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