In Exercise 10.19, you showed that o could be written as a linear function of independent

In Exercise 10.19, you showed that β̂_o could be written as a linear function of independent random variables. Use Theorem 6.8 to show that

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THEOREM 6.8 The Expected Value E(C) and variance V(O* of a Linear Function of Y, Y2, .. Yn Suppose the means and variances of Y1, Y2. ... Y4 are (u1, oi). (u2, oi)... (Hn, o). respectively. If f = a,Y, + azY, + .. + a„Y, then E(e) = ajui + azu2 +. + anea and oi = v(€) = ajoi + ažoi + + ao + 2aja,Cov(y1. y2) + 2aja;Cov(y1, y3) + ... + 2aja,Cov(yi. Ya) + 2azazCov(y2. y3) +... + 2aza,Cov(y2. Ya) + ... + 2a,-19,Cov(ya-1. Ya) Note: If Y1. Y2. .... Y, are independent, then oi = V(e) = afaf + ažaž + ... + ao Proof of Theorem 6.8 By Theorem 6.3, we know E(€) = E(a, Y1) + E(a,Y2) + ..+ E(a,Y,) %3D Then, by Theorem 6.2, E(€) = a,E(Y1) + azE(Y,) + + a,E(Y,) %3D = aiui + azuz +*+ anfln Similarly. V(e) = E{[{ – E(()F} %3D = E[(a,Y, + azY2 +...+ a, Y, - aji - azl2 %3D -.. . = E{[(a(Y1 - H1) + az(Y2 - p2) + . + aa(Ya - Hn)l} = E[af(Y - p) + a(Y, - p2)? + ...+ (Y - H)? + 2ajaz(Y, - Hi)(Y, – p2) + 2aja3( Y1 - Hi)(Y - u3) +... + 2a4-19a(Ya-1 - Ha-1)(Y - H)] %3D + ... + a,E[ (Ya - Ha)²] + 2ajazE[ (Y - H1)(Y - 2)]+ 2aja;E[ (Y, - Hi)(Y3 - H3)] +... + 2an-14,E[(Y-1 - Ha-1)(Yn - Ha)l %3D By the definitions of variance and covariance, we have E[ (Y; - H;)1 = o and E[(Y; - Hi)(Y; - M)) = Cov(Y, Y;) %3! Therefore. V(€) = afof + ažož + ... + ao + 2a;a,Cov(Y1. Y2) + 2aja;Cov(Y, Y3) + ... + 2azaz Cov (Y2, Y3) + +2an-1a,Cov(Y-1, Ya)