例 3.4.9

hard一级题目发布者: ai-batch

题干

二维正态分布 $N(\mu_1,\mu_2,\sigma_1^2,\sigma_2^2,\rho)$ 的相关系数就是 $\rho$ 。

答案点击展开后可查看解析

解析

下面先求 $\mathrm{Cov}(X,Y)$ 。

\mathrm{Cov}(X,Y) = E[(X-E(X))(Y-E(Y))]

= \frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x-\mu_1)(y-\mu_2) \cdot \exp\left\{-\frac{1}{2(1-\rho^2)}\left[\frac{(x-\mu_1)^2}{\sigma_1^2} - 2\rho\frac{(x-\mu_1)(y-\mu_2)}{\sigma_1\sigma_2} + \frac{(y-\mu_2)^2}{\sigma_2^2}\right]\right\}\mathrm{d}x\mathrm{d}y.

先将上式中方括号内化成

\left(\frac{x-\mu_1}{\sigma_1} - \rho\frac{y-\mu_2}{\sigma_2}\right)^2 + \left(\sqrt{1-\rho^2}\,\frac{y-\mu_2}{\sigma_2}\right)^2,

再作变量变换

\begin{cases} u = \dfrac{1}{\sqrt{1-\rho^2}}\left(\dfrac{x-\mu_1}{\sigma_1} - \rho\dfrac{y-\mu_2}{\sigma_2}\right), \\[6pt] v = \dfrac{y-\mu_2}{\sigma_2}, \end{cases}

则

\begin{cases} x-\mu_1 = \sigma_1(u\sqrt{1-\rho^2}+\rho v), \\ y-\mu_2 = \sigma_2 v, \end{cases}

\mathrm{d}x\mathrm{d}y = |J|\,\mathrm{d}u\mathrm{d}v = \sigma_1\sigma_2\sqrt{1-\rho^2}\,\mathrm{d}u\mathrm{d}v.

由此得

\mathrm{Cov}(X,Y) = \frac{\sigma_1\sigma_2}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(uv\sqrt{1-\rho^2}+\rho v^2)\exp\left\{-\frac{1}{2}(u^2+v^2)\right\}\mathrm{d}u\mathrm{d}v.

上式右端积分可以分为两个积分之和，其中

\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}uv\exp\left\{-\frac{1}{2}(u^2+v^2)\right\}\mathrm{d}u\mathrm{d}v = 0,

\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}v^2\exp\left\{-\frac{1}{2}(u^2+v^2)\right\}\mathrm{d}u\mathrm{d}v = 2\pi.

从而

\mathrm{Cov}(X,Y) = \frac{\sigma_1\sigma_2}{2\pi}\cdot\rho\cdot 2\pi = \rho\sigma_1\sigma_2,

\mathrm{Corr}(X,Y) = \frac{\mathrm{Cov}(X,Y)}{\sigma_1\sigma_2} = \rho.