[1]
尤度関数は
\(f(\boldsymbol{x})=\displaystyle \left(\frac{1}{\theta}\right)^n\)(ただし任意の\(i\)について\(0<X_i \leq \theta\))
任意の\(i\)について\(0<X_i \leq \theta\)より\(\theta\)の範囲は、\(X_{max} \leq \theta\)となり、尤度関数は\(\theta\)の減少関数なので\(\theta\)が最小の時に最大となることから示された。
[2]
\(E[\theta’]\)\( \displaystyle =E\left[\frac{2}{n}\sum_{i=1}^nX_i\right]\)
\(\displaystyle =2E[X_i]\)
\(E[X]\)\(\displaystyle = \int^\theta_0 x\frac{1}{\theta}dx\)
\(\displaystyle =\frac{1}{2}\theta\)
より、
\(E[\theta’]=\theta\)
[3]
\(P(X_{max} \leq x)=P(\forall i,X_i \leq x)\)(全ての\(i\)に対して、\(x_i\)が\(x\)となる確率)
\(\displaystyle =\left(\frac{x}{\theta}\right)^n\)
\(f_{X_{max}}(x)=\displaystyle \frac{d}{dx}P(X_{max} \leq x)\)
\(\displaystyle =n\left(\frac{x}{\theta}\right)^{n-1} \)
次に、
\(E[\theta”]\)\(\displaystyle =E\left[\frac{n+1}{n}X_{max}\right]\)
\(\displaystyle =\frac{n+1}{n}E[X_{max}]\)
\(\displaystyle =\frac{n+1}{n}\int^\theta_0 xn\left(\frac{x}{\theta}\right)^{n-1}dx\)\(=\theta\)
[4]
\(V[\theta’]\)\(\displaystyle =V\left[\frac{2}{n}\sum_{i=1}^nX_i\right]\)
\(\displaystyle =\frac{4}{n}V[X_i]\)
\(\displaystyle =\frac{4}{n}\left(\int^\theta_0x^2\frac{1}{\theta}dx-\left(\frac{1}{2}\theta\right)^2\right)\)
\(\displaystyle =\frac{\theta^2}{3n}\)
\(V[\theta”]\)\(\displaystyle =V\left[\frac{n+1}{n}X_{max}\right]\)
\(\displaystyle =\frac{(n+1)^2}{n^2}V[X_{max}]\)
\(\displaystyle =\frac{(n+1)^2}{n^2}\left(\int^\theta_0x^2n\left(\frac{x}{\theta}\right)^{n-1}dx\right. \)\(\displaystyle \left. -\left(\frac{n}{n+1}\theta\right)^2\right)\)
\(\displaystyle =\frac{\theta^2}{n(n+2)}\)
\(V[\theta’]>V[\theta”]\)なので、\(\theta”\)が良い。
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