Duplicate of the stats stack exchange question here; however, I need some help with some of the steps in the accepted answer.

A uniform distribution on the unit sphere $\mathbb{S}^{D-1}$ is obtained by normalizing a $D$-variate normal distribution and the dot product $t$ of normalized vectors is their correlation coefficient

Method: The exact distribution of the dot product of unit vectors is easily obtained geometrically, because this is the component of the second vector in the direction of the first. Since the second vector is independent of the first and is uniformly distributed on the unit sphere, its component in the first direction is distributed the same as any coordinate of the sphere. (Notice that the distribution of the first vector does not matter.)

Finding the Density Letting that coordinate be the last, the density at $t\in[-1,1]$ is therefore proportional to the surface area lying at a height between $t$ and $t+dt$ on the unit sphere. That proportion occurs within a belt of height $dt$ and radius $\sqrt{1-t^2}$, which is essentially a conical frustum constructed out of an $\mathbb{S}^{D-2}$ of radius $\sqrt{1-t^2}$, of height $dt$, and slope $1/\sqrt{1-t^2}$. Whence the probability is proportional to $\dfrac{\sqrt{1-t^2}^{D-2}}{\sqrt{1-t^2}}dt=(1-t^2)^{(D-3)/2}dt$.

I understand that the method detailed above results in a spherical cap and that $t:=cos\theta$, i.e. the dot product between two unit vectors.

Using the Pythagorean theorem, we know that the radius of the base of the cap is $sin\theta=\sqrt{1-cos^2\theta}=\sqrt{1-t^2}$.

1) Now is the slope/slant of the "conical frustum": $\dfrac{\textrm{radius of the unit. sphere}}{\textrm{radius of the cap base}}$, i.e. $\dfrac{1}{\sqrt{1-t^2}}$?

2) And is the thickness of the conical frustum $R\cdot dt=dt$, where $R=1$ (unit $n$-sphere radius)?

3) I understand that the surface area of $\mathbb{S}^{D-1}$ with radius $R$ is proportional to $R^{D-1}$. But, how do we get the proportional probability?

  • $\begingroup$ I lost you at "We have the spherical cap area:" none of those equalities seems to comport with the laws of Euclidean geometry. Could you perhaps provide explanations for the steps you are taking? $\endgroup$ – whuber Mar 16 at 22:18
  • $\begingroup$ @whuber I've updated my question $\endgroup$ – patagonia89 Mar 17 at 13:08

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