確率変数の変数変換 Z=c1X+c2Y / 正規分布の定数倍同士の和

確率変数の変数変換 \(Z=c_1X+c_2Y\) / 正規分布の定数倍同士の和

確率変数の変数変換 \(Z=c_1X+c_2Y\)

$$ \begin{eqnarray} p_{c_1X+c_2Y}(z) &=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \delta(z-(c_1x+c_2y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/delta-function.html}{\int_{-\infty}^{\infty}\delta(x)f(x)\mathrm{d}x=f(0)} \\&&\;\cdots\;X=x,Y=yの同時確率f(x)g(x)のうちz-\left(c_1x+c_2y\right)=0を満たすものだけを足し合わせる. \\&=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \delta(z-(c_1x+c_2y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/delta-function.html}{\delta(u(x))=\sum_{\alpha\in u^{-1}(0)}\frac{1}{\left|u^{\prime}(\alpha)\right|}\delta\left(x-\alpha\right)} \\&&\;\cdots\;u(x)=z-(c_1x+c_2y) \\&&\;\cdots\;u(x=\alpha)=0,\;\alpha=\frac{z-c_2y}{c_1} \\&&\;\cdots\;u^\prime=\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\left(z-(c_1x+c_2y)\right)=-c_1 \\&&\;\cdots\;\delta\left(z-(c_1x+c_2y)\right) =\frac{1}{|u^\prime(\alpha)|}\delta\left(x-\alpha\right) =\frac{1}{\left|-c_1\right|}\delta\left(x-\frac{z-c_2y}{c_1}\right) =\frac{1}{\left|c_1\right|}\delta\left(x-\frac{z-c_2y}{c_1}\right) \\&=&\int_{-\infty}^{\infty}\frac{1}{\left|c_1\right|}\delta\left(x-\frac{z-c_2y}{c_1}\right)f_X\left(x\right)g_Y(y)\mathrm{d}y \\&=&\int_{-\infty}^{\infty}\frac{1}{\left|c_1\right|}\delta\left(\frac{z-c_2y}{c_1}-\frac{z-c_2y}{c_1}\right)f_X\left(\frac{z-c_2y}{c_1}\right)g_Y(y)\mathrm{d}y \;\cdots\;z=c_1x+c_2y,\;x=\frac{z-c_2y}{c_1} \\&=&\frac{1}{\left|c_1\right|}\int_{-\infty}^{\infty}\delta(0)f_X\left(\frac{z-c_2y}{c_1}\right)g_Y(y)\mathrm{d}y \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{\left|c_1\right|}\int_{-\infty}^{\infty}f_X\left(\frac{z-c_2y}{c_1}\right)g_Y(y)\mathrm{d}y \end{eqnarray} $$

正規分布の定数倍同士の和の例

$$ \begin{eqnarray} f_X(x)&=&\frac{1}{\sqrt{2\pi\sigma_1^2}}\mathrm{e}^{\frac{-(x-\mu_1)^2}{2\sigma_1^2}} \;\cdots\;\mathrm{N}(\mu_1,\sigma_1^2)(\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\g_Y(y)&=&\frac{1}{\sqrt{2\pi\sigma_2^2}}\mathrm{e}^{\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \;\cdots\;\mathrm{N}(\mu_2,\sigma_2^2)(\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\p_{c_1X+c_2Y}(z) &=&\frac{1}{\left|c_1\right|}\int_{-\infty}^{\infty}f_X\left(\frac{z-c_2y}{c_1}\right)g_Y(y)\mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma_1^2}}\mathrm{e}^{\frac{-\left\{\left(\frac{z-c_2y}{c_1}\right)-\mu_1\right\}^2}{2\sigma_1^2}}\frac{1}{\sqrt{2\pi\sigma_2^2}}\mathrm{e}^{\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{\sqrt{2\pi\sigma_1^2}}\frac{1}{\sqrt{2\pi\sigma_2^2}}\int_{-\infty}^{\infty}\mathrm{e}^{\frac{-\left\{\left(\frac{z-c_2y}{c_1}\right)-\mu_1\right\}^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}}\int_{-\infty}^{\infty}\mathrm{e}^{\frac{-\left\{\left(\frac{z-c_2y}{c_1}\right)-\mu_1\right\}^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{f_1(c_1,c_2,y,z,\mu_1,\mu_2,\sigma_1,\sigma_2)} \mathrm{d}y \;\cdots\;f_1はyが引数にある． \end{eqnarray} $$

