# Distribution of the maximum of two correlated normal variables

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 $\max(X_1, X_2)$?

• Jan 6, 2018 at 20:16

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.

$\hskip2in$ • What does this look like if $r = 0$ (no correlation at all)? I'm having trouble visualizing it. Feb 25, 2015 at 0:53
• I added a figure visualizing the distribution. It looks like a squeezed Gaussian slightly skewed to the right. Feb 25, 2015 at 9:31
• That is precisely the skew normal distribution with $\alpha = \frac{1-r}{\sqrt{1-r^2}}$. Mar 6, 2020 at 23:29
• 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 Sep 18, 2021 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

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

### Recapitulation

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 $$\rho.$$ 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 $$|Z|.$$

### Edit 2022-06-22

The same method described here leads to the density of the maximum of any Binormal variable $$(X,Y):$$ you add the contributions from the vertical and horizontal strips; the tiny square where they overlap is negligible. On each strip there is a width factor given by a Normal density and a height factor given by a Normal distribution function. See https://stats.stackexchange.com/a/579392/919 for a statement of the result.

• Can this be extended to more than two standard normal variables with given correlation matrix? Aug 28, 2018 at 14:36
• @A.Donda Yes--but the expression gets more complicated. With each new dimension comes the need to integrate once more.
– whuber
Aug 28, 2018 at 14:39
• @whuber are there any known results we could look into for the expressions in the extended cases?
– runr
Jul 8, 2021 at 19:24
• @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 ;-).
– whuber
Jul 8, 2021 at 19:31
• @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.
– whuber
Sep 18, 2021 at 14:17