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
$$ \begin{eqnarray} f_X(x)&=&\lambda e^{-\lambda x} (x \geq 0, \lambda \gt 0) \;\cdots\;Exp(\lambda)\;(\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布}) \\g_Y(y)&=&\lambda e^{-\lambda y} (x \geq 0, \lambda \gt 0) \;\cdots\;Exp(\lambda)\;(\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布}) \\p_{X+Y}(z)&=&\int \int \delta(z-(x+y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(z-(x+y))\lambda e^{-\lambda x} \lambda e^{-\lambda y} \mathrm{d}x\mathrm{d}y \\&=&\int_0^z \lambda e^{-\lambda (z-y)} \lambda e^{-\lambda y}\mathrm{d}y \;\cdots\;z=x+y,x=z-y \\&=&\lambda^2\int_0^z e^{-\lambda (z-y)} e^{-\lambda y}\mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\lambda^2\int_0^z e^{-\lambda (z-y)-\lambda y}\mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\lambda^2\int_0^z e^{-\lambda (z-y+y)}\mathrm{d}y \\&=&\lambda^2\int_0^z e^{-\lambda z}\mathrm{d}y \\&=&\lambda^2 e^{-\lambda z}\int_0^z\mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\lambda^2 e^{-\lambda z}\left[y\right]_0^z \\&=&\lambda^2 e^{-\lambda z}\left[z-0\right] \\&=&\lambda^2 z e^{-\lambda z} \\&=&\Gamma(2, \lambda) \end{eqnarray} $$ $$ \begin{eqnarray} \Gamma(\alpha, \lambda)&=&\frac{1}{\Gamma(\alpha)}\lambda^{\alpha} x^{\alpha-1} e^{-\lambda x}\;(\href{https://shikitenkai.blogspot.com/2020/05/n.html}{ガンマ分布;\;gamma\;disribution}) \\\Gamma(1, \lambda)&=&\frac{1}{\Gamma(1)}\lambda^1 x^{1-1} e^{-\lambda x}=\frac{1}{1}\lambda e^{-\lambda x}=\lambda e^{-\lambda x} \\&&\;\cdots\;\Gamma(1)=(1-1)!=0!=1 \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布} \\\Gamma(2, \lambda)&=&\frac{1}{\Gamma(2)}\lambda^2 x^{2-1} e^{-\lambda x}=\frac{1}{1}\lambda^2xe^{-\lambda x}=\lambda^2xe^{-\lambda x} \\&&\;\cdots\;\Gamma(2)=(2-1)!=1!=1 \\&&\;\cdots\;指数分布に従う確率変数同士の和の分布,\;\alpha=2のガンマ分布 \end{eqnarray} $$n回の事象が発生するまでの間隔の確率密度分布 (ガンマ分布)
\(n\)回の事象が発生するまでの間隔(時間)の確率\(Y_n\)について考える．
$$
\begin{eqnarray}
F_n(t)
&=&P(Y_n\leq t)
\;\cdots\;F_nをY_nの累積分布凾数(cumulative\;distribution\;function, CDF)とする．(少なくともtまでにn回の事象が発生している)\\
&=&P(X_t\geq n)
\;\cdots\;n回以上の事象が発生するのが時刻t以降になる確率とも言える\\
&=&\sum_{x=n}^\infty P_o(x;\lambda t)
\;\cdots\;事象の発生回数の確率を\lambda tをパラメータとするポアソン分布とする．ポアソン分布: P_o(x;\lambda)= \mathrm{e}^{-\lambda} \frac{\left(\lambda\right)^x}{x!}\\
&=&\sum_{x=n}^\infty\mathrm{e}^{-\lambda t} \frac{\left(\lambda t\right)^x}{x!}\;\cdots\;ポアソン分布は期間内で何回起こるか？の確率分布(指数分布は次の発生までの期間の確率分布)\\
&=&1-\sum_{x=0}^{n-1}\mathrm{e}^{-\lambda t} \frac{\left(\lambda t\right)^x}{x!}
