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!


  • 2
    $\begingroup$ 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? $\endgroup$ Commented Feb 15, 2022 at 7:17

1 Answer 1


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


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

  • 1
    $\begingroup$ You may proceed analogously using mass functions if they are discrete $\endgroup$ Commented Feb 15, 2022 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.