Does $X-\operatorname{E}(X \mid Z)$ vary more or less then $X$? 
Does $X-\operatorname{E}(X \mid Z)$ vary more or less then $X$?

I've tried the following:
$$\begin{align}
\operatorname{Var}(\color{green}{X-\operatorname{E}(X \mid Z)})
&= \operatorname{E}\left[\operatorname{Var}(\color{green}{X-\operatorname{E}(X \mid Z)} \mid Z)\right] + \operatorname{Var}\left[\operatorname{E}(\color{green}{X-\operatorname{E}(X\mid Z)}\mid Z)\right]\\
&= \operatorname{E}\left[\operatorname{Var}(X\mid Z)\right] + \operatorname{Var}\left[\operatorname{E}(X\mid Z)-\operatorname{E}\left\{\operatorname{E}(X\mid Z)\mid Z\right\}\right]\\
&= \operatorname{E}\left[\operatorname{Var}(X \mid Z)\right]\end{align}$$
I don't see how I can go on from here.
 A: $\newcommand{\Var}{\operatorname{Var}}
\newcommand{\Expect}{\operatorname{E}}$
The law of total variance tells us that, if $X$ and $Y$ are r.v.s on the same space, and $X$ has finite variance, then
$$\Var(X) = \Expect\Var[X \mid Z]) + \Var(\Expect[X \mid Z]).$$
(We can assume that $X$ has finite variance, because if it doesn’t, then the answer is trivial.)
Proof is in the Wikipedia entry, so I won't repeat it here.
Dropping that into the equation above, we have
$$
\begin{align*}
\Var(X - \Expect[X \mid Z]) = \cdots
&= \Expect(\Var[X \mid Z]) \\
&= \Var(X) - \Var(\Expect[X \mid Z])
\end{align*}
$$
Since $\Expect[X \mid Z]$ is a random variable in its own right, it has non-negative variance. It follows that
$$\Var(X - \Var[X \mid Z]) \leq \Var(X).$$
Further, we can deduce an equality condition: equality if and only if $\Var(\Expect[X \mid Z]) = 0$, which occurs if $X$ is a function of $Z$.
A: By Cauchy-Schwarz inequality, $$E(E(X\mid Z)^2)\geqslant E(E(X\mid Z))^2=E(X)^2,$$ hence $$E(X^2)-E(E(X\mid Z)^2)\leqslant E(X^2)-E(X)^2,$$ that is, $$\mathrm{Var}(X-E(X\mid Z))\leqslant \mathrm{Var}(X).$$
