In article <8a1r90$qgs$[EMAIL PROTECTED]>,
Enfilade <[EMAIL PROTECTED]> wrote:
>Hi,
> Many thanks to Herman Rubin for his comment on my earlier post. I
>may not have fully understood him, so I'm restating the problem with a
>little more detail. I also progressed a bit further, and am wondering
>if it is, in fact, as far as I can go.
> The problem is to derive an expression for E(X*Z^2). If you are
>wondering why, it is part of the formula for E(dx*dz^2), where dx=X - E
>(X), and dz = Z - E(Z). I know that if BOTH X and Z are continuous and
>distributed bivariate normal, E(dx*dz^2) = 0. I'm trying to determine
>what this odd moment will be if X is binary 0/1 with proportion (px)
>free to vary, and Z distributed univariate normal. (Note: I will be
>very pleased if someone tells me E(dx*dz^2)=0 here, too. I believe it
>is the case with px=.5, but only that case.)
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.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
[EMAIL PROTECTED] Phone: (765)494-6054 FAX: (765)494-0558
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