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$$ \begin{eqnarray} f_1(c_1,c_2,y,z,\mu_1,\mu_2,\sigma_1,\sigma_2) &=&\frac{-\left\{\left(\frac{z-c_2y}{c_1}\right)-\mu_1\right\}^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2} \\&=&-\frac{1}{2} \left\{ \frac{\left(\frac{1}{c_1}z-\frac{c_2}{c_1}y-\mu_1\right)^2}{\sigma_1^2} +\frac{(y-\mu_2)^2}{\sigma_2^2} \right\} \\&=&-\frac{1}{2} \left\{ \frac{\sigma_2^2\left(\frac{1}{c_1}z-\frac{c_2}{c_1}y-\mu_1\right)^2 +\sigma_1^2(y-\mu_2)^2}{\sigma_1^2\sigma_2^2} \right\} \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \sigma_2^2\left(\frac{1}{c_1}z-\frac{c_2}{c_1}y-\mu_1\right)^2 +\sigma_1^2(y-\mu_2)^2 \right\} \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left[ \sigma_2^2\left\{\left(\frac{1}{c_1}z\right)^2-2\left(\frac{1}{c_1}z\right)\left(\frac{c_2}{c_1}y\right)-2\left(\frac{1}{c_1}z\right)\mu_1+\left(\frac{c_2}{c_1}y\right)^2+2\left(\frac{c_2}{c_1}y\right)\mu_1+\mu_1^2\right\} +\sigma_1^2(y^2-2y\mu_2+\mu_2^2) \right] \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \frac{1}{c_1^2}\sigma_2^2z^2 -2\frac{c_2}{c_1^2}\sigma_2^2zy -2\frac{1}{c_1}\sigma_2^2z\mu_1 +\frac{c_2^2}{c_1^2}\sigma_2^2y^2 +2\frac{c_2}{c_1}\sigma_2^2y\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2y^2 -2\sigma_1^2y\mu_2 +\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{c_1^2} \left\{ \sigma_2^2z^2 -2c_2\sigma_2^2zy -2c_1\sigma_2^2z\mu_1 +c_2^2\sigma_2^2y^2 +2c_1c_2\sigma_2^2y\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2y^2 -2c_1^2\sigma_1^2y\mu_2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right)y^2 +\left( -2c_2\sigma_2^2z +2c_1c_2\sigma_2^2\mu_1 -2c_1^2\sigma_1^2\mu_2 \right)y +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right)y^2 -2\left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)y +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right) \left( y^2 -2\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2}y \right) +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=& \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right) \left( y^2 -2\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2}y \right) +A-A +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&&\;\cdots\;平方完成させるための項Aを考える \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right) \left( y^2 -2\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2}y +\frac{1}{c_1^2\sigma_1^2 +c_2^2\sigma_2^2}A \right) -A +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left[ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right) \left\{ y^2 -2\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2}y +\frac{1}{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \right\} -\frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right] \\&&\;\cdots\;A=\frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 } { c_1^2\sigma_1^2 +c_2^2\sigma_2^2 } \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right) \left( y^2 -\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \right)^2 -\frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=& -\frac{\left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right)}{2c_1^2\sigma_1^2\sigma_2^2} \left( y^2 -\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \right)^2 -\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ -\frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=& -f_2(c_1,c_2,\sigma_1^2,\sigma_2^2) \left\{ y-f_3(c_1,c_2,z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2) \right\}^2 -f_4(c_1,c_2,z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2) \;\cdots\;f_2,f_3,f_4はyを引数としない. \\&=& -f_2\left(y-f_3\right)^2-f_4 \;\cdots\;yが平方完成の中の一つだけになっている. \\ \\f_2(c_1,c_2,\sigma_1^2,\sigma_2^2)&=& \frac{\left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right)}{2c_1^2\sigma_1^2\sigma_2^2} \\ \\f_3(c_1,c_2,z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2)&=&\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \\ \\f_4(c_1,c_2,z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2) &=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ -\frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \frac{ -\left( c_2\sigma_2^2 \left(z -c_1\mu_1 \right) +c_1^2\sigma_1^2\mu_2 \right)^2 +\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right) \left(\sigma_2^2\left( z-c_1\mu_1 \right)^2 +c_1^2\sigma_1^2\mu_2^2\right) }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \right\} \\&=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \frac{ -\left( c_2\sigma_2^2 B +c_1^2\sigma_1^2\mu_2 \right)^2 +\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right) \left(\sigma_2^2B^2 +c_1^2\sigma_1^2\mu_2^2\right) }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \;\cdots\;B=z-c_1\mu_1 \\&=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2(c_1^2\sigma_1^2 +c_2^2\sigma_2^2)} \left\{ -\left( c_2\sigma_2^2 B +c_1^2\sigma_1^2\mu_2 \right)^2 +\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right) \left(\sigma_2^2B^2 +c_1^2\sigma_1^2\mu_2^2\right) \right\} \\&=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2(c_1^2\sigma_1^2 +c_2^2\sigma_2^2)} \left( \color{red}{-c_2^2\sigma_2^4B^2} \color{black}{-2c_1^2c_2\sigma_1^2\sigma_2^2\mu_2B} \color{green}{-c_1^4\sigma_1^4\mu_2^2} \color{black}{+c_1^2\sigma_1^2\sigma_2^2B^2} \color{red}{+c_2^2\sigma_2^4B^2} \color{green}{+c_1^4\sigma_1^4\mu_2^2} \color{black}{+c_1^2c_2^2\sigma_1^2\sigma_2^2\mu_2^2} \right) \\&=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2(c_1^2\sigma_1^2 +c_2^2\sigma_2^2)} c_1^2\sigma_1^2 \sigma_2^2\left\{ -2 c_2 \mu_2 B + B^2 + c_2^2 \mu_2^2 \right\} \\&=&\frac{1}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}\left(B-c_2\mu_2\right)^2 \\&=&\frac{\left(z-c_1\mu_1-c_2\mu_2\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)} \;\cdots\;B=z-c_1\mu_1 \\&=&\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)} \end{eqnarray} $$

