Given a correlation between A and B, what are the best bounds on the product of the AC and BC correlations? Let's say I have three vectors (or random variables) $A, B,$ and $C.$ I can of course calculate the correlation between any of them (and have these numbers). However what I'm interested in is if there's a bound on the product of the correlations, $r_{A,C}$ $\cdot$ $r_{B,C}$, given $r_{A,B}$.
For example, let's say the correlation between $A$ and $B$ is 0.5.  It's obvious that one couldn't get $r_{,}\,r_{,}=1$, since $C$ cannot be perfectly correlated with both B and C. Does anyone know how to calculate this? I have found similar questions on StackExchange but nothing exactly like this.
 A: Because the correlation matrix must be positive semidefinite, then as shown at https://stats.stackexchange.com/a/5753/919 the three correlations $\tau=\rho_{AB},$ $\sigma=\rho_{AC},$ and $\rho=\rho_{BC}$ must satisfy the inequality
$$1 + 2(\rho\sigma)\tau - (\rho^2 + \sigma^2 + \tau^2) \ge 0\tag{*}$$
and, of course, all three values must be in the range $[-1,1].$
The question asks how to maximimize and minimize $\rho\sigma,$ given a value of $\tau$ and the constraint $(*).$  The following solution is elementary.  Alternatively, it can be derived using Lagrange multipliers.
Setting $\phi = (\rho+\sigma)/2$ and $\delta = \rho-\phi$ (whence $\sigma-\phi = -\delta$) we find, with easy algebra, that
$$\rho\sigma=\phi^2 - \delta^2 \le \frac{1 - \tau^2 - 4\delta^2}{2(1-\tau)}$$
(provided only that $\tau \ne \pm 1$).  Clearly the upper limit for $\rho\sigma$ on the right hand side can be made as large as possible only by setting $\delta=0,$ which means $\rho\sigma$ is maximized when $\delta=0;$ that is, $\rho=\sigma.$  In this case the problem of maximizing $\rho\sigma$ is very simple (because $(*)$ becomes linear in $\phi^2$), with solution

$$\rho\sigma \le \frac{1+\tau}{2}.$$

To minimize $\rho\sigma$ we deduce, in the same manner, that we wish to make $\delta$ as large as possible.  This leads to $\phi=0,$ whence $\rho\sigma=-\delta^2.$  Again the problem becomes simple, with solution

$$\rho\sigma \ge \frac{\tau-1}{2}.$$


One way to visualize (and check via brute force) these results is to pick $\tau=\rho_{AB},$ plot the region in the $(\rho,\sigma) = (\rho_{AC}, \rho_{BC})$ plane where $(*)$ holds, and overplot contour lines of the function $f(\rho,\sigma)=\rho\sigma.$  The best bounds for $\rho\sigma$ are the lowest and highest contour levels that intersect the region plot.
Here are three examples for typical values of $\tau=\rho_{AB},$ including the case posited by the question, $\rho_{AB}=0.5,$ at the right.  The solid regions denote the loci of solutions of $(*):$ that is, the mathematically possible values of the two other correlation coefficients.

The contour lines at levels $(\rho_{AB}-1)/2$ and $(\rho_{AB}+1)/2$ are highlighted: you can see how they just barely skim the vertices of the ellipses determined by $(*),$ osculating precisely at the points $\rho_{AC}=\pm\rho_{BC}$ as deduced in this solution.
It should now be clear that although $\tau=\pm 1$ might have caused algebraic difficulties in the derivation, they aren't really special: the same formulas for the bounds will apply.
A: A geometric perspective makes this surprisingly easy and provides detailed information about how the extreme products are reached.
Abstractly, $A, B,$ and $C$ are Euclidean vectors and their correlations are the cosines of the angles between them or, equivalently, their inner products.  We may simplify the question with a few observations:

*

*$C$ can be expressed as the sum of its component in the $AB$ plane and the orthogonal component (this is just regression of $C$ on $A$ and $B$).  Because the orthogonal component contributes nothing to the inner products, its presence can only decrease the magnitudes of those products.  Thus, we may with no loss of generality assume $C$ lies in the $AB$ plane.


*Because only angles are under consideration, normalize all vectors to have unit length.  (The possibility that any vector can be zero is excluded from the question because the correlation with it is undefined.)


*We may therefore adopt a polar coordinate system in which $A=(1,0)$ and $B$ lies in the upper half plane at, say, angle $\alpha$ with $A.$  Thus, $\rho_{AB} = \cos(\alpha)$ is the correlation between $A$ and $B.$

This figure depicts unit vectors with arrows.  Dotted lines show their projections onto each other.  The projections themselves are highlighted in black.  The correlations $\rho_{**}$ are signed: in this image, $\rho_{AB}$ is negative (approximately $-0.5$).
As a result, $C$ is a unit vector in the plane having some polar coordinate $\theta$ and its correlations with $A$ and $B$ are the cosines of the angles it makes with them: that is,
$$\rho_{AC}\rho_{BC} = \cos(\theta)\cos(\theta-\alpha) = \frac{1}{2}\left(\cos(2\theta-\alpha) + \cos(\alpha)\right).$$
The part of the right hand side that varies with $\theta$ is a multiple of a cosine wave, which has maxima when its argument is an even multiple of $\pi$ that is, $2n\pi=2\theta-\alpha$ for integral $n;$ and minima at odd multiples of $\pi;$ that is, $(2n+1)\pi=2\theta-\alpha.$ Solving for $\theta$ gives the answer:

$\rho_{AC}\rho_{BC}$ is maximized when $2\theta=\alpha+2n\pi;$ that is, $C$ bisects $AB$ and therefore the maximum is $$\cos(\alpha/2)\cos(-\alpha/2)=\cos^2(\alpha/2) = \frac{1+\cos(\alpha)}{2} = \frac{1 + \rho_{AB}}{2}.$$ $\rho_{AC}\rho_{BC}$ is minimized when $2\theta=\alpha+(2n+1)\pi,$ with a minimum $$\cos(\pi/2+\alpha/2)\cos(\pi/2-\alpha/2) = -\sin^2(\alpha/2) = \cos^2(\alpha/2)-1 = \frac{-1 + \rho_{AB}}{2}.$$

Furthermore, observation $(1)$ shows that when this product can attain any value, it can also attain any fraction of that value down to zero simply by pulling $C$ sufficiently far out of the $AB$ plane.  Thus, every number in the unit-length interval $[(-1+\rho_{AB})/2, (1+\rho_{AB})/2]$ is a possible value of the product $\rho_{AC}\rho_{BC}.$
