Page 137 - Textos de Matemática Vol. 47

STONGLY DEPENDENT GAUSSIAN RANDOM FIELDS because1+↵< 2 .WhenA2⇥B1 isconsidere!d,weget
1+
127
nq2 XX ! un,m
nm |ri,j ⇢n,m|exp 1+wi,j i=p+1 j=1
u2n,m XnXq
nmexp 1+ (p,1) i=p+1j=1|ri,j ⇢n,m|
u2 nm exp n,m
!! 2 1+ (p,1)
Xn Xq ⇢

+ + (2.4)
Now, since lnn+lnm  lnn⇥lnm, for n,m 7, and (p,1)  k , with lnp
k > 0, we obtain, for n and m large enough,
ri,j ln(ij) ln(nm↵) ln(nm↵) ln(nm)
2
⇥
i=p+1 j=1
ln(ij)
" !#2
n2m u2 1+ (p,1) n2m ln p 2 ln p
exp n,m  K
lnn 2 lnn
(ln(nm))k+ln p (nm) k+ln p
K 1 n2 2↵lnn m1 2↵lnn (lnn+lnm) ↵lnn
1 k+↵ ln n k+↵ ln n k+↵ ln n lnn
1 k+↵lnn k+↵lnn Kn 2k (lnn) k
(lnm)↵lnn m1 2 ↵lnn k+↵lnn k+↵lnn
(2.5)
exp⇣ 2klnn ⌘
⇣k+↵lnn⌘(lnm) ↵lnn m1 2 ↵lnn
=K
K (lnm) ↵lnn m1 2 ↵lnn .
1
2 k+↵ ln n k+↵ ln n
k+↵ ln n k+↵ ln n
exp klnlnn k+↵lnn
Furthermore, it can be shown that
lnn Xn Xq ri,j  lnn Xn Xq 1 |ri,j ln(ij) |
n i=p+1 j=1 ln(i j) n i=p+1 j=1 ln(i j)
 1 m↵ sup |ri,j ln(i j) |. (2.6)
k+↵lnn
1+↵ 2↵ ln n < 0, for n large, for the first sum in the right hand side of (2.4), k+↵lnn
↵ i 1,j 1
Thus, due to (2.5) and (2.6) and taking into account that ↵ ln n < 1 and