# Claims and questions regarding $n$-ball distribution?

CONTEXT

In my research, I am utilizing an $$n$$-ball distributions along with two related distributions. I'd like to make certain I have a firm handle on the way to describe my three distributions. I have been looking for reference books on the subject , I've got some guidance, but have not been able to find what I'm looking for yet.

In this post, I attempt to deal with the simplest case that i am concerned with. I'm modelling my approach to this based on the format I find on wikipedia (e.g. )

QUESTIONS

What is the characteristic function of the uniform distribution on a ball in $$\mathbb{R}^𝑛$$? [edit: solution offered by @whuber in the below solution.]

What is the entropy of the uniform distribution on a ball in $$\mathbb{R}^𝑛$$? [edit: solution offered by @whuber in the below comment.]

MY UNDERSTANDING

The $$n$$-ball distribution here is a generalization of the uniform distribution.

Parameters

By $$n\in \mathbb{N}$$ I denote the dimension of the ball.

By $$R\in \mathbb{R}, R>0$$ I denote the radius of the $$n$$-ball.

By $$\gamma$$ I denote a parametrization of the $$n$$-ball given as $$\gamma: (0,R) \times \left[0, \pi\right) \times \cdots \times \left[0, \pi\right) \times \left[0, 2\pi\right) \rightarrow \mathbb{R}^n$$, which is defined by: $$\gamma\begin{pmatrix}r\\\\ \phi_1 \\\\ \vdots \\\\ \phi_{n-1}\end{pmatrix} \rightarrow \begin{bmatrix} r \cos{(\phi_{1})} \prod\limits_{i=1}^{1-1} \sin{(\phi_{i })} \\\\ r \cos{(\phi_{2})} \prod\limits_{i=1}^{2-1} \sin{(\phi_{i })} \\\\ \vdots \\\\ r \cos{(\phi_{n-1 })} \prod\limits_{i=1}^{n-1-1} \sin{(\phi_{i })} \\\\ r \prod\limits_{i=1}^{n-1} \sin{(\phi_{i })} \end{bmatrix}.$$

Support $$i = {1, 2, \ldots, n}$$ $$x_i = (-R,R)$$ $$0\leq \sum_{i=1}^{n}x_i^2 < R^2$$

Probability Density Function

With respect to the indicator for the $$n$$-ball, $$\mathcal{I}(\left\|\textbf{x}\right\|_2 , the probability density is $$\frac{\Gamma\left(\frac{n}{2} + 1\right)}{\pi^\frac{n}{2} R^n}\,\mathcal{I}(\left\|\textbf{x}\right\|_2

Mean

$$E(X_i) = 0$$

Variance

$$\textrm{Var}(X_i) = \dfrac{1}{n+2} R^2$$

$$\textrm{Cov}(X_i,X_j) = 0\quad\quad i \neq j$$

Entropy

$$\log\left(\frac{\Gamma\left(\frac{n}{2} + 1\right)}{\pi^\frac{n}{2} R^n}\right)$$

Characteristic function

$$\phi_n(t) = e^{-i\frac{|t|}{R}} \,_1F_1\left(\frac{n+1}{2};n+1; i\frac{2|t|}{R}\right).$$

BIBLIOGRAPHY

• Aug 26, 2019 at 15:09
• Those are great leads. Aug 26, 2019 at 15:30
• There's no such thing as a PGF for these continuous distributions and the MGF is just the CF evaluated at purely imaginary arguments, so what looks like a large battery of questions could be fruitfully narrowed to a single one: what is the characteristic function of the uniform distribution on a ball in $\mathbb{R}^n$?
– whuber
Aug 26, 2019 at 15:47
• Re entropy: Perhaps surprisingly, that's trivial. The reason is that for the uniform distribution over any region $\mathcal B$ the density equals $1/|\mathcal B|$ (the reciprocal volume) throughout $\mathcal B,$ whence the entropy is $\log|\mathcal B|.$ There is a simple formula for the volume of an $n$-ball.
– whuber
Oct 8, 2019 at 13:21
• Re the cf: You might want to replace $|t|$ by $|t|/R$ on the right hand side so that the formula is correct for (origin-centered) balls of radius $R$ to match all the other formulas. (2) It might also be worthwhile multiplying your formula for the pdf by the indicator function of the ball, $\mathcal{I}(|x|\le R).$ (3) The left hand side of the inequality for the support is useless; you can replace $-R^2$ by $0.$
– whuber
Oct 8, 2019 at 13:49

The question asks for the characteristic function of the uniform distribution on a ball.

Let's begin with definitions and simplifications, because it turns out that's all the computation we will need.

### Definitions

The characteristic function of a density $$\mathrm{d}\mu$$ on $$\mathbb{R}^n$$ is the function of the $$n$$-vector $$t=(t_1,t_2,\ldots, t_n)$$ defined by

$$\phi_{\mathrm{d}\mu}(t) = \int \cdots \int e^{it\cdot x}\mathrm{d}\mu(x)$$

where $$t\cdot x = t_1x_1 + t_2 x_2 + \cdots t_n x_n$$ is the Euclidean dot product. (This dot product determines the Euclidean length $$|t|^2 = t\cdot t.$$) Because $$e^0=1,$$ note that $$\phi_{\mathrm{d}\mu}(0) = \int\cdots\int \mathrm{d}\mu(x)$$ is just the integral of the density.

