t分布の期待値と分散

t分布の期待値

$$\begin{array}{rcl}\mathrm{E}\left[X\right]& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x\cdot \frac{1}{\sqrt{n}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\mathrm{d}x\dots t\mathrm{分}\mathrm{布}\mathrm{の}\mathrm{確}\mathrm{率}\mathrm{密}\mathrm{度}\mathrm{凾}\mathrm{数}:\frac{1}{\sqrt{n}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\\ & =& \frac{1}{\sqrt{n}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\mathrm{d}x\end{array}$$ $$\begin{array}{rcl}\mathrm{彼}\mathrm{積}\mathrm{分}\mathrm{凾}\mathrm{数}(x)& =& x{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\\ \mathrm{彼}\mathrm{積}\mathrm{分}\mathrm{凾}\mathrm{数}(-x)& =& (-x){(1+\frac{(-x{)}^2}{n})}^{-\frac{n+1}{2}}\\ & =& -x{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\\ & =& -\mathrm{彼}\mathrm{積}\mathrm{分}\mathrm{凾}\mathrm{数}(x)\dots \mathrm{奇}\mathrm{凾}\mathrm{数}\end{array}$$ $$\begin{array}{rcl}\mathrm{E}\left[X\right]& =& \frac{1}{\sqrt{n}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\mathrm{d}x\\ & =& \frac{1}{\sqrt{n}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}\cdot 0\\ & & \dots \mathrm{奇}\mathrm{凾}\mathrm{数}\mathrm{の}\mathrm{上}\mathrm{端}\mathrm{と}\mathrm{下}\mathrm{端}\mathrm{の}\mathrm{絶}\mathrm{対}\mathrm{値}\mathrm{が}\mathrm{等}\mathrm{し}\mathrm{い}\mathrm{定}\mathrm{積}\mathrm{分}\mathrm{は}0．\\ & & \dots {\int }_{-a}^{a}f(x)\mathrm{d}x=0,f(x)\mathrm{が}\mathrm{奇}\mathrm{凾}\mathrm{数}\mathrm{の}\mathrm{場}\mathrm{合}\\ & =& 0\end{array}$$

