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Suppose every point in the plane undergoes brownian motion for a time t. What is the probability n particles ended up at 0? For n finite, countable or uncountable?

What proportion of the plane does not have a particle on it after time t? Ie. pick n random points inside an open disc, as n approaches infinity, what fraction of those n points will have k particles on it?

Edit: Cut the plane into regions of equal area, let each region undergo brownian motion for t approaching infinity, what is the resulting distribution for the number of overlapping areas at origo?

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If the regions are anything like a square grid, then the distribution of the count approaches a Poisson distribution. – Douglas Zare Aug 4 '14 at 3:42
As the answers point out, this question is incompletely posed, since you have not stated whether there is to be any dependence between these Brownian motions. (As pointed out by Robert Israel, if you want them all to be independent, you have measurability problems.) For the question in your edit, do you mean that each region should be translated according to an independent Brownian motion? I don't see how you can get a concrete answer without specifying the shapes of the regions. – Nate Eldredge Aug 4 '14 at 13:19
@NateEldredge Independent motions. Are you saying the answer depends on the shapes of the regions? – JGH Aug 4 '14 at 15:38
You might find this paper interesting, though it's about percolation properties of a disk packing in the plane, rather than what you've specifically asked about: – j.c. Aug 13 '14 at 15:00
Regarding the question in the edit: I interpreted it as asking for how many of the regions contain the origin. In that case, it tends to Poisson(1) regardless of the shapes. – Ori Gurel-Gurevich Aug 13 '14 at 18:35
up vote 6 down vote accepted

Presumably you mean you have continuum-many independent Brownian motions, one (call it $W_p(t)$) with $W_p(0) = p$ for each $p$. Unfortunately I'm pretty sure the number of $p$ for which $W_p(t) = 0$ is not a measurable function, so your question does not have an answer.

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