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!



1 Answer 1


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.


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