Closed-form CDF for bivariate normal distribution in point $(\Phi^{-1}(p),\,\Phi^{-1}(p))$

Question

Let $\Phi(x)$ be a CDF of standard normal distribution and $\Phi^{-1}(p),\,p\in(0,1)$ its inverse.

It is evident that $$ \mathbb{P}(X<\Phi^{-1}(p))=\Phi(\Phi^{-1}(p))=p, $$ where $X\sim N(0,1)$.

Is there possible to get any simplified expression of $$ \mathbb{P}(X<\Phi^{-1}(p),\,Y<\Phi^{-1}(p))= \int_{-\infty}^{\Phi^{-1}(p)}\varphi(y)\Phi\left(\frac{-\varrho y+\Phi^{-1}(p)}{\sqrt{1-\varrho^2}}\right)\,dy, $$ where $X,\,Y\sim N(0,1)$, $\textrm{cov}(X,Y)=\varrho$ and $\varphi(y)=\frac{1}{\sqrt{2\pi}}e^{-y^2/2}$?

Answer 1

If $\varrho$ is not zero there isn't much you can do. If you have an implementation of the bi-variate normal CDF $\Phi_2(x,y,\varrho)$ you can "simplify" it to $$ \mathbb P\Big(X<\Phi^{-1}(p),Y<\Phi^{-1}(p)\Big)=\Phi_2\Big(\Phi^{-1}(p),\Phi^{-1}(p),\varrho\Big)\,. $$ This RHS is known as Gauss copula. If $p=1$ and $\Phi^{-1}(p)=+\infty$ the RHS is one.

Needless to say that for $\varrho=0$ the RHS is $\Phi(\Phi^{-1}(p))\,\Phi(\Phi^{-1}(p))=p^2\,.$