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$$ \begin{eqnarray} p_{c_1X+c_2Y}(z)&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{f_1(c_1,c_2,y,z,\mu_1,\mu_2,\sigma_1,\sigma_2)} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{-f_2\left(y-f_3\right)^2-f_4} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{-f_2\left(y-f_3\right)^2} \mathrm{e}^{-f_4} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \int_{-\infty}^{\infty} \mathrm{e}^{-f_2\left(y-f_3\right)^2} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \int_{-\infty}^{\infty} \mathrm{e}^{-u^2} \frac{1}{\sqrt{f_2}}\mathrm{d}u \\&&\;\cdots\;u=\sqrt{f_2}\left(y-f_3\right),\;\frac{\mathrm{d}u}{\mathrm{d}y}=\sqrt{f_2},\;\mathrm{d}y=\frac{1}{\sqrt{f_2}}\mathrm{d}u \\&&\;\cdots\;y:-\infty \rightarrow \infty,\;u:-\infty \rightarrow \infty \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \frac{1}{\sqrt{f_2}} \int_{-\infty}^{\infty} \mathrm{e}^{-u^2} \mathrm{d}u \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \frac{1}{\sqrt{f_2}} \sqrt{\pi} \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/gaussian-integral.html}{\int_{-\infty}^{\infty}\mathrm{e}^{-u^2}\mathrm{d}u=\sqrt{\pi}} \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \frac{\sqrt{\pi}}{\sqrt{f_2}} \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \sqrt{\frac{\pi}{f_2}} \mathrm{e}^{-f_4} \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \sqrt{\frac{\pi}{\frac{\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right)}{2c_1^2\sigma_1^2\sigma_2^2}}} \mathrm{e}^{-\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}} \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \sqrt{\frac{2\pi c_1^2\sigma_1^2\sigma_2^2}{\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right)}} \mathrm{e}^{-\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}} \\&=&\frac{1}{\left|c_1\right|}\frac{\sqrt{c_1^2}}{\sqrt{2\pi\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right)}} \mathrm{e}^{-\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}} \\&=&\frac{1}{\left|c_1\right|}\frac{\left|c_1\right|}{\sqrt{2\pi\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right)}} \mathrm{e}^{-\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}} \;\cdots\;\sqrt{A^2}= \left\{ \begin{array} \\A&(A\geq0) \\-A&(A\lt0) \end{array} \right. =\left|A\right| \\&=&\frac{1}{\sqrt{2\pi\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right)}} \mathrm{e}^{-\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}} \\&\sim&\mathrm{N}\left(c_1\mu_1+c_2\mu_2,\;c_1^2\sigma_1^2+c_2^2\sigma_2^2\right) \end{eqnarray} $$

確率変数の変数変換 Z=X+Y その2 / 正規分布の再生性

確率変数の変数変換 Z=X+Y その2 / 正規分布の再生性

Z=X+Y 正規分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\frac{1}{\sqrt{2\pi\sigma_1^2}}\mathrm{e}^{\frac{-(x-\mu_1)^2}{2\sigma_1^2}} \;\cdots\;\mathrm{N}(\mu_1,\sigma_1^2)(\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\g_Y(x)&=&\frac{1}{\sqrt{2\pi\sigma_2^2}}\mathrm{e}^{\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \;\cdots\;\mathrm{N}(\mu_2,\sigma_2^2)(\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\p_{X+Y}(z)&=&\href{https://shikitenkai.blogspot.com/2020/09/zxy.html}{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \delta(z-(x+y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y} \\&=&\int_{-\infty}^{\infty}\delta(z-((z-y)+y))f_X(z-y)g_Y(y)\mathrm{d}y \;\cdots\;z=x+y,\;x=z-y \\&=&\int_{-\infty}^{\infty}\delta(0)f_X(z-y)g_Y(y)\mathrm{d}y \\&=&\int_{-\infty}^{\infty}f_X(z-y)g_Y(y)\mathrm{d}y \\&=&\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma_1^2}}\mathrm{e}^{\frac{-(z-y-\mu_1)^2}{2\sigma_1^2}}\frac{1}{\sqrt{2\pi\sigma_2^2}}\mathrm{e}^{\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \\&=&\frac{1}{\sqrt{2\pi\sigma_1^2}}\frac{1}{\sqrt{2\pi\sigma_2^2}}\int_{-\infty}^{\infty}\mathrm{e}^{\frac{-(z-y-\mu_1)^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \\&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}}\int_{-\infty}^{\infty}\mathrm{e}^{\frac{-(z-y-\mu_1)^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \\&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{f_1(y,z,\mu_1,\mu_2,\sigma_1,\sigma_2)} \mathrm{d}y \;\cdots\;f_1はyが引数にある． \end{eqnarray} $$

