1.5 Poisson representation
We decompose the path of S under P
∗ µ,Π
according to excursions of the total mass of ρ above its
past minimum, see Section 4.2.3 in [22]. More precisely, let a
i
, b
i
, i ∈ K be the excursion intervals of X
− I above 0 under P
∗ µ,Π
. For every i ∈ K , we define h
i
= H
a
i
and ¯ S
i
= ¯ ρ
i
, ¯ m
i
by the formulas: for t
≥ 0 and f ∈ B
+
R
+
, 〈 ¯
ρ
i t
, f 〉 =
Z
h
i
,+ ∞
f x − h
i
ρ
a
i
+t∧b
i
d x 22
〈 ¯ m
a t
i
, f 〉 =
Z
h
i
,+ ∞
f x − h
i
m
a a
i
+t∧b
i
d x, a
∈ {nod, ske}, 23
with ¯ m
i
= ¯ m
nod i
, ¯ m
ske i
. We set ¯ σ
i
= inf{s 0; 〈ρ
i s
, 1 〉 = 0}. It is easy to adapt Lemma 4.2.4. of
[22] to get the following Lemma.
Lemma 1.6. Let µ, Π ∈ S. The point measure
X
i ∈K
δ
h
i
, ¯ S
i
is under P
∗ µ,Π
a Poisson point measure with intensity
µd rN[dS ].
1.6 The dual process and representation formula
We shall need the M
f
R
+
-valued process η = η
t
, t ≥ 0 defined by
η
t
d r = β1
[0,H
t
]
r d r + X
s≤t
X
s −
I
s t
X
s
− I
s t
δ
H
s
d r. The process
η is the dual process of ρ under N see Corollary 3.1.6 in [22]. The next Lemma on time reversibility can easily be deduced from Corollary 3.1.6 of [22] and the
construction of m.
Lemma 1.7. Under N, the processes ρ
s
, η
s
, 1
{m
s
=0}
, s ∈ [0, σ] and η
σ−s−
, ρ
σ−s−
,
1
{m
σ−s−
=0}
, s ∈ [0, σ] have the same distribution. We present a Poisson representation of
ρ, η, m under N. Let N d x dℓ du, N
1
d x dℓ du and N
2
d x be independent Poisson point measures respectively on [0, +∞
3
, [0, + ∞
3
and [0, + ∞
with respective intensity d x
ℓπ dℓ 1
[0,1]
udu, d x
ℓπ
1
dℓ 1
[0,1]
udu and
α
1
d x .
For every a 0, let us denote by M
a
the law of the pair µ, ν, m
nod
, m
ske
of measures on R
+
with finite mass defined by: for any f
∈ B
+
R
+
〈µ, f 〉 = Z
N d x dℓ du + N
1
d x dℓ du 1
[0,a]
xuℓ f x + β Z
a
f r d r ,
24 〈ν, f 〉 =
Z N
d x dℓ du + N
1
d x dℓ du
1
[0,a]
x1 − uℓ f x + β Z
a
f r d r ,
25 〈m
nod
, f 〉 =
Z N
1
d x dℓ du 1
[0,a]
xuℓ f x and
〈m
ske
, f 〉 =
Z N
2
d x 1
[0,a]
x f x. 26
1442
Remark 1.8. In particular
µd r + νd r is defined as 1
[0,a]
rdξ
r
, where ξ is a subordinator with
Laplace exponent ψ
′
− α. We finally set M =
R
+∞
d a e
−αa
M
a
. Using the construction of the snake, it is easy to deduce from Proposition 3.1.3 in [22], the following Poisson representation.
Proposition 1.9. For every non-negative measurable function F on M
f
R
+ 4
, N
Z
σ
F ρ
t
, η
t
, m
t
d t
= Z
Mdµ dν d mF µ, ν , m, where m = m
nod
, m
ske
and σ = inf{s 0; ρ
s
= 0} denotes the length of the excursion.
2 The pruned exploration process
We define the following continuous additive functional of the process ρ
t
, m
t
, t ≥ 0: A
t
= Z
t
1
{m
s
=0}
ds for t
≥ 0. 27
Lemma 2.1. We have the following properties.
i For λ 0, N[1 − e
−λA
σ
] = ψ
−1
λ. ii N-a.e. 0 and σ are points of increase for A. More precisely, N-a.e. for all ǫ 0, we have A
ǫ
and A
σ
− A
σ−ǫ∨0
0. iii If
lim
λ→∞
φ
1
λ = +∞, then N-a.e. the set {s; m
s
6= 0} is dense in [0, σ]. Proof.
