-1
$\begingroup$

Let $H$ be a separable Hilbert space and let $\mu$ be an arbitrary probability measure on $H$. I would like to approximate this distribution by by a finitely supported discrete distribution is the following sense:

For every $\varepsilon>0$ I would like to have a finitely supported probability measure $\nu_{\varepsilon}$ so that for every open ball $B$ of diameter $1$ we have $$|\mu(B)-\nu(B)|<\varepsilon.$$

Is such a statement true? Is it true at least in $\mathbb{R}^d$?

Comment: The question has been changed due to a comment by Brendan Mckay's example that refuted the stronger version of the question.

$\endgroup$
6
  • 1
    $\begingroup$ Take $\mu$ to be the normal distribution on the real line and $\nu$ to be supported on a finite set $X$. The complement $A$ of $X$ is open yet $\mu(A)=1$ and $\nu(A)=0$. $\endgroup$ May 22 at 9:35
  • $\begingroup$ Thank you, I will mend the question as I need a weaker statement, but hoped that something stronger could be true, but it turned out to be naive. $\endgroup$
    – TOM
    May 22 at 10:14
  • $\begingroup$ (For second version) Partition $\mathbb{R}^d$ into cubes of side $d^{-1/2}$. Choose cubes in decreasing order of $\mu$ measure until their total measure exceeds $1-\varepsilon$. That only takes a finite number of choices, since the sequence converges. Now support your discrete measure on the centres of the selected cubes. $\endgroup$ May 22 at 11:16
  • $\begingroup$ Maybe smaller cubes are needed so that balls centred nearby will cover them. Hopefully you can work out the details. $\endgroup$ May 22 at 11:50
  • $\begingroup$ Thank you for your answers - it is most useful! $\endgroup$
    – TOM
    May 26 at 0:53
1
$\begingroup$

This is much too ambitious—it won‘t even work in $[0,1]$ but it is true if in any completely regular space if you replace „open“ by „compact“. This is under the assumption that you are talking about tight measures. So it will work for $\sigma$-additive measures in your situation, since these are tight when the underlying space is polish as are separable Hilbert spaces.

$\endgroup$

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.