UA MATH566 统计理论 QE练习题2.2
UA MATH566 統(tǒng)計(jì)理論 QE練習(xí)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(shí)在是不想算了，棄療，下面是答案。個(gè)人建議考試就四個(gè)小時(shí)還是放棄這種思路不難但計(jì)算很麻煩的題吧，反正六選五。
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