t分布の分散

$$\begin{array}{rcl}\mathrm{E}\left[X^2\right]& =& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^2\cdot \frac{1}{\sqrt{n}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\mathrm{d}x\\ & =& \frac{1}{\sqrt{n}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^2{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\mathrm{d}x\end{array}$$ $$\begin{array}{rcl}\mathrm{彼}\mathrm{積}\mathrm{分}\mathrm{凾}\mathrm{数}(x)& =& x^2{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\\ \mathrm{彼}\mathrm{積}\mathrm{分}\mathrm{凾}\mathrm{数}(-x)& =& (-x{)}^2{(1+\frac{(-x{)}^2}{n})}^{-\frac{n+1}{2}}\\ & =& x^2{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\\ & =& \mathrm{彼}\mathrm{積}\mathrm{分}\mathrm{凾}\mathrm{数}(x)\dots \mathrm{偶}\mathrm{凾}\mathrm{数}\end{array}$$ $$\begin{array}{rcl}\mathrm{E}\left[X^2\right]& =& \frac{1}{\sqrt{n}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^2{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\mathrm{d}x\\ & =& 2{\int }_0^{\mathrm{\infty }}{x}^2\cdot \frac{1}{\sqrt{n}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}\mathrm{d}x\\ & & \dots \mathrm{偶}\mathrm{凾}\mathrm{数}\mathrm{の}\mathrm{上}\mathrm{端}\mathrm{と}\mathrm{下}\mathrm{端}\mathrm{の}\mathrm{絶}\mathrm{対}\mathrm{値}\mathrm{が}\mathrm{等}\mathrm{し}\mathrm{い}\mathrm{定}\mathrm{積}\mathrm{分}\mathrm{は}2{\int }_0^{a}f(x)\mathrm{d}x．\\ & & \dots {\int }_{-a}^{a}f(x)\mathrm{d}x=2{\int }_0^{a}f(x)\mathrm{d}x,f(x)\mathrm{が}\mathrm{偶}\mathrm{凾}\mathrm{数}\mathrm{の}\mathrm{場}\mathrm{合}\\ & =& 2{\int }_1^0\left(n(t^{-1}-1)\right)\cdot \frac{1}{\sqrt{n}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}{(1+\frac{n(t^{-1}-1)}{n})}^{-\frac{n+1}{2}}{n}^{\frac{1}{2}}\frac{1}{2}{(t^{-1}-1)}^{-\frac{1}{2}}(-t^{-2})\mathrm{d}t\\ & & \dots x^2=n(t^{-1}-1)\\ & & \dots x=n^{\frac{1}{2}}{(t^{-1}-1)}^{\frac{1}{2}}=n^{\frac{1}{2}}{s}^{\frac{1}{2}}\dots s=t^{-1}-1\\ & & \dots x:0\to \mathrm{\infty },t:1\to 0\\ & & \dots \frac{\mathrm{d}x}{\mathrm{d}t}=\frac{\mathrm{d}x}{\mathrm{d}s}\frac{\mathrm{d}s}{\mathrm{d}t}=n^{\frac{1}{2}}\frac{1}{2}{s}^{\frac{1}{2}-1}\frac{\mathrm{d}s}{\mathrm{d}t}=n^{\frac{1}{2}}\frac{1}{2}{(t^{-1}-1)}^{-\frac{1}{2}}(-t^{-2})\\ & =& \cancel{2}\frac{1}{\cancel{\sqrt{n}}}\frac{1}{\beta (\frac{1}{2},\frac{n}{2})}\cancel{\frac{1}{2}}n\cancel{n^{\frac{1}{2}}}(-1){\int }_1^0{(1+\frac{\cancel{n}(t^{-1}-1)}{\cancel{n}})}^{-\frac{n+1}{2}}{(t^{-1}-1)}^{1-\frac{1}{2}}{t}^{-2}\mathrm{d}t\\ & =& \frac{n}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_0^1{(\cancel{1}+t^{-1}\cancel{-1})}^{-\frac{n+1}{2}}{(t^{-1}-1)}^{\frac{1}{2}}{t}^{-2}\mathrm{d}t\\ & =& \frac{n}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_0^1{t}^{\frac{n+1}{2}}{(t^{-1}-1)}^{\frac{1}{2}}{t}^{-2}\mathrm{d}t\\ & =& \frac{n}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_0^1{t}^{\frac{n}{2}-\frac{3}{2}}{\left(\frac{t}{t}(t^{-1}-1)\right)}^{\frac{1}{2}}\mathrm{d}t\\ & =& \frac{n}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_0^1{t}^{\frac{n}{2}-\frac{3}{2}}{\left(\frac{1}{t}(1-t)\right)}^{\frac{1}{2}}\mathrm{d}t\\ & =& \frac{n}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_0^1{t}^{\frac{n}{2}-\frac{3}{2}}{\left(\frac{1}{t}\right)}^{\frac{1}{2}}{(1-t)}^{\frac{1}{2}}\mathrm{d}t\\ & =& \frac{n}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_0^1{t}^{\frac{n}{2}-\frac{3}{2}}{t}^{-\frac{1}{2}}{(1-t)}^{\frac{1}{2}}\mathrm{d}t\\ & =& \frac{n}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_0^1{t}^{\frac{n}{2}-\frac{4}{2}}{(1-t)}^{\frac{1}{2}}\mathrm{d}t\\ & =& \frac{n}{\beta (\frac{1}{2},\frac{n}{2})}{\int }_0^1{t}^{\frac{n}{2}-1-1}{(1-t)}^{\frac{3}{2}-1}\mathrm{d}t\\ & =& \frac{n}{\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\mathrm{\Gamma }(\frac{1}{2}+\frac{n}{2})}}\frac{\mathrm{\Gamma }(\frac{n}{2}-1)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }(\frac{n}{2}-1+\frac{3}{2})}\dots \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\\ & =& n\frac{\cancel{\mathrm{\Gamma }(\frac{1}{2}+\frac{n}{2})}}{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{\mathrm{\Gamma }(\frac{n}{2}-1)\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\cancel{\mathrm{\Gamma }(\frac{n}{2}+\frac{1}{2})}}\\ & =& n\frac{1}{\cancel{\mathrm{\Gamma }\left(\frac{1}{2}\right)}(\frac{n}{2}-1)\cancel{\mathrm{\Gamma }(\frac{n}{2}-1)}}\frac{\cancel{\mathrm{\Gamma }(\frac{n}{2}-1)}\frac{1}{2}\cancel{\mathrm{\Gamma }\left(\frac{1}{2}\right)}}{1}\dots \mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }\left(z\right)\\ & =& n\frac{1}{(\frac{n}{2}-1)}\frac{\frac{1}{2}}{1}\\ & =& \frac{n}{n-2}\end{array}$$ $$\begin{array}{rcl}\mathrm{V}\left[X\right]& =& \mathrm{E}\left[X^2\right]-\mathrm{E}{\left[X\right]}^2\\ & =& \frac{n}{n-2}-0^2\\ & =& \frac{n}{n-2}\end{array}$$

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}$$