Penalisations of Brownian motion with its maximum and minimum processes as weak forms of Skorokhod embedding
We develop a Brownian penalisation procedure related to weight processes (Ft) of the type: Ft := f(It, St) where f is a bounded function with compact support and St (resp. It) is the one-sided maximum (resp. minimum) of the Brownian motion up to time t. Two main cases are treated: either Ft is the i...
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irk-123456789-45582009-12-07T12:00:33Z Penalisations of Brownian motion with its maximum and minimum processes as weak forms of Skorokhod embedding Roynette, B. Vallois, P. Yor, M. We develop a Brownian penalisation procedure related to weight processes (Ft) of the type: Ft := f(It, St) where f is a bounded function with compact support and St (resp. It) is the one-sided maximum (resp. minimum) of the Brownian motion up to time t. Two main cases are treated: either Ft is the indicator function of {It ≥ α, St ≤ β} or Ft is null when {St − It > c} for some c > 0. Then we apply these results to some kind of asymptotic Skorokhod embedding problem. 2008 Article Penalisations of Brownian motion with its maximum and minimum processes as weak forms of Skorokhod embedding / B. Roynette, P. Vallois, M. Yor // Theory of Stochastic Processes. — 2008. — Т. 14 (30), № 2. — С. 116–138. — Бібліогр.: 25 назв.— англ. 0321-3900 http://dspace.nbuv.gov.ua/handle/123456789/4558 519.21 en Інститут математики НАН України |
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We develop a Brownian penalisation procedure related to weight processes (Ft) of the type: Ft := f(It, St) where f is a bounded function with compact support and St (resp. It) is the one-sided maximum (resp. minimum) of the Brownian motion up to time t. Two main cases are treated: either Ft is the indicator function of {It ≥ α, St ≤ β} or Ft is null when {St − It > c} for some c > 0. Then we apply these results to some kind of asymptotic Skorokhod embedding problem. |
| format |
Article |
| author |
Roynette, B. Vallois, P. Yor, M. |
| spellingShingle |
Roynette, B. Vallois, P. Yor, M. Penalisations of Brownian motion with its maximum and minimum processes as weak forms of Skorokhod embedding |
| author_facet |
Roynette, B. Vallois, P. Yor, M. |
| author_sort |
Roynette, B. |
| title |
Penalisations of Brownian motion with its maximum and minimum processes as weak forms of Skorokhod embedding |
| title_short |
Penalisations of Brownian motion with its maximum and minimum processes as weak forms of Skorokhod embedding |
| title_full |
Penalisations of Brownian motion with its maximum and minimum processes as weak forms of Skorokhod embedding |
| title_fullStr |
Penalisations of Brownian motion with its maximum and minimum processes as weak forms of Skorokhod embedding |
| title_full_unstemmed |
Penalisations of Brownian motion with its maximum and minimum processes as weak forms of Skorokhod embedding |
| title_sort |
penalisations of brownian motion with its maximum and minimum processes as weak forms of skorokhod embedding |
| publisher |
Інститут математики НАН України |
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2008 |
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http://dspace.nbuv.gov.ua/handle/123456789/4558 |
| citation_txt |
Penalisations of Brownian motion with its maximum and minimum processes as weak forms of Skorokhod embedding / B. Roynette, P. Vallois, M. Yor // Theory of Stochastic Processes. — 2008. — Т. 14 (30), № 2. — С. 116–138. — Бібліогр.: 25 назв.— англ. |
| work_keys_str_mv |
AT roynetteb penalisationsofbrownianmotionwithitsmaximumandminimumprocessesasweakformsofskorokhodembedding AT valloisp penalisationsofbrownianmotionwithitsmaximumandminimumprocessesasweakformsofskorokhodembedding AT yorm penalisationsofbrownianmotionwithitsmaximumandminimumprocessesasweakformsofskorokhodembedding |
| first_indexed |
2025-07-02T07:46:23Z |
| last_indexed |
2025-07-02T07:46:23Z |
| _version_ |
1836520449376256000 |
| fulltext |
Theory of Stochastic Processes
Vol. 14 (30), no. 2, 2008, pp. 116–138
UDC 519.21
B. ROYNETTE, P. VALLOIS, AND M. YOR
PENALISATIONS OF BROWNIAN MOTION WITH
ITS MAXIMUM AND MINIMUM PROCESSES AS
WEAK FORMS OF SKOROKHOD EMBEDDING
We develop a Brownian penalisation procedure related to weight processes (Ft) of
the type: Ft := f(It, St) where f is a bounded function with compact support and
St (resp. It) is the one-sided maximum (resp. minimum) of the Brownian motion
up to time t. Two main cases are treated: either Ft is the indicator function of
{It ≥ α, St ≤ β} or Ft is null when {St − It > c} for some c > 0. Then we apply
these results to some kind of asymptotic Skorokhod embedding problem.
1. Introduction
In a series of papers [18], [17], [16], [23], [24], [21], [22] (see also the surveys [19] and
[20] and the monograph [25] we have introduced a penalisation procedure of Brownian
paths and applied it to many settings. To present the aim of this paper, we first introduce
a few notations.
Let
(
Ω = C(R+,R), (Xt)t≥0, (Ft)t≥0
)
be the canonical space, where (Xt)t≥0 denotes
the coordinate maps: Xt(ω) = ω(t), for any t ≥ 0. Let (Px)x∈R be the family of Wiener
probability measures on Ω : under Px, (Xt)t≥0 is a standard one-dimensional Brownian
motion started at x.
Next, we consider a stochastic process (Ft)t≥0 which takes its values in [0,∞[ and
satisfies:
(1.1) 0 < E0(Ft) <∞ ∀t ≥ 0.
We shall say that the penalisation procedure (associated with the weight process (Ft))
holds if:
(1.2) QF
0 (Λs) := lim
t→∞
E0[1Λs Ft]
E0[Ft]
exists for any s ≥ 0 and Λs ∈ Fs.
We briefly recall (see Theorem 3.6 in [17], [16] and [23]) that the penalisation procedure
holds with Ft = ϕ(St) where:
1. ϕ : [0,∞[→ [0,∞[ is a Borel function such that
∫ ∞
0
ϕ(x)dx = 1;
2. (St)t≥0 is the one-sided maximum process associated with (Xt)t≥0:
(1.3) St := max
u≤t
Xu ; t ≥ 0.
According to Theorem 4.6 of [17], under QF
0 , the r.v. S∞ is finite and admits ϕ as a
density function. Consequently, the procedure (1.2) forces Brownian motion to have a
2000 AMS Mathematics Subject Classification. Primary 60G17, 60G40, 60G44, 60H10, 60J25, 60J60,
60J65.
Key words and phrases. Skorokhod’s problem, penalisation, one-sided maximum and minimum,
Laplace’s method.
116
PENALISATIONS OF BROWNIAN MOTION 117
finite one-sided total maximum with the given probability density ϕ. This result presents
some analogy with Skorokhod’s embedding problem.
1.1. On Skorokhod’s problem for linear Brownian motion. Let μ be a probability
measure (p.m.) on R, such that:
(1.4)
∫
R
|y|μ(dy) <∞ and
∫
R
y μ(dy) = 0.
A number of constructions of (finite) stopping times T such that (i) T is standard, i.e.
(Xs∧T ; s ≥ 0) is a uniformly integrable martingale, and (ii) the distribution of XT ,
under P0 is μ, have been made by many authors, see e.g. Ob�lój’s thorough survey [11] of
the subject. We briefly recall the particular construction given by Azéma and Yor ([2],
[1]): there exists a non-decreasing function φμ : [0,+∞[→ R such that for :
(1.5) Tμ := inf
{
t ≥ 0, Xt ≤ φμ(St)
}
,
Tμ is standard and XTμ ∼ μ (under P0).
Precisely, φμ is the right-continuous inverse of ψμ where
ψμ(x) =
1
μ
(
]x,∞[
) ∫
]x,∞[
ydμ(y).
1.2. An asymptotic resolution of Skorokhod’s problem for diffusions. Similarly
to the Skorokhod embedding problem for Brownian motion, let us start with a target
p.m. μ on R. For simplicity we suppose that μ admits a density function μ0 which is
supposed to be positive and of class C1
b : μ(dx) = μ0(x)dx.
