies that the probability must be $1$ for this event.

In English, if an event does not happen infinitely often, then it will eventually cease to happen, hence its complement will always happen.


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.