# A question about conditional expectation involving independence

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!

Help

We have the following definitions: \begin{align} E(u|x,v) &= \int u \: p(u|x,v) \: du\\ E(u|v) &= \int u \: p(u|v) \: du. \end{align} I.e., if we could show that for $$p(u|x,v) = p(u|v)$$, we were done. Now we have: \begin{align} p(u|xv) &= \frac{p(u,x|v)}{p(x|v)}\\ &= \frac{p(u|v)p(x|v)}{p(x|v)}\\ &= p(u|v). \end{align} The first equality is the definition of conditional probability, the second equality is because of independence. And we are done.
Maybe the second equality needs some clarification: We want to show that under the presumption of independence between $$(u,v)$$ and $$x$$, i.e. $$p(u,v,x) = p(u,v)p(x)$$, it follows that $$p(u,x|v) = p(u|v)p(x|v)$$. This can be seen as follows: \begin{align} p(u,x|v) &= \frac{p(u,v,x)}{p(v)}\\ &= \frac{p(u,v)p(x)}{p(v)}\\ &= p(u|v)\frac{p(x)p(v)}{p(v)}\\ &= p(u|v)\frac{p(x,v)}{p(v)}\\ &= p(U|v)p(x|v), \end{align} and we are done.