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$$ \begin{eqnarray} f_1(y,z,\mu_1,\mu_2,\sigma_1,\sigma_2) &=&\frac{-(z-y-\mu_1)^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2} \\&=&-\frac{1}{2} \left( \frac{(z-y-\mu_1)^2}{\sigma_1^2} +\frac{(y-\mu_2)^2}{\sigma_2^2} \right) \\&=&-\frac{1}{2} \left( \frac{\sigma_2^2(z-y-\mu_1)^2 +\sigma_1^2(y-\mu_2)^2}{\sigma_1^2\sigma_2^2} \right) \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left( \sigma_2^2(z-y-\mu_1)^2 +\sigma_1^2(y-\mu_2)^2 \right) \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \sigma_2^2(z^2-2zy-2z\mu_1+y^2+2y\mu_1+\mu_1^2) +\sigma_1^2(y^2-2y\mu_2+\mu_2^2) \right\} \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \sigma_2^2z^2 -2\sigma_2^2zy -2\sigma_2^2z\mu_1 +\sigma_2^2y^2 +2\sigma_2^2y\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2y^2 -2\sigma_1^2y\mu_2 +\sigma_1^2\mu_2^2 \right\} \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \left( \sigma_1^2 +\sigma_2^2 \right)y^2 +\left( -2\sigma_2^2z +2\sigma_2^2\mu_1 -2\sigma_1^2\mu_2 \right)y +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \left( \sigma_1^2 +\sigma_2^2 \right)y^2 -2\left( \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 \right)y +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \left( \sigma_1^2 +\sigma_2^2 \right) \left(y^2 -2\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2} y \right) +A -A +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&&\;\cdots\;平方完成させるための項Aを考える \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \left( \sigma_1^2 +\sigma_2^2 \right) \left(y^2 -2\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2} y +\frac{1}{\sigma_1^2+\sigma_2^2}A \right) -A +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left[ \left( \sigma_1^2 +\sigma_2^2 \right) \left\{y^2 -2\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2} y +\frac{1}{\sigma_1^2+\sigma_2^2} \frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2 +\sigma_2^2} \right\}-\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2 +\sigma_2^2} +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right] \\&&\;\cdots\;A=\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2 +\sigma_2^2} \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \left( \sigma_1^2 +\sigma_2^2 \right) \left(y-\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2} \right)^2 -\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2 +\sigma_2^2} +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{\sigma_1^2+\sigma_2^2}{2\sigma_1^2\sigma_2^2} \left(y-\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2} \right)^2 -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ -\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2+\sigma_2^2} +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=& -f_2(\sigma_1^2,\sigma_2^2) \left\{ y-f_3(z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2) \right\}^2 -f_4(z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2) \;\cdots\;f_2,f_3,f_4はyを引数としない. \\&=& -f_2\left(y-f_3\right)^2-f_4 \;\cdots\;yが平方完成の中の一つだけになっている. \\ \\f_2(\sigma_1^2,\sigma_2^2)&=& \frac{\sigma_1^2+\sigma_2^2}{2\sigma_1^2\sigma_2^2} \\ \\f_3(z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2)&=&\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2 } \\ \\f_4(z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2)&=&\frac{1}{2\sigma_1^2\sigma_2^2}\left\{ -\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2 +\sigma_2^2} +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=&\frac{1}{2\sigma_1^2\sigma_2^2}\left\{ -\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 +(\sigma_1^2+\sigma_2^2)(\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2) }{\sigma_1^2+\sigma_2^2} \right\} \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{\sigma_1^2+\sigma_2^2} \left\{ -(\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 +(\sigma_1^2+\sigma_2^2)(\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2) \right\} \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{\sigma_1^2+\sigma_2^2} \left\{ -\sigma_2^4z^2 -\sigma_2^4\mu_1^2 -\sigma_1^4\mu_2^2 +2\sigma_2^4z\mu_1 -2\sigma_1^2\sigma_2^2z\mu_2 +2\sigma_1^2\sigma_2^2\mu_1\mu_2 +(\sigma_1^2+\sigma_2^2)\sigma_2^2z^2 -2(\sigma_1^2+\sigma_2^2)\sigma_2^2z\mu_1 +(\sigma_1^2+\sigma_2^2)\sigma_2^2\mu_1^2 +(\sigma_1^2+\sigma_2^2)\sigma_1^2\mu_2^2 \right\} \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{\sigma_1^2+\sigma_2^2} \left( \color{red }{-\sigma_2^4 z^2} \color{green}{-\sigma_2^4 \mu_1^2} \color{blue }{-\sigma_1^4 \mu_2^2} \color{cyan }{+2\sigma_2^4 z\mu_1} \color{black}{-2\sigma_1^2 \sigma_2^2z\mu_2} \color{black}{+2\sigma_1^2 \sigma_2^2\mu_1\mu_2} \color{black}{+ \sigma_1^2 \sigma_2^2 z^2} \color{red }{+ \sigma_2^4 z^2} \color{black}{-2\sigma_1^2 \sigma_2^2 z\mu_1} \color{cyan}{-2\sigma_2^4 z\mu_1} \color{black}{+ \sigma_1^2 \sigma_2^2 \mu_1^2} \color{green}{+ \sigma_2^4 \mu_1^2} \color{blue }{+ \sigma_1^4 \mu_2^2} \color{black}{+ \sigma_1^2 \sigma_2^2 \mu_2^2} \right) \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{\sigma_1^2+\sigma_2^2} \left( \color{black}{-2\sigma_1^2 \sigma_2^2z\mu_2} \color{black}{+2\sigma_1^2 \sigma_2^2\mu_1\mu_2} \color{black}{+ \sigma_1^2 \sigma_2^2 z^2} \color{black}{-2\sigma_1^2 \sigma_2^2 z\mu_1} \color{black}{+ \sigma_1^2 \sigma_2^2 \mu_1^2} \color{black}{+ \sigma_1^2 \sigma_2^2 \mu_2^2} \right) \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{\sigma_1^2+\sigma_2^2} \left(\sigma_1^2 \sigma_2^2\right) \left( \color{black}{-2z\mu_2} \color{black}{+2\mu_1\mu_2} \color{black}{+ z^2} \color{black}{-2z\mu_1} \color{black}{+ \mu_1^2} \color{black}{+ \mu_2^2} \right) \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2} \left(z-\mu_1-\mu_2\right)^2 \\&=&\frac{\left(z-\mu_1-\mu_2\right)^2}{2\left(\sigma_1^2+\sigma_2^2\right)} \\&=&\frac{\left\{z-\left(\mu_1+\mu_2\right)\right\}^2}{2\left(\sigma_1^2+\sigma_2^2\right)} \end{eqnarray} $$

