Download (535Kb) - LSE Research Online - London School of ...

set of =(V1:::Vk) such thatVs 2V0, s =1:::k. By Gaussianity, we can write, using diagram
formalism (see e.g. Brillinger (1975), p.21),
E m = X
Q (7.21)
where
Q =
mX
kY
2;
(
j1:::jk=l p=1
a (p)
jp )qV1 :::qV k (7.22)
where, for Vs =((p v) (p0v0 )), qVs qVs(jpjp0)=E[ pv(jp) p0v0(jp0)]: Set
Clearly
q Vs
jQ j
ja (p)
jp a(p0 )
j p 0 j1=2 jqVs(jpjp 0)j:
mX
j1:::j k=l
q V1 :::q Vk : (7.23)
Each argument j1:::jk in (7.23) belongs to exactly two functions q . Therefore, by the Cauchy
Vs
inequality, weget
where
We nowshow that
jQ j
jjq Vs jj2 =(
mX
j1:::j k=l
mX
jk=l
which together with (7.24), (7.21) implies (7.19).
We estimate
jjq Vs jj 2
2
mX
j=l
q V1 :::q Vk jjq V1 jj2 :::jjq Vk jj2 (7.24)
fq Vs (j k)g 2 ) 1=2 s =1:::k:
jjq Vs jj2 C(jja (p) jj2jja (p0 ) jj2) 1=2 (7.25)
ja (p)
j a(p0 )
j jjqVs(j j)j 2 + X
=: jjq Vs1 jj 2
l kj m:k6=j
2 + jjqVs2jj 2
2
ja (p)
k a(p0 )
j jjqVs(j k)j 2
From (a), (b) of Lemma 2.2 or Lemma 2.1 it follows that jqVs(j j)j C. Thus
jjq Vs1 jj 2
2
With tapering, from (c) and (d) of Lemma 2.2 it follows that
Therefore
C
mX
j=l
: (7.26)
ja (p)
j a(p0 )
j j Cjja (p) jj2jja (p0 ) jj2: (7.27)
jqVs(k j)j C((m=n) jj ; kj ;2 +(min(k j)) ;1 jj ; kj) ;3=2 ) l k 6= j m:
jjq Vs2 jj 2
2
C X
l kj m:k6=j
ja (p)
j a(p0 )
k j (m=n) 2 jj ; kj ;4 +(min(k j)) ;2 jj ; kj) ;3
35