Then, there exists a p.m. Q0 on (Ω,F∞) such that :
(1.6) Xt = Bt +
1
2
∫ t
0
μ′
0(Xs)
μ0(Xs)
ds, t ≥ 0
and (Bt)t≥0 is a Q0-Brownian motion started at 0.
Moreover, under Q0, Xt converges in distribution to μ, as t→ ∞. In other words, we
have introduced a diffusion process, whose limit distribution is the given p.m. μ, which
may be considered as an asymptotic kind of resolution of Skorokhod’s problem.
1.3. Solving Skorokhod’s problem for (St, Xt) . The solution of Skorokhod’s prob-
lem given by Azéma and Yor [2] suggested to consider the embedding problem for the
two-dimensional process (St, Xt)t≥0. A complete answer has been given by Rogers [13]
(see also [15]). Starting with a p.m. ν on R+ × R, assumed to satisfy:
(1.7)
∫
R+×R
|y| ν(dx, dy) <∞,
∫
R+×R
y ν(dx, dy) = 0
(1.8) a ν
(
[a,∞[×R
)
=
∫
R+×R
1{x≥a} y ν(dx, dy), ∀a ≥ 0,
it is shown that there exists a finite standard stopping time T such that (ST , XT ) ∼ ν
(under P0).
Note that (1.7) and (1.8) correspond to
(1.9) E
[
|XT |
]
<∞, E
[
XT
]
= 0,
resp.
(1.10) aP (ST ≥ a) = E
[
XT |ST ≥ a
]
, ∀ a ≥ 0.
118 B. ROYNETTE, P. VALLOIS, AND M. YOR
1.4. The (St, Xt) asymptotic resolution of Skorokhod’s problem for diffusion
processes. Comparing Subsections 1.2 and 1.3, we may ask the following question : for
which class of p.m.’s ν on R+ × R, does there exist a p.m. Q1 on (Ω,F∞) under which:
(1.11) Xt = Bt +
∫ t
0
b(u,X·)du
where
(1.12) (Bt)t≥0 is a Q1-standard Brownian motion with, B0 = 0,
(1.13)
(
b(t,X·)
)
t≥0
is an (Ft)-adapted process,
(1.14) and the couple (St, Xt) converges in distribution towards ν, as t→ ∞.
Our approach is based on a Brownian penalisation procedure. This method is well
fitted for our purpose since it permits to obtain Markov processes whose distributions
are locally equivalent to that of Brownian motion and enjoy new path properties (for
instance a finite total unilateral maximum, see the beginning of Introduction).
Unfortunately we have not been able to solve completely this question. In Section 6,
we only give a class of p.m.’s ν verifying (1.11)-(1.14).
1.5. Organisation of the paper. Section 2 is a short survey of Brownian penalisa-
tions. To show that the penalisation procedure (see (1.2) or Section 2 for more details)
associated with a weight process (Ft) holds, we need to be able to determine the rates
of decay of E0[Ft] and E[Ft|Fs
]
as t → ∞. In this paper, we consider penalisations
with Ft = f(It, St) where f : ] −∞, 0] × [0,∞[→]0,∞[ is a Borel bounded function with
compact support and
It := inf
u≤t
Xu (t ≥ 0).
To determine the rate of decay of t !→ E
[
f(It, St)
]
as t → ∞, we have been led to
consider two classes of functions f . These developments are given in Section 3. With
these estimates at hand, we shall show in Section 4 that the associated penalisation
procedures hold. These schemes give rise to two families of p.m. on the canonical space
(Ω,F∞). In Section 5 we determine the law of the process (Xt) under each of these
new p.m.’s. Finally, in Section 6 we apply our results to the question discussed above in
subsection 1.4.
2. A survey of Brownian penalisations
We keep notations from the Introduction.
2.1. The goal. Penalisations provide a method to define P0(· |A) for A a negligible
event in F∞ := ∨
t≥0
Ft, i.e. P0(A) = 0.
This question arises naturally in probability theory and especially in the study of
stochastic processes. Let us give a few explicit examples :
(2.1) 1) A1 = {Xt ≥ 0 ; ∀t ≥ 0}
(2.2) 2) A2 = {sup
t≥0
Xt ≤ a}, (a > 0)
(2.3) 3) A3 = {inf
t≥0
Xt ≥ α, sup
t≥0
Xt ≤ β} (α < 0, β > 0).
Conditioning by A1 may be treated by h-Doob’s transforms (see for instance Section 4,
Chap. V in [4]). As for (2.2), it is proved in [9], [10] that for any 0 < t1 < · · · < tn, the
PENALISATIONS OF BROWNIAN MOTION 119
conditional distribution of the random vector
(
Xt1 , · · · , Xtn
)
given {St ≤ a} converges
as t→ ∞.
The third case is the subject of our study. It is then demanded that (Xt)t≥0 stays in
the strip [α, β].
2.2. A solution via approximation. Given a decreasing family (At)t≥0 of events in
F∞ such that P0(At) > 0 , ∀ t ≥ 0, we set A = ∩
t≥0
At.
As an example, the set A3 given by (2.3) satisfies : A3 = ∩
t≥0
A3,t, with A3,t := {It ≥
α, St ≤ β}, where (It)t≥0 is the one-sided minimum process, i.e.
(2.4) It := inf
u≤t
Xu, t ≥ 0.
Going back to the the general case of the family (At)t≥0, we would like to define :
(2.5) PA
0 (Λ) := lim
t→∞P0(Λ|At),
for Λ ∈ F∞ such that the limit exists.
At this stage, three questions arise immediately :
(2.6) 1) for which Λ, does PA
0 (Λ) exist ?
(2.7) 2) Can PA
0 be extended to a p.m. on (Ω,F∞) ?
(2.8) 3) How does PA
0 depend on the family (At)t≥0 ?
2.3. A penalisation procedure. It is actually easier to generalize the previous ap-
proach by replacing (At)t≥0 by a stochastic process (Ft)t≥0 which takes its values in
[0,∞[ and satisfies (1.1).
Our penalisation procedure is the following : the assumptions of the next theorem
have been shown to be satisfied for a large number of such weight processes
(
Ft
)
t≥0
; see
[18], [17], [16], [23], [24] and [22].
Theorem 2.1. Let (Ft)t≥0 as above. Assume that:
(2.9)
E0[Ft|Fs]
E0(Ft)
a.s.−→
t→∞MF
s ∀s ≥ 0
and
(2.10) E0(MF
s ) = 1 ∀s ≥ 0.
Then:
1. (MF
s ; s ≥ 0) is a non-negative P0-martingale.
2. For any s ≥ 0 and Λs ∈ Fs :
lim
t→∞
E0[1Λs Ft]
E0[Ft]
= QF
0 (Λs)
and
QF
0 (Λs) = E0[1Λs M
F
s ].
3. QF
0 extends as a p.m. on (Ω,F∞).
Note that if we choose for the weight process (Ft)t≥0 : Ft = 1At , where (At)t≥0 satisfies
(??)-(??), then assuming that (2.9) and (2.10) hold in this framework, we get:
QF
0 (Λs) = lim
t→∞P0
(
Λs|At
)
.
120 B. ROYNETTE, P. VALLOIS, AND M. YOR
Consequently QF
0 agrees with PA
0 , as tentatively defined by (2.5). Morever a solution to
the questions (2.6) and (2.7) has been given.
3. Preliminary results
In subsection 7 below we shall study in a general framework the asymptotic behavior
of t !→ E[f(It, St)] as t→ ∞, where f : ] −∞, 0] × [0,∞[→ [0,∞[ is a bounded function
with compact support. However, in order to obtain an explicit rate of convergence, we
will need to impose some restrictions on f , see subsection 3.2.
3.1. A general result. For any Borel function f : ] − ∞, 0] × [0,∞[→ [0,∞[ with
compact support, let us define :
(3.1) Kf = sup{β − α ; f(α, β) > 0}.
This means that the support of f is included in the triangle with vertices (−Kf , 0), (0, 0)
and (0,Kf ).
Let us state the main result of this subsection.
