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