[R] Warning - Naive Question Alert

[Prof Brian D Ripley]

On Sun, 27 Aug 2000, Bill Venables wrote:

> Extending this to the multivariate case should not be all that difficult.
> One way you could do it is generate any old multivariate normal sample of
> the right dimension, "standardize" it by subtracting the mean and
> multiplying by the inverse of a symmetric square root of the variance
> matrix, then multiply by the same sort of square root of the given summary
> statistic variance matrix and add back on the given sample mean.  [Warning:
> I am doing this at the keyboard, ideally I would like a bit more time to
> think about it...but I don't have that sort of time right now.  Caveat
> emptor!]

Here's the code I promised, simplified:

mvrnorm <- function (n = 1, mu, Sigma, tol = 1e-06, empirical=FALSE)
{
    p <- length(mu)
    if (!all(dim(Sigma) == c(p, p))) stop("incompatible arguments")
    eS <- eigen(Sigma, sym = TRUE)
    ev <- eS$values
    if (!all(ev >= -tol * abs(ev[1]))) stop("Sigma is not positive definite")
    X <- matrix(rnorm(p * n), n)
    if(empirical) {
        X <- scale(X, T, F) # remove means
        X <- X %*% svd(X, nu = 0)$v # rotate to PCs
        X <- scale(X, F, T) # rescale PCs to unit variance 
    }
    X <- mu + eS$vectors %*% diag(sqrt(pmax(ev, 0))) %*% t(X)
    nm <- names(mu)
    if (is.null(nm) && !is.null(dn <- dimnames(Sigma))) nm <- dn[[1]]
    dimnames(X) <- list(nm, NULL)
    if (n == 1) drop(X) else t(X)
}

With empirical=TRUE you get the exact mu and Sigma in the sample.

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272860 (secr)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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