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}
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