Proposition 3.1. Let f : ]−∞, 0]× [0,∞[→ [0,∞[, bounded with compact support, then
(3.2) E[f(It, St)] = Δt(fa
(t)
0 ) + tΔt(fa
(t)
1 ) + t2Δt(fa2) +Rt(f) ; t ≥ 1
where
1. Δt is the linear operator acting on functions g : ] −∞, 0] × [0,∞[→ [0,∞[ :
(3.3) Δt(g) :=
∫
]−∞,0]×[0,∞[
g(α, β)
(β − α)6
exp
{
− π2t
2(β − α)2
}
dαdβ
2. a(t)
0 and a(t)
1 are two continuous functions defined on ] −∞, 0] × [0,∞[ satisfying:
(3.4) |a(t)
i (α, β)| ≤ C(1 +K4
f ) i = 0, 1
and
(3.5) a2(α, β) := 4π3 sin
(
πβ
β − α
)
·
3. Rt(f) is a remainder term, which satisfies :
(3.6) |Rt(f)| ≤ C(1 +K4
f )
(
sup
α,β
f(α, β)
)
exp
{
− 8π2t
K2
f
}
; t ≥ 1.
To prove Proposition 3.1, three Lemmas are required.
Lemma 3.2. For any t > 0, α < 0 and β > 0, we have :
(3.7) P0(It > α, St < β) =
4
π
∑
k≥0
1
2k + 1
sin
((2k + 1)πβ
β − α
)
exp
{
− (2k + 1)2π2
2(β − α)2
t
}
Proof. According to ([3] section 11 of chap 2 ; [12], ex. 3.15, chap. III) we have:
P0(It > α, St < β, Xt ∈ dx) =
(3.8) 1[α,β](x)dx ×
∑
k∈Z
pt
(
x+ 2k(β − α)
)
− pt
(
2β − x+ 2k(β − α)
)
where pt(x) is the density function of the Gaussian distribution with mean 0 and variance
t :
(3.9) pt(x) =
1√
2πt
exp
{
− x2
2t
}
·
PENALISATIONS OF BROWNIAN MOTION 121
Using the Poisson summation formula (see for instance [7], Chap. XIX, p. 630) we get :∑
k∈Z
pt
(
x+ 2k(β − α)
)
− pt
(
2β − x+ 2k(β − α)
)
=
1
β − α
∑
k≥1
[
cos
(
kπx
β − α
)
− cos
(
kπ(2β − x
β − α
)]
exp
{
− k2π2t
2(β − α)2
}
Since cos a− cos b = −2 sin
(a+ b
2
)
sin
(a− b
2
)
, we obtain:
(3.10) P0
(
It > α, St < β, Xt ∈ dx
)
= 1[α,β](x)dx
× 2
β − α
∑
k≥1
sin
( kπβ
β − α
)
sin
(kπ(β − x)
β − α
)
exp
{
− k2π2t
2(β − α)2
}
·
Integrating (3.10) over [α, β] we easily obtain the announced result. �
Lemma 3.3. Let h1 : [0,∞[×] −∞, 0]×]0,∞[→ R be the function:
(3.11) h1(t, α, β) :=
4
π
sin
( πβ
β − α
)
exp
{
− π2t
2(β − α)2
}
·
Then
(3.12)
∂2h1
∂α∂β
(t, α, β) =
(
b0(t, α, β) + b1(t, α, β)t+ b2(α, β)t2
)
exp
{
− π2t
2(β − α)2
}
where b0(t, ·) and b1(t, ·) are of C∞ class (in the variables α and β) except at 0, and:
(3.13) |bi(t, α, β)| ≤ C
(β − α)6
(
1 + (β − α)4
)
, i = 0, 1
and
(3.14) b2(α, β) := − 4π3
(β − α)6
sin
(
πβ
β − α
)
Proof. We first observe that h1(t, α, β) is the term which is obtained by taking k = 0 in
the series (3.7).
We begin with the α-partial derivative of h1:
∂h1
∂α
(t, α, β) =
4
π
[
πβ
(β − α)2
cos
(
πβ
β − α
)
− π2t
(β − α)3
sin
(
πβ
β − α
)]
exp
{
− π2t
2(β − α)2
}
·
Taking the β-partial derivative in the above expression, it is clear that (3.12) holds. Then
(3.14) and (3.13) follow after straightforward calculations and estimates. �
Lemma 3.4. Let h2 be the function:
(3.15) h2(t, α, β) :=
4
π
∑
k≥2
1
2k + 1
sin
(
(2k + 1)πβ
β − α
)
exp
{
− (2k + 1)2π2t
2(β − α)2
}
·
Then
(3.16)
∣∣∣∣ ∂2h2
∂α∂β
(t, α, β)
∣∣∣∣ ≤ C
(
1 + (β − α)4
)
t
exp
{
− 8π2t
(β − α)2
}
·
Proof. We proceed as in the proof of Lemma 3.3:∣∣∣∣ ∂2h2
∂α∂β
(t, α, β)
∣∣∣∣ ≤ C t2
(
1 + (β − α)4
)⎛⎝∑
k≥2
(2k + 1)3
(β − α)6
exp
{
− (2k + 1)2π2t
2(β − α)2
}⎞⎠
122 B. ROYNETTE, P. VALLOIS, AND M. YOR
Since k ≥ 2, we have :
(2k + 1)2 = (2k − 3 + 4)2 ≥ (2k − 3)2 + 16
As a result: ∣∣∣∣ ∂2h2
∂α∂β
(t, α, β)
∣∣∣∣ ≤ C t2
(
1 + (β − α)4
)
⎛⎝∑
k≥2
(2k + 1)3
(β − α)6
exp
{
− (2k − 3)2π2t
2(β − α)2
}⎞⎠× exp
{
− 8π2t
(β − α)2
}
·
Let C := sup
x≥0
x3e−x. Then:
x3e−ax ≤ C
a3
, ∀x ≥ 0.
Taking x =
1
(β − α)2
and a =
(2k − 3)2π2t
2
in the above inequality, we get:
1
(β − α)6
exp
{
− (2k − 3)2π2t
2(β − α)2
}
≤ C
(2k − 3)6t3
·
This implies (3.16). �
Proof of Proposition 3.1. It is clear that the definition of h1(t, α, β)
(
resp. h2(t, α, β)
)
given by (3.11)
(
resp. (3.15)
)
implies that
P (It > α, St < β) = h1(t, α, β) + h2(t, α, β), α < 0, β > 0.
Consequently:
E0
[
f(It, St)
]
= −
∫
]−∞,0]×[0,∞[
f(α, β)
∂2h1
∂α∂β
(t, α, β) dαdβ +Rt(f)
where
Rt(f) := −
∫
]−∞,0]×[0,∞[
f(α, β)
∂2h2
∂α∂β
(t, α, β) dαdβ.
It is obvious that (3.16) implies (3.6).
From (3.13) and (3.14), we may deduce:
E0
[
f(It, St)
]
= −
1∑
i=0
ti
∫
]−∞,0]×[0,∞[
bi(t, α, β)(β−α)6
f(α, β)
(β − α)6
exp
{
− π2t
(β − α)2
}
dαdβ
+t2
∫
]−∞,0]×[0,∞[
b2(α, β)(β − α)6
f(α, β)
(β − α)6
exp
{
− π2t
2(β − α)2
}
dαdβ +Rt(f).
Setting:
a
(t)
i (α, β) = −(β − α)6 bi(t, α, β) (i = 0, 1)
leads to (3.2).
Obviously, (3.4) is a consequence of (3.13). �
PENALISATIONS OF BROWNIAN MOTION 123
3.2. Applications. Let f : ] − ∞, 0] × [0,∞[→ [0,∞[ be a bounded function with
compact support. Recall that Kf has been defined by (3.1). This quantity and the set:
Sf :=
{
β ∈]0, Kf [; f(β −Kf , β) > 0
}
will play an important role in our study.
One aim of our paper is to show that the penalisation procedure holds with the weight
process Ft := f(It, St). We briefly detail our approach. Formula (2.9) shows that it is
natural to first investigate the asymptotic behavior of t !→ E
[
f(It, St)
]
, as t → ∞.
Roughly speaking (3.2) tells us that the dominant term is Δt(fa2). We observe that:
(3.17) sup
α<0,β>0,f(α,β)>0
exp
{
− π2t
2(β − α)2
}
= exp
{
− π2t
2K2
f
}
,
and the above maximum is achieved at any point of the type (β−Kf , β) where β ∈ Sf .
