$\operatorname{Cov} \, (A,B)\geq 0$ and $\operatorname{Cov}(B,C)\geq 0\Rightarrow \operatorname{Cov}(A,C)\geq 0$? 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 $\cov(A,C)\geq 0$?
 A: 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 $\text{cov} \, (A,C) \geq 0$.
A: 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: 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 $A = \{1\}, B = \{1, 2\}, C = \{2\}$.
