# Samples in the convex body vs. samples on the convex surface

Let $K$ be a bounded convex body in $\mathbb{R}^n$. Suppose we have a sampler $\mathcal{S}_1$ that can generate points uniformly distributed in $\mathrm{int}K$, and another sampler $\mathcal{S}_2$ that can generate points uniformly distributed on $\partial K$.

The figure below illustrates two samplers.

Let $\{\mathbf{x}^j\}_{j=1}^M$ be $M$ samples generated by $\mathcal{S_1}$ and $\{\mathbf{y}^j\}_{j=1}^M$ by $M$ samples generated by $\mathcal{S_2}$. Denote the empirical centroid $\mathbf{u}$, $\mathbf{v}$ as $\mathbf{u}=\frac{1}{M}\sum_j \mathbf{x}^j$ and $\mathbf{v}=\frac{1}{M}\sum_j \mathbf{y}^j$, respectively. My questions are:

1. Are $\mathbf{u}$ and $\mathbf{v}$ same or not?
2. If not, can anyone provide a lower and upper bound of $\|\mathbf{u}-\mathbf{v}\|$?

Thanks

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1. Not necessarily.
2. Unfortunately, the value of $||{\bf u} - {\bf v}||$ can be anywhere from $0$ to $\infty$.

This holds true even if the empirical measures capture the actual centroid of $K$ and the actual centroid of the boundary of $K$, so let's look at that situation.

Take the simple case of an isosceles triangle. Let the vertices of the triangle be at $(0,x)$, $(1,0)$ and $(-1,0)$. The centroid of the triangle has barycentric coordinates in a $1:1:1$ ratio, which means: $$\text{The centroid lies at } \left(0,\frac{x}{3}\right).$$

The centroid of the boundary of the triangle is called the Spieker center of the triangle and has barycentric coordinates in a $b+c:a+c:a+b$ ratio (see X(10) in the Encyclopedia of Triangle Centers; also, $a, b, c$ are the lengths of the sides opposite vertices $A, B, C$, respectively). Since the side lengths of our example triangle are $2, \sqrt{x^2+1}, \sqrt{x^2+1}$: $$\text{The Spieker center lies at }\left(0, \frac{2x\sqrt{x^2+1}}{4+4\sqrt{x^2+1}}\right) = \left(0, \frac{x}{2+2/\sqrt{x^2+1}}\right).$$

If the triangle is equilateral, $x = \sqrt{3}$, and the centroid and Spieker center coincide. On the other hand, the distance between the centroid and the Spieker center is on the order of $\frac{x}{6}$ and so can be made arbitrarily large by increasing $x$.

This also points to the general case that the distance between the centroid of the region and the centroid of the perimeter can be made arbitrarily large: Moving a vertex further and further away from the other vertices has more effect on the centroid of the perimeter than it does on the centroid of the region, as this process pulls away from the main part of the region a larger percentage of the boundary than of the area.

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Thanks @MikeSpivey for this nice and clear answer. The example is really convincing. –  han Nov 5 '11 at 20:23
@han: You're welcome. I'm glad it was helpful. –  Mike Spivey Nov 5 '11 at 20:57