二変量正規分布のモーメント母関数の導出

モーメント母関数の定義式 $M_{X,Y}\left(\mathit{\theta }\right)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{\mathit{\theta }\mathit{x}}\cdot f(x,y)dxdy$ より、 $$\begin{array}{rl}{M}_{X,Y}\left(\mathit{\theta }\right)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{{\theta }_{1}x+{\theta }_{2}y}\cdot \frac{1}{2\pi {\sigma }_X{\sigma }_Y\sqrt{1-{\rho }^2}}\exp \{-\frac{1}{2(1-{\rho }^2)}Q(x,y)\}dxdy\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{2\pi {\sigma }_X{\sigma }_Y\sqrt{1-{\rho }^2}}\exp \{{\theta }_{1}x+{\theta }_{2}y-\frac{1}{2(1-{\rho }^2)}Q(x,y)\}dxdy\end{array}$$ ここで、以下のように変数変換すると、 $$\begin{array}{c}\{\begin{array}{c}u=\frac{x-{\mu }_X}{{\sigma }_X}\\ v=\frac{y-{\mu }_Y}{{\sigma }_Y}\end{array}\Rightarrow \{\begin{array}{c}x=u{\sigma }_X+{\mu }_X\\ y=v{\sigma }_Y+{\mu }_Y\end{array}\\ \begin{array}{c}x:-\mathrm{\infty }\to \mathrm{\infty }\\ y:-\mathrm{\infty }\to \mathrm{\infty }\end{array}\Rightarrow \begin{array}{c}u:-\mathrm{\infty }\to \mathrm{\infty }\\ v:-\mathrm{\infty }\to \mathrm{\infty }\end{array}\\ \left|J\right|={\sigma }_X{\sigma }_Y\end{array}$$ また、指数部分を完全平方にすると、 $$\begin{array}{c}{\theta }_{1}x+{\theta }_{2}y-\frac{1}{2(1-{\rho }^2)}Q(x,y)={\theta }_1{\mu }_X+{\theta }_2{\mu }_Y-\frac{1}{2(1-{\rho }^2)}R(u,v)\\ R(u,v)={\{u-\rho v-(1-{\rho }^2){\theta }_1{\sigma }_X\}}^2+(1-{\rho }^2){(v-\rho {\theta }_1{\sigma }_X-{\theta }_2{\sigma }_Y)}^2-(1-{\rho }^2)({\theta }_1^2{\sigma }_X^2+2\rho {\theta }_1{\theta }_2{\sigma }_X{\sigma }_Y+{\theta }_2^2{\sigma }_Y^2)\end{array}$$ したがって、置換積分法により、 $$\begin{array}{rl}{M}_{X,Y}\left(\mathit{\theta }\right)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{2\pi {\sigma }_X{\sigma }_Y\sqrt{1-{\rho }^2}}\exp \{{\theta }_1{\mu }_X+{\theta }_2{\mu }_Y-\frac{1}{2(1-{\rho }^2)}R(u,v)\}\cdot {\sigma }_X{\sigma }_{Y}dudv\\ & =\exp \{{\theta }_1{\mu }_X+{\theta }_2{\mu }_Y\}\cdot {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{2\pi \sqrt{1-{\rho }^2}}\exp \{-\frac{1}{2(1-{\rho }^2)}R(u,v)\}dudv\end{array}$$ ここで、以下のように変数変換すると、 $$\begin{array}{c}\{\begin{array}{c}w=\frac{u-\rho v-(1-{\rho }^2){\theta }_1{\sigma }_X}{\sqrt{1-{\rho }^2}}\\ z=v-\rho {\theta }_1{\sigma }_X-{\theta }_2{\sigma }_Y\end{array}\Rightarrow \{\begin{array}{c}u=\sqrt{1-{\rho }^2}w+\rho v+(1-{\rho }^2){\theta }_1{\sigma }_X\\ v=z+\rho {\theta }_1{\sigma }_X+{\theta }_2{\sigma }_Y\end{array}\\ \begin{array}{c}u:-\mathrm{\infty }\to \mathrm{\infty }\\ v:-\mathrm{\infty }\to \mathrm{\infty }\end{array}\Rightarrow \begin{array}{c}w:-\mathrm{\infty }\to \mathrm{\infty }\\ z:-\mathrm{\infty }\to \mathrm{\infty }\end{array}\end{array}$$ 変数変換のヤコビアンは、 $$\begin{array}{rl}\left|J\right|& =\left|\begin{array}{cc}\frac{\partial u}{\partial w}& \frac{\partial v}{\partial w}\\ \frac{\partial u}{\partial z}& \frac{\partial v}{\partial z}\end{array}\right|\\ & =\left|\begin{array}{cc}\sqrt{1-{\rho }^2}& 0\\ 0& 1\end{array}\right|\\ & =\sqrt{1-{\rho }^2}\end{array}$$ したがって、置換積分法により、 $$\begin{array}{rl}{M}_{X,Y}\left(\mathit{\theta }\right)& =\exp \{{\theta }_1{\mu }_X+{\theta }_2{\mu }_Y+\frac{1}{2}({\theta }_1^2{\sigma }_X^2+2\rho {\theta }_1{\theta }_2{\sigma }_X{\sigma }_Y+{\theta }_2^2{\sigma }_Y^2)\}\cdot {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{2\pi \sqrt{1-{\rho }^2}}\exp (-\frac{w^2}{2}-\frac{z^2}{2})\sqrt{1-{\rho }^2}dwdz\\ & =\exp \{{\theta }_1{\mu }_X+{\theta }_2{\mu }_Y+\frac{1}{2}({\theta }_1^2{\sigma }_X^2+2\rho {\theta }_1{\theta }_2{\sigma }_X{\sigma }_Y+{\theta }_2^2{\sigma }_Y^2)\}\cdot {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{2\pi }\exp (-\frac{w^2}{2}-\frac{z^2}{2})dwdz\\ & =\exp \{{\theta }_1{\mu }_X+{\theta }_2{\mu }_Y+\frac{1}{2}({\theta }_1^2{\sigma }_X^2+2\rho {\theta }_1{\theta }_2{\sigma }_X{\sigma }_Y+{\theta }_2^2{\sigma }_Y^2)\}\cdot {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}\exp (-\frac{w^2}{2})dw\cdot {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}\exp (-\frac{z^2}{2})dz\\ & =\exp \{{\theta }_1{\mu }_X+{\theta }_2{\mu }_Y+\frac{1}{2}({\theta }_1^2{\sigma }_X^2+2\rho {\theta }_1{\theta }_2{\sigma }_X{\sigma }_Y+{\theta }_2^2{\sigma }_Y^2)\}\end{array}$$ $◼$