66 and then
∂ϕµ ∂µ
1
= log µ
1
− 1
2 µ
2 1
k
X
ℓ=2
µ
2 ℓ
+ f
µ
2
, . . . , µ
k
, 5.13
where f : R
k−1
→ R is an analytical function to be determined. Note that since
µ
1
0 then log µ
1
and 1 2µ
2 1
in 5.13 are well-defined. Derivative of 5.13 with respect to
µ
ℓ
gives ∂
2
ϕ ∂µ
1
µ
ℓ
= − µ
ℓ
µ
2 1
+ ∂ f µ
2
, . . . , µ
k
∂µ
ℓ
. 5.14
Expression 5.14 is equal to the 1, ℓth element of
h
V
F
µ i
−1
in 5.11, that is −
µ
ℓ
µ
2 1
+ ∂ f µ
2
, . . . , µ
k
∂µ
ℓ
= − µ
ℓ
µ
2 1
; therefore,
∂ f µ
2
, . . . , µ
k
∂µ
ℓ
= 0 for all ℓ ∈ {2,...,k} implies f µ
2
, . . . , µ
k
= c
1
a real constant. Thus, 5.13 becomes
∂ϕ ∂µ
1
= log µ
1
− 1
2 µ
2 1
k
X
ℓ=2
µ
2 ℓ
+ c
1
5.15 and its primitive can be written as
ϕµ = µ
1
log µ
1
− µ
1
+ 1
2 µ
1 k
X
ℓ=2
µ
2 ℓ
+ c
1
µ
1
+ h
µ
2
, . . . , µ
k
, 5.16
where h : R
k−1
→ R is an analytical function to be determined. From now on, complete informations of the model i.e. normal components begin to show
itself. The two first derivatives of 5.16 with respect to µ
ℓ
give, respectively, ∂ϕµ
∂µ
ℓ
= µ
ℓ
µ
1
+ ∂hµ
2
, . . . , µ
k
∂µ
ℓ
, ∀ℓ ∈ {2,...,k}
5.17 and
∂
2
ϕµ ∂µ
2 ℓ
= 1
µ
1
+ ∂h
2
µ
2
, . . . , µ
k
∂µ
2 ℓ
, ∀ℓ ∈ {2,...,k}.
5.18 Expression 5.18 is equal to the diagonal
ℓ, ℓth element of h
V
F
µ i
−1
in 5.11 for all
ℓ ∈ {2,...,k}, hence we have 1
µ
1
+ ∂
2
h µ
2
, . . . , µ
k
∂µ
2 ℓ
= 1
µ
1
. Consequently,
∂
2
h µ
2
, . . . , µ
k
∂µ
2 ℓ
= 0 and ∂hµ
2
, . . . , µ
k
∂µ
ℓ
= c
ℓ
a real con-
67 stant for all
ℓ ∈ {2,...,k}. Then, equation 5.17 becomes ∂ϕµ
∂µ
ℓ
= µ
ℓ
m
1
+ c
ℓ
∀ℓ ∈ {2,...,k}. 5.19
Using equation 5.15 and 5.19 one obtains
θµ =
logµ
1
− 1
2 µ
2 1
k
X
ℓ=2
µ
2 ℓ
, µ
2
m
1
, . . .
, µ
k
µ
1
⊤
+ c
1
, . . . , c
k ⊤
or
θµ =
θ
1
= log µ
1
− 1
2 µ
2 1
P
k ℓ=2
µ
2 ℓ
+ c
1
θ
ℓ
= µ
ℓ
µ
1
+ c
ℓ
, ℓ = 2, . . . , k.
