I want to know the formula of the volume of the multivariate normal distribution in the positive orthant. For example, when there is only one variable, like $X\sim N(0,1)$, then $P(X>0)=\frac{1}{2}$. Similarly, when $(X,Y)\sim N(0,0;1,1;\rho)$, we have $P(X>0,Y>0)=\frac{1}{4}+\frac{1}{2\pi}\arcsin\rho$, which I have solved (If you are interested in it, keep reading and I will prove it in the final part).
When it comes to three variables, like $$\begin{pmatrix}X\\ Y\\ Z\end{pmatrix} \sim N\left( \begin{pmatrix}0\\ 0\\ 0\end{pmatrix} ,\begin{pmatrix}1&\rho_{1} &\rho_{2} \\ \rho_{1} &1&\rho_{3} \\ \rho_{2} &\rho_{3} &1\end{pmatrix} \right) $$, just use the principle of inclusion-exclusion, we can get $P(X>0,Y>0,Z>0)=\frac{1}{8}+\frac{1}{4\pi}\sum_{i=1}^{3}\arcsin\rho_i$.
Well, it seems too complex to continue. So I guess the formula of this question, that is $P(X_1>0,X_2>0,\cdots, X_n>0)$, may be $\frac{1}{2^n}+\frac{1}{2^{n-1}\pi}\sum_i\rho_i$(for each correlation).
If it is correct, how to prove it? What is the correct formula if not?
Here is the proof of $P(X>0,Y>0)$
$$U=\frac{Y-\rho X}{\sqrt{1-\rho^{2} } } $$, then $(X,U)\sim N(0,0;1,1;0)$ $$P(X>0,Y>0)=\frac{1}{2}P(\frac{Y}{X}>0)=\frac{1}{2}P(\frac{U}{X}>\frac{-\rho}{\sqrt{1-\rho^2}})$$
where $U/X\sim Cauchy(0,1)$, that is $$\frac{1}{2}\int_{\frac{-\rho}{\sqrt{1-\rho^2}}}^{+\infty}\frac{1}{\pi(1+t^2)}dt=\frac{1}{4}+\frac{1}{2\pi}\arcsin\rho$$