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$$ \begin{eqnarray} p_{X+Y}(z)&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{f_1(y,z,\mu_1,\mu_2,\sigma_1,\sigma_2)} \mathrm{d}y \\&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty}\mathrm{e}^{ -f_2 \left( y-f_3 \right)^2 -f_4} \mathrm{d}y \\&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty}\mathrm{e}^{ -f_2 \left( y-f_3 \right)^2 } \mathrm{e}^{-f_4} \mathrm{d}y \;\cdots\;A^{B+C}=A^BA^C \\&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \int_{-\infty}^{\infty}\mathrm{e}^{ -f_2 \left( y-f_3 \right)^2 } \mathrm{d}y \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \int_{-\infty}^{\infty}\mathrm{e}^{-f_2\left(y-f_3\right)^2} \mathrm{d}y \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \int_{-\infty}^{\infty}\mathrm{e}^{-u^2} \left(\frac{1}{\sqrt{f_2}}\right)\mathrm{d}u \\&&\;\cdots\;u=\sqrt{f_2}\left(y-f_3\right),\;\frac{\mathrm{d}u}{\mathrm{d}y}=\sqrt{f_2},\;\mathrm{d}y=\frac{1}{\sqrt{f_2}}\mathrm{d}u, \\&&\;\cdots\;y:-\infty \rightarrow \infty,\;u:-\infty \rightarrow \infty \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \frac{1}{\sqrt{f_2}} \int_{-\infty}^{\infty}\mathrm{e}^{-u^2}\mathrm{d}u \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \frac{1}{\sqrt{f_2}} \mathrm{e}^{-f_4} \sqrt{\pi} \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/gaussian-integral.html}{\int_{-\infty}^{\infty}\mathrm{e}^{-u^2}\mathrm{d}u=\sqrt{\pi}} \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \frac{1}{\sqrt{\frac{\sigma_1^2+\sigma_2^2}{2\sigma_1^2\sigma_2^2}}} \mathrm{e}^{-\frac{\left\{z-\left(\mu_1+\mu_2\right)\right\}^2}{2\left(\sigma_1^2+\sigma_2^2\right)}} \sqrt{\pi} \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \sqrt{\frac{2\pi\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2}} \mathrm{e}^{-\frac{\left\{z-\left(\mu_1+\mu_2\right)\right\}^2}{2\left(\sigma_1^2+\sigma_2^2\right)}} \\&=& \frac{1}{\sqrt{2\pi\left(\sigma_1^2+\sigma_2^2\right)}} \mathrm{e}^{-\frac{\left\{z-\left(\mu_1+\mu_2\right)\right\}^2}{2\left(\sigma_1^2+\sigma_2^2\right)}} \;\cdots\;\mathrm{N}(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2) (再生性) \end{eqnarray} $$ 再生性は，同じ確率分布族に含まれる確率分布\(F_1\),\(F_2\)に対して \(X_1\sim F_1\),\(X_2\sim F_2\)とする互いに独立な確率変数に対して， \(X_1+X_2\)がやはり同一の確率分布族に含まれる性質．

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} $$

確率変数の変数変換 Z=XY

確率変数の変数変換 Z=XY

Z=XY

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

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_{XY}(z)&=&\int \int \delta(z-xy)f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(z-xy)\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} \frac{1}{|y|} \frac{1}{\sqrt{2\pi}}e^{-\frac{\left(\frac{z}{y}\right)^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;z=xy,x=\frac{z}{y} \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{|y|} e^{-\frac{\left(\frac{z}{y}\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} \frac{1}{|y|} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \mathrm{d}y \\&=&\frac{1}{2\pi}\left\{ \int_{-\infty}^{0} \frac{1}{|y|} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \mathrm{d}y + \int_{0}^{\infty} \frac{1}{|y|} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \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} -\frac{1}{y} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \mathrm{d}y + \int_{0}^{\infty} \frac{1}{y} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \mathrm{d}y \right\} \\&&\;\cdots\;\frac{1}{|y|}=-\frac{1}{y}\;(y\lt0),\frac{1}{|y|}=\frac{1}{y}\;(0\lt y) \\&=&\frac{1}{2\pi}2\int_{0}^{\infty} \frac{1}{y} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \mathrm{d}y \\&&\;\cdots\;y=\frac{1}{x}が奇凾数でy=e^{-\frac{1}{2}\left(\frac{z^2}{x^2}+x^2\right)}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_{条件付きで偶凾数}(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} \color{red}{\frac{1}{y}} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \color{red}{\mathrm{d}y} \\&=&\frac{1}{\pi} \int_{0}^{\infty} e^{-\frac{1}{2}\left\{\frac{z^2}{y^2}+y^2\right\}} \frac{\mathrm{d}t}{2t} \\&&\;\cdots\;t=\frac{y^2}{2},y^2=2t,\;y:0\rightarrow\infty,t:0\rightarrow\infty \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}y}=y=\frac{y^2}{y}=\frac{2t}{y},\frac{\mathrm{d}y}{y}=\frac{\mathrm{d}t}{2t} \\&=&\frac{1}{\pi}\frac{1}{2}\int_{0}^{\infty} \frac{1}{t} e^{-\frac{1}{2}\left\{\frac{z^2}{2t}+2t\right\}} \mathrm{d}t \\&=&\frac{1}{2\pi}\int_{0}^{\infty} \frac{1}{t} e^{-t-\frac{z^2}{4t}} \mathrm{d}t \\&&\;\cdots\;K_n(z)=\frac{1}{2}\left(\frac{z}{2}\right)^n\int_0^{\infty} t^{-n-1} e^{-t-\frac{z^2}{4t}} \mathrm{d}t\;(第二種変形ベッセル凾数) \\&&\;\cdots\;K_0(z)=\frac{1}{2}\int_0^{\infty} t^{-1} e^{-t-\frac{z^2}{4t}} \mathrm{d}t \\&=&\frac{1}{\pi}K_0(z) \end{eqnarray} $$

