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

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