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

Lemma 2.5. Let a and b be two real numbers such that a < b and let f : R → R be a function which can be expressed as a difference of two non-decreasing functions f₁ and f₂. Then

sup |f (s)| 6 |f (a)| + |f (b)| + |f2 (b) − f2 (a)| .

(2.65)

s∈[a,b]
Proof. Let s ∈ [a, b]. Then by non-decreasingness of f₁ and f₂,

• 
  (s) = f₁ (s)−f₂ (s) 6 f₁ (b)−f₂ (a) = f (b)+f₂ (b)−f₂ (a) 6 |f (a)|+|f (b)|+|f₂ (b) − f₂ (a)| .
  

Moreover,

f (s) = f₁ (s) − f₂ (s) > f₁ (a) − f₂ (b) = f (a) + f₂ (a) − f₂ (b) > − |f (a)| − |f₂ (b) − f₂ (a)| ,
which allows to conclude.
In the expression sup_s∈Ik sup_06t61 |R_n (s, t)|, the supremum over t is actually a maximum; for the supremum over s, we will apply Lemma 2.5 in the following setting: for a fixed t ∈ [0, 1],

[nt]		n
f1 (s) =		(1 {g (Xi, Xj) 6 s} + P{g (Xi, Xj) 6 s}) ;							(2.66)
i=1		nt
X		j=[^X]+1
^f2	[nt]n		h1,s (Xi) + h2,s (Xj)			,		(2.67)
	^(^s^) = i=1 j=[nt]+1
	X X		g	g

where hg_1,s (u) = P{g (u, X₂) 6 s} and hg_2,s (v) = P{g (X₁, v) 6 s}. In view of (2.64), we derive that

sup

sup

|

R

(s, t)

| 6

2 max sup

|

R

(a

, t)

|

−R6s6R 06t61

n

^16^k^6^Bn 06t61

n

k

2,ak ^( j^)

2,ak−1 ^(

j^)

^+ 3/2

sup

1

k

Bn

[nt]

n

1,ak ^(

i^)

1,ak−1 ^( i^) +

n

1

06t61

6 6

−

−

i=1 j=[nt]+1

max

^X

X

h^] X

h^^ X

h^] X

h^^ X

(2.68)
and after having rearranged the second term of the right hand side of (2.68), we end up with the estimate

P

−R6s6R 06t61

|

R

n

|

6 P

16k6Bn 06t61 ^|

R

n

k

|

sup sup

(s, t)

> 4ε

max

sup

(a

, t)

> ε

^+ ^P(√n 16k6Bn i=1 ^1^,ak ^(

i) − 1,ak−1 ( i)

)

1

n

max

X

h^] X

h^^ X

> ε

P

1 k Bn

n

2,ak ^( j^)

2,ak−1 ^( j^)

1

6 6

j=1

−

√n

X

h^] X

h^^ X

+

max

> ε . (2.69)

14 HEROLD DEHLING, DAVIDE GIRAUDO AND OLIMJON SHARIPOV

Let us estimate the first term of the right hand side of (2.69). The use of (2.62) and a union bound yields

P

^16^k^6^Bn 06t61

|

n

k

|

6

n exp

−

max

sup

R

(a

, t)

> ε

CB

cε^√n

.

(2.70)

Now, using the assumptions (A.1) and assumption (A.2), we derive that

n

q,ak ^( i^) ^−

q,ak−1 ^(

i)

6

16k6Bn

q ^( k ^−

k−1^) ^6

q n

∈{1 2}

16k6Bn i=1

X

h^] X

h^^ X

max

max

n

M

a

a

M nδ , q

,

hence the condition

(2.71)

√

lim

(2.72)

nδn = 0

n→+∞

guarantees the convergence in probability of the last two terms of (2.69).

We thus need

√

log n

lim

lim

= 0,

(2.73)

nδn = 0;

n→+∞

n→+∞ nδn

which can be done by choosing δ_n = n^−3/4. This ends the proof of Theorem 1.1.
2.2.