Let us suppose that T = 1 for simplicity. Using invariance of the Haar measure H by multiplication, it is plain to see that E[X xX y] is of the form F
| y − x| where F : R
+
→ R
+
∪ {∞}. We have the scaling relation F a b = ln1
a + F b if a ≤ 1 and b ≤ 1 see also lemma 4.8 below which entails that for
|x| ≤ T : F
|x| = ln1|x| − Z
G
ln |e
m 1
|Hdm. where e = 1, 0, . . . , 0 is the first vector of the canonical basis. As a corollary, we get the existence
of some constant C take C = −
R
G
ln |e
m 1
|Hdm such that ln1|x| + C is positive definite as a tempered distribution in a neighborhood of 0. This easily implies that ln1
|x| is positive definite in a neighborhood of 0: to our knowledge, this result is new. This contrasts with the fact that
ln
+
1|x| is positive definite in dimension d ≤ 3 but is not positive definite for d ≥ 4 see [15].
Remark 3.1. It is easy to see that 1 − |x|
α
is positive definite in a neighborhhod of 0 as a function defined in R if
α ≤ 2. Therefore, one can consider the isotropic and positive definite function in a neighborhhood of 0 in R
d
defined by: F
|x| = Z
G
1 − |x
m 1
|
α
Hd m. By scaling, one can see that F
|x| = 1 − C|x|
α
for some C 0. It is easy to see that this entails that
1 − |x|
α
is positive definite in a neighborhood V of 0. Using the main theorem in [13], one can extend 1
− |x|
α
defined in V in an isotropic and positive definite function defined in R
d
. This is in contrast with the so called Kuttner-Golubov problem which is to determine the
α, κ 0 such that 1 − |x|
α κ
+
is positive definite in R
d
. It is known see [8] for instance that for α 0 the function 1 − |x|
α +
is not positive definite in R
d
if d ≥ 3.
Finally, let us mention that the field X with correlations given by 4 exhibits long range correlations as soon as d
≥ 2. More precisely, we have:
Lemma 3.2. If d ≥ 2, we get the following equivalence:
E [X xX y]
∼
|x− y|→∞
2Γd 2
Γ12Γd − 12 r
T |x − y|
. where Γ is the standard gamma function: Γx =
R
∞
t
x −1
e
−t
d t.
4 Proofs
4.1 Proof of lemma 1.1
We will consider the case d = 1 for simplicity the general case is an adaptation of the one dimen- sional case. With no restriction, we can suppose that the
α in lemma 1.1 belongs to ]0, 1[. For s, t
0, we have the following inequalities by using the assumptions on M and the concavity of
248
x 7→ x
α
: E
[M [0, t + s]
α
] = t + s
α
E [
t t + s
M [0, t] t
+ s
t + s M [t, t + s]
s
α
] ≥ t + s
α
E [
t t + s
M [0, t] t
α
+ s
t + s M [t, t + s]
s
α
] = t + s
α
t t + s
E [
M [0, t] t
α
] + s
t + s E
[ M [t, t + s]
s
α
] = t + s
α
E [M [0, 1]
α
] = E[M [0, t + s]
α
]. Therefore, the above inequalities are in fact equalities and, since x
7→ x
α
is strictly concave, this shows that
M [0,t] t
and
M [t,t+s] s
are equal almost surely. Let us consider the random non decreasing function f t = M [0, t]. The function f satisfies for all s, t
0: f t
t =
f t + s − f t
s .
Letting s go to 0 in the above equality, we conclude that f has a derivative f
′
which satisfies: f
′
t = f t
t .
Therefore, f t is of the form C t where C is some constant.
4.2 Proof of Theorem 2.7
The relation E[e
ω
l
x
] = e
ψ1H⊗θ C
l
x
= 1 and the fact that for l
′
l, ω
l
′
x − ω
l
x is indepen- dent of
F
l
ensures that for each Borelian subset A ⊂ R
d
, the process M
l
A is a martingale with respect to
F
l
. Existence of the MRM then results from [10]. Properties i and ii result from Fatou’s lemma.
