Marginal distribution of uniform distribution over sphere

Question

Let $(x_1,…,x_n)$ be a random vector uniformly distributed on the $n$-dimensional unit sphere.

Is there a closed form solution for the joint distribution of $P(x_1, x_2)$?

Answer 1

I want to flesh out John L's idea. Let $d\gt 2$ be the dimension of the space in which we will be working. (When $d=2$ the marginal is the uniform distribution on the circle -- that fully determines it, but it has no density function.) As you go through, note that the same analysis applies mutatis mutandis to finding the distribution of any proper subset of the coordinates, from $1$ through $d-1$ of them.

1. The uniform distribution on the surface of the unit $d-1$ sphere $S^{d-1}\subset\mathbb{R}^d$ is the radial projection of the standard $d$-variate Normal distribution in $\mathbb{R}^d.$ See How to generate uniformly distributed points on the surface of the 3-d unit sphere?.

2. Writing $(Z_1,Z_2,\ldots,Z_d)$ for such a Normal variate and $|Z| = \sqrt{Z_1^2+\cdots + Z_d^2}$ for its norm, $(1)$ means $(X_1,\ldots, X_d) = (Z_1/|Z|,\ldots,Z_d/|Z|)$ has a uniform distribution on $S^{d-1}.$

3. By definition, $U^2=Z_1^2+Z_2^2$ has a $\chi^2(2)$ distribution and $V^2=Z_3^2 + \cdots + Z_d^2$ has a $\chi^2(d-2)$ distribution. (This is one place we must require $d\gt 2.$)

4. Because the $Z_i$ are independent, $(Z_1,Z_2)$ is independent of $(Z_3,\ldots,Z_d),$ whence $U^2$ and $V^2$ are independent.

5. The ratio $\frac{U^2}{U^2+V^2} = X_1^2+X_2^2$ has a Beta$(1,d/2-1)$ distribution.

6. Because $(Z_1,\ldots,Z_d)$ is spherically symmetric, so is $(X_1,\ldots,X_d),$ whence (via projection) $(X_1,X_2)$ is circularly symmetric.

7. $(5)$ and $(6)$ say that in polar coordinates $(R,\Theta)$ with $X_1=R\cos\Theta$ and $X_2=R\sin\Theta,$ $R^2$ has a Beta distribution and $\Theta$ is independently uniformly distributed on (say) the interval $[0,2\pi).$

8. Conditional on $(X_1,X_2),$ the remaining coordinates $(X_3,\ldots,X_d)$ must be uniformly distributed on the slice of the sphere determined by $(X_1,X_2).$ That slice has radius $\sqrt{1-R^2}.$

We can immediately write down expressions for the distribution. Since the density of a Beta distribution is $f(t;\alpha,\beta) = t^{\alpha-1}(1-t)^{\beta-1}/B(\alpha,\beta),$ setting $t=r^2$ gives the density for $R$ as

$$f_R(r;d) = \frac{2}{B\left(1,\frac{d}{2}-1\right)}\,r(1-r^2)^{d/2-2} = (d-2)\,r\,(1-r^2)^{d/2-2}$$

for $0\le r\le 1$ and $d\gt 2.$

The joint density $f_{R,\Theta;d}$ is the product of $f_R$ and the density of $\Theta,$ given by $1/(2\pi)$ on the interval $[0,2\pi).$ Changing back to rectangular coordinates gives

$$f_{x_1,x_2}(x_1,x_2;d) = \frac{d-2}{2\pi}\,(1-x_1^2-x_2^2)^{d/2-2}\tag{*}$$

for $x_1^2+x_2^2\le 1.$ (See the example at https://stats.stackexchange.com/a/154298/919 for details.)

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As an example, I generated a million draws from the uniform distribution on $S^{11}$ (so that $d=12$). Here is a scatterplot matrix of the first five components (showing just the first thousand observations for clarity).

The circular symmetry of each pair is apparent.

