t分布の導出
\( X=\frac{Y}{\sqrt{\frac{Z}{n}}} \), \(t\)分布の確率変数
\(t\)分布は次の確率変数\(X\)を考えます． $$\begin{eqnarray} X&=&\frac{Y}{\sqrt{\frac{Z}{n}}}\;\cdots\;\frac{Y}{\mathrm{RMS}\left[Y\right]} \\Y&:&f_Y(y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \;\cdots\;z分布\left(標準正規分布\right) \\Z&:&f_Z(z)=\frac{1}{2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}}e^{-\frac{z}{2}}z^{\frac{n}{2}-1} \;\cdots\;\chi^2分布\left(Z=\sum_{i=1}^n Y_i^2\right) \\&&\end{eqnarray}$$\( f_{YZ}(y,z) \), \(Y\)と\(Z\)の同時確率
$$\begin{eqnarray} f_{YZ}(y,z)&=&f_Y(y)\cdot f_Z(z)\;\cdots\;YとZは独立 \end{eqnarray}$$変数変換, \(Y\)と\(Z\)から\(X\)と\(U\)へ
$$\begin{eqnarray} x&=&\frac{y}{\sqrt{\frac{z}{n}}}&より&y&=&x\sqrt{\frac{z}{n}} \end{eqnarray}$$ $$ \left\{ \begin{eqnarray} z&=&u \\y&=&x\sqrt{\frac{u}{n}} \end{eqnarray} \right.$$ \begin{eqnarray} J&=&\left|\frac{\partial(y,z)}{\partial(x,u)}\right| \\&=&\left|\begin{matrix} \frac{\partial y}{\partial x}&\frac{\partial y}{\partial u} \\\frac{\partial z}{\partial x}&\frac{\partial z}{\partial u} \end{matrix}\right| \\&=&\left|\begin{matrix} \sqrt{\frac{u}{n}}&\frac{x}{\sqrt{n}}\frac{1}{2}u^{-\frac{1}{2}} \\0&1 \end{matrix}\right| \\&=&\sqrt{\frac{u}{n}}\cdot 1 - \frac{x}{\sqrt{n}}\frac{1}{2}u^{-\frac{1}{2}} \cdot 0 \\&=&\sqrt{\frac{u}{n}} \end{eqnarray} \begin{eqnarray} &&\int_{0}^{\infty}\int_{\infty}^{-\infty}f_Y(y)f_Z(z)\mathrm{d}y\mathrm{d}z\;\cdots\;Y,Zの同時確率の全事象 \\&=&\int_{0}^{\infty}\int_{\infty}^{-\infty}f_Y\left(x\sqrt{\frac{u}{n}}\right)f_Z(u)\sqrt{\frac{u}{n}}\mathrm{d}x\mathrm{d}u \\&&\;\cdots\;変数変換:y=x\sqrt{\frac{u}{n}},\;z=u,J=\sqrt{\frac{u}{n}} \\&&\;\cdots\;x=\frac{y}{\sqrt{\frac{u}{n}}}なので,\;y:-\infty\rightarrow\infty,\;x:-\infty\rightarrow\infty \\&&\;\cdots\;u=zなので,\;z:0\rightarrow\infty,\;u:0\rightarrow\infty \\&=&\int_{0}^{\infty}\int_{\infty}^{-\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{\left(x\sqrt{\frac{u}{n}}\right)^2}{2}} \frac{1}{2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}}e^{-\frac{u}{2}}u^{\frac{n}{2}-1} \sqrt{\frac{u}{n}} \mathrm{d}x\mathrm{d}u \\&=& \frac{1}{\sqrt{2\pi}} \frac{1}{2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \frac{1}{\sqrt{n}} \int_{0}^{\infty}\int_{\infty}^{-\infty} e^{-\frac{x^2u}{2n}} e^{-\frac{u}{2}} u^{\frac{n}{2}-1} \sqrt{u} \mathrm{d}x\mathrm{d}u \\&=& \frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \int_{0}^{\infty}\int_{\infty}^{-\infty} e^{-\frac{x^2u}{2n}-\frac{u}{2}} u^{\frac{n}{2}-1+\frac{1}{2}} \mathrm{d}x\mathrm{d}u \\&=& \frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \int_{0}^{\infty}\int_{\infty}^{-\infty} e^{-\frac{u}{2}\left(\frac{x^2}{n}+1\right)} u^{\frac{n+1}{2}-1} \mathrm{d}x\mathrm{d}u \\&=&1\;\cdots\;全事象 \\&=&F_{XU}\left(x,u\right) \end{eqnarray}\(f_X\left(x\right)\), \(X\)の確率密度凾数
\begin{eqnarray} f_{XU}\left(x,u\right) &=&\frac{\partial^2}{\partial x \partial u}F_{XU}\left(x,u\right) \\&=&\frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} e^{-\frac{u}{2}\left(1+\frac{x^2}{n}\right)} u^{\frac{n+1}{2}-1} \end{eqnarray} \(X\)の確率密度凾数\(f_X\left(x\right)\)がすなわち\(t\)分布なので，uについて積分することで周辺確率を求めます． \begin{eqnarray} t(x)=f_X(x)&=&\int_{0}^{\infty}f_{XU}\left(x,u\right)\mathrm{d}u \\&=&\frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \int_{0}^{\infty} e^{-\frac{u}{2}\left(\frac{x^2}{n}+1\right)} u^{\frac{n+1}{2}-1} \mathrm{d}u \\&=&\frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \int_{0}^{\infty} e^{-t} \cdot\left(\frac{2t}{\frac{x^2}{n}+1}\right)^{\frac{n+1}{2}-1} \cdot\frac{2}{\frac{x^2}{n}+1}\mathrm{d}t \\&&\;\cdots\;t=\frac{u}{2}\left(\frac{x^2}{n}+1\right) \\&&\;\cdots\;u:0\rightarrow\infty,t:0\rightarrow\infty \\&&\;\cdots\;u=\frac{2t}{\frac{x^2}{n}+1} \\&&\;\cdots\;\frac{\mathrm{d}u}{\mathrm{d}t}=\frac{2}{\frac{x^2}{n}+1} \\&&\;\cdots\;\mathrm{d}u=\frac{2}{\frac{x^2}{n}+1}\mathrm{d}t \\&=&\frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \frac{2}{\frac{x^2}{n}+1} \left(\frac{2}{\frac{x^2}{n}+1}\right)^{\frac{n+1}{2}-1} \int_{0}^{\infty} e^{-t} t^{\frac{n+1}{2}-1} \mathrm{d}t \\&=&\frac{1}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} \left(\frac{2}{\frac{x^2}{n}+1}\right)^{\frac{n+1}{2}} \Gamma{\left(\frac{n+1}{2}\right)} \\&=&\frac{\Gamma{\left(\frac{n+1}{2}\right)}}{\sqrt{2\pi n}\;2^{\frac{n}{2}}\Gamma{\left(\frac{n}{2}\right)}} 2^{\frac{n+1}{2}} \left(\frac{x^2}{n}+1\right)^{-\frac{n+1}{2}} \\&=&\frac{\Gamma{\left(\frac{n+1}{2}\right)}}{\sqrt{\cancel{2}\pi n}\;\cancel{2^{\frac{n}{2}}}\Gamma{\left(\frac{n}{2}\right)}} \cancel{2^{\frac{n}{2}}}\cancel{2^{\frac{1}{2}}}\left(\frac{x^2}{n}+1\right)^{-\frac{n+1}{2}} \\&=&\frac{\Gamma{\left(\frac{n+1}{2}\right)}}{\sqrt{\pi n}\Gamma{\left(\frac{n}{2}\right)}} \left(\frac{x^2}{n}+1\right)^{-\frac{n+1}{2}} \\&=&\frac{1}{\sqrt{n}}\frac{\Gamma{\left(\frac{n+1}{2}\right)}}{\sqrt{\pi}\;\Gamma{\left(\frac{n}{2}\right)}} \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\&=&\frac{1}{\sqrt{n}} \frac{ \Gamma{ \left(\frac{n+1}{2}\right) } }{ \Gamma{ \left(\frac{1}{2}\right) } \Gamma{ \left(\frac{n}{2}\right) } } \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \;\cdots\;\Gamma{ \left(\frac{1}{2}\right) }=\sqrt{\pi} \\&=&\frac{1}{\sqrt{n}} \frac{ \Gamma{ \left(\frac{1}{2}+\frac{n}{2}\right) } }{ \Gamma{ \left(\frac{1}{2}\right) } \Gamma{ \left(\frac{n}{2}\right) } } \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\&=&\frac{1}{\sqrt{n}} \frac{1} { \beta{ \left(\frac{1}{2},\frac{n}{2}\right) } } \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{ \beta{\left(a,b\right)}=\frac{ \Gamma{\left(a\right)} \Gamma{\left(b\right)} }{ \Gamma{ \left(a+b\right) } }=\int_0^1 x^{a-1}(1-x)^{b-1}\mathrm{d}x} \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
$$ \begin{eqnarray} f_X(x)&=&\lambda e^{-\lambda x} (x \geq 0, \lambda \gt 0) \;\cdots\;Exp(\lambda)\;(\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布}) \\g_Y(y)&=&\lambda e^{-\lambda y} (x \geq 0, \lambda \gt 0) \;\cdots\;Exp(\lambda)\;(\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布}) \\p_{X+Y}(z)&=&\int \int \delta(z-(x+y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(z-(x+y))\lambda e^{-\lambda x} \lambda e^{-\lambda y} \mathrm{d}x\mathrm{d}y \\&=&\int_0^z \lambda e^{-\lambda (z-y)} \lambda e^{-\lambda y}\mathrm{d}y \;\cdots\;z=x+y,x=z-y \\&=&\lambda^2\int_0^z e^{-\lambda (z-y)} e^{-\lambda y}\mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\lambda^2\int_0^z e^{-\lambda (z-y)-\lambda y}\mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\lambda^2\int_0^z e^{-\lambda (z-y+y)}\mathrm{d}y \\&=&\lambda^2\int_0^z e^{-\lambda z}\mathrm{d}y \\&=&\lambda^2 e^{-\lambda z}\int_0^z\mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\lambda^2 e^{-\lambda z}\left[y\right]_0^z \\&=&\lambda^2 e^{-\lambda z}\left[z-0\right] \\&=&\lambda^2 z e^{-\lambda z} \\&=&\Gamma(2, \lambda) \end{eqnarray} $$ $$ \begin{eqnarray} \Gamma(\alpha, \lambda)&=&\frac{1}{\Gamma(\alpha)}\lambda^{\alpha} x^{\alpha-1} e^{-\lambda x}\;(\href{https://shikitenkai.blogspot.com/2020/05/n.html}{ガンマ分布;\;gamma\;disribution}) \\\Gamma(1, \lambda)&=&\frac{1}{\Gamma(1)}\lambda^1 x^{1-1} e^{-\lambda x}=\frac{1}{1}\lambda e^{-\lambda x}=\lambda e^{-\lambda x} \\&&\;\cdots\;\Gamma(1)=(1-1)!=0!=1 \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布} \\\Gamma(2, \lambda)&=&\frac{1}{\Gamma(2)}\lambda^2 x^{2-1} e^{-\lambda x}=\frac{1}{1}\lambda^2xe^{-\lambda x}=\lambda^2xe^{-\lambda x} \\&&\;\cdots\;\Gamma(2)=(2-1)!=1!=1 \\&&\;\cdots\;指数分布に従う確率変数同士の和の分布,\;\alpha=2のガンマ分布 \end{eqnarray} $$二項分布(binomial distribution)の積率母凾数(moment-generating function)と期待値(expected value)・分散(variance)
二項分布(binomial distribution)の積率母凾数(moment-generating function)と期待値(expected value)・分散(variance)
二項分布
$$ \begin{eqnarray} X&\sim&B(n, p) \\f(X=x)&=& \begin{cases} _n\mathrm{C}_x\;p^x(1-p)^{n-x} & x \in \left\{0,1,2, \dotsc ,n\right\} \\0 & x \notin \left\{0,1,2, \dotsc ,n\right\} \end{cases}\;\cdots\;確率密度凾数 \end{eqnarray} $$積率母凾数
$$ \begin{eqnarray} M_X(t)&=&\mathrm{E}\left[e^{tx}\right] \\&=&\sum_{k=0}^n e^{tx}\;_n\mathrm{C}_x\;p^x\left(1-p\right)^{n-x} \\&=&\sum_{k=0}^n\;_n\mathrm{C}_x\;\left( e^{t} p \right)^x \left(1-p\right)^{n-x} \\&=&\;_n\mathrm{C}_0\;\left( e^{t} p \right)^0 \left(1-p\right)^{n-0} +\;_n\mathrm{C}_1\;\left( e^{t} p \right)^1 \left(1-p\right)^{n-1} +\;_n\mathrm{C}_2\;\left( e^{t} p \right)^2 \left(1-p\right)^{n-2} +\cdots +\;_n\mathrm{C}_n\;\left( e^{t} p \right)^n \left(1-p\right)^{n-n} \\&=&\left\{e^{t} p + \left(1-p\right) \right\}^n \\&&\;\cdots\;(A+B)^D=\;_D\mathrm{C}_0\;A^0B^{D-0}+\;_D\mathrm{C}_1\;A^1B^{D-1}+\cdots+\;_D\mathrm{C}_D\;A^DB^{D-D}=\sum_{k=0}^D \;_D\mathrm{C}_k\;A^kB^{D-k}\;(二項関係) \\&=&\left(e^{t}p - p + 1\right)^n \end{eqnarray} $$(原点周りの)一次モーメント = 期待値
