相关系数范围

相关矩阵必须半正定。设 $\rho=\mathrm{Corr}(X,Z)$，则

$$\det\begin{pmatrix} 1&5/13&\rho\\ 5/13&1&12/13\\ \rho&12/13&1 \end{pmatrix} =\frac{120}{169}\rho-\rho^2\ge 0.$$

因此 $0\le \rho\le \frac{120}{169}$，故 $b=\frac{120}{169}$。

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Original Explanation

The key idea here is that the correlation matrix for any collection of random variables must be positive semi-definite. Denoting $\text{Corr}(X,Z) = \rho$, the correlation matrix for $(X,Y,Z)$, say $C$, is a $3 \times 3$ matrix

$$C = \begin{bmatrix} 1 & \frac{5}{13} & \rho \\ \frac{5}{13} & 1 & \frac{12}{13} \\ \rho & \frac{12}{13} & 1 \end{bmatrix}$$

A condition to show that a matrix is positive definite is that each of the sub-matrices of order $1 \times 1, 2 \times 2, \ldots, n \times n$, where $n$ is a size of the matrix, originates from the top left corner having a non-negative determinant. In this case, it is easy to see that the top left $2 \times 2$ matrix has determinant

$$C_{11}C_{22} - C_{12}C_{21} = \dfrac{144}{169} > 0$$

so we only need to check that the entire matrix $C$ has non-negative determinant.

Evaluating this determinant as a function of $\rho$, we obtain that it is

$$\dfrac{120}{169}\rho - \rho^2$$

Setting this equal to $0$, we obtain that $\rho = 0,\dfrac{120}{169}$. As this polynomial had a negative leading coefficient for $\rho^2$, it follows that the parabola (i.e. determinant) must have been positive between the two roots, so $\rho^* = \dfrac{120}{169}$ is the largest value where this determinant is non-negative.