t分布の期待値と分散

t分布の期待値

\begin{eqnarray} \mathrm{E}\left[X\right]&=&\int_{-\infty}^{\infty}x\cdot\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}} \mathrm{d}x \;\ldots\;\href{https://shikitenkai.blogspot.com/2022/10/t.html}{t分布の確率密度凾数:\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}}} \\&=&\frac{1}{\sqrt{n}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \end{eqnarray} \begin{eqnarray} 彼積分凾数(x)&=&x\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\彼積分凾数(-x)&=&(-x)\left(1+\frac{(-x)^2}{n}\right)^{-\frac{n+1}{2}} \\&=&-x\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\&=&-彼積分凾数(x)\;\ldots\;奇凾数 \end{eqnarray} \begin{eqnarray} \mathrm{E}\left[X\right] &=&\frac{1}{\sqrt{n}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \\&=&\frac{1}{\sqrt{n}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}}\cdot0 \\&&\;\ldots\;奇凾数の上端と下端の絶対値が等しい定積分は0． \\&&\;\ldots\;\int_{-a}^{a}f(x)\mathrm{d}x=0,f(x)が奇凾数の場合 \\&=&0 \end{eqnarray}

t分布の分散

\begin{eqnarray} \mathrm{E}\left[X^2\right] &=&\int_{-\infty}^{\infty}x^2\cdot\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}} \mathrm{d}x \\&=&\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x^2 \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \end{eqnarray} \begin{eqnarray} 彼積分凾数(x)&=&x^2\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\彼積分凾数(-x)&=&(-x)^2\left(1+\frac{(-x)^2}{n}\right)^{-\frac{n+1}{2}} \\&=&x^2\left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \\&=&彼積分凾数(x)\;\ldots\;偶凾数 \end{eqnarray} \begin{eqnarray} \mathrm{E}\left[X^2\right]&=&\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{-\infty}^{\infty}x^2 \left(1+\frac{x^2}{n}\right)^{-\frac{n+1}{2}} \mathrm{d}x \\&=&2\int_{0}^{\infty}x^2\cdot\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}} \mathrm{d}x \\&&\;\ldots\;偶凾数の上端と下端の絶対値が等しい定積分は2\int_{0}^{a}f(x)\mathrm{d}x． \\&&\;\ldots\;\int_{-a}^{a}f(x)\mathrm{d}x=2\int_{0}^{a}f(x)\mathrm{d}x,f(x)が偶凾数の場合 \\&=&2\int_{1}^{0}\left(n\left(t^{-1}-1\right)\right)\cdot\frac{1}{\sqrt{n}} \frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \left(1+\frac{n\left(t^{-1}-1\right)}{n}\right)^{-\frac{n+1}{2}} n^{\frac{1}{2}}\frac{1}{2}\left(t^{-1}-1\right)^{-\frac{1}{2}}\left(-t^{-2}\right) \mathrm{d}t \\&&\;\ldots\;x^2=n\left(t^{-1}-1\right) \\&&\;\ldots\;x=n^{\frac{1}{2}}\left(t^{-1}-1\right)^{\frac{1}{2}}=n^{\frac{1}{2}}s^{\frac{1}{2}}\;\ldots\;s=t^{-1}-1 \\&&\;\ldots\;x:0\rightarrow\infty, t:1\rightarrow0 \\&&\;\ldots\;\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}\left(t^{-1}-1\right)^{-\frac{1}{2}}\left(-t^{-2}\right) \\&=&\cancel{2}\frac{1}{\cancel{\sqrt{n}}}\frac{1}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}}\cancel{\frac{1}{2}} n\cancel{n^{\frac{1}{2}}}(-1) \int_{1}^{0} \left(1+\frac{\cancel{n}\left(t^{-1}-1\right)}{\cancel{n}}\right)^{-\frac{n+1}{2}} \left(t^{-1}-1\right)^{1-\frac{1}{2}} t^{-2} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} \left(\cancel{1}+t^{-1}\cancel{-1}\right)^{-\frac{n+1}{2}} \left(t^{-1}-1\right)^{\frac{1}{2}} t^{-2} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n+1}{2}} \left(t^{-1}-1\right)^{\frac{1}{2}} t^{-2} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} \left(\frac{t}{t}\left(t^{-1}-1\right)\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} \left(\frac{1}{t}\left(1-t\right)\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} \left(\frac{1}{t}\right)^\frac{1}{2}\left(1-t\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{3}{2}} t^{-\frac{1}{2}}\left(1-t\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-\frac{4}{2}} \left(1-t\right)^{\frac{1}{2}} \mathrm{d}t \\&=&\frac{n}{\beta{\left(\frac{1}{2},\frac{n}{2}\right)}} \int_{0}^{1} t^{\frac{n}{2}-1-1} \left(1-t\right)^{\frac{3}{2}-1} \mathrm{d}t \\&=&\frac{n}{ \frac{ \Gamma{\left(\frac{1}{2}\right)} \Gamma{\left(\frac{n}{2}\right)} }{ \Gamma{ \left(\frac{1}{2}+\frac{n}{2}\right) } }} \frac{ \Gamma{\left(\frac{n}{2}-1\right)} \Gamma{\left(\frac{3}{2}\right)} }{ \Gamma{ \left(\frac{n}{2}-1+\frac{3}{2}\right) } } \;\ldots\;\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} \\&=&n \frac{ \cancel{\Gamma{ \left(\frac{1}{2}+\frac{n}{2}\right) }} }{ \Gamma{\left(\frac{1}{2}\right)} \color{green}{\Gamma{\left(\frac{n}{2}\right)}} } \frac{ \Gamma{\left(\frac{n}{2}-1\right)} \color{blue}{\Gamma{\left(\frac{3}{2}\right)}} }{ \cancel{\Gamma{ \left(\frac{n}{2}+\frac{1}{2}\right) }} } \\&=&n \frac{1}{ \cancel{\Gamma{\left(\frac{1}{2}\right)}} \color{green}{\left(\frac{n}{2}-1\right)\cancel{\Gamma{\left(\frac{n}{2}-1\right)}}} } \frac{ \cancel{\Gamma{\left(\frac{n}{2}-1\right)}} \color{blue}{\frac{1}{2}\cancel{\Gamma{\left(\frac{1}{2}\right)}}} }{1} \;\ldots\;\href{https://shikitenkai.blogspot.com/2020/08/s1ss.html}{\Gamma\left(z+1\right)=z\Gamma\left(z\right)} \\&=&n \frac{1}{ \left(\frac{n}{2}-1\right) } \frac{ \frac{1}{2} }{1} \\&=&\frac{n}{n-2} \end{eqnarray} \begin{eqnarray} \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{eqnarray}