UA MATH566 統計理論 QE練習 位置變換后的指數分布

• 2016年1月第六題
• 2018年5月第六題

2016年1月第六題

Part a
Joint likelihood is
$$L(\theta )=\exp \left(?\sum _{i=1}^n(X_1?\theta )\right)=\exp \left(n\theta ?\sum _{i=1}^n{X}_1\right)I(X_{(1)}\ge \theta )$$

Compute likelihood ratio
$$\frac{L(\theta |\text{X})}{L(\theta |\text{Y})}=\frac{\exp \left(n\theta ?{\sum }_{i=1}^n{X}_1\right)I(X_{(1)}\ge \theta )}{\exp \left(n\theta ?{\sum }_{i=1}^n{Y}_1\right)I(Y_{(1)}\ge \theta )}=\exp \left(\sum _{i=1}^n{Y}_i?\sum _{i=1}^n{X}_i\right)\frac{I(X_{(1)}\ge \theta )}{I(Y_{(1)}\ge \theta )}$$

To make likelihood ration independent of $\theta$,
$$X_{(1)}=Y_{(1)}$$

So $X_{(1)}$ is minimal sufficient statistics.

Part b
$$EX={\int }_{\theta }^{\infty }x{e}^{?(x?\theta )}dx=\theta +1=\bar{X}?{\hat{\theta }}_{MME}=\bar{X}?1$$

Part c
Notice if $\theta >X_{(1)}$, $L(\theta )=0$. To make $L(\theta )$ as greater as possible, $\theta \le X_{(1)}$. Since $L(\theta )$ is increasing on $\theta$, when ${\hat{\theta }}_{MLE}=X_{(1)}$

Part d
Compute
$$\begin{gathered} E[{\hat{\theta }}_{MME}]=E[\bar{X}?1]=\theta \\ Var({\hat{\theta }}_{MME})=Var(\bar{X})=\frac{1}{n} \end{gathered}$$

By the property of order statistics, density of $X_1$ is
$$f_{X_{(1)}}=n{e}^{?n(x?\theta )}$$

Compute
$$\begin{gathered} E{X}_{(1)}={\int }_{\theta }^{\infty }nx{e}^{?n(x?\theta )}dx=\frac{n\theta +1}{n} \\ E{X}_{(1)}^2={\int }_{\theta }^{\infty }n{x}^2{e}^{?n(x?\theta )}dx=\frac{n^2{\theta }^2+2n\theta +2}{n^2} \\ Var(X_{(1)})=E{X}_{(1)}^2?E{X}_{(1)}=\frac{1}{n^2} \\ MSE(X_{(1)})=bia{s}^2+\frac{1}{n^2}=\frac{2}{n^2} \end{gathered}$$

Part e
By Lehmann-Scheffe theorem, $E[\bar{X}?1|X_{(1)}]$ will be better. Or $X_{(1)}?1/n$.

2018年5月第六題

Part a
Joint likelihood function is
$$L(\lambda ,\theta )=\prod _{i=1}^{n}f(X_i|\lambda ,\theta )={\lambda }^n\exp \left(?\lambda \sum _{i=1}^n(X_i?\theta )\right)I(X_{(1)}>\theta )$$
Consider two samples $\{X_i{\}}_{i=1}^n$ and $\{Y_i{\}}_{i=1}^n$,

$$\begin{gathered} \frac{L(\lambda ,\theta |\text{X})}{L(\lambda ,\theta |\text{Y})}=\frac{{\lambda }^n\exp \left(?\lambda {\sum }_{i=1}^n(X_i?\theta )\right)I(X_{(1)}>\theta )}{{\lambda }^n\exp \left(?\lambda {\sum }_{i=1}^n(Y_i?\theta )\right)I(Y_{(1)}>\theta )} \\ =\frac{I(X_{(1)}>\theta )}{I(Y_{(1)}>\theta )}\exp \left(?\lambda \sum _{i=1}^n(Y_i?X_i)\right) \end{gathered}$$

To make this likelihood ratio independent of parameters,
$$X_{(1)}=Y_{(1)},\sum _{i=1}^n{X}_i=\sum _{i=1}^n{Y}_i$$

Let $T_1(X)=X_{(1)}$, $T_2(X)={\sum }_{i=1}^n{X}_i$ and they are minimal sufficient statistics.

