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相关系数范围

Correlation Ranges

专题
Probability / 概率
难度
L4

题目详情

随机变量 X,Y,ZX,Y,Z 满足 Corr(X,Y)=513\mathrm{Corr}(X,Y)=\frac{5}{13} 且 Corr(Y,Z)=1213\mathrm{Corr}(Y,Z)=\frac{12}{13}。

Corr(X,Z)\mathrm{Corr}(X,Z) 的可取范围是形如 [0,b][0,b] 的区间，其中 bb 为最简分数。求 bb。

英文原题

Suppose that X,Y,X,Y, and ZZ are three random variables. We know that Corr(X,Y)=513\text{Corr}(X,Y) = \frac{5}{13} and Corr(Y,Z)=1213\text{Corr}(Y,Z) = \frac{12}{13}. The range of possible values for Corr(X,Z)\text{Corr}(X,Z) is an interval in the form [0,b][0,b], where bb is a fraction in fully reduced form. Find bb.

解析

设 ρ=Corr(X,Z)\rho=\mathrm{Corr}(X,Z)。

相关矩阵必须半正定：

R=(15/13ρ5/13112/13ρ12/131)⪰0.R=\begin{pmatrix} 1&5/13&\rho\\ 5/13&1&12/13\\ \rho&12/13&1 \end{pmatrix}\succeq 0.

因此其行列式非负：

det⁡(R)=1+2⋅513⋅1213ρ−(513)2−(1213)2−ρ2≥0.\det(R)=1+2\cdot\frac{5}{13}\cdot\frac{12}{13}\rho-\left(\frac{5}{13}\right)^2-\left(\frac{12}{13}\right)^2-\rho^2\ge 0.

化简得到

−ρ2+120169ρ≥0⇒0≤ρ≤120169.-\rho^2+\frac{120}{169}\rho\ge 0 \Rightarrow 0\le \rho\le \frac{120}{169}.

所以 b=120169b=\boxed{\frac{120}{169}}。

英文解析

Set ρ=Corr(X,Z)\rho=\mathrm{Corr}(X,Z).

Correlation matrix must be semi-positive:

R=(15/13ρ5/13112/13ρ12/131)⪰0.R=\begin{pmatrix} 1&5/13&\rho\\ 5/13&1&12/13\\ \rho&12/13&1 \end{pmatrix}\succeq 0.

So its determinant is not negative:

det⁡(R)=1+2⋅513⋅1213ρ−(513)2−(1213)2−ρ2≥0.\det(R)=1+2\cdot\frac{5}{13}\cdot\frac{12}{13}\rho-\left(\frac{5}{13}\right)^2-\left(\frac{12}{13}\right)^2-\rho^2\ge 0.

Simplify to get

−ρ2+120169ρ≥0⇒0≤ρ≤120169.-\rho^2+\frac{120}{169}\rho\ge 0 \Rightarrow 0\le \rho\le \frac{120}{169}.

So b=120169b=\boxed{\frac{120}{169}}.