$$ \begin{eqnarray} \mathrm{E}\left[X\right]&=&M^{(1)}_X(t) \\&=&\left.\frac{\mathrm{d} M_X(t)}{\mathrm{d}t}\right|_{t=0} \\&=&\left.\frac{\mathrm{d}}{\mathrm{d}t} \left(e^{t}p - p + 1\right)^n \right|_{t=0} \\&=&\left.\frac{\mathrm{d}}{\mathrm{d}u} u^n \frac{\mathrm{d}u}{\mathrm{d}t} \right|_{t=0} \;\cdots\;u=e^{t}p - p + 1,\frac{\mathrm{d}u}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{t}p - p + 1\right)=e^{t}p \\&=&\left.nu^{n-1} e^{t}p \right|_{t=0} \\&=&\left.n\left( e^{t}p - p + 1 \right)^{n-1} e^{t}p \right|_{t=0} \\&=&\left.npe^{t} \left( e^{t}p - p + 1 \right)^{n-1} \right|_{t=0} \\&=&npe^{0} \left( e^{0}p - p + 1 \right)^{n-1} \\&=&np\cdot1 \left( 1\cdot p - p + 1 \right)^{n-1} \\&=&np \left(1\right)^{n-1} \\&=&np\;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/blog-post_30.html}{Xの期待値の定義から求めた二項分布の期待値}と同じ \end{eqnarray} $$(原点周りの)二次モーメント
$$ \begin{eqnarray} \mathrm{E}\left[X^2\right]&=&M^{(2)}_X(t) \\&=&\left.\frac{\mathrm{d}^2 M_X(t)}{\mathrm{d}t^2}\right|_{t=0} \\&=&\left.\frac{\mathrm{d}^2}{\mathrm{d}t^2} \left(e^{t}p - p + 1\right)^n \right|_{t=0} \\&=&\left.\frac{\mathrm{d}}{\mathrm{d}t} npe^{t} \left(e^{t}p - p + 1\right)^{n-1} \right|_{t=0} \\&=&\left.np\frac{\mathrm{d}}{\mathrm{d}t} e^{t} \left(e^{t}p - p + 1\right)^{n-1} \right|_{t=0} \;\cdots\;\frac{\mathrm{d}}{\mathrm{d}x}cf(x)=c\frac{\mathrm{d}}{\mathrm{d}x}f(x)\;(c:定数) \\&=&\left.np\frac{\mathrm{d}}{\mathrm{d}t} uv \right|_{t=0} \;\cdots\;u=e^t,\;v= \left(e^{t}p - p + 1\right)^{n-1} \\&=&\left.np\left\{\left(\frac{\mathrm{d}}{\mathrm{d}t}u\right)v+u\left(\frac{\mathrm{d}}{\mathrm{d}t}v\right) \right\}\right|_{t=0} \\&=&\left.np\left[\left(e^{t}\right)v+u\left\{\left(n-1\right)\left(e^{t}p - p + 1\right)^{n-2}pe^t\right\} \right]\right|_{t=0} \;\cdots\;\frac{\mathrm{d}u}{\mathrm{d}t}=e^{t},\frac{\mathrm{d}v}{\mathrm{d}t}=\left(n-1\right)\left(e^{t}p - p + 1\right)^{n-2}pe^t \\&=&\left.np\left[\left(e^{t}\right)\left(e^{t}p - p + 1\right)^{n-1}+e^t\left\{\left(n-1\right)\left(e^{t}p - p + 1\right)^{n-2}pe^t\right\} \right]\right|_{t=0} \;\cdots\;u=e^t,\;v= \left(e^{t}p - p + 1\right)^{n-1} \\&=&\left.npe^{t}\left(e^{t}p - p + 1\right)^{n-1}+n\left(n-1\right)p^2e^{2t}\left(e^{t}p - p + 1\right)^{n-2} \right|_{t=0} \\&=&npe^{0}\left(e^{0}p - p + 1\right)^{n-1}+n\left(n-1\right)p^2e^{2\cdot0}\left(e^{0}p - p + 1\right)^{n-2} \\&=&np\cdot1\cdot\left(1\cdot p - p + 1\right)^{n-1}+n\left(n-1\right)p^2\cdot1\cdot\left(1\cdot p - p + 1\right)^{n-2} \\&=&np\left(1\right)^{n-1}+n\left(n-1\right)p^2\left(1\right)^{n-2} \\&=&np+n\left(n-1\right)p^2 \end{eqnarray} $$二次の中心(化)モーメント / 母平均周りの二次モーメント = 分散
$$ \begin{eqnarray} \mathrm{V}\left[X\right]&=&\mathrm{E}\left[\left(X-\mathrm{E}\left[X\right]\right)^2\right] \\&=&\mathrm{E}\left[X^2\right]-\mathrm{E}\left[X\right]^2 \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-variance.html}{\mathrm{E}\left[\left(X-\mathrm{E}\left[X\right]\right)^2\right]=\mathrm{E}\left[X^2\right]-\left[X\right]^2} \\&=&M^{(2)}_X(t) -\left(M^{(1)}_X(t)\right)^2 \\&=&np+n\left(n-1\right)p^2 - (np)^2 \\&=&np+n^2p^2-np^2 - n^2p^2 \\&=&np-np^2 \\&=&np(1-p)\;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/blog-post_75.html}{X(X-1)の期待値を利用した二項分布の分散}と同じ \end{eqnarray} $$標本平均の分布の尖度
標本平均\(\overline{X}\)の尖度\(\beta_2\left(\overline{X}\right)\)
$$ \begin{eqnarray} \beta_2\left(\overline{X}\right) &=&\frac{\mathrm{E}\left[(\overline{X}-\mu)^4\right]}{\mathrm{V}\left[\overline{X}\right]^{\frac{4}{2}}}-3 \;\cdots\; \href{https://shikitenkai.blogspot.com/2020/08/blog-post_27.html}{\beta_2=\beta_2\left(X\right)=\frac{\mathrm{E}\left[(X-\mu)^4\right]}{\mathrm{V}\left[X\right]^{\frac{4}{2}}}-3=\frac{\mu_4}{\sigma^4}-3\;:尖度} \\&=&\frac{\frac{1}{n^3}\left(\mu_4+3(n-1)\sigma^4\right)}{\left(\frac{\sigma^2}{n}\right)^2}-3 \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/4-4.html}{\mathrm{E}\left[(\overline{X}-\mu)^4\right]=\frac{1}{n^3}\left(\mu_4+3(n-1)\sigma^4\right) \;:標本平均の母平均まわりの4次モーメント (標本平均の4次の中心(化)モーメント) } \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable_3.html}{\mathrm{V}\left[\overline{X}\right]=\frac{\sigma^2}{n}\;:標本平均の分散} \\&=&\frac{n^2}{n^3}\frac{\mu_4+3(n-1)\sigma^4}{\sigma^4}-3 \\&=&\frac{1}{n}\left(\frac{\mu_4}{\sigma^4}+3(n-1)\right)-3 \\&=&\frac{1}{n}\left(\frac{\mu_4}{\sigma^4}+3n-3\right)-3 \\&=&\frac{\frac{\mu_4}{\sigma^4}-3}{n}+3-3 \\&=&\frac{\frac{\mu_4}{\sigma^4}-3}{n} \\&=&\frac{\beta_2}{n} \;\cdots\;\beta_2=\frac{\mu_4}{\sigma^4}-3 \end{eqnarray} $$ よって標本数\(n\)を増やすことで\(\beta_2\left(\overline{X}\right)\)は\(0\)に近づいていく．カイ二乗分布の期待値と分散
カイ二乗分布の期待値と分散
カイ二乗分布の期待値(一次モーメント)
$$ \begin{eqnarray} \mathrm{E}\left[x\right]&=&\int_0^\infty x \chi^2(x) \mathrm{d}x \\&=&\int_0^\infty x \frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \mathrm{d}x \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\int_0^\infty x e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \mathrm{d}x \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\int_0^\infty e^{-\frac{x}{2}}x^{\frac{n}{2}} \mathrm{d}x \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\int_0^\infty e^{-\frac{x}{2}}x^{\frac{n}{2}} \color{red}{2^{\frac{n}{2}}\left(\frac{1}{2}\right)^{\frac{n}{2}}} \color{black}{\mathrm{d}x} \\&=&\frac{1}{\color{red}{2^{\frac{n}{2}}}\color{black}{\Gamma\left(\frac{n}{2}\right)}}\color{red}{2^{\frac{n}{2}}}\color{black}{\int_0^\infty e^{-\frac{x}{2}}x^{\frac{n}{2}} \left(\frac{1}{2}\right)^{\frac{n}{2}}\mathrm{d}x} \\&=&\frac{1}{\Gamma\left(\frac{n}{2}\right)}\int_0^\infty e^{-\frac{x}{2}}\left(\frac{x}{2}\right)^{\frac{n}{2}}\mathrm{d}x \\&=&\frac{1}{\Gamma\left(\frac{n}{2}\right)}\int_0^\infty e^{-t}t^{\frac{n}{2}}\;2\mathrm{d}t \;\cdots\;t=\frac{x}{2},\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{2},\mathrm{d}x=2\mathrm{d}t \\&=&\frac{1}{\Gamma\left(\frac{n}{2}\right)}2\int_0^\infty e^{-t}t^{\frac{n}{2}}\;\mathrm{d}t \;\cdots\;\int cf(x) \mathrm{d}x=c\int f(x) \mathrm{d}x \\&=&\frac{1}{\Gamma\left(\frac{n}{2}\right)}2\int_0^\infty e^{-t}t^{\frac{n}{2}\color{red}{+1-1}}\;\mathrm{d}t \\&=&\frac{1}{\Gamma\left(\frac{n}{2}\right)}2\Gamma\left(\frac{n}{2}+1\right) \;\cdots\;\Gamma\left(s\right)=\int_0^\infty e^{-t}t^{s-1}\mathrm{d}t \\&=&\frac{1}{\Gamma\left(\frac{n}{2}\right)}2\frac{n}{2}\Gamma\left(\frac{n}{2}\right) \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/s1ss.html}{\Gamma(s+1)=\int_0^\infty e^{-t}t^{s}\;\mathrm{d}t=s\Gamma(s)} \\&=&n \end{eqnarray} $$カイ二乗分布の二次モーメント
$$ \begin{eqnarray} \mathrm{E}\left[x^2\right]&=&\int_0^\infty x^2 \chi^2(x) \mathrm{d}x \\&=&\int_0^\infty x^2 \frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \mathrm{d}x \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\int_0^\infty x^2 e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \mathrm{d}x \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\int_0^\infty e^{-\frac{x}{2}}x^{\frac{n}{2}+1} \mathrm{d}x \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\int_0^\infty e^{-\frac{x}{2}}x^{\frac{n}{2}+1} \color{red}{2^{\frac{n}{2}+1} \left(\frac{1}{2}\right)^{\frac{n}{2}+1} } \color{black}{\mathrm{d}x} \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}2^{\frac{n}{2}+1}\int_0^\infty e^{-\frac{x}{2}}x^{\frac{n}{2}+1} \left(\frac{1}{2}\right)^{\frac{n}{2}+1}\mathrm{d}x \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}2^{\frac{n}{2}}2\int_0^\infty e^{-\frac{x}{2}}x^{\frac{n}{2}+1} \left(\frac{1}{2}\right)^{\frac{n}{2}+1}\mathrm{d}x \\&=&\frac{1}{\color{red}{2^{\frac{n}{2}}}\color{black}{\Gamma\left(\frac{n}{2}\right)}}\color{red}{2^{\frac{n}{2}}}\color{black}{2\int_0^\infty e^{-\frac{x}{2}}x^{\frac{n}{2}+1} \left(\frac{1}{2}\right)^{\frac{n}{2}+1}\mathrm{d}x} \\&=&\frac{2}{\Gamma\left(\frac{n}{2}\right)}\int_0^\infty e^{-\frac{x}{2}}\left(\frac{x}{2}\right)^{\frac{n}{2}+1}\mathrm{d}x \\&=&\frac{2}{\Gamma\left(\frac{n}{2}\right)}\int_0^\infty e^{-t}t^{\frac{n}{2}+1}\;2\mathrm{d}t \;\cdots\;t=\frac{x}{2},\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{2},\mathrm{d}x=2\mathrm{d}t \\&=&\frac{2}{\Gamma\left(\frac{n}{2}\right)}2\int_0^\infty e^{-t}t^{\frac{n}{2}+1}\;\mathrm{d}t \;\cdots\;\int cf(x) \mathrm{d}x=c\int f(x) \mathrm{d}x \\&=&\frac{2}{\Gamma\left(\frac{n}{2}\right)}2\int_0^\infty e^{-t}t^{\frac{n}{2}+1\color{red}{+1-1}}\;\mathrm{d}t \\&=&\frac{2}{\Gamma\left(\frac{n}{2}\right)}2\int_0^\infty e^{-t}t^{\frac{n}{2}+2-1}\;\mathrm{d}t \\&=&\frac{2}{\Gamma\left(\frac{n}{2}\right)}2\Gamma\left(\frac{n}{2}+2\right) \;\cdots\;\Gamma\left(s\right)=\int_0^\infty e^{-t}t^{s-1}\mathrm{d}t \\&=&\frac{2}{\Gamma\left(\frac{n}{2}\right)}2\left(\frac{n}{2}+1\right)\frac{n}{2}\Gamma\left(\frac{n}{2}\right) \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/s1ss.html}{\Gamma\left(s+2\right)=(s+1)\Gamma(s+1)=(s+1)s\Gamma(s)} \\&=&2n\left(\frac{n}{2}+1\right) \\&=&n(n+2) \end{eqnarray} $$カイ二乗分布の分散(二次の中心モーメント)
