Let Xh,2,it be the hth element of Xh,2,it. The required result follows if we can
show that
sup
i,T
√T X (xh,2,it
T t
- Exh,2,it)
< M for all h.
4
(41)
The proof of (41) is similar to the proof of Lemma 1 of Andrews (1991). This
proof relies on the following α- mixing inequality presented in Hall and Heyde
(1980, p. 278). Suppose that Y and W are random variables that are G-
measurable and H- measurable, respectively, with E ∣Y ∣p < ∞ and E ∣W∣q <
∞, where p,q > 1 with 1/p + 1/q < 1. Then,
∣E (Y - EY ) (W - EW ) ∣ ≤ 8 H Y ∣∣p ∣∣ W ∣∣g [α (G, H)]1-1/p-1/q , (42)
where α (G, H) is the α-mixing coefficient between the sigma fields G and H.
Now, let Xit = xh,2,it - Exh,2,it. Notice that
4
p e .. Σ X«)
sup T χ x x x∣e (Xi,XisXipXik )∣
i,τ t=1 s = 1 p=1 k= 1
T -tT-sT -p
⅛ 5⅛2χχχΣ∣E (XitXi,t+sXi,t+s+pXi,t+s+p+k) ∣
i,T t s=0 p=0 k=0
4!siuτp T X
Σ
0≤p,k≤s
0≤p+k+s≤T-t
∣ E (Xit (Xi,t+sXi,t+s+pXi,t+s+p+k)) ∣
+4!sup ɪ X X
i,T t 0≤s,k≤p
0≤p+k+s≤T-t
+4!sup .=2 X X
i,i t 0≤s,k≤p
0≤p+k+s≤T-t
1
+4!siup .χ
x
0≤s,p≤k
0≤p+k+s≤T-t
= I + II + III + IV, say.
E [(Xit
Xi,t+s) (Xi,t+s+pXi,t+s+p+k)]
-E (XitXi,t+s) E (Xi,t+s+pXi,t+s+p+k )
∣ E (XitXi,t+s) E (Xi,t+s+pXi,t+s+p+k) ∣
E ((Xit
Xi,t+sXi,t+s+p') Xi,t+s+p+k) ∣
By applying the inequality of (42) to Xi,t+sXi,t+s+pXi,t+s+p+k and Xit and
then by the Holder inequality, we have
1T
i ≤ 4!8sup .2 HXitk4q HXi,t+sH4q Il Xi,t+s+p ∣∣ 4q
i,T t=10≤p,k≤s≤T-t
×∣∣Xi,t+s+p+k∣∣4q αi (s)q-1
35