The following is Sheldon Ross's definition:

We say that the random variables $X,Y$ have a bivariate normal distribution if, for some constants $\mu_x,\mu_y,\sigma_x>0,\sigma_y>0, -1<\rho < 1$, their joint density function is given, for all $-\infty < x,y < \infty$, by $$f(x,y)=\frac{\exp\left(-\frac1{2(1-\rho^2)}\left(\left(\frac{x-\mu_x}{\sigma_x}\right)^2+\left(\frac{y-\mu_y}{\sigma_y}\right)^2-2\rho\frac{(x-\mu_x)(y-\mu_y)}{\sigma_x\sigma_y}\right)\right)}{2\pi\sigma_x\sigma_y\sqrt{1-\rho^2}}$$

Is there a combinatorial/intuitive meaning of this definition?


I don't have a combinatorial meaning, but you can think of it as follows. $(X,Y)$ is the result of applying an affine transformation to a pair $(W,Z)$ of independent standard normal random variables. Many such transformations exist, and one in particular is

$$\begin{align*} X &= \mu_x + \sigma_x W\\ Y &= \mu_y + \rho \sigma_y W + \sqrt{1-\rho^2} \sigma_y Z \end{align*}$$

See for example this set of slides. The contours of the joint density (points at equal height above the $x$-$y$ plane) are ellipses centered at $(\mu_x,\mu_y)$.

  • $\begingroup$ @Didier Thanks for inserting the missing $\sigma_y$. $\endgroup$ – Dilip Sarwate Apr 16 '12 at 14:29

There are several equivalent definitions of a random vector being multivariate normal. Every characterization of a multivariate normal distribution is of course important. However sometimes it happens that one of the characterizations has an greater appeal from an intuitive standpoint.

One such characterization is the following:

One can show that a random vector $(X,Y)$ is bivariate normal iff $(a_1,a_2)(X,Y)$ is normal for every vector $(a_1,a_2)$.

This result generalizes to higher dimensions, i.e. for random vectors of dimension $n$.

The nice thing about this definition of a multivariate normal variable is that many results become almost trivial to prove. For example it follows immediately that if X is multivariate normal, any marginal distribution of X is (multivariate) normal.

  • $\begingroup$ This also points up the fact that Ross's definition is incomplete in the sense that $(X,aX+b)$ is also called a bivariate normal vector though it does not have the bivariate joint density function given by Ross (which applies only when $|\rho| < 1$) $\endgroup$ – Dilip Sarwate Apr 16 '12 at 12:18
  • $\begingroup$ That, or a Dirac mass at (0,0). $\endgroup$ – Did Apr 16 '12 at 13:08

Another characterization of a multivariate normal is: a random vector ${\bf X}$ is multivariate normal iff its density has the form

$$ f_{\bf X}({\bf x}) = \alpha \; e^{-g\,({\bf x})}$$

with $g\,({\bf x})$ being a positive definite quadratic form (i.e., it can be written as $g\,({\bf x})={\bf x}^T {\bf A} {\bf x} + {\bf b}^T {\bf x} + c$ with ${\bf A}$ positive definite, ${\bf b}$ and $c$ arbitrary), and $\alpha$ the appropiate normalization constant. Applying this to the two dimensional case, and expressing the degrees of freedom as function of the probabilistic parameters $\mu_x$, $\mu_y$ , $\sigma_x^2$ $\sigma_y^2$ and $\rho$ you get the above formula.

  • $\begingroup$ Could you please define what you mean by the degrees of freedom here? I have not seen this phrase used in this context before. Thanks. $\endgroup$ – Dilip Sarwate Apr 16 '12 at 15:38
  • $\begingroup$ @DilipSarwate: by degrees of freedom I simply mean parameters -as scalars; eg, an arbitrary quadratic in 2D has 6 parameters (3 for the matrix A, 2 for b , 1 for c) but one of them is absorbed into the normalizacion constant. $\endgroup$ – leonbloy Apr 16 '12 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.