指数分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\lambda e^{-\lambda x} (x \geq 0, \lambda \gt 0) \\g_Y(y)&=&\lambda e^{-\lambda y} (x \geq 0, \lambda \gt 0) \\p_{XY}(z)&=&\int \int \delta(x-xy)f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(x-xy)\lambda e^{-\lambda x} \lambda e^{-\lambda y} \mathrm{d}x\mathrm{d}y \\&=&\int_0^{\infty} \frac{1}{|y|}\lambda e^{-\lambda \left(\frac{z}{y}\right)} \lambda e^{-\lambda y}\mathrm{d}y \;\cdots\;z=xy,x=\frac{z}{y} \\&=&\lambda^2\int_0^{\infty} \frac{1}{|y|} e^{-\lambda \left(\frac{z}{y}\right)} e^{-\lambda y} \mathrm{d}y \;\cdots\;\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x \\&=&\lambda^2\int_0^{\infty} \frac{1}{|y|} e^{-\lambda \left(\frac{z}{y}\right)-\lambda y} \mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\lambda^2\int_0^{\infty} \frac{1}{|y|} e^{-\lambda \left(\frac{z}{y}+y\right)} \mathrm{d}y \\&=&\lambda^2\int_0^{\infty} \frac{1}{|\frac{t}{\lambda}|} e^{-\lambda \left(\frac{z}{\frac{t}{\lambda}}+\frac{t}{\lambda}\right)} \frac{1}{\lambda}\mathrm{d}t \\&&\;\cdots\;t=\lambda y, y=\frac{t}{\lambda}, \frac{\mathrm{d}y}{\mathrm{d}t}=\frac{1}{\lambda},\mathrm{d}y=\frac{1}{\lambda}\mathrm{d}t,\;y:0\rightarrow\infty, t:0\rightarrow\infty \\&=&\lambda^2\int_0^{\infty} \frac{|\lambda|}{|t|} e^{-\lambda^2\frac{z}{t}-t} \frac{1}{\lambda}\mathrm{d}t \\&=&\lambda^2\int_0^{\infty} \frac{|\lambda|}{\lambda}\frac{1}{|t|} e^{-\lambda^2\frac{z}{t}-t} \mathrm{d}t \\&=&\lambda^2\int_0^{\infty} \frac{1}{|t|} e^{-\lambda^2\frac{z}{t}-t} \mathrm{d}t \;\cdots\;\lambda \gt 0 \\&=&\lambda^2\int_0^{\infty} \frac{1}{t} e^{-\lambda^2\frac{z}{t}-t} \mathrm{d}t \;\cdots\;t \gt 0\;(積分範囲より) \\&=&2\lambda^2K_0(2\lambda\sqrt{z}) \\&&\;\cdots\;K_n(z)=\frac{1}{2}\left(\frac{z}{2}\right)^n\int_0^{\infty} t^{-n-1} e^{-t-\frac{z^2}{4t}} \mathrm{d}t\;(第二種変形ベッセル凾数) \\&&\;\cdots\;K_0(z)=\frac{1}{2}\int_0^{\infty} t^{-1} e^{-t-\frac{z^2}{4t}} \mathrm{d}t \end{eqnarray} $$

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

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

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\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\g_Y(y)&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \;\cdots\;N(0,1)\;(標準\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\p_{X+Y}(z)&=&\int\int \delta(z-(x+y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int\int \delta(z-(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} \frac{1}{\sqrt{2\pi}}e^{-\frac{(z-y)^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;z=x+y,x=z-y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{(z-y)^2}{2}}e^{-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\frac{1}{2\pi}\int e^{-\frac{(z-y)^2}{2}-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{(z-y)^2+y^2}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2-2yz+y^2+y^2}{2}}\mathrm{d}y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2-2yz+2y^2}{2}}\mathrm{d}y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2}{2}+yz-y^2}\mathrm{d}y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2}{2}\color{red}{+\frac{z^2}{4}-\frac{z^2}{4}}\color{black}{+yz-y^2}\mathrm{d}y} \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2}{4}-(\frac{z^2}{4}-yz+y^2)}\mathrm{d}y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2}{4}}e^{-(\frac{z^2}{4}-yz+y^2)}\mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}\int_{-\infty}^{\infty} e^{-(\frac{z}{2}-y)^2}\mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}\int_{\infty}^{-\infty} e^{-t^2}(-1)\mathrm{d}t \\&&\;\cdots\;t=\frac{z}{2}-y,\frac{\mathrm{d}t}{\mathrm{d}y}=-1,\mathrm{d}y=-\mathrm{d}t,\;y:-\infty\rightarrow\infty, t:\infty\rightarrow -\infty \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}(-1)\int_{\infty}^{-\infty} e^{-t^2}\mathrm{d}t \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}(-1)\left(-\int_{-\infty}^{\infty} e^{-t^2}\mathrm{d}t\right) \;\cdots\;\int_{A}^{B} f(x)\mathrm{d}x=-\int_{B}^{A} f(x)\mathrm{d}x \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}\int_{-\infty}^{\infty} e^{-t^2}\mathrm{d}t \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}\sqrt{\pi} \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/gaussian-integral.html}{\int_{-\infty}^{\infty} e^{-t^2}\mathrm{d}y=\sqrt{\pi}} \\&=&\frac{1}{2\sqrt{\pi}}e^{-\frac{z^2}{4}} \\&=&\frac{1}{\sqrt{2\pi\cdot2}}e^{-\frac{z^2}{2\cdot2}} \;\cdots\;N(0,2) \end{eqnarray} $$