$$ \begin{eqnarray} \mathrm{V}\left[x^2\right]&=&\mathrm{E}\left[(x-\mathrm{E}\left[x\right])^2\right] \\&=&\mathrm{E}\left[x^2\right]-\mathrm{E}\left[x\right]^2 \\&=&n(n+2)-n^2 \\&=&n^2+2n-n^2 \\&=&2n \end{eqnarray} $$カイ二乗分布の導出
カイ二乗\(\chi^2\)分布の導出
\(k=1\)のXの確率密度凾数
$$ \begin{eqnarray} X&=&Z_1^2\;\cdots\;Z_1:標準正規分布に従う確率変数 \\F_1(X=t)&=&\int_0^t f_1(x) \mathrm{d}x \;\cdots\;k=1のXの累積分布凾数F_1,確率密度凾数f_1 \\F_1(X=t)&=&\int_{-\sqrt{t}}^{+\sqrt{t}}g(z_1)\mathrm{d}z_1 \;\cdots\;Z_1=\pm\sqrt{X},x:0\rightarrow t,z_1:0\rightarrow\pm\sqrt{t}\;標準正規分布の確率密度凾数g \\&=&2\int_{0}^{\sqrt{t}}g(z_1)\mathrm{d}z_1 \;\cdots\;標準正規分布は偶凾数(y軸(x＝0)で左右対称),\;Z_1\geq0のみ考えればよくする \\&=&2\int_{0}^{t}g(\sqrt{x})\frac{1}{2}x^{-\frac{1}{2}}\mathrm{d}x \\&&\;\cdots\;Z_1=\sqrt{X}(Z_1\geq0)より\frac{\mathrm{d}z_1}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x}=\frac{\mathrm{d}}{\mathrm{d}x}x^{\frac{1}{2}}=\frac{1}{2}x^{-\frac{1}{2}},\mathrm{d}z_1=\frac{1}{2}x^{-\frac{1}{2}}\mathrm{d}x \\&&\;\cdots\;z_1:0\rightarrow\sqrt{t},\;x:0\rightarrow t \\&=&\int_{0}^{t}g(\sqrt{x})x^{-\frac{1}{2}}\mathrm{d}x \\\int_0^t f_1(x) \mathrm{d}x &=&\int_{0}^{t}g(\sqrt{x})x^{-\frac{1}{2}}\mathrm{d}x \;\cdots\;累積分布凾数同士なので等しい(0から任意のt(\gt0)までの定積分が等しい) \\f_1(x)&=&g(\sqrt{x})x^{-\frac{1}{2}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/blog-post_25.html}{0から任意のt(\gt0)までの定積分が等しい\rightarrow0からtの範囲で等しい\rightarrow各tでの微分が等しい\rightarrow被積分凾数同士も等しい} \\&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{\sqrt{x}^2}{2}}\;x^{-\frac{1}{2}} \;\cdots\;標準正規分布の確率密度凾数:g(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \\&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{x}{2}}\;x^{-\frac{1}{2}} \end{eqnarray} $$\(k=2\)のXの確率密度凾数
\(k=2\) $$ \begin{eqnarray} X&=&Z_1^2+Z_2^2\;\cdots\;Z_1,Z_2:互いに独立に標準正規分布に従う確率変数 \\Y&=&Z_1^2 \\X-Y&=&Z_2^2 \\f_2(x) &=&\int_{0}^{\infty}f_1(y)f_1(x-y)\mathrm{d}y \;\cdots\;k=2のXの確率密度凾数f_2,\;Z_1とZ_2は独立なので同時確率はf_1(Z_1)とf_1(Z_2)の積 \\&=&\int_{0}^{x} \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{y}{2}}\;y^{-\frac{1}{2}} \right\} \;\cdot\; \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \right\} \mathrm{d}y \\&=&\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}}e^{-\frac{(x-y)}{2}}\;y^{-\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi} \int_{0}^{x} e^{-\frac{y}{2}-\frac{(x-y)}{2}}\;y^{-\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{x}{2}} \int_{0}^{x} y^{-\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{x}{2}}\pi \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/1xa-x.html}{\int_{0}^{a}\frac{1}{\sqrt{x}\sqrt{a-x}}\mathrm{d}x=\pi} \\&=&\frac{1}{2}e^{-\frac{x}{2}} \end{eqnarray} $$\(k=3\)のXの確率密度凾数
$$ \begin{eqnarray} X&=&Z_1^2+Z_2^2+Z_3^2\;\cdots\;Z_1,Z_2,Z_3:互いに独立に標準正規分布に従う確率変数 \\Y&=&Z_1^2+Z_2^2 \\X-Y&=&Z_3^2 \\f_3(x) &=&\int_{0}^{\infty}f_2(y)f_1(x-y)\mathrm{d}y \;\cdots\;k=3のXの確率密度凾数f_3,\;YとZ_3は独立なので同時確率はf_2(Y)とf_1(Z_3)の積 \\&=&\int_{0}^{x} \left\{ \frac{1}{2}e^{-\frac{y}{2}} \right\} \;\cdot\; \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \right\} \mathrm{d}y \\&=&\frac{1}{2}\frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}}e^{-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\sqrt{2\pi}}e^{-\frac{x}{2}} \int_{0}^{x} (x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\sqrt{2\pi}}e^{-\frac{x}{2}} 2x^{\frac{1}{2}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/1a-x.html}{\int_{0}^{a}\frac{1}{\sqrt{a-x}}\mathrm{d}x=2a^{\frac{1}{2}}} \\&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{x}{2}} x^{\frac{1}{2}} \end{eqnarray} $$\(k=4\)のXの確率密度凾数
$$ \begin{eqnarray} X&=&Z_1^2+Z_2^2+Z_3^2+Z_4^2\;\cdots\;Z_1,Z_2,Z_3,Z_4:互いに独立に標準正規分布に従う確率変数 \\Y&=&Z_1^2+Z_2^2+Z_3^2 \\X-Y&=&Z_4^2 \\f_4(x) &=&\int_{0}^{\infty}f_3(y)f_1(x-y)\mathrm{d}y \;\cdots\;k=4のXの確率密度凾数f_4,\;YとZ_4は独立なので同時確率はf_3(Y)とf_1(Z_4)の積 \\&=&\int_{0}^{x} \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{y}{2}}y^{\frac{1}{2}} \right\} \;\cdot\; \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \right\} \mathrm{d}y \\&=&\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}}e^{-\frac{(x-y)}{2}}\;y^{\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi} \int_{0}^{x} e^{-\frac{y}{2}-\frac{(x-y)}{2}}\;y^{\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{x}{2}} \int_{0}^{x} y^{\frac{1}{2}}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{x}{2}} \frac{x\pi}{2} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/xa-x.html}{\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{a-x}}\mathrm{d}x=\frac{a\pi}{2}} \\&=&\frac{1}{4}e^{-\frac{x}{2}} x \end{eqnarray} $$\(k=5\)のXの確率密度凾数
$$ \begin{eqnarray} X&=&Z_1^2+Z_2^2+Z_3^2+Z_4^2+Z_5^2\;\cdots\;Z_1,Z_2,Z_3,Z_4,Z_5:互いに独立に標準正規分布に従う確率変数 \\Y&=&Z_1^2+Z_2^2+Z_3^2+Z_4^2 \\X-Y&=&Z_5^2 \\f_5(x) &=&\int_{0}^{\infty}f_4(y)f_1(x-y)\mathrm{d}y \;\cdots\;k=5のXの確率密度凾数f_5,\;YとZ_4は独立なので同時確率はf_4(Y)とf_1(Z_5)の積 \\&=&\int_{0}^{x} \left\{ \frac{1}{4}e^{-\frac{y}{2}} y \right\} \;\cdot\; \left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{(x-y)}{2}}\;(x-y)^{-\frac{1}{2}} \right\} \mathrm{d}y \\&=&\frac{1}{4}\frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}}e^{-\frac{(x-y)}{2}}\;y(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{4\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{y}{2}-\frac{(x-y)}{2}}\;y(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{4\sqrt{2\pi}}e^{-\frac{x}{2}} \int_{0}^{x} y(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{4\sqrt{2\pi}}e^{-\frac{x}{2}}\frac{4}{3}x^{\frac{3}{2}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/xa-x_22.html}{\int_{0}^{a}\frac{x}{\sqrt{a-x}}\mathrm{d}x=\frac{4}{3}a^{\frac{3}{2}}} \\&=&\frac{1}{3\sqrt{2\pi}}e^{-\frac{x}{2}} x^{\frac{3}{2}} \end{eqnarray} $$\(k=n\)の\(X\)の確率密度凾数の類推
$$ \begin{array}[crcccc] \\k=1&:&f_1(x)=&\frac{1}{\sqrt{2\pi}}&e^{-\frac{x}{2}}&x^{-\frac{1}{2}} \\k=2&:&f_2(x)=&\frac{1}{2}&e^{-\frac{x}{2}}&x^0 \\k=3&:&f_3(x)=&\frac{1}{\sqrt{2\pi}}&e^{-\frac{x}{2}}&x^{\frac{1}{2}} \\k=4&:&f_4(x)=&\frac{1}{4}&e^{-\frac{x}{2}}&x^1 \\k=5&:&f_5(x)=&\frac{1}{3\sqrt{2\pi}}&e^{-\frac{x}{2}}&x^{\frac{3}{2}} \\k=n&:&f_n(x)=&\alpha_n&e^{-\frac{x}{2}}&x^{\frac{n}{2}-1} \end{array} $$係数\(\alpha_n\)を求める
$$ \begin{eqnarray} F_n(t) &=&\int^{t}_0 f_n(x) \mathrm{d}x \;\cdots\;任意のnにおけるXの累積分布凾数F_n,確率密度凾数f_n \\F_n(t)&=&\int^{t}_0 \alpha_ne^{-\frac{x}{2}}x^{\frac{n}{2}-1} \mathrm{d}x \;\cdots\;前述の類推より \\&=&\alpha_n\int^{t}_0e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \mathrm{d}x\;\cdots\;\alpha_nはxによらない定数,\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\alpha_n\int^{t}_0e^{-t}(2u)^{\frac{n}{2}-1} 2\mathrm{d}u \;\cdots\;u=\frac{x}{2},x=2u,\frac{dx}{du}=2,x:0\rightarrow\infty,u:0\rightarrow\infty \\&=&\alpha_n\int^{t}_0e^{-t}2^{\frac{n}{2}-1}t^{\frac{n}{2}-1} 2\mathrm{d}t \\&=&\alpha_n\int^{t}_0e^{-t}2^{\frac{n}{2}}t^{\frac{n}{2}-1}\mathrm{d}t \\&=&\alpha_n2^{\frac{n}{2}}\int^{t}_0e^{-t}t^{\frac{n}{2}-1}\mathrm{d}t \\F_n(t=\infty)&=&\int^{\infty}_0 f_n(x) \mathrm{d}x \\&=&\alpha_n2^{\frac{n}{2}}\int^{\infty}_0e^{-t}t^{\frac{n}{2}-1}\mathrm{d}t \\&=&\alpha_n2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)\;\cdots\;\Gamma\left(s\right)=\int^{\infty}_0e^{-t}t^{s-1}\mathrm{d}t \\&=&\alpha_n2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right) \\&=&1\;\cdots\;全事象は1 \\\alpha_n &=&\frac{F_n(t=\infty)}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)} \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)} \end{eqnarray} $$係数\(\alpha_k\)の確認
$$ \begin{array}[crlllll] \\\alpha_1&=&\frac{1}{2^{\frac{1}{2}}\Gamma\left(\frac{1}{2}\right)}&=&\frac{1}{2^{\frac{1}{2}}\;\sqrt{\pi}} &=&\frac{1}{\sqrt{2\pi}} \\\alpha_2&=&\frac{1}{2^{\frac{2}{2}}\Gamma\left(\frac{2}{2}\right)}&=&\frac{1}{2^1 \; 1} &=&\frac{1}{2} \\\alpha_3&=&\frac{1}{2^{\frac{3}{2}}\Gamma\left(\frac{3}{2}\right)}&=&\frac{1}{2^{\frac{3}{2}}\;\frac{\sqrt{\pi}}{2}} &=&\frac{1}{\sqrt{2\pi}} \\\alpha_4&=&\frac{1}{2^{\frac{4}{2}}\Gamma\left(\frac{4}{2}\right)}&=&\frac{1}{2^2 \; 1} &=&\frac{1}{4} \\\alpha_5&=&\frac{1}{2^{\frac{5}{2}}\Gamma\left(\frac{5}{2}\right)}&=&\frac{1}{2^{\frac{5}{2}}\;\frac{3\sqrt{\pi}}{4}}&=&\frac{1}{3\sqrt{2\pi}} \end{array} $$累積分布凾数の等式より\(k=n\)の\(X\)の確率密度凾数を求める
