The maximum of $n$ normal i.i.d. random variables $$Y=\max\{x_1,...,x_n\},$$ $$x_i \sim N[0,1]$$

has the CDF $$P(Y\le y)=\Phi(y)^n $$

but how does the CDF look like, if the variables are identically correlated? $$\text{corr}(x_i, x_j)=\rho,$$ $$0\le\rho\le1.$$

I am looking for the analytical formula $$P(Y\le y)=f(y,n,\rho).$$


2 Answers 2


If $X_i$ are exchangeable multivariate Normal variables you can write $X_i= Y_i +Z$ where $Y_i$ are independent Normal$(0,\sigma^2)$ and $Z$ is Normal$(0,\tau^2)$, independent of the $Y_i$ (and $\rho^2=\tau^2/(\tau^2+\sigma^2)$)

So, $\max X_i=\max(Y_i+Z)= Z+ \max Y_i$ and the distribution of the maximum is the convolution of the distribution of $Z$ and the distribution of $n$ independent Normals.

That is, the exponent doesn't change. The distribution is the sum of a part with exponent $n$ (that vanishes at $\rho=1$) and a part with exponent 1 (that vanishes at $\rho=0$)

If $f_z$ is the density of $z$ and $f_y$ is the density of of $\max Y_i$ (which you already know), the density $f_x$ of $\max X_i$ is $$f_x(x)=\int_{-\infty}^\infty f_z(z)f_y(x-z)\,dz.$$ It probably doesn't have a simple closed form.

  • $\begingroup$ thank you @thomas-lumley, I am probably not enough sophisticated, but I struggle to derive from your reasoning a formula like $𝑃(𝑌≤𝑦)=f(y,n,\rho)$. Could you kindly elaborate further? $\endgroup$
    – elemolotiv
    Sep 17, 2021 at 14:37

This answer is essentially the same method as the answer by Thomas Lumley, but I'll give you a bit more detail on the derivations. Suppose we take independent values:

$$\tilde{X}_i \sim \text{N}(0, 1-\rho^2) \quad \quad \quad \quad \quad Z \sim \text{N}(0, \rho^2).$$

Setting $X_i = \tilde{X}_i+Z$ gives a random vector containing equicorrelated values with unit variance:

$$\mathbf{X} \sim \text{N}(\mathbf{0}, \mathbf{\Sigma}) \quad \quad \quad \quad \quad \mathbf{\Sigma} = \begin{bmatrix} 1 & \rho & \cdots & \rho \\ \rho & 1 & \cdots & \rho \\ \vdots & \vdots & \ddots & \vdots \\ \rho & \rho & \cdots & 1 \\ \end{bmatrix}.$$

Let $\phi$ and $\Phi$ denote the PDF and CDF of the standard normal distribution. We can then write the CDF of the random variable of interest as:

$$\begin{align} F_Y(y) &\equiv \mathbb{P}(Y \leqslant y) \\[16pt] &= \mathbb{P}(\max (X_1,...,X_n) \leqslant y) \\[16pt] &= \mathbb{P}(Z + \max (\tilde{X}_1,...,\tilde{X}_n) \leqslant y) \\[12pt] &= \int \limits_{-\infty}^\infty \mathbb{P}(Z + \max (\tilde{X}_1,...,\tilde{X}_n) \leqslant y |Z=z) \cdot \text{N}(z|0,\rho^2) \ dz \\[6pt] &= \int \limits_{-\infty}^\infty \mathbb{P}(\max (\tilde{X}_1,...,\tilde{X}_n) \leqslant y-z) \cdot \text{N}(z|0,\rho^2) \ dz \\[6pt] &= \int \limits_{-\infty}^\infty \mathbb{P}\Bigg( \frac{\max (\tilde{X}_1,...,\tilde{X}_n)}{\sqrt{1-\rho^2}} \leqslant \frac{y-z}{\sqrt{1-\rho^2}} \Bigg) \cdot \text{N}(z|0,\rho^2) \ dz \\[6pt] &= \int \limits_{-\infty}^\infty \mathbb{P} \Bigg( \frac{\tilde{X}_1}{\sqrt{1-\rho^2}} \leqslant \frac{y-z}{\sqrt{1-\rho^2}}, ..., \frac{\tilde{X}_n}{\sqrt{1-\rho^2}} \leqslant \frac{y-z}{\sqrt{1-\rho^2}} \Bigg) \cdot \text{N}(z|0,\rho^2) \ dz \\[6pt] &= \int \limits_{-\infty}^\infty \Phi \bigg( \frac{y-z}{\sqrt{1-\rho^2}} \bigg)^n \cdot \text{N}(z|0,\rho^2) \ dz \\[6pt] &= \frac{1}{|\rho|} \int \limits_{-\infty}^\infty \Phi \bigg( \frac{y-z}{\sqrt{1-\rho^2}} \bigg)^n \cdot \phi \bigg( \frac{z}{\rho} \bigg) \ dz. \\[6pt] \end{align}$$

The corresponding density is:

$$\begin{align} f_Y(y) &= \frac{dF_Y}{dy}(y) \\[16pt] &= \frac{1}{|\rho|} \frac{d}{dy} \int \limits_{-\infty}^\infty \Phi \bigg( \frac{y-z}{\sqrt{1-\rho^2}} \bigg)^n \cdot \phi \bigg( \frac{z}{\rho} \bigg) \ dz \\[6pt] &= \frac{1}{|\rho|} \int \limits_{-\infty}^\infty \frac{\partial}{\partial y} \Phi \bigg( \frac{y-z}{\sqrt{1-\rho^2}} \bigg)^n \cdot \phi \bigg( \frac{z}{\rho} \bigg) \ dz \\[6pt] &= \frac{n}{|\rho| \sqrt{1-\rho^2}} \int \limits_{-\infty}^\infty \Phi \bigg( \frac{y-z}{\sqrt{1-\rho^2}} \bigg)^{n-1} \cdot \phi \bigg( \frac{y-z}{\sqrt{1-\rho^2}} \bigg) \cdot \phi \bigg( \frac{z}{\rho} \bigg) \ dz. \\[6pt] \end{align}$$

This function has no closed form expression (except in the trivial case where $n=1$) but it can be evaluated using numerical methods.

  • $\begingroup$ thank you @Ben ! Just a clarification about your notation in the workings of the CDF. What do you mean by $N(z|0,\rho^2)$ inside the integral, is that the normal pdf $\phi(z,0,\sigma^2)$? - Where does that $\frac{1}{|\rho|} $ come from? $\endgroup$
    – elemolotiv
    Sep 22, 2021 at 18:25
  • 1
    $\begingroup$ I have used $\phi$ to refer to the density for the standard normal distribution, so $\phi(x) = \text{N}(x|0,1)$. The $1/|\rho|$ term comes from converting the density to the standard normal (see the density of the normal distribution here). $\endgroup$
    – Ben
    Sep 22, 2021 at 19:36

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