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If the vector $(u,v)$ is independent of the vector $x$, then I would like to show that $$E(u|x,v)= E(u|v)$$

The only thing I can derive from the definitions is that if $(u,v)$ is independent of $x$, then $E( (u,v) | x)= E((u,v))$.

I can no longer attack this problem!

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    $\begingroup$ What level do you want your answer at? Discrete random variables? Everything with densities? The Kolmogorov formulation of a conditional expectation as a Radon-Nikodym derivative? $\endgroup$
    – Michael Greinecker
    Feb 15, 2022 at 7:17

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Well assuming the random variables are absolutely continuous, you can use densities:

$$f(u|x,v)=\frac{f(u,v,x)}{f(x,v)}=\frac{f(u,v)f(x)}{f(x,v)}=\frac{f(u|v)f(v)f(x)}{f(x,v)}=f(u|v),$$

where the second and last equalities use independence between $x$ and $(u,v)$.

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    $\begingroup$ You may proceed analogously using mass functions if they are discrete $\endgroup$
    – Golden_Ratio
    Feb 15, 2022 at 14:57

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