7
$\begingroup$

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?

$\endgroup$
7
  • 1
    $\begingroup$ $(X,Y)$ is bivariate normal? $\endgroup$ Jan 12, 2018 at 19:50
  • $\begingroup$ I would prefer to avoid that assumption if possible. $\endgroup$
    – dimitriy
    Jan 12, 2018 at 20:09
  • $\begingroup$ Is there a reason why you can't make that assumption? $\endgroup$
    – Jon
    Jan 12, 2018 at 20:16
  • 2
    $\begingroup$ How did you work out the pdf without making some assumption about the joint distribution? $\endgroup$
    – jbowman
    Jan 12, 2018 at 22:13
  • 1
    $\begingroup$ Let's assume bivariate normality if that makes things easier. $\endgroup$
    – dimitriy
    Jan 12, 2018 at 22:59

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
2
  • 3
    $\begingroup$ (+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. $\endgroup$ Jan 13, 2018 at 3:15
  • 1
    $\begingroup$ @AlecosPapadopoulos I didn't know that that's called $\Phi_2$, thanks for the info. $\endgroup$ Jan 13, 2018 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.