$$ \begin{eqnarray} \\\int^{t}_0 f_n(x) \mathrm{d}x &=&\int^{t}_0 \frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \mathrm{d}x \;\cdots\;累積分布凾数同士なので等しい(0から任意のt(\gt0)までの定積分が等しい) \\f_n(x) &=& \frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/blog-post_25.html}{0から任意のt(\gt0)までの定積分が等しい\rightarrow0からtの範囲で等しい\rightarrow各tでの微分が等しい\rightarrow被積分凾数同士も等しい} \end{eqnarray} $$数学的帰納法による証明
\(f_1(x),f_n(x)\)を認めた上で，\(f_{n+1}(x)\)が\(f_n(x)\)の\(n\)を\(n+1\)とした式になるかを確認する． $$ \begin{eqnarray} f_1(x) &=& \frac{1}{\sqrt{2\pi}}e^{-\frac{x}{2}}\;x^{-\frac{1}{2}} \\f_{n}(x) &=& \frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}e^{-\frac{x}{2}}x^{\frac{n}{2}-1} \\f_{n+1}(x) &=& \frac{1}{2^{\frac{\color{red}{n+1}}{2}}\Gamma\left(\frac{\color{red}{n+1}}{2}\right)}e^{-\frac{x}{2}}x^{\frac{\color{red}{n+1}}{2}-1}\;\cdots\;f_n(x)のnをn+1とした式,上2式を認めた上でこの式が導出できるか． \end{eqnarray} $$ $$ \begin{eqnarray} X&=&Z_1^2+\cdots+Z_n^2+Z_{n+1}^2\;\cdots\;Z_1,\cdots,Z_n,Z_{n+1}:互いに独立に標準正規分布に従う確率変数 \\Y&=&Z_1^2+\cdots+Z_{n} \\X-Y&=&Z_{n+1}^2 \\f_{n+1}(x) &=& \int_0^\infty f_n(y)f_1(x-y) \mathrm{d}y \\&=& \int_0^\infty \left\{ \frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}e^{-\frac{y}{2}}y^{\frac{n}{2}-1} \right\}\left\{ \frac{1}{\sqrt{2\pi}}e^{-\frac{x-y}{2}}\;(x-y)^{-\frac{1}{2}} \right\} \mathrm{d}y \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{2\pi}} \int_0^\infty e^{-\frac{y}{2}}e^{-\frac{x-y}{2}} y^{\frac{n}{2}-1}(x-y)^{-\frac{1}{2}} \mathrm{d}y \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{2}\sqrt{\pi}} \int_0^\infty e^{-\frac{y}{2}-\frac{x-y}{2}} y^{\frac{n}{2}-1}(x-y)^{-\frac{1}{2}} \mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{2^{\frac{1}{2}}\sqrt{\pi}} \int_0^\infty e^{-\frac{x}{2}} y^{\frac{n}{2}-1}(x-y)^{-\frac{1}{2}} \mathrm{d}y \\&=&\frac{1}{2^{\frac{n}{2}}2^{\frac{1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} \int_0^\infty y^{\frac{n}{2}-1}(x-y)^{-\frac{1}{2}} \mathrm{d}y \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} \int_0^1 (xu)^{\frac{n}{2}-1}(x-xu)^{-\frac{1}{2}} x \mathrm{d}u \;\cdots\; y=xu, \frac{\mathrm{d}y}{\mathrm{d}u}=x, x:0\rightarrow\infty, u:0\rightarrow1 \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} \int_0^1 (xu)^{\frac{n}{2}-1}\left\{x(1-u)\right\}^{-\frac{1}{2}} x \mathrm{d}u \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} \int_0^1 x^{\frac{n}{2}-1}u^{\frac{n}{2}-1}x^{-\frac{1}{2}}(1-u)^{-\frac{1}{2}} x \mathrm{d}u \;\cdots\;(AB)^C=A^CB^C \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{\frac{n}{2}-1} x^{-\frac{1}{2}} x \int_0^1 u^{\frac{n}{2}-1}(1-u)^{-\frac{1}{2}} \mathrm{d}u \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n}{2}-1-\frac{1}{2}+1} \int_0^1 u^{\frac{n}{2}-1}(1-u)^{-\frac{1}{2}} \mathrm{d}u \;\cdots\;A^BA^C=A^{B+C} \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \int_0^1 u^{\frac{n}{2}-1}(1-u)^{-\frac{1}{2}} \mathrm{d}u \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} B\left(\frac{n}{2}-1, -\frac{1}{2}\right) \;\cdots\;B(p,q)=\int_0^1 x^{p}(1-x)^{q} \mathrm{d}x \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\left(\frac{n}{2}-1\right)!\left(-\frac{1}{2}\right)!}{\left(\frac{n}{2}-1-\frac{1}{2}+1\right)!} \;\cdots\;B(p,q)=\frac{p!q!}{\left(p+q+1\right)!} \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\left(\frac{n}{2}-1\right)!\left(-\frac{1}{2}\color{red}{+1-1}\right)!} {\left(\frac{n}{2}+\frac{1}{2}-1\right)!} \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\left(\frac{n}{2}-1\right)!\left(\frac{1}{2}-1\right)!} {\left(\frac{n+1}{2}-1\right)!} \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\Gamma\left(\frac{n}{2}\right)\Gamma\left(\frac{1}{2}\right)} {\Gamma\left(\frac{n+1}{2}\right)} \;\cdots\;\Gamma(n)=\left(n-1\right)! \\&=&\frac{1}{2^{\frac{n+1}{2}}\color{red}{\Gamma\left(\frac{n}{2}\right)}} \frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\color{red}{\Gamma\left(\frac{n}{2}\right)}\color{black}{\Gamma\left(\frac{1}{2}\right)}} {\Gamma\left(\frac{n+1}{2}\right)} \\&=&\frac{1}{2^{\frac{n+1}{2}}}\frac{1}{\sqrt{\pi}} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \frac{\Gamma\left(\frac{1}{2}\right)} {\Gamma\left(\frac{n+1}{2}\right)} \\&=&\frac{1}{2^{\frac{n+1}{2}}} \color{red}{ \frac{1}{\sqrt{\pi}} } \color{black}{e^{-\frac{x}{2}}} x^{ \frac{n+1}{2}-1} \frac{\color{red}{\sqrt{\pi}}} {\Gamma\left(\frac{n+1}{2}\right)} \;\cdots\;\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi} \\&=&\frac{1}{2^{\frac{n+1}{2}}\Gamma\left(\frac{n+1}{2}\right)} e^{-\frac{x}{2}} x^{ \frac{n+1}{2}-1} \end{eqnarray} $$ 任意の\(n(n\geq1)\)において上記式が成り立つことを証明できた．以上により\(\chi^2\)分布の確率密度凾数が導出された．
ベータ分布の期待値(平均)と分散
$$
\begin{eqnarray}
f(x;m,n)&=&\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{\frac{x^{(m-1)}(1-x)^{(n-1)}}{B(m,n)}\;\cdots\;ベータ分布}
\\B(m,n)&=&\int_0^1x^{(m-1)}(1-x)^{(n-1)} \mathrm{d}x\;\cdots\;ベータ凾数
\\&=&\frac{(m-1)!\;(n-1)!}{\left\{(m-1)+(n-1)+1\right\}!}
\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_22.html}{\int_\alpha^\beta(x-\alpha)^p(\beta-x)^q \mathrm{d}x=\frac{p!\;q!}{(p+q+1)!}(\beta-\alpha)^{(p+q+1)}(第一種オイラー積分)}
\\&=&\frac{(m-1)!\;(n-1)!}{(m+n-1)!}
\end{eqnarray}
$$
ベータ分布の期待値(一次モーメント・平均)
$$ \begin{eqnarray} \mathbf{E}[X]&=&\int_0^1xf(x)\mathrm{d}x \\&=&\int_0^1x\; \frac{x^{(m-1)}(1-x)^{(n-1)}}{B(m,n)} \;\mathrm{d}x \\&=&\frac{1}{B(m,n)} \int_0^1x^{m}(1-x)^{(n-1)}\;\mathrm{d}x \\&=&\frac{(m+n-1)!}{(m-1)!\;(n-1)!}\frac{m!\;(n-1)!}{(m+n)!} \\&=&\frac{(m+n-1)!}{(m+n)!}\frac{m!}{(m-1)!} \\&=&\frac{1}{m+n}\frac{m}{1} \\&=&\frac{m}{m+n}\;\cdots\;ベータ分布の平均 \end{eqnarray} $$ベータ分布の二次モーメント
$$ \begin{eqnarray} \mathbf{E}[X^2]&=&\int_0^1x^2f(x)\mathrm{d}x \\&=&\int_0^1x^2\; \frac{x^{(m-1)}(1-x)^{(n-1)}}{B(m,n)} \;\mathrm{d}x \\&=&\frac{1}{B(m,n)} \int_0^1x^{(m+1)}(1-x)^{(n-1)}\;\mathrm{d}x \\&=&\frac{(m+n-1)!}{(m-1)!\;(n-1)!}\frac{(m+1)!\;(n-1)!}{((m+1)+n)!} \\&=&\frac{(m+n-1)!}{(m-1)!\;(n-1)!}\frac{(m+1)!\;(n-1)!}{(m+n+1)!} \\&=&\frac{(m+n-1)!}{(m+n+1)!}\frac{(m+1)!}{(m-1)!} \\&=&\frac{1}{(m+n+1)(m+n)}\frac{(m+1)m}{1} \\&=&\frac{(m+1)m}{(m+n+1)(m+n)} \end{eqnarray} $$ベータ分布の分散(二次の中心モーメント
$$ \begin{eqnarray} \mathbf{V}[X]&=&\mathbf{E}[X^2]-\mathbf{E}[X]^2 \\&=&\frac{(m+1)m}{(m+n+1)(m+n)}-\left\{\frac{m}{m+n}\right\}^2 \\&=&\frac{(m+1)m}{(m+n+1)(m+n)}-\frac{m^2}{\left(m+n\right)^2} \\&=&\frac{(m+n)(m+1)m-(m+n+1)m^2}{(m+n+1)(m+n)^2} \\&=&\frac{(m^2+m+mn+n)m-(m^3+m^2n+m^2)}{(m+n+1)(m+n)^2} \\&=&\frac{m^3+m^2+m^2n+mn-m^3-m^2n-m^2}{(m+n+1)(m+n)^2} \\&=&\frac{mn}{(m+n+1)(m+n)^2}\;\cdots\;ベータ分布の分散 \end{eqnarray} $$第一種オイラー積分 / ベータ凾数
第一種オイラー積分 / ベータ凾数
\(p,q\geq0\)の整数の元で以下の式を第一種オイラー積分という． $$ \begin{eqnarray} \int_\alpha^\beta(x-\alpha)^p(\beta-x)^q \mathrm{d}x \end{eqnarray} $$ 部分積分を一回適用すると以下のように\(p+1\),\(q-1\)の第一種オイラー積分がでてくる． $$ \begin{eqnarray} \int_\alpha^\beta(x-\alpha)^p(\beta-x)^q \mathrm{d}x &=&\int_\alpha^\beta\left[\frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{p+1}(x-\alpha)^{(p+1)}\right\}\right](\beta-x)^q \mathrm{d}x \;\cdots\;\frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{p+1}(x-\alpha)^{(p+1)}\right\}=(x-\alpha)^p\\ &=&\left[\frac{1}{p+1}(x-\alpha)^{(p+1)}(\beta-x)^q \right]_\alpha^\beta -\int_\alpha^\beta\frac{1}{p+1}(x-\alpha)^{(p+1)}\left\{q(\beta-x)^{(q-1)}(-1)\right\} \mathrm{d}x \;\cdots\;\int f'g\;\mathrm{d}x= fg - \int fg' \mathrm{d}x\\ &=&0-\int_\alpha^\beta\frac{-q}{p+1}(x-\alpha)^{(p+1)}(\beta-x)^{(q-1)} \mathrm{d}x \;\cdots\;(\alpha-\alpha)=0, (\beta-\beta)=0\\ &=&\frac{q}{p+1}\int_\alpha^\beta(x-\alpha)^{(p+1)}(\beta-x)^{(q-1)} \mathrm{d}x\\ \end{eqnarray} $$ 第一種オイラー積分の乗数\(p,q\)に着目し元の式を\(I_{p,q}\),部分積分によって出てきた積分を\(I_{p+1,q-1}\)と表すと以下の様になる． $$ \begin{eqnarray} I_{p,q}&=&\frac{q}{p+1}I_{p+1,q-1} \;\cdots\;I_{a,b}=\int_\alpha^\beta(x-\alpha)^a(\beta-x)^b \mathrm{d}x\\ &=&\frac{q}{p+1}\frac{q-1}{p+2}I_{p+2,q-2}\\ &=&\frac{q}{p+1}\frac{q-1}{p+2}\cdots \frac{2}{p+(q-1)}\frac{1}{p+q}I_{p+q,0}\\ I_{p+q,0} &=&\int_\alpha^\beta(x-\alpha)^{(p+q)}(\beta-x)^0 \mathrm{d}x\\ &=&\int_\alpha^\beta(x-\alpha)^{(p+q)} \mathrm{d}x\\ &=&\left[\frac{1}{p+q+1}(x-\alpha)^{\left\{(p+q)+1\right\}}\right]_\alpha^\beta\\ &=&\frac{1}{p+q+1}(\beta-\alpha)^{(p+q+1)}\\ &=&\frac{(\beta-\alpha)^{(p+q+1)}}{p+q+1}\\ I_{p,q} &=&\frac{q}{p+1}\frac{q-1}{p+2}\cdots \frac{2}{p+q-1}\frac{1}{p+q}\frac{(\beta-\alpha)^{(p+q+1)}}{p+q+1}\\ &=&\frac{q(q-1)\cdots 2\;1}{(p+1)(p+2)\cdots (p+q-1)(p+q)(p+q+1)}(\beta-\alpha)^{(p+q+1)}\\ &=&\frac{q!