- 正規性の検討
- p.202下：自由度\(\nu\)の\(t\)分布の尖度の導出
- p.203上：一様分布の尖度の導出
統計学基礎の行間埋め 第６章
-
導出の参考
自由度\(\nu\)の\(t\)分布の確率密度関数はこちらなど参考 \begin{eqnarray} f(t)&=&\frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})}(1+\frac{t}{\nu})^{-\frac{\nu+1}{2}} \end{eqnarray} 書ける。\(\nu\gt 4\)のとき \begin{eqnarray} \int_{-\infty}^{\infty}dtt^2f(t) &=& \int_{-\infty}^{\infty}dtt^4\frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})}(1+\frac{t^2}{\nu})^{-\frac{\nu+1}{2}} \\ \\ &=& 2\int_{0}^{\infty}dtt^4\frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})}(1+\frac{t^2}{\nu})^{-\frac{\nu+1}{2}}&...&\text{偶関数であるから} \\ \\ &=& 2\frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})}\int_{0}^{\infty}dtt^4(1+\frac{t^2}{\nu})^{-\frac{\nu+1}{2}}\\ \\ &=& 2\frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})}\int_{1}^{0}\left(-\frac{1}{2}\nu^{\frac{1}{2}}x^{-2}(\frac{1}{x}-1)^{-\frac{1}{2} }dx\right)\left(\nu(\frac{1}{x}-1)\right)^2x^{\frac{\nu+1}{2}}&...&\frac{1}{x}=1+\frac{t^2}{\nu},dt=-\frac{1}{2}\nu^{\frac12}x^{-2}(\frac1x-1)^{-\frac12}\text{を利用}\\ \\ &=& {\color{red}-\frac{1}{2}}2\frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})}\int_{1}^{0}dx\nu^{\frac{5}{2}}\left(\frac{1}{x}-1\right)^{\frac{3}{2}}x^{\frac{\nu-3}{2}}&\\ \\ &=& \frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})}\int_{\color{red}0}^{\color{red}1}dx\nu^{\frac{5}{2}}\left(\frac{1}{x}-1\right)^{\frac{3}{2}}x^{\frac{\nu-3}{2}}&\\ \\ &=& \frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})}\nu^{\frac{5}{2}}\int_{0}^{1}dx\frac{1}{x^{\frac32}}\left(1-x\right)^{\frac{3}{2}}x^{\frac{\nu-3}{2}}&\\ \\ &=& \frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu\pi}\Gamma(\frac{\nu}{2})}\nu^{\frac{5}{2}}\int_{0}^{1}dx\left(1-x\right)^{\frac{3}{2}}x^{\frac{\nu-6}{2}}&\\ \\ &=& \frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\pi}\Gamma(\frac{\nu}{2})}\nu^2\int_{0}^{1}dx\left(1-x\right)^{\frac{5}{2}-1}x^{\frac{\nu-4}{2}-1}&\\ \\ &=& \frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\pi}\Gamma(\frac{\nu}{2})}\nu^2Beta(\frac{5}{2},\frac{\nu-4}{2})&...&\text{(1)}\\ \\ &=& \frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\pi}\Gamma(\frac{\nu}{2})}\nu^2\frac{\Gamma(\frac{5}{2})\Gamma(\frac{\nu-4}{2})}{\Gamma(\frac{5}{2}+\frac{\nu-4}{2})}&...&\text{(1)}\\ \\ &=& \frac{\Gamma(\frac{\nu+1}{2})}{\sqrt{\pi}\frac{\nu-2}{2}\frac{\nu-4}{2}\Gamma(\frac{\nu-4}{2})}\nu^2\frac{\Gamma(\frac{5}{2})\Gamma(\frac{\nu-4}{2})}{\Gamma(\frac{\nu+1}{2})}&\\ \\ &=& \frac{\Gamma(\frac{5}{2})}{\sqrt{\pi}\frac{\nu-2}{2}\frac{\nu-4}{2}}\nu^2&\\ \\ &=& \frac{\frac{3}{2}\frac{1}{2}\sqrt{\pi}}{\sqrt{\pi}\frac{\nu-2}{2}\frac{\nu-4}{2}}\nu^2&\\ \\ &=& \frac{3}{(\nu-2)(\nu-4)}\nu^2&\\ \\ \end{eqnarray} が得られる。(1)ではベータ関数(参考
)を用いた。ここで、\(s^2\)は2次のモーメントなので、\(s^2=\frac{\nu}{\nu-2}\)となる(参考サイトより)。p.202の標本尖度より、 \begin{eqnarray} \frac{\frac{3}{(\nu-2)(\nu-4)}\nu^2}{s^4}-3 &=& \frac{\frac{3}{(\nu-2)(\nu-4)}\nu^2}{\frac{\nu^2}{(\nu-2)^2}}-3 \\ \\ &=& \frac{3(\nu-2)}{(\nu-4)}-3 \\ \\ &=& \frac{6}{(\nu-4)} \end{eqnarray} と導出できる。
- 一様分布を
\begin{eqnarray}
f(t)&=&\frac{1}{b-a}...(a\leq t\leq b)
\end{eqnarray}
とする。期待値は式(2.8.2)より\(\frac{a+b}{2}\)なので、
4次のモーメントは
\begin{eqnarray}
\int_{a}^{b}dt(t-\frac{a+b}{2})^4f(t)
&=&
\int_{a}^{b}dt(t-\frac{a+b}{2})^4\frac{1}{b-a} \\ \\
&=&
\left[\frac{(t-\frac{a+b}{2})^5}{5}\right]_{a}^{b}\frac{1}{b-a} \\ \\
&=&
\left[\frac{(b-\frac{a+b}{2})^5}{5}-\frac{(a-\frac{a+b}{2})^5}{5}\right]\frac{1}{b-a} \\ \\
&=&
\left[\frac{(\frac{b-a}{2})^5}{5}-\frac{(\frac{-a+b}{2})^5}{5}\right]\frac{1}{b-a} \\ \\
&=&
\left[\frac{(\frac{b-a}{2})^5}{5}+\frac{(\frac{b-a}{2})^5}{5}\right]\frac{1}{b-a} \\ \\
&=&
\left[\frac{2(\frac{b-a}{2})^5}{5}\right]\frac{1}{b-a} \\ \\
&=&
\frac{2\frac{(b-a)^4}{2^5}}{5} \\ \\
&=&
\frac{(b-a)^4}{80} \\ \\
\end{eqnarray}
が得られる。式(2.8.3)より二次のモーメントは\(\frac{(b-a)^2}{12}\)であるから、尖度は
\begin{eqnarray}
\frac{\frac{(b-a)^4}{80}}{(\frac{(b-a)^2}{12})^2}-3
&=&
\frac{144}{80}-3 \\ \\
&=&
-1.2
\end{eqnarray}
と導出できる。