Integral Equations for Rost’s Reversed
Barriers: Existence and Uniqueness Results
Tiziano De Angelis & Yerkin Kitapbayev
First version: 15 September 2015
Research Report No. 5, 2015, Probability and Statistics Group
School of Mathematics, The University of Manchester
Integral equations for Rost’s reversed
barriers: existence and uniqueness results
Tiziano De Angelis∗ and Yerkin Kitapbayev†
August 24, 2015
Abstract
We establish that the boundaries of the so-called Rost’s reversed barrier are the
unique couple of left-continuous monotonic functions solving a suitable system of
nonlinear integral equations of Volterra type. Our result holds for any atom-less
target distribution µ of the related Skorokhod embedding problem.
The integral equations we obtain here generalise the ones often arising in optimal
stopping literature and our proof of the uniqueness of the solution goes beyond the
existing results in the field.
MSC2010: 60G40, 60J65, 60J55, 35R35, 45D05.
Key words: Skorokhod embedding, Rost’s reversed barriers, optimal stopping, freeboundary problems, Volterra integral equations.
1
Introduction
The celebrated Skorokhod embedding problem formulated in the 1960’s by Skorokhod [20]
can be stated as follows: given a probability law µ, find a stopping time τ of a standard
Brownian motion W such that Wτ is distributed according to µ. A solution of this
problem, among many others (see [15] for a survey), was found by Rost [18] and can be
given in terms of first hitting times of the time-space process (t, Wt )t≥0 to a set, depending
on µ, usually called reversed barrier [3]. The boundaries of this set can be expressed as two
monotonic functions of time (see [4]) and we denote them t 7→ s± (t) (here we understand
s± (t) = s± (t; µ) depending on µ but prefer to adopt a simpler notation as no confusion
should arise).
For any atom-less target distribution µ (hence also singular ones like Cantor distribution) we use purely probabilistic methods to characterise the couple (s+ , s− ) as the
unique solution of suitable nonlinear integral equations of Volterra type. To achieve that
we use a known connection between Rost’s solution of the embedding problem and an
∗
School of Mathematics, University of Manchester, Oxford Rd. M13 9PL Manchester, UK;
[email protected]
†
Questrom School of Business, Boston University, 595 Commonwealth Avenue, 02215, Boston, MA,
USA; [email protected]
1
Integral equations for Rost’s barriers
2
optimal stopping problem (see for instance [5], [12] and [6]) so that our contribution becomes twofold: on the one hand we provide a new way of characterising and evaluating s±
and on the other hand we extend in several directions the existing literature concerning
uniqueness of solutions to certain integral equations.
To the best of our knowledge until now there exist essentially two ways of computing
Rost’s reversed barriers: a constructive way and a PDE way. The former approach was
proposed by Cox and Peskir [4] who approximate the boundaries s± via a sequence of
piecewise constant boundaries (in our notation (sn± )n≥0 ). They prove suitable convergence
of sn± to s± as n → ∞ and for each n they also provide equations that can be solved
numerically to find sn± (however no equations could be provided for s± in the limit). In
practice for n large enough one can compute an accurate approximation of s± . The PDE
approach instead relies upon the aforementioned connection to optimal stopping as proved
in [5] through viscosity solution of variational inequalities. The idea is that one should
numerically solve a variational inequality to obtain the value function of the stopping
problem and then use it to compute the stopping set. The boundaries of such stopping
set indeed coincide with s± up to a suitable time reversal. An approach of this kind was
used in Section 4 of [9], although in the different context of Root’s solutions to Skorokhod
embedding.
Here we do use the connection to optimal stopping but instead of employing PDE
theory we only rely upon stochastic calculus. Our work offers a point of view different to
the one of [4] and [5] since we directly characterise the boundaries s± of Rost’s barriers
as the unique solutions to specific equations. We do not develop a constructive approach
nor we use an indirect approach via variational inequalities. The computational effort
required to evaluate our equations numerically is rather modest and hinges on a simple
algorithm which is widely used in the optimal stopping literature (see for example Remark
4.2 in [7]). It is worth mentioning that McConnell [12] provided a family of uncountably
many integral equations for s± which appear to be very different from ours as they involve
at the same time the boundaries and (in our language) the value function of the optimal
stopping problem. Solvability of these equations seems to be left as an open question and
we are not aware of any further