Volume computation using probabilistic approach

Question

Let $\mathbb{S}^{d-1}=\{v\in\mathbb{R}^d:\|v\|_2=1\}$, namely $d-$dimensional sphere. It is well-known that if a random vector $X$ is distributed uniformly on $\mathbb{S}^{d-1}$, then there exists i.i.d. standard normal random variables $N_1,\dots,N_d$ such that $$ X \stackrel{d}{=}(N_1/N,\cdots,N_d/N), $$ where $N=\sqrt{N_1^2+\cdots+N_d^2}$.

Now, my question is the following. Suppose that I am interested in understanding the size of a certain subset $S$ of $\mathbb{S}^{d-1}$ (with respect to $(d-1)-$dimensional Lebesgue measure). One can then set $$ \mathbb{P}(X\in S) = {\rm Size}(S)/{\rm Size}(\mathbb{S}^{d-1}). $$ Here, the presence of Gaussianity often makes the computation simple, thereby suggesting a potential approach for computing the size of $S$. Is this type of an approach well-known? Namely, is there any work on calculating the sizes of complicated sets using this type of probabilistic reasoning?

Answer 1

This approach is of course well known. Clearly, it just says that $$P(X\in A)=P((N_1,\dots,N_d)\in C_A),$$ where $A$ is a Borel subset of the unit sphere $S^{d-1}$ and $C_A:=\mathbb R_+A$ is the corresponding cone.

The hard part is to compute the Gaussian measure, $P((N_1,\dots,N_d)\in C_A)$, of the cone $C_A$.

This is hard even when $A$ is a spherical simplex. The work on this begins with Schläfli (1858), whose paper (I think) is hard to read. A more general result was obtained by Plackett (1954), who gave a recursive formula (see formulas (7) and (6) there) for the centered Gaussian measure of a simplicial cone in $\mathbb R^d$ (not necessarily with the vertex at the origin) in terms of a certain integral functional of certain partial derivatives of the Gaussian measure of a varying simplicial cone in $\mathbb R^{d-2}$. See also references in Plackett's paper, including the one to Schläfli. (Plackett is actually dealing with the obviously equivalent problem, where the simplicial cone is just $c+\mathbb R_+^d$ for some $c\in\mathbb R^d$, but the Gaussian distribution is not necessarily the standard one.)