Lonnie Hamm wrote:
> Suppose I want to generate a vector of n normal deviates
> with the standard deviation the same for all elements and
> the covariances are zero. Since the covariances are zero,
> is there such a thing as a multivariate normal deviate
> in this case
Yes (see below).
> or can I just generate a univariate normal for
> each of the n components?
Yes, after which the set of n outcomes will follow the special case of
the n-variate normal distribution in which all n(n-1)/2 covariance
(i.e., correlation) parameters are equal to zero. This follows directly
from the fact that uncorrelated *normal* random variables are
independent (which can be proven by examining the form of the general
multivariate normal density function when its covariance matrix is diagonal).
Charles Metz
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