We first prove i. Let λ 0. Before computing v = N[1 − exp −λA
σ
], notice that A
σ
≤ σ implies, thanks to 15, that v
≤ N[1 − exp −λσ] = ψ
−1
λ +∞. We have v =
λN Z
σ
dA
t
e
−λ R
σ t
dA
u
= λN
Z
σ
dA
t
E
∗ ρ
t
,0
[e
−λA
σ
]
, where we replaced e
−λ R
σ t
dA
u
in the last equality by E
∗ ρ
t
,m
t
[e
−λA
σ
], its optional projection, and used that dA
t
-a.e. m
t
= 0. In order to compute this last expression, we use the decomposition of S under P
∗ µ
according to excursions of the total mass of ρ above its minimum, see Lemma 1.6. Using
the same notations as in this lemma, notice that under P
∗ µ
, we have A
σ
= A
∞
= P
i ∈K
¯ A
i ∞
, where for every T
≥ 0, ¯
A
i T
= Z
T
1
{ ¯ m
i t
=0}
d t .
28 By Lemma 1.6, we get
E
∗ µ
[e
−λA
σ
] = e
−〈µ,1〉N[1−exp −λA
σ
]
= e
−v〈µ,1〉
.
1443
We have v =
λN h Z
σ
dA
t
e
−v〈ρ
t
,1 〉
i = λN
h Z
σ
d t 1
{m
t
=0}
e
−v〈ρ
t
,1 〉
i 29
= λ Z
+∞
d a e
−αa
M
a
[1
{m=0}
e
−v〈µ,1〉
] = λ
Z
+∞
d a e
−αa
exp −α
1
a −
Z
a
d x Z
1
du Z
0,∞
ℓ
1
π
1
dℓ
1
exp n
− β va − Z
a
d x Z
1
du Z
0,∞
ℓ π
dℓ
1 − e
−vuℓ
o
= λ Z
+∞
d a exp
−a Z
1
du ψ
′
uv 30
= λ v
ψ v
, 31
where we used Proposition 1.9 for the third and fourth equalities, and for the last equality that α
= α + α
1
+ R
0,∞
ℓ
1
π
1
dℓ
1
as well as ψ
′
λ = α +
Z
0,∞
π dℓ
ℓ 1 − e
−λℓ
. 32
Notice that if v = 0, then 30 implies v = λψ
′
0, which is absurd since ψ
′
0 = α 0 thanks to
19. Therefore we have v ∈ 0, ∞, and we can divide 31 by v to get ψ
v = λ. This proves i. Now, we prove ii. If we let
λ goes to infinity in i and use that lim
r →∞
ψ r = +∞, then we
get that N[A
σ
0] = +∞. Notice that for µ, Π ∈ S, we have under P
∗ µ,Π
, A
∞
≥ P
i ∈K
¯ A
i ∞
, with ¯
A
i
defined by 28. Thus Lemma 1.6 implies that if µ 6= 0, then P
∗ µ,Π
-a.s. K is infinite and A
∞
0. Using the Markov property at time t of the snake under N, we get that for any t 0, N-a.e. on
{σ t}, we have A
σ
− A
t
0. This implies that σ is N-a.e. a point of increase of A. By time reversibility, see Lemma 1.7, we also get that N-a.e. 0 is a point of increase of A. This gives ii.
If α
1
0 then the snake ρ
s
, W
s
, s ≥ 0 is non trivial. It is well known that, since the Lévy process X
is of infinite variation, the set {s; ∃t ∈ [0, H
s
, W
s
t 6= 0} is N-a.e. dense in [0, σ]. This implies that
{s; m
s
6= 0} is N-a.e. dense in [0, σ]. If
α
1
= 0 and π
1
0, ∞ = ∞, then the set J
1
of jumping time of X is N-a.e. dense in [0, σ]. This also implies that
{s; m
s
6= 0} is N-a.e. dense in [0, σ]. If
α
1
= 0 and π
1
0, ∞ ∞, then the set J
1
of jumping time of X is N-a.e. finite. This implies that
{s; m
s
6= 0} ∩ [0, σ] is N-a.e. a finite union of intervals, which, thanks to i, is not dense in [0, σ].
To get iii, notice that lim
λ→∞
φ
1
λ = ∞ if and only if α
1
0 or π
1
0, ∞ = ∞. We set C
t
= inf{r 0; A
r
t} the right continuous inverse of A, with the convention that inf ; = ∞. From excursion decomposition, see Lemma 1.6, ii of Lemma 2.1 implies the following Corollary.
Corollary 2.2. For any initial measures µ, Π ∈ S, P
µ,Π
-a.s. the process C
t
, t ≥ 0 is finite. If m
= 0, then P
µ,Π
-a.s. C = 0.
1444
We define the pruned exploration process ˜ ρ = ˜
ρ
t
= ρ
C
t
, t ≥ 0 and the pruned marked exploration
process ˜ S = ˜
ρ, ˜ m
, where ˜ m = m
C
t
, t ≥ 0 = 0. Notice C
t
is a F -stopping time for any t ≥ 0 and
is finite a.s. from Corollary 2.2. Notice the process ˜ ρ, and thus the process ˜
S , is càd-làg. We also set ˜
H
t
= H
C
t
and ˜ σ = inf{t 0; ˜
ρ
t
= 0}. Let ˜
F = ˜ F
t
, t ≥ 0 be the filtration generated by the pruned marked exploration process ˜
S com- pleted the usual way. In particular ˜
F
t
⊂ F
C
t
, where if τ is an F -stopping time, then F
τ
is the σ-field associated with τ.