4.3 Characteristic function of
ω
l
x
As in [1], the crucial point is to compute the characteristic function of ω
l
x. We consider
x
1
, . . . , x
q
∈ R
d q
and λ
1
, . . . , λ
q
∈ R
q
and we have to compute φλ = E
e
i λ
1
ω
l
x
1
+···+iλ
q
ω
l
x
q
. Let us denote by
S
q
the permutation group of the set {1; . . . ; q}. For a generic element σ ∈ S
q
, we define
B
σ
= {m ∈ G; x
σ1,m 1
· · · x
σq,m 1
}. Finally, given x, z
∈ R
d
, we define cone like subset product C
σ l
x = C
l
x ∩ B
σ
= {m, t, y ∈ B
σ
× S; t, y ∈ A
l
x
m 1
} and
C
σ l
x, z = {m, t, y ∈ B
σ
× S; t, y ∈ A
l
x
m 1
∩ A
l
z
m 1
}. 249
Lemma 4.4. The characteristic function of the vector ω
l
x
i 1
≤i≤q
exactly matches E
h exp
i λ
1
ω
l
x
1
+ · · · + iλ
q
ω
l
x
q
i = exp
X
σ∈S
q
q
X
j=1 j
X
k=1
α
σ
j, kρ
σ l
x
σk
− x
σ j
where ρ
σ l
x = H ⊗ θ C
σ l
0, x, and α
σ
j, k = ϕr
σ k, j
+ ϕr
σ k+1, j
−1
− ϕr
σ k, j
−1
− ϕr
σ k+1, j
r
σ k, j
=
j
X
i=k
λ
σi
or 0 if k j. Moreover
q
X
j=1 j
X
k=1
α
σ
j, k = ϕ
q
X
k=1
λ
k
.
Proof. Without loss of generality, we assume x
i
6= x
j
for i 6= j. We point out that the family B
σ σ∈S
d
is a partition of G up to a set of null H-measure. The function φ breaks down as
φλ =E e
i λ
1
µC
l
x
1
+···+iλ
q
µC
l
x
q
=E e
P
σ∈Sd
i λ
1
µC
σ l
x
1
+···+iλ
q
µC
σ l
x
q
= Y
σ∈S
d
E e
i λ
1
µC
σ l
x
1
+···+iλ
q
µC
σ l
x
q
Let us fix σ ∈ S
q
. We focus on the term φ
σ
λ = E e
i λ
1
µC
σ l
x
1
+···+iλ
q
µC
σ l
x
q
= Ee
i λ
σ1
µC
σ l
x
σ1
+···+iλ
σq
µC
σ l
x
σq
. Given
σ ∈ S
d
and p ≤ q, we further define
φ
σ
λ, p = E e
i λ
σ1
µC
σ l
x
σ1
+···+iλ
σp
µC
σ l
x
σp
. From now on, we adapt the proof of [1, Lemma 1] and proceed recursively. We define
Y
σ q
=
q
X
k=1
λ
σk
µC
σ l
x
σk
\ C
l
x
σq
, which stands for the contribution of the points of the above sum that do not belong to C
l
x
σq
. Moreover, the points in the set C
l
x
σq
can be grouped into the disjoint sets C
l
x
σk
, x
σq
\ C
l
x
σk−1
, x
σq
. We stress that the latter assertion is valid since, for m
∈ B
σ
, the coordinates are suitably sorted, that is: x
σ1,m 1
x
σ2,m 2
· · · x
σq,m 1
. We define X
σ k,q
= µ C
l
x
σk
, x
σq
\ C
l
x
σk−1
, x
σq
250
with the convention C
l
x
σk
, x
σ0
= C
l
x
σ0
, x
σk
= ;, in such a way that one has λ
σ1
µC
σ l
x
σ1
+ · · · + λ
σq
µC
σ l
x
σq
= Y
σ q
+
q
X
k=1
r
σ k,q
X
σ k,q
. Furthermore, since the variable Y
q
and X
σ k,q
k
are mutually independent, we get the following decomposition:
φ
σ
λ = E e
iY
σ q
q
Y
k=1
E e
i r
σ k,q
X
σ k,q
. 5
Similarly, one can prove φ
σ
λ, q − 1 = E e
iY
σ q
q
Y
k=1
E e
i r
σ k,q
−1
X
σ k,q
. 6
Gathering 5 and 6 yields φ
σ
λ, q = φ
σ
λ, q − 1
q
Y
k=1
E e
i r
σ k,q
X
σ k,q
E e
i r
σ k,q
−1
X
σ k,q
. For any m
∈ B
σ
, one has x
σk−1,m 1
x
σk,m 1
x
σq,m 1
and therefore E
e
i αX
σ k,q
=e
ϕαH⊗θ C
l
x
σk
,x
σq
\C
l
x
σk−1
,x
σq
=e
ϕα H⊗θ C
l
x
σk
,x
σq
−H⊗θ C
l
x
σk−1
,x
σq
Note that H
⊗ θ C
l
x
σi
, x
σ j
= Z
B
σ
θ A
l
x
σi,m 1
∩ A
l
x
σ j,m 1
Hdm =
Z
B
σ
θ A
l
0 ∩ A
l
x
σi,m 1
− x
σ j,m 1
Hdm =ρ
σ l
x
σi
− x
σ j
The proof can now be completed recursively. For further details, the reader is referred to [1].
4.5 Homogenity and isotropy