指数分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\lambda e^{-\lambda x} (x \geq 0, \lambda \gt 0) \;\cdots\;Exp(\lambda)\;(\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布}) \\g_Y(y)&=&\lambda e^{-\lambda y} (x \geq 0, \lambda \gt 0) \;\cdots\;Exp(\lambda)\;(\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布}) \\p_{X+Y}(z)&=&\int \int \delta(z-(x+y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(z-(x+y))\lambda e^{-\lambda x} \lambda e^{-\lambda y} \mathrm{d}x\mathrm{d}y \\&=&\int_0^z \lambda e^{-\lambda (z-y)} \lambda e^{-\lambda y}\mathrm{d}y \;\cdots\;z=x+y,x=z-y \\&=&\lambda^2\int_0^z e^{-\lambda (z-y)} e^{-\lambda y}\mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\lambda^2\int_0^z e^{-\lambda (z-y)-\lambda y}\mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\lambda^2\int_0^z e^{-\lambda (z-y+y)}\mathrm{d}y \\&=&\lambda^2\int_0^z e^{-\lambda z}\mathrm{d}y \\&=&\lambda^2 e^{-\lambda z}\int_0^z\mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\lambda^2 e^{-\lambda z}\left[y\right]_0^z \\&=&\lambda^2 e^{-\lambda z}\left[z-0\right] \\&=&\lambda^2 z e^{-\lambda z} \\&=&\Gamma(2, \lambda) \end{eqnarray} $$ $$ \begin{eqnarray} \Gamma(\alpha, \lambda)&=&\frac{1}{\Gamma(\alpha)}\lambda^{\alpha} x^{\alpha-1} e^{-\lambda x}\;(\href{https://shikitenkai.blogspot.com/2020/05/n.html}{ガンマ分布;\;gamma\;disribution}) \\\Gamma(1, \lambda)&=&\frac{1}{\Gamma(1)}\lambda^1 x^{1-1} e^{-\lambda x}=\frac{1}{1}\lambda e^{-\lambda x}=\lambda e^{-\lambda x} \\&&\;\cdots\;\Gamma(1)=(1-1)!=0!=1 \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布} \\\Gamma(2, \lambda)&=&\frac{1}{\Gamma(2)}\lambda^2 x^{2-1} e^{-\lambda x}=\frac{1}{1}\lambda^2xe^{-\lambda x}=\lambda^2xe^{-\lambda x} \\&&\;\cdots\;\Gamma(2)=(2-1)!=1!=1 \\&&\;\cdots\;指数分布に従う確率変数同士の和の分布,\;\alpha=2のガンマ分布 \end{eqnarray} $$

デルタ凾数(delta function)

デルタ凾数(delta function)

以下の式を満たす\(\delta(x)\)をデルタ凾数と呼ぶ． $$ \begin{eqnarray} \int_{-\infty}^{\infty}\delta(x)f(x)\mathrm{d}x&=&f(0) \end{eqnarray} $$ これは\(f(x)\)の積分の中から\(x=0\;(\delta(0))\)となる時のf(x)のみを取り出すことになる.

上記式を満たすため，\(\delta(x)\)のみでの積分値は1でとされる． $$ \begin{eqnarray} \int_{-\infty}^{\infty}\delta(x)\mathrm{d}x&=&1\;\cdots\;前述の式においてf(x)=1の場合と同じ \end{eqnarray} $$

デルタ凾数の性質

\(\delta(x-\alpha)\)の場合

足し合わせたいxの値を動かす．これは\(f(x)\)の積分の中から\(x=\alpha\;(\delta(\alpha-\alpha)=\delta(0))\)となる時の\(f(\alpha)\)のみを取り出すことになる. $$ \begin{eqnarray} \\\int_{-\infty}^{\infty}\delta(x-\alpha)f(x)\mathrm{d}x &=&\int_{-\infty}^{\infty}\delta(t)f(t+\alpha)\mathrm{d}t \\&&\;\cdots\;t=x-\alpha,x=t+\alpha,\frac{\mathrm{d}x}{\mathrm{d}t}=1,\mathrm{d}x=\mathrm{d}t \\&&\;\cdots\;x:-\infty\rightarrow \infty, t:-\infty\rightarrow \infty \\&=&f(\alpha) \end{eqnarray} $$

