[1]
\(\displaystyle E[T]=\int^\infty_0 tf(t)dt\)
\(=\mu\)
\(\displaystyle P(T=\tau|T>t)=\frac{P(T=\tau>t)}{P(T>t)}\)
\(\displaystyle =\frac{f(\tau)}{\int^\infty_t f(t)dt}\)
\(\displaystyle =\frac{\frac{1}{\mu}e^{-\tau/\mu}}{e^{- t/\mu}}\)
\(\displaystyle =\frac{1}{\mu}e^{-(\tau -t)/\mu}\)
\(\displaystyle E[T|T>t]=\int^\infty_t \tau \frac{1}{\mu}e^{-(\tau -t)/\mu} d\tau\)
\(\displaystyle =\frac{1}{\mu}e^{t/\mu}\int^\infty_t \tau e^{-\tau/\mu}d\tau\)
\(=\mu+t\)
[2]
\(\displaystyle l_1=\ln \frac{1}{\mu}e^{-t_1/\mu}\frac{1}{\mu}e^{-t_2/\mu}\)
\(\displaystyle =\ln \frac{1}{\mu^2}e^{-\frac{t_1+t_2}{\mu}}\)
\(\displaystyle =-\frac{t_1+t_2}{\mu}-2\ln \mu\)
\(\displaystyle \frac{\partial l_1}{\partial \mu}=\frac{t_1+t_2}{\mu^2}-\frac{2}{\mu}\)\(=0\)
として、
\(\hat{\mu}=\displaystyle \frac{t_1+t_2}{2}\)
[3]
\( \displaystyle l_2=\ln \frac{1}{\mu}e^{-t_1/\mu}\int^\infty_t \frac{1}{\mu}e^{-t/\mu}dt\)
\( \displaystyle =\ln \frac{1}{\mu}e^{-t_1/\mu}e^{-t/\mu}\)
\(\displaystyle =\ln \frac{1}{\mu}e^{-\frac{t_1+t}{\mu}}\)
\(\displaystyle =-\ln \mu -\frac{t_1+t}{\mu}\)
\(\displaystyle \frac{\partial l_2}{\partial \mu}=-\frac{1}{\mu}+\frac{t_1+t}{\mu^2}\)\(=0\)
として、
\(\displaystyle \hat{\mu}=t_1+t\)
[4]
\(T_2=\xi^{(k)}\)
\(=E[T_2|\mu^{(k)},T_2>t]\)
\(=\mu^{(k)}+t\)([1]より)
\(T_1=t_1,T_2=\mu^{(k)}+t\)が観測されたとき、\(\mu\)の推定値は[2]より、
\(\displaystyle \mu^{(k+1)}=\frac{t_1+\mu^{(k)}+t}{2}\)
[5]
漸化式の特性方程式を用いて、
\(\displaystyle \mu^{(k+1)}-(t_1+t)=\frac{1}{2}(\mu^{(k)}-(t_1+t))\)
\(\displaystyle \mu^{(k)}=(\mu^{(0)}-(t_1+t))\left(\frac{1}{2}\right)^k+t_1+t\)
\(k \to \infty\)として、
\(\displaystyle \mu^{(\infty)}=t_1+t\)
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