# CDF of maximum of $n$ correlated normal random variables

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

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.

• 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? 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.

• 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? Sep 22, 2021 at 18:25
• 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).
– Ben
Sep 22, 2021 at 19:36