確率変数の変数変換 Z=X/Y

確率変数の変数変換 Z=X/Y

Z=X/Y

$$ \begin{eqnarray} p_{X/Y}(z) &=&\int \int \delta\left(z-\frac{x}{y}\right)f(x)g(y)\mathrm{d}x\mathrm{d}y \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/delta-function.html}{\int_{-\infty}^{\infty}\delta(x)f(x)\mathrm{d}x=f(0)} \\&&\;\cdots\;X=x,Y=yの同時確率f(x)g(x)のうちz-\left(\frac{x}{y}\right)=0を満たすものだけを足し合わせる. \\&=&\int \int \left|y\right|\delta\left(x-yz\right)\;f\left(x\right)g(y)\mathrm{d}x\mathrm{d}y \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/delta-function.html}{\delta(u(x))=\sum_{\alpha\in u^{-1}(0)}\frac{1}{\left|u^{\prime}(\alpha)\right|}\delta\left(x-\alpha\right)} \\&&\;\cdots\;u(x)=z-\frac{x}{y} \\&&\;\cdots\;u(x=\alpha)=0,\;\alpha=yz \\&&\;\cdots\;u^\prime=\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(z-\frac{x}{y})=-\frac{1}{y} \\&&\;\cdots\;\delta\left(z-\frac{x}{y}\right) =\frac{1}{|u^\prime(\alpha)|}\delta\left(x-\alpha\right) =\frac{1}{\left|-\frac{1}{y}\right|}\delta\left(x-yz\right) =\frac{1}{\left|\frac{1}{y}\right|}\delta\left(x-yz\right) =\left|y\right|\delta\left(x-yz\right) \\&=&\int |y|\delta\left(yz-yz\right)f\left(yz\right)g(y)\mathrm{d}y \;\cdots\;z=\frac{x}{y},x=yz \\&=&\int |y|\delta\left(0\right)f\left(yz\right)g(y)\mathrm{d}y \\&=&\int |y|f(yz)g(y)\mathrm{d}y \;\cdots\;\delta凾数の引数が全域で0なので，全域で|y|f(yz)g(y)を足し合わせる積分になる. \end{eqnarray} $$

標準正規分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \;\cdots\;N(0,1)\;(標準正規分布) \\g_Y(y)&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \;\cdots\;N(0,1)\;(標準正規分布) \\p_{X/Y}(z)&=&\int \int \delta(z-\frac{x}{y})f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(z-\frac{x}{y})\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\mathrm{d}x\mathrm{d}y \\&=&\int_{-\infty}^{\infty} \left|y\right| \frac{1}{\sqrt{2\pi}}e^{-\frac{\left(yz\right)^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;z=\frac{x}{y},x=yz \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} \left|y\right| e^{-\frac{\left(yz\right)^2}{2}}e^{-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} \left|y\right| e^{-\frac{\left(yz\right)^2}{2}-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} \left|y\right| e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}\left\{ \int_{-\infty}^{0} \left|y\right| e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y + \int_{0}^{\infty} \left|y\right| e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y \right\} \\&&\;\cdots\;\int_{A}^{B} f(x)\mathrm{d}x=\int_{A}^{C} f(x)\mathrm{d}x+\int_{C}^{B} f(x)\mathrm{d}x\;(ただしA\leq C \leq B) \\&=&\frac{1}{2\pi}\left\{ \int_{-\infty}^{0} -y e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y + \int_{0}^{\infty} y e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y \right\} \;\cdots\;\left|A\right|=\left\{ \begin{array} \\A&(A\geq0) \\-A&(A\lt0) \end{array} \right. \\&=&\frac{1}{2\pi}2\int_{0}^{\infty} y e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y \\&&\;\cdots\;y=xが奇凾数でy=e^{-\frac{x^2\left(z^2+1\right)}{2}}が偶凾数なのでその積は奇凾数. \\&&\;\cdots\;負の区間(-\infty,0)と正の区間(0,\infty)で符号が逆なので全体としては偶凾数と同じに扱える. \\&&\;\cdots\;-f_{奇凾数(-\infty,0)}(x) + f_{奇凾数(0,\infty)}(x) = f_{偶凾数}(x)\mathrm{d}x \\&&\;\cdots\;偶凾数の(-a,a)での積分は正の区間(0,a)の2倍として計算できる. \\&&\;\cdots\;\int_{-a}^a f_{奇凾数}(x) \mathrm{d}x=2\int_0^a f_{偶凾数}(x)\mathrm{d}x \\&=&\frac{1}{\pi}\int_{0}^{\infty} y e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y \\&=&\frac{1}{\pi}\int_{0}^{\infty} \color{red}{y} e^{-t}\cdot\frac{1}{\left(z^2+1\right)\color{red}{y}}\mathrm{d}t \\&&\;\cdots\;t=\frac{y^2\left(z^2+1\right)}{2},\frac{\mathrm{d}t}{\mathrm{d}y}=\frac{\left(z^2+1\right)}{2}2y=\left(z^2+1\right)y,\mathrm{d}y=\frac{1}{\left(z^2+1\right)y}\mathrm{d}t \\&=&\frac{1}{\pi}\frac{1}{\left(z^2+1\right)}\int_{0}^{\infty} e^{-t} \mathrm{d}t \;\cdots\;\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x \\&=&\frac{1}{\pi}\frac{1}{\left(z^2+1\right)}\cdot1 \;\cdots\;\int_{0}^{\infty} e^{-t} \mathrm{d}t=\left[-e^{-t}\right]_{0}^{\infty}=\left[\left(-e^{-\infty}\right)-\left(-e^0\right)\right]=\left[0+1\right]=1 \\&=&\frac{1}{\pi}\frac{1}{\left(z^2+1\right)}\;\cdots\;標準コーシー分布に等しい. \\f(x;x_0, \gamma)&=&\frac{1}{\pi}\frac{\gamma}{(x-x_0)^2+\gamma^2}\;\cdots\;コーシー分布(Cauchy\;distribution)の確率密度凾数 \\f(x;0, 1)&=&\frac{1}{\pi}\frac{1}{(x-0)^2+1^2}\;\cdots\;標準コーシー分布の確率密度凾数 \\&=&\frac{1}{\pi}\frac{1}{x^2+1} \end{eqnarray} $$

