# 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.

1. 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)

2. 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)

3. 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.

4. 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.

• Erm... Out of $\frac D2$ and $\frac{D-1}2$ at most one is an integer, so such a straightforward interpretation may be just impossible. – fedja Jun 12 at 2:20

## 1 Answer

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 ...

• The question is how to relate this idea with the probability $1- \left( \frac{x_0}{R} \right)^2$ of having $\frac{D-1}{2}$ successes out of $\frac{D}{2}$ trials. – Puco4 Jun 12 at 20:41
• yeah ... surprising perhaps that the 2 exponent is $D$-independent ... – phdmba7of12 Jun 13 at 14:19