Show that with suitable smoothness assumptions on r(x), 2 in equation 2

(5.89) is a consistent estimator of .

86 5. Nonparametric Regression where T2 v= tr(L), [= tr(L*L)=E|| (z)|l?. i=1 If r is sufficiently smooth, v = o(n) and = o(n) then ? is a consistent estimator of o. We will now outline the proof of this result. Recall that if Y is a random vector and Q is a symmetric matrix then YTQY is called a quadratic form and it is well known that E(YfQY) = tr(QV) + *Qu (5.87) where V = V(Y) is the covariance matrix of Y and = E(Y) is the mean vector. Now, Y r = Y - LY = (I LY and so YTAY tr(A) (5.88) where A = (1-L)"(1 - L). Hence, rT Ar E() E(YTAY) tr(A) n-2v + Assuming that v and 7 do not grow too quickly, and that r is smooth, the last term is small for large n and hence E(a) = 0 . Similarly, one can show that v(a) - 0. Here is another estimator, due to Rice (1984). Suppose that the e;s are ordered. Define (5.89) (+1 - Y;)2. 2(n-1) The motivation for this estimator is as follows. Assuming r() is smooth, we have r(21+1)-(2)0 and hence = 0? + n-1 1 i=1 Yi+1 - Y, = [(x+1) + 6i+1] - [r(?:) +] - 6+1 - 6 and hence (Yi+1 - Y) - 4+1+? - 2+1. Therefore, E(Yi+1 - Y)2 El&+1)+E(?) 2E(1+1)E(e) E(+1)+E(C) = 20 (5.90)