2
n8>><>>:
iX
j=k0
M(Yj,ˆ j, j)9>>=>>;
2
for constants c and C in R. As in the research of Belitser and van de Geer [number],
we choose the level of backward recursion k0 as follows:
k0 = k0(k) = k0(k, n) = bk − C0
−1
n log nc
Hence,
C0
−1
n log n  k − k0  C0
−1
n log n + 1
and therefore k0(k) are properly defined for those k which satisfy
k 2 {i 2 N : C0
−1
n log n  i  n} = Jn
Now, let us derive a bound for (1 −
nG1)k−k0 :
(1 −
nG1)k−k0  C(1 −
nG1)C0
−1
n log n  Ce−2 log n = Cn−2
for sufficiently large n as long as C0 satisfies C0G1  2.
Moreover:
max
k0ik{
iX
j=k0

j}2  max
k0ik
L2(xi − xk0 )2  c(k − k0)2/n2
54
uniformly in  2 . In order to evaluate C
2
n{Pi
j=k0 M(Yj,ˆ j, j)}2, observe that
{M(Yj,ˆ j, j)}kj
=k0
is a martingale difference with respect to the natural filtration
{Aj}kj
=k0
, where Aj = (Y0, Y1, . . . , Yj), the -algebra generated by Y0, Y1, . . . , Yj.
Lets first remind the definition of martingale:
Definition: A stochastic process M = (Mn)n0 is called a martingale with respect
to the filtration F if it satisfies the following conditions:
• (1)M is adapted to F
• (2)E(|Mn|) < 1 for all n  0
• (3)E(Mn+1|Fn) = Mn a.s. 8n
First, observe from the definition of M and the bound of |ˆk| that |M(Yj,ˆ j, j)| 
2 max{1, H + L + h + b}, almost surely. Moreover,
E(M(Yj+1,ˆ j+1, j+1)|Aj) = E(S (Yj+1,ˆ j+1)|Aj) − E(g(ˆ j+1, j+1)|Aj)
= 0, j = 1, 2, . . . , n − 1. (3.7)
Therefore, {Pm
j=k0 M(Yj,ˆ j, j)}k
m=k0
is a martingale. Now, remember Doob’s Submartingale
Inequality:
Doob’s Submartingale Inequality: Let Z be a martingale or non-negative submartingale.
Then, for c > 0 and p > 1,
P(max
kn
Zk  c)  c−1E(Zn) (3.8)
and
|| max
kn
Zk||p 
p
p − 1||Zn||p (3.9)
The special case where p = 2 gives
E[(max
kn
Zk)2]  4E[Z2
n ]
55
Applying Doob’s Submartingale Inequality to Pkj
=k0 M(Yj,ˆ j, j), we obtain:
E{ max
k0ik
[
iX
j=k0
M(Yj,ˆ j, j)]2}  4E{
kX
j=k0
M(Yj,ˆ j, j)}2 = C(k − k0)
uniformly in  2  and over k 2 J for some constant C.
From this point on, following the same steps as Belitser and de Geer, one concludes:
n[(1 −
nG1)k−k0 + max
k0ik |
iX
j=k0

j|] 
Cn
n2 +
Ln(k − k0)
n

c
(log n)3/2+ (3.10)
uniformly in  2  and over k 2 Jn, where n = n/(2+1)(log n)−3/2+.
Furthermore, using the Azuma-Hoeffding inequality, one has the following bound:
P{cn
n max
k0ik |
iX
j=k0
M(Yj,ˆ j, j)| > /2}  2 exp{−
Cn2/(2+1)(log n)1+2
k − k0 }
 2 exp{−c(log n)1+2}  Cn−2(3.11)
uniformly in  2  and over k 2 Kn. Uniformness of the above results lead us to
1 Xn=1
P{n|kn | > } < 1
uniformly in  2  for any sequence {kn} such that kn 2 Kn. The result then follows.

Remark
As Belitser and de Geer[2000]([5]) state, the results hold if we allow k’s to have
different distributions, all satisfying the conditions (A1) - (A4). Furthermore, defining
the class P = P(, p, M) of densities satisfying (A1) - (A4) with the condition
supu2R |f (u)|  M, one can show that all the results hold uniformly overall joint
distributions of (0, 1, . . . , n), for all independent  with densities f 2 P.
56
3.5.2 Estimation of
n
Estimation of
n is a nonlinear optimization problem. One has to choose an accuracy
measure to minimize in order to decide on
n. In this study we chose this
measure as the mean square error.
Experiments showed the estimations using a parameter depending on the number
of observations, n, does not lead to accurate results. (The optimization, estimation
results and the conclusion will be written later.)
We try to improve our model parameters.
　
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