Finding correlation between CDF of two normal distributions Suppose that $X\sim N(0,1)$, $Y\sim N(0,1)$ with correlation $(X, Y) =ρ$ where $ρ ∈ (−1, 1)$.
Show the following, 
Correlation $(Φ(X),Φ(Y))=\dfrac6π \arcsin\dfrac ρ2 $.
Here $Φ(X)$, and $Φ(Y)$ denote the CDF of Random variables $X$ and $Y$ respectively.
What I know so far:
$$\text{Cov}( Φ(X),Φ(Y)) =E(Φ(X)Φ(Y)) - E(Φ(X)) ×E(Φ(Y)) $$
We also know since  $X \sim N(0,1)$ and $Y\sim N(0,1)$; 
$Φ(X)\sim\text{unif}(0,1)$, $Φ(Y)\sim\text{unif}(0,1)$.
Hence
$$E(Φ(X)) =E(Φ(Y)) =\frac12.$$
I am stuck in finding $E(Φ(X)Φ(Y))$ [I tried to find it using double expectation].
What I have written may be the wrong way of attempting the question. So I am thankful for any help.
 A: Assuming $(X,Y)$ is jointly normal with zero means, unit variances and $\text{Corr}(X,Y)=\rho$, having joint density $f_{X,Y}$.
Define $$(X',Y')\stackrel{\text{i.i.d}}{\sim} N(0,1)$$
such that $(X',Y')$ is independent of $(X,Y)$. 
So $(X',Y')$ is (trivially) jointly normal with correlation zero. 
This implies $$(U,V)=\left(\frac{X-X'}{\sqrt 2},\frac{Y-Y'}{\sqrt 2}\right)$$ is also jointly normal with zero means and unit variances and $\text{Corr}(U,V)=\rho/2$.
Then,
\begin{align}
E(\Phi(X)\Phi(Y))&=\int_{\mathbb R}\int_{\mathbb R}\Phi(x)\Phi(y)f_{X,Y}(x,y)\,dx\,dy
\\&=\int_{\mathbb R}\int_{\mathbb R}P(X'\leqslant x,Y'\leqslant y)f_{X,Y}(x,y)\,dx\,dy
\\&=\int_{\mathbb R}\int_{\mathbb R}P(X'\leqslant x,Y'\leqslant y\mid X=x,Y=y)f_{X,Y}(x,y)\,dx\,dy
\\\\&=P(X'\leqslant X,Y'\leqslant Y)
\\\\&=P(X-X'\geqslant 0,Y-Y'\geqslant 0)
\\\\&=P\left(\frac{X-X'}{\sqrt 2}\geqslant 0,\frac{Y-Y'}{\sqrt 2}\geqslant 0\right)
\\\\&=P(U\geqslant 0,V\geqslant 0)
\\\\&=\frac{1}{4}+\frac{1}{2\pi}\arcsin\left(\frac{\rho}{2}\right)
\end{align}
In the last line, we used a popular result for the probability that two jointly normal variables both lie in the first quadrant. Proofs can be found here and here.
Finally, 
\begin{align}
\text{Corr}(\Phi(X),\Phi(Y))&=\frac{\frac{1}{4}+\frac{1}{2\pi}\arcsin\left(\frac{\rho}{2}\right)-\frac{1}{4}}{\frac{1}{12}}
\\\\&=\frac{6}{\pi}\arcsin\left(\frac{\rho}{2}\right)
\end{align}