\(\delta(\alpha x)\)の場合

\(f(x)\)と\(\delta(x)\)でスケールが\(\alpha\)倍異なる場合の性質を求める． $$ \begin{eqnarray} \int_{-\infty}^{\infty}\color{red}{\delta(\alpha x)}\color{black}{f(x)\mathrm{d}x} &=&\int_{-\infty}^{\infty}\delta(t)f\left(\frac{t}{\alpha}\right)\frac{\mathrm{d}t}{\alpha} \\&&\;\cdots\;t=\alpha x,x=\frac{t}{\alpha}\;\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{\alpha},\mathrm{d}x=\frac{\mathrm{d}t}{\alpha} \\&&\;\cdots\;x:-\infty\rightarrow \infty, t:-\infty\rightarrow \infty \\&=&\begin{cases} \int_{-\infty}^{\infty}\delta(t)f\left(\frac{t}{\alpha}\right)\frac{\mathrm{d}t}{|\alpha|}&(\alpha \gt 0)\;\cdots\;a=|a|\;(a\gt0) \\ \int_{\infty}^{-\infty}\delta(t)f\left(\frac{t}{\alpha}\right)\frac{\mathrm{d}t}{|\alpha|}&(\alpha \lt 0)\;\cdots\;a=-|a|\;(a\lt0) \end{cases} \\&=&\int_{-\infty}^{\infty}\delta(t)f\left(\frac{t}{\alpha}\right)\frac{\mathrm{d}t}{|\alpha|} \\&&\;\cdots\;t=0以外は\delta(t)=0なので積分範囲が逆転しても\delta(t)=0での \\&&\;\cdots\;f(t)の値以外関係なく,\alphaの絶対値をとれば\alphaの正負によらない式にできる． \\&=&\frac{1}{|\alpha|}\int_{-\infty}^{\infty}\delta(t)f\left(\frac{t}{\alpha}\right)\mathrm{d}t \\&=&\frac{1}{|\alpha|}f\left(0\right) \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\int_{-\infty}^{\infty} \color{red}{\frac{1}{|\alpha|}\delta(t)}\color{black}{f\left(x\right)\mathrm{d}t} \;\cdots\;最初の式と比較しやすいように並べ直してみる． \\\delta(\alpha x)&=&\frac{1}{|\alpha|}\delta(t) \end{eqnarray} $$

\(\delta(u(x))\)の場合

足し合わせたいxの値を動かす．\(u(x)=0\)を満たす\(x\)を\(x=\alpha_1,\cdots,\alpha_k,\cdots,\alpha_n\)とするとこれは\(f(x)\)の積分の中から\(x=\alpha_k\;(\delta(\alpha_k-\alpha_k)=\delta(0))\)となる時の\(f(\alpha_k)\)のみを取り出すことになり，積分はその足し合わせとなる.

\(u(x)=0\)の解\(x=\alpha_k\)が区間にただ一つとなるように\(\epsilon\)を十分小さくとるとする． $$ \begin{eqnarray} \int_{\alpha_k-\epsilon}^{\alpha_k+\epsilon}\color{red}{\delta\left(u(x)\right)}\color{black}{f(x)\mathrm{d}x} &=&\int_{\alpha_k-\epsilon}^{\alpha_k+\epsilon}\delta\left(u(\alpha_k)+u^{\prime}(\alpha_k)\left(x-\alpha_k\right)+\frac{1}{2!}u^{\prime\prime}(\alpha_k)\left(x-\alpha_k\right)^2+\cdots\right)f(x)\mathrm{d}x \\&&\;\cdots\;u(x)のx=\alpha周りでのテイラー展開を行った． \\&&\;\cdots\;a周りでのf(x)のテイラー展開: f(x)=f^{}(a)+f^{\prime}(a)\left(x-a\right)+\frac{1}{2}f^{\prime\prime}(a)\left(x-a\right)^2+\cdots \\&=&\int_{\alpha_k-\epsilon}^{\alpha_k+\epsilon}\delta\left(0+u^{\prime}(\alpha_k)\left(x-\alpha_k\right)+0\right)f(x)\mathrm{d}x \\&&\;\cdots\;u(\alpha_k)は\alpha_kの定義より0である． \\&&\;\cdots\;(x-\alpha_k)=\pm\epsilonであり二次以上の項は\epsilon^2以下の数を掛けることになる． \\&&\;\cdots\;これは無視できるほど小さく,0として扱える. \\&=&\int_{\alpha_k-\epsilon}^{\alpha_k+\epsilon}\delta\left(u^{\prime}(\alpha_k)\left(x-\alpha_k\right)\right)f(x)\mathrm{d}x \\&=&\int_{\alpha_k-\epsilon}^{\alpha_k+\epsilon}\color{red}{\frac{1}{\left|u^{\prime}(\alpha_k)\right|}\delta\left(x-\alpha_k\right)}\color{black}{f(x)\mathrm{d}x} \;\cdots\;\delta(\alpha x)=\frac{1}{|\alpha|}\delta(t)(前述) \end{eqnarray} $$ よって全区間では以下のようになる． $$ \begin{eqnarray} \int_{-\infty}^{\infty}\color{red}{\delta\left(u(x)\right)}\color{black}{f(x)\mathrm{d}x} &=&\sum_{k=1}^{n} \int_{\alpha_k-\epsilon}^{\alpha_k+\epsilon}\delta\left(u(x)\right)f(x)\mathrm{d}x \\&=&\sum_{k=1}^{n} \int_{\alpha_k-\epsilon}^{\alpha_k+\epsilon}\frac{1}{\left|u^{\prime}(\alpha_k)\right|}\delta\left(x-\alpha_k\right)f(x)\mathrm{d}x \\&=&\int_{-\infty}^{\infty}\left\{\color{red}{\sum_{k=1}^{n}\frac{1}{\left|u^{\prime}(\alpha_k)\right|}\delta\left(x-\alpha_k\right)}\right\}\color{black}{f(x)\mathrm{d}x} \end{eqnarray} $$ $$ \begin{eqnarray} \\\delta(u(x))&=&\sum_{k=1}^{n}\frac{1}{\left|u^{\prime}(\alpha_k)\right|}\delta\left(x-\alpha_k\right)(ただし\alpha_kはu(x)=0の解) \\&=&\sum_{\alpha\in u^{-1}(0)}\frac{1}{\left|u^{\prime}(\alpha)\right|}\delta\left(x-\alpha\right) \end{eqnarray} $$