关键点：用matrix notation对矩阵迹进行变换
$hat{y}=X{(X^TX)}^{-1}X^Ty=Hy$
$sum_i^n {(y_i-hat{y}_i)}^2={(y-Hy)}^T(y-Hy)=y^Ty-y^THy-y^THy+y^TH^THy\ =y^Ty-y^THy$ 其中$H^TH=H$
$E(y^Ty)=E(sum_i^ny_i^2)=sum_i^n (sigma^2+C_i)=nsigma^2+sum_i C_i^2=nsigma^2+C^TC$ 其中$E(y_i)=C_i$
$E(y^THy)=E(sum_isum_jy_iy_jH_{ij})=E(sum_i y_i^2H_{ii}+sum_isum_{j eq i}y_iy_jH_{ij})\ =sigma^2sum_i H_{ii}+sum_i C_i^2H_{ii}+sum_isum_{j eq i}C_iC_jH_{ij}\ =sigma^2sum_i H_{ii}+sum_isum_jC_iC_jH_{ij}\ =sigma^2sum_i H_{ii}+C^THC$
$C=Xeta$
$C^THC=eta^TX^TX{(X^TX)}^{-1}X^TXeta=C^TC$
$sum_iH_{ii}=sum_ileft[X{(X^TX)}^{-1}X^T ight]_{ii}=sum_i sum_j left[X{(X^TX)}^{-1} ight]_{ij}X_{ij}\ =sum_i sum_j sum_k X_{ik}{(X^TX)}_{kj}^{-1}X_{ij}\ =sum_jsum_k left[X^TX ight]_{kj}{(X^TX)}_{kj}^{-1}=sum_jleft[X^TX{(X^TX)}^{-1} ight]_{jj}\ =sum_j I_{jj}=p+1$
所以$E(sum_i^n {(y_i-hat{y}_i)}^2)=(n-p-1)sigma^2$