4
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $A = \{1\}, B = \{1, 2\}, C = \{2\}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.