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


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Cov (W (s, t) , W (s0, t0)) = t (1 − t) (1 − t0) C1,1 (s, s0) − 2t (1 − t) t0 (1 − t0) C1a (s, s0)




        • 2t2 (1 − t) (1 − t0) C1a (s0, s) + 4t2 (1 − t) t0 (1 − t0) Ca (s, s0)




          • (t0t) t (1 − t0) C2,1 (s, s0) − 2 (t0t) tt0 (1 − t0) C2a (s, s0)




  • 2 (t0t) t (1 − t) (1 − t0) C1a (s0, s) + 4 (t0t) t (1 − t) t0 (1 − t0) Ca (s, s0)




            • (1 − t0) tt0C2,2 (s, s0) − 2 (1 − t0) tt0 (1 − t0) C2a (s, s0)




    • 2 (1 − t0) t (1 − t) t0C2a (s0, s) + 4 (1 − t0) t (1 − t) t0 (1 − t0) Ca (s, s0)




      • 2t2 (1 − t) t0 (1 − t0) C1a (s, s0) − 2 (1 − t0) t2tt0 (1 − t0) C2a (s, s0)




+ 4t (1 − t) t0 (1 − t0) Ca (s, s0) +

4

t (1

t) t0 (1 − t0) Cs,sa0. (2.96)

3

Simplifying this expression leads to the covariance mentioned in (1.17).

18 HEROLD DEHLING, DAVIDE GIRAUDO AND OLIMJON SHARIPOV


2.2.3. Convergence of the linear part. We also use Theorem 2.3. The convergence of (Wn0)n>1 also hold to a process having continuous paths. Define























ξn (s, t) := Wn0 (−R + 2Rs, t) , s, t [0, 1],



















(2.97)

where Wn0 is defined by (2.75). Observe that
































































ξn (s, t) = Wn (−R + 2Rs, t)

(




− )

1,R+2Rs ( i)




(










− 1)




2,R+2Rs (




j)




(2.98)

+

n3/2










n























































[nt] (n − [nt]) 1

n−1




n

i h






















X










n




j




h










X




.



















X








































X














































































































































2

i=1








































j=2








































Now, it suffices to prove that
























































































ξ




s, t

[nt] (n − [nt])







1







n−1




n i h













X













and













X































n,1

(

) :=






















n













− )










1,R+2Rs (




i)










(2.99)













n3/2










(















































































2







i=1























































ξ













s, t

[nt] (n − [nt]) 1







n

j







h













X





































X








































n,2 (




) :=


































− 1)







2,R+2Rs (




j)










(2.100)






































































n3/2

n







(















































































2







j=2



















































satisfy the conditions 2, 3 and 4. Since the treatment of ξn,1 is completely analoguous to that of ξn,2, we will do it only for the latter. By writing [nt] (n − [nt]) and h2,R+2Rs (Xj) as a difference of non-decreasing function, we can see that 2 is satisfied with










[nt]

1




n




[nt]2 1




n













X







X




where





































ξn,2

(s, t) = n1/2




n




j=2

(j − 1) bs (Xj) +

n3/2 n




(j − 1) θ2,s (Xj) ,

(2.101)










2







2




j=2




bs (v) = P{g (X1, v) 6 R + 2Rs} ; θs = P{g (X1, X2) 6 R + 2Rs} .

(2.102)

To do, so rewrite (j − 1) θ2,s (Xj) in terms of partial sums of θ2,s (Xj), use triangle inequality for the L2p-norm, then we apply Proposition A.4.

Similar estimates as those who led to (2.55) give



06j1,j26n

n,2

n

n



n,2

n n




6

32n−1/2.

(2.103)

max




ξ

j1 , j2

+ 1




ξ




j1 , j2



































































































































































2.2.4. Negligibility of the degenerated part. In view of (2.4) and (2.76), it suffices to prove that




















[nt] (n − [nt]) n




−1










h




X
















.




s

[







R,R] 06t61




























2



















1

i












































n











































































3/2


































6X6

3,s (

i)







0 in probability

(2.104)




sup sup















































































































































































[nt](n [nt])




n




1







23/22

















































Bounding




n3/2




2




by 2n




, it

suffices to show that for all positive ε,














































P




sup

























h3,s

(Xi)




> ε



0.

(2.105)

























s




3/2




1











































∈ −










1













6

6






























































































X













































































































































































































































































































































This is done in the same way as in the proof of Proposition 2.4: we cut the interval [−R, R] into intervals of length δn = n−3/4 and do the same estimate. This ends the proof of Theo-rem 1.3.


19
2.3.


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