# Is this possible that Cor(X,Y)=0.99Cor(X, Y)=0.99, Cor(Y,Z)=0.99Cor(Y, Z)=0.99 but Cor(X,Z)=0Cor(X, Z)=0?

Let $$XX$$, $$YY$$, and $$ZZ$$ are three random variables. Intuitively, I think that it is impossible to have $$Cor(X,Y)=0.99Cor(X, Y)=0.99$$, $$Cor(Y,Z)=0.99Cor(Y, Z)=0.99$$ but $$Cor(X,Z)=0Cor(X, Z)=0$$. My intuitive thought is that $$XX$$ and $$ZZ$$ are nearly linearly correlated to $$YY$$. Hence, they are more or less linearly correlated, which makes the last equality impossible.

I pose this question because of the question and the comments (include my comments) here.

In general, as some others point out, I agree that it is possible that for some $$ρ>0\rho>0$$ we may have $$Cor(X,Y)=ρ,Cor(Y,Z)=ρ and Cor(X,Z)=0(1).Cor(X, Y)=\rho, Cor(Y, Z)=\rho \mbox{ and } Cor(X, Z)=0 \qquad (1).$$

My questions are:

1. Do you think that (1) is wrong when $$ρ\rho$$ is close to 1, e.g., 0.99?
2. If (1) is wrong when $$ρ\rho$$ is close to 1, what is the maximum value of $$ρ\rho$$ so that (1) can be correct?

$$|1−λρρρ1−λ0ρ01−λ|=(1−λ)((1−λ)2−2ρ2))=0 \left| \begin{matrix} 1-\lambda & \rho & \rho \\ \rho & 1-\lambda & 0 \\ \rho & 0 & 1-\lambda \end{matrix} \right| =(1-\lambda)\big((1-\lambda)^2-2\rho^2)\big) =0$$
so the eigenvalues are $$11$$ and $$1±√2ρ1\pm\sqrt{2}\rho$$. These are all non-negative for
$$−1√2≤ρ≤1√2. -\frac1{\sqrt{2}} \le \rho \le \frac1{\sqrt{2}}.$$