of the support [a,b of Xt,we have Eal-g0=7∑Eke-x)-g T t=1 E [Kn (-X)]-g(x)(by identical distribution) k(（) g(y)dy-g(x) (b-x)/h J(a-x)/h (d)(y chue of variabie) K(w)ode halde-) g(r)K(u)du-g(z) +he国 uK(u)du +号r四ge+nh Cx((+m)(K(w)du 2cxe+ad的 where the second term re+回--rK@i-0 as h-0 by Lebesgue's dominated convergence theorem,and the boundedness and continuity of g"()and2K(u)du<oo. Therefore,for the point x in the interior region a+h,b-h,the bias of g(x)is proportional to h2.Thus,we must let h0as Too in order to have the bias vanish to zero as T→o. The above result for the bias is obtained under the identical distribution assumption on {Xt.It is irrelevant to whether {Xt}is IID or serially dependent.In other words, it is robust to serial dependence in [X}. Question:What happens to the bias of g(r)if x is outside the interior region [a+h,b- ? 17of the support [a; b] of Xt ; we have E [^g(x)] ￾ g(x) = 1 T X T t=1 EKh(x ￾ Xt) ￾ g(x) = E [Kh (x ￾ Xt)] ￾ g(x) (by identical distribution) = Z b a 1 h K  x ￾ y h  g(y)dy ￾ g(x) = Z (b￾x)=h (a￾x)=h K(u)g(x + hu)du ￾ g(x) (by change of variable y ￾ x h = u) = Z 1 ￾1 K(u)g(x + hu)du ￾ g(x) = g(x) Z 1 ￾1 K(u)du ￾ g(x) +hg0 (x) Z 1 ￾1 uK(u)du + 1 2 h 2 Z 1 ￾1 u 2K(u)g 00(x + hu)du = 1 2 h 2CKg 00(x) + 1 2 h 2 Z 1 ￾1 [g 00(x + hu) ￾ g 00(x)]u 2K(u)du = 1 2 h 2CKg 00(x) + o(h 2 ) where the second term Z 1 ￾1 [g 00(x + hu) ￾ g 00(x)]u 2K(u)du ! 0 as h ! 0 by Lebesgueís dominated convergence theorem, and the boundedness and continuity of g 00(
) and R 1 ￾1 u 2K(u)du < 1: Therefore, for the point x in the interior region [a + h; b ￾ h]; the bias of g^(x) is proportional to h 2 : Thus, we must let h ! 0 as T ! 1 in order to have the bias vanish to zero as T ! 1. The above result for the bias is obtained under the identical distribution assumption on fXtg. It is irrelevant to whether fXtg is IID or serially dependent. In other words, it is robust to serial dependence in fXtg: Question: What happens to the bias of g^(x) if x is outside the interior region [a+h; b￾ h]? 17