[1]
\(f(x) = \displaystyle \frac{d}{dx}F(x)\)
\(\displaystyle =\left \{ \begin{array}{} \frac{1}{\gamma}x^{-\frac{1}{\gamma}-1} & (x>1) \\ 0 & (x \leq 1)\end{array} \right.\)
\(f_{\gamma=1}(x) \)\(\displaystyle =\left \{ \begin{array}{} x^{-2} & (x>1) \\ 0 & (x \leq 1)\end{array} \right.\)
[2]
\(E[X]\)
\(E[X] \displaystyle = \int^\infty_1 x \frac{1}{\gamma}x^{-\frac{1}{\gamma}-1}dx\)
\(\displaystyle = \int^\infty_1 \frac{1}{\gamma}x^{-\frac{1}{\gamma}}dx\)
\(\displaystyle =\left \{ \begin{array}{} \frac{1}{\gamma}[\ln x]^\infty _ 1 & (\gamma = 1) \\ \frac{1}{\gamma-1}\left[x^{1-\frac{1}{\gamma}}\right]^\infty _ 1 & (\gamma \neq 1)\end{array} \right.\)
\(\displaystyle =\left \{ \begin{array}{} \infty & (\gamma = 1) \\ \frac{1}{\gamma-1}\left[x^{\frac{\gamma-1}{\gamma}}\right]^\infty _ 1 & (\gamma > 1)\\ \frac{1}{\gamma-1}\left[\frac{1}{x^{\frac{1-\gamma}{\gamma}}}\right]^\infty _ 1 & (0<\gamma < 1)\end{array} \right.\)
\(\displaystyle =\left \{ \begin{array}{} \infty & (\gamma = 1) \\ \infty & (\gamma > 1)\\ \frac{1}{1-\gamma} & (0<\gamma < 1)\end{array} \right.\)
\(\displaystyle =\left \{ \begin{array}{} \frac{1}{1-\gamma} & (0<\gamma < 1) \\ \infty & (\gamma \geq 1) \end{array} \right.\)
\(V[X]\)
\(E[X]\)が収束する\(0<\gamma < 1\)の範囲で考える。
\(E[X^2] \displaystyle = \int^\infty_1 x^2 \frac{1}{\gamma}x^{-\frac{1}{\gamma}-1}dx\)
\(\displaystyle = \int^\infty_1 \frac{1}{\gamma}x^{\frac{\gamma-1}{\gamma}}dx\)
\(\displaystyle =\left \{ \begin{array}{} \frac{1}{2}[\ln x]^\infty _ 1 & \left(\gamma = \frac{1}{2}\right) \\ \frac{1}{2\gamma-1}\left[x^{\frac{2\gamma-1}{\gamma}}\right]^\infty _ 1 & \left(\gamma \neq \frac{1}{2}\right)\end{array} \right.\)
\(\displaystyle =\left \{ \begin{array}{} \infty &\left(\gamma = \frac{1}{2}\right) \\ \frac{1}{2\gamma-1}\left[x^{\frac{2\gamma-1}{\gamma}}\right]^\infty _ 1 & \left(\frac{1}{2}<\gamma<1\right)\\ \frac{1}{2\gamma-1}\left[\frac{1}{x^{\frac{1-2\gamma}{\gamma}}}\right]^\infty _ 1 & \left(0<\gamma<\frac{1}{2}\right)\end{array} \right.\)
\(\displaystyle =\left \{ \begin{array}{} \infty & \left(\frac{1}{2} \leq \gamma<1\right)\\ \frac{1}{1-2\gamma} & \left(0<\gamma<\frac{1}{2}\right)\end{array} \right.\)
ここで、公式から
\(V[X] = E[X^2]-E[X]^2\)
\(\displaystyle = \frac{1}{1-2\gamma} - \left( \frac{1}{1-\gamma}\right)^2\)\(\displaystyle = \frac{\gamma^2}{(1-2\gamma)(1-\gamma)^2} \)
[3]
尤度\(L\)は
\(\displaystyle L = \prod_{i=1}^n f(x_i)\)\(\displaystyle = \frac{1}{\gamma^n}\prod_{i=1}^n x_i^{-\frac{1}{\gamma}-1}\)
\(\displaystyle \ln L = -n \ln \gamma +\left(-\frac{1}{\gamma}-1\right)\sum _{i=1}^n \ln x_i \)
\(\displaystyle \frac{ \partial \ln L}{\partial \gamma} = -\frac{n}{\gamma} +\frac{\sum _{i=1}^n \ln x_i}{\gamma ^2} \)\(=0\)として、
\(\displaystyle \hat{\gamma} = \frac{1}{n}\sum _{i=1}^n \ln x_i\)
[4]
\(t = \displaystyle \frac{1}{\gamma}\ln x\)として、
\(x = e^{\gamma t}\)
\(dx = \gamma e^{\gamma t}dt\)と、[1]で求めた\(f(x)\)から、
\(T\)の確率密度関数\(g(t)\)は、
\(\displaystyle g(t) = \frac{1}{\gamma}e^{-\gamma t-t}\times \gamma e^{\gamma t}\)
\(\displaystyle = e^{-t} \) (\(T\)は指数分布に従う)
[5]
\(E[\hat{\gamma}]\)
\(\displaystyle E[\hat{\gamma}] = E\left[\frac{1}{n}\sum _{i=1}^n \ln x_i\right]\)
\(\displaystyle = \frac{1}{n}\sum _{i=1}^nE\left[ \ln x_i\right]\)
\(\displaystyle = \frac{1}{n}\sum _{i=1}^nE\left[\gamma T\right]\)
\(\displaystyle =\gamma E\left[ T\right] = \gamma\) (\(T\)は期待値\(1\)の指数分布に従うため)
\(V[\hat{\gamma}]\)
\(\displaystyle V[\hat{\gamma}] = V\left[\frac{1}{n}\sum _{i=1}^n \ln x_i\right]\)
\(\displaystyle = \frac{1}{n^2}\sum _{i=1}^nV\left[ \ln x_i\right]\)
\(\displaystyle = \frac{1}{n^2}\sum _{i=1}^nV\left[\gamma T\right]\)
\(\displaystyle =\frac{\gamma^2}{n}V\left[ T\right] \)\(\displaystyle =\frac{\gamma^2}{n}\) (\(T\)は分散\(1\)の指数分布に従うため)
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