Range of a Third Correlation Given Two Correlations of 0.8
Three random variables $x$, $y$, $z$ satisfy $$\operatorname{corr}(x, y) = 0.8 \qquad \text{and} \qquad \operatorname{corr}(x, z) = 0.8.$$ What are the largest and smallest possible values of $\operatorname{corr}(y, z)$?
Solve it geometrically, by viewing the (centered) random variables as vectors and correlations as cosines of the angles between them.
Hints
- For centered random variables (or their sample vectors), $\operatorname{corr}(u, v) = \dfrac{\langle u, v\rangle}{\|u\|\,\|v\|} = \cos\theta_{uv}$, the cosine of the angle between them.
- Both $y$ and $z$ make the same angle $\theta$ with $x$, where $\cos\theta = 0.8$ (so $\sin\theta = 0.6$). Picture $y$ and $z$ on a cone of half-angle $\theta$ around $x$. How small and how large can the angle between $y$ and $z$ be?
- The angle between $y$ and $z$ ranges from $0$ (same direction) to $2\theta$ (opposite sides of $x$ in one plane). So $\operatorname{corr}(y,z)$ ranges from $\cos 2\theta = \cos^2\theta - \sin^2\theta$ up to $1$.
Worked Solution
How to Think About It: After centering, random variables live in an inner-product space where covariance is the inner product, standard deviation is the norm, and correlation is the cosine of the angle. The question becomes: two unit vectors each make angle $\theta$ with a third unit vector; what angles can they make with each other?
Quick Estimate: Since $y$ and $z$ are both strongly aligned with $x$, they should be at least mildly positively correlated with each other. Two vectors at $\approx 37^\circ$ from $x$ can be at most $\approx 74^\circ$ apart, and $\cos 74^\circ \approx 0.28$.
Formal Solution:
*Step 1 -- Correlation as a cosine.* Let $\tilde{x} = x - E[x]$, etc. Then $\operatorname{cov}(u, v) = \langle \tilde u, \tilde v\rangle$ and $\sigma_u = \|\tilde u\|$, so $$\operatorname{corr}(u, v) = \frac{\langle \tilde u, \tilde v\rangle}{\|\tilde u\|\|\tilde v\|} = \cos\theta_{uv}.$$
*Step 2 -- Set up the angles.* Let $\theta$ be the angle between $\tilde x$ and $\tilde y$, and also between $\tilde x$ and $\tilde z$: $\cos\theta = 0.8$, hence $\sin\theta = 0.6$. The vector $\tilde y$ lies on a cone of half-angle $\theta$ around $\tilde x$, and so does $\tilde z$.
*Step 3 -- Extreme configurations.* The angle $\varphi$ between $\tilde y$ and $\tilde z$ is smallest ($\varphi = 0$) when $\tilde z$ points along $\tilde y$: then $\operatorname{corr}(y, z) = 1$ (e.g. $z = y$). It is largest when $\tilde y$ and $\tilde z$ lie in a common plane with $\tilde x$ on opposite sides of it, giving $\varphi = 2\theta$; every intermediate angle is attainable by rotating $\tilde z$ around the cone (the spherical triangle inequality $\varphi \le \theta + \theta$ shows nothing larger is possible).
*Step 4 -- Compute the minimum.* $$\min \operatorname{corr}(y, z) = \cos 2\theta = \cos^{2}\theta - \sin^{2}\theta = 0.8^{2} - 0.6^{2} = 0.64 - 0.36 = 0.28.$$
*Step 5 -- Explicit witnesses.* Write $y = 0.8x + 0.6u$ and $z = 0.8x - 0.6u$ with $x, u$ independent standard normals: each has unit variance, correlation $0.8$ with $x$, and $\operatorname{corr}(y, z) = 0.64 - 0.36 = 0.28$. Taking $z = y$ instead gives $1$.
Answer: $\operatorname{corr}(y, z)$ can be as large as $1$ and as small as $0.28$; the attainable range is $[0.28, 1]$.
Intuition
Correlation is a cosine, so correlations obey triangle-like constraints: if $y$ and $z$ each sit at angle $\theta = \arccos 0.8 \approx 36.9^\circ$ from $x$, they can be at most $2\theta \approx 73.7^\circ$ apart, hence $\operatorname{corr}(y,z) \ge \cos 2\theta = 0.28$. Traders meet this when two assets are each strongly correlated with a benchmark: they must be positively correlated with each other, and any hedge assuming otherwise is inconsistent with a valid (positive semidefinite) correlation matrix.