}{\frac{(p+q+1)!}{p!}}(\beta-\alpha)^{(p+q+1)}\\ &=&\frac{p!\;q!}{(p+q+1)!}(\beta-\alpha)^{(p+q+1)}\\ \end{eqnarray} $$ 特に\(\alpha=0,\beta=1\)の時 $$ \begin{eqnarray} \int_\alpha^\beta(x-\alpha)^p(\beta-x)^q \mathrm{d}x &=&\int_0^1(x-0)^p(1-x)^q \mathrm{d}x\\ &=&\int_0^1x^p(1-x)^q \mathrm{d}x\\ &=&\frac{p!\;q!}{(p+q+1)!}(1-0)^{(p+q+1)} \;\cdots\;\href{}{\int_\alpha^\beta(x-\alpha)^p(\beta-x)^q \mathrm{d}x=\frac{p!\;q!}{(p+q+1)!}(\beta-\alpha)^{(p+q+1)}}\\ &=&\frac{p!\;q!}{(p+q+1)!}\\ \end{eqnarray} $$ 特に\(\alpha=0,\beta=1,p=m-1,q=n-1\)の時 $$ \begin{eqnarray} \int_\alpha^\beta(x-\alpha)^p(\beta-x)^q \mathrm{d}x &=&\int_0^1x^{(m-1)}(1-x)^{(n-1)} \mathrm{d}x\\ &=&\frac{(m-1)!\;(n-1)!}{((m-1)+(n-1)+1)!} \;\cdots\;\href{}{\int_0^1x^p(1-x)^q \mathrm{d}x=\frac{p!\;q!}{(p+q+1)!}}\\ &=&\frac{(m-1)!\;(n-1)!}{(m+n-1)!}\\ &=&\frac{\Gamma(m)\;\Gamma(n)}{\Gamma(m+n)}\;\cdots\;\Gamma(a)=(a-1)!\\ &=&B(m,n)\;\cdots\;ベータ凾数\\ \end{eqnarray} $$ 逆を言えば上記積分の値を1にしたい時はベータ凾数で割れば(逆数を掛ければ)よい(確率密度凾数との関係性)． $$ \begin{eqnarray} \frac{1}{B(m,n)}\int_0^1x^{(m-1)}(1-x)^{(n-1)} \mathrm{d}x&=&1\\ f(x;m,n)&=&\frac{x^{(m-1)}(1-x)^{(n-1)}}{B(m,n)}\;\cdots\;ベータ分布\\ \end{eqnarray} $$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)．
$$
\begin{eqnarray}
\frac{\mathrm{e}^{-\beta t}}{\Gamma(\alpha)}\beta^\alpha t^\left(\alpha-1\right)
&=&\frac{\mathrm{e}^{-\beta t}}{\Gamma(1)}\beta^1 t^\left(1-1\right)\;\cdots\;\alpha=1\\
&=&\frac{\mathrm{e}^{-\beta t}}{1}\beta t^\left(0\right)\;\cdots\;\Gamma(n)=(n-1)!,\Gamma(1)=(1-1)!=0!=1\\
&=&\mathrm{e}^{-\beta t}\beta\;\cdots\;t^0=1\\
&=&\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{\lambda\mathrm{e}^{-\lambda t}}\;\cdots\;\beta=\lambda\\
\end{eqnarray}
$$
事象の発生する回数についての確率密度分布 (ポアソン分布)
単位時間当たりの事象の発生確率を\(\lambda\)とし，t時間の間にk回事象が発生する確率を考える．
tまでをn等分した時間を\(\Delta t=\frac{t}{n}\)として二項分布により以下のように表すことができる．
$$
\begin{eqnarray}
p(k;\lambda,\Delta t)&=&
_nC_k
\left(\lambda\Delta t\right)^k
\left(1-\left(\lambda\Delta t\right)\right)^{\left(n-k\right)}\\
\end{eqnarray}
$$
二項分布からポアソン分布を導いた時と同様に\(\lambda\)を\(\lambda \Delta t\)として計算を進めると以下のようになる．
$$
\begin{eqnarray}
p(k;\lambda,\Delta t)
&=&_nC_k\left(\lambda\Delta t\right)^k\left(1-\left(\lambda\Delta t\right)\right)^{\left(n-k\right)}\\
&=&\frac{n!}{k!(n-k)!}\left(\lambda\Delta t\right)^k\left(1-\left(\lambda\Delta t\right)\right)^{\left(n-k\right)}\\
&=&\frac{1}{k!}\left\{n(n-1)\cdots(n-k+1)\right\}\left(\lambda\Delta t\right)^k\left(1-\left(\lambda\Delta t\right)\right)^{\left(n-k\right)}\\
&=&\frac{1}{k!}\left\{1(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})\;n^k\right\}\left(\lambda\Delta t\right)^k\left(1-\left(\lambda\Delta t\right)\right)^{\left(n-k\right)}\\
&=&\frac{1}{k!}\left\{1(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})\right\}\left(n\lambda\Delta t\right)^k\left(1-\left(\lambda\Delta t\right)\right)^{\left(n-k\right)}\\
&=&\frac{1}{k!}\left\{1(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})\right\}\left(n\lambda\frac{t}{n}\right)^k\left(1-\left(\lambda\frac{t}{n}\right)\right)^{n}\left(1-\left(\lambda\frac{t}{n}\right)\right)^{-k}
\;\cdots\;\Delta t=\frac{t}{n}\\
&=&\frac{1}{k!}\left\{1(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})\right\}\left(\lambda t\right)^k\left(1-\left(\frac{\lambda t}{n}\right)\right)^{n}\left(1-\left(\frac{\lambda t}{n}\right)\right)^{-k}\\
&=&\frac{1}{k!}\left\{1(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})\right\}\left(\lambda t\right)^k
\left(1+\frac{1}{\frac{-n}{\lambda t}}\right)^{n}
\left(1-\left(\frac{\lambda t}{n}\right)\right)^{-k}
\\
&=&\frac{1}{k!}\left\{1(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})\right\}\left(\lambda t\right)^k
\left\{\left(1+\frac{1}{\frac{-n}{\lambda t}}\right)^{\frac{-n}{\lambda t}\frac{-\lambda t}{n}}\right\}^{n}
\left(1-\left(\frac{\lambda t}{n}\right)\right)^{-k}
\\
&=&\frac{1}{k!}\left\{1(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})\right\}\left(\lambda t\right)^k
\left\{\left(1+\frac{1}{\frac{-n}{\lambda t}}\right)^{\frac{-n}{\lambda t}}\right\}^ {\frac{-\lambda t}{n}n}
\left(1-\left(\frac{\lambda t}{n}\right)\right)^{-k}
\\
&=&\frac{1}{k!}\left\{1(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})\right\}\left(\lambda t\right)^k
\left\{\left(1+\frac{1}{\frac{-n}{\lambda t}}\right)^{\frac{-n}{\lambda t}}\right\}^ {-\lambda t}
\left(1-\left(\frac{\lambda t}{n}\right)\right)^{-k}
\\
\end{eqnarray}
$$
ここでnを\(\infty\)に極限をとるとnを分母にもつ分数は0へ収束することや，収束数列によるネイピア数の定義により以下のようになる．
$$
\begin{eqnarray}
&&\lim_{n\to\infty}\frac{1}{k!}\left\{1(1-\frac{1}{n})\cdots(1-\frac{k-1}{n})\right\}\left(\lambda t\right)^k
\left\{\left(1+\frac{1}{\frac{-n}{\lambda t}}\right)^{\frac{-n}{\lambda t}}\right\}^ {-\lambda t}
\left(1-\left(\frac{\lambda t}{n}\right)\right)^{-k}\\
&=&\frac{1}{k!}\left\{1(1-0)\cdots(1-0)\right\}\left(\lambda t\right)^k
\mathrm{e}^{-\lambda t}
\left(1-0\right)^{-k}\\
&=&\frac{1}{k!}\left(\lambda t\right)^k\mathrm{e}^{-\lambda t}\\
&=&\frac{\left(\lambda t\right)^k}{k!}\mathrm{e}^{-\lambda t}\\
\end{eqnarray}
$$
標本平均の分布の歪度
標本平均(\(\overline{X}\))の分布の歪度
$$\begin{array}{rcl} \displaystyle \beta_1(\overline{X}) &=&\displaystyle \frac{E\left[\left(\overline{X}-\mu\right)^3\right]}{\left(V\left[\overline{X}\right]^{\frac{1}{2}}\right)^3}\\ &=&\displaystyle \frac{\frac{\mu_3}{n^2}}{\left\{ \left( \frac{\sigma^2}{n} \right) ^{\frac{1}{2}} \right\}^3} &&\displaystyle\,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/overlinexmu3.html}{E\left[\left(\overline{X}-\mu\right)^3\right]=\mu_3\left(\overline{X}\right)=\frac{\mu_3}{n^2}} ,\,\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable_3.html}{V\left[\overline{X}\right]=\frac{\sigma^2}{n}}\\ &=&\displaystyle \frac{\frac{\mu_3}{n^2}}{\left( \frac{\sigma}{\sqrt{n}} \right)^3}\\ &=&\displaystyle \frac{\frac{\mu_3}{n^2}}{ \frac{\sigma^3}{n\sqrt{n}} }\\ &=&\displaystyle \frac{\mu_3}{n^2} \frac{n\sqrt{n}}{\sigma^3}\\ &=&\displaystyle \frac{\mu_3}{\sigma^3} \frac{1}{\sqrt{n}}\\ &=&\displaystyle \frac{\beta_1}{\sqrt{n}} &&\displaystyle\,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/mu-sigma2beta1.html}{\frac{\mu_3}{\sigma^3}=\beta_1}\\ \end{array}$$ よって標本数\(n\)を増やすことで\(\beta_1\left(\overline{X}\right)\)は\(0\)に近づいていく．連続型確率変数(continuous random variable) の一様分布(uniform distribution)の分散(variance)
$$\begin{array}{rcl}
\displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\
&=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\
&=&\displaystyle E[X^m]\\
\end{array}$$
積率母凾数の二階微分
$$\begin{array}{rcl} \displaystyle M_X^{(2)} &=& \displaystyle \frac{\mathrm{d}^2}{\mathrm{d}t^2}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform.html}{\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t(b-a)}} \displaystyle \right\}\\ &=& \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform_8.html}{\frac{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}}{t^2(b-a)}} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (t^{-2})'\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\}' \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[\left\{(tb-1)\mathrm{e}^{tb}\right\}'-\left\{(ta-1)\mathrm{e}^{ta}\right\}'\right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[ \left\{ (tb-1)'\mathrm{e}^{tb}+(tb-1)(\mathrm{e}^{tb})' \right\} -\left\{ (ta-1)'\mathrm{e}^{ta}+(ta-1)(\mathrm{e}^{ta})' \right\} \right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[ \left\{ b\mathrm{e}^{tb}+(tb-1)b\mathrm{e}^{tb} \right\} -\left\{ a\mathrm{e}^{ta}+(ta-1)a\mathrm{e}^{ta} \right\} \right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[\frac{ \displaystyle -2\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +t\left[ \left\{ b\mathrm{e}^{tb}+(tb-1)b\mathrm{e}^{tb} \right\} -\left\{ a\mathrm{e}^{ta}+(ta-1)a\mathrm{e}^{ta} \right\} \right]}{t^3} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left[ \displaystyle -2(tb-1)\mathrm{e}^{tb}+2(ta-1)\mathrm{e}^{ta} \displaystyle +tb\mathrm{e}^{tb}+tb(tb-1)\mathrm{e}^{tb} \displaystyle -ta\mathrm{e}^{ta}-ta(ta-1)\mathrm{e}^{ta} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left[ \displaystyle \left\{-2(tb-1)+tb+tb(tb-1)\right\} \mathrm{e}^{tb} \displaystyle +\left\{2(ta-1)-ta-ta(ta-1)\right\} \mathrm{e}^{ta} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left\{ \displaystyle ((tb-1)^2+1) \mathrm{e}^{tb} \displaystyle -((ta-1)^2+1) \mathrm{e}^{ta} \displaystyle \right\}\\ &=& \displaystyle \frac{((tb-1)^2+1) \mathrm{e}^{tb}-((ta-1)^2+1) \mathrm{e}^{ta}}{t^3(b-a)}\\ \end{array}$$原点周りの二次モーメント