Next is a histogram of all million values of $R$ on which the root-Beta density function $f_R$ is plotted, with excellent agreement:

The $(X_i+1)/2$ have Beta$((d-1)/2,(d-1)/2)$ distributions, as shown at Distribution of scalar products of two random unit vectors in $D$ dimensions. This indeed is what one obtains from $(*)$ by integrating out one of the variables. Here is a histogram from the simulation with that Beta density overplotted:

Answer 2

I don't think there is a closed form, but there are some observations below.

Let $Z_1,...,Z_d$ be iid standard normal.
Then, $\left(\frac{Z_1}{\sqrt{Z_1^2+...+Z_d^2}},...,\frac{Z_d}{\sqrt{Z_1^2+...+Z_d^2}} \right)$ is uniform on the $d$-dimensional unit sphere.
We want to find the joint distribution of $(X_1,X_2)=\left(\frac{Z_1}{\sqrt{Z_1^2+...+Z_d^2}},\frac{Z_2}{\sqrt{Z_1^2+...+Z_d^2}} \right)$.

First notice that $\left(\frac{1}{X_1^2},\frac{1}{X_2^2} \right)=\left(\frac{Z_1^2+...+Z_d^2}{Z_1^2},\frac{Z_1^2+...+Z_d^2}{Z_2^2} \right)=\left(1+\frac{Z_2^2+Z_3^2...+Z_d^2}{Z_1^2},1+\frac{Z_1^2+Z_3^2...+Z_d^2}{Z_2^2} \right)$.

Thus, $\left(\frac{1}{d-1}\left( \frac{1}{X_1^2}-1 \right),\frac{1}{d-1}\left( \frac{1}{X_2^2}-1 \right)\right)$ has the same distribution as $\left(\frac{(V_2+V_3)/(d-1)}{V_1},\frac{(V_1+V_3)/(d-1)}{V_2} \right)$ where $V_1,V_2,V_3$ are independent chi-square random variables with degrees of freedom 1, 1, and $d-2$ respectively.

$\frac{(V_2+V_3)/(d-1)}{V_1}$ and $\frac{(V_1+V_3)/(d-1)}{V_2}$ are dependent and each has an F-distribution with $d-1$ (numerator) and 1 (denominator) degrees of freedom. There are some other definitions of the bivariate F-distribution that have closed forms, but I don't think this one does. The density for $X_1$ or $X_2$ is $$f(x)=\frac{((d-2)(1 - x^2))^{d/2}(d-2 + x^2)^{(1 - d)/2}\Gamma((1 + d)/2)}{\sqrt{d-1}\sqrt{\pi}(1 - x^2)^2\Gamma(d/2)}$$ for $x \in (-1,1)$. I don't think it is possible to find the distribution of $X_2$ given $X_1$ in an easy form.

The following R functions generate random numbers and calculate the joint distribution function by numerical integration.

library(cubature)

rX1X2=function(n,d) {
  z1=rnorm(n)
  z2=rnorm(n)
  v3=rchisq(n,d-2)
  return(matrix(c(z1,z2)/sqrt(z1^2+z2^2+v3),byrow=F,ncol=2))
}

pX1X2=function(x1,x2,d) {
  f1=function(x,d,x1,x2) {
    den=sqrt(x[1]^2+x[2]^2+x[3])
    return(ifelse((x[1]/den)<x1 & (x[2]/den)<x2,
                     dnorm(x[1])*dnorm(x[2])*dchisq(x[3],d-2),0))
  }
 return(adaptIntegrate(f1,lowerLimit=c(-Inf,-Inf,0),upperLimit =c(Inf,Inf,Inf), 
                          d=d,x1=x1,x2=x2,maxEval=10000)$integral)
}

pX1X2(-0.2,0.3,5) #estimated probability that X1<-0.2 and X2<0.3
x=rX1X2(100000,5)
mean(x[,1]<(-0.2) & x[,2]<0.3) #estimated probability from simulation

y=(1/x[,1]^2-1)/4
plot(log(quantile(y,c(1:99)/100)),
       log(qf(c(1:99)/100,4,1))) #verify that y has an F(4,1) distribution

Q-Q plot (log-scale) verifying that the transformed variable has an F-distribution.