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Let $X$ be a Gaussian vector in $\mathbb{R}^n$ with $\mathbb{E}[X]=0$ and $\mathbb{E}[X X^\intercal]=I_n$. Let $\mathbf{S}$ be a convex polytope in $\mathbb{R}^n$ defined as the intersection of $m$ $(m<n)$ half spaces, $\{ x\in\mathbb{R}^n\mid \langle x,\theta_i\rangle\ge 0 \}_{1\le i\le m}$. How do we compute $\mathbb{E}[|X|^2 \mathbb{1}_{X\in \mathbf{S}}]$?

I believe when $m=2$, there is a clean explicit formula for the expectation and it should depend on $n$ and the angle between $\theta_1$ and $\theta_2$. When $n$ is larger, can we still get an explicit formula for this expectation? If we can, how does it depend on $n$ and the angles?

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  • $\begingroup$ You write "hyperplanes" but then describe half spaces. Since you talk about defining a polytope as their intersection, I assume you meant "half spaces"? Please clarify your question either way :-) $\endgroup$
    – Max Horn
    Commented Aug 24, 2023 at 7:22
  • $\begingroup$ Thanks for the comment. Yes, I mean intersection of half spaces. $\endgroup$
    – Ye He
    Commented Aug 24, 2023 at 14:57
  • $\begingroup$ I doubt this has a closed form even in the case of the intersection of two half-planes when $n=2.$ If the two half-planes are orthogonal to each other, then it's reducible to two instances of the one-dimensional case. But even the one-dimensional case my require numerical methods: What is $\operatorname E(Z^2\mid Z>c)$ when $Z\sim\operatorname N(0,1)\text{?} \qquad$ $\endgroup$ Commented Aug 24, 2023 at 17:23
  • $\begingroup$ @MichaelHardy : An expression in elementary functions exists for $n\le3$ -- see my answer (mathoverflow.net/a/453379/36721). The hyperplanes are through the origin; so, your $c$ must be $0$. $\endgroup$ Commented Aug 24, 2023 at 18:02
  • $\begingroup$ @IosifPinelis : oh . . . . Somehow I missed the part about the boundaries going straight through the origin. I think that makes it far simpler. $\endgroup$ Commented Aug 25, 2023 at 19:29

1 Answer 1

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$\newcommand\R{\mathbb R}$Let $S:=\mathbf S$. Let $X$ be any random vector in $\R^n$ with a spherical symmetric distribution (say such that $P(X=0)=0$). Then $|X|$ is independent of $U:=X/|X|$ and $U$ is uniformly distributed on the unit sphere. So, $$E|X|^2\,1(X\in S)=E|X|^2\,1(U\in S)=E|X|^2\,P(U\in S)=E|X|^2\,p_{n,S},$$ where $p_{n,S}:=P(U\in S)$; the displayed equalities actually hold even if $P(X=0)\ne0$. (If $X$ is a standard Gaussian vector in $\R^n$, then of course $E|X|^2=n$.)

As noted by Ruben, p. 213, "For dimensionality greater than three (spherical tetrahedra, spherical pentahedra, etc.) the areas [that are the probabilities $p_{n,S}$ -- I.P.] can no longer be expressed in terms of elementary functions." For such dimensions, there are only recursive formulas expressing $p_{n,S}$ in terms of integrals of $p_{n-2,S'}$ for some polyhedral cones $S'$ in $\R^{n-2}$. These recursive formulas were first obtained by Schläfli in 1858; see e.g. Plackett for a generalization and further references.

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  • $\begingroup$ This is helpful. It is great to know $p_{n,s}$ can be computed recursively. Thanks! $\endgroup$
    – Ye He
    Commented Aug 24, 2023 at 15:22
  • $\begingroup$ If $X$ is a random vector with isotropic log-concave density, is it possible to get a recursive formula(might be inequality) for the corresponding $p_{n,S}$? $\endgroup$
    – Ye He
    Commented Aug 24, 2023 at 18:17
  • $\begingroup$ Thanks! The follow-up question has been removed. $\endgroup$
    – Ye He
    Commented Aug 24, 2023 at 18:34
  • $\begingroup$ @YeHe : All right. Are you satisfied with the answer? $\endgroup$ Commented Aug 24, 2023 at 18:51
  • $\begingroup$ For now, yes. I need to read Schläfli's and Plackett's papers to convince myself it is hard to work on non-gaussian densities. $\endgroup$
    – Ye He
    Commented Aug 24, 2023 at 19:05

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