UA MATH566 統計理論 QE練習題2.2

• 第五題

這是2014年5月的5題。

第五題

Part (a)
Joint density of the bivariate normal disrtribution is
$$f_{Y_1,Y_2}(y_1,y_2)=\frac{1}{2\pi {\sigma }^2\sqrt{1?{\rho }^2}}\exp \left(?\frac{1}{2(1?{\rho }^2)}\left[\frac{(Y_1?\mu {)}^2}{{\sigma }^2}+\frac{(Y_2?\mu {)}^2}{{\sigma }^2}?\frac{2\rho (Y_1?\mu )(Y_2?\mu )}{{\sigma }^2}\right]\right)$$

Joint likelihood for random sample is
$$\begin{gathered} L(\alpha ,\beta ,{\sigma }^2,\rho )=\prod _{i=1}^n{f}_{Y_1,Y_2}(Y_{1i},Y_{2i})=(2\pi {)}^n{\sigma }^{2n}(1?{\rho }^2{)}^{n/2}\times \\ \exp \left(?\frac{1}{2{\sigma }^2(1?{\rho }^2)}\left[\sum _{i=1}^n(Y_{1i}?\alpha ?\beta x_i{)}^2+\sum _{i=1}^n(Y_{2i}?\alpha ?\beta x_i{)}^2?2\rho \sum _{i=1}^n(Y_{1i}?\alpha ?\beta x_i)(Y_{2i}?\alpha ?\beta x_i)\right]\right) \end{gathered}$$

Let’s compute
$$\begin{gathered} \left[\sum _{i=1}^n(Y_{1i}?\alpha ?\beta x_i{)}^2+\sum _{i=1}^n(Y_{2i}?\alpha ?\beta x_i{)}^2?2\rho \sum _{i=1}^n(Y_{1i}?\alpha ?\beta x_i)(Y_{2i}?\alpha ?\beta x_i)\right] \\ =\sum _{i=1}^n{Y}_{1i}^2?2\sum _{i=1}^n{Y}_{1i}(\alpha +\beta x_i)+\sum _{i=1}^n(\alpha +\beta x_i{)}^2+\sum _{i=1}^n{Y}_{2i}^2?2\sum _{i=1}^n{Y}_{2i}(\alpha +\beta x_i) \\ +\sum _{i=1}^n(\alpha +\beta x_i{)}^2?2\rho \left[\sum _{i=1}^n{Y}_{1i}{Y}_{2i}?\sum _{i=1}^n(\alpha +\beta x_i)Y_{1i}?\sum _{i=1}^n(\alpha +\beta x_i)Y_{2i}+\sum _{i=1}^n(\alpha +\beta x_i{)}^2\right] \\ =\sum _{i=1}^n{Y}_{1i}^2+\sum _{i=1}^n{Y}_{2i}^2+2n{\alpha }^2?2\alpha \sum _{i=1}^n\left(Y_{1i}+Y_{2i}\right)?2\beta \sum _{i=1}^n{x}_i(Y_{1i}+Y_{2i})+2\alpha \beta \sum _{i=1}^n{x}_i+{\beta }^2\sum _{i=1}^n{x}_i^2 \\ ?2\rho \left[\sum _{i=1}^n{Y}_{1i}{Y}_{2i}?\alpha \sum _{i=1}^n\left(Y_{1i}+Y_{2i}\right)?\beta \sum _{i=1}^n{x}_i(Y_{1i}+Y_{2i})+n{\alpha }^2+2\alpha \beta \sum _{i=1}^n{x}_i+\beta \sum _{i=1}^n{x}_i^2\right] \end{gathered}$$

Define $T_1(Y)={\sum }_{i=1}^n\left(Y_{1i}+Y_{2i}\right)$，$T_2(Y)={\sum }_{i=1}^n{x}_i(Y_{1i}+Y_{2i})$，$T_3(Y)={\sum }_{i=1}^n{Y}_{1i}{Y}_{2i}$. By Neyman-Fisher theorem, they are sufficient statistics. Consider another group of random sample $\{(Z_{1i},Z_{2i})\}$,
$$\begin{gathered} \frac{L(\alpha ,\beta ,{\sigma }^2,\rho |\text{Y})}{L(\alpha ,\beta ,{\sigma }^2,\rho |\text{Y})}=\exp (\sum _{i=1}^n(Y_{1i}^2+Y_{2i}^2?Z_{1i}^2?Z_{2i}^2)+2\alpha (T_1(Z)?T_1(Y))+ \\ 2\beta (T_2(Z)?T_2(Y))?2\rho [T_3(Y)?T_3(Z)?\alpha (T_1(Y)?T_1(Z))?\beta (T_2(Y)?T_2(Z))]) \end{gathered}$$

only if $T_1(Y)=T_1(Z),T_2(Y)=T_2(Z),T_3(Y)=T_3(Z)$, this likelihood ratio is independent of parameters. So $T_1,T_2,T_3$ are minimal sufficient statistics.

Part (b)
Log-likelihood function is
$$\begin{gathered} l(\alpha ,\beta ,{\sigma }^2,\rho )=n\log (2\pi )+n\log ({\sigma }^2)+\frac{n}{2}\log (1?{\rho }^2)? \\ \frac{1}{2{\sigma }^2(1?{\rho }^2)}\left[\sum _{i=1}^n(Y_{1i}?\alpha ?\beta x_i{)}^2+\sum _{i=1}^n(Y_{2i}?\alpha ?\beta x_i{)}^2?2\rho \sum _{i=1}^n(Y_{1i}?\alpha ?\beta x_i)(Y_{2i}?\alpha ?\beta x_i)\right] \end{gathered}$$

$$\frac{?l}{?\beta }=?\frac{1}{2{\sigma }^2(1?{\rho }^2)}\left[\sum _{i=1}^n?2x_i(Y_{1i}?\alpha ?\beta x_i)+\sum _{i=1}^n?2x_i(Y_{2i}?\alpha ?\beta x_i)?2\rho \sum _{i=1}^n(Y_{1i}?\alpha ?\beta x_i)(Y_{2i}?\alpha ?\beta x_i)\right]$$

這道題，我實在是不想算了，棄療，下面是答案。個人建議考試就四個小時還是放棄這種思路不難但計算很麻煩的題吧，反正六選五。

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