Hello Petr and thanks for your help! Thanks also for the correction on the
code, of cause it is better to use the real mean and covariance than those
estimated by mean() and cov(). What I am after is that if I have the two
two-dimensional probability density functions of the distribution of my
parameters, these functions should have an intersect. And if I know the
intersect I should be able to say that a certain "volume" under both functions
is shared by both comparable to the surface shared by two one-dimensional
normal distributions. From there on I was hoping to be able to say something
about how high the probability is that my samples come from two different
populations (in the logic of a t-test).
Would this be possible or reasonable at all?
Thank again, I really appreciate your help!
best
Fabian
Would it be possible, too to
On 26 Apr 2012, at 10:56, Petr Savicky wrote:
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.
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Fabian Roger, Ph.D. student
Dept of Biological and Environmental Sciences
University of Gothenburg
Box 461
SE-405 30 Göteborg
Sweden
Tel. +46 31 786 2933
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