5.20
From 5.20, each θ
ℓ
belongs to R for ℓ ∈ {1,2,...,k} because µ
1
0 and µ
ℓ
∈ R for
ℓ ∈ {2,...,k}. Thus, one has ΘM
F
=: Θ ⊆ R
k
and also µ
1
= exp
θ
1
− c
1
+ 1
2
k
X
ℓ=2
θ
ℓ
− c
ℓ 2
, 5.21
µ
ℓ
= θ
ℓ
− c
ℓ
exp
θ
1
− c
1
+ 1
2
k
X
ℓ=2
θ
ℓ
− c
ℓ 2
. 5.22
Since µ =
∂K
ν
θ ∂θ
, then using 5.21 one can obtain K
ν
θ as follow:
K
ν
θ = Z
∂K
′ ν
θ ∂θ
1
d θ
1
= exp
θ
1
− c
1
+ 1
2
k
X
ℓ=2
θ
ℓ
− c
ℓ 2
+ gθ
2
, . . . , θ
k
, 5.23
where g : R
k−1
→ R is an analytical function to be determined. Again, derivative of 5.23 with respect to
θ
ℓ
produces
∂K
ν
θ ∂θ
ℓ
= θ
ℓ
− c
ℓ
exp
θ
1
− c
1
+ 1
2
k
X
ℓ=2
θ
ℓ
− c
ℓ 2
+ ∂gθ
2
, . . . , θ
k
∂θ
ℓ
which is equal to 5.22; then, one gets ∂gθ
2
, . . . , θ
k
∂θ
ℓ
= 0 for all ℓ ∈ {2,...,k}
implying g θ
2
, . . . , θ
k
= C a real constant. Finally, it ensues from it that we have
K
ν
θ = exp
θ
1
− c
1
+ 1
2
k
X
ℓ=2
θ
ℓ
− c
ℓ 2
+C.
68
By Proposition 5.2.1 one can see that, up to affinity, this K
ν
is a normal- Poisson
1
cumulant function as given in 5.2 with t = 1 on its corresponding support 5.3. Theorem 5.2.1 is therefore proven by using the analytical
property of K
ν
. 5.2.2.
Characterization by Generalized Variance Function
Before stating our next result, let us briefly recall that, for an unknown
smooth function K : Θ ⊆ R
k
→ R, k 2, the Monge–Ampère equation is de- fined by
det K
′′
θ = gθ 5.24
where K
′′
= D
2 ij
K
i ,j=1,...,k
denotes the Hessian matrix of K and g is a given
positive function see e.g. Gutierrez 2001. The class of equation 5.24 given g has been a source of intense investigations which are related to
many areas of mathematics. Note also that explicit solutions of 5.24, even if in particular situations of g, remain generally challenging problems. We
can refer to Kokonendji and Seshadri 1996; Kokonendji and Masmoudi 2013, 2006 for some details and handled particular cases.
We now state the next result in the following sense.
Theorem 5.2.2. Let F
t;j
= F
ν
t;j
be an infinitely divisible NEF on R
k
k 1 for
given j ∈ {1,2,...,k} such that 1. Θ
ν
t;j
= R
k
, and
2. det K
′′ ν
t;j
θ = t exp k × θ
⊤
˜θ
c j
for θ and ˜θ
c j
given as in 5.4. Then F
t;j
is of normal-Poisson
j
type. The proof of Theorem 5.2.2 is given below. A reformulation of this
theorem, by changing the canonical parameterization into the mean param- eterization, is stated in the following theorem without proof.
Theorem 5.2.3. Let j ∈ {1,2,...,k} fixed and F
t;j
= F
ν
t;j
be an infinitely divisible NEF on R
k
such that
1. M
F
t;j
= n
µ ∈ R
k
; µ
j
0 and µ
ℓ
∈ Rwithℓ , j o
, and
2. det V
F
t;j
µ = µ
k j
. Then F
t;j
is of normal-Poisson
j
type. Theorem 5.2.2 can be viewed as the solution to a particular Monge-
Ampère equation. Whereas Theorem 5.2.3 is interesting for generalized variance estimations.
69 Proof. To proof Theorem 5.2.2 is to solve the Monge-Ampère equation prob-
lem of normal-Poisson models item 2 of the theorem. For that purpose, we need three propositions which are already used in Kokonendji and Mas-
moudi 2006 and Kokonendji and Masmoudi 2013 and we provide the propositions below for making the paper as self-contained as possible.
Proposition 5.2.2. If
ν is an infinitely divisible measure on R
k
, then there exist a symmetric non-negative definite d × d matrix Σ with rank r 6 k and a positive
measure ψ on R
k
such that
K
′′ ν
θ = Σ + Z
R
k
xx
⊤
expθ
⊤
x ψdx.
See, e.g., Gikhman and Skorokhod 2004, page 342.
The above expression of K
′′ ν
θ is an equivalent of the Lévy-Khinchine for- mula see e.g. Bertoin 1996; Kokonendji and Khoudar 2006; Sato 1999; thus,
Σ
comes from a Brownian part and the rest L
′′ ψ
θ := R
R
k
xx
⊤
expθ
⊤
x ψdx
corresponds to jumps part of the associated Lévy process through the Lévy measure
ψ.
Proposition 5.2.3. Let A and B be two k × k matrices. Then
detA + B =
X
S⊂{1,2,...,k}
det A
S
′
det B
S
, with S
′
= {1,2,...,k} \ S and A
S
= a
ij i
,j∈S
2
for A = a
ij i
,j∈{1,2,...,k}
2
. See Muir 1960.