We are now able to restate precisely Theorem 0.3. Let ρ
be the exploration process of a Lévy process with Laplace exponent
ψ .
Theorem 2.3. For every finite measure
µ, the law of the pruned process ˜ ρ under P
µ,0
is the law of the exploration process
ρ associated with a Lévy process with Laplace exponent
ψ under P
µ
. The proof of this Theorem is given at the end of Section 4.
3 A special Markov property
In order to define the excursions of the marked exploration process away from {s ≥ 0; m
s
= 0}, we define O as the interior of
{s ≥ 0, m
s
6= 0}. We shall see that the complementary of O has positive Lebesgue measure.
Lemma 3.1. i If the set
{s ≥ 0, m
s
6= 0} is non empty then, N-a.e. O is non empty. ii If we have
lim
λ→∞
φ
1
λ = ∞, then N-a.e. the open set O is dense in [0, σ]. Proof.
For any element s
′
in {s ≥ 0, m
s
6= 0}, there exists u ≤ H
s
′
such that m
s
′
[0, u] 6= 0 and ρ
s
′
[u, ∞ 0. Then we consider τ
s
′
= inf{t s
′
, ρ
t
[u, ∞ = 0}. By the right continuity of ρ, we have
τ
s
′
s
′
and the snake property 21 implies that N-a.e. s
′
, τ
s
′
⊂ O. Use iii of Lemma 2.1 to get the last part.
We write O = S
i ∈I
α
i
, β
i
and say that α
i
, β
i i
∈I
are the excursions intervals of the marked exploration process
S = ρ, m away from {s ≥ 0, m
s
= 0}. For every i ∈ I , let us define the measure-valued process
S
i
= ρ
i
, m
i
. For every f ∈ B
+
R
+
, t ≥ 0, we set 〈ρ
i t
, f 〉 =
Z
[H
αi
,+ ∞
f x − H
α
i
ρ
α
i
+t∧β
i
d x 〈m
a i
t
, f 〉 =
Z
H
αi
,+ ∞
f x − H
α
i
m
a α
i
+t∧β
i
d x with a
∈ {nod, ske} 33
and m
i t
= m
nod i
t
, m
ske i
t
. Notice that the mass located at H
α
i
is kept, if there is any, in the definition of
ρ
i
whereas it is removed in the definition of m
i
. In particular, if ∆
α
i
0, then ρ
i
= ∆
α
i
δ and for every t
β
i
− α
i
, the measure ρ
i t
charges 0. On the contrary, as m
i
= 0, we have, for every t
β
i
− α
i
, m
i t
{0} = 0.
1445
Let ˜ F
∞
be the σ-field generated by
˜ S = ρ
C
t
, m
C
t
, t ≥ 0. Recall that P
∗ µ,Π
dS denotes the law of the marked exploration process
S started at µ, Π ∈ S and stopped when ρ reaches 0. For ℓ ∈ 0, +∞, we will write P
∗ ℓ
for P
∗ ℓδ
,0
. If Q is a measure on S and ϕ is a non-negative measurable function defined on the measurable space
R
+
× S × S, we denote by Q[
ϕu, ω, ·] = Z
S
ϕu, ω, S QdS . In other words, the integration concerns only the third component of the function
ϕ. We can now state the Special Markov Property.
Theorem 3.2 Special Markov property. Let ϕ be a non-negative measurable function defined on
R
+
× M
f
R
+
× S. Then, we have P-a.s. E
exp
− X
i ∈I
ϕA
α
i
, ρ
α
i
−
, S
i
˜ F
∞
= exp
− Z
∞
du α
1
N
1 − e
−ϕu,µ,·
|µ= ˜ ρ
u
exp −
Z
∞
du Z
0,∞
π
1
dℓ
1 − E
∗ ℓ
[e
−ϕu,µ,·
]
|µ= ˜ ρ
u
. 34
In other words, the law under P of the excursion process X
i ∈I
δ
A
αi
, ρ
αi −
, S
i
du dµ dS , given ˜ F
∞
, is the law of a Poisson point measure with intensity
1
{u≥0}
du δ
˜ ρ
u
dµ α
1
Nd S +
Z
0,∞
π
1
dℓP
∗ ℓ
dS .
Informally speaking, this Theorem gives the distribution of the marked exploration process “above” the pruned CRT. The end of this section is now devoted to its proof.
Let us first remark that, if lim
λ→+∞
φ
1
λ +∞, we have α
1
= 0 and π
1
is a finite measure. Hence, there is no marks on the skeleton and the number of marks on the nodes is finite on every bounded
interval of time. The proof of Theorem 3.2 in that case is easy and left to the reader. For the rest of this Section, we assume that lim
λ→+∞
φ
1
λ = +∞.
3.1 Preliminaries
Kategori