How to get $E(f(X)\mid X_1,\dots, X_{k-1})$?

Question

Let $X_i$ be a sequence of independent r.v.-s and consider $f(X)=f(X_1, \dots, X_n)$ for some function $f: \mathbb{R}^n\to \mathbb{R}$. Why does the following identity hold? $$ E[E(f(X)\mid X_1,\dots, X_{k-1})\mid X_1,\dots, X_k]=E(f(X)\mid X_1,\dots, X_{k-1})? $$

Answer 1

If $Y$ is independent of $Z$, then $E[Y|Z]=E[Y]$. Notice that $E[f(X)|X_1,\ldots,X_{k-1},X_{k+1},\ldots,X_n]$ is independent of $X_k$. It follows that $$E[E[f(X)|X_1,\ldots,X_{k-1},X_{k+1},\ldots,X_n]|X_1,\ldots,X_k] = E[E[f(X)|X_1,\ldots,X_{k-1},X_{k+1},\ldots,X_n]|X_1,\ldots,X_{k-1}]$$ By the tower property, the r.h.s. equals to $E[f(X)|X_1,\ldots,X_{k-1}]$, which completes the proof.