Page 113 - Textos de Matemática Vol. 47

ESTIMATION OF THE CONDITIONAL QUANTILE FUNCTION 103
Before stating the next lemma, we must fix some additional notations. For i = 0, 1, put ⇣ . ⌘
W(i)(x,y)=Z(i)(x,y) E Z(i)(x,y) Ft , t,T t,T t,T
where ✓ ◆✓ ◆
Z(i)(x,y)= 1 Gi y Yt K x Xt , y2E, x2Rd.
t , T T h dT h T h T DenotingbyntheintegerpartofT,n=[T],T 1,let besuchthat
n =T⌘Tn (andso1 <2).Wethendefine,fork2{0,...,n}, Tk =k and Fk, = {(Xs,Ys):0sTk}.
Lemma 4.2. IfZK is a bounded kernel, we have, f or all n 2 N and ✏ > 0, Tn(i) ! 22d
P Wt,Tn(x,y)dt >✏ 4 exp C1✏ Tn hTn
0 S with C1 a positive constant.
n
Proof. Fix i in {0, 1}. Writing [0, Tn] = [Tk 1, Tk], we get
Z T n
W (i)
k=1
Xn (x, y) dt =
Xn
U (i) k,Tn
V(i) (x,y)=Z Tl ⇣E⇣Z(i) (x,y).Fl 1, ⌘ E⇣Z(i) (x,y).Ft ⌘⌘dt. l 1,Tn t,Tn t,Tn
(x, y) +
with ZTk⇣ ⇣.⌘⌘
0 t,Tn
U(i) (x,y)= Z(i) (x,y) E Z(i) (x,y) Fk 1, dt,
k,Tn t,Tn t,Tn Tk 1
Note that U(i)
k,Tn k=1,...,n ` 1,Tn `=2,...,n
respectively, verifying
k=1
l=2
⇣Tl 1 ⌘ ⇣ ⌘
V (i) l 1,Tn
and V(i) (x,y) are martingale
(x,y)
di↵erences with respect to the fields (Fk, )k=1,...,n and (F` 1, )`=2,...,n,
(x, y)
and
U(i) (x,y)  2k⇤ 1 , k , T n T n h dT n
V (i) (x, y)  2 k⇤ 1 , ` 1,Tn Tn hdTn
k = 1,...,n, ` = 2, . . . , n.