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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?

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    $\begingroup$ $(X,Y)$ is bivariate normal? $\endgroup$ – kjetil b halvorsen Jan 12 '18 at 19:50
  • $\begingroup$ I would prefer to avoid that assumption if possible. $\endgroup$ – Dimitriy V. Masterov Jan 12 '18 at 20:09
  • $\begingroup$ Is there a reason why you can't make that assumption? $\endgroup$ – Jon Jan 12 '18 at 20:16
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    $\begingroup$ How did you work out the pdf without making some assumption about the joint distribution? $\endgroup$ – jbowman Jan 12 '18 at 22:13
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    $\begingroup$ Let's assume bivariate normality if that makes things easier. $\endgroup$ – Dimitriy V. Masterov Jan 12 '18 at 22:59
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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.

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    $\begingroup$ (+1) This integral equals $\Phi_2 (a,a;\rho)$, i.e. the bivariate cumulative distribution function, where both variables are evaluated at $a$, and having correlation $\rho$. The bivariate normal integral is available as a ready-made special function in almost all software, so there is really no need to go into quadrature ourselves. $\endgroup$ – Alecos Papadopoulos Jan 13 '18 at 3:15
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    $\begingroup$ @AlecosPapadopoulos I didn't know that that's called $\Phi_2$, thanks for the info. $\endgroup$ – Bjørn Kjos-Hanssen Jan 13 '18 at 8:03

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