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

Theorem 1.3. Let (X_i)_i∈Z be a strictly stationary sequence. Let e⁰_n be the two-sample U-statistics empirical process with kernel g : R × R → R defined for n > 1, 0 6 t 6 1 and s ∈ R

by

e0	(s, t) :=	1
n	n3/2

[nt] n
X X
i=1 j=[nt]+1

• 
  X
  

1 {g (X_i, X_j) 6 s} − _n 1 {g (X_i0, X_j0) 6 s} .

• 
  16i⁰⁰6n
  

(1.15)

Suppose that the assumptions (A.1), (A.2), (A.3) and (A.4) hold. Then for all R,
en^0 (s, t) → W (s, t) in distribution in D ([−R, R] × [0, 1]) ,	(1.16)

where (W (s, t) , s ∈ R, t ∈ [0, 1]) is a centered Gaussian process, with covariance given for 0 6 t 6 t⁰ 6 1 and s, s⁰ ∈ R by the following formula:

Cov (W (s, t) , W (s⁰, t⁰)) = t (1 − t) (1 − t⁰) C_1,1 (s, s⁰) + (t⁰ − t) t (1 − t⁰) C_2,1 (s, s⁰)

• 
  (1 − t⁰) tt⁰C_2,2 (s, s⁰) − 2tt⁰ 1 − t² (1 − t⁰) C₁^a (s, s⁰)
  

• 
  2t (1 − t) t⁰ (1 − t⁰) (s⁰, s) C₁^a (s⁰, s) − 2tt⁰ (1 − t⁰) 2 + t⁰ − 2t − t² C₂^a (s, s⁰)
  

− 2t (1 − t) t^0 (1 − t^0) C2^a (s^0, s) + 4t (1 − t) t^0 (1 − t^0) t^2 + t +

3

C^a (s, s^0) ,

(1.17)

10

where

h1,s (u) = P{g (u, X1) 6 s} − P{g (X1, X2) 6 s} ,

(1.18)

h2,s (v) = P{g (X1, v) 6 s} − P{g (X1, X2) 6 s} ,

(1.19)

for i, j = 1; 2,

^as ^= ^h1,s ^− ^h2,s^,

(1.20)

X

(1.21)

Ci,j (s, s^0) =

Cov (hi,s (X0) , hj,s0 (Xk)) ,

k∈Z

Ci^a (s, s^0) =

X

Cov (hi,s (X0) , as0 (Xk))

(1.22)

k∈Z

C^a (s, s^0) =

X

(1.23)

Cov (as (X0) , as0 (Xk)) .

k∈Z
Remark 1.4. When g is symmetric, h_1,s = h_2,s and a_s = 0 hence the covariance admits the simpler form

Cov (W (s, t) , W (s^0, t^0)) = t (1 − t^0) (1 − 2t + 2t^0) C1,1 (s, s^0) , t 6 t^0, s ∈ R,

(1.24)

in particular, we get the same limiting process as in the centering of the indicator by their expectation (see Remark 1.2).

• 
  HEROLD DEHLING, DAVIDE GIRAUDO AND OLIMJON SHARIPOV
  

Let us give examples where the assumptions (A.1) and (A.2) are satisfied. let g₁ and g₂ be function defined from R to itself. Assume that g₁ (X₁) has a density f₁ and g₂ (X₁) has a density f₂, where f₁ and f₂ are bounded.

(1) Let g : (u, v) 7→g₁ (u) + g₂ (v). Then g (u, X₁) has density f_1,u where f_1,u (x) =
f₂ (x − g₁ (u)) hence sup_x,u∈R f_1,u (x) = sup_x∈R f₂ (x) < +∞ and similarly, sup_x,u∈R f_2,u (x) = sup_x∈R f₁ (x) < +∞.

