Compute $E[a\cdot b|a \geq 0, b \geq 0]$ if $(a,b)$ is centered, jointly normal, with known covariance

Question

Suppose there exists a bivariate normal distribution where $a \sim \mathcal{N}(0, \sigma_a)$ and $a \sim \mathcal{N}(0, \sigma_b)$ with correlation $\rho$. What is the value of $$ E[a\cdot b|a \geq 0, b \geq 0] $$

Answer 1

Note that $a=\sigma_aX$ and $b=\sigma_b(\cos(u) X+\sin(u)Y)$ where $(X,Y)$ is i.i.d. standard normal, with $$\rho=\cos(u)\qquad 0\leqslant u\leqslant\pi$$ Thus, $(X,Y)=(R\cos(T),R\sin(T))$ where $(R,T)$ is independent with known marginals. This implies that $$ab=\sigma_a\sigma_b R^2\cos(T)\cos(T-u)$$ and that the event $A=[a\geqslant0,b\geqslant0]$ is $$A=[\cos(T)\geqslant0,\cos(T-u)\geqslant0]$$ Since $E(R^2)=E(X^2)+E(Y^2)=2$ and $2\cos(T)\cos(T-u)=\cos(2T-u)+\cos(u)$, one gets $$E(ab\mid a\geqslant0,b\geqslant0]=\sigma_a\sigma_b E(\cos(2T-u)\mid A)+\sigma_a\sigma_b \cos(u)$$ One knows that $T$ can be chosen as being uniform on any interval of length $2\pi$ so we decide that $T$ is uniform on $(-\pi,\pi]$. Then $$A=[-\tfrac\pi2\leqslant T\leqslant\tfrac\pi2,-\tfrac\pi2\leqslant T-u\leqslant \tfrac\pi2]=[u-\tfrac\pi2\leqslant T\leqslant \tfrac\pi2]$$ hence $$E(\cos(2T-u)\mid A)=\int_{u-\pi/2}^{\pi/2}\cos(2t-u)\frac{dt}{\pi-u}=\frac{\sin u}{\pi-u}$$ Finally, $$E(ab\mid a\geqslant0,b\geqslant0]=\sigma_a\sigma_b \left(\rho+\frac{\sqrt{1-\rho^2}}{\pi-\arccos(\rho)}\right)$$