0
$\begingroup$

Let $X_n$ be a sequence of independent real valued random variables on the same event space, with the same (finite) mean $\mu$.

Suppose that for almost every couple of points $(\omega,\omega')$ in the squared space of events, we have $$ X_n(\omega) - X_n(\omega') \to 0. $$

Is it true that $X_n$ converges to $\mu$ almost surely?

And if it is false, is it true at least that there exists a random variable $X$ that is an almost surely limit for the sequence?


Another equivalent way we can formulate the question is as follows:

consider two sequences of independent random variables $X_n$ and $Y_n$ where for every $n$, $X_n\equiv Y_n$ (meaning they have the same distribution), and all the variables have the same finite mean $\mu$.

If $X_n-Y_n\to 0$ almost surely, is it true that $X_n\to\mu$ almost surely? or that there exists a random variable $X$ that is an almost sure limit for $X_n$?


The idea here should be that every $X_n$ tends to be a constant function as $n$ goes to infinity, since the couples $(\omega,\omega')$ where $X_n$ differs more than $\varepsilon$ are few.

Since the mean is the same for every $X_n$, then the constant it converges to should be the mean.

The weak point is that the mean can actually be manipulated by big deviation on very small events, but it should be possible to handle from the fact that the variables are independent.

$\endgroup$
  • $\begingroup$ p.s. I think I have a counterexample, but I'm not totally sure of it $\endgroup$ – Exodd May 31 at 9:36
  • $\begingroup$ This is true but its proof requires a knowledge in the induced measures on $\mathbb R^{\infty}$ by the two sequences. You have to apply Fubini's Theorem on $\mathbb R^{\infty} \times \mathbb R^{\infty}$ $\endgroup$ – Kavi Rama Murthy May 31 at 9:42
  • $\begingroup$ @KaviRamaMurthy could you post the proof? I am courios to see why my counterexample does not work $\endgroup$ – Exodd May 31 at 9:45
  • $\begingroup$ I have posted a proof of a weaker result. You should post your example also. $\endgroup$ – Kavi Rama Murthy May 31 at 9:55
  • $\begingroup$ @KaviRamaMurthy on it $\endgroup$ – Exodd May 31 at 9:57
1
$\begingroup$

What we can say is that $X_n-y_n$ converges almost surely for some sequence $(y_n)$ of real numbers. (To say that $(X_n)$ itself converges you need some addition hypothesis).

Sketch of proof: Let $X=(X_n)$ and $Y=(Y_n)$. These are independent $\mathbb R^{\infty}$ valued random vectors. Let $E$ be the set of all Cauchy sequences in $\mathbb R^{\infty}$. The $E$ is a measurable set and we have $P(X-Y\in E)=1$. Fubini's Theorem shows that there exists some sequence $y=(y_n)$ such that $P(X-y\in E)=1$. Hence $X_n-y_n$ converges almost surely.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm not sure how do you use Fubini here. $\endgroup$ – Exodd May 31 at 10:07
  • $\begingroup$ oh, wait. Are you saying that $y_n = Y_n(\omega)$ for some $\omega$ because if that's not true then for every $\omega$ you have a positive prob. of $X-Y$ not being convergent, and the integral of strictly positive is strictly positive? $\endgroup$ – Exodd May 31 at 10:11
  • $\begingroup$ And.. Let me guess.. The missing condition for it to be true is that $X_n-y_n$ is dominated by an integrable function (so that $\mu -y_n\to 0$), and so, it is practically equivalent to say that $X_n$ is dominated? $\endgroup$ – Exodd May 31 at 10:19
  • $\begingroup$ @Exodd Yes. some condition lie that is needed. $\endgroup$ – Kavi Rama Murthy May 31 at 11:41
0
$\begingroup$

Here's a (I hope) counterexample.

Suppose $X_n:[0,1]\to \mathbb R$ with $$ X_n|_{[0,n^{-2}]} \equiv n^3-n , \qquad X_n|_{(n^{-2},1]} \equiv -n $$ and notice that its mean is $(n^3-n)/n^2 - n(n^2-1)/n^2 = 0$ for every $n$.

The space of couple of points $(x,y)$ where $X_n(x)-X_n(y)\ne 0$ is also the space where $|X_n(x)-X_n(y)| = n^3$. This space has probability $2(1-n^{-2})n^{-2}\le 2/n^2$. As a consequence, we can use Borel-Cantelli to say that, almost surely $$ X_n(x)-X_n(y)\to 0. $$

Since $X_n$ do not converge a.s. to anything, this should be a valid counterexample.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Looks fine to me. $\endgroup$ – Kavi Rama Murthy May 31 at 11:44

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.