Intuitive explanation of CDF of a Binomial distribution in the volume of a Hyperspherical Cap Note: This is my first question ever in stackexchange, I apologize for any mistakes in formatting, on the appropriateness of the question and tags.


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*From Wikipedia, I know the regularized incomplete beta function is related to the CDF of a random variable $X$ from a Binomial distribution:
$$\mathcal{F} \left(k; n, p\right) = 1 - I_{p} \left( k+1, n-k \right). $$
(https://en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function)

*Also from Wikipedia, the expression of a hyperspherical cap in $D$ dimensions is given by
$$ V_{D\text{-cap}} = \frac{1}{2}V_{D\text{-ball}} \, I_{(2Rh-h^2)/R^2} \left(\frac{D+1}{2}, \frac{1}{2} \right),$$
where $V_{D\text{-ball}}$ is the the volume of the the $D$-ball with radius $R$ and $h$ is its height.
(https://en.wikipedia.org/wiki/Spherical_cap)  

*I re-expressed the previous expression to incorporate heights larger than $R$ and considered the cap is cut at the hyperplane $x=0$ and the ball is centered at $x=x_0$ (See image).
Then, $R = h-x_0$, and
$$ \frac{V_{D\text{-cap}} (x_0)}{V_{D\text{-ball}}} = -\frac{1}{2} \text{sgn}\left(\frac{x_0}{R} \right) I_{1-\left( \frac{x_0}{R} \right)^2} \left( \frac{D+1}{2}, \frac{1}{2} \right) + \Theta \left(\frac{x_0}{R}\right),$$
where $\text{sgn}(x)$ is the sign operator, $I_x \left(a,b \right)$ the regularized incomplete beta function and $\Theta(x)$ the Heaviside function.

*Considering the property from point 1 and $\Theta (x) = \frac{1}{2} + \frac{1}{2} \text{sgn} (x)$, this can be re-expressed as:
$$\frac{V_{D\text{-cap}} (x_0)}{V_{D\text{-ball}}} =\frac{1}{2} + \frac{1}{2} \text{sgn}\left(\frac{x_0}{R}\right) \mathcal{F} \left( \frac{D-1}{2}; \frac{D}{2}, 1-\left( \frac{x_0}{R} \right)^2 \right).$$
So here comes my question: What is the meaning of the CDF of the Binomial distribution in this context? That is, which random variable is associated to this problem that has probability $1- \left( \frac{x_0}{R} \right)^2$ of having $\frac{D-1}{2}$ successes out of $\frac{D}{2}$ trials?

EDIT:
After some plots I see the expression from step 1 is never true for $k=\frac{D-1}{2}$ and $n=\frac{D}{2}$. For odd $D$ the Binomial CDF appears to not be defined, I guess a fractional number of experiments $n$ is not well-defined; and for even $D$ the two sides give different values, I imagine a fractional number of successes $k$ is neither well-defined. So as pointed by @fedja this interpretation may be ill-posed.
 A: 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.
A: I'd think the volume ratio is tied to Binomial distribution if you consider computing it via a Monte Carlo method where you pick random points inside a hypercube completely containing the sphere (and thus the cap) and tally up how many such points lie in the cap region vs entire sphere itself
like computing pi by filling up a square with random points and recording which fraction lie inside a circle ...
