In article <8a2sv8$[EMAIL PROTECTED]>,
[EMAIL PROTECTED] (Herman Rubin) wrote:
>
> Not even here. To show the problem, let me assume that Z
> is normal with mean 0 and variance 1, and P(X=1) = .5 and X
> is uncorrelated with Z. This is not enough to determine
> E(X*Z^2).
>
> To see this, note that if X and Z are independent, the
> expectation is .5. For another model, assume that
> P(X=1|Z) = exp(-1.5*Z^2). This makes Z normal with
> mean 0 and variance .25 conditional on X=1, and also
> P(X=1) = .5. However, E(X*Z^2) = .125.
>
> It is the multidimensional normal which is the unusual case.
> --
Thanks. I think I undertand it now. To determine (X*Z^2), it
appears I either have to specify the variance of Z in both categories
of X (getting overall V(Z) from the variance formula for mixture
models), or assume they are equal. Leaving in two variance terms
doesn't complicate the expression too much, so I'll probably just let
the user make the homogeneity assumption if they wish.
-E
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