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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 $x$.

$n=1$

(the circle)

$x_1$ follows the arcsine distribution.

$n=2$

(the sphere)

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

$n>2$

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!

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  • $\begingroup$ 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 at 14:48
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$\def\d{\mathrm{d}}$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.

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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 $$f_{X_1}(x) = \frac{1}{B \bigl( \frac{n}{2}, \frac{1}{2} \bigr)}(1-x^2)^{(n/2-1)}, \quad x\in[-1,1]. $$


Edit:
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,

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

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