曝光台 注意防骗
网曝天猫店富美金盛家居专营店坑蒙拐骗欺诈消费者
– if |u − k| ) p G(u, k) sup|x| f(x)
– if |u − k| > ) p/2(H + L) + h G(u, k) 1/
Now we can take
G1 = min{p, p/(2(H + L) + h), 1/2}
and
G2 = max{1/, sup
|x|
f(x), 1 + [(H + L)/h]}
We proved the lemma.
Now we are ready to state and prove the theorem which specifies the rate of convergence
of the estimator ˆk.
Theorem 3.5.1.2 Suppose the conditions (A1) - (A4) are satisfied. Let Kn = {k 2
N : C0n2 /(2 +1) k n}, where C0 is defined by (algorithm). Then, for some
51
positive constant C1, any fixed , and any sequence {kn} such that kn 2 Kn, the
relations
1 Xn=1
P{n /(2 +1)(log n)(−3/2+ )|ˆkn − kn | > } < 1
lim sup
n1
max
k2Kn
n2 /(2 +1)
(log n)2
E(ˆk − k)2 C1
hold uniformly over 2
Proof. Introduce the following identities for convenience:
for any sequence {bi}, Pm
i=m+1 bi = 0, and Qmi
=m+1 bi = 1.
Now define the differences k = k,n = ˆk − k and k = k − k+1. Denote the difference
S (Yk, ˆ k) − g( ˆ k, k) by M(Xk,ˆk, k). We can, now, rewrite the (algorithm):
ˆ
k = ˆk−1 +
nS (Yk−1,ˆk−1) = (k − k−1) +ˆk − k = (ˆk−1 − k−1) +
nS (Yk−1,ˆk−1)
) k = k−1 +
n(M(Yk−1,ˆk−1, k−1) + g(ˆk−1, k−1)) + k−1
Using the relation, g(ˆk, k) = G(ˆk, k)k from the LEMMA:
k = k−1(1 −
nG(ˆk−1, k−1)) +
nM(Yk−1,ˆk−1, k−1) + k−1
k = 1, . . . , n. Observe that, for any k0, 0 k0 k, iterating the above algorithm
gives:
k = (k−2(1−
nG(ˆk−2, k−2))+
nM(Yk−2,ˆk−2, k−2))(1−
nG(ˆk−1, k−1))+
n+k−1
= ((k−3(1 −
nG(ˆk−3, k−3)) +
nM(Yk−3,ˆk−3, k−3) + k−3)(1 −
nG(ˆk−2, k−2))
+
nM(Yk−2,ˆk−2, k−2)+k−2)(1−
nG(ˆk−1, k−1))+
nM(Yk−1,ˆk−1, k−1)+k−1
.
.
.
52
= k0
k−1Yi=k0
[1 −
nG(ˆi, i)] +
k−1Xi=k0
{[i +
nM(Yi,ˆi, i)]
k−1Yj=i+1
[1 −
nG(ˆ j, j)]}
(3.5)
Now, remember summation by parts formula, which asserts:
nX
k=1
xkzk = Xnzn −
n−1Xk=1
Xk(zk+1 − zk)
where Xk = x1 + x2 + . . . + xk.
Denoting Ai = Pi
j=k0 aj, with aj = j +
nM(Yj, ˆj, j), we can apply the summation
by parts formula to the second part of the equation 3.4. Using the conventions
we set, we get:
k−1Xi=k0
{[i +
nM(Yi,ˆi, i)]
k−1Yj=i+1
[1 −
nG(ˆ j, j)]} =
k−1Xi=k0
ai
k−1Yj=i+1
[1 −
nG(ˆ j, j)]
= Ak−1 −
k−2Xi=k0
Ai
nG(ˆi+1, i+1)
k−2Yj=i+2
[1 −
nG(ˆ j, j)] (3.6)
Now, we can rewrite k as:
k = k0
k−1Yi=k0
[1 −
nG(ˆi, i)] + Ak−1 −
k−2Xi=k0
Ai
nG(ˆi+1, i+1)
k−2Yj=i+2
[1 −
nG(ˆ j, j)]
The relation 3.5 holds for any sequence {ai}, which allows us to take ak0 = 1 and
aj = 0 for all j > k0 and Ai = Pi
j=k0 aj = 1 for all i, so:
k−2Xi=k0
nG(ˆi+1, i+1)
k−1Yj=i+2
[1 −
nG(ˆ j, j)] = 1 −
k−1Yi=k0+1
[1 −
nG(ˆi, i)] 1
for sufficiently large n, by Lemma
As Belitser and de Geer pointed out,
|ˆk| H + L + h + b, b = sup
u1
u−2 /(2 +1) log u
Because ˆk and k are bounded as we showed, |k| is bounded uniformly in k and
2 as well.
53
Now we can write the relation for |k| as follows:
|k| |k0 |(1 −
nGk−k0
1 + max
k0ik |Ai|
for sufficiently large n, uniformly in 2 , where G1 is as defined in the proof of
the LEMMA. The above inequality is equalivant to:
2k
2k
0(1 −
nG1)2(k−k0) + 2k0(1 −
nG1)k−k0 max
k0ik |Ai| + max
k0ik
A2i
Using the definition of Ai:
中国航空网 www.aero.cn
航空翻译 www.aviation.cn
本文链接地址：航空资料31(22)
