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 [1], 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. [2])


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.]


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


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, $\mathcal{I}(\left\|\textbf{x}\right\|_2 <R)$, the probability density is $$\frac{\Gamma\left(\frac{n}{2} + 1\right)}{\pi^\frac{n}{2} R^n}\,\mathcal{I}(\left\|\textbf{x}\right\|_2 <R)$$


$$E(X_i) = 0$$


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

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


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

Characteristic function

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


[1] Reference books on uniform spherical distributions in multiple dimensions

[2] https://en.wikipedia.org/wiki/Multinomial_distribution

  • $\begingroup$ Those are great leads. $\endgroup$ Aug 26, 2019 at 15:30
  • 1
    $\begingroup$ 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 $\mathbb{R}^n$? $\endgroup$
    – whuber
    Aug 26, 2019 at 15:47
  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Oct 8, 2019 at 13:21
  • 1
    $\begingroup$ 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 $0.$ $\endgroup$
    – whuber
    Oct 8, 2019 at 13:49

1 Answer 1


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.


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.


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 $a=(n+1)/2.$


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,

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


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.


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