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