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.)
Here is how I have proceeded:
Because of the 0/1 coding for X:
E(X*Z^2) = px*E(Z^2|X=1)
E(Z^2|X=1) can be expressed in terms of conditional variance:
E(Z^2|X=1) = V(Z)|X=1 + E^2(Z)|Z=1
Given C(X,Z) = E(XZ) - E(X)E(Z),
and in this case, E(XZ) = px*E(Z)|X=1,
I solve for E(Z)|X=1 and obtain:
E(Z)|X=1 = (C(X,Z) + px*E(Z)) / px
Substitution back into my first expression gives:
E(X*Z^2) = (V(Z)|X=1 + C^2(X,Z)+2*C(X,Z)*px*E(Z)+px^2*E(Z)^2) / px
This expression verifies to be correct using real data. I haven't
gotten beyond this point. The V(Z)|X=1 term still needs to be removed
if I want this in terms of px,E(z),V(x),V(z),and C(x,z). I thought
about Herman's comment, and wondered whether knowledge of V(z), px, and
C(x,z) determine V(Z)|X=1. I played with the mixture formula:
V(Z) = px * ( V(Z)|X=1 + (E(Z)|X=1 - E(Z))^2) +
(1-px) * (V(Z)|X=0 + (E(Z)|X=0 - E(Z))^2)
but this doesn't appear to help me (still playing with it).
Is what I'm trying to do even possible? It seems all I need is
another expression in terms of V(Z), V(Z)|X=1, and V(Z)|X=0. Of
course, I have E(Z)|X=1 and E(Z)|X=0 tied to E(Z), but doubt that helps
me.
Any help will be appreciated. I stopped when my symbolic math
package said: "E(X*Z^2) = E(X*Z^2)". :)
-E
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