[1]
\(E[S^2]\)\(=\displaystyle E\left[\frac{1}{n-1}\sum^n_{i=1}(X_i-\bar{X})^2\right]\)
\(\displaystyle =\frac{1}{n-1}E\left[\sum^n_{i=1}(X_i-\bar{X})^2\right]\)
\(\displaystyle =\frac{1}{n-1}E\left[\sum^n_{i=1}\{(X_i-\mu) -(\bar{X}-\mu)\}^2\right]\)
\(\displaystyle =\frac{1}{n-1}\left(E\left[\sum^n_{i=1}(X_i-\mu)^2\right]\right.\)\(\displaystyle -2E\left[\sum^n_{i=1}(X_i-\mu)(\bar{X}-\mu)\right]\)\(\displaystyle \left. +E\left[\sum^n_{i=1}(\bar{X}-\mu)^2\right]\right)\)
括弧内の1項目は、
\(\displaystyle E\left[\sum^n_{i=1}(X_i-\mu)^2\right]\)
\(\displaystyle =\sum^n_{i=1}E\left[(X_i-\mu)^2\right]\)
\(\displaystyle =\sum^n_{i=1}\sigma^2\)
\(\displaystyle =n\sigma^2\)
2項目は、
\(\displaystyle E\left[\sum^n_{i=1}(X_i-\mu)(\bar{X}-\mu)\right]\)
\(\displaystyle =E\left[n(\bar{x}-\mu)(\bar{X}-\mu)\right]\)
\(\displaystyle =nE\left[(\bar{x}-\mu)^2\right]\)
\(\displaystyle =n\frac{\sigma^2}{n}\)
\(=\sigma^2\)
3項目は、
\(\displaystyle E\left[\sum^n_{i=1}(\bar{X}-\mu)^2\right]\)
\(\displaystyle =\sum^n_{i=1}E\left[(\bar{X}-\mu)^2\right]\)
\(\displaystyle =\sum^n_{i=1}\frac{\sigma^2}{n}\)
\(=\sigma^2\)
よって、
\(E[S^2]=\sigma^2\)
[2]
\(E[Y]=\displaystyle \int^\infty_0 yg(y)dy\)
\(\displaystyle =\frac{\left(\frac{1}{2}\right)^{\frac{n-1}{2}}}{\Gamma \left(\frac{n-1}{2}\right)}\int^\infty_0 y^{\frac{n-1}{2}}e^{-\frac{y}{2}}dy\)
\(\displaystyle \frac{y}{2}=t\)と変数変換すると、
\(\displaystyle =\frac{\left(\frac{1}{2}\right)^{\frac{n-1}{2}}}{\Gamma \left(\frac{n-1}{2}\right)}\int^\infty_0 (2t)^{\frac{n-1}{2}}e^{-t}2dt\)
\(\displaystyle =\frac{2}{\Gamma \left(\frac{n-1}{2}\right)}\Gamma \left(\frac{n-1}{2}+1\right)\)
\(\displaystyle =\frac{2}{\Gamma \left(\frac{n-1}{2}\right)}\frac{n-1}{2}\Gamma \left(\frac{n-1}{2}\right)\)
\(=n-1\)
\(E[Y^2]=\displaystyle \int^\infty_0 y^2g(y)dy\)
\(\displaystyle =\frac{\left(\frac{1}{2}\right)^{\frac{n-1}{2}}}{\Gamma \left(\frac{n-1}{2}\right)}\int^\infty_0 y^{\frac{n-1}{2}+1}e^{-\frac{y}{2}}dy\)
\(\displaystyle \frac{y}{2}=t\)と変数変換すると、
\(\displaystyle =\frac{\left(\frac{1}{2}\right)^{\frac{n-1}{2}}}{\Gamma \left(\frac{n-1}{2}\right)}\int^\infty_0 (2t)^{\frac{n-1}{2}+1}e^{-t}2dt\)
\(\displaystyle =\frac{2^2}{\Gamma \left(\frac{n-1}{2}\right)}\Gamma \left(\frac{n-1}{2}+2\right)\)
\(\displaystyle =\frac{2^2}{\Gamma \left(\frac{n-1}{2}\right)}\left(\frac{n-1}{2}+1\right)\)\(\displaystyle \frac{n-1}{2}\Gamma \left(\frac{n-1}{2}\right)\)
\(=(n+1)(n-1)\)
よって、
\(V[Y]=E[Y^2]-E[Y]^2\)\(=2(n-1)\)
[3]
\(E[\sqrt{Y}]=\displaystyle \int^\infty_0 \sqrt{y}g(y)dy\)
