Explain why $Cov(X-E(X\mid Y), Y) =0 $

Question

$$Cov(X-E(X\mid Y), Y) =0 $$

Above $Cov$ denotes Covariance.

My textbook says above identity always holds, but I can't figure why it is so.

Any hint or advice?

Also, if $Cov(X-E(X\mid Y), Y) =0 $ always holds, Does $Var(X-E(X\mid Y), Y) =0 $ always hold?

Answer 1

By definition, \begin{align} \operatorname{Cov}(X-\mathsf E(X|Y),Y)&=\mathsf E(X-\mathsf E(X|Y)-\mathsf EX+\mathsf E\mathsf E(X|Y))(Y-\mathsf EY)\\ &=\mathsf E(X-\mathsf E(X|Y))(Y-\mathsf EY)\\ &=\mathsf E(XY)-\mathsf E(Y\mathsf E(X|Y)) -\mathsf EX\cdot\mathsf EY +\mathsf EY\cdot\mathsf E\mathsf E(X|Y)\\ &=\mathsf E(XY)-\mathsf E(XY) -\mathsf EX\cdot\mathsf EY +\mathsf EY\cdot\mathsf EX\\ &=0 \end{align} Here several properties of conditional expectations are used. For example, $\mathsf E\mathsf E(X|Y)=X$ and $Y\mathsf E(X|Y)=\mathsf E(XY|Y)$.

Answer 2

Note that in the law of total expectation $$\operatorname{E}[\operatorname{E}[X \mid Y]] = \operatorname{E}[X]$$ the nested expectations on the LHS are such that the innermost expectation is with respect to $X$ for a fixed $Y$, and the outer is taken with respect to $Y$. That is to say, $\operatorname{E}[X \mid Y]$ is a random variable that is a function of $Y$. For the sake of conceptual convenience, call this function $g_X(Y) = \operatorname{E}[X \mid Y]$, and it obeys the property that $\operatorname{E}[g_X(Y)] = \operatorname{E}[X]$.

With this in mind, $$\begin{align*} \operatorname{Cov}[(X-\operatorname{E}[X \mid Y]), Y] &= \operatorname{E}[X - g_X(Y), Y] \\ &= \operatorname{E}[(X - g_X(Y))Y] - \operatorname{E}[X - g_X(Y)]\operatorname{E}[Y] \\ &= \operatorname{E}[XY - g_X(Y)Y] - (\operatorname{E}[X] - \operatorname{E}[g_X(Y)])\operatorname{E}[Y] \\ &= \operatorname{E}[XY] - \operatorname{E}[g_X(Y)Y] - (\operatorname{E}[X] - \operatorname{E}[X])\operatorname{E}[Y] \\ &= \operatorname{E}[XY] - \operatorname{E}[g_X(Y)Y]. \\ \end{align*}$$ From here, it seems that no further simplification is possible. But since $$g_X(Y)Y = Y \operatorname{E}[X \mid Y] = \operatorname{E}[XY \mid Y],$$ it follows that $$\operatorname{E}[g_X(Y) Y] = \operatorname{E}[\operatorname{E}[XY \mid Y]] = \operatorname{E}[XY],$$ where the rightmost expectation is with respect to $X$ and $Y$ jointly. Thus the covariance is zero as claimed.