A ball $$B(y,r)$$ for $$y\in\mathbb{R}^n$$ and $$r \ge 0$$ is the set of points within distance $$r$$ of $$y;$$ that is, $$x\in B(y,r)$$ if and only if $$|x-y| \le r.$$

The uniform distribution on any set $$\mathcal{B}\subset \mathbb{R}^n$$ with finite (Lebesgue) integral, such as a ball, has a density that is a constant multiple of Lebesgue measure on $$\mathcal{B}$$ and otherwise zero. The constant is adjusted to make a unit integral.

### Simplifications

Given $$B(y,r)$$ and a vector $$t,$$ we may translate the ball by $$-y,$$ scale it by $$1/r,$$ and rotate it to make $$t=(0,0,\ldots,0,|t|).$$ The translation multiplies its characteristic function $$\phi$$ by $$e^{-it\cdot y};$$ the scaling changes $$\phi(t)$$ to $$\phi(tr);$$ and because the ball is spherically symmetric, the rotation doesn't change its characteristic function at all.

This reduces the problem to that of finding

$$\phi_n(t) = \int \cdots \int_{B(0,1)} e^{i |t| x_n}\, \mathrm{d} x_1\cdots \mathrm{d} x_n,\tag{1}$$

after which we may replace $$|t|$$ by $$|t|/r$$ and multiply the result by $$e^{it\cdot y}$$ to obtain the characteristic function of $$B(y,r).$$

The strategy to minimize computation is to compute this integral up to a multiplicative constant and then discovering that constant from the fact that $$\phi_n(0)=1$$ because the density must integrate to unity.

The integral $$(1)$$ slices the unit $$n$$-ball into horizontal $$n-1$$-balls of radii $$\sqrt{1-x_n^2}$$ (from the Pythagorean Theorem). Being $$n-1$$-dimensional, such balls have $$n-1$$-volumes proportional to the $$n-1$$ power of their radii,

$$\left(\sqrt{1-x_n^2}\right)^{n-1} = (1-x_n)^{(n+1)/2-1}\,(1+x_n)^{(n+1)/2-1}.$$

By Cavalieri's Principle the integral therefore is proportional to

$$\phi_n(t) \propto \int_{-1}^1 e^{i|t|x_n}\, (1-x_n)^{(n+1)/2-1}\, (1+x_n)^{(n+1)/2-1} \, \mathrm{d}x_n\tag{2}.$$

For convenience, write $$a=(n+1)/2.$$

### Calculation

The substitution $$1+x=2u$$ entails $$\mathrm{d}x = 2\mathrm{d}u$$ with $$0\le u\le 1.$$ Observing that $$1-x = 2-(1+x) = 2-2u,$$ $$(2)$$ has become

$$\phi_n(t)\propto \int_0^1 e^{i|t|(2u-1)} (2u)^{a-1}(2-2u)^{a-1}\,2\mathrm{d}u \propto e^{-i|t|} \int_0^1 e^{i(2|t|)u} u^{a-1}(1-u)^{a-1}\,\mathrm{d}u .$$

The integral is explicitly the value of the characteristic function at $$2|t|$$ of the univariate density

$$F_{a,a}(u) \propto u^{a-1}(1-u)^{a-1},$$

which we immediately recognize as the Beta$$(a,a)$$ distribution. Its characteristic function is given by the confluent hypergeometric function $$_1F_1$$ with parameters $$a,2a,$$ whence

$$\phi_n(t) \propto e^{-i|t|} \,_1F_1(a;2a; 2i|t|).\tag{3}$$

Indeed, since $$_1F_1$$ is a characteristic function, $$_1F_1(a,2a; 0) = 1$$ and obviously $$e^{-i|0|}=1.$$ Accordingly, formula $$(3)$$ already is normalized: the constant of proportionality is $$1.$$ (That's why no calculations are needed.) Thus,

$$\phi_n(t) = e^{-i|t|} \,_1F_1\left(\frac{n+1}{2};n+1; 2i|t|\right).$$

### Implications

Most people are unfamiliar with hypergeometric functions. They actually are very tractable. One definition is in terms of power series:

\eqalign{ _1F_1(a;b; z) &= \sum_{n=0}^\infty \frac{a^{(n)}}{b^{(n)}} \frac{z^n}{n!} \\ &= 1 + \frac{a}{b}z + \frac{a(a+1)}{b(b+1)}\frac{z^2}{2!} + \cdots + \frac{a(a+1)\cdots(a+n-1)}{b(b+1)\cdots(b+n-1)}\frac{z^n}{n!} + \cdots,}

from which we may read off the moments $$a^{(n)}/b^{(n)}.$$ For integral $$a$$ (the dimension $$n$$ is odd) these are linear combinations of exponentials with rational coefficients; for half-integral $$a$$ (even dimension $$n$$) they are rational linear combinations of Bessel functions $$J_0,$$ $$J_1,$$ through $$J_{\lfloor a \rfloor}.$$ For instance,

$$\phi_1(t) = e^{-i|t|}\,_1F_1 (1;2;2i|t|) = e^{-i|t|}\left(\frac{e^{2i|t|} - 1}{2i|t|}\right) = \frac{\sin|t|}{|t|}$$

is the characteristic function of the unit ball in one dimension: the interval $$[-1,1]$$ and

$$\phi_2(t) = e^{-i|t|}\,_1F_1 (3/2;3;2i|t|) = 2\frac{J_1(|t|)}{|t|}$$

is the characteristic function of the unit disk in the plane.