# 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.

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?

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: $$E[Y(X_1-Y)]=E[E[Y(X_1-Y)|X_2]]=E[YE(X_1|X_2)-Y^2]=E[Y^2-Y^2]=0.$$

There is a much simpler way to consider this. We make use of the following formulae:

$$(1) \space Cov[X,Y] = E[XY]-E[X]E[Y]$$ $$\text{and (2) the Tower Law } E[X]=E[E[X|Y]]$$

We now apply this to your problem and see what happens.

Step 1

We take 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:

$$(a) \quad E[X_1E[X_1|X_2]-E[X_1|X_2]^2]$$

and

$$(b) \quad E[E[X_1|X_2]]E[X_1-E[X_1|X_2]]$$

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.

• Oh my god I made such a stupid mistake that putting $E(X_1,X_2)$ as just $(X_1,X_2$.. Thank you!! Commented Apr 21, 2022 at 11:09
• No problem. Glad I could help @RyanBertrand Commented Apr 21, 2022 at 11:11