\;\cdots\;時刻tまでにn回未満の事象しか発生しない場合の余事象\\
&=&1-\mathrm{e}^{-\lambda t}\left\{
\frac{\left(\lambda t\right)^0}{0!}
+ \frac{\left(\lambda t\right)^1}{1!}
+ \frac{\left(\lambda t\right)^2}{2!}
+ \dots
+ \frac{\left(\lambda t\right)^\left(n-2\right)}{\left(n-2\right)!}
+ \frac{\left(\lambda t\right)^\left(n-1\right)}{\left(n-1\right)!}
\right\}\\
&=&1-\mathrm{e}^{-\lambda t}A
\;\cdots\;A=\left\{
\frac{\left(\lambda t\right)^0}{0!}
+ \frac{\left(\lambda t\right)^1}{1!}
+ \frac{\left(\lambda t\right)^2}{2!}
+ \dots
+ \frac{\left(\lambda t\right)^\left(n-2\right)}{\left(n-2\right)!}
+ \frac{\left(\lambda t\right)^\left(n-1\right)}{\left(n-1\right)!}
\right\}\\
Y_n=\frac{\partial F_n(t)}{\partial t}
&=&0-\left[(\mathrm{e}^{-\lambda t})'A
+\mathrm{e}^{-\lambda t}(A)'
\right]
\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/02/blog-post.html}{(fg)'=f'g+fg'}
\;\cdots\;累積分布凾数の微分は確率密度凾数(probability\;density\;function、PDF)\\
(\mathrm{e}^{-\lambda t})'A
&=&\mathrm{e}^{-\lambda t}(-\lambda)A\\
&=&\mathrm{e}^{-\lambda t}(-\lambda)\left\{
\frac{\left(\lambda t\right)^0}{0!}
+ \frac{\left(\lambda t\right)^1}{1!}
+ \frac{\left(\lambda t\right)^2}{2!}
+ \dots
+ \frac{\left(\lambda t\right)^\left(n-2\right)}{\left(n-2\right)!}
+ \frac{\left(\lambda t\right)^\left(n-1\right)}{\left(n-1\right)!}
\right\}\\
&=&\mathrm{e}^{-\lambda t}\left\{
- \frac{\lambda}{0!}
- \frac{\lambda^2 t}{1!}
- \frac{\lambda^3 t^2}{2!}
- \dots
- \frac{\lambda^\left(n-1\right) t^\left(n-2\right)}{\left(n-2\right)!}
- \frac{\lambda^n t^\left(n-1\right)}{\left(n-1\right)!}
\right\}\\
\\
\mathrm{e}^{-\lambda t}(A)'
&=&\mathrm{e}^{-\lambda t}\left(\left\{
\frac{\left(\lambda t\right)^0}{0!}
+ \frac{\left(\lambda t\right)^1}{1!}
+ \frac{\left(\lambda t\right)^2}{2!}
+ \dots
+ \frac{\left(\lambda t\right)^\left(n-2\right)}{\left(n-2\right)!}
+ \frac{\left(\lambda t\right)^\left(n-1\right)}{\left(n-1\right)!}
\right\}\right)'\\
&=&\mathrm{e}^{-\lambda t}\left\{
\left(\frac{1}{0!}\right)'
+ \left(\frac{\lambda t}{1!}\right)'
+ \left(\frac{\lambda^2 t^2}{2!}\right)'
+ \dots
+ \left(\frac{\lambda^\left(n-2\right) t^\left(n-2\right)}{\left(n-2\right)!}\right)'
+ \left(\frac{\lambda^\left(n-1\right) t^\left(n-1\right)}{\left(n-1\right)!}\right)'
\right\}\\
&=&\mathrm{e}^{-\lambda t}\left\{
0
+ \frac{\lambda}{1!}
+ \frac{2\lambda^2 t}{2!}
+ \dots
+ \frac{(n-2)\lambda^{(n-2)} t^\left(n-3\right)}{\left(n-2\right)!}
+ \frac{(n-1)\lambda^{(n-1)} t^\left(n-2\right)}{\left(n-1\right)!}
\right\}\\
