Thanks for the help. The answer should have been obvious to me if I would
have bothered to actually look at the form of the multivariate normal pdf.
Lonnie
"Charles Metz" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> 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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