[1]
\(\mu=\ln \theta\)なので、\(X \sim N(\ln \theta, 1)\)で、
\(L(\theta)=\displaystyle \prod_{i=1}^n f(x_i)\)
\(\displaystyle = \prod_{i=1}^n \frac{1}{\sqrt{2\pi}}e^{-\frac{(x_i-\ln \theta)^2}{2}}\)
\(\displaystyle = (2\pi)^{-n/2}\exp\left[-\frac{ \sum_{i=1}^n(x_i-\ln \theta)^2}{2}\right]\)
\(\displaystyle \ln L(\theta)=\ln (2\pi)^{-n/2}\exp\left[-\frac{ \sum_{i=1}^n(x_i-\ln \theta)^2}{2}\right]\)
\(\displaystyle =-\frac{n}{2}\ln (2\pi) \)\(\displaystyle -\frac{ \sum_{i=1}^n(x_i-\ln \theta)^2}{2}\)
\(\displaystyle \frac{\partial \ln L(\theta)}{\partial \theta}=\frac{1}{\theta} \left(\sum_{i=1}^nx_i-n\ln \theta \right)\)\(=0\)
として、
\(\displaystyle \hat{\theta}=\exp\left[\frac{1}{n}\sum_{i=1}^n x_i\right]\)\(=e^{\bar{X}}\)
[2]
\(バイアス=E[\hat{\theta}]-\theta\)
\(=E[e^{\bar{X}}]-\theta\)
ここで、\(\bar{X}\sim N(\ln \theta,1/n)\)なので、
\(\displaystyle M_{\bar{X}}(t)=E[e^{t\bar{X}}]\)\(\displaystyle =\exp\left[ (\ln \theta) t+\frac{1}{2n}t^2 \right]\)
これを使うと、
\(バイアス=E[e^{\bar{X}}]-\theta\)(\(M_{\bar{X}}(t)\)で\(t=1\))
\(\displaystyle =\exp\left[ \ln \theta+\frac{1}{2n} \right]\)\(-\theta\)
\(\displaystyle =\theta (e^{\frac{1}{2n}}-1)>0\)
よって、バイアスは正。
また、
\(\displaystyle E[\tilde{\theta}]=E[e^{\bar{X}-\frac{1}{2n}}]\)
\(\displaystyle =e^{-\frac{1}{2n}}E[e^{\bar{X}}]\)
\(\displaystyle =e^{-\frac{1}{2n}}\exp\left[ \ln \theta+\frac{1}{2n} \right] \)\(=\theta\)
[3]
\(MSE[\hat{\theta}]=E[(\hat{\theta}-\theta)^2]\)
\(=E[\hat{\theta}^2-2\theta\hat{\theta}+\theta^2]\)
\(=E[\hat{\theta}^2]-2\theta E[\hat{\theta}]+\theta^2\)
\(=E[e^{2\bar{X}}]-2\theta E[e^{\bar{X}}]+\theta^2\)
\(\displaystyle =e^{2\ln \theta +\frac{2}{n}}-2\theta e^{\ln \theta +\frac{1}{2n}}+\theta^2\)
\(\displaystyle =\theta^2(e^{\frac{2}{n}}-2e^{\frac{1}{2n}}+1)\)
よって、
\(\displaystyle \lim_{n \to \infty} MSE[\hat{\theta}]=0\)
[4]
\(I(\theta)\)
[1]より、
\(\displaystyle \frac{d \ln L(\theta)}{d\theta}=\frac{1}{\theta} \left(\sum_{i=1}^nx_i-n\ln \theta \right) \)
\(\displaystyle =\frac{\bar{X}-\ln \theta}{\theta/n}\)
\(\displaystyle I(\theta)=E\left[\left(\frac{\bar{X}-\ln \theta}{\theta/n}\right)^2\right]\)
\(\displaystyle =V\left[\frac{\bar{X}-\ln \theta}{\theta/n}\right]\)\(\displaystyle +E\left[\frac{\bar{X}-\ln \theta}{\theta/n}\right]^2\)
ここで、\(\bar{X}\sim N(\ln \theta,1/n)\)より、\(\displaystyle \frac{\bar{X}-\ln \theta}{\theta/n} \sim N\left(0,\frac{n}{\theta^2}\right)\)なので、
\(\displaystyle =\frac{n}{\theta^2}\)
\(V[\tilde{\theta}]\)
\(\displaystyle V[\tilde{\theta}]=V[e^{\bar{X}-\frac{1}{2n}}]\)
\(\displaystyle =e^{-\frac{1}{n}}V[e^{\bar{X}}]\)
\(\displaystyle =e^{-\frac{1}{n}}(E[e^{2\bar{X}}]-E[e^{\bar{X}}]^2)\)
\(M_{\bar{X}}(t)\)を用いて、
\(\displaystyle =e^{-\frac{1}{n}}(e^{2\ln \theta +\frac{2}{n}}-(e^{\ln \theta +\frac{1}{2n}})^2)\)
\(\displaystyle = \theta^2 (e^{\frac{1}{n}}-1)\)
よって、\(V[\tilde{\theta}]\)と\(\displaystyle \frac{1}{I(\theta)}\)は不一致。
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