UA MATH566 统计理论 QE练习题2.2
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
fY1,Y2(y1,y2)=12πσ21?ρ2exp?(?12(1?ρ2)[(Y1?μ)2σ2+(Y2?μ)2σ2?2ρ(Y1?μ)(Y2?μ)σ2])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)fY1?,Y2??(y1?,y2?)=2πσ21?ρ2?1?exp(?2(1?ρ2)1?[σ2(Y1??μ)2?+σ2(Y2??μ)2??σ22ρ(Y1??μ)(Y2??μ)?])
Joint likelihood for random sample is
L(α,β,σ2,ρ)=∏i=1nfY1,Y2(Y1i,Y2i)=(2π)nσ2n(1?ρ2)n/2×exp?(?12σ2(1?ρ2)[∑i=1n(Y1i?α?βxi)2+∑i=1n(Y2i?α?βxi)2?2ρ∑i=1n(Y1i?α?βxi)(Y2i?α?βxi)])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)L(α,β,σ2,ρ)=i=1∏n?fY1?,Y2??(Y1i?,Y2i?)=(2π)nσ2n(1?ρ2)n/2×exp(?2σ2(1?ρ2)1?[i=1∑n?(Y1i??α?βxi?)2+i=1∑n?(Y2i??α?βxi?)2?2ρi=1∑n?(Y1i??α?βxi?)(Y2i??α?βxi?)])
Let’s compute
[∑i=1n(Y1i?α?βxi)2+∑i=1n(Y2i?α?βxi)2?2ρ∑i=1n(Y1i?α?βxi)(Y2i?α?βxi)]=∑i=1nY1i2?2∑i=1nY1i(α+βxi)+∑i=1n(α+βxi)2+∑i=1nY2i2?2∑i=1nY2i(α+βxi)+∑i=1n(α+βxi)2?2ρ[∑i=1nY1iY2i?∑i=1n(α+βxi)Y1i?∑i=1n(α+βxi)Y2i+∑i=1n(α+βxi)2]=∑i=1nY1i2+∑i=1nY2i2+2nα2?2α∑i=1n(Y1i+Y2i)?2β∑i=1nxi(Y1i+Y2i)+2αβ∑i=1nxi+β2∑i=1nxi2?2ρ[∑i=1nY1iY2i?α∑i=1n(Y1i+Y2i)?β∑i=1nxi(Y1i+Y2i)+nα2+2αβ∑i=1nxi+β∑i=1nxi2]\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][i=1∑n?(Y1i??α?βxi?)2+i=1∑n?(Y2i??α?βxi?)2?2ρi=1∑n?(Y1i??α?βxi?)(Y2i??α?βxi?)]=i=1∑n?Y1i2??2i=1∑n?Y1i?(α+βxi?)+i=1∑n?(α+βxi?)2+i=1∑n?Y2i2??2i=1∑n?Y2i?(α+βxi?)+i=1∑n?(α+βxi?)2?2ρ[i=1∑n?Y1i?Y2i??i=1∑n?(α+βxi?)Y1i??i=1∑n?(α+βxi?)Y2i?+i=1∑n?(α+βxi?)2]=i=1∑n?Y1i2?+i=1∑n?Y2i2?+2nα2?2αi=1∑n?(Y1i?+Y2i?)?2βi=1∑n?xi?(Y1i?+Y2i?)+2αβi=1∑n?xi?+β2i=1∑n?xi2??2ρ[i=1∑n?Y1i?Y2i??αi=1∑n?(Y1i?+Y2i?)?βi=1∑n?xi?(Y1i?+Y2i?)+nα2+2αβi=1∑n?xi?+βi=1∑n?xi2?]
Define T1(Y)=∑i=1n(Y1i+Y2i)T_1(Y) = \sum_{i=1}^n \left( Y_{1i} + Y_{2i}\right)T1?(Y)=∑i=1n?(Y1i?+Y2i?)，T2(Y)=∑i=1nxi(Y1i+Y2i)T_2(Y) =\sum_{i=1}^n x_i(Y_{1i} + Y_{2i})T2?(Y)=∑i=1n?xi?(Y1i?+Y2i?)，T3(Y)=∑i=1nY1iY2iT_3(Y) = \sum_{i=1}^n Y_{1i}Y_{2i}T3?(Y)=∑i=1n?Y1i?Y2i?. By Neyman-Fisher theorem, they are sufficient statistics. Consider another group of random sample {(Z1i,Z2i)}\{(Z_{1i},Z_{2i})\}{(Z1i?,Z2i?)},
L(α,β,σ2,ρ∣Y)L(α,β,σ2,ρ∣Y)=exp?(∑i=1n(Y1i2+Y2i2?Z1i2?Z2i2)+2α(T1(Z)?T1(Y))+2β(T2(Z)?T2(Y))?2ρ[T3(Y)?T3(Z)?α(T1(Y)?T1(Z))?β(T2(Y)?T2(Z))])\frac{L(\alpha,\beta,\sigma^2,\rho|\textbf{Y})}{L(\alpha,\beta,\sigma^2,\rho|\textbf{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))])L(α,β,σ2,ρ∣Y)L(α,β,σ2,ρ∣Y)?=exp(i=1∑n?(Y1i2?+Y2i2??Z1i2??Z2i2?)+2α(T1?(Z)?T1?(Y))+2β(T2?(Z)?T2?(Y))?2ρ[T3?(Y)?T3?(Z)?α(T1?(Y)?T1?(Z))?β(T2?(Y)?T2?(Z))])
only if T1(Y)=T1(Z),T2(Y)=T2(Z),T3(Y)=T3(Z)T_1(Y) = T_1(Z),T_2(Y)=T_2(Z),T_3(Y) = T_3(Z)T1?(Y)=T1?(Z),T2?(Y)=T2?(Z),T3?(Y)=T3?(Z), this likelihood ratio is independent of parameters. So T1,T2,T3T_1,T_2,T_3T1?,T2?,T3? are minimal sufficient statistics.
Part (b)
Log-likelihood function is
l(α,β,σ2,ρ)=nlog?(2π)+nlog?(σ2)+n2log?(1?ρ2)?12σ2(1?ρ2)[∑i=1n(Y1i?α?βxi)2+∑i=1n(Y2i?α?βxi)2?2ρ∑i=1n(Y1i?α?βxi)(Y2i?α?βxi)]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]l(α,β,σ2,ρ)=nlog(2π)+nlog(σ2)+2n?log(1?ρ2)?2σ2(1?ρ2)1?[i=1∑n?(Y1i??α?βxi?)2+i=1∑n?(Y2i??α?βxi?)2?2ρi=1∑n?(Y1i??α?βxi?)(Y2i??α?βxi?)]
?l?β=?12σ2(1?ρ2)[∑i=1n?2xi(Y1i?α?βxi)+∑i=1n?2xi(Y2i?α?βxi)?2ρ∑i=1n(Y1i?α?βxi)(Y2i?α?βxi)]\frac{\partial l}{\partial \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]?β?l?=?2σ2(1?ρ2)1?[i=1∑n??2xi?(Y1i??α?βxi?)+i=1∑n??2xi?(Y2i??α?βxi?)?2ρi=1∑n?(Y1i??α?βxi?)(Y2i??α?βxi?)]
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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