&=&\mathrm{e}^{-\lambda t}\left\{
\frac{\lambda}{1!}
+ \frac{\lambda^2 t}{1!}
+ \dots
+ \frac{\lambda^{(n-2)} t^\left(n-3\right)}{\left(n-3\right)!}
+ \frac{\lambda^{(n-1)} t^\left(n-2\right)}{\left(n-2\right)!}
\right\}
\;\cdots\;\frac{n}{n!}=\frac{n}{n\times(n-1)\times\dots\times 1}=\frac{1}{(n-1)\times\dots\times 1}=\frac{1}{(n-1)!}\\
(\mathrm{e}^{-\lambda t})'A+\mathrm{e}^{-\lambda t}(A)'
&=&\mathrm{e}^{-\lambda t}\left\{
- \frac{\lambda^1}{0!}
- \frac{\lambda^2 t}{1!}
- \frac{\lambda^3 t^2}{2!}
- \dots
- \frac{\lambda^\left(n-1\right) t^\left(n-2\right)}{\left(n-2\right)!}
- \frac{\lambda^n t^\left(n-1\right)}{\left(n-1\right)!}
\right\}
+ \mathrm{e}^{-\lambda t}\left\{
\frac{\lambda}{1!}
+ \frac{\lambda^2 t}{1!}
+ \dots
+ \frac{\lambda^{(n-2)} t^\left(n-3\right)}{\left(n-3\right)!}
+ \frac{\lambda^{(n-1)} t^\left(n-2\right)}{\left(n-2\right)!}
\right\}\\
&=&\mathrm{e}^{-\lambda t}\left\{
\left(- \frac{\lambda}{0!}+ \frac{\lambda}{1!}\right)
+ \left(- \frac{\lambda^2 t}{1!}+ \frac{\lambda^2 t}{1!}\right)
+ \dots
+ \left(- \frac{\lambda^\left(n-1\right) t^\left(n-2\right)}{\left(n-2\right)!}+ \frac{\lambda^{(n-1)} t^\left(n-2\right)}{\left(n-2\right)!}\right)
- \frac{\lambda^n t^\left(n-1\right)}{\left(n-1\right)!}
\right\}\\
&=&\mathrm{e}^{-\lambda t}\left\{
- \frac{\lambda^n t^\left(n-1\right)}{\left(n-1\right)!}
\right\}\\
Y_n=\frac{\partial F_n(t)}{\partial t}
&=&-\left[(\mathrm{e}^{-\lambda t})'A
+\mathrm{e}^{-\lambda t}(A)'
\right]\\
&=&-\left[\mathrm{e}^{-\lambda t}\left\{
- \frac{\lambda^n t^\left(n-1\right)}{\left(n-1\right)!}
\right\}\right]\\
&=&\mathrm{e}^{-\lambda t}\frac{\lambda^n t^\left(n-1\right)}{\left(n-1\right)!}\\
&=&\mathrm{e}^{-\beta t}\frac{\beta^\alpha t^\left(\alpha-1\right)}{\left(\alpha-1\right)!}
\;\cdots\;\alpha=n, \beta=\lambdaとする.\\
&=&\frac{\mathrm{e}^{-\beta t}}{\left(\alpha-1\right)!}\beta^\alpha t^\left(\alpha-1\right)\\
&=&\frac{\mathrm{e}^{-\beta t}}{\Gamma(\alpha)}\beta^\alpha t^\left(\alpha-1\right)
\;\cdots\;ガンマ分布, \Gamma(n)=(n-1)!\\
\end{eqnarray}
$$
\(\alpha=1\)の時，指数分布となる(事象と事象の間隔の確率分布)．
$$
\begin{eqnarray}
\frac{\mathrm{e}^{-\beta t}}{\Gamma(\alpha)}\beta^\alpha t^\left(\alpha-1\right)
&=&\frac{\mathrm{e}^{-\beta t}}{\Gamma(1)}\beta^1 t^\left(1-1\right)\;\cdots\;\alpha=1\\
&=&\frac{\mathrm{e}^{-\beta t}}{1}\beta t^\left(0\right)\;\cdots\;\Gamma(n)=(n-1)!,\Gamma(1)=(1-1)!=0!=1\\
&=&\mathrm{e}^{-\beta t}\beta\;\cdots\;t^0=1\\
&=&\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{\lambda\mathrm{e}^{-\lambda t}}\;\cdots\;\beta=\lambda\\
