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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?

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

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  • $\begingroup$ If $a=b+c$ then $|a| \ge |b|-|c|$ and $|a| \ge |c| - |b|$. I am not sure you can do any better, since (as far as I can tell) all three terms could be any real number. $\endgroup$ – angryavian May 16 at 20:33
  • $\begingroup$ Yeah, something better would be needed. The inequalities in my original question would hold if $\mathbb{E}(\text{Cov}(X,Y \mid Z))$ and $\text{Cov}(\mathbb{E}(X \mid Z),\mathbb{E}(Y \mid Z))$ both had the same sign. I'm not sure when they have different signs though $\endgroup$ – John Doe May 16 at 22:00
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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

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