Law of total covariance inequality

Question

The law of total variance: $$ \text{Var}(X) = \mathbb{E}(\text{Var}(X\mid Y)) + \text{Var}(\mathbb{E}(X\mid Y)).$$

There is also something called the law of total covariance: $$ \text{Cov}(X,Y) = \mathbb{E}(\text{Cov}(X,Y\mid Z)) + \text{Cov}(\mathbb{E}(X\mid Z),\mathbb{E}(Y\mid Z)).$$

The law of total variance gives us the following two inequalities, since the variance is non-negative: $$\text{Var}(X) \geq \mathbb{E}(\text{Var}(X\mid Y))$$ $$\text{Var}(X) \geq \text{Var}(\mathbb{E}(X\mid Y)).$$

Are the following true with the covariances like when dealing with variances?

$$ |\text{Cov}(X,Y)| \geq |\mathbb{E}(\operatorname{Cov}(X,Y\mid Z))|$$ $$ |\operatorname{Cov}(X,Y)| \geq |\operatorname{Cov}(\mathbb{E}(X\mid Z),\mathbb{E}(Y\mid Z))|$$

Answer 1

If $\text{Cov}(X,Y)=0$ and the others are non-zero then this would be a counter-example to your covariance conjectures. For example if there are four equally likely combinations of values:

 x    y    z 
+1   -2   -5
-2   -1   -5
-1   +2   +5
+2   +1   +5 

these are in fact $(x,y)$ vertices of an oblique square with $z=x+3y$ but that is not significant

Then

• $\text{Cov}(X,Y) = 0$
• $\mathbb{E}(\text{Cov}(X,Y\mid Z)) = -\frac34$
• $\text{Cov}(\mathbb{E}(X\mid Z),\mathbb{E}(Y\mid Z)) =+\frac34$