Let $X_1,...,X_n$ be independent Random Variables and let $Z:=f(X_1,...,X_n)$, where f is a function $ f: \Omega^n \rightarrow \mathbb{R} $. Let $Z \in \mathcal{L}^2$ and denote $E^{(i)}[Z] := E[Z \vert X_1,...,X_{i-1},X_{i+1},...,X_n] $ and $E_{(i)}[Z] := E[Z \vert X_1,...,X_i] $.
I already proved: $E^{(i)}[E_{(i)}[Z]]=E_{(i-1)}[Z]$. with this i need to prove: \begin{equation*} E[Var^{(i)}(E_{(i)}[Z])]=E[E_{(i)}[Z]^2-E_{(i-1)}[Z]^2] \end{equation*}
My Attempt was: \begin{align*} E[Var^{(i)}(E_{(i)}[Z])]&=E[E^{(i)}[E_{(i)}[Z]^2]-E^{(i)}[E_{(i)}[Z]]^2] \\ &= E[E^{(i)}[E_{(i)}[Z]^2]-E_{(i-1)}[Z]^2] \end{align*} but now i dont know what i can do with this $E^{(i)}[E_{(i)}[Z]^2]$ Term. I tried to show with the defintion of the conditional expectation that: $E^{(i)}[E_{(i)}[Z]^2]=E_{(i)}[Z]^2$ but i failed. Can someone help?
For an integrable random variable $Y$ and a $\sigma$-algebra $\mathcal F$, $$ \mathbb E\left[\mathbb E\left[X\mid \mathcal F\right]\right]=\mathbb E\left[X\right]. $$ Apply this to $$ X= \left(E_{(i)}[Z]\right)^2;\quad \mathcal F:=\sigma\left( X_1,...,X_{i-1},X_{i+1},...,X_n\right). $$