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

1 {u 6 k} h_{1,s`</sub> (X<sub>i</sub>) + t<sub>k</sub>1 {u > k + 1} h<sub>2,s`} (X_i))

Zi,u :=

(2.21)

k,`=1

we have |Yn,i − Zi,u| 6 K/n for a constant K independent of n and i hence

lim

1

E

n

^Yn,i

2

= d+1

lim

1

E

^[^ntu^]−^[^ntu−1^] Zi

2

.

(2.22)

n→+∞ n

X

!

n→+∞ n

X

i=1

u=1

i=1

By expanding the square and using stationarity, it follows that

n→+∞ n ^E

2

=

(2.23)

n

n,i^!

^X

lim ^1

Y

σ2

i=1

where

d+1

X

σ^2 =

X

(tu − tu−1)

Cov (Z0,u, Zi,u) .

(2.24)

u=1

i∈Z

This is the variance of

d

a

N

N

d

covariance

^Pk,`=1

k,`

k,`, where (

k,`)k,`=1 is a centered Gaussian vector having

d+1

E^hZk,`,^(^u^)0, Zk^(^u0,`^)

0

,i^i

,

(2.25)

^Cov (^Nk,`^, Nk^0,`^0) = u=1 ^(^tu − ^tu−1^) i∈Z

where

X

X

Zk,`,i^(^u^) = (1 − tk) 1 {u 6 k} h1,s` (Xi) + tk1 {u > k + 1} h2,s` (Xi) .

(2.26)

We can also check by spliting the sum over u that for k 6 k^0,

Cov (Nk,`, Nk0,`0) = Cov (W (s`, tk) , W (s`0, tk0)) .

(2.27)

• 
  HEROLD DEHLING, DAVIDE GIRAUDO AND OLIMJON SHARIPOV
  

Now, it remains to check the third condition of Theorem A.1. We take a_k := α (k) and since Y_n,i is a function of X_i, the inequality α_n (k) 6 α (k) holds.

This ends the proof of Proposition 2.1.

2.1.3. Convergence of the linear part.