In this general setting, it seems difficult to obtain an equivalent for E
[
f(It, St)
]
as
t→ ∞. This led us to consider two extreme cases: either Sf reduces to a single point or
Sf =]0, Kf [. The two corresponding prototypes of functions f are either:
(3.18) f(α, β) = 1{α≥α0, β≤β0} where α0 < 0, β0 > 0
or
(3.19) f(α, β) = 1{β−α≤c} where c > 0.
More generally we have been able to deal with the two following cases.
Case 1.
(3.20) f(α, β) = Φ(α, β) 1{α≥α0, β≤β0}
where α0 < 0, β0 > 0 and Φ : [α0, 0] × [0, β0] → R+ is continuous and
(3.21) Φ(α0, β0) > 0.
Note that then Kf = β0−α0, Sf = {β0}, and (α0, β0) is the unique point which achieves
the maximum in (3.17). It seems reasonable to believe that:
(3.22) E
[
f(It, St)
]
∼
t→∞Cα0,β0 Φ(α0, β0) exp
{
− π2t
2(β0 − α0)2
}
.
Case 2.
(3.23) f(α, β) = Φ(α, β) 1{β−α≤c}
where c > 0, Φ :
{
(α, β); α < 0, β > 0, β − α ≤ c
}
→ R+ is continuous and
(3.24)
∫ c
0
Φ(β − c, β)dβ > 0.
In this case Kf = c and Sf =
{
β ∈]0, c[; Φ(β − c, β) > 0
}
. Therefore the maximum
in (3.17) is achieved at any (β − c, β), where β belongs to ]0, c[. The expected behavior
would be:
(3.25) E
[
f(It, St)
]
∼
t→∞C(Φ, t) exp
{
− π2t
2c2
}
,
where t !→ C(Φ, t) has a polynomial rate of decay in t and depends on all the values of
Φ over the segment in R2 with endpoints (−c, 0) and (0, c).
These heuristic arguments will be justified in the remainder of the section.
124 B. ROYNETTE, P. VALLOIS, AND M. YOR
Proposition 3.5. 1. In Case 1, we have:
(3.26) E
[
f(It, St)
]
∼
t→∞
8
π
Φ(α0, β0) sin
(
πβ0
β0 − α0
)
exp
{
− π2t
2(β0 − α0)2
}
2. In Case 2, we have:
(3.27) E
[
f(It, St)
]
∼
t→∞
4π
c2
(∫ 1
0
Φ
(
c(r − 1), cr
)
sin(πr)dr
)
t exp
{
−π2t
2c2
}
·
Remark 3.6. Suppose that ϕ : [0,∞[→ [0,∞[ is a Borel function such that
∫ ∞
0
ϕ(y)dy <
∞. Note that the rate of decay of t !→ E
[
ϕ(St)
]
as t→ ∞ is very different from that of
E
[
f(It, St)
]
. Indeed, (Lemma 3.8 of [17]) it is easy to prove that:
E
[
ϕ(St)
]
∼
t→∞
√
2
π
∫ ∞
0
ϕ(y)dy
1√
t
.
Our proof of Proposition 3.5 requires 3 steps. First, we specify the Δi
t(f) introduced
in Proposition 3.1, in Cases 1 and 2.
Lemma 3.7. Let a :] −∞, 0] × [0,∞[→ R be a continuous function. Assume that f is
a function which satisfies either (3.20) or (3.23). Then
(3.28) Δt(fa) =
1
2
∫ ∞
1/K2
f
Λ(Φa)(x) x exp
{
− π2t x
2
}
dx
where
(3.29) Λ(g)(x) =
⎧⎪⎪⎪⎨⎪⎪⎪⎩
∫ inf{1, β0
√
x}
(1+α0
√
x)+
g
(r − 1√
x
,
r√
x
)
in Case 1
∫ 1
0
g
(r − 1√
x
,
r√
x
)
dr in Case 2
Proof. We only consider Case 2. According to (3.3) we have
Δt(fa) =
∫
]−∞,0]×[0,∞[
Φ(α, β)
a(α, β)
(β − α)6
exp
{
− π2t
2(β − α)2
}
1{β−α≤c}dαdβ
Setting x =
1
(β − α)2
(β fixed), we get:
Δt(fa) =
1
2
∫ ∞
1/c2
x3/2 exp
{
− π2t x
2
}(∫ 1/
√
x
0
Φ
(
β − 1√
x
, β
)
a
(
β − 1√
x
, β
)
dβ
)
dx.
The change of variable β = r/
√
x leads directly to (3.29). �
The proof of (3.26) and (3.27) is based on Laplace’s method, whose main result we
briefly recall (see for instance [6], chap. IV). We consider:
I(t) =
∫ ∞
0
g(x)et h(x)dx,
where g, h :]0,∞[→ R are continuous and satisfy the two following properties:
(3.30)
∫ ∞
0
|g(x)| et h(x)dx <∞, ∀ t > 0
(3.31) ∃δ0 > 0, such that h(x) ≤ h(δ) for any x ≥ δ and 0 < δ < δ0.
PENALISATIONS OF BROWNIAN MOTION 125
Proposition 3.8. Suppose that the functions g and h satisfy:
(3.32) g(x) ∼
x→0+
g0x
ρ
(3.33) h(x) = h0 − h1 x
τ + o(xτ ), x→ 0
for some:
(3.34) g0 �= 0, ρ > −1, h1 ≥ 0, τ > 0.
Then:
(3.35) I(t) ∼
t→∞
g0
τ
Γ
(ρ+ 1
τ
)
(h1t)−
ρ+1
τ eh0t.
As an application of the previous instance of Laplace’s method, we obtain the following
asymptotics.
Lemma 3.9. Let a :] −∞, 0] × [0,∞[→ R be a continuous function.
1. In Case 1, we have:
(3.36) Δt(fa) ∼
t→∞
2
π4
(Φa)(α0, β0)
1
t2
exp
{
− π2t
2K2
f
}
when a(α0, β0) �= 0.
2. In Case 2, we have
(3.37) Δt(fa) ∼
t→∞
1
π2K2
f
(∫ 1
0
(Φa)
(
(r − 1)Kf , rKf
)
dr
)
1
t
exp
{
− π2t
2K2
f
}
where it is assumed that
∫ 1
0
(Φa)
(
(r − 1)Kf , rKf
)
dr �= 0.
Proof. According to (3.28), we have
Δt(fa) =
1
2
∫ ∞
1/K2
f
Λ(Φa)(y) y exp
{
− π2ty
2
}
dy.
Setting y = x+ 1/K2
f we get:
(3.38) Δt(fa) =
1
2
(∫ ∞
0
Λ(Φa)
(
x+
1
K2
f
)(
x+
1
K2
f
)
exp
{
− π2tx
2
}
dx
)
exp
{
− π2t
2K2
f
}
.
a) We begin with Case 2 which is easier.
From (3.29) we deduce:
Λ(Φa)
(
x+
1
K2
f
)
=
∫ 1
0
(Φa)
⎛⎝ r − 1√
x+ 1/K2
f
,
r√
x+ 1/K2
f
⎞⎠ dr.
Since Φ and a are continuous, we obtain:
(3.39) lim
x→0
Λ(Φa)
(
x+
1
K2
f
)
=
∫ 1
0
(Φa)
(
(r − 1)Kf , rKf
)
dr.
b) Next, we deal with Case 1. Due to (3.29) we have:
Λ(Φa)
(
x+
1
K2
f
)
=
∫ τ1(x)
τ0(x)
(Φa)
( r − 1√
x+ 1/K2
f
,
r√
x+ 1/K2
f
)
dr
126 B. ROYNETTE, P. VALLOIS, AND M. YOR
with
τ0(x) =
(
1 + α0
√
x+ 1/K2
f
)
+
, τ1(x) = inf
{
1, β0
√
x+ 1/K2
f
}
.
We observe that
lim
x→0+
τ0(x) =
(
1 +
α0
Kf
)
+
=
(
1 +
α0
β0 − α0
)
+
=
β0
β0 − α0
lim
x→0+
τ1(x) = inf
{
1,
β0
Kf
}
= inf
{
1,
β0
β0 − α0
}
=
β0
β0 − α0
.
This implies that:
Λ(Φa)
(
x+
1
K2
f
)
∼
x→0+
(
τ1(x) − τ0(x)
)
(Φa)(α0, β0).
