Can we express the distribution of a coordinate of the $n$-sphere in any known distribution?

In formal terms, consider $S^n = \{x\in\mathbb{R}^{n + 1}: \|x\|=1\}$ (i.e. the usual $n$-sphere). If we sample $x$ uniformly from $S^n$ what is the distribution of $x_1$?

By "sampling uniformly" I mean that any point in $S^n$ has the same value for the density probability function. And $x_1$ means the first coordinate of vector $x$.


(the circle)

$x_1$ follows the arcsine distribution.


(the sphere)

Thanks to Archimedes we know that $x_1$ follows the Uniform distribution.


Do we know?


I know that this is equivalent to ask the distribution of the dot product of two random points on the $n$-sphere. But I also do not know that!

  • $\begingroup$ I'd say it is the normal distribution $N(0,1/m)$, and this guess is correct up to within Kolmogorov distance of $\mathcal O(1/m^c)$. $\endgroup$
    – dohmatob
    Jun 1 '20 at 14:48

$\def\d{\mathrm{d}}$This is a direct application of the formula for the surface area of the hyperspherical cap. Denoting by $I(x; a, b)$ the regularized beta function, the surface area of an $(n + 1)$-dimensional hyperspherical cap with height $x \leqslant 1$ and radius $1$ is$$ A_n(x) = \frac{1}{2} A_n(2) I\left( x(2 - x); \frac{n}{2}, \frac{1}{2} \right), $$ thus for $-1 \leqslant x \leqslant 0$,$$ P(X_1 \leqslant x) = \frac{A_n(x + 1)}{A_n(2)} = \frac{1}{2} I\left( 1 - x^2; \frac{n}{2}, \frac{1}{2} \right). $$

Since $\dfrac{∂I}{∂x}(x; a, b) = \dfrac{1}{B(a, b)} x^{a - 1} (1 - x)^{b - 1}$, then for $-1 < x < 0$,$$ f_{X_1}(x) = \frac{\d}{\d x} P(X_1 \leqslant x) = \frac{(1 - x^2)^{\frac{n}{2} - 1}}{B\left( \dfrac{n}{2}, \dfrac{1}{2} \right)}. $$ By symmetry,$$ f_{X_1}(x) = \frac{(1 - x^2)^{\frac{n}{2} - 1}}{B\left( \dfrac{n}{2}, \dfrac{1}{2} \right)}. \quad \forall -1 < x < 1 $$ Indeed, for $n = 2$ this is a uniform distribution.


I came across this answer previously. The original question was about the marginals of a point on the unit sphere in $\mathbb{R}^3$, but the poster added the case for the sphere in $\mathbb{R}^d$ (see Added section).

His method is summarized as follows (I have replaced $d$ in his answer with $n+1$):

1) Let $Z$ be a multivariate Gaussian r.v. with $(n+1)$ components, then $ \frac{Z}{\lVert Z \rVert}$ is uniformly distributed on the unit sphere $S^n$. Thus, we set $ X= \frac{Z}{\lVert Z \rVert}$.

2) Rewrite the CDF of $X_1 = \frac{Z_1}{\lVert Z \rVert}$ in terms of $\frac{Z_1^2}{Z_2^2 + \dots Z_n^2}$.

3) The ratio $\frac{n\cdot Z_1^2}{Z_2^2 + \dots Z_n^2}$ follows an $F_{1,n}$ distribution. Thus the CDF of $ X_1 = \frac{Z_1}{\lVert Z \rVert}$ can be determined from (2). Differentiating gives $$f_{X_1}(x) = \frac{1}{B \bigl( \frac{n}{2}, \frac{1}{2} \bigr)}(1-x^2)^{(n/2-1)}, \quad x\in[-1,1]. $$

Therefore, for $n=1$, $$f_{X_1}(x) \propto (1-x^2)^{-1/2} = \frac{1}{\sqrt{1+x}\sqrt{1-x}},$$ which is the arcsine distribution with support $x\in[-1,1]$.

For $n=2$, $$f_{X_1}(x) \propto 1 ,$$ which is the uniform distribution.

For a general $n$, $$f_{X_1}(x) \propto (1+x)^{(n/2-1)} (1-x)^{(n/2-1)},$$

which resembles a $\text{Beta}\left(\tfrac{n}{2},\tfrac{n}{2}\right)$ distribution, but rescaled so that its support is in $[-1,1]$.
That is,

$$\frac{X_1 + 1}{2} \sim \text{Beta}\left(\frac{n}{2}, \frac{n}{2}\right).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.