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
Henry wrote:
> On Fri, 13 Apr 2001 15:50:55 -0500, Charles Metz <[EMAIL PROTECTED]>
> wrote:
> > 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 fu
On Fri, 13 Apr 2001 15:50:55 -0500, Charles Metz <[EMAIL PROTECTED]>
wrote:
> 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
On Thu, 12 Apr 2001 16:11:48 -0500, "Lonnie Hamm"
<[EMAIL PROTECTED]> 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
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).
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 or can I just generate a univariate normal for each of t
