Prove that $EX=E(E(X|Y))$

Prove that $EX=E(E(X|Y))$

I know that I should prove it from definition of conditional distribution and conditional expected value, but I don't know how.

I have also looked at theorem("Total expectation") which should be connected with the proof.

If someone would tell me, which properties of expectation I should use to prove it, I would be really glad.

Solution 1:

\begin{eqnarray*} \mathbb{E}(\mathbb{E}(X|Y)) & = & \int_{-\infty}^{\infty} \mathbb{E}(X|Y = y) f_Y(y) dy \\ & = & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}xf_{X|Y}(x|y) dx \ f_Y(y) dy \\ & = & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}xf_{X|Y}(x|y) f_Y(y) dx dy \\ & = & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}xf_{X,Y}(x,y) dx dy \\ & = & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}xf_{X,Y}(x,y) dy dx \\ & = & \int_{-\infty}^{\infty} x\int_{-\infty}^{\infty}f_{X,Y}(x,y) dy dx \\ & = & \int_{-\infty}^{\infty} xf_{X}(x) dx \\ & = & \mathbb{E}(X) \end{eqnarray*}

Solution 2:

By definition $E(X\mid Y)$ is random variable which satisfies: $$ E(X \cdot 1_G) = E\bigl(E(X\mid Y)\cdot 1_G\bigr), \quad \text{for all} \ G \in \sigma(Y). $$ Simply take $G=\Omega$ such that $1_G=1$ and you get the desired result.