Can we always write a random variable as conditional expectation plus error?

Question

Consider the random variables $Y,X$. I believe that we can always write $$ Y=E(Y|X)+\epsilon $$ with $E(\epsilon|X)=0$.

Question: Is the above true regardless whether $Y$ is a discrete or continuous random variable?

---

My thoughts: I believe that the answer is yes.

For example, let $Z$ be a discrete random variable and $Y\equiv \mathbb{1}\{Z=3\}$, where $\mathbb{1}\{\cdots\}$ is the indicator function taking value $1$ if the condition inside is satisfied and zero otherwise. We can write $$ \mathbb{1}\{Z=3\}= E(\mathbb{1}\{Z=3\}| X)+ \epsilon \quad E(\epsilon|X)=0 $$ that is $$ \mathbb{1}\{Z=3\}= Pr(Z=3|X)+ \epsilon \quad E(\epsilon|X)=0 $$

Answer 1

Yes this always possible, as long as the expectations exist:

$$E(Y|X) = E(Y|X).$$ $$E(Y|X) + Y= E(Y|X) + Y.$$ $$Y= E(Y|X) + Y - E(Y|X).$$ $$Y= E(Y|X) + \epsilon.$$

We then have $E(\epsilon) = E(Y - E(Y|X)) = E(Y) - E(E(Y|X)) = E(Y) - E(Y) = 0$ by linearity of expectations and the law of iterated expectations.

Note that in general and especially when $Y$ is discrete, $E(Y|X)$ will be non-linear in $X$.