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}$$ $$\begin{array}{rcl}\mathrm{V}\left[X\right]& =& \mathrm{E}\left[X^2\right]-\mathrm{E}{\left[X\right]}^2\\ & =& \frac{n}{n-2}-0^2\\ & =& \frac{n}{n-2}\end{array}$$