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Let $Z$ be a random variable on a probability space. Is it true that we can always find $X,Y$ two random variables on the same space such that $E[Y|X]=Z$ almost surely ?

What about this weaker (is it really weaker ? It feels so, but I am not sure) statement: Let $\mu$ be a probability measure on $\mathbb R$. Is it true that we can always find $X,Y$ such that $E[Y|X]\sim \mu$ ?

Any solution is welcome, but extra points if you can "build" the random variables $X,Y$.

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  • $\begingroup$ Doesn't $X=Y=Z$ work? $\endgroup$
    – Karl
    May 15, 2021 at 17:42
  • $\begingroup$ I should have added that the r.v have to be different on a set of probability bigger than $0$ $\endgroup$
    – W. Volante
    May 15, 2021 at 20:52

1 Answer 1

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Let $(X, \Omega, \mu) \in \mathcal{P}$ a probability space, $\mathcal{B} \subseteq \Omega$ a sub $\sigma$-field of $\Omega$, $Y : X \rightarrow \mathbb{R}$ a $\Omega$-measurable function of finite expectation. Conditional expectation naturally extends itself to the question of existance of a $\mathcal{B} $-measurable function $Z$ on $X$ such that : $$ \forall A \in \mathcal{B}, \int_A ZdP = \int_A Y dP $$ If $Y$ has a finite expectation (and we need it for the very definition of conditional expectation) such a random variable $Z$ does exists (We define $m$ a measure on $\mathcal{B}$ by $ \forall A \in \mathcal{B}, m(A) = \int_A YdP $ and apply Radon-Nykodym theorem to get a $\mathcal{B} $-measurable function $Z$ such that $ \forall A \in \mathcal{B}, m(A) = \int_A ZdP $) A good intuition on this for $\mathcal{L}^2$ functions is that expectation is just the (linear) projection from $\Omega$-measurable functions onto $\mathcal{B} $-measurable function.

Now for the second question, by definition of $Z$ you have that the answer is positive.

Now if we drop the assumption of finite expectation it becomes a bit more complex as ( R. E. Strauch. "Conditional Expectations of Random Variables Without Expectations." Ann. Math. Statist. 36 (5) 1556 - 1559, October, 1965. ) in short, you could define a new conditional expectation to be $$ E(Y^+ \mid \mathcal{B}) - E(Y^- \mid \mathcal{B}) $$ provided this difference is defined almost everywhere. The paper goes on about pathologies of such a definition.

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