二変量正規分布の期待値・分散・共分散の導出

（i）期待値
二変量正規分布のモーメント母関数の公式より、 $$\begin{array}{r}{M}_{X,Y}\left(\mathit{\theta }\right)=\exp \{{\theta }_1{\mu }_X+{\theta }_2{\mu }_Y+\frac{1}{2}({\theta }_1^2{\sigma }_X^2+2\rho {\theta }_1{\theta }_2{\sigma }_X{\sigma }_Y+{\theta }_2^2{\sigma }_Y^2)\}\end{array}$$ モーメント母関数の ${\theta }_1$ による1階微分を求めると、合成関数の微分法より、 $$\begin{array}{rl}\frac{\partial }{\partial {\theta }_1}{M}_{X,Y}\left(\mathit{\theta }\right)& =M_{X,Y}\left(\mathit{\theta }\right)\cdot \frac{\partial }{\partial {\theta }_1}\{{\theta }_1{\mu }_X+{\theta }_2{\mu }_Y+\frac{1}{2}({\theta }_1^2{\sigma }_X^2+2\rho {\theta }_1{\theta }_2{\sigma }_X{\sigma }_Y+{\theta }_2^2{\sigma }_Y^2)\}\\ & =({\mu }_X+{\theta }_1{\sigma }_X^2+\rho {\theta }_2{\sigma }_X{\sigma }_Y)\exp \{{\theta }_1{\mu }_X+{\theta }_2{\mu }_Y+\frac{1}{2}({\theta }_1^2{\sigma }_X^2+2\rho {\theta }_1{\theta }_2{\sigma }_X{\sigma }_Y+{\theta }_2^2{\sigma }_Y^2)\}\end{array}$$ 1次モーメントとモーメント母関数の関係 $\frac{\partial }{\partial {\theta }_1}{M}_{X,Y}\left(\mathbf{0}\right)=E\left(X\right)$ より、 $$\begin{array}{rl}E\left(X\right)& =({\mu }_X+0+0)\exp \{0+0+\frac{1}{2}(0+0+0)\}\\ & ={\mu }_X\end{array}$$ $◼$

（ii）分散
モーメント母関数の ${\theta }_1$ による2階微分を求めると、積の微分公式と合成関数の微分法より、 $$\begin{array}{rl}\frac{{\partial }^2}{\partial {\theta }_1^2}{M}_{X,Y}\left(\mathit{\theta }\right)& =M_{X,Y}\left(\mathit{\theta }\right)\cdot \frac{\partial }{\partial {\theta }_1}({\mu }_X+{\theta }_1{\sigma }_X^2+\rho {\theta }_2{\sigma }_X{\sigma }_Y)+({\mu }_X+{\theta }_1{\sigma }_X^2+\rho {\theta }_2{\sigma }_X{\sigma }_Y)\cdot \frac{\partial }{\partial {\theta }_1}{M}_{X,Y}\left(\mathit{\theta }\right)\\ & =M_{X,Y}\left(\mathit{\theta }\right)\cdot {\sigma }_X^2+({\mu }_X+\rho {\theta }_2{\sigma }_X{\sigma }_Y+2{\theta }_1{\sigma }_Y^2)\cdot \frac{\partial }{\partial {\theta }_1}{M}_{X,Y}\left(\mathit{\theta }\right)\end{array}$$ 2次モーメントとモーメント母関数の関係 $\frac{{\partial }^2}{\partial {\theta }_1^2}{M}_{X,Y}\left(\mathbf{0}\right)=E\left(X^2\right)$ より、 $$\begin{array}{rl}E\left(X^2\right)& =\exp \{0+0+\frac{1}{2}(0+0+0)\}\cdot {\sigma }_X^2+({\mu }_X+0+0)\cdot {\mu }_X\\ & ={\mu }_X^2+{\sigma }_X^2\end{array}$$ 分散の公式 $V\left(X\right)=E\left(X^2\right)-{\left\{E\left(X\right)\right\}}^2$ より、 $$\begin{array}{r}V\left(X\right)={\mu }_X^2+{\sigma }_X^2-{\mu }_X^2={\sigma }_X^2\end{array}$$ $◼$

（iii）共分散
モーメント母関数を ${\theta }_1,{\theta }_2$ で1回ずつ微分すると、合成関数の微分法より、 $$\begin{array}{rl}\frac{{\partial }^2}{\partial {\theta }_1\partial {\theta }_2}{M}_{X,Y}\left(\mathit{\theta }\right)& =M_{X,Y}\left(\mathit{\theta }\right)\cdot \frac{\partial }{\partial {\theta }_2}({\mu }_X+{\theta }_1{\sigma }_X^2+\rho {\theta }_2{\sigma }_X{\sigma }_Y)+({\mu }_X+{\theta }_1{\sigma }_X^2+\rho {\theta }_2{\sigma }_X{\sigma }_Y)\cdot \frac{\partial }{\partial {\theta }_2}{M}_{X,Y}\left(\mathit{\theta }\right)\\ & =M_{X,Y}\left(\mathit{\theta }\right)\cdot \left(\rho {\sigma }_X{\sigma }_Y\right)+({\mu }_X+{\theta }_1{\sigma }_X^2+\rho {\theta }_2{\sigma }_X{\sigma }_Y)\cdot ({\mu }_Y+\rho {\theta }_1{\sigma }_X{\sigma }_Y+{\theta }_2{\sigma }_Y^2)\cdot M_{X,Y}\left(\mathit{\theta }\right)\\ & =M_{X,Y}\left(\mathit{\theta }\right)\cdot \{\rho {\sigma }_X{\sigma }_Y+({\mu }_X+{\theta }_1{\sigma }_X^2+\rho {\theta }_2{\sigma }_X{\sigma }_Y)\cdot ({\mu }_Y+\rho {\theta }_1{\sigma }_X{\sigma }_Y+{\theta }_2{\sigma }_Y^2)\}\end{array}$$ 積の期待値とモーメント母関数の関係 $\frac{{\partial }^2}{\partial {\theta }_1\partial {\theta }_2}{M}_{X,Y}\left(\mathbf{0}\right)=E\left(XY\right)$ より、 $$\begin{array}{rl}E\left(XY\right)& =\exp \{0+0+\frac{1}{2}(0+0+0)\}\cdot \{\rho {\sigma }_X{\sigma }_Y+({\mu }_X+0+0)\cdot ({\mu }_Y+0+0)\}\\ & =\rho {\sigma }_X{\sigma }_Y+{\mu }_X{\mu }_Y\end{array}$$ 共分散の公式 $\mathrm{Cov}(X,Y)=E\left(XY\right)-E\left(X\right)E\left(Y\right)$ より、 $$\begin{array}{r}\mathrm{Cov}(X,Y)=\rho {\sigma }_X{\sigma }_Y+{\mu }_X{\mu }_Y-{\mu }_X{\mu }_Y=\rho {\sigma }_X{\sigma }_Y\end{array}$$ 相関係数の定義式 ${\rho }_{XY}=\frac{\mathrm{Cov}(X,Y)}{{\sigma }_X{\sigma }_Y}$ より、 $$\begin{array}{r}{\rho }_{XY}=\frac{\rho {\sigma }_X{\sigma }_Y}{{\sigma }_X{\sigma }_Y}=\rho \end{array}$$ $◼$