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


Download 380.03 Kb.
bet20/37
Sana16.11.2020
Hajmi380.03 Kb.
#146511
1   ...   16   17   18   19   20   21   22   23   ...   37
































































X






































































































































































Bounding n



[nt] and [nt] by n, we derive by






































































stationarity that








































E ξn (s0

, t)



ξn (s, t)

|

2p

6 CpnpE




[nt] (as

(Xi)



as0 (Xi) + ms0



ms)

2p






















h|
















i













i=1










































































































X

















































































































































































































































n







[nt]








































2p































































(bs (Xi) bs0

(Xi) + bs0







bs)










(2.46)


































+ CpnpE














































































































































































































i=1



























































































X

































































































































































































































































































































































































and taking into account the inequalities |as (x) − as0 (x)| 6 2RM |ss0| and |bs (x) − bs0 (x)| 6 2RM |ss0|, we derive that

h

i




E |ξn (s0, t) − ξn (s, t)|2p

6 C (s0s)p ,

(2.47)

showing (2.36).
Now we show (2.37). Going back to the expression of ξn given by (2.34), we derive that

|ξn (s, t) − ξn (s0, t)| 6

1













n

(|as (Xi) − as0 (Xi)| + |bs (Xi) − bs0 (Xi)|) + 2

























|msms0|
















n

n3/2




i=1

6 8R




X














































(2.48)




|s0s|








































n








































hence

6n

n

n




n



n

n n

6

n







06j1,j2




(2.49)




max







ξ

j1 + 1

, j2




ξ

j1 , j2







8R ,



















































































































































































































giving (2.37).
Finally, we show (2.38). For 0 6 t 6 t0 6 1 and s ∈ [0, 1], denoting cs following inequalities hold:




n (

)




n

(




) 6

3/2




s (

i)




3/2







s (




i) + 2




s







1/2

|

ξ

s, t



ξ




s, t0

|




[nt0

]

[nt0] c

X






[nt]

[nt]

c




X




m




[nt0

] − [nt]
















n




i=1







n

i=1
















n





































X













X






































































































































































































































































:= as + bs, the



(2.50)




[nt0] [nt] [nt]

c







X













[nt0]

[nt0]

c

X

m




[nt0] − [nt]




6





































s (




i) +






















s ( i) + 2




s













(2.51)







n

3/2










i=1







n

3/2



















n

1/2






































































nt
































































X





































i=[X]+1





































1
















nt0







nt










nt













nt0




[nt0] − [nt]

























6

4 n3/2 ([




] − [




]) ([




] + [




])+8




























(2.52)































n1/2



















6

8




(

n

(

t0

t

)+1)+8

n (t0 t) + 1




























(2.53)




























































n3/2
















n1/2




























6 16n1/2 (t0t) + 16n−1/2.














































(2.54)


12 HEROLD DEHLING, DAVIDE GIRAUDO AND OLIMJON SHARIPOV


As a consequence, we derive that



06j1

,j2

6n

n

n

n



n

n n

max

ξ

j1 , j2

+ 1




ξ

j1 , j2



























and (2.38) follows. This ends the proof of Proposition 2.2.

2.1.4. Negligibility of the degenerated part.

6 32n−1/2, (2.55)




Download 380.03 Kb.

Do'stlaringiz bilan baham:
1   ...   16   17   18   19   20   21   22   23   ...   37




Ma'lumotlar bazasi mualliflik huquqi bilan himoyalangan ©fayllar.org 2024
ma'muriyatiga murojaat qiling