When x is small, we have
τ1(x) − τ0(x) = β0
√
x+ 1/K2
f − 1 − α0
√
x+ 1/K2
f =
√
xK2
f + 1 − 1.
As a result:
(3.40) Λ(Φa)
(
x+
1
K2
f
)
∼
x→0+
K2
f
2
(Φa)(α0, β0) x.
c) We apply Laplace’s method with
g(x) =
(
x+
1
K2
f
)
Λ(Φa)
(
x+
1
K2
f
)
, and h(x) = −π
2x
2
·
It is clear that (3.30), (3.31) and (3.33) hold with
h0 = 0, h1 =
π2
2
and τ = 1.
In Case 2 (resp. Case 1), relation (3.39)
(
resp. (3.40)
)
implies that (3.32) is satisfied
with g0 =
1
K2
f
∫ 1
0
(Φa)
(
(r − 1)Kf , rKf
)
dr and ρ = 0
(
resp. g0 =
1
2
(Φa)(α0, β0) and
ρ = 1
)
.
Lemma 3.9 is a direct consequence of (3.35) and (3.38). �
Proof of Proposition 3.5. 1) We begin with Case 2. It is clear that (3.37) and (3.5)
imply:
(3.41) Δt(fa2) ∼
t→∞
4π
K2
f
(∫ 1
0
Φ
(
(r − 1)Kf , rKf
)
sin(πr)dr
)
1
t
exp
{
− π2t
2K2
f
}
.
Let i = 0, 1. From (3.4) we have:
(3.42) |Δt(fa
(t)
i )| ≤ C(1 +K4
f ) Δt(f).
Using (3.37), we get:
(3.43) Δt(f) ∼
t→∞
μ
t
exp
{
− π2t
2K2
f
}
(for some μ > 0).
Applying (3.2), (3.6) and (3.41)–(3.43) shows (3.27).
2) Similarly to Case 2, in Case 1, the main term of E
[
f(It, St)
]
is Δt(fa). As a result,
(3.26) follows from (3.36). �
PENALISATIONS OF BROWNIAN MOTION 127
4. Penalisation with the maximum and the minimum
4.1. Penalisation with f(It, St), Case 1. In this subsection we suppose that f satisfies
(3.20).
Theorem 4.1. The Brownian penalisation procedure holds with the weight process Ft =
f(It, St), i.e.
1. Property (2.9) holds:
(4.1) lim
t→∞
E0
[
f(It, St)|Fu
]
E0
[
f(It, St)
] = Mα0,β0
u , for any u ≥ 0
where
(4.2) Mα0,β0
u = Nα0,β0(u ∧ Tα0 ∧ Tβ0) ; u ≥ 0
(4.3) Nα0,β0
u =
1
sin
(
πβ0
β0−α0
) sin
(
π(β0 −Xu)
β0 − α0
)
exp
{
π2u
2(β0 − α0)2
}
u ≥ 0.
(4.4) Tx = inf{t ≥ 0, Xt = x}.
2. Moreover (2.10) is satisfied:
(4.5) E0[Mα0,β0
u ] = 1 ∀u ≥ 0.
Remark 4.2. 1. According to Theorem 2.1:
(a) (Mα0,β0
u ; u ≥ 0) is a non-negative P0-martingale, which converges to 0.
(b) For any u ≥ 0 and Λu ∈ Fu,
(4.6) Qα0,β0
0 (Λu) := lim
t→∞
E0
[
1ΛuΦ(It, St)1{It≥α0,St≤β0}
]
E0
[
Φ(It, St)1{It≥α0,St≤β0}
]
exists and Qα0,β0
0 is a p.m. on (Ω,F∞) which satisfies:
(4.7) Qα0,β0
0 (Λu) = E0[1ΛuM
α0,β0
u ] u ≥ 0, Λu ∈ Fu.
2. In the particular case: f = 1 and β0 = −α0, then
Ft = 1{It≥−β0,St≤β0} = 1{X∗
t ≤β0},
where X∗
t = St ∨ (−It) = max
u≤t
|Xu|. Moreover:
M−β0,β0
u = cos
(πX(u ∧ T ∗
β0
)
2β0
)
exp
{π2(u ∧ T ∗
β0
)
2β2
0
}
,
where T ∗
β0
= inf{t ≥ 0, |Xt| = β0}.
3. Note that in [17], a penalisation procedure has been considered with weight pro-
cesses:
Ft :=
∫
]−∞,0]×[0,∞]
1{It≥α,St≤β} exp
{1
2
( 1
β
− 1
α
)
L0
t
}
ν(dα, dβ)
where ν is a p.m. on ]−∞, 0]× [0,∞] and (L0
t ) is the local time process at 0 associated
with (Xt).
It has been proved (cf Theorem 3.18) that the martingales generated by this penal-
isation are functions of the triplet (It, St, L
0
t ). Consequently, they are not of the form(
M−β0,β0
t
)
.
Proof of Theorem 4.1. To show that the penalisation procedure holds with Ft = f(It, St)
we need to prove (4.1) and (4.6). Observe that relation (3.26) in Proposition 3.5 gives the
128 B. ROYNETTE, P. VALLOIS, AND M. YOR
rate of decay of t → E0
[
f(It, St)
]
, t → ∞. Next, we need to determine the asymptotic
behavior of t→ E
[
f(It, St)|Fu
]
. We will prove in step 1 below that the rate of decay of
E
[
f(It, St)|Fu
]
may be deduced from (3.26). In step 2 we will prove (4.4).
1) Proof of (4.1)
Let u > 0 be fixed. We introduce X ′
t = Xt+u −Xu, t ≥ 0.
Under P0, (X ′
t)t≥0 is a Brownian motion started at 0 and independent from Fu.
Moreover:
Su+h = Su ∨ (Xu + S′
h), Iu+h = Iu ∧ (Xu + I ′h), h > 0
where S′
h = max
v≤h
X ′
v and I ′h = min
v≤h
X ′
v.
This implies that
(4.8) E0
[
f(Iu+v, Su+v)|Fu
]
= H(Iu, Su, Xu, v) ; v ≥ 0
where
(4.9) H(a, b, x, v) := E0
[
f
(
a ∧ (x+ I ′v), b ∨ (x+ S′
v)
)]
.
Suppose that a, b, x are fixed, a ≤ x, b ≥ x. We introduce:
(4.10) f̃(α, β) = f
(
a ∧ (x+ α), b ∨ (x+ β)
)
α ≤ 0, β ≥ 0.
Since:
a ∧ (x+ α) ≥ α0 ⇔ a ≥ α0 and α ≥ α0 − x
and
b ∨ (x+ β) ≤ β0 ⇔ b ≤ β0 and β ≤ β0 − x
then (3.20) implies that
H(a, b, x, v) = 1{a≥α0,β≤β0}E0
[
Φ
(
a ∧ (x+ Iv), b ∨ (x+ Sv)
)
1{Iv≥α0−x, Sv≤β0−x}
]
.
We may apply (3.26):
H(a, b, x, v) ∼
v→∞
8
π
Φ(α0, β0) sin
(
π(β0 − x)
β0 − α0
)
exp
{
− π2v
2(β0 − α0)2
}
1{a≥α0, b≤β0}.
Consequently:
E0
[
f(Iu+v , Su+v)|Fu
]
∼
v→∞
8
π
Φ(α0, β0) sin
(
π(β0 −Xu)
β0 − α0
)
exp
{
− π2v
2(β0 − α0)2
}
(4.11) ×1{Iu≥α0, Su≤β0}.
Recall that
E0
[
f(Iu+v, Su+v)
]
∼
v→∞
8
π
Φ(α0, β0) sin
(
πβ0
β0 − α0
)
exp
{
− π2(u+ v)
2(β0 − α0)2
}
.
This proves (4.1) because
Nα0,β0(Tα0) = Nα0,β0(Tβ0) = 0
and
{Iu ≥ α0} = {u ≤ Tα0}, {Su ≤ β0} = {u ≤ Tβ0}.
2) Proof of (4.5)
PENALISATIONS OF BROWNIAN MOTION 129
(Nα0,β0
u , u ≥ 0) is a continuous local martingale as combination of exponential mar-
tingales. Itô’s formula confirms this:
(4.12) dNα0,β0
u = − π
β0 − α0
1
sin
(
πβ0
β0−α0
) cos
(
π(β0 −Xu)
β0 − α0
)
exp
{ π2u
2(β0 − α0)2
}
dXu.
