平成29年損保数理過去問　問題文中の積分

下記を積分する。
$$\begin{array}{r}I={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Phi }(a+bx)\varphi (x)dx\end{array}$$

ただし、
$$\begin{array}{rl}\varphi (x)& =\frac{1}{\sqrt{2\pi }}exp(-\frac{1}{2}{x}^2)\\ \mathrm{\Phi }(x)& ={\int }_{-\mathrm{\infty }}^x\frac{1}{\sqrt{2\pi }}exp(-\frac{1}{2}{t}^2)\end{array}$$
である。

積分

$$\begin{array}{rl}I& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Phi }(a+bx)\varphi (x)dx\\ & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx{\int }_{-\mathrm{\infty }}^{a+bx}dyexp[-\frac{1}{2}(x^2+y^2)]\\ & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx{\int }_{-\mathrm{\infty }}^{bx}dyexp[-\frac{1}{2}{x}^2+(y+a{)}^2](\because \mathrm{平}\mathrm{行}\mathrm{移}\mathrm{動})\\ & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx{\int }_{-\mathrm{\infty }}^{0}dyexp[-\frac{1}{2}(xcos\theta +ysin\theta {)}^2+(-xsin\theta +ycos\theta +a{)}^2]\\ & (\because \mathrm{座}\mathrm{標}\mathrm{軸}\mathrm{の}\mathrm{回}\mathrm{転}:\theta =arctan(b))\\ & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx{\int }_{-\mathrm{\infty }}^{0}dyexp[-\frac{1}{2}(x-asin\theta {)}^2+(y+acos\theta {)}^2]\\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}exp[-\frac{1}{2}(x-asin\theta {)}^2]dx{\int }_{-\mathrm{\infty }}^0\frac{1}{\sqrt{2\pi }}exp[-\frac{1}{2}(y+acos\theta {)}^2]\\ & =1\mathrm{・}\mathrm{\Phi }(acos\theta )\\ & =\mathrm{\Phi }(\frac{a}{\sqrt{1+b^2}})\end{array}$$

まとめ

$$\begin{array}{r}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Phi }(a+bx)\varphi (x)dx=\mathrm{\Phi }(\frac{a}{\sqrt{1+b^2}})\end{array}$$

例

$a=\frac{1}{\sqrt{2}}$、$b=1$のとき、

$$\begin{array}{rl}I& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Phi }(\frac{1}{\sqrt{2}}+x)\varphi (x)dx\\ & =\mathrm{\Phi }(\frac{\frac{1}{\sqrt{2}}}{\sqrt{1+1^2}})\\ & =\mathrm{\Phi }(\frac{1}{2})\\ & =0.6915\end{array}$$