3

Consider the unit $n$-sphere and $l$ points distributed uniformly randomly inside its hypervolume. What is the expectation of the smallest distance from a point to the sphere centre?

Steps done so far are: Find distribution function of distance from random point to zero: $P(\xi < x) = x^n$, and the distribution function of the closest to zero point is $P(\xi < x) = 1 - (1-x^n)^l$. Then considering E(x) = $\int_{-\infty}^{+\infty} x \cdot d(F(x))$, the problem is how to calculate it.

Parcly Taxel
  • 103,344
NotImplemented
  • 33
  • 5
  • 1
    It cannot be true that $X_k\sim U(-1,1)$ for each $k$ if the point is restricted inside the unit ball. – Ѕᴀᴀᴅ Mar 10 '18 at 10:18
  • Thank you, fixed this. – NotImplemented Mar 10 '18 at 10:21
  • well, but still you have to tell how the $l$ points are distributed on the ball: uniform on the spherical coordinates $\rho,\theta\phi$ ? – G Cab Mar 10 '18 at 10:27
  • $$E(\xi)=\int_0^1P(\xi>x)dx=\int_0^1(1-x^n)^\ell dx=\int_0^1(1-t)^\ell \frac1nt^{1/n-1}dt=\frac1n\mathrm{B}\left(\ell+1,\frac1n\right)$$ where $\mathrm B$ denotes the Beta numbers, defined as $$\mathrm{B}(u,v)=\frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)}$$ – Did Mar 10 '18 at 10:28
  • The points are uniformly distributed over the interior of an n-dimensional hypersphere of radius 1 centered at the origin. – NotImplemented Mar 10 '18 at 10:50

1 Answers1

0

The cdf of the random variable describing the distance $r$ from a random point in the unit $n$-sphere to the centre is $r^n$, so the pdf is its derivative or $nr^{n-1}$. Now we are looking for the first order statistic of a sample of $l$ points from this distribution, whose pdf computes to (using the formula given on MathWorld) $$l(1-r^n)^{l-1}nr^{n-1}$$ Thus the desired expectation is $$E(n,l)=\int_0^1rl(1-r^n)^{l-1}nr^{n-1}\,dr$$ Substitute $s=r^n$: $$=l\int_0^1s^{1/n}(1-s)^{l-1}\,ds=l\mathrm B\left(1+\frac1n,l\right)=\frac{l\Gamma(1+1/n)\Gamma(l)}{\Gamma(1+1/n+l)}=\frac{l!}{\prod_{i=0}^{l-1}(1+1/n+i)}$$ $$E(n,l)=\prod_{i=1}^l\frac i{1/n+i}$$

Parcly Taxel
  • 103,344