0
$\begingroup$

Let $x_{1},x_{2},\cdots,x_{n}$ and $y_{1},y_{2},\cdots,y_{n}$ be correlated normal random variables the covariance between two arbitrary random variables is known. In other words, let $X=\left[x_{1},\cdots,x_{n}\right]^{T}\in\mathbb{R}^{n}$ and $Y=\left[y_{1},\cdots,y_{n}\right]^{T}\in\mathbb{R}^{n}$, we have \begin{equation} \left[\begin{array}{c} X\\ Y \end{array}\right]\sim\mathcal{N}\left(\left[\begin{array}{c} \mu_{X}\\ \mu_{Y} \end{array}\right],\left[\begin{array}{cc} \Sigma_{X} & \Sigma_{XY}\\ \Sigma_{YX} & \Sigma X \end{array}\right]\right) \end{equation}

where the mean vectors $\mu_{X}$ and $\mu_{Y}$ and the covariance matrices $\Sigma_{X}$, $\Sigma_{Y}$, $\Sigma_{XY}=\Sigma_{YX}$ are known. The problem is to find the distribution of a random variable $z$ defined by \begin{equation} z=\max_{1\leq k\leq n}x_{k}+\max_{1\leq k\leq n}y_{k} \end{equation} Could you please give me any suggestion?

$\endgroup$

2 Answers 2

2
$\begingroup$

Even the distribution of the maximum of two correlated normal random variables is a rather complicated thing. Do you really think this has a closed form?

$\endgroup$
1
  • $\begingroup$ I need to find a method for calculate the c.d.f of the variable $z$ numerically. For example, for the maximum of a normal random sequence, let $z=\max_{1\leq k\leq n}x_{k}$, then \begin{equation}\mathbb{P}\left(z<a\right) = \mathbb{P}\left(\max_{1\leq k\leq n}x_{k}<a\right)=\mathbb{P}\left(\bigcap_{k=1}^{n}\left\{ x_{k}<a\right\} \right)\end{equation} Then, the c.d.f. $\mathbb{P}\left(z<a\right)$ can be calculated if we know the mean vector $\mu_{X}$ and the covariance matrix $\Sigma_{X}$ by using the mvncdf function in Matlab. Correct me if i was wrong? $\endgroup$ Jun 19, 2014 at 7:20
0
$\begingroup$

We have:

$F(z)=P(X+Y\leq z)=\int_{-\infty}^{\infty}\int_{-\infty}^{z-x}f_{x,y}(x,y)\,dx\,dy$

(see CDF of sum of dependent random variables)

The marginal distributions:

$F(x)=P(\max x_k\leq x)=P(x_1\leq x,...,x_n \leq x)=\Phi(x)\cdots\Phi(x)$

(same for y-max, also see http://ocw.mit.edu/courses/civil-and-environmental-engineering/1-151-probability-and-statistics-in-engineering-spring-2005/lecture-notes/app11_max.pdf).

The joint distribution $f(x,y)$ can be expressed using the Copula concept:

$f(x,y)=f(x)f(y)\,c^N(F(x),F(y))$

(see slide 10 http://www-1.ms.ut.ee/tartu07/presentations/liebscher.pdf)

The Gaussian Copula density (from Normal marginals) is:

$c^N(u,v)=\frac{\phi(\Phi^{-1}(u,v))}{\phi(\Phi^{-1}(u))\phi(\Phi^{-1}(u))}$

Plugging these into $F(z)$ would give the distribution, but I also doubt it would have closed form; numerically you can of course find the Gausscopula pdf and Normal distribution in MATLAB and approximate the integral.

$\endgroup$
1
  • $\begingroup$ The marginal distributions would be parametrized by $(\mu,\Sigma)$ as above. $\endgroup$
    – emcor
    Jun 19, 2014 at 9:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .