$\operatorname{Cov} \, (A,B)\geq 0$ and $\operatorname{Cov}(B,C)\geq 0\Rightarrow \operatorname{Cov}(A,C)\geq 0$?

Question

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 $\cov(A,C)\geq 0$?

Answer 1

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 $\text{cov} \, (A,C) \geq 0$.

Answer 2

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.