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


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In order to get rid of terms of the form bntkc and reduce the dependence in n, we will define for iIn,u the random variable


d

X


Yn,i =




ak,` (1 tk) h1,s` (Xi) 1 {u 6 k}




























k,`=1































2




d










d
















X







X


















(ni)

ak,`tk (1 tk) h1,s` (Xi) +







ak,`tkh2,s` (Xi) 1 {u > k}




n



















k,`=1







k,`=1































2




d



























i

X

ak,`tk (1 tk) h2,s` (Xi) . (2.80)































n




























k,`=1










Since |An,iYn,i| 6 1/n, it suffices to show that n−1/2







n







d







i=1 Yn,i =: n−1/2Sn




k,`=1 ak,`W 0 (s`, tk)

in distribution. Here again, will use Theorem A.1.

The first condition can be easily checked







P







P




by bounding |hq,s` (Xi)| by 2 and the terms i/n by 1. For the third condition, we also use ak = α (k), since each random variable Yni is a function of Xi. It remains to compute the limit of the sequence n−1 Var (Sn) n>1.
By similar argument as those who gave (2.13), this reduces to compute for all 1 6 u 6 d + 1 the limit

2


lim

1







X













(2.81)

E







Yn,i







.

n→+∞ n











































i In,u

For convenience, we write







d







Zi :=

ak,` (1 tk) h1,s` (Xi) 1 {u 6 k}







k,`=1










X

d

d







X

X




+

ak,`tkh2,s` (Xi) 1 {u > k} − 2

ak,`tk (1 tk) h1,s` (Xi) ; (2.82)







k,`=1

k,`=1




16







HEROLD DEHLING, DAVIDE GIRAUDO AND OLIMJON SHARIPOV































2




d


















































































X






































































Zn,i0

=







i

ak,`tk (1 tk) (h1,s` (Xi) h2,s` (Xi)) ,







(2.83)













n





































k,`=1
























































































d









































































Zi00

:= 2

X

ak,`tk (1 tk) (h1,s` (Xi) h2,s` (Xi)) .







(2.84)




















































k,`=1


























































Then Yn,i = Zi + Zi0 and consequently

























Zn,i0




+ E




Zn,i0










E




Yn,i




2 = E







Zi




2

+ 2E













Zi



















2










X





































X
















X

























X




























X













































































































iIn,u










iIn,u
















iIn,u




i0In,u










iIn,u

(2.
































































85)

For the first term, we get, by stationarity, that




























































































































2


























































lim




1













X































X






































































= ( u













0 i)




(2.86)













n +




n E










i










u−1) Cov (






















→ ∞














































i∈Z














































iIn,u










































































































































The second and third term cannot be treated similarly because of the terms i/n. Neverthe-less, we can use the following lemma.


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