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 $P(x_1, x_2)$?


2 Answers 2


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.)

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).

Figure 1

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:

Figure 2

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:

Figure 3

  • $\begingroup$ Hi, thank you for your detailed response! I actually realized that my question is a special case of mathoverflow.net/questions/359643/… With this said, interestingly, it looks like the constants differ in your and Iosif's answer (the key dependence on $(1 - x_1^2 - x_2^2)^{d / 2 - 2}$ is the same). And I was wondering if there might be a difference in the derivation that lead to this? $\endgroup$
    – student_t
    Commented Apr 30, 2021 at 19:06
  • $\begingroup$ @Student_t The Mathematics post doesn't explicitly give the constant, so there is no difference to note. The constant in my answer here is correct: look at the example and estimate the area under the curve. Clearly it's close to $1,$ as it should be. The fact that the curve agrees with the histogram shows the constant is correct. $\endgroup$
    – whuber
    Commented Apr 30, 2021 at 19:11
  • $\begingroup$ Thanks for your fast reply! Sorry for the confusion and just to clarify, I was referring to the third answer to that question, and not the first, where there is an explicit form at the end of the answer. $\endgroup$
    – student_t
    Commented Apr 30, 2021 at 20:10
  • $\begingroup$ @student_t Unfortunately, the order of answers varies for each user, so "third answer" doesn't identify any of them. Nevertheless, if you detect any difference in the constant, just perform the kinds of checks I have done here: integrate the density (numerically if you have to) and also compare it to simulated data. $\endgroup$
    – whuber
    Commented Apr 30, 2021 at 21:00
  • 1
    $\begingroup$ I agree that's odd--and it's a good check to make. A review of the analysis indicates it applies only to $d\ge 3.$ I have now noted that in the answer. $\endgroup$
    – whuber
    Commented May 23, 2021 at 18:38

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.


rX1X2=function(n,d) {

pX1X2=function(x1,x2,d) {
  f1=function(x,d,x1,x2) {
    return(ifelse((x[1]/den)<x1 & (x[2]/den)<x2,
 return(adaptIntegrate(f1,lowerLimit=c(-Inf,-Inf,0),upperLimit =c(Inf,Inf,Inf), 

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

       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.

enter image description here

  • $\begingroup$ Spherical geometry will reveal a simple closed formula. $\endgroup$
    – whuber
    Commented Apr 20, 2021 at 18:15
  • 1
    $\begingroup$ Here's another hint: $r^2 = Z_1^2+Z_2^2$ has a Beta distribution and $\arctan(Z_2/Z_1)$ has a uniform distribution. That fully describes the joint distribution. $\endgroup$
    – whuber
    Commented Apr 20, 2021 at 20:02
  • 3
    $\begingroup$ @whuber OK, I can see that. I don't think I would have ever thought of using those transformations. $\endgroup$
    – John L
    Commented Apr 20, 2021 at 20:33

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