Let $X, Y$ be continuous, real valued random variables, and let $f$ be a measurable function such that $f(X)$ is again a random variable.
EDITED:
How would the conditional expectation $\mathbb{E}[Y|f(X)=f(x)]=\int y d\mathbb{P}_{Y|f(X)}(f(x))$ relate to $\mathbb{E}[Y|X=x]=\int y d\mathbb{P}_{Y|X}(x)$? Which are the minimum conditions that they are equal? How can I get from the one integral representation to the other and which properties must $f$ fullfil to do so?
Best Answer
The conditional expectation $\mathbb{E}[Y|X]$ is $\sigma (X)$-measurable, so there exists a measurable function $\varphi$ such that $\mathbb{E}[Y|X]=\varphi(X)$. Likewise, there exists a measurable function $\psi$ such that $\mathbb{E}[Y|f(X)]=\psi(f(X))$.
The sigma algebra $\sigma(f(X))$ generated by $f(X)$ is contained in the sigma algebra $\sigma(X)$ generated by $X$ (because $f(X)^{-1}(C)=X^{-1}(f^{-1}(C))\in\sigma(X)$ for every Borel set $C$). Hence, by the tower property for conditional expectations, $$\mathbb{E}[\varphi(X)|f(X)]=\psi(f(X))$$ This means that $$\mathbb{E}[Y|f(X)=f(x)]=\mathbb{E}[\mathbb{E}[Y|X]|f(X)=f(x)]$$
So, if $f$ is injective then $\mathbb{E}[Y|f(X)=f(x)]=\mathbb{E}[Y|X=x]$. In general, $\mathbb{E}[Y|f(X)=f(x)]$ is an integral of $\mathbb{E}[Y|X]$ over all $x'$ such that $f(x')=f(x)$.