Show that $E(E(Y|X)=E(Y)$

Question

$X,Y$ discrete, integrable Random variables on $(\Omega,A,P)$.

Show that $E(E(Y|X))=E(Y)$.

First of all in another task it was to show that $E(Y|X)$ is a discrete Random variable. So it is $$ E(E(Y|X))=\sum_{z}zP(E(Y|X)=z), $$ Moreover it is $$ E(Y|X)(\omega):=\sum_x E(Y|X=x)1_{X=x}(\omega) $$ and $$ E(Y|X=x):=\sum_y y P(Y=y|X=x). $$

So putting this all together I get $$ E(E(Y|X))=\sum_z zP\left(\sum_x\sum_y yP(Y=y|X=x)1_{X=x}=z\right) $$

Oh my god. What can I do now to get the result?

Answer 1

A general thing to remember is not to introduce $\omega$ unless you absolutely have to.

Just stay with $$ E(Y|X=x)=\sum_y y P(Y=y|X=x). $$ Use $$ P(Y=y|X=x)P(X=x) = {P(Y=y,X=x)}$$ Now remember that $E(Y|X)$ is a function of $X(\omega)$, so $$ E[E(Y|X)] = \sum P(X=x) E(Y|X=x) $$ and you will be fine.