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Problem instance: A closed convex body $B\subset {\Bbb R}^n$ of volume 1; a point $p\in B$; and a real number $v\in(0,1)$.

Objective: Find the probability $P(B,v,p)$ that $p\in B'$, for $B'$ a randomly chosen closed convex subset of $B$ of volume $v$.

Of course to make this problem to make sense, I must define "choosing $B'$ at random."

I'll propose a definition or two myself. Still, I'd be interested in any solution to this problem relative to any reasonable definition.

So I propose taking $P(B,v,p)$ as the limit of probabilities $P_m(B,v,p)$ defined by restricting the range of $B'$. For $P_m(B,v,p)$, we pick $m$ points at random from $B$, get $B'$ by taking their convex hull, but condition on getting the correct volume $v$. By using Lebesgue measure on ${\Bbb R}^n$ to pick points at random from $B$, this all makes sense.

Alternatively (if it makes a difference?), we pick $m$ points at random from $B$, take their convex hull to get $B'$ and condition on getting volume $v$ and also on having none of the $m$ points wind up in the interior of the hull.

The sort of answer I'm looking for would single out the function $P(B,v,\cdot)$ in some reasonably nice way but not necessarily explicit way, as the solution of some differential or integral equation perhaps.

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One definition of "randomly chosen convex subset of volume $v$" is to choose a direction at random, then the hyperplane perpendicular to that direction which cuts off a piece of $B$ of volume $v$. Maybe this is too simple, but if it isn't satisfactory, what other properties do you want? – Douglas Zare Jun 21 '12 at 18:38
Thanks Douglas At the very least, given bodies $B_I \subset B_O$ of volumes $v-\epsilon$ and $v+\epsilon$, we should get $B_I \subset B'\subset B_O$ with positive probability. Also the probability shouldn't vary when we translate this picture within $B$. Both my proposals clearly do satisfy these properties. – David Feldman Jun 21 '12 at 19:05
One might take an $\varepsilon$-lattice $L_\varepsilon$ and consider only convex polygons with the vertices in $L$ with counted measure. Then you pass to the limit as $\varepsilon\to0$. In this case I bet that "random" convex sets will be congruent with probability 1... – Anton Petrunin Jun 21 '12 at 19:20
Another nice property of your definition is that it is affine invariant. However, it seems hard to calculate it even for the simplest examples, say for a disk and its center. – Douglas Zare Jun 21 '12 at 19:50
>it seems hard to calculate it even for the simplest examples, say for a disk and its center Agreed, but the whole function must be rotationally symmetric, so it might actually be easier to characterize the function and then calculate the value at the center than solve for the central value directly. – David Feldman Jun 21 '12 at 20:35

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