It is clear that Nα0,β0
u is uniformly bounded on any interval [0, T ], T fixed. This shows
that (Nα0,β0
u , u ≥ 0) is a martingale. Applying the Doob’s optional stopping theorem we
get:
E0[Mα0,β0
u ] = E
[
Nα0,β0(u ∧ Tα0 ∧ Tβ0)
]
= Nα0,β0(0) = 1.
�
4.2. Penalisation with f(It, St), Case 2 . We study penalisation in Case 2. Namely
f satisfies (3.23):
f(α, β) = Φ(α, β)1{β−α≤c}.
Theorem 4.3. The Brownian penalisation holds with Ft = f(It, St), f satisfying (3.23).
1. Property (2.9) holds:
(4.13) lim
t→∞
E0
[
f(It, St)|Fu
]
E0
[
f(It, St)
] = MΦ,c
u , ∀u ≥ 0
where
(4.14) MΦ,c
u = NΦ,c
(
u ∧ θ(c)
)
; u ≥ 0
(4.15) NΦ,c(u) =
c+ Iu − Su
cρ(F )
{∫ 1
0
[
Φ
(
Su − c+ r(c+ Iu − Su), Su + r(c+ Iu − Su)
)
× sin
(π
c
(
Su −Xu + r(c + Iu − Su)
))]
dr
}
exp
{π2u
2c2
}
(4.16) =
1
ρ(Φ)
(∫ (Iu−Xu+c)/c
(Su−Xu)/c
Φ(Xu + c(r − 1), Xu + rc) sin(πr)dr
)
× exp
{π2u
2c2
}
(4.17) ρ(Φ) =
∫ 1
0
Φ
(
c(r − 1), cr
)
sin(πr)dr
(4.18) θ(c) = inf{t ≥ 0; St − It = c}
2. Moreover (2.10) holds:
(4.19) E0[MΦ,c(u)] = 1.
Remark 4.4. 1. Applying Theorem 2.1, we may deduce that:
(a) (MΦ,c
u ; u ≥ 0) is a non-negative P0-martingale, which converges to 0.
(b) For any u ≥ 0 and Λu ∈ Fu,
QΦ,c
0 (Λu) := lim
t→∞
E0
[
1ΛuΦ(It, St)1{St−It≤c}
]
E0
[
Φ(It, St)1{St−It≤c}
]
exists.
Moreover QΦ,c
0 is a p.m. on (Ω,F∞) and
(4.20) QΦ,c
0 (Λu) = E0[1ΛuM
Φ,c
u ] u ≥ 0, Λu ∈ Fu.
130 B. ROYNETTE, P. VALLOIS, AND M. YOR
2. When Φ(α, β) = Φ0(β − α) then f(α, β) = Φ0(β − α)1{β−α≤c} and
(4.21) NΦ,c
u =
1
2
[
cos
(π(Su −Xu)
c
)
+ cos
(π(Iu −Xu)
c
)]
exp
{π2u
2c2
}
.
Recall that MΦ,c
u = NΦ,c(u ∧ θ(c)) u ≥ 0.
Proof of Theorem 4.3. Our proof of Theorem 4.3 is close to that of Theorem 4.1. The
details are left to the reader. However we would like to explain how the martingale
(MΦ,c
t )t≥0 appears. We go back to the proof of Theorem 4.1. Obviously (4.8)-(4.10) are
still valid. We have to make f̃ explicit in Case 2. From (3.23):
f̃(α, β) = Φ
(
a ∧ (x+ α), b ∨ (x + β)
)
1{b∨(x+β)−a∧(x+α)≤c}
Since:
b ∨ (x+ β) − a ∧ (x+ α) ≤ c ⇔ b− a ≤ c, b− x− α ≤ c, x+ β − a ≤ c, β − α ≤ c
then
f̃(α, β) = 1{b−a≤c} Φ̃(α, β) 1{β−α≤c}
where
Φ̃(α, β) = Φ
(
a ∧ (x + α), b ∨ (x+ β)
)
1{b−x−α≤c, x+β−a≤c}.
Consequently, K �f = Kf = c and
Φ̃
(
c(r − 1), cr
)
= Φ
(
a ∧ (x+ cr − c), b ∨ (x+ cr)
)
1{b−x≤cr≤c+a−x}
= Φ(x+ cr − c, x+ cr) 1{b−x≤cr≤c+a−x}.
Applying (3.27) leads to (4.13) and (4.16). �
5. The law of (Xt) under Qα0,β0
0 and QΦ,c
0
We first consider the distribution of the canonical process (Xt) under Qα0,β0
0 . Let
α0 < 0 and β0 > 0 be two fixed real numbers. Recall the definition of the p.m. Qα0,β0
0
on (Ω,F∞) :
(5.1) Qα0,β0
0 (Λu) = E0[1ΛuM
α0,β0
u ] Λu ∈ Fu
where (Mα0,β0
u )u≥0 is the P0-martingale defined by (4.2).
Theorem 5.1. Under Qα0,β0
0 :
1. (Xt) is a diffusion process solving:
(5.2) Xt = Bt −
π
β0 − α0
∫ t
0
cot
(
π(β0 −Xu)
β0 − α0
)
du ; t ≥ 0,
where (Bt)t≥0 is a Qα0,β0
0 -Brownian motion started at 0.
2. (Xt) has the following path properties :
(5.3) α0 < Xt < β0 ∀t ≥ 0
(5.4) S∞ = sup
t≥0
Xt = β0, I∞ = inf
t≥0
Xt = α0.
3. Xt converges in distribution, as t→ ∞ to the p.m. pα0,β0(x)dx on R, with :
(5.5) pα0,β0(x) :=
2
β0 − α0
sin2
(
π(β0 − x)
β0 − α0
)
1(α0,β0)(x).
PENALISATIONS OF BROWNIAN MOTION 131
Remark 5.2. 1. Property (5.3) follows intuitively from our penalisation procedure and
the fact that the support of the p.m. on Fu:
Λu !→
E0
[
1ΛuΦ(It, St)1{It≥α0,St≤β0}
]
E0
[
Φ(It, St)1{It≥α0,St≤β0}
] (t > u)
is included in
{
Iu ≥ α0, Su ≤ β0
}
.
2. Note that Qα0,β0
0 do not depend on Φ.
Proof of Theorem 5.1.
a) (Mα0,β0
t ; t ≥ 0) is a non-negative Qα0,β0
0 -martingale and can be written as:
Mα0,β0
t = E(J)t := exp
{∫ t
0
Ju dXu − 1
2
∫ t
0
J2
udu
}
for any t < Tα0 ∧ Tβ0 . Indeed, from (4.12) and (4.2) we have:
Ju = − π
β0 − α0
cot
(
π(β0 −Xu)
β0 − α0
)
, 0 ≤ u < Tα0 ∧ Tβ0 .
Then Girsanov’s theorem implies (5.2).
b) To investigate the path properties of (Xt), we define Yt = β0 −Xt, then
(5.6) Yt = β0 − Bt +
π
β0 − α0
∫ t
0
cot
(
π Yu
β0 − α0
)
du
Obviously (Yt)t≥0 is a one-dimensional diffusion. Let S
(
resp. m(dy)
)
denote its scale
function (resp. speed measure), see for instance: (Section 1, chap. 4 of [8]), (Section 1,
chap II of [4]). Using standard calculations (i.e. [14] (Section 52, chap V), [5] (Section
12, chap 16))) we easily get:
S(y) = − cot
(
πy
β0 − α0
)
, y ∈]0, β0 − α0[
m
(
[0, y]
)
=
(β0 − α0)
π
y − (β0 − α0)2
2π2
sin
(
2πy
β0 − α0
)
, y ∈]0, β0 − α0[.
According to the classification of boundary points of a linear diffusion (cf Section 1,
chap II of [4]), we have:
i) 0 is not an exit point since :
(5.7)
∫
(0,x)
m
(
[y, z]
)
S′(y)dy = ∞ (z ∈]0, β0 − α0[).
ii) 0 is an entrance point since:
(5.8)
∫
(0,z)
(
S(z) − S(y)
)
m(dy) <∞ (z ∈]0, β0 − α0[).
