Say I have two standard normal random variables $X_1$ and $X_2$ that are jointly normal with correlation coefficient $r$.

What is the distribution function of $\max(X_1, X_2)$?


According to Nadarajah and Kotz, 2008, Exact Distribution of the Max/Min of Two Gaussian Random Variables, the PDF of $X = \max(X_1, X_2)$ appears to be

$$f(x) = 2 \cdot \phi(x) \cdot \Phi\left( \frac{1 - r}{\sqrt{1 - r^2}} x\right),$$

where $\phi$ is the PDF and $\Phi$ is the CDF of the standard normal distribution.

$\hskip2in$enter image description here

  • $\begingroup$ What does this look like if $r = 0$ (no correlation at all)? I'm having trouble visualizing it. $\endgroup$
    – Mitch
    Feb 25 '15 at 0:53
  • 3
    $\begingroup$ I added a figure visualizing the distribution. It looks like a squeezed Gaussian slightly skewed to the right. $\endgroup$
    – Lucas
    Feb 25 '15 at 9:31
  • 6
    $\begingroup$ That is precisely the skew normal distribution with $\alpha = \frac{1-r}{\sqrt{1-r^2}}$. $\endgroup$
    – corey979
    Mar 6 '20 at 23:29
  • $\begingroup$ thank you @Lucas any hint on how to integrate the PDF $f(x) = 2 \cdot \phi(x) \cdot \Phi\left( \frac{1 - r}{\sqrt{1 - r^2}} x\right)$ into the CDF? It's basically $\int f'(x)f(\alpha x) dx$, but I can't get any further $\endgroup$
    – elemolotiv
    Sep 18 at 8:21

Let $f_\rho$ be the bivariate Normal PDF for $(X,Y)$ with standard marginals and correlation $\rho$. The CDF of the maximum is, by definition,

$$\Pr(\max(X, Y)\le z) = \Pr(X\le z,\ Y\le z) = \int_{-\infty}^z\int_{-\infty}^z f_\rho(x,y)dy dx.$$

The bivariate Normal PDF is symmetric (via reflection) around the diagonal. Thus, increasing $z$ to $z+dz$ adds two strips of equivalent probability to the original semi-infinite square: the infinitesimally thick upper one is $(-\infty, z]\times (z, z+dz]$ while its reflected counterpart, the right-hand strip, is $(z, z+dz]\times (-\infty, z]$.


The probability density of the right-hand strip is the density of $X$ at $z$ times the total conditional probability that $Y$ is in the strip, $\Pr(Y\le z\,|\, X=z)$. The conditional distribution of $Y$ is always Normal, so to find this total conditional probability we only need the mean and variance. The conditional mean of $Y$ at $X$ is the regression prediction $\rho X$ and the conditional variance is the "unexplained" variance $\text{var}(Y) - \text{var}(\rho X) = 1-\rho^2$.

Now that we know the conditional mean and variance, the conditional CDF of $Y$ given $X$ can be obtained by standardizing $Y$ and applying the standard Normal CDF $\Phi$:

$$\Pr(Y \le y\,|\, X) = \Phi\left(\frac{y-\rho X}{\sqrt{1-\rho^2}}\right).$$

Evaluating this at $y=z$ and $X=z$ and multiplying by the density of $X$ at $z$ (a standard Normal pdf $\phi$) gives the probability density of the second (right-hand) strip

$$\phi(z)\Phi\left(\frac{z-\rho z}{\sqrt{1-\rho^2}}\right) = \phi(z)\Phi\left(\frac{1-\rho}{\sqrt{1-\rho^2}}z\right).$$

Doubling this accounts for the equi-probable upper strip, giving the PDF of the maximum as

$$\frac{d}{dz}\Pr(\max(X,Y)\le z) = \color{blue}{2}\color{black}{\phi(z)}\color{darkred}{\Phi\left(\frac{1-\rho}{\sqrt{1-\rho^2}}z\right)}.$$


I have colored the factors to signify their origins: $\color{blue}2$ for the two symmetrical strips; $\color{black}{\phi(z)}$ for the infinitesimal strip widths; and $\color{darkred}{\Phi\left(\cdots\right)}$ for the strip lengths. The argument of the latter, $\frac{1-\rho}{\sqrt{1-\rho^2}}z$, is just a standardized version of $Y=z$ conditional on $X=z$.

Here are plots of this density for a range of values of $\rho.$

Figure showing density functions

Two dotted black curves in this figure provide references for the extreme cases:

For extremely positive $\rho,$ the density must approximate a Standard Normal distribution, because we are (to a good approximation) computing the larger of a standard normal variable $Z$ and a copy of itself.

For extremely negative $\rho,$ the density must approximate a Standard Half-Normal distribution, because the maximum of $Z$ and $-Z$ is $|Z|.$

  • $\begingroup$ Can this be extended to more than two standard normal variables with given correlation matrix? $\endgroup$
    – A. Donda
    Aug 28 '18 at 14:36
  • 3
    $\begingroup$ @A.Donda Yes--but the expression gets more complicated. With each new dimension comes the need to integrate once more. $\endgroup$
    – whuber
    Aug 28 '18 at 14:39
  • $\begingroup$ @whuber are there any known results we could look into for the expressions in the extended cases? $\endgroup$
    – runr
    Jul 8 at 19:24
  • 1
    $\begingroup$ @runr They exist: I ran across them just a few months ago -- but I can't remember where! However, a site search turns up this useful post: stats.stackexchange.com/a/311444/919. The formula should look familiar ;-). $\endgroup$
    – whuber
    Jul 8 at 19:31
  • 1
    $\begingroup$ @elemolotiv Some methods for the definite integral are given at stats.stackexchange.com/questions/61080--perhaps they will suggest useful approaches to the indefinite integral. $\endgroup$
    – whuber
    Sep 18 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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