Determine whether or not the following result is True or False:

$$Cov(E(X_1|X_2), X_1 - E(X_1|X_2)) = 0$$


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?


2 Answers 2


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:

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


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

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.

  • 1
    $\begingroup$ Oh my god I made such a stupid mistake that putting $E(X_1,X_2)$ as just $(X_1,X_2$.. Thank you!! $\endgroup$ Commented Apr 21, 2022 at 11:09
  • $\begingroup$ No problem. Glad I could help @RyanBertrand $\endgroup$
    – FD_bfa
    Commented Apr 21, 2022 at 11:11

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