Similarly β0 − α0 is not an exit point and is an entrance one. This shows (5.3).
c) The diffusion (Yt)t≥0 which takes its values in (0, β0−α0) is recurrent. This implies
(5.4). Let p be the density of its invariant p.m. From the Fokker Planck equation, p
solves:
(5.9)
1
2
p′′(y) − π
β0 − α0
(
cot
( π, y
β0 − α0
)
p(y)
)′
= 0.
It is easy to verify that
p(y) =
2
β0 − α0
sin2
(
πy
β0 − α0
)
y ∈ (0, β0 − α0)
132 B. ROYNETTE, P. VALLOIS, AND M. YOR
is the unique density function solving (5.9). �
To study the law of (Xt)t≥0 under QΦ,c
0 it is more convenient to express the p.m. QΦ,c
0
via the family of p.m.’s
{
Qβ−c,β
0 ; 0 < β < c
}
(see Theorem 5.3 below). This result will
allow to determine easily the distribution of (Xt) under QΦ,c
0 .
Theorem 5.3. Let c > 0, and Φ :
{
(α, β) ; α < 0, β > 0, β − α ≤ c
}
→ R+ be a
continuous function satisfying (3.24). Then:
(5.10) QΦ,c
0 (·) =
1
cρ(Φ)
∫ c
0
Φ(β − c, β) sin
(
πβ
c
)
Qβ−c,β
0 (·)dβ.
Proof. a) Assume for a while that the followings holds:
(5.11) EΦ,c
0
[
1Λuh(Su, S∞)
]
=
1
cρ(Φ)
∫ c
0
Φ(v − c, v) sin
(πv
c
)
Ev−c,v
0
[
1Λuh(Su, v)
]
dv
for any h : R2
+ → R+ Borel, u > 0, Λu ∈ Fu, and that EΦ,c
0 (resp. Eα0,β0
0 ) stands for the
expectation under QΦ,c
0 (resp. Qα0,β0
0 ).
Taking h = 1 in (5.1) implies that (5.10) holds on Fu. Since the p.m.’s in (5.10) are
defined on F∞ they coincide on F∞.
b) The remainder is devoted to the proof of (5.11).
i) We claim that:
(5.12) QΦ,c
0 (S∞ − I∞ ≤ c) = 1.
Indeed, from the definition of QΦ,c
0 we have:
QΦ,c
0 (St − It > c) = EΦ,c
0 [1{St−It>c}M
Φ,c
t ] = EΦ,c
0 [1{t>θ(c)}M
Φ,c
t ]
From Doob’s optional stopping theorem and the fact that MΦ,c
θ(c) = 0, we get:
QΦ,c
0 (St − It > c) = EΦ,c
0 [1{t>θ(c)}M
Φ,c
θ(c)] = 0.
Taking t→ ∞, we obtain (5.12).
ii) Due to the monotone class theorem and (5.12), it is sufficient to show (5.11) with
h(x, y) := 1[0,β](x)1]β′,c](y) (0 < β < β′ < c)
In this case, (5.11) reduces to:
(5.13) A =
1
cρ(Φ)
∫ c
β′
Φ(v − c, v) sin
(πv
c
)
P v−c,v
0
(
Λu ∩ {Su ≤ β}
)
dv,
where
(5.14) A := PΦ,c
0
(
Λu ∩ {Su ≤ β, S∞ > β′}
)
.
It is clear that
A = lim
t→∞A(t), with A(t) := PΦ,c
0
(
Λu ∩ {Su ≤ β, St > β′}
)
.
Then, for t > u, we have:
A(t) = E0[1{Λu∩{Su≤β} 1{St>β′}M
Φ,c
t ]
= E0[1Λu∩{Su≤β} 1{Tβ′<t}M
Φ,c
t
]
Due to Doob’s optional stopping theorem we get:
A(t) = E0
[
1{Λu∩{Su≤β} 1{Tβ′<t}MΦ,c(Tβ′)
]
PENALISATIONS OF BROWNIAN MOTION 133
Taking the limit t→ ∞, we obtain:
A = E0
[
1∧u∩{Su≤β}MΦ,c(Tβ′)
]
.
Using (4.14) and (4.16) we have:
A =
1
ρ(Φ)
E0
[
1Λu∩{Su≤β}
(∫ 1
0
1{rc<I(Tβ′)−β′+c} sin(πr)Φ
(
β′ + c(r − 1), β′ + rc
)
dr
)
× 1{β′−I(Tβ′)<c} exp
{π2Tβ′
2c2
}]
.
Observe that 0 > I(Tβ′) > rc + β′ − c implies that I(Tβ′) > β′ − c and β′ + rc < c.
Making the change of variable v = β′ + rc leads to
(5.15) A =
1
cρ(Φ)
∫ c
β′
Φ(v − c, v) sin
(
π(v − β′)
c
)
A1(v)dv
where
A1(v) := E0
[
1Λu∩{Su≤β} 1{I(Tβ′ )>v−c} exp
{π2Tβ′
2c2
}]
.
Since u ≤ Tβ < Tβ′ , we get, by using the Markov property at time u:
E0
[
1{I(Tβ′)>v−c} exp
{π2Tβ′
2c2
}∣∣∣Fu
]
= 1{Iu>v−c}A2(Xu) exp
{π2u
2c2
}
with:
A2(x) := E
[
1{I(Tβ′−x)>v−c−x} exp
{π2Tβ′−x
2c2
}]
.
Let α1 < 0 < β1 and assume that β1 − α1 < c. Then, using the fact that
Zt := sin
(π(Xt − α1)
c
)
exp
{π2t
2c2
}
, t ≥ 0
is a martingale, and Doob’s optional stopping theorem at Tβ1 ∧ Tα1 lead to:
(5.16) E0
[
1{I(Tβ1)>α1} exp
{π2Tβ1
2c2
}]
= −
sin
(
πα1
c
)
sin
(
π
c (β1 − α1)
) .
Consequently, we obtain successively:
A2(x) =
sin
(
π (v−x)
c
)
sin
(π(v−β′)
c
)
A =
1
cρ(Φ)
∫ c
β′
Φ(v − c, v)A3(v)dv,
with
A3(v) := E0
[
1Λu∩{Su≤β, Iu>v−c} sin
(π(v −Xu)
c
)
exp
{π2u
2c2
}]
.
Note that Su ≤ β < β′ < v ; then according to (4.2) and (4.3) we have:
A2(v) = sin
(πv
c
)
E0
[
1Λu∩{Su≤β}Mv−c,v
u
]
.
Finally
A =
1
cρ(Φ)
∫ c
β′
Φ(v − c, v) sin
(πv
c
)
E0
[
1Λu M
v−c,v
u 1{Su≤β}
]
dv.
=
1
cρ(Φ)
∫ c
β′
Φ(v − c, v) sin
(πv
c
)
P v−c,v
0
(
Λu ∩ {Su ≤ β}
)
dv.
This shows (5.13). �
134 B. ROYNETTE, P. VALLOIS, AND M. YOR
We will deduce from Theorem 5.3 two main consequences (see Theorems 5.4 and 5.5
below). We first interpret the identity (5.10) in a more probabilistic way.
Theorem 5.4. 1. Conditionally on S∞ = v, (It, St, Xt)t≥0 is distributed under QΦ,c
0 as
the three dimensional process (It, St, Xt)t≥0 under Qv−c,v
0 .
2. The density function of S∞ under QΦ,c
0 is
1
cρ(Φ)
Φ(v − c, v) sin
(πv
c
)
1(0,c)(v).
Proof. Let u > 0 and Λu ∈ Fu. Let us apply (5.11) with h(x, y) = H(y), for H : [0,∞[→
[0,∞[:
EΦ,c
0
[
1ΛuH(S∞)
]
=
1
cρ(Φ)
∫ c
0
Φ(v − c, v) sin
(πv
c
)
H(v)Qv−c,v
0 (Λu)dv
If we take in particular Λu = Ω, we get
EΦ,c
0
[
H(S∞)
]
=
1
cρ(Φ)
∫ c
0
H(v)Φ(v − c, v) sin
(πv
c
)
dv.
This shows ii).
Then i) follows from
EΦ,c
0
[
1ΛuH(S∞)
]
= EΦ,c
0
[
H(S∞)PΦ,c
0 (Λu|S∞)
]
=
1
cρ(Φ)
∫ c
0
H(v)PΦ,c
0 (Λu|S∞ = v)Φ(v − c, v) sin
(πv
c
)
dv.
