2
$\begingroup$

We were taught in class that :

A sequence of random variables $(X_n)_{n\geqslant 1}$ converges to $X$ almost surely (or, almost everywhere), if $~\exists~$ an event $A$ such that $\mathbb{P}(A)=0$ , and $\forall~\omega\in\Omega\setminus A$ , $X_n(\omega) \longrightarrow X(\omega)$ pointwise.

Now, my question is, for a sequence $(X_n)_{n\geqslant 1}$ of random variables, if it's given that $X_n\longrightarrow X$ almost surely, then is it true that for all events $E$ such that $\mathbb{P}(E)=0$ and $\forall~\omega\in\Omega\setminus E$ , $X_n(\omega) \longrightarrow X(\omega)$ pointwise ?

If the answer to my doubt is NO, then, in general, is there a common way to show that a sequence of random variables does NOT converge almost surely ?

$\endgroup$

1 Answer 1

3
$\begingroup$

Suppose $X_n\to X$ almost surely. By definition you are guaranteed a null event $A$ such that $X_n(\omega)\to X(\omega)$ (pointwise) for all $\omega\in A^c.$ But if you take an event $B\subsetneq A,$ and $\omega\in A\setminus B$ then obviously you are not guaranteed that$X_n(\omega)\to X(\omega).$ As long as your measure space is complete and $|A|>1,$ You can always find such events. So in general, as you expected the answer to your first question is ‘No’.

For the second question, I don’t know of a universal technique to show that $X_n$ does not converge to $X.$ But here are a few tricks, I have come across.

If $X_n\to X$ a.s., then $X_n\to X$ in probability. It is easier usually to show that $X_n$ does not converge in probability. Of course, a sequence can converge in probability but may fail to converge almost surely and this method would not work.

Another way would be to try and identify $X^1:=\limsup X_n$ and $X_{1}:=\liminf X_n.$ The good thing is that they are always guaranteed to exist. If you could show that $X^1\neq X_{1}$ then you are done.

If you have added assumptions (like boundedness) that combined with almost sure convergence would give you the convergence of the expectations, then it might be useful to check that the expectations does not converge (this may itself be very tricky sometimes).

A naive trick is to find two subsequences which converge almost surely but to two different limits. This tells you that the original sequence can not converge almost surely.

$\endgroup$
2
  • $\begingroup$ Thanks a lot for your answer. However, if I mention in my "definition" or "question", that $E$ is non-null, is the answer YES by any chance ? You've written something regarding measure space, but sadly, measure theory is completely beyond my scope till now. $\endgroup$
    – Kolmogorov
    Nov 3, 2020 at 19:34
  • $\begingroup$ By the way, this doubt occured to me while solving a problem in which I have to show that a certain sequence $(X_n)_{n\geqslant 1}$ of random variables converge to some random variable $X$ in $\mathcal{L}_2$ , but it doesn't converge almost surely. $\endgroup$
    – Kolmogorov
    Nov 3, 2020 at 19:36

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.