Consider the random variables $X$ and $Y$ defined on the same probability space $(\Omega, \mathcal{F}, \mathbb{P})$ taking values respectively in $\mathbb{R}^d$ and $\mathbb{R}^p$. Let $f:\mathbb{R}^d\rightarrow \mathbb{R}^m$. Let $g:\mathbb{R}^m\rightarrow \mathbb{R}^n$.
(1) What is the relation between $E(Y|X)$ and $E(Y|f(X))$?
(2) What is the relation between $E\Big (g(f(X))|X \Big)$ and $E\Big (g(f(X))|f(X) \Big)$?
Let $x$ denotes a realisation of $X$. I am confused by the fact that $E(Y|X=x)$ may be different from $E(Y|f(X)=f(x))$ unless $f$ is injective. Similarly for $E\Big (g(f(X))|X=x \Big)$ and $E\Big (g(f(X))|f(X)=f(x) \Big)$.
On the other hand $\sigma(X)\supseteq \sigma(f(X))$, where $\sigma$ denotes the sigma-algebra symbol.
Could you help me to clarify?
(1) If $f$ is injective, then $X$ and $f(X)$ span the same $\sigma$-algebra, so $E(Y\mid X) = E(Y\mid f(X))$.
If not, you can say that $E(E(Y\mid X)\mid f(X)) = E(Y\mid f(X))$ by tower property, but practically nothing beyond that.
(2) $E(g(f(X)\mid X) = E(g(f(X)\mid f(X)) = g(f(X))$, since it is measurable with respect to $\sigma(f(X))$.