For d ≤ 2 assume that there exist 0 > 0, a nonnegative function g such that for all x ∈ Rd, and 0 <>
µ(Sx,) > g(x) d (6.3)
and
1 g(x)2/d µ(dx)
Prove the rate of convergence given in Theorem 6.2.
Hint: Prove that under the conditions of the problem
E{X(1,n)(X) − X2 } ≤ c˜ n2/d .
Formula (6.3) implies that for almost all x mod µ and 0
µ(Sx,) ≥ µ(Sx,0 ) ≥ g(x) d 0 ≥ g(x) 0 L d d,
hence we can assume w.l.o.g. that (6.3) holds for all 0 0,
P{X(1,n)(X) − X> } = E{(1 − µ(SX,))n}
≤ E e −nµ(SX,)
≤ E e −ng(X)d ,
therefore, E{X(1,n)(X) − X2 } =
L2 0 P{X(1,n)(X) − X > √} d
≤
L2 0 E e −ng(X)d/2 d
≤
∞ 0 e −ng(x)d/2 d µ(dx)
=
1 n2/dg(x)2/d
∞ 0 e −zd/2 dz µ(dx)
= c˜ n2/d .