Conditional expectation and variable decomposition Suppose that $X$ and $Y$ have an uknown joint distribution $f_{XY}$.
How can I formally demostrate that it always exists a unique decomposition of the form :
$$ Y = E[Y|X] +\epsilon $$ 
without assuming any explicity form of $f_{XY}$?
 A: This follows from the linearity of expectation and law of total expectation
$X$ and $Y$ have an unknown joint distribution $F_{XY}$, and the distribution of $Y\mid X$  is some unknown $F_{Y|X}$. Suppose the mean of $Y|X$ is $\mu(X)$. 
Consider the random variable $\epsilon = Y - \mu(X)$. Then $\epsilon$ is a mean 0 random variable. To see this,
\begin{align*}
E(\epsilon)& = E(Y - \mu(X))\\
& = E(E(Y - \mu(X) \mid X))\\
& = E\left[ E(Y\mid X) - \mu(X) \right]\\
& = E(0)\\
& = 0\,.
\end{align*}
Thus, we can always write 
$$Y = \mu(X) + \epsilon  = E(Y\mid X) + \epsilon \,.$$

Uniqueness:
Assume that there exists a $\delta(X)$ and  $\eta$, such that
$$Y = \delta(x) + \eta. $$
Uniqueness holds under two constraints.


*

*$\eta$ is independent of $X$

*$E(\eta) = 0$


Taking expectation with respect to $F_{Y|X}$,
\begin{align*}
E(Y\mid X) & = E( \delta(X) \mid X) + E(\eta \mid X)\\
\Rightarrow \mu(X) & = \delta(X) + 0\,.
\end{align*}
Thus, we obtain that $\delta(X) = \mu(X)$, which implies $\eta = \epsilon$.
Note that the two constraints are crucial.
