# 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

• 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