\(\displaystyle =\frac{\left(\frac{1}{2}\right)^{\frac{n-1}{2}}}{\Gamma \left(\frac{n-1}{2}\right)}\int^\infty_0 y^{\frac{n}{2}-1}e^{-\frac{y}{2}}dy\)
\(\displaystyle \frac{y}{2}=t\)と変数変換すると、
\(\displaystyle =\frac{\left(\frac{1}{2}\right)^{\frac{n-1}{2}}}{\Gamma \left(\frac{n-1}{2}\right)}\int^\infty_0 (2t)^{\frac{n}{2}-1}e^{-t}2dt\)
\(\displaystyle =\frac{\sqrt{2}\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n-1}{2}\right)}\)
\(\displaystyle \frac{(n-1)S^2}{\sigma^2}=\sum^n_{i=1}\left(\frac{X_i-\bar{X}}{\sigma}\right)^2\)は、\(\chi^2(n-1)\)に従うので、\(Y=\displaystyle \frac{(n-1)S^2}{\sigma^2}\)すなわち、\(S=\displaystyle \frac{\sigma}{\sqrt{n-1}}\sqrt{Y}\)で、
\(E[S]=\displaystyle \frac{\sigma}{\sqrt{n-1}}E[\sqrt{Y}]\)
\(\displaystyle =\sigma \sqrt{\frac{2}{n-1}}\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n-1}{2}\right)}\)
[4]
スターリングの公式を用いると、
\(\displaystyle E[S]=\sigma \sqrt{\frac{2}{n-1}}\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n-1}{2}\right)}\)
\(\displaystyle = \sigma \sqrt{\frac{2}{n-1}}\frac{\sqrt{2 \pi} e^{-\frac{n}{2}}\left(\frac{n}{2}\right)^{\frac{n-1}{2}}}{\sqrt{2 \pi} e^{-\frac{n-1}{2}}\left(\frac{n-1}{2}\right)^{\frac{n-2}{2}}}\)
\(\displaystyle =\sigma e^{-\frac{1}{2}}\left(\frac{n}{n-1}\right)^{\frac{n-1}{2}}\)
\(\displaystyle f(n)=\left(\frac{n}{n-1}\right)^{\frac{n-1}{2}}\)
とすると、
\(f(\infty)=\displaystyle \lim_{n\to \infty}\left(\frac{n}{n-1}\right)^{\frac{n-1}{2}}\)
\(=\displaystyle \lim_{n\to \infty}\left(\left(1+\frac{1}{n-1}\right)^{n-1}\right)^{\frac{1}{2}}\)
\(\displaystyle =e^{\frac{1}{2}}\)
\(f(n)\)に対して、\(t=\displaystyle \frac{1}{n}\)で変数変換をしてマクローリン展開をする。
\(\displaystyle f(n)=\left(\frac{1}{1-t}\right)^{\frac{1-t}{2t}}\)\(\equiv h(t)\)とすると、
\(h(0)\)\(=f(\infty)\)\(=\displaystyle e^{\frac{1}{2}}\)
\(\displaystyle \ln h(t)=\frac{1-t}{2t}\ln \frac{1}{1-t}\)
\(\displaystyle 2\ln h(t)=\left(1-\frac{1}{t}\right)\ln (1-t)\)
両辺を\(t\)で微分して、
\(\displaystyle 2\frac{h'(t)}{h(t)}=\frac{t+\ln (1-t)}{t^2}\)
\(t \to 0\)の時に不定形なので、ロピタルの定理を利用して、
\(\displaystyle \lim_{t \to 0}\frac{t+\ln (1-t)}{t^2}\)\(\displaystyle =\lim_{t \to 0}\frac{1-\frac{1}{1-t}}{2t}\)
\(\displaystyle =\lim_{t \to 0 }-\frac{1}{2(1-t)}\)
\(=\displaystyle -\frac{1}{2}\)
よって、
\(\displaystyle \lim_{t \to 0}h'(t)=-\frac{e^{\frac{1}{2}}}{4}\)
以上より1次までのマクローリン展開は、
\(\displaystyle h(t)\approx e^{\frac{1}{2}}-\frac{e^{\frac{1}{2}}}{4}t\)
\(n\)に戻して、
\(\displaystyle f(n)\approx e^{\frac{1}{2}}-\frac{e^{\frac{1}{2}}}{4}\frac{1}{n}\)
よって、
\(\displaystyle E[S] \approx \sigma -\frac{\sigma}{4n}\)
なので、
\(\displaystyle E[S]-\sigma \approx -\frac{\sigma}{4n}\)
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