-3
$\begingroup$

Let $(\Omega,\Sigma,\mathbb{P})$ be a complete probability space, $B\subseteq X$ be a non-empty Borel subset of a polish space $X$, $A$ be an uncountable indexing set, and $\{X_{\alpha,n}\}_{a \in A, n \in \mathbb{N}}$ a set of $X$-valued random elements with the following properties:

  • For every pair of distinct $\alpha,\alpha'\in A$ and every $\omega \in \Omega$ there does not exist $n,n'\in\mathbb{N}$ satisfying $$ X_{\alpha,n}(\omega) = X_{\alpha',n'}(\omega). $$
  • For every $\alpha \in A$, and every non-empty open subset $O'\subseteq X$, $0<\mathbb{P}\left(\bigcup_{n \in \mathbb{N}}\{X_{\alpha,n} \in O'\}\right)$.

Then can we conclude that $1=\mathbb{P}\left(\bigcup_{n \in \mathbb{N},\alpha \in A}\{X_{\alpha,n} \in B\}\right)$?

$\endgroup$
7
  • $\begingroup$ consider any system $\Omega, P, X,\{X_{\alpha,n}\}$ like this, and then double it: $\Omega'=\Omega\sqcup \Omega$, $X'=X\sqcup X$ and so on (probability should be divided by 2). Take $B$ to be one of the two copies of $X$. Then your set in the end will be of measure 1/2. Am i missing something? $\endgroup$
    – erz
    Jun 2, 2020 at 12:08
  • $\begingroup$ Actually, I was missing something. Namely that the second point should hold for all non-empty Borel subsets. It's a form of ergodicity. $\endgroup$
    – ABIM
    Jun 2, 2020 at 12:46
  • $\begingroup$ isn't my comment still valid though? $\endgroup$
    – erz
    Jun 2, 2020 at 12:56
  • 1
    $\begingroup$ Your second point can never be satisfied if $X$ is uncountable. Fix $\alpha$ once and for all. For every $x$, taking $B' = \{x\}$, you demand there exists $n$ such that $\mathbb{P}(X_{\alpha, n}=x) > 0$. By pigeonhole there must exist an $n$ such that $\mathbb{P}(X_{\alpha, n}=x) > 0$ for uncountably many $x$, which is impossible. $\endgroup$ Jun 2, 2020 at 13:10
  • $\begingroup$ Right, if I replace B by an open set then this resolves that issue. $\endgroup$
    – ABIM
    Jun 2, 2020 at 13:41

2 Answers 2

1
$\begingroup$

Here is a "very regular" counterexample:

Let $X=\mathbb R$, $A:=(0,1)=:B$, and $X_{a,n}:=Z+a$ for all $a\in A$ and $n\in \mathbb R$, where $Z\sim N(0,1)$. Then all your conditions hold. However, $$P\Big(\bigcup_{n\in\mathbb N,a\in A}\{X_{a,n}\in B\}\Big)=P(|Z|<1)\ne1.$$

$\endgroup$
0
1
$\begingroup$

An attempt at a counterexample. Pick any $(\Omega,\Sigma,\mathbb{P})$. Pick $X = \{0,1\}^{\omega}$ with standard Borel structure, and $B = \{0^\omega\}$. Pick $A = \mathbb{R}$. Have each $X_{\alpha,n}$ be the constant function at $x_{\alpha,n}$ for all $\alpha,n$, and have $(x_{\alpha,n})_n$ enumerate a dense set in $X$ for each $\alpha$. It's easy to pick the sequences $(x_{\alpha,n})_n$ so they are all disjoint (note that each $X_{\alpha,n}$ is measurable trivially as it's constant, so it's just about cardinality), and you can pick them so that their values are not equal to $0^\omega$. Now $\mathbb{P}(\bigcup_{n \in \mathbb{N}} \{X_{\alpha, n} \in O'\}) = 1$ for all $\alpha$ and open $O'$ (if I'm interpreting it correctly). But no values are in $B$ so the other probability is zero.

$\endgroup$
3
  • $\begingroup$ I didn't notice the comments. If $B$ is open you can do the same with two-valued random variables. Throw a global coin, and set $X_{\alpha,n} = x_{\alpha,n}$ if heads, $X_{\alpha,n} = 1^\omega$ if tails (for all $\alpha, n$ at once). If $B$ does not contain $1^\omega$ we're good. $\endgroup$
    – Ville Salo
    Jun 2, 2020 at 14:57
  • $\begingroup$ Do you have an idea of a "regularity requirement" on $\{X_\alpha\}$ which evadues this type of construction? $\endgroup$
    – ABIM
    Jun 2, 2020 at 15:13
  • $\begingroup$ I do not. It would be a bit surprising to me if there is a true statement even in the spirit of what you ask for; of course all the more interesting if you find one. $\endgroup$
    – Ville Salo
    Jun 2, 2020 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.