Suppose that $K$ and $L$ are two compact convex sets in space. Suppose that $K$ contains $L$. Now consider two quantities

• $X$ is the average volume of a simplex forms by four points in $K$ drawn uniformly at random.
• $Y$ is the average volume of a simplex forms by four points in $L$ drawn uniformly at random.

### TYI: Is it always the case that X ≥ Y?

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### 12 Responses to Test Your intuition 51

1. DS says:

No. The number 4 seems arbitrary here, I did it first with simplices formed by 2 points a.k.a. segments 🙂

2. ava says:

It seems a triangular bipyramid with its heights to the diagonal face 1 and eps works. If we cut a small portion of points around the obtuse corner (corresponding to the eps height) the average volume increases.

3. Johan Aspegren says:

My intuition says that a cube containing the maximal ellipsoid is a counter example if there is one.

4. Johan Aspegren says:

But I can’t say if that is an extremal case. There may be at least nonsymmetric cases when the claim fails even more. I just thought that for that pair the claim is true asymptotically e.g when the dimension grows. The number 4 is not arbitary because that’s the minimal number of points you need to form the simplex in space. Anyway, you can solve these questions for (at least 1-unconditional) pairs just by figuring out the average distance from the origin , when both bodies have the (different) densities that integrate to the unit. Because we assumed that the bodies are 1-uncondional it suffices to consider projections to the axes. If I figured this right, for the cube this is solved from 1/2 = x and for the ball from 1/2 = 4\pi*\int_0^1 r^2dr.

5. No, for the ball it seems to be 1/2 = 4/pi\int_0^1 r^3 dr.

• Johan Aspegren says:

And I forgot to put the right densities for the cube and the ball. Now to sleep.

• Johan Aspegren says:

OK, a new day. For the maximal ball in the unit cube the formula would be 3/8 =\int_0^(1/2) 8*3*r^3 dr is the expectation of the distance from the origin, so the expectation of the projection to an axis would be \sqrt(3)/8 for the inscribed ball and in general \sqrt(n)/(2n+2). For the cube this would be \sqrt(3)/4 and in general \sqrt(n)/(n+1). It seems that this is again one of those questions very related to isotropic constants, because you might as well consider seconds moments. It’s interesting that the analogous claim seems to fail in the plane also.

6. Johan Aspegren says:

No, for the cube of unit volume the general formula is \sqrt(n)/4. These things are so counterintuitive.

• And I didn’t find a counterexample after all.

7. pxjsmmu says:

X=Y?

8. George Beck says:

No, because we can take K to be very long, forcing the tetrahedra to often be thin and so of small volume on average.

• Why this order Y > X is not preserved by a linear transformation, if you transform K to less thin?