# Does $Cov(X,Y)<0$ and $Cov(X,Z)>0$ imply $Cov(Y,Z)<0$?

Consider three normally distributed random variables, $$X,Y,Z$$ where $$Cov(X,Y)<0$$ and $$Cov(X,Z)>0$$. Can we say anything about the sign of $$Cov(Y,Z)$$?

Intuitively, $$Y$$ goes down when $$X$$ goes up. Next, $$Z$$ goes up when $$X$$ goes up. I would guess $$Cov(Y,Z)<0$$.

Set $$\rho=Corr(X,Y)<0$$. Then, $$Y=\rho X+\sqrt{1-\rho^2}\epsilon$$ where $$\epsilon\sim N(0,1)$$ is orthogonal to $$X$$. Then, \begin{align} Cov(Y,Z) &= Cov\left(\rho X+\sqrt{1-\rho^2}\epsilon,Z\right) \\ &=\underbrace{\rho Cov(X,Z)}_{<0} + \sqrt{1-\rho^2}Cov(\epsilon,Z). \end{align} Is there an argument why $$Cov(\epsilon,Z)$$ should be zero or negative? Or are additional assumptions necessary?

Let's write $$Z=r X+\sqrt{1-r^2}\epsilon_Z$$ where $$\epsilon_Z\sim N(0,1)$$ is orthogonal to $$X$$.. Then, \begin{align} Cov(\epsilon,Z) = \sqrt{1-r^2}Cov(\epsilon,\epsilon_Z). \end{align} I cannot see why $$Cov(\epsilon,\epsilon_Z)$$ should be zero/negative.

• The answer is definitely yes. See stats.stackexchange.com/questions/72790 for one particular case and how it can be analyzed. A correct answer has been posted after you accepted an incorrect one. Please reconsider your decision.
– whuber
Dec 19, 2021 at 15:39
• Oh, yes means here “yes, we can say something about the sign of the cov(y,z)”, which is the answer for the question in the body, and I think Jarle’s answer is the better one. I, on the other hand, despite reading the whole text, answered the question directly in the title, i.e. “Does cov(x,y) < 0 and cov(x,z) > 0 imply cov(y,z) < 0”, and the answer is no, generally speaking. Dec 19, 2021 at 16:47

If the correlations between the three variables are $$a$$, $$b$$ and $$c$$, the eigenvalues of the correlation matrix satisfies $$\left|\begin{matrix} 1-\lambda & a & c \\ a & 1-\lambda & b \\ c & b & 1-\lambda \end{matrix}\right|=0$$ which after some algebra simplifies to $$(1-\lambda)^3 - (1-\lambda)(a^2+b^2+c^2)+2abc=0.$$ For given values of $$a$$ and $$b$$, the correlation matrix is positive semi-definite when $$c$$ lies in a closed interval. At the endpoints of this interval, one of the eigenvalues are zero implying that $$1-a^2-b^2-c^2+2abc=0.$$ Solving this quadratic equation for $$c$$, we find that the endpoints of the interval of possible values of $$c$$ (plotted below) for given values of $$a$$ and $$b$$ are $$ab\pm \sqrt{(a^2-1)(b^2-1)}.$$ If $$a$$ and $$b$$ are of opposite sign, the whole interval for $$c$$ thus contains only negative values if $$ab+\sqrt{(a^2-1)(b^2-1)}<0.$$ Moving $$ab$$ to the right hands side and squaring both sides yields $$(a^2-1)(b^2-1) which simplifies to the condition $$1 Thus, to summarise, if $$a=\corr(X,Y)$$ and $$b=\corr(X,Z)$$ are of opposite signs and $$(a,b)$$ lies outside the unit circle, then $$c=\corr(Y,Z)$$ is necessarily negative.
No (to the title question). Any positive semi-definite matrix is also a covariance matrix. So, if there exists a $$\rho$$ that the following matrix is PSD, then you've a contradiction:
$$\Sigma=\begin{bmatrix}1&-\rho&\rho\\-\rho &1&\rho\\\rho&\rho&1\end{bmatrix}$$
This occurs for $$\rho=0.1$$ for example. The conditions on when it implies are correctly laid out in Jarle’s answer.