[1]
\(\displaystyle E[X]=\int^\infty_0 xf(x)dx\)
\(\displaystyle = \int^\infty_0 \frac{x^\alpha e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)}dx\)
\(\displaystyle = \frac{\beta^{\alpha+1} \Gamma(\alpha+1)}{\beta^\alpha \Gamma(\alpha)}\int^\infty_0 \frac{x^\alpha e^{-x/\beta}}{\beta^{\alpha+1} \Gamma(\alpha+1)}dx\)
\(\Gamma(\alpha+1)=\alpha\Gamma(\alpha)\)より、
\(=\alpha\beta\)
\(\displaystyle E[X^2]=\int^\infty_0 x^2f(x)dx\)
\(\displaystyle = \int^\infty_0 \frac{x^{\alpha+1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)}dx\)
\(\displaystyle = \frac{\beta^{\alpha+2} \Gamma(\alpha+2)}{\beta^\alpha \Gamma(\alpha)}\int^\infty_0 \frac{x^{\alpha+1} e^{-x/\beta}}{\beta^{\alpha+2} \Gamma(\alpha+2)}dx\)
\(\Gamma(\alpha+2)=\alpha(\alpha+1)\Gamma(\alpha)\)より、
\(=\alpha(\alpha+1)\beta^2\)
\(V[X]=E[X^2]-E[X]^2\)\(=\alpha\beta^2\)
[2]
\(l\)\(\displaystyle =\ln \prod^n_{i=1}\frac{x_i^{\alpha-1}e^{-x_i/\beta}}{\beta^\alpha\Gamma(\alpha)}\)
\(\displaystyle =\sum^n_{i=1}\ln \frac{x_i^{\alpha-1}e^{-x_i/\beta}}{\beta^\alpha\Gamma(\alpha)}\)
\(\displaystyle =(\alpha-1)\sum^n_{i=1} \ln x_i\)\(\displaystyle -\frac{1}{\beta}\sum^n_{i=1}x_i\)\(\displaystyle -n\alpha\ln\beta\)\(\displaystyle -n\ln\Gamma(\alpha)\)
[3]
\(l\)\(\displaystyle =(\alpha-1)\sum^n_{i=1} \ln x_i\)\(\displaystyle -\frac{1}{\beta}\sum^n_{i=1}x_i\)\(\displaystyle -n\alpha\ln\beta\)\(\displaystyle -n\ln\Gamma(\alpha)\)
\(\displaystyle =(\alpha-1)n \ln \tilde{x_i}\)\(\displaystyle -\frac{1}{\beta}n\bar{x_n}\)\(\displaystyle -n\alpha\ln\beta\)\(\displaystyle -n\ln\Gamma(\alpha)\)
偏微分して、
\(\displaystyle \frac{\partial l}{\partial \alpha}=n \ln \tilde{x_i}-n\ln\beta-n\frac{\Gamma(\alpha)’}{\Gamma(\alpha)}\)
\(\displaystyle =n \ln \tilde{x_i}-n\ln\beta-n\psi(\alpha)\)
\(\displaystyle \frac{\partial l}{\partial \beta}=\frac{n\bar{x_n}}{\beta^2}-\frac{n\alpha}{\beta}\)
\(\displaystyle \frac{\partial l}{\partial \beta}=0\)として、\(\displaystyle \beta=\frac{\bar{x_n}}{\alpha}\)
\(\displaystyle \frac{\partial l}{\partial \alpha}\)に代入して\(=0\)とすると、
\(\displaystyle \ln \tilde{x_i}-\ln\frac{\bar{x_n}}{\alpha}-\psi(\alpha)=0\)
整理して、
\(\displaystyle \psi(\alpha)-\ln\alpha=\ln\frac{\tilde{x_i}}{\bar{x_n}}\)
[4]
\(\displaystyle \ln \frac{\tilde{x_i}}{\bar{x_n}}=\ln \tilde{x_i}-\ln\bar{x_n}\)
\(\displaystyle =\frac{63.6}{10}-\ln \frac{10000}{10}\)
\(\displaystyle =6.36-3\ln 10\)
\(\displaystyle =6.36-3\times 2.3026=-0.5478\)
グラフで\(\eta(\alpha)\)が\(-0.55\)の時の\(\alpha\)を見ると、\(\alpha=1\)
また\(\beta=\displaystyle \frac{\bar{x_n}}{\alpha}=\frac{10000}{1}\)\(=10000\)
[5]
\(\psi(\alpha)-\ln\alpha\)が単調増加関数なので、\(\displaystyle \ln \frac{\tilde{x_i}}{\bar{x_n}}\)に対して、一意の\(\alpha\)が存在する。また\(\beta\)の推定値は\(\displaystyle \frac{\bar{x_n}}{\alpha}\)なので、これも一意。
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