Consider $X,Y,Z$ all standard normal random variables. Now I also want to have some correlation between them.
Lets denote these $\rho_{XY},\rho_{YZ}$ and $\rho_{XZ}$. Surely I cannot choose them freely (on [-1,1]). Is there a necessary and sufficient condition on, say, $\rho_{XY}$ given the others? I've found the necessary condition that must lie between
$$\rho_{XY}\rho_{YZ} \pm \sqrt{1-\rho_{XY}^2}\sqrt{1-\rho_{YZ}},$$ but (simulation indicates) it is not sufficient.
My findings are that if $(\rho_{yz}, \rho_{zx}, \rho_{xy})$ lies inside the region enclosed by the symmetric parametric surface $$(x,y,z) = (\sin (\theta + \phi), \cos(\theta - \phi), \sin 2 \phi), \quad \theta \in [0, 2\pi), \quad \phi \in [-\pi/4, \pi/4],$$ then the eigenvalues of the correlation matrix $$\boldsymbol \Sigma = \begin{bmatrix} 1 & \rho_{xy} & \rho_{zx} \\ \rho_{xy} & 1 & \rho_{yz} \\ \rho_{zx} & \rho_{yz} & 1 \end{bmatrix}$$ will all be nonnegative, thus $(X, Y, Z)$ will be multivariate normal.
In other words, if, say, $\rho_{xy} \in [-1,1]$ is given, then $(\rho_{yz}, \rho_{zx})$ will be constrained to the interior of an ellipse whose boundary is given by the parametric equation $$(\sin(\theta + \sin^{-1}(\rho_{xy})/2), \cos (\theta - \sin^{-1}(\rho_{xy})/2), \quad \theta \in [0,2\pi).$$ You can further solve this to get an interval for one of the correlation coefficients in terms of the other two.