$$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{((tb-1)^2+1) \mathrm{e}^{tb}-((ta-1)^2+1) \mathrm{e}^{ta}}{t^3(b-a)} \displaystyle \right\}\,\dotso\,0を代入すると分母が0になってしまうので極限で考える．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle ((tb-1)^2+1) \mathrm{e}^{tb} - ((ta-1)^2+1) \mathrm{e}^{ta} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle ((tb-1)^2+1) \left(\frac{(tb)^0}{0!}+\frac{(tb)^1}{1!}+\frac{(tb)^2}{2!}+\frac{(tb)^3}{3!}\right) \displaystyle -((ta-1)^2+1) \left(\frac{(ta)^0}{0!}+\frac{(ta)^1}{1!}+\frac{(ta)^2}{2!}+\frac{(ta)^3}{3!}\right) \displaystyle \right\} \displaystyle \right]\\ && \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\mathrm{e}^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb} (マクローリン展開), t^3が分母にあるのでt^4の項以上は分子にtが残ることになるのでt^3の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle (t^2b^2-2tb+2) \left(1+tb+t^2\frac{b^2}{2}+t^3\frac{b^3}{6}\right) \displaystyle -(t^2a^2-2ta+2) \left(1+ta+t^2\frac{a^2}{2}+t^3\frac{a^3}{6}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left[ \left\{ \left( t^2b^2 +t^3b^3 +t^4\frac{b^4}{2} +t^5\frac{b^5}{6} \right) +\left( -2tb -2t^2b^2 -t^3b^3 -t^4\frac{b^4}{3} \right) +\left(2+2tb + t^2b^2 +t^3\frac{b^3}{3} \right) \right\} -\left\{ \left( t^2a^2 +t^3a^3 +t^4\frac{a^4}{2} +t^5\frac{a^5}{6} \right) +\left( -2ta -2t^2a^2 -t^3a^3 -t^4\frac{a^4}{3} \right) +\left(2+2ta + t^2a^2 +t^3\frac{a^3}{3} \right) \right\} \displaystyle \right] \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \left( 2+t^3\frac{b^3}{3}+t^4\frac{b^4}{6}+t^5\frac{b^5}{6} \right) -\left( 2+t^3\frac{a^3}{3}+t^4\frac{a^4}{6}+t^5\frac{a^5}{6} \right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ t^3\frac{b^3-a^3}{3} +t^4\frac{b^4-a^4}{6} +t^5\frac{b^5-a^5}{6} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ t^3\frac{(b-a)(b^2+ab+a^2)}{3} +t^4\frac{(b-a)(b+a)(a^2+b^2)}{6} +t^5\frac{(b-a)\frac{b^5-a^5}{b-a}}{6} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{b^2+ab+a^2}{3} +t \frac{(b+a)(a^2+b^2)}{6} +t^2\frac{\left(\frac{b^5-a^5}{b-a}\right)}{6} \displaystyle \right\}\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}=\frac{a^2+ab+b^2}{3} \,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$分散(二次の中心モーメント)
$$\begin{array}{rcl} \displaystyle V[X] &=&\displaystyle E[X^2]-E[X]^2\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}-\left(\href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform_8.html}{\frac{b+a}{2}}\right)^2\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}-\frac{b^2+2ab+a^2}{4}\\ &=&\displaystyle \frac{4(b^2+ab+a^2)-3(b^2+2ab+a^2)}{12}\\ &=&\displaystyle \frac{4b^2+4ab+4a^2-3b^2-6ab-3a^2}{12}\\ &=&\displaystyle \frac{b^2-2ab+a^2}{12}\\ &=&\displaystyle \frac{(b-a)^2}{12}\\ \end{array}$$連続型確率変数(continuous random variable) の一様分布(uniform distribution)の期待値(expected value)
$$\begin{array}{rcl}
\displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\
&=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\
&=&\displaystyle E[X^m]\\
\end{array}$$
積率母凾数の一階微分
$$\begin{array}{rcl} \displaystyle M_X^{(1)} &=& \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform.html}{\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t(b-a)}} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left\{ \displaystyle \left(t^{-1}\right)'\left(\mathrm{e}^{tb}-\mathrm{e}^{ta}\right) \displaystyle +\left(t^{-1}\right)\left(\mathrm{e}^{tb}-\mathrm{e}^{ta}\right)' \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left\{ \displaystyle \left(-t^{-2}\right)\left(\mathrm{e}^{tb}-\mathrm{e}^{ta}\right) \displaystyle +\left(t^{-1}\right)\left(b\mathrm{e}^{tb}-a\mathrm{e}^{ta}\right) \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left( \displaystyle -\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t^2} \displaystyle +\frac{b\mathrm{e}^{tb}-a\mathrm{e}^{ta}}{t} \displaystyle \right)\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left\{ \displaystyle -\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t^2} \displaystyle +\frac{(b\mathrm{e}^{tb}-a\mathrm{e}^{ta})t}{t^2} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left( \displaystyle \frac{tb\mathrm{e}^{tb}-\mathrm{e}^{tb}-ta\mathrm{e}^{ta}+\mathrm{e}^{ta}}{t^2} \displaystyle \right)\\ &=& \displaystyle \frac{1}{t^2(b-a)} \displaystyle \left\{ \displaystyle (tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta} \displaystyle \right\}\\ &=& \displaystyle \frac{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}}{t^2(b-a)}\\ \end{array}$$原点周りの一次モーメント=期待値
$$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}}{t^2(b-a)} \displaystyle \right\}\,\dotso\,0を代入すると分母が0になってしまうので極限で考える．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle (tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle (tb-1)\left(\frac{(tb)^0}{0!}+\frac{(tb)^1}{1!}+\frac{(tb)^2}{2!}\right) \displaystyle -(ta-1)\left(\frac{(ta)^0}{0!}+\frac{(ta)^1}{1!}+\frac{(ta)^2}{2!}\right) \displaystyle \right\} \displaystyle \right]\\ && \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\mathrm{e}^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\dotsb} (マクローリン展開), t^2が分母にあるのでt^3の項以上は分子にtが残ることになるのでt^2の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle (tb-1)\left(1+tb+t^2\frac{b^2}{2}\right) \displaystyle -(ta-1)\left(1+ta+t^2\frac{a^2}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left[ \displaystyle \left\{\left(tb+t^2b^2+t^3\frac{b^3}{2}\right)-\left(1+tb+t^2\frac{b^2}{2}\right)\right\} \displaystyle -\left\{\left(ta+t^2a^2+t^3\frac{a^3}{2}\right)-\left(1+ta+t^2\frac{a^2}{2}\right)\right\} \displaystyle \right] \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle \left(-1+t(b-b)+t^2(b^2-\frac{b^2}{2})+t^3\frac{b^3}{2}\right) \displaystyle -\left(-1+t(a-a)+t^2(a^2-\frac{a^2}{2})+t^3\frac{a^3}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle \left(-1+t^2\frac{b^2}{2}+t^3\frac{b^3}{2}\right) \displaystyle -\left(-1+t^2\frac{a^2}{2}+t^3\frac{a^3}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{1}{t^2(b-a)}\left( \displaystyle t^2\frac{b^2-a^2}{2} \displaystyle +t^3\frac{b^3-a^3}{2} \displaystyle \right) \displaystyle \right\}\\ &=&\displaystyle \lim_{t \to 0} \left[\frac{1}{t^2(b-a)}\left\{ \displaystyle t^2 \frac{ (b-a)(b+a) }{2} \displaystyle +t^3 \frac{ (b-a)(b^2+ab+a^2) }{2} \displaystyle \right\}\right] \,\dotso\,b^2-a^2=(b-a)(b+a),\,b^3-a^3=(b-a)(b^2+ab+a^2)\\ &=&\displaystyle \lim_{t \to 0} \displaystyle \left\{ \displaystyle \frac{ (b+a) }{2} \displaystyle +t \frac{ (b^2+ab+a^2) }{2} \displaystyle \right\}\\ &=&\displaystyle \frac{b+a}{2}=\frac{a+b}{2} \,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$連続型確率変数(continuous random variable) の一様分布(uniform distribution)の積率母凾数(moment-generating function)
$$f_X(x) =
\begin{cases}
\displaystyle \frac{1}{b-a} & \quad \left\{ a\leq x \leq b\right\}\\
\displaystyle 0 & \quad \left\{ x\lt a,\,b\lt x\right\}\\
\end{cases}
$$
$$\begin{array}{rcl}
\displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\
&=&\displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{tx}f_X(x)\mathrm{d}x\\
&=&\displaystyle \int_{a}^{b}\mathrm{e}^{tx}\left(\frac{1}{b-a}\right)\mathrm{d}x\\
&=&\displaystyle \frac{1}{b-a}\int_{a}^{b}\mathrm{e}^{tx}\mathrm{d}x\\
&=&\displaystyle \frac{1}{b-a}\int_{ta}^{tb}\mathrm{e}^{s}\frac{1}{t}\mathrm{d}s\,\dotso\,s=tx,\,\frac{\mathrm{d}s}{\mathrm{d}x}=t,\, \mathrm{d}x=\frac{1}{t}\mathrm{d}s,\,a\to ta,\,b\to tb\\
&=&\displaystyle \frac{1}{b-a}\frac{1}{t}\int_{ta}^{tb}\mathrm{e}^{s}\mathrm{d}s\\
&=&\displaystyle \frac{1}{t(b-a)}\left[\mathrm{e}^{s}\right]_{ta}^{tb}\\
&=&\displaystyle \frac{1}{t(b-a)}\left[\mathrm{e}^{tb}-\mathrm{e}^{ta}\right]\\
&=&\displaystyle \frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t(b-a)}\\
\end{array}$$
離散型確率変数(discrete random variable) の一様分布(uniform distribution)の分散(variance)
$$\begin{array}{rcl}
\displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\
&=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\
&=&\displaystyle E[X^m]\\
\end{array}$$
積率母凾数の二階微分
