5
$\begingroup$

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 $$P(Y\le y)=f(y,n,\rho).$$

$\endgroup$
2

2 Answers 2

11
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
4
+100
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

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.