# Probability of a Gaussian random vector in a cone

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?

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 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, $$$$V_n:=\sqrt2\,(\sqrt Y_n-\sqrt{n-2})\to V$$$$ in distribution, where $$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 $$V$$.

• 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? Commented Mar 16, 2022 at 21:40
• @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. Commented Mar 17, 2022 at 3:56
• @HaoHe : Also, I have just recalled one of my old results, which implies the just added nice lower bound on the probability in question. Commented Mar 17, 2022 at 4:21
• 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/….
– Boby
Commented Mar 18, 2022 at 14:10
• @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. Commented Mar 18, 2022 at 14:41