ECON 710 Midterm: ECON 710 UW Madison Midterm 2015a
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= e(cid:2)y2(cid:28) (x)(cid:3) (cid:0) 2e [yx0(cid:28) (x)] (cid:18) + (cid:18)0e [xx0(cid:28) (x)] (cid:18) Setting it equal to zero and solving for (cid:18) T ((cid:18)) = 2e [xy(cid:28) (x)] (cid:18) + 2e [xx0(cid:28) (x)] (cid:18) (cid:18) = (e [xx0(cid:28) (x)]) (cid:0)1. E [xy(cid:28) (x): since e = y (cid:0) x0(cid:18), E [xe(cid:28) (x)] = e [xy(cid:28) (x)] (cid:0) e [xx0(cid:28) (x)] (cid:18) = e [xy(cid:28) (x)] (cid:0) e [xx0(cid:28) (x)] (e [xx0(cid:28) (x)]) = e [xy(cid:28) (x)] (cid:0) e [xy(cid:28) (x)] For this result, you do not need an additional assumption. For example, e(ejx) = 0 is not needed: if the conditional mean is linear e(yjx) = x0(cid:12) then (cid:18) = (e [xx0(cid:28) (x)]) (cid:0)1. E (x(cid:28) (x)x0(cid:12)) so that (cid:18) = (cid:12). When the conditional mean is linear it equals the best linear predictor, and thus (cid:18) equals the best linear predictor as well. A common answer was: (cid:147)when (cid:28) (x) = (cid:28) is independent of x(cid:148).
