How to prove $Cov(E(X_1|X_2), X_1 - E(X_1|X_2)) = 0$ Question
Determine whether or not the following result is True or False:
$$Cov(E(X_1|X_2), X_1 - E(X_1|X_2)) = 0$$
Attempt
I tried at first:
$=E[((X_1|X_2)(X_1-E(X_1|X_2)) - E(X_1|X_2)E(X_1-E(X_1|X_2))]$
$E[(X_1(X_1|X_2) - (X_1|X_2)E(X_1|X_2)+E(X_1|X_2)^2-E(X_1|X_2)E(X_1)]$
$E(X_1)^2 - E(X_1|X_2)E(X_1) + E(E(X_1|X_2)^2) - E(X_1)^2$
$-E(X_1|X_2)E(X_1) + E(E(X_1|X_2)^2)$
From here I am lost.
Further Comments
I know $E(E(X_1|X_2)^2) = E(X_1)^2$ But what about $E(X_1|X_2)E(X_1)$?
Is there way to simplify it?
 A: There is a much simpler way to consider this. We make use of the following formulae:
($1$) $Cov[X,Y] = E[XY]-E[X]E[Y]$ and ($2$) the Tower Law $E[X]=E[E[X|Y]]$
We now apply this to your problem and see what happens.
Step 1
We have the desired quantity $Cov ($$E(X_1|X_2),$ $X_1 - E(X_1|X_2)$$)$ and we first expand this using formula $(1)$ from above:
$= E$$[X_1E[X_1|X_2]-E[X_1|X_2]^2]$ - $E[E[X_1|X_2]]E[X_1-E[X_1|X_2]]$
Step 2
At the moment this looks very untidy, but this will simplify very easy when we start applying the tower property. Let's break up the above expression into:

*

*a) $E$$[X_1E[X_1|X_2]-E[X_1|X_2]^2]$
and

*

*b) $E[E[X_1|X_2]]E[X_1-E[X_1|X_2]]$
We can then look at them both separately and take the difference between the two terms in the final step.
Step 3
Let us first consider result a) and see if we can simplify this. If we apply the tower property here, we reduce this down to the much simplify result: $E(X_1)^2-E(X_1)^2$ which, of course, is equal to $0$.
Step 4
Now let us consider b). This simplifies down $E[X_1]$ ($E[X_1]-E[X_1])$ which is also equal to $0$.
Step 5
Therefore, by substituting these two values back into our original expression (found in step 1). We get the result: $Cov (E(X_1|X_2),$ $X_1 - E(X_1|X_2)$$)$ $=0-0=0$. This is exactly what we need.
Therefore, the statement is True.
A: Let $Y=E(X_1|X_2)$. Then: $E(Y)=E[E(X_1|X_2)]=E(X_1)$. And so:
$$
E(X_1-Y)=E(X_1)-E(Y)=0\implies \text{Cov}(Y,X_1-Y)=E[Y(X_1-Y)]
$$
To finish:
$$
E[Y(X_1-Y)]=E[E[Y(X_1-Y)|X_2]]=E[Y^2-YE(X_1|X_2)]=E[Y^2-Y^2]=0.
$$
