Convergence of the empirical two-sample -statistics with -mixing data

Proposition 2.4. Let R_n be defined by (2.4). Under the conditions of Theorem 1.1, the following convergence holds:
sup sup |R_n (s, t)| → 0 in probability. (2.56)

−R6s6R 06t61

As a first step, we can look to the control of the supremum on t for a fixed s. Then

sup

n ^(

) =

^3^/^2 1 ` n 1

`

n

s

i

j

06t61 ^|

|

n

1

6 6 − i=1 j=`+1

R

s, t

max

X X

h

(X,X)

.

(2.57)

The problem is that the index ` over which we take

the maximum appear

in both sums and

we cannot directly apply maximal inequality in Lemma A.5.

Here is a way to overcome this issue. Denote hi,j

:= h3,s (Xi, Xj) and for a fixed n, S` :=

`

n

i,j^, ^1 ^6

6

^−`

1

n

0

`−1 n

6

6

^Pi=1

^Pj=`+1

h

`

n

S

`

n,

and

= 0. Then for 2

^S` − ^S`−1 ^=

X X

^hi,j −

X X

(2.58)

^hi,j

i=1 j=`+1

i=1 j=`

`−1 n

n

`−1 n

`−1

=

X X

^hi,j ^+

X

X X

X

(2.59)

^h`,j −

^hi,j −

^hi,`

i=1 j=`+1

j=`+1

i=1 j=`+1

i=1

n

`−1

=

X

^h`,j −

X

^hi,`

(2.60)

j=`+1

i=1

hence

k

n

k

k

n

6^X6

X X

^hi,j ^=

X

X X

^h`,j −

(2.61)

^Sk

=

(S` − S`−1) =

^hi,`^.

i=1 j=k+1

`=1

`=1 j=`+1

1 i<`

k

Therefore, the maximum over k can be treated by using the maximum of degenerated U-

statistic for the term

16i<`6k ^hi,`^.

The other one can be reduced to a similar contribution

data and (X

, . . . , X ) instead of (X , . . . , X

).

by using this time the^P

n

1

1

n

Using Lemma A.5, we derive that for each fixed s,

P 06t61

|

n ^(

)|

6

n

sup

R

s, t

> ε

Cε−2

log n

,

(2.62)

where the constant C depends only on (β (k))_k>1.

We divide the interval [−R, R] in intervals of length δ_n, where δ_n with be specified later.

Let ak := −R + 2Rkδn and the interval
Ik := [ak−1, ak], 1 6 k 6 [1/δn] + 1 =: Bn	(2.63)
(here for simplicity, we ommit the dependence in n for ak and Ik).

											13
Then
sup sup	|	R		(s, t)	| 6	max sup sup	|	R	s, t	.	(2.64)
−R6s6R 06t61			n			^16^k^6^Bn s∈Ik 06t61		n ^(	)|

In order to handle the supremum on I_k, we need the following lemma: