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． $$\begin{array}{rcl}X& =& \frac{Y}{\sqrt{\frac{Z}{n}}}\cdots \frac{Y}{\mathrm{RMS}\left[Y\right]}\\ Y& :& f_Y(y)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{y^2}{2}}\cdots z\mathrm{分}\mathrm{布}\left(\mathrm{標}\mathrm{準}\mathrm{正}\mathrm{規}\mathrm{分}\mathrm{布}\right)\\ Z& :& f_Z(z)=\frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{e}^{-\frac{z}{2}}{z}^{\frac{n}{2}-1}\cdots {\chi }^2\mathrm{分}\mathrm{布}(Z=\sum _{i=1}^n{Y}_i^2)\\ & & \end{array}$$

$f_{YZ}(y,z)$, $Y$と$Z$の同時確率

$$\begin{array}{rcl}{f}_{YZ}(y,z)& =& f_Y(y)\cdot f_Z(z)\cdots Y\mathrm{と}Z\mathrm{は}\mathrm{独}\mathrm{立}\end{array}$$

変数変換, $Y$と$Z$から$X$と$U$へ

$$\begin{array}{rclrclr}x& =& \frac{y}{\sqrt{\frac{z}{n}}}& \mathrm{よ}\mathrm{り}& y& =& x\sqrt{\frac{z}{n}}\end{array}$$ $$\{\begin{array}{rcl}z& =& u\\ y& =& x\sqrt{\frac{u}{n}}\end{array}$$ $$\begin{array}{rcl}J& =& \left|\frac{\partial (y,z)}{\partial (x,u)}\right|\\ & =& \left|\begin{array}{cc}\frac{\partial y}{\partial x}& \frac{\partial y}{\partial u}\\ \frac{\partial z}{\partial x}& \frac{\partial z}{\partial u}\end{array}\right|\\ & =& \left|\begin{array}{cc}\sqrt{\frac{u}{n}}& \frac{x}{\sqrt{n}}\frac{1}{2}{u}^{-\frac{1}{2}}\\ 0& 1\end{array}\right|\\ & =& \sqrt{\frac{u}{n}}\cdot 1-\frac{x}{\sqrt{n}}\frac{1}{2}{u}^{-\frac{1}{2}}\cdot 0\\ & =& \sqrt{\frac{u}{n}}\end{array}$$ $$\begin{array}{rcl}& & {\int }_0^{\mathrm{\infty }}{\int }_{\mathrm{\infty }}^{-\mathrm{\infty }}{f}_Y(y)f_Z(z)\mathrm{d}y\mathrm{d}z\cdots Y,Z\mathrm{の}\mathrm{同}\mathrm{時}\mathrm{確}\mathrm{率}\mathrm{の}\mathrm{全}\mathrm{事}\mathrm{象}\\ & =& {\int }_0^{\mathrm{\infty }}{\int }_{\mathrm{\infty }}^{-\mathrm{\infty }}{f}_Y\left(x\sqrt{\frac{u}{n}}\right)f_Z(u)\sqrt{\frac{u}{n}}\mathrm{d}x\mathrm{d}u\\ & & \cdots \mathrm{変}\mathrm{数}\mathrm{変}\mathrm{換}:y=x\sqrt{\frac{u}{n}},z=u,J=\sqrt{\frac{u}{n}}\\ & & \cdots x=\frac{y}{\sqrt{\frac{u}{n}}}\mathrm{な}\mathrm{の}\mathrm{で},y:-\mathrm{\infty }\to \mathrm{\infty },x:-\mathrm{\infty }\to \mathrm{\infty }\\ & & \cdots u=z\mathrm{な}\mathrm{の}\mathrm{で},z:0\to \mathrm{\infty },u:0\to \mathrm{\infty }\\ & =& {\int }_0^{\mathrm{\infty }}{\int }_{\mathrm{\infty }}^{-\mathrm{\infty }}\frac{1}{\sqrt{2\pi }}{e}^{-\frac{{\left(x\sqrt{\frac{u}{n}}\right)}^2}{2}}\frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{e}^{-\frac{u}{2}}{u}^{\frac{n}{2}-1}\sqrt{\frac{u}{n}}\mathrm{d}x\mathrm{d}u\\ & =& \frac{1}{\sqrt{2\pi }}\frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{n}}{\int }_0^{\mathrm{\infty }}{\int }_{\mathrm{\infty }}^{-\mathrm{\infty }}{e}^{-\frac{x^{2}u}{2n}}{e}^{-\frac{u}{2}}{u}^{\frac{n}{2}-1}\sqrt{u}\mathrm{d}x\mathrm{d}u\\ & =& \frac{1}{\sqrt{2\pi n}{2}^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{\int }_{\mathrm{\infty }}^{-\mathrm{\infty }}{e}^{-\frac{x^{2}u}{2n}-\frac{u}{2}}{u}^{\frac{n}{2}-1+\frac{1}{2}}\mathrm{d}x\mathrm{d}u\\ & =& \frac{1}{\sqrt{2\pi n}{2}^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{\int }_{\mathrm{\infty }}^{-\mathrm{\infty }}{e}^{-\frac{u}{2}(\frac{x^2}{n}+1)}{u}^{\frac{n+1}{2}-1}\mathrm{d}x\mathrm{d}u\\ & =& 1\cdots \mathrm{全}\mathrm{事}\mathrm{象}\\ & =& F_{XU}(x,u)\end{array}$$

$f_X\left(x\right)$, $X$の確率密度凾数

$$\begin{array}{rcl}{f}_{XU}(x,u)& =& \frac{{\partial }^2}{\partial x\partial u}{F}_{XU}(x,u)\\ & =& \frac{1}{\sqrt{2\pi n}{2}^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{e}^{-\frac{u}{2}(1+\frac{x^2}{n})}{u}^{\frac{n+1}{2}-1}\end{array}$$ $X$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{array}{rcl}t(x)=f_X(x)& =& {\int }_0^{\mathrm{\infty }}{f}_{XU}(x,u)\mathrm{d}u\\ & =& \frac{1}{\sqrt{2\pi n}{2}^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{u}{2}(\frac{x^2}{n}+1)}{u}^{\frac{n+1}{2}-1}\mathrm{d}u\\ & =& \frac{1}{\sqrt{2\pi n}{2}^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\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}(\frac{x^2}{n}+1)\\ & & \cdots u:0\to \mathrm{\infty },t:0\to \mathrm{\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}}\mathrm{\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^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{n+1}{2}-1}\mathrm{d}t\\ & =& \frac{1}{\sqrt{2\pi n}{2}^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\left(\frac{2}{\frac{x^2}{n}+1}\right)}^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n+1}{2}\right)\\ & =& \frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{\sqrt{2\pi n}{2}^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{2}^{\frac{n+1}{2}}{(\frac{x^2}{n}+1)}^{-\frac{n+1}{2}}\\ & =& \frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{\sqrt{\cancel{2}\pi n}\cancel{2^{\frac{n}{2}}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\cancel{2^{\frac{n}{2}}}\cancel{2^{\frac{1}{2}}}{(\frac{x^2}{n}+1)}^{-\frac{n+1}{2}}\\ & =& \frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{\sqrt{\pi n}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{(\frac{x^2}{n}+1)}^{-\frac{n+1}{2}}\\ & =& \frac{1}{\sqrt{n}}\frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{\sqrt{\pi }\mathrm{\Gamma }\left(\frac{n}{2}\right)}{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\\ & =& \frac{1}{\sqrt{n}}\frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{n}{2}\right)}{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\cdots \mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }\\ & =& \frac{1}{\sqrt{n}}\frac{\mathrm{\Gamma }(\frac{1}{2}+\frac{n}{2})}{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{n}{2}\right)}{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\\ & =& \frac{1}{\sqrt{n}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\cdots \beta (a,b)=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}={\int }_0^1{x}^{a-1}(1-x{)}^{b-1}\mathrm{d}x\end{array}$$