Let's say we have three random variables $A$, $B$ and $C$. I know that $\DeclareMathOperator{cov}{Cov} \cov(A,B)\geq 0$ and $\cov(B,C)\geq 0$ .

Then is it true that $\cov(A,C)\geq 0$?


No, this is in general not true. Just consider any random variable $X \in L^2$, $X \neq 0$, $\mathbb{E}X=0$, and choose $$A := X \qquad B := 0 \qquad C := -X.$$ Then $$\text{cov} \, (A,B) = \text{cov} \, (B,C) =0,$$ but $$\text{cov} \, (A,C) = - \mathbb{E}(X^2)<0.$$

Remark: It is also possible to construct counterexamples if the inequalities are strict, i.e. $\text{cov} \, (A,B) >0$, $\text{cov} \, (B,C)>0$ does not imply $\text{cov} \, (A,C) \geq 0$.

  • $\begingroup$ This would not hold even if the inequalities are strict? $\endgroup$ – Klein Gordon Apr 16 '15 at 6:42
  • $\begingroup$ @KleinGordon Yeah, I think it's also possible to construct a counterexample if the inequalities are strict. $\endgroup$ – saz Apr 16 '15 at 6:45

Another counterexample:

Let $X$ and $Y$ be independent $N(0,1)$ random variables. Define $V$ and $W$ as $V=X+Y$ and $W=Y-aX$ for some constant $a$ in $(0,1)$. Then it is readily verified that the correlation coefficients between $X$ and $V$ and between $V$ and $W$ are both positive (note that $\hbox{cov}(V, W)>0$ requires $0<a<1$), but the correlation coefficient between $X$ and $W$ is negative, showing that correlation need not be transitive.


A simple counter-example for the case where the inequalities are strict.

It is sometimes easier to visualize things when the random variables are indicators. Lets allow a slightly abuse of notation and let $A$ stand for both an event and its indicator variable. We want:

  • $Cov(A,B) > 0 \iff P(A\cap B) > P(A)P(B)$

  • $Cov(B,C) > 0 \iff P(B\cap C) > P(B)P(C)$

  • $Cov(A,C) < 0 \iff P(A\cap C) < P(A)P(C)$

Now all we need is to visualize some Venn diagram where the overlaps satisfy the above. Here is a specific example. Roll a $6$-sided die and:

  • $A = \{1,2,3\}, B = \{2,3,4\}, C=\{3,4,5\}$

  • $P(A)=P(B)=P(C) = \frac12$

  • $P(A\cap B) = P(B \cap C) = \frac13 > \frac12 \times \frac12$

  • But $P(A\cap C) = \frac16 < \frac12 \times \frac12$.

A perhaps even simpler example is $A = \{1\}, B = \{1, 2\}, C = \{2\}$.


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