$$\begin{array}{rcl} \displaystyle M_X^{(2)} &=&\displaystyle \frac{\mathrm{d^2}}{\mathrm{d}t^2}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/uniform-distribution.html}{\frac{1}{n}\frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)}} \displaystyle \right\}\\ &=&\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/discrete-random-variable-uniform.html}{\frac{\mathrm{e}^{t}}{n}\frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2}} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \mathrm{e}^{t} \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \mathrm{e}^{t} (n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ (\mathrm{e}^{t})'(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)'(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left((\mathrm{e}^{t}-1)^{-2}\right)' \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ \mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}((n\mathrm{e}^{(n+1)t})'-((n+1)\mathrm{e}^{nt})'+(1)')(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left(-2\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-3}\right) \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ \mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt})+0)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left(-2\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-3}\right) \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{ (n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt}))(\mathrm{e}^{t}-1)^{-2} +(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left(-2\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-3}\right) \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ (n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1) +((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt}))(\mathrm{e}^{t}-1) -2\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1) }{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ (n\mathrm{e}^{(n+2)t}-(n+1)\mathrm{e}^{(n+1)t}+\mathrm{e}^{t})-(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1) +((n(n+1)\mathrm{e}^{(n+2)t})-(n(n+1)\mathrm{e}^{(n+1)t}))-((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt})) -(2n\mathrm{e}^{(n+2)t}-2(n+1)\mathrm{e}^{(n+1)t}+2\mathrm{e}^{t}) }{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ n\mathrm{e}^{(n+2)t}-(n+1)\mathrm{e}^{(n+1)t}+\mathrm{e}^{t}-n\mathrm{e}^{(n+1)t}+(n+1)\mathrm{e}^{nt}-1 +n(n+1)\mathrm{e}^{(n+2)t}-n(n+1)\mathrm{e}^{(n+1)t}-n(n+1)\mathrm{e}^{(n+1)t}+n(n+1)\mathrm{e}^{nt} -2n\mathrm{e}^{(n+2)t}+2(n+1)\mathrm{e}^{(n+1)t}-2\mathrm{e}^{t} }{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ (n+n(n+1)-2n)\mathrm{e}^{(n+2)t} +(-(n+1)-n-n(n+1)-n(n+1)+2(n+1))\mathrm{e}^{(n+1)t} +((n+1)+n(n+1))\mathrm{e}^{nt} +(1-2)\mathrm{e}^{t} -1}{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ n^2\mathrm{e}^{(n+2)t} -(2n^2+2n-1)\mathrm{e}^{(n+1)t} +(n^2+2n+1)\mathrm{e}^{nt} -\mathrm{e}^{t} -1}{(\mathrm{e}^{t}-1)^{3}} \right\}\\ \end{array}$$原点周りの二次モーメント
$$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\mathrm{e}^{t}}{n} \left\{\frac{ n^2\mathrm{e}^{(n+2)t} -(2n^2+2n-1)\mathrm{e}^{(n+1)t} +(n^2+2n+1)\mathrm{e}^{nt} -\mathrm{e}^{t} -1}{(\mathrm{e}^{t}-1)^{3}} \right\} \right]\,\dotso\,0を代入すると分母が0になってしまうので極限で考える．\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\mathrm{e}^{t}}{n} \left\{\frac{ n^2\mathrm{e}^{(n+2)t} -(2n^2+2n-1)\mathrm{e}^{(n+1)t} +(n^2+2n+1)\mathrm{e}^{nt} -\mathrm{e}^{t} -1}{ \mathrm{e}^{3t} -3\mathrm{e}^{2t} +3\mathrm{e}^{t} -1} \right\} \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}\right)}{n} \left\{\frac{ n^2\left( \frac{((n+2)t)^0}{0!}+\frac{((n+2)t)^1}{1!}+\frac{((n+2)t)^2}{2!}+\frac{((n+2)t)^3}{3!}+\frac{((n+2)t)^4}{4!} \right) -(2n^2+2n-1)\left( \frac{((n+1)t)^0}{0!}+\frac{((n+1)t)^1}{1!}+\frac{((n+1)t)^2}{2!}+\frac{((n+1)t)^3}{3!}+\frac{((n+1)t)^4}{4!} \right) +(n^2+2n+1)\left( \frac{(nt)^0}{0!}+\frac{(nt)^1}{1!}+\frac{(nt)^2}{2!}+\frac{(nt)^3}{3!}+\frac{(nt)^4}{4!} \right) -\left( \frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!} \right) -1}{ \left( \frac{(3t)^0}{0!}+\frac{(3t)^1}{1!}+\frac{(3t)^2}{2!}+\frac{(3t)^3}{3!}+\frac{(3t)^4}{4!} \right) -3\left( \frac{(2t)^0}{0!}+\frac{(2t)^1}{1!}+\frac{(2t)^2}{2!}+\frac{(2t)^3}{3!}+\frac{(2t)^4}{4!} \right) +3\left( \frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!} \right) -1} \right\} \right]\\ && \displaystyle \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\mathrm{e}^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dotsb} (マクローリン展開), ひとまず4乗の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{n} \left\{\frac{ n^2\left( 1 +(n+2)t +\frac{(n+2)^2}{2}t^2 +\frac{(n+2)^3}{6}t^3 +\frac{(n+2)^4}{24}t^4 \right) -(2n^2+2n-1)\left( 1 +(n+1)t +\frac{(n+1)^2}{2}t^2 +\frac{(n+1)^3}{6}t^3 +\frac{(n+1)^4}{24}t^4 \right) +(n^2+2n+1)\left( 1 +nt +\frac{n^2}{2}t^2 +\frac{n^3}{6}t^3 +\frac{n^4}{24}t^4 \right) -\left( 1 +t +\frac{1}{2}t^2 +\frac{1}{6}t^3 +\frac{1}{24}t^4 \right) -1}{ 1 +3t +\frac{9}{2}t^2 +\frac{27}{6}t^3 +\frac{81}{24}t^4 -3 -6t -\frac{12}{2}t^2 -\frac{24}{6}t^3 -\frac{48}{24}t^4 +3 +3t +\frac{3}{2}t^2 +\frac{3}{6}t^3 +\frac{3}{24}t^4 -1} \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{n} \left\{\frac{ n^2 \left(1+(n+2)t +\frac{(n+2)^2}{2}t^2 +\frac{(n+2)^3}{6}t^3 +\frac{(n+2)^4}{24}t^4 \right) -(2n^2+2n-1)\left(1+(n+1)t +\frac{(n+1)^2}{2}t^2 +\frac{(n+1)^3}{6}t^3 +\frac{(n+1)^4}{24}t^4 \right) +(n^2+2n+1) \left(1+nt +\frac{n^2}{2}t^2 +\frac{n^3}{6}t^3 +\frac{n^4}{24}t^4 \right) - \left(1+t +\frac{1}{2}t^2 +\frac{1}{6}t^3 +\frac{1}{24}t^4 \right) -1}{t^3+\frac{3}{2}t^4} \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ n^2 \left(1+(n+2)t +\frac{(n+2)^2}{2}t^2 +\frac{(n+2)^3}{6}t^3 +\frac{(n+2)^4}{24}t^4 \right) -(2n^2+2n-1)\left(1+(n+1)t +\frac{(n+1)^2}{2}t^2 +\frac{(n+1)^3}{6}t^3 +\frac{(n+1)^4}{24}t^4 \right) +(n^2+2n+1) \left(1+nt +\frac{n^2}{2}t^2 +\frac{n^3}{6}t^3 +\frac{n^4}{24}t^4 \right) - \left(1+t +\frac{1}{2}t^2 +\frac{1}{6}t^3 +\frac{1}{24}t^4 \right) -1 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ n^2 +n^2(n+2)t +\frac{n^2(n+2)^2}{2}t^2 +\frac{n^2(n+2)^3}{6}t^3 +\frac{n^2(n+2)^4}{24}t^4 -(2n^2+2n-1) -(2n^2+2n-1)(n+1)t -\frac{(2n^2+2n-1)(n+1)^2}{2}t^2 -\frac{(2n^2+2n-1)(n+1)^3}{6}t^3 +\frac{(2n^2+2n-1)(n+1)^4}{24}t^4 +(n^2+2n+1) +(n^2+2n+1)nt +\frac{(n^2+2n+1)n^2}{2}t^2 +\frac{(n^2+2n+1)n^3}{6}t^3 +\frac{(n^2+2n+1)n^4}{24}t^4 -1 -t -\frac{1}{2}t^2 -\frac{1}{6}t^3 +\frac{1}{24}t^4 -1 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ (n^2 -(2n^2+2n-1) +(n^2+2n+1) -1 -1) +(n^2(n+2) -(2n^2+2n-1)(n+1) +(n^2+2n+1)n -1 )t +(\frac{n^2(n+2)^2}{2} -\frac{(2n^2+2n-1)(n+1)^2}{2} +\frac{(n^2+2n+1)n^2}{2} -\frac{1}{2} )t^2 +(\frac{n^2(n+2)^3}{6} -\frac{(2n^2+2n-1)(n+1)^3}{6} +\frac{(n^2+2n+1)n^3}{6} -\frac{1}{6} )t^3 +(\frac{n^2(n+2)^4}{24} -\frac{(2n^2+2n-1)(n+1)^4}{24} +\frac{(n^2+2n+1)n^4}{24} -\frac{1}{24} )t^4 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ (0) +(0)t +(0)t^2 +\frac{n(n+1)(2n+1)}{6}t^3 +\frac{n(n+1)(3n^2+5n+1)}{12}t^4 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} nt^3 \left\{ \frac{(n+1)(2n+1)}{6} +\frac{(n+1)(3n^2+5n+1)}{12}t \right\}\right]\\ &&\,\dotso\,分母はt^3の項からが残っている．t^4以上の項はtが残るのでマクローリン展開はt^4で十分となる\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{1+\frac{3}{2}t} \left\{ \frac{(n+1)(2n+1)}{6} +\frac{(n+1)(3n^2+5n+1)}{12}t \right\}\right]\\ &=&\displaystyle \frac{(n+1)(2n+1)}{6}\\ &&\,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$分散
$$\begin{array}{rcl} \displaystyle V[X]&=&\displaystyle E[X^2]-E[X]^2\\ &=&\displaystyle \frac{(n+1)(2n+1)}{6}-\left(\href{https://shikitenkai.blogspot.com/2019/07/discrete-random-variable-uniform.html}{\frac{n+1}{2}}\right)^2\\ &=&\displaystyle \frac{(n+1)(2n+1)}{6}-\frac{(n+1)^2}{4}\\ &=&\displaystyle \frac{2(n+1)(2n+1)-3(n+1)^2}{12}\\ &=&\displaystyle \frac{(n+1)(2(2n+1)-3(n+1))}{12}\\ &=&\displaystyle \frac{(n+1)(4n+2-3n-3)}{12}\\ &=&\displaystyle \frac{(n+1)(n-1)}{12}\\ &=&\displaystyle \frac{n^2-1}{12}\\ \end{array}$$離散型確率変数(discrete random variable) の一様分布(uniform distribution)の期待値(expected value)
$$\begin{array}{rcl}
\displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\
&=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\
&=&\displaystyle E[X^m]\\
\end{array}$$
積率母凾数の一階微分