(2) Let g : (u, v) 7→g|1 (u) − g2 (v)|. Then
	f1,u (x) = f2 (g1 (u) + x) + f2 (g1 (u) − x)						(1.25)
hence
sup f1,u (x) 6 sup f2 (g1 (u) + x) +			sup f2 (g1 (u) − x) 6 2 sup f2 (x) < +∞				(1.26)
x,u∈R	x,u∈R	x,u∈R		x∈R
and by a similar reasoning, we also derive
	sup f2,u (x) 6 2 sup f1 (x) < +∞.						(1.27)
	x,u∈R	x∈R

1.2. Application. The following is a consequence of Theorems 1.1 and 1.3.

Corollary 1.5. Let µ be a finite measure on the Borel subsets of R. Then under the conditions of Theorems 1.1 and 1.3 the following convergences in distribution take place

06t61 ^ZR

e

n (

)

^2 dµ

(

^) ^→ 06t61

ZR

(

)

^2 dµ

(

)

(1.28)

sup

s, t

s

sup

W

s, t

s

06t61 ^Z

R

n

(

)

^2 dµ

(

) →

06t61

ZR

(

s, t

)

^2 dµ

( )

.

(1.29)

sup

e0

s, t

s

sup

W 0

s

2. 
   Proof
   

The proof of Theorems 1.1 and 1.3 will be done according to the following steps.

1. 
   Let h_s : R² → R be the kernel defined by h_s (u, v) = 1{g (u, v) 6 s}. The Hoeffding’s decomposition of this kernel gives a spliting of the empirical two-sample U-statistics into a linear part and a degenerated part.
   

2. 
   We prove the convergence of the finite dimensional distributions of the linear part to the corresponding ones of the process W .
   

3. 
   Then we prove that the process associated to the linear part converges to W in D ([−R, R] × [0, 1]) for all R > 0.
   

4. 
   Finally, we show the negligibility of the contribution of the degenerated part.
   

We do it first in the context of Theorem 1.1. The proof of Theorem 1.3 is closely related.

Consequently, we will only mention the required modifications.

2.1. Proof of Theorem 1.1.

5
2.1.1. Hoeffding’s decomposition. Let h_s : R² → R be defined by h_s (u, v) = 1 {g (u, v) 6 s}. Let us do the Hoeffding’s decomposition of h_s for each fixed s. Let θ_s := P{g (X₁, X₂) 6 s},

h1,s (u) = P{g (u, X1) 6 s} − θs, h2,s (v) = P{g (X1, v) 6 s} − θs,

(2.1)

and

Then
1

^en ^{(s, t) =} _n3/2

h_3,s (u, v) = h_s (u, v) − h_1,s (u) − h_2,s (v) − θ_s.
[nt] n

X X

(h_s (X_i, X_j) − θ_s)

i=1 j=[nt]+1
(2.2)

=
1

[nt]n	1	[nt]n
^X X h3,s (Xi, Xj) +		^X X h1,s (Xi) +

1	[nt]n
	^X X h2,s (Xj)

_n3/2

i=1 j=[nt]+1	n3/2
	i=1 j=[nt]+1

_n3/2

i=1 j=[nt]+1

=
[nt]

1 X

n	n − [nt]	[nt]	[nt]	n
^X h3,s (Xi, Xj) +		^X h1,s (Xi) +		^X h2,s (Xj)

_n3/2

i=1
or in other words,

where

j=[nt]+1	n3/2	n3/2
	i=1	j=[nt]+1
	en (s, t) = Rn (s, t) + Wn (s, t) ,	(2.3)

Rn (s, t) =

1

[nt]

n

h3,s (Xi, Xj) ,

(2.4)

n

3/2

i=1

nt

X

j=[^X]+1

Wn (s, t) =

n − [nt]

[nt]

^h1,s

(Xi) +

[nt] ^n

h2,s (Xj) .

(2.5)

n3/2

n3/2

i=1

nt

X

j=[^X]+1

Moreover, observe that by the assumpions (A.1) and (A.2), there exists a constant M such that for all i ∈ {1; 2; 3},

sup sup	|hi,s0 (x) − hi,s (x)|	6 M.	(2.6)
	s^0 − s
x∈R s^0

2.1.2. Convergence of the finite dimensional distributions of the linear part.