\end{eqnarray}
$$
事象の発生する間隔についての確率密度分布 (指数分布)
ある事象が起こってから次の事象が起きるまでの時間をTとする．
事象が起こってからt時間までに次の事象が起きる確率は，tまでをn等分した時間を\(\Delta t=\frac{t}{n}\)として，
“事象が\((k-1)\)回の\(\Delta t\)の間に発生しなかったのち次の\(k\)回目の\(\Delta t\)の間に発生する確率”の\(k=1\)から\(n\)までの和によって以下のように表すことができる
(独立同分布の和によって決まる過程，単位時間当たりの事象の発生確率(個々の分布)を\(\lambda\)とする)．
$$
\begin{eqnarray}
P(T\leq t)
&=&\lambda \Delta t
+ (1-\lambda \Delta t )\Delta t
+ (1-\lambda \Delta t )^2\Delta t
+ \cdots
+ (1-\lambda \Delta t )^{(n-1)}\Delta t\\
&=&\sum_{k=1}^n (\lambda \Delta t)(1-\lambda \Delta t )^{k-1}\\
\lambda\Delta t&&\Delta tの間に発生する確率\\
(1-\lambda \Delta t )\lambda\Delta t&&最初の\Delta t間に発生せず，次の\Delta tの間に発生する確率\\
(1-\lambda \Delta t )^2\lambda\Delta t&&2回の\Delta t間に発生せず，次の\Delta tの間に発生する確率\\
(1-\lambda \Delta t )^{n-1}\lambda\Delta t&&n-1回の\Delta t間に発生せず，次の\Delta tの間に発生する確率\\
\end{eqnarray}
$$
初項を\(\lambda \Delta t\)，公比を\((1-\lambda \Delta t )\)とする等比級数として上式は以下のようになる．
$$
\begin{eqnarray}
P(T\leq t)
&=&\lambda \Delta t
+ (1-\lambda \Delta t )\Delta t
+ (1-\lambda \Delta t )^2\Delta t
+ \cdots
+ (1-\lambda \Delta t )^{(n-1)}\Delta t\\
&=&\lambda \Delta t\frac{1-(1-\lambda \Delta t )^n}{1-(1-\lambda \Delta t )}
\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_17.html}{\sum_{k=1}^{n}ar^{(k-1)}=\frac{a\left(1-r^n\right)}{1-r}}\\
&=&1-\left(1-\lambda \Delta t \right)^n\\
&=&1-\left(1-\lambda \frac{t}{n} \right)^n\;\cdots\;\Delta t=\frac{t}{n}\\
&=&1-\left(1+\frac{1}{-\frac{n}{\lambda t}} \right)^n\\
&=&1-\left\{\left(1+\frac{1}{\frac{-n}{\lambda t}} \right)^{\frac{-n}{\lambda t}\frac{\lambda t}{-n}}\right\}^n\\
&=&1-\left\{\left(1+\frac{1}{\frac{-n}{\lambda t}} \right)^{\frac{-n}{\lambda t}}\right\}^{\frac{\lambda t}{-n} n}\\
&=&1-\left\{\left(1+\frac{1}{\frac{-n}{\lambda t}} \right)^{\frac{-n}{\lambda t}}\right\}^{-\lambda t}
\end{eqnarray}
$$
ここでnを\(\infty\)に極限をとると収束数列によるネイピア数の定義より以下のようになる．
$$
\begin{eqnarray}
\lim_{n\to\infty} 1-\left\{\left(1+\frac{1}{\frac{-n}{\lambda t}} \right)^{\frac{-n}{\lambda t}}\right\}^{-\lambda t}
&=&1-\mathrm{e}^{-\lambda t}
\;\cdots\; \lim_{n\to+\infty} \left(1+\frac{1}{n}\right)^n=\mathrm{e}\\
\end{eqnarray}
$$
上式は時刻tまでの累積分布凾数\(F(t)\)となる．
累積分布凾数を微分することで確率密度凾数を得ることができる．
$$
\begin{eqnarray}
F(t)&=&1-\mathrm{e}^{-\lambda t}\\
f(t)
&=&\frac{\mathrm{d}F(t)}{\mathrm{d}{t}}\\
&=&0-\left(
-\lambda \mathrm{e}^{-\lambda t}
\right)\\
&=&\lambda \mathrm{e}^{-\lambda t}
\;\cdots\;指数分布
\end{eqnarray}
$$
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