i have on Friday a presentation and i have one small problem. Let $ X_1,...,X_n$ be independent Random Variables. Let $f: \Omega^n \rightarrow \mathbb{R}$ and define $Z:=f(X_1,...,X_n)$. Define also: \begin{align*} E_{(i)}[Z]&:=E[Z \vert X_1,...,X_i], \\ E^{(i)}[Z]&:=E[Z \vert X_1,...,X_{i-1},X_{i+1},...,X_n]. \end{align*} Now let $T$ be $(X_1,...,X_i)$ measureable. Then following equation holds: \begin{equation*} E[Cov^{(i)}(T,Z)]=E[Z(T-E^{(i)}[T]] \end{equation*} I dont know why this should be true.
I wanted to show this with the defintion: \begin{align*} E[Cov^{(i)}(T,Z)]=E[E^{(i)}[ZT]-E^{(i)}[Z]E^{(i)}[T]] \end{align*} now when i compare this with the result, i see $E^{(i)}[ZT]=ZT$ and $ E^{(i)}[Z]=Z$. Hence $Z$ and $ZT$ must be $(X_1,...,X_{i-1},X_{i+1},...,X_n)$ measureable but i dont know why this should be true. I would really appreciate some help. Thanks alot.
Observe that $$ E[E^{(i)}[ZT]]=E[ZT] $$ hence it suffices to check that $$ E\left[ E^{(i)}[Z]E^{(i)}[T]\right]= E\left[ ZE^{(i)}[T]\right]. $$ Denoting $\mathcal F$ as the $\sigma$-algebra generated by $X_j$, $j\in\{1,\dots,n\}\setminus \{i\}$, we get that $$ E\left[ E^{(i)}[Z]E^{(i)}[T]\right]=E[E[Z\mid \mathcal F]E[T\mid \mathcal F]] $$ and since $E[Z\mid \mathcal F]$ is $\mathcal F$-measurable, we get that $$ E[ZE[T\mid \mathcal F]\mid \mathcal F] $$ and the result follows by taking the expectation.