证明二维正态分布中的参数ρ为相关系数

在二维正态分布中，参数$\rho$被定义为随机变量$X,Y$的相关系数，即$\rho =r(x,y)$。由

$$\begin{gathered} r(x,y)=\frac{Cov(x,y)}{\sqrt{Var[x]Var[y]}} \\ Cov(x,y)=E[XY]-E[X]E[Y] \end{gathered}$$

即可证明。由于$E[XY]$的求解较为繁琐，本文记录此过程。

假设随机变量$(X,Y)\sim N({\mu }_1,{\mu }_2,{\sigma }_1^2,{\sigma }_2^2,\rho )$，求$E[XY]$。

解：

$$\begin{array}{rl}E[XY]& ={\int }_{-\infty }^{\infty }\text{d}x{\int }_{-\infty }^{\infty }\frac{xy}{2\pi {\sigma }_1{\sigma }_2\sqrt{1-{\rho }^2}}\exp \left\{-\frac{1}{2(1-{\rho }^2)}\left(\frac{(x-{\mu }_1{)}^2}{{\sigma }_1^2}-2\rho \frac{(x-{\mu }_1)(y-{\mu }_2)}{{\sigma }_1{\sigma }_2}+\frac{(y-{\mu }_2{)}^2}{{\sigma }_2^2}\right)\right\}\text{d}y\\ & =\frac{1}{2\pi {\sigma }_1{\sigma }_2\sqrt{1-{\rho }^2}}{\int }_{-\infty }^{\infty }x\exp \left\{-\frac{1}{2(1-{\rho }^2)}\left(\frac{(x-{\mu }_1^2)}{{\sigma }_1^2}\right)\right\}\text{d}x{\int }_{-\infty }^{\infty }y\exp \left\{-\frac{1}{2(1-{\rho }^2)}\left(-2\rho \frac{(x-{\mu }_1)(y-{\mu }_2)}{{\sigma }_1{\sigma }_2}+\frac{(y-{\mu }_2{)}^2}{{\sigma }_2^2}\right)\right\}\text{d}y\end{array}$$

作代换$x\to {\sigma }_{1}x+{\mu }_1,y\to {\sigma }_{2}y+{\mu }_2$，则

$$\begin{array}{rl}E[XY]& =\frac{1}{2\pi \sqrt{1-{\rho }^2}}{\int }_{-\infty }^{\infty }({\sigma }_{1}x+{\mu }_1)\exp \left\{-\frac{1}{2(1-{\rho }^2)}{x}^2\right\}\text{d}x{\int }_{-\infty }^{\infty }({\sigma }_{2}y+{\mu }_2)\exp \left\{-\frac{1}{2(1-{\rho }^2)}\left(-2\rho xy+y^2\right)\right\}\text{d}y\end{array}$$

由于

$$\begin{array}{rl}{\int }_{-\infty }^{\infty }\exp \left(\frac{-x^2+2kx}{l^2}\right)\text{d}x& ={\int }_{-\infty }^{\infty }\exp \left[\frac{-(x-k{)}^2+k^2}{l^2}\right]\text{d}x\\ & =\exp \frac{k^2}{l^2}{\int }_{-\infty }^{\infty }\exp \left(-\frac{u^2}{l^2}\right)\text{d}u\\ & =l\exp \frac{k^2}{l^2}{\int }_{-\infty }^{\infty }\exp \left(-v^2\right)\text{d}v\\ & =l\exp \frac{k^2}{l^2}\sqrt{\pi }\\ {\int }_{-\infty }^{\infty }x\exp \left(\frac{-x^2+2kx}{l^2}\right)\text{d}x& ={\int }_{-\infty }^{\infty }x\exp \left[\frac{-(x-k{)}^2+k^2}{l^2}\right]\text{d}x\\ & =\exp \frac{k^2}{l^2}{\int }_{-\infty }^{\infty }(u+k)\exp (-\frac{u^2}{l^2})\text{d}u\\ & =\exp \frac{k^2}{l^2}l{\int }_{-\infty }^{\infty }(vl+k)\exp (-v^2)\text{d}v\\ & =kl\exp \frac{k^2}{l^2}\sqrt{\pi }\end{array}$$

因此

$$\begin{array}{rl}E[XY]& =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }({\sigma }_{1}x+{\mu }_1)({\sigma }_2\rho x+{\mu }_2)\exp \left\{-\frac{x^2}{2}\right\}\text{d}x\\ & ={\mu }_1{\mu }_2+\rho {\sigma }_1{\sigma }_2\end{array}$$