$$\begin{array}{rcl} \displaystyle M_X^{(1)} &=& \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/uniform-distribution.html}{\frac{1}{n}\frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)}} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{n} \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{n} \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \mathrm{e}^{t}(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1)^{-1} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ (\mathrm{e}^{t})'(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(\mathrm{e}^{nt}-1)'(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(\mathrm{e}^{nt}-1)((\mathrm{e}^{t}-1)^{-1})' \right\}\\ &&\,\dotso\,(uvw)'=u'(vw)+u(vw)'=u'(vw)+u(v'w+vw')=u'vw+uv'w+uvw'\\ &=&\displaystyle \frac{1}{n} \left\{ \mathrm{e}^{t}(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(n\mathrm{e}^{nt})(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(\mathrm{e}^{nt}-1)(-\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-2}) \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)} + \frac{\mathrm{e}^{t}(n\mathrm{e}^{nt})}{(\mathrm{e}^{t}-1)} + \frac{\mathrm{e}^{t}\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{-(\mathrm{e}^{t}-1)^2} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{ \frac{(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)} + \frac{(n\mathrm{e}^{nt})}{(\mathrm{e}^{t}-1)} + \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{-(\mathrm{e}^{t}-1)^2} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1) + (n\mathrm{e}^{nt})(\mathrm{e}^{t}-1) - \mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{(\mathrm{e}^{nt}\mathrm{e}^{t}-\mathrm{e}^{nt}-\mathrm{e}^{t}+1) + (n\mathrm{e}^{nt}\mathrm{e}^{t}-n\mathrm{e}^{nt}) - (\mathrm{e}^{nt}\mathrm{e}^{t}-\mathrm{e}^{t})}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{\mathrm{e}^{nt}\mathrm{e}^{t}-\mathrm{e}^{nt}-\mathrm{e}^{t}+1 + n\mathrm{e}^{nt}\mathrm{e}^{t}-n\mathrm{e}^{nt} - \mathrm{e}^{nt}\mathrm{e}^{t}+\mathrm{e}^{t}}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{\mathrm{e}^{nt}(\mathrm{e}^{t} - 1 +n\mathrm{e}^{t} -n-\mathrm{e}^{t}) +1}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{\mathrm{e}^{nt}(n\mathrm{e}^{t}-(n+1))+1}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2}\\ \end{array}$$原点周りの一次モーメント=期待値
$$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\mathrm{e}^{t}}{n} \displaystyle \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2} \right\}\,\dotso\,0を代入すると分母が0になってしまうので極限で考える．\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\mathrm{e}^{t}}{n} \displaystyle \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{\mathrm{e}^{2t}-2\mathrm{e}^{t}+1} \right\}\\ &=&\displaystyle \lim_{t \to 0} \left\{\frac{\left(\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}\right)}{n} \frac{n\left(\frac{((n+1)t)^0}{0!}+\frac{((n+1)t)^1}{1!}+\frac{((n+1)t)^2}{2!}+\frac{((n+1)t)^3}{3!}\right) -(n+1)\left(\frac{(nt)^0}{0!}+\frac{(nt)^1}{1!}+\frac{(nt)^2}{2!}+\frac{(nt)^3}{3!}\right) +1}{ \left(\frac{(2t)^0}{0!}+\frac{(2t)^1}{1!}+\frac{(2t)^2}{2!}+\frac{(2t)^3}{3!}\right) -2 \left(\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}\right) +1} \right\}\\ && \displaystyle \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\mathrm{e}^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb} (マクローリン展開), ひとまずt^3の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{n} \frac{n\left(1+(n+1)t+\frac{(n+1)^2}{2}t^2+\frac{(n+1)^3}{6}t^3\right) -(n+1)\left(1+nt+\frac{n^2}{2}t^2+\frac{n^3}{6}t^3\right) +1}{ \left(1+(2t)+\frac{(2t)^2}{2}+\frac{(2t)^3}{6}\right) -2 \left(1+(t)+\frac{t^2}{2}+\frac{t^3}{6}\right) +1} \right\}\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{n} \frac{n\left(1+(n+1)t+\frac{(n+1)^2}{2}t^2+\frac{(n+1)^3}{6}t^3\right) -(n+1)\left(1+nt+\frac{n^2}{2}t^2+\frac{n^3}{6}t^3\right) +1}{ 1+2t+2t^2+\frac{4}{3}t^3 -2-2t-t^2-\frac{1}{3}t^3 +1} \right\}\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{n} \frac{n\left(1+(n+1)t+\frac{(n+1)^2}{2}t^2+\frac{(n+1)^3}{6}t^3\right) -(n+1)\left(1+nt+\frac{n^2}{2}t^2+\frac{n^3}{6}t^3\right) +1}{t^2(1+t)} \right\}\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} \left\{n+n(n+1)t+\frac{n(n+1)^2}{2}t^2+\frac{n(n+1)^3}{6}t^3 -(n+1)-(n+1)nt-\frac{(n+1)n^2}{2}t^2-\frac{(n+1)n^3}{6}t^3 +1\right\} \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} \left\{(n-(n+1)+1) +(n(n+1)-(n+1)n)t +(\frac{n(n+1)^2}{2}-\frac{(n+1)n^2}{2})t^2 +(\frac{n(n+1)^3}{6}-\frac{(n+1)n^3}{6})t^3 \right\} \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} \left\{(0) +(0)t +(\frac{n+1}{2})nt^2 +(\frac{(n+1)(2n+1)}{6})nt^3 \right\} \right]\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} nt^2\left(\frac{n+1}{2}+\frac{(n+1)(2n+1)}{6}t\right) \right\}\\ &&\,\dotso\,分母はt^2の項からが残っている．t^3以上の項はtが残るのでマクローリン展開はt^3で十分となる．\\ &=&\displaystyle \lim_{t \to 0}\left\{ \left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right) \left(\frac{n+1}{2}+\frac{(n+1)(2n+1)}{6}t\right) \right\}\\ &=&\displaystyle \frac{n+1}{2}\\ &&\,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$離散型確率変数(discrete random variable) の一様分布(uniform distribution)の積率母凾数(moment-generating function)
$$f_X(x) =
\begin{cases}
\displaystyle \frac{1}{n} & \quad x \in \left\{1,2, \dots ,n\right\}\\
\displaystyle 0 & \quad x \notin \left\{1,2, \dots ,n\right\}
\end{cases}
$$
$$\begin{array}{rcl}
\displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\
&=&\displaystyle \sum_{x=1}^{n}\mathrm{e}^{tx}\left(\frac{1}{n}\right)\\
&=&\displaystyle \frac{1}{n} \sum_{x=1}^{n}\mathrm{e}^{tx}\\
&=&\displaystyle \frac{1}{n} \left(\mathrm{e}^t+\mathrm{e}^{2t}+\dotsb+\mathrm{e}^{nt}\right)\\
&=&\displaystyle \frac{1}{n} \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{\mathrm{e}^{t}-1}
\,\dotso\,a+ar+ar^2+\dotsb+ar^{n-1}=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}\;(r \neq 1)\\
&=&\displaystyle \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{n(\mathrm{e}^{t}-1)}\\
\end{array}$$
ポアソン分布(Poisson distribution)の積率母凾数(moment-generating function)と期待値(expected value)・分散(variance)
ポアソン分布
$$\begin{array}{rcl} \displaystyle Po(\lambda)&=&\displaystyle \frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\ \end{array}$$積率母凾数
$$\begin{array}{rcl} \displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\ &=&\displaystyle \sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{\mathrm{e}^{tx}\lambda^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{(\mathrm{e}^{t}\lambda)^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\mathrm{e}^{\mathrm{e}^{t}\lambda} \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\sum_{x=0}^{\infty}\frac{a^{x}}{x!}=\mathrm{e}^a}\\ &=&\displaystyle \mathrm{e}^{\mathrm{e}^{t}\lambda-\lambda}=\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)} \end{array}$$期待値・分散
$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_x(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0} \,\dotso\,s=\lambda(\mathrm{e}^{t}-1),\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}\\ &=&\displaystyle \left\{ (\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)})(\lambda\mathrm{e}^{t}) \right\}|_{t=0} \,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }\mathrm{e}^{x}=\mathrm{e}^{x}\\ &=&\displaystyle \left\{ \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0}\\ &=&\displaystyle \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+0})\\ &=&\displaystyle \lambda\mathrm{e}^0 \,\dotso\,a^0=1\\ &=&\lambda \,\dotso\,a^0=1\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d}^2 }{ \mathrm{d} t^2 }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0} \,\dotso\,E[X]の展開から．\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\lambda\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0} \,\dotso\,s=\lambda(\mathrm{e}^{t}-1)+t,\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}+1\\ &=&\displaystyle \left\{ (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+1})(\lambda\mathrm{e}^{t}+1) \right\}|_{t=0} \,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }C\mathrm{e}^{x}=C\mathrm{e}^{x}\\ &=&\displaystyle (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+1})(\lambda\mathrm{e}^{0}+1)\\ &=&\displaystyle \lambda\mathrm{e}^0(\lambda+1) \,\dotso\,a^0=1\\ &=&\lambda(\lambda+1) \,\dotso\,a^0=1\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle V[X]&=&\displaystyle E[X^2]-E[X]^2\\ &=&\lambda(\lambda+1)-\lambda^2\\ &=&\lambda^2+\lambda-\lambda^2\\ &=&\lambda\\ \end{array}$$二項分布(binomial distribution)
$$B(n, p) = f_X(x) =
\begin{cases}
\displaystyle _nC_x\,p^x(1-p)^{n-x} & \quad x \in \left\{0,1,2, \dotsc ,n\right\}\\
\displaystyle 0 & \quad x \notin \left\{0,1,2, \dotsc ,n\right\}
\end{cases}
$$
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