指数分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\lambda e^{-\lambda x} (x \geq 0, \lambda \gt 0) \\g_Y(y)&=&\lambda e^{-\lambda y} (x \geq 0, \lambda \gt 0) \\p_{X/Y}(z)&=&\int \int \delta(x-\frac{z}{y})f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(x-\frac{z}{y})\lambda e^{-\lambda x} \lambda e^{-\lambda y} \mathrm{d}x\mathrm{d}y \\&=&\int_0^{\infty} \left|y\right|\lambda e^{-\lambda yz} \lambda e^{-\lambda y}\mathrm{d}y \;\cdots\;z=\frac{x}{y},x=yz \\&=&\lambda^2 \int_0^{\infty} \left|y\right|e^{-\lambda yz} e^{-\lambda y}\mathrm{d}y \;\cdots\;\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x \\&=&\lambda^2 \int_0^{\infty} \left|y\right|e^{-\lambda yz-\lambda y}\mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\lambda^2 \int_0^{\infty} \left|y\right|e^{-\lambda y\left(z+1\right)}\mathrm{d}y \\&=&\lambda^2 \int_0^{\infty} ye^{-\lambda y\left(z+1\right)}\mathrm{d}y \;\cdots\;y \gt 0\;(積分範囲より) \\&=&\lambda^2 \int_0^{\infty} \frac{t}{\lambda\left(z+1\right)}e^{-t}\frac{1}{\lambda\left(z+1\right)}\mathrm{d}t \\&&\;\cdots\;t=\lambda\left(z+1\right)y, y=\frac{t}{\lambda\left(z+1\right)} \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}y}=\lambda\left(z+1\right),\mathrm{d}y=\frac{1}{\lambda\left(z+1\right)}\mathrm{d}t \\&=&\lambda^2 \left\{\frac{1}{\lambda\left(z+1\right)}\right\}^2\int_0^{\infty} te^{-t}\mathrm{d}t \;\cdots\;\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x \\&=&\frac{1}{\left(z+1\right)^2}\int_0^{\infty} te^{-t}\mathrm{d}t \\&=&\frac{1}{\left(z+1\right)^2}\int_0^{\infty} t\left(-e^{-t}\right)^\prime\mathrm{d}t \;\cdots\;\left(-e^{-t}\right)^\prime=e^{-t} \\&=&\frac{1}{\left(z+1\right)^2}\left\{\left[-te^{-t}\right]_0^{\infty}-\int_0^{\infty} -e^{-t}\mathrm{d}t\right\} \;\cdots\;\int_a^b f^\prime(x)g(x) \mathrm{d}x=\left[f(x)g(x)\right]_a^b -\int_a^b f(x)g^\prime(x)\mathrm{d}x \\&=&\frac{1}{\left(z+1\right)^2}\left\{\left[-(\infty)e^{-\infty}-\left(-0e^{-0}\right)\right]+\int_0^{\infty} e^{-t}\mathrm{d}t\right\} \\&=&\frac{1}{\left(z+1\right)^2}\left(0+1\right) \;\cdots\;\int_{0}^{\infty} e^{-t} \mathrm{d}t=\left[-e^{-t}\right]_{0}^{\infty}=\left[\left(-e^{-\infty}\right)-\left(-e^0\right)\right]=\left[0+1\right]=1 \\&=&\frac{1}{\left(z+1\right)^2} \end{eqnarray} $$