sorry, my mistake...
since I build up a correlation matrix, I forgot the fact that the diagonal
should be one....
NR=2
CORRELATION=matrix(c(0.4,-0.25,
                     -0.25,0.3),NR,NR)
REGION=sample(1:NR,size=n,replace=TRUE)
SIGMA=CORRELATION[REGION,REGION]
diag(SIGMA)=1
> min(eigen(SIGMA)$values)
[1] 0.6


2011/5/6 Arthur Charpentier <arthur.charpentier@gmail.com>

> thanks for the tip
> actually, I know that the covariance matrix has rank 2, but it should still
> be definite positive (not strictly positive, but positive)
> my problem is that Cholesky needs a positive matrix...
> my concern is that I have
>
> > min(eigen(SIGMA)$values)
> [1] -2.109071e-17
> while theoretically it should be 0 (if it was [1] 2.109071e-17 I guess it
> would work, the problem is the minus sign)
> Arthur
>
> 2011/5/6 Petr Savicky <savicky@cs.cas.cz>
>
>> On Thu, May 05, 2011 at 02:31:59PM -0400, Arthur Charpentier wrote:
>> > I do have some trouble with matrices. I want to build up a covariance
>> matrix
>> > with a hierarchical structure). For instance, in dimension n=10, I have
>> two
>> > subgroups (called REGION).
>> >
>> > NR=2; n=10
>> > CORRELATION=matrix(c(0.4,-0.25,
>> >                      -0.25,0.3),NR,NR)
>> > REGION=sample(1:NR,size=n,replace=TRUE)
>> > R1=REGION%*%t(rep(1,n))
>> > R2=rep(1,n)%*%t(REGION)
>> > SIGMA=matrix(NA,n,n)
>> >
>> > for(i in 1:NR){
>> > for(j in 1:NR){
>> > SIGMA[(R1==i)&(R2==j)]=CORRELATION[i,j]
>> > }}
>> >
>> > If I run quickly some simulations, I build up the following matrix
>> >
>> > > CORRELATION
>> >       [,1]  [,2]
>> > [1,]  0.40 -0.25
>> > [2,] -0.25  0.30
>> > > REGION
>> >  [1] 2 2 1 1 2 1 2 1 1 2
>> > > SIGMA
>> >        [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8]  [,9] [,10]
>> >  [1,]  0.30  0.30 -0.25 -0.25  0.30 -0.25  0.30 -0.25 -0.25  0.30
>> >  [2,]  0.30  0.30 -0.25 -0.25  0.30 -0.25  0.30 -0.25 -0.25  0.30
>> >  [3,] -0.25 -0.25  0.40  0.40 -0.25  0.40 -0.25  0.40  0.40 -0.25
>> >  [4,] -0.25 -0.25  0.40  0.40 -0.25  0.40 -0.25  0.40  0.40 -0.25
>> >  [5,]  0.30  0.30 -0.25 -0.25  0.30 -0.25  0.30 -0.25 -0.25  0.30
>> >  [6,] -0.25 -0.25  0.40  0.40 -0.25  0.40 -0.25  0.40  0.40 -0.25
>> >  [7,]  0.30  0.30 -0.25 -0.25  0.30 -0.25  0.30 -0.25 -0.25  0.30
>> >  [8,] -0.25 -0.25  0.40  0.40 -0.25  0.40 -0.25  0.40  0.40 -0.25
>> >  [9,] -0.25 -0.25  0.40  0.40 -0.25  0.40 -0.25  0.40  0.40 -0.25
>> > [10,]  0.30  0.30 -0.25 -0.25  0.30 -0.25  0.30 -0.25 -0.25  0.30
>>
>> Hi.
>>
>> If X is a random vector from the 2 dimensional normal distribution
>> with the covariance matrix
>>
>>        [,1]  [,2]
>>  [1,]  0.40 -0.25
>>  [2,] -0.25  0.30
>>
>> then the vector X[REGION], which consists of replicated components
>> of X, has the expanded covariance matrix n times n, which you ask
>> for. Since the mean and the covariance matrix determine the distribution
>> uniquely, this is also a description of the required distribution.
>>
>> The distribution is concentrated in a 2 dimensional subspace, since
>> the covariance matrix has rank 2.
>>
>> Hope this helps.
>>
>> Petr Savicky.
>>
>> ______________________________________________
>> R-devel@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-devel
>>
>
>

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