Correlation Bounds From Positive Semidefiniteness
Three random variables $x$, $y$, $z$ satisfy $\operatorname{corr}(x, y) = 0.8$ and $\operatorname{corr}(x, z) = 0.8$. Let $\rho = \operatorname{corr}(y, z)$.
Using only the fact that a correlation matrix must be positive semidefinite, find the maximum and minimum possible values of $\rho$.
Hints
- Write the $3 \times 3$ correlation matrix $P$ of $(x, y, z)$ with unit diagonal, $0.8$ in the $(x,y)$ and $(x,z)$ slots, and $\rho$ in the $(y,z)$ slot.
- $P$ is a covariance matrix of standardized variables, so $w^{T}Pw = \operatorname{Var}(w_1 x + w_2 y + w_3 z) \ge 0$ for all $w$: $P$ is positive semidefinite, and in particular $\det P \ge 0$.
- Expand $\det P$ as a quadratic in $\rho$ and find where it is nonnegative. It factors nicely.
Worked Solution
How to Think About It: Every valid correlation matrix must be positive semidefinite, because quadratic forms in it are variances of linear combinations. For a $3 \times 3$ matrix with unit diagonal, the binding constraint is that its determinant is nonnegative; solving $\det P \ge 0$ for the unknown entry gives the attainable range.
Quick Estimate: If $\rho = 0$, the portfolio $w = (1, -\tfrac12, -\tfrac12)$ on standardized variables has variance $1 + \tfrac14 + \tfrac14 - 0.8 - 0.8 + 0 = -0.1 < 0$, impossible. So $\rho = 0$ is out and the minimum must be positive; the geometric argument ($\cos 2\theta$ with $\cos\theta = 0.8$) suggests $0.28$.
Formal Solution:
*Step 1 -- Correlation matrix.* $$P = \begin{pmatrix} 1 & 0.8 & 0.8 \\ 0.8 & 1 & \rho \\ 0.8 & \rho & 1 \end{pmatrix}.$$
*Step 2 -- Why PSD.* For standardized variables $\tilde x, \tilde y, \tilde z$ and any $w \in \mathbb{R}^3$, $w^{T}Pw = \operatorname{Var}(w_1\tilde x + w_2\tilde y + w_3\tilde z) \ge 0$. A symmetric matrix is PSD iff all its principal minors are nonnegative; the $1 \times 1$ minors are $1$, the $2 \times 2$ minors are $1 - 0.64 = 0.36$ (twice) and $1 - \rho^2$, so the only nontrivial condition is $\det P \ge 0$ together with $|\rho| \le 1$.
*Step 3 -- Determinant.* Expanding along the first row, $$\det P = 1\cdot(1 - \rho^{2}) - 0.8\,(0.8 - 0.8\rho) + 0.8\,(0.8\rho - 0.8) = 1 - \rho^{2} - 0.64 + 0.64\rho + 0.64\rho - 0.64 = -\rho^{2} + 1.28\rho - 0.28.$$
*Step 4 -- Solve $\det P \ge 0$.* $$-\rho^{2} + 1.28\rho - 0.28 = -(\rho - 0.28)(\rho - 1) \ge 0 \quad \Longleftrightarrow \quad 0.28 \le \rho \le 1.$$ (Check the factorization: $(\rho - 0.28)(\rho - 1) = \rho^2 - 1.28\rho + 0.28$.)
*Step 5 -- Attainability.* At $\rho = 1$, $P$ is PSD with a zero eigenvalue ($z = y$). At $\rho = 0.28$, $\det P = 0$ and $P$ is singular but PSD: the witness $y = 0.8x + 0.6u$, $z = 0.8x - 0.6u$ with independent standard normals $x, u$ realizes it. Every $\rho$ between is attainable by continuity (e.g. $z = 0.8x + 0.6(\cos\alpha\,u + \sin\alpha\,v)$ with $v$ another independent normal). In general, with $\operatorname{corr}(x,y) = a$ and $\operatorname{corr}(x,z) = b$, the same computation gives $ab - \sqrt{(1-a^2)(1-b^2)} \le \rho \le ab + \sqrt{(1-a^2)(1-b^2)}$; here $0.64 \pm 0.36$.
Answer: $\rho = \operatorname{corr}(y, z)$ must satisfy $0.28 \le \rho \le 1$: the maximum is $1$ and the minimum is $0.28$.
Intuition
A correlation matrix is a Gram matrix, so it is automatically positive semidefinite, and that single fact constrains every entry given the others: here $\det P = -(\rho - 0.28)(\rho - 1) \ge 0$ pins $\rho$ to $[0.28, 1]$. This is why risk systems reject hand-edited correlation matrices and why stress tests that shock one pairwise correlation must repair the matrix (nearest-PSD projection) before it can be used for portfolio variance.