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?



Probability of a Gaussian random vector in a cone
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](https://en.wikipedia.org/wiki/Delta_method#Univariate_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)](https://projecteuclid.org/journals/annals-of-statistics/volume-22/issue-1/Extremal-Probabilistic-Problems-and-Hotellings-T2-Test-Under-a-Symmetry/10.1214/aos/1176325373.full), which means that $V\_n$ is strictly stochastically greater than $V$.


