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

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 $$\varphi(y)=\frac{1}{\sqrt{2\pi}}e^{-y^2/2}$$?

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 $$\Phi(\Phi^{-1}(p))\,\Phi(\Phi^{-1}(p))=p^2\,.$$