Let $x$ be a $n$-dimensional Gaussian random vector, i.e., $x \sim \mathcal{N}(0,\sigma^2 I_n)$. What is its probability of falling in a cone? Say a cone $C = \{ x \in \mathbb R^n: \frac{\langle x - v, -v \rangle}{\|x-v\|_2 \|v\|_2} \leq \cos \theta \}$ is parameterized by $v \in \mathbb R^n$ and $\theta \in (0,\pi / 2)$. What is $\mathbb P(x \in C)$? Is there a closed form?


1 Answer 1


Let \begin{equation*} Z=(Z_1,\dots,Z_n):=x/\sigma\sim N(0,I_n), \end{equation*} \begin{equation*} Y_n:=Z_2^2+\dots+Z_n^2\sim\chi^2_{n-1}, \end{equation*} \begin{equation*} c:=\|v\|_2/\sigma>0,\quad t:=\cos\theta\in(0,1),\quad u:=\frac t{\sqrt{1-t^2}}=\cot\theta\in(0,\infty). \end{equation*} By the spherical symmetry, without loss of generality $v=c\sigma(1,0,\dots,0)$. So, \begin{equation*} \begin{aligned} P(x\in C)&=1-P(c-Z_1>t\sqrt{(Z_1-c)^2+Y_n}) \\ &=1-P(Z_1<c-u\sqrt Y_n). \end{aligned} \tag{1}\label{1} \end{equation*} Note that the random variables (r.v.'s) $Z_1$ and $Y_n$ are independent. So, \begin{equation*} P(x\in C)=1 -\frac{2^{(1-n)/2}} {\Gamma ((n-1)/2)} \int_0^\infty \Phi(c-u\sqrt y) e^{-y/2} y^{(n-3)/2}\,dy, \tag{2}\label{2} \end{equation*} where $\Phi$ is the standard normal cdf.

Mathematica cannot do anything with the latter integral. So, it is unlikely that it can be expressed in closed form.

However, using \eqref{1} or \eqref{2}, one can easily find various approximations to $P(x\in C)$, depending on how $n,c,u$ vary.

For instance, suppose that $u$ is fixed and $n\to\infty$. Then, by the central limit theorem and the delta method, \begin{equation} V_n:=\sqrt2\,(\sqrt Y_n-\sqrt{n-2})\to V \end{equation} in distribution, where $V$ is a standard normal r.v., which let us choose to be independent of $Z_1$. Then, by \eqref{1}, \begin{equation*} \begin{aligned} P(x\in C)&=P(Z_1\ge c-u\sqrt Y_n) \\ &=P(Z_1+\tfrac u{\sqrt2}\,V_n\ge c-u\sqrt{n-2}) \\ &\to P(Z_1+\tfrac u{\sqrt2}\,V\ge c_0) \\ &=1-\Phi\Big(\frac{c_0}{\sqrt{1+u^2/2}}\Big) \end{aligned} \tag{3}\label{3} \end{equation*} if $c$ varies with $n$ so that $c-u\sqrt{n-2}$ converges to some real $c_0$. Similarly, if $u$ and $c$ are fixed whereas $n\to\infty$, then $P(x\in C)\to1$.

Of course, one can also use various asymptotic expansions to obtain more detailed asymptotics.

In fact, we have a lower bound on $P(x\in C)$ that is essentially the same as the limit in \eqref{3}:
\begin{equation*} \begin{aligned} P(x\in C) &=P(Z_1+\tfrac u{\sqrt2}\,V_n\ge c-u\sqrt{n-2}) \\ &>P(Z_1+\tfrac u{\sqrt2}\,V\ge c-u\sqrt{n-2}) \\ &=1-\Phi\Big(\frac{c-u\sqrt{n-2}}{\sqrt{1+u^2/2}}\Big). \end{aligned} \tag{4}\label{4} \end{equation*} The inequality in \eqref{4} follows by formula (2.6), which means that $V_n$ is strictly stochastically greater than $V$.

  • $\begingroup$ Thanks a lot for your great answer! As I look the equation 2, I am wondering is it possible to approximate (or lower/upper-bound) the $\Phi$ function such that the integral can have a closed-form? $\endgroup$
    – Hao He
    Commented Mar 16, 2022 at 21:40
  • $\begingroup$ @HaoHe : I tried to differentiate the integral in (2) with respect to $c$, to get $\varphi(c-u\sqrt y)$ in place of $\Phi(c-u\sqrt y)$, where $\varphi:=\Phi'$, but Mathematica cannot take that simpler integral either. However, you can of course analyze the integral by the Laplace method (en.wikipedia.org/wiki/Laplace%27s_method). You can also use asymptotic expansions in the delta method, used in the above answer. $\endgroup$ Commented Mar 17, 2022 at 3:56
  • $\begingroup$ @HaoHe : Also, I have just recalled one of my old results, which implies the just added nice lower bound on the probability in question. $\endgroup$ Commented Mar 17, 2022 at 4:21
  • $\begingroup$ Hi Iosif, you seem to be an expert in how to manipulate gaussian. I was wondering if you can take a look at the question I asked here: mathoverflow.net/questions/418204/…. $\endgroup$
    – Boby
    Commented Mar 18, 2022 at 14:10
  • $\begingroup$ @Boby : I have seen that question and tried to play with it, unsuccessfuly. You are asking completeness-/characterization-type questions, which can be very hard (if at all possible) to answer. $\endgroup$ Commented Mar 18, 2022 at 14:41

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