# Prove or disprove $E[(X-E[X|Y,Z])^2]\leq E[(X-E[X|Y])^2]$

Prove or disprove $E[(X-E[X|Y,Z])^2]\leq E[(X-E[X|Y])^2]$ if $X,Y,Z$ are real random variables with finite second moments.

I try to claim the statement is true, i,e, showing the inequality. I noticed that the equality holds if $X,Y$ are independent, and $X,Z$ are independent (both side are variance of $X$ since we know that $E[X|F]=E[X]$ if $X,F$ are independent. )

However, for the case that $X,Y$ are independent, but $X,Z$ are not independent, we know that $E[X|Y,Z]=E[X|Z],$ and $E[X|Y]=E[X]$. I can't go further from here, and I have no idea how to deal with the case $Y,Z$ are both not independent of $X$. Could anyone give me some hints to work it. Thanks!

Essentially, the statement boils down to the fact $E[Var(U \mid V)] \le Var(U)$ from the law of total variance.

You want to show $E[Var(X \mid Y,Z)] \le E[Var(X \mid Y)]$.

By the tower property, the left-hand side is $E[E[Var(X \mid Y,Z) \mid Y]]$. So, it suffices to show $E[Var(X \mid y,Z)] \le Var(X \mid y)$ for each $y$. This follows from the law of total variance.

I found another way to solve it by using the following equality: If $X\in L^2$ and $G\subset F$ are $\sigma$-fields, then $$E[(X-E[X|F])^2]+E[(E[X|F]-E[X|G])^2]=E[(X-E[X|G])^2].$$ Thus, we have $$E[(X-E[X|F])^2]\leq E[(X-E[X|G])^2],$$ Which prove our goal by consider $F=\sigma(Y,Z)$ and $G=\sigma(Y)$.

Therefore, it suffices to show that $E[(X-E[X|F])^2]+E[(E[X|F]-E[X|G])^2]=E[(X-E[X|G])^2].$

We note that

\begin{align} E[(X-E[X|G])^2]&=E[\{(X-E[X|F])+(E[X|F]-E[X|G])\}^2] \\ &=E[(X-E[X|F])^2]+E[(E[X|F]-E[X|G])^2]+2E[(X-E[X|F])(E[X|F]-E[X|G])] \\ \end{align}

Hence, we need to showing that $E[(X-E[X|F])(E[X|F]-E[X|G])]=0$. Observe that $$(X-E[X|F])E[X|F]=XE[X|F]-E[X|F]E[X|F]=XE[X|F]-E[E[X|F]X|F].$$ Thus, $E[ (X-E[X|F])E[X|F] ]=0$.

Also, we note that $E[X|G]$ is $F$-measurable since $G\subset F$, so $$(X-E[X|F])E[X|G]]=XE[X|G]-E[X|F]E[X|G]=XE[X|G]-E[E[X|G]X|F].$$ Therefore, we also have $E[(X-E[X|F])E[X|G]]=0$.