Intuitive explanation of CDF of a Binomial distribution in the volume of a Hyperspherical Cap

Solution 1:

Assume vlog that $R=1$. The problem is equivalent to proving that the probability that the random point inside a hypersphere has $x$-coefficient less than $x_0$ is equal to: $$\mathcal{F} \left( \frac{D-1}{2}; \frac{D}{2}, 1-x_0^2 \right)=I_{x_0^2}(\frac{1}{2},\frac{D+1}{2})$$ To do this we have to ask how we can draw random points from the inside of a hypersphere uniformly at random. Notice that if $N_1$, $N_2$, ..., $N_D$ are independent standard normal random variables, the vector: $$(N_1, N_2, ..., N_D)$$ Points to a point on a hemisphere uniformly at random. The same would be true for: $$(\frac{N_1}{X}, \frac{N_2}{X}, ..., \frac{N_D}{X})$$ For any random variable $X$. If we choose: $$X=\sqrt{N_1^2+N_2^2+...+N_D^2+Y}$$ Where $Y \sim Exp(2)=\Gamma(1,2)$ we see that the radius squared is equal to: $$r^2=\frac{N_1^2+N_2^2+...+N_D^2}{N_1^2+N_2^2+...+N_D^2+Y}=\frac{Z}{Z+Y}$$ Where $Z \sim \Gamma(\frac{D}{2},1)$. It follows that $r^2 \sim B(\frac{D}{2},1)$ which is exactly the distribution of a squared distance of a uniformly random point inside a hypersphere (which can be easily seen from the CDF of a radius which is $P(r \le x)=x^D$). So the distribution of the square of the first coordinate of a random point inside a hypersphere is the same as the distribution of: $$\frac{N_1^2}{X^2}=\frac{N_1^2}{N_1^2+N_2^2+...+N_D^2+Y}=\frac{A}{A+B}$$ Where $A \sim \Gamma(\frac{1}{2},2)$ and $B \sim \Gamma(\frac{D+1}{2})$ from which it follows that $x_0^2$ has a distribution $B(\frac{1}{2},\frac{D+1}{2})$ which concludes the proof.