Conditional expectations question

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The discussion centers on the existence of a random variable V, jointly distributed with X, such that W equals the conditional expectation E[X|V]. The participants explore the relationship between the densities of the random variables involved, specifically focusing on the equations q(x) = ∫ f(x,y) dy and p(x) = ∫ x f(x,y) dy. The conclusion drawn indicates that a necessary condition for the existence of V is that q(x) must equal p(x)/x. This highlights a critical dependency in the joint distribution of the random variables.

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Obie
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I have a simple-seeming question on conditional expectation. It seems simple, but it has eluded my attempt to answer.

Suppose that two jointly distributed random variables (X,Y) exist with support on positive real line.

From these, it is possible to construct a new random variable, Z=E[X|Y]. Of course, by Bayes rule, E[Z]=E[E[X|Y]]=E[X]. Support of Z is a subset of positive real line.

Suppose now that you are given some third random variable W, and you know that E[W]=E[X]. W hassupport on positive part of real line. Does there exist a random variable V, jointly distributed with X for which W=E[X|V].

Any help is appreciated...
 
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Obie said:
. Does there exist a random variable V, jointly distributed with X for which W=E[X|V].

That is an interesting question. My attempt to translate it to calculus is:

Give the random variable W with density p(x) and the random variable X with density q(x), find f(x,y) satisfying

[tex]q(x) = \int f(x,y) dy[/tex]
[tex]p(x) = \int x f(x,y) dy[/tex]

The last equation implies [tex]\frac {p(x)}{x} = \int f(x,y) dy[/tex]

So are we out of luck unless [itex]q(x) = \frac{p(x)}{x}[/itex] ?
 

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