[R] comparison of bivariate normal distributions

[Petr Savicky]

On Wed, Apr 25, 2012 at 08:43:34PM +0000, Fabian Roger wrote:
> sorry for cross-posting
> 
> Dear all,
> 
> I have tow (several) bivariate distributions with a known mean and variance-covariance structure (hence a known density function) that I would like to compare in order to get an intersect that tells me something about "how different" these distributions are (as t-statistics for univariate distributions).
> 
> In order to visualize what I mean hear a little code example:
> 
> ########################################
> library(mvtnorm)
> 
> c<-data.frame(rnorm(1000,5,sd=1),rnorm(1000,6,sd=1))
> c2<-data.frame(rnorm(1000,10,sd=2),rnorm(1000,7,sd=1))
> 
> xx=seq(0,20,0.1)
> yy=seq(0,20,0.1)
> xmult=cbind(rep(yy,201),rep(xx,each=201))
> dens=dmvnorm(xmult,mean(c),cov(c))
> dmat=matrix(dens,ncol=length(yy),nrow=length(xx),byrow=F)
> 
> dens2=dmvnorm(xmult,mean(c2),cov(c2))
> dmat2=matrix(dens2,ncol=length(yy),nrow=length(xx),byrow=F)
> contour(xx,yy,dmat,lwd=2)
> contour(xx,yy,dmat2,lwd=2,add=T)
> ##############################################
> 
> Is their an easy way to do this (maybe with dmvnorm()?) and could I interpret the intersect ("shared volume") in the sense of a t-statistic?

Hello:

I am not sure, what is exactly the question. The parameters
of a bivariate normal distribution are the covariance matrix and
the mean vector. For the distributions above, these are

  mean1 <- c(5, 6)
  cov1 <- diag(c(1, 1))

  mean2 <- c(10, 7)
  cov2 <- diag(c(2, 1)^2)

These parameters may be used in the code above instead of mean(c), cov(c)
and mean(c2), cov(c2).

The curves of equal density are ellipses, whose equations may be derived
from the mean vector mu and covariance matrix Sigma using the formula for
the exponent in the bivariate density of normal distribution. For any
fixed value of the density, the formula has the form

  (x - mu)' Sigma^(-1) (x - mu) = const

where (x - mu)' is the transpose of (x - mu), (x - mu) is a column
vector and const is some constant. The value of const may be derived
using the full formula for the bivariate density, which is at

  http://en.wikipedia.org/wiki/Multivariate_normal_distribution

In order to compute the area of this ellipse, we have to specify a required
density, more exactly a lower bound on the density. The area of an ellipse
is \pi a b, where a, b, are its axes. If we have two such ellipses, it is
possible to compute the area of their intersection, but again, for each
ellipse, a lower bound on the density is needed.

Is the area of the intersection of two ellipses for some specified lower
bounds on the density, what you want to compute?

Petr Savicky.