Part b
If $\lambda =1$,
$$f_X(x)=e^{?(x?\theta )},x>\theta ,F_X(x)={\int }_{\theta }^x{e}^{?(s?\theta )}ds=1?e^{?(x?\theta )},x>\theta$$

Compute
$$\begin{gathered} P(X_{(1)}\le x)=P(\min (X_1,?,X_n)\le x)=1?P(\min (X_1,?,X_n)>x)=1?[1?F(x){]}^n \\ {f}_{X_{(1)}}(x)=n[1?F(x){]}^{n?1}f(x)=n{e}^{?(n?1)(x?\theta )}{e}^{?(x?\theta )}=n{e}^{?n(x?\theta )} \end{gathered}$$

This means $X_{(1)}～\theta +Gamma(1,n)$

Part c
Define $Q=2n(X_{(1)}?\theta )$. By this location-scale transformation, $Q～{\chi }_2^2{=}_d\Gamma (1,\frac{1}{2})$. Let ${\chi }_{y,2}^2$ and ${\chi }_{1?\alpha +y,2}^2$ denote $y$ and $1?\alpha +y$ quantile of ${\chi }_2^2$.
$$\begin{gathered} P({\chi }_{y,2}^2\le Q\le {\chi }_{1?\alpha +y,2}^2)=1?\alpha \\ P\left(X_{(1)}?\frac{{\chi }_{y,2}^2}{2n}\le \theta \le X_{(1)}?\frac{{\chi }_{1?\alpha +y,2}^2}{2n}\right)=1?\alpha \end{gathered}$$

Notice the length of confidential interval is
$$L=\frac{{\chi }_{1?\alpha +y,2}^2?{\chi }_{y,2}^2}{2n}$$

Let $Z$ denote standard normal variable. By normal approximation of chi-square distribution (see UA MATH564 概率論VI 數理統計基礎3 卡方分布的正態近似)
$$L\approx \frac{2+2Z_{1?\alpha +y}?(2+2Z_y)}{2n}=\frac{Z_{1?\alpha +y}?Z_y}{n}$$

Notice standard normal distribution is symmetric on y-axis, so the shortest length should be $y=\frac{\alpha }{2}$. Hence the shortest confidential interval is
$$P\left(X_{(1)}?\frac{{\chi }_{\frac{\alpha }{2},2}^2}{2n}\le \theta \le X_{(1)}?\frac{{\chi }_{1?\frac{\alpha }{2},2}^2}{2n}\right)=1?\alpha$$

Part d
Posterior kernel of $\theta$ is
$$\pi (\theta |\text{X})\propto \exp \left(?\sum _{i=1}^n(X_i?\theta )\right)$$

Compute
$$\begin{gathered} {\int }_0^1\theta \exp \left(?\sum _{i=1}^n(x_i?\theta )\right)d\theta =\frac{1}{n}{\int }_0^1\theta d\exp \left(?\sum _{i=1}^n(x_i?\theta )\right) \\ =\frac{1}{n}\theta \exp \left(?\sum _{i=1}^n(x_i?\theta )\right){|}_0^1?\frac{1}{n}{\int }_0^1\exp \left(?\sum _{i=1}^n(x_i?\theta )\right)d\theta \\ =\frac{n?1}{n^2}\exp \left(n?\sum _{i=1}^n{X}_i\right)+\frac{1}{n^2}\exp \left(?\sum _{i=1}^n{X}_i\right) \end{gathered}$$

Marginal density of $X$ is
$$m(x)={\int }_0^1\exp \left(?\sum _{i=1}^n(x_i?\theta )\right)d\theta =\frac{1}{n}\exp \left(n?\sum _{i=1}^n{X}_i\right)?\frac{1}{n}\exp \left(?\sum _{i=1}^n{X}_i\right)$$

So the estimator is
$$\hat{\theta }=\frac{\frac{n?1}{n^2}\exp \left(n?{\sum }_{i=1}^n{X}_i\right)+\frac{1}{n^2}\exp \left(?{\sum }_{i=1}^n{X}_i\right)}{\frac{1}{n}\exp \left(n?{\sum }_{i=1}^n{X}_i\right)?\frac{1}{n}\exp \left(?{\sum }_{i=1}^n{X}_i\right)}=\frac{(n?1)e^n+1}{n{e}^n?n}$$

總結

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