Enfilade,
Are you making any assumptions about the joint distribution of X and Z?
If they are independent, or even uncorrelated, E(dx*dz^2)= Edx*Edz^2 = 0.
I assume you are not making that assumption or it would be too easy :).
On the other hand, let Z be standard normal and c such that
Pr(abs(Z)>c)=1/2.
Let X = 1 if abs(Z)>c, 0 otherwise . Then
E(dx*dz^2)= 1/2*(1-1/2) E(Z|Z^2>c) + 1/2*(0-1/2) E(Z|Z^2<=c) which is
positive
even with px=.5.
To get a nice expression, it may be necessary to make some additonal
assumptions or settle for a linear approximation.
Regards,
Ellen Hertz
Enfilade 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.)
>
> 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
>
> Sent via Deja.com http://www.deja.com/
> Before you buy.
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