R中的多元偏态正态

Question

I'm trying to generate random numbers with a multivariate skew normal distribution using the rmsn command from the sn package in R. I would like, ideally, to be able to get three columns of numbers with a specified variances and covariances, while having one column strongly skewed. But I'm struggling to achieve both goals simultaneously.

The post at skew normal distribution was related and useful (and the source of some of the code below), but hasn't completely clarified the issue for me.

I've been trying:

a <- c(5, 0, 0) # set shape parameter
s <- diag(3) # create variance-covariance matrix
w <- sqrt(1/(1-((2*(a^2)/(1 + a^2))/pi))) # determine scale parameter to get sd of 1
xi <- w*a/sqrt(1 + a^2)*sqrt(2/pi) # determine location parameter to get mean of 0

apply(rmsn(n=1000, xi=c(xi), Omega=s, alpha=a), 2, sd)
colMeans(rmsn(n=1000, xi=c(xi), Omega=s, alpha=a))

The columns means and SDs are correct for the second and third columns (which have no skew) but not the first (which does). Can anyone clarify where my code above, or my thinking, has gone wrong? I may be misunderstanding how to use rmsn , or the output. Any assistance would be appreciated.

Answer 1

The location is not the mean (except when there is no skew). From the documentation:

Notice that the location vector 'xi' does not represent the mean vector of the distribution (which in fact may not even exist if 'df <= 1'), and similarly 'Omega' is not the covariance matrix of the distribution

And you may want to replace Omega=s with Omega=w . And this is supposed to be a variance matrix: there should be no square root.