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?
 A: 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 $V$.
