CHi nt e R s a hi o nre n Ry ne Rt re y a t He r he t fo ow b Co(e,b压)=Ee(bRB)压 E(IRP)eeX( XE I(IRPX(XX Cov(e,b)=0 dd d、1c、ofex(b,(b、 过刀)) fiLC 0, of e. SnpT or BR12 bk R B RK&/n R 8(X'X nR K d gr、、off、d N(0.1 i tk (0, d duc Rt IR 1)of丘、don,E、00 tRd. r b FBk F bR+ ta/2Sbg (ii)FRtest Co.、dr Ho: RB JTK x R fu row r T FR for uld、fi,,dCHAPTER 4 FINITE—SAMPLE PROPERTIES OF THE LSE 5 This follows because Cov (e, b|X) = E  e (b − β) ′ |X  = E
(I − P) ee′X (X ′X) −1 |X = σ 2 I (I − P) X (X ′X) −1 = 0 which implies Cov (e, b) = 0 and independence of e and b, and because s 2 is a function of e. (See also Theorem B−12). Therefore, tk =  bk − β 0 k  / σ 2 (X′X) −1 kk (n − K) s 2 σ2 / (n − K) = bk − β 0 k s 2 (X′X) −1 kk has Student’s t−distribution with n − K degrees of freedom. Recall N (0, 1)  χ 2 k /k ∼ tk when N (0, 1) and χ 2 k are independent. We deduce from the distribution of tk P  bk − tα/2sbk ≤ βk ≤ bk + tα/2sbk  = 1 − α where sbk = s 2 (X′X) −1 kk and tα/2 is the critical value from the t−distribution with (n − K) degrees of freedom. The 100(1-α)% confidence interval for βk is bk − tα/2sbk ≤ βk ≤ bk + tα/2sbk  . (ii) F−test Consider the null hypothesis H0 : Rβ = r where the J × K matrix R has full row rank. The F−test for this null is defined as F = (Rb − r) ′
R (X ′X) −1 R ′ −1 (Rb − r) /Js2