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 $C = \{ x \in \mathbb R^n: \frac{\langle x  v, v \rangle}{\xv\_2 \v\_2} \leq \cos \theta \}$ 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
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_{n1}, \end{equation*} \begin{equation*} c:=\v\_2/\sigma>0,\quad t:=\cos\theta\in(0,1),\quad u:=\frac t{\sqrt{1t^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)&=1P(cZ_1>t\sqrt{(Z_1c)^2+Y_n}) \\ &=1P(Z_1<cu\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^{(1n)/2}} {\Gamma ((n1)/2)} \int_0^\infty \Phi(cu\sqrt y) e^{y/2} y^{(n3)/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{n2})\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 cu\sqrt Y_n) \\ &=P(Z_1+\tfrac u{\sqrt2}\,V_n\ge cu\sqrt{n2}) \\ &\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 $cu\sqrt{n2}$ 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 cu\sqrt{n2}) \\
&>P(Z_1+\tfrac u{\sqrt2}\,V\ge cu\sqrt{n2}) \\
&=1\Phi\Big(\frac{cu\sqrt{n2}}{\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/upperbound) the $\Phi$ function such that the integral can have a closedform? $\endgroup$– Hao HeCommented Mar 16, 2022 at 21:40

$\begingroup$ @HaoHe : I tried to differentiate the integral in (2) with respect to $c$, to get $\varphi(cu\sqrt y)$ in place of $\Phi(cu\sqrt y)$, 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/…. $\endgroup$– BobyCommented Mar 18, 2022 at 14:10

$\begingroup$ @Boby : I have seen that question and tried to play with it, unsuccessfuly. You are asking completeness/characterizationtype questions, which can be very hard (if at all possible) to answer. $\endgroup$ Commented Mar 18, 2022 at 14:41