1
$\begingroup$

I got two sequences of random variables $(X_n)_{n \in \mathbb N}$ and $(Y_n)_{n \in \mathbb N}$ and know that

1) $X_n \to 0, \, n \to \infty,$ almost surely 2) For all $n \in \mathbb N$ the sequences are equal in distribution: $X_n \stackrel{d}{=} Y_n$

Now I'm wondering, does this imply that also $Y_n \to 0, n \to \infty,$ almost surely?

$\endgroup$
3
$\begingroup$

I don't have a simple counterexample at the moment but you can argue using Skorohod's Theorem: let $(Y_n)$ tend to $0$ is probability but not almost surely. Since $Y_n \to 0$ in distribution Skorohod's Theorem tells us that there exits random variable $X_1,X_2,...$ such that $X_n \to 0$ almost surely and $X_n$ has same distribution as $Y_n$ for each $n$.

Reference: https://eventuallyalmosteverywhere.wordpress.com/2014/10/13/skorohod-representation-theorem/

A better example: consider $[0,1)$ with Lebesgue measure. Arrange the intervals $[\frac {i-1} {2^{n}},\frac i {2^{n}})$, $1 \leq i \leq 2^{n}$, $n \geq 1$ in as sequence using the 'natural' ordering. Let $Y_1,Y_2,..$ be the indicator functions of these intervals. Now form $X_1,X_2,...$ by replacing $[\frac {i-1} {2^{n}},\frac i {2^{n}})$ by $[0,\frac 1 {2^{n}})$ for each $i$. Then $X_n \to 0$ almost surely, $Y_n$ does not tend to $0$ almost surely and $X_n$ has the same distribution as $Y_n$ for each $n$.

$\endgroup$
  • $\begingroup$ Thanks. So if I get this right, this means the statement does not hold on the original probability space that $(X_n)_n$ and $(Y_n)_n$ are defined on, but using Skorohods theorem I can define those sequences on some other space in which the almost surve convergence of $(Y_n)$ holds? $\endgroup$ – Bazzan Jul 19 at 10:14
  • $\begingroup$ Yes, it may be possible to give a better example but this is what I have at the moment. $\endgroup$ – Kavi Rama Murthy Jul 19 at 10:19
  • $\begingroup$ @Bazzan I have now given a better example. $\endgroup$ – Kavi Rama Murthy Jul 19 at 10:36
  • $\begingroup$ thank you, this makes it clear $\endgroup$ – Bazzan Jul 19 at 10:53
  • 1
    $\begingroup$ @Ak19 Thank you very much. $\endgroup$ – Kavi Rama Murthy Jul 19 at 11:54

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.