Prove (5.2). Hint:
Step (a).
E(mn(x) − m(x))2 ≥ E(mn(x) − E{mn(x)|X1,...Xn}) 2 + (E{mn(x)} − m(x))2 .
Step (b).
E(mn(x) − E{mn(x)|X1,...Xn}) 2
= 1 n{µ([x − hn, x + hn]) P{µn([x − hn, x + hn]) > 0}2 .
Step (c).
E
1 0 (mn(x) − E{mn(x)|X1,...Xn}) 2 µ(dx)
≥
1−hn hn 1 2nhn (1 − (1 − 2hn) n) 2 dx
≥ 1 + o(1) 2nhn .
Step (d).
(E{mn(x)} − m(x))2
= nE X1I{|X1−x|≤hn} E 1 1 + n i=2 I{|Xi−x|≤hn} − x 2 .
Step (e). Fix 0 ≤ x ≤ hn/2,
then E X1I{|X1−x|≤hn} = (x + hn) 2 2 ,
Step (f).
E 1 1 + n i=2 I{|Xi−x|≤hn} ≥ 1 1+(n − 1)(x + hn) .
Step (g). For large nhn,
nE X1I{|X1−x|≤hn} E 1 1 + n i=2 I{|Xi−x|≤hn} ≥ x + hn 3 ≥ x.
Step (h). For large nhn,
1 0 (E{mn(x)} − m(x))2 µ(dx) ≥
hn/2 0 x + hn 3 − x 2 dx = h3 n 54 .