�
We are now able to present a few path properties of (Xt) under QΦ,c
0 .
Theorem 5.5. Under QΦ,c
0 :
1. We have:
(5.17) It > I∞ and St < S∞ for any t ≥ 0
(5.18) S∞ − I∞ = c.
2. When t goes to infinity, the couple (St, Xt) converges in distribution to the p.m.
on R+ × R:
2
c2ρ(Φ)
sin2
(
π(x− y)
c
)
sin
(πx
c
)
Φ(x− c, x)1{0<x<c, x−c<y<x}dxdy.
Remark 5.6. 1. Property (5.18) can be deduced intuitively from our penalisation proce-
dure and the fact that the support of the p.m. on Fu:
Λu !→
E0
[
1ΛuΦ(It, St)1{St−It≤c}
]
E0
[
Φ(It, St)1{St−It≤c}
] (t > u)
is included in
{
Su − Iu ≤ c
}
.
2. Recall that the p.m. Qα0,β0
0 which arises from penalisation with
Φ(It, St)1{It≥α0,St≤β0}
does not depend on the values of Φ. Therefore penalisation associated with
Φ(It, St)1{St−It≤c}
is very different since the p.m. QΦ,c
0 depends on the values of Φ over the segment{
(β − c, β); β ∈ [0, c]
}
.
PENALISATIONS OF BROWNIAN MOTION 135
Proof of Theorem 5.5.
It is easy to deduce (5.17)
(
resp. (5.18)
)
from (5.3)
(
resp. (5.4)
)
and Theorem 5.4.
The details are left to the reader. Let g : R+ ×R be a continuous and bounded function.
According to Theorem 5.4 we have:
EΦ,c
0
[
g(St, Xt)
]
=
1
cρ(Φ)
∫ c
0
Φ(v − c, v) sin
(πv
c
)
Ev−c,v
0
[
g(St, Xt)
]
dv.
Applying Theorem 5.1, we get:
lim
t→∞Ev−c,v
0
[
g(St, Xt)
]
=
2
c
∫ v
v−c
g(v, y) sin2
(π(v − y)
c
)
dy.
Point 2) of Theorem 5.5 is a direct consequence of the dominated convergence theorem.
�
We formulate differently item 2) of Theorem 5.5.
Corollary 5.7. The pair
(St + It
2
, Xt −
St + It
2
)
converges in distribution as t → ∞
to:(
1
cρ(Φ)
Φ
(
x− c
2
, x+
c
2
)
cos
(πx
c
)
1[− c
2 , c
2
](x)dx
)
×
(
2
c
cos2
(πy
c
)
1[− c
2 , c
2
](y)dy
)
hence, its two components are asymptotically independent.
Proof. This is a direct consequence of 2) of Theorem 5.4, (5.18) and simple changes of
variables. �
Remark 5.8. Using Girsanov’s theorem it may be proved that (Xt) solves:
(5.19) Xt = Bt +
∫ t
0
∂ΓΦ,c
∂x
ΓΦ,c
(Iu, Su, Xu)1{Su−Iu<c}du
where (Bt)t≥0 is a QΦ,c
0 Brownian motion, and
(5.20) ΓΦ,c(a, b, x) =
∫ (a−x+c)/c
(b−x)/c
Φ(x+ c(r − 1), x+ rc) sin(πr)dr
∂ΓΦ,c
∂x
(a, b, x) = −π
c
∫ (a−x+c)/c
(b−x)/c
Φ(x+ c(r − 1), x+ rc) cos(πr)dr.
In the particular case Φ = 1 (i.e. penalisation with St − It ≤ c) then:
Γ1,c(a, b, x) =
1
π
(
cos
(
π(b − x)
c
)
+ cos
(
π(a− x)
c
))
This implies that (Xt) solves:
(5.21) Xt = Bt +
π
c
∫ t
0
tan
(
π
c
(
Su + Iu
2
−Xu
))
1{Su−Iu<c}du.
136 B. ROYNETTE, P. VALLOIS, AND M. YOR
6. Application to diffusions
Our approach is based on Theorem 5.1. Let μ be a p.m. on ] −∞, 0] × [0,∞[ which
does not charge (0, 0). Let us consider the associated p.m. Qμ
0 on (Ω,F∞) :
(6.1) Qμ
0 (·) =
∫
]−∞,0]×[0,∞[
Qα,β
0 (·)μ(dα, dβ)
where Qα,β
0 has been defined by (5.1). Note that from Theorem 5.3, the p.m. QΦ,c
0 is
equal to Qμ
0 where
(6.2) μ(dα, dβ) = μΦ,c(dα, dβ) :=
1
cρ(Φ)
Φ(β − c, β) sin
(
πβ
c
)
δβ−c(dα)1[0,c](β)dβ.
Proposition 6.1. Under Qμ
0 :
1. (It, St) converges a.s. as t→ ∞ to (I∞, S∞) and the distribution of (I∞, S∞) is μ.
2. (It, St, Xt) converges in distribution, as t→ ∞ to the p.m. on ]−∞, 0]× [0,∞[×R:
(6.3) λ(dα, dβ, dx) = pα,β(x)μ(dα, dβ)dx,
where the density function pα,β(x) has been defined by (5.5).
In particular, (St, Xt) converges in law, as t→ ∞, to
(6.4) ν(dβ, dx) :=
(∫ 0
−∞
pα,β(x)μ(dα, dβ)
)
dx.
Proof.
Proposition 6.1 is a direct consequence of Theorem 5.1. �
Remark 6.2. 1. It seems difficult to characterize all possible p.m.’s ν obtained by this
randomization procedure, i.e. to describe the set of p.m.’s ν which are defined by (6.4),
μ varying in the set of p.m.’s on ] −∞, 0] × [0,∞].
2. It can be proved that if μ(dα, dβ) satisfies:
(6.5)
∫
]−∞,0]×[0,∞[
(β − α)μ(dα, dβ) <∞,
(6.6)
∫
]−∞,0]×[0,∞[
(β + α)μ(dα, dβ) = 0,
(6.7) 2 a μ
(
] −∞, 0] × [a,∞[
)
=
∫
]−∞,0]×[0,∞[
(α+ β)μ(dα, dβ).
Then, the Rogers conditions (1.7) and (1.8) hold.
3. The p.m. Qμ
0 is locally absolutely continuous with respect to the Wiener measure
P0, namely:
(6.8) Qμ
0 (Λt) = E0[1ΛtM
μ
t ], Λt ∈ Ft
where
(6.9) Mμ
t =
∫
]−∞,0]×[0,∞[
Mα,β
t μ(dα, dβ),
and (Mα,β
t )t≥0 is the P0-martingale introduced in Theorem 4.1. Moreover
(6.10) Mμ
t = Γμ(It, St, Xt, t)
PENALISATIONS OF BROWNIAN MOTION 137
where
(6.11) Γμ(a, b, x, t) :=
∫
]−∞,0]×[0,∞[
1
sin
(
πβ
β−α
) sin
(
π(β − x)
β − α
)
exp
{
π2t
2(β − α)2
}
×1{α≥a, β≤b} μ(dα, dβ).
Due to Girsanov’s theorem, it may be infered that Xt solves:
(6.12) Xt = Bt +
∫ t
0
∂
∂x Γμ
Γμ
(Iu, Su, Xu, u) du
where (Bt)t≥0 is a Qμ
0 -Brownian motion started at 0 and
(6.13)
∂
∂x
Γμ(a, b, x, t) = −π
∫
]−∞,0]×[0,∞[
1
(β − α) sin
(
πβ
β−α
) cos
(
π(β − x)
β − α
)
(6.14) × exp
{
π2t
2(β − α)2
}
1{α≥a, β≤b} μ(dα, dβ)
Consequently, (It, St, Xt) is a non-homogeneous Markov process.
4. There exists a p.m. μ satisfying (6.5)-(6.7). It is easy to show that these conditions
hold with
(6.15) μ(dα, dβ) =
(
2c2
(β + c)3
1{β>0}dβ
)
μ1(dα)
where μ1(dα) is a p.m. on ] −∞, 0], such that −
∫ 0
−∞
αμ1(dα) = c ∈]0,∞[.
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