Percentiles of $Z=\max(X,Y)$, where $X$ and $Y$ are correlated bivariate normal random variables Suppose $X \sim N(\mu_x,\sigma_x^2)$, $Y \sim N(\mu_y,\sigma_y^2)$, and $\operatorname{Corr}(X,Y)=\rho$. I am interested in calculating percentiles of $Z = \max(X,Y)$. We can assume bivariate normality.
I know how to find the pdf, mean, and variance of $Z$, but I am having trouble solving or finding an approximation for the percentiles. Has this been worked out somewhere in the literature?     
 A: You can compute this numerically. As for theoretical results, I don't have a reference to the literature, but here's a calculation of how your problem is related to the standard normal CDF $\Phi$.
The joint pdf is
$$f(x_1,x_2)=\frac1{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\exp\left[-\frac{z}{2(1-\rho^2)}\right]$$
where
$$z=\frac{(x_1-\mu_1)^2}{\sigma_1^2}-\frac{2\rho(x_1-\mu_1)(x_2-\mu_2)}{\sigma_1\sigma_2}+\frac{(x_2-\mu_2)^2}{\sigma_2^2}.$$
For simplicity I'll assume $\mu_1=\mu_2=0$, $\sigma_1=\sigma_2=1$:
$$f(x_1,x_2)=\frac1{2\pi\sqrt{1-\rho^2}}\exp\left[-\frac{z}{2(1-\rho^2)}\right],\qquad z=x_1^2 - 2\rho x_1x_2 + x_2^2.$$
Now we have, using $x^2-2\rho xy = (x-\rho y)^2-\rho^2y^2$, that
$$\Pr(\max(X,Y)\le a)=\int_{-\infty}^a\int_{-\infty}^a f(x,y)\,dx\,dy=$$
$$\frac1{2\pi\sqrt{1-\rho^2}}\int_{-\infty}^a\exp\left(-\frac{y^2}{2(1-\rho^2)}\right)\int_{-\infty}^a \exp\left(-\frac{x^2-2\rho xy}{2(1-\rho^2)}\right)\,dx\,dy$$
$$=\frac1{2\pi\sqrt{1-\rho^2}}\int_{-\infty}^a\exp\left(-\frac{y^2}{2}\right)\int_{-\infty}^a \exp\left(-\frac{(x-\rho y)^2}{2(1-\rho^2)}\right)\,dx\,dy$$
Let $W$ be normal with mean $\rho y$ and variance $1-\rho^2$. Then
$$\Pr(W\le a)=\Pr\left((W-\rho y)/\sqrt{1-\rho^2}\le (a-\rho y)/\sqrt{1-\rho^2}\right)$$
$$=\Phi\left((a-\rho y)/\sqrt{1-\rho^2}\right).$$
So we get

$$\Pr(\max(X,Y)\le a)=\frac1{\sqrt{2\pi}}\int_{-\infty}^a \exp\left(-y^2/2\right)\Phi\left(\frac{a-\rho y}{\sqrt{1-\rho^2}}\right)\,dy.$$

You can see that if $\rho=0$ then this is just $\Phi(a)^2$, as it should be.