Proposition 5.2.4. Let f : R
k
→ R be a C
2
map. Then, f is an affine polynomial if and only if
∂
2
f θ∂θ
2 i
= 0 ,
for i = 1 , . . . , k.
See Bar-Lev, et al. 1994, Lemma 4.1. Without loss of generality, we assume t = 1 in Theorem 5.2.2. Letting
F
j
= F
ν
j
in Theorem 5.2.2 for fixed j ∈ {1,2,...,k}, we have to solve the following equation with respect to
ν
j
or its characteristic function:
det K
′′ ν
j
θ = exp
k ·
θ
j
+ 1
2 X
ℓ,j
θ
2 ℓ
, ∀θ ∈ R
k
. 5.25
From Proposition 5.2.2 relative to the representation of infinitely divisible distribution, the unknown left member of Equation 5.25 can be written as
det K
′′ ν
j
θ = det Σ
+ Z
R
k
xx
⊤
expθ
⊤
x ψdx
= det h
Σ + L
′′ ψ
θ i
. 5.26
70 For S = {i
1
, i
2
, . . . , i
ℓ
}, with 1 6 i
1
i
2
··· i
ℓ
6 k, a non-empty subset of
{1,2,...,k}, and τ
S
: R
k
→ R
ℓ
the map defined by τ
S
x = x
i
1
, x
i
2
, . . . , x
i
ℓ
⊤
, we define
ψ
S
the image measure of H
ℓ
dx
1
, . . . , dx
ℓ
= 1
ℓ det [
τ
S
x
1
. . . τ
S
x
ℓ
]
2
ψdx
1
. . . ψdx
ℓ
by ψ
ℓ
: R
k ℓ
→ R
k
, x
1
, . . . , x
ℓ
7→ x
1
+ x
2
+ ··· + x
ℓ
. By Proposition 5.2.3 and Expression 5.26 the modified Lévy measure
ρν in 5.1.1 can be expressed as
ρν
j
= det Λ δ
+ X
∅,S⊂{1,2,...,k}
det Λ
S
′
ψ
S
, 5.27
where Λ is a diagonal representation of Σ in an orthonormal basis e = e
i i=1
,...,k
see, e.g., Hassairi 1999, page 384. Since Σ is the Brownian part, then it corresponds to the k − 1 normal components from the right member
of 5.25; that implies r = rankΣ = k − 1 and detΣ = 0. Therefore detΛ = 0 with Λ = diag
λ
1
, λ
2
, . . . , λ
k
such that λ
j
= 0 and λ
ℓ
0 for all ℓ , j. For all non-empty subsets S of {1,2,...,k} there exist real numbers α
S
0 such that det Λ
S
′
ψ
S
=
Y
iS
λ
i
ψ
S
= α
S
h δ
e
j
∗ N0,1e
c j
i
∗k
, 5.28
where e
c j
= e
1
, . . . , e
j−1
, e
j+1
, . . . , e
k
denotes the induced orthonormal basis of
e without component e
j
; i.e. k − 1 is the dimension of e
c j
. With respect to Kokonendji and Masmoudi 2006, Lemma 7 for making
precise the measure ν of 5.28, it is easy to see that S
= {j} is a singleton i.e.
set with exactly one element such that, for x = x
1
e
1
+ ··· + x
k
e
k
, x
2 j
ψdx = βδ
ae
j
, with
β 0 and a , 0. Consequently, we have the following complementary set S
′
= {1,2,...,k} \ {j}. So, from 5.28 we have kth power of convolution of
only one Poisson at the jth component e
j
and k −1-variate standard normal.
That means K
′′ ν
j
θ = K
ν
j
θ ˜θ
c j
˜θ
c⊤ j
+ I
j
k
, with notations of 5.4. Let Bθ =
exp θ
j
+
1 2
P
ℓ,j
θ
2 ℓ
from 5.25. Since we check that ∂
2
K
ν
j
−Bθ∂θ
2 i
= 0 for all i = 1
, . . . , k, Proposition 5.2.4 allows that K
ν
j
− Bθ is an affine function
on R
k
and therefore
K
ν
j
θ = exp
θ
j
+ 1
2 X
ℓ,j
θ
2 ℓ
+u
⊤
θ + b,
for u , b ∈ R
k
×R. Hence F
j
= F
ν
j
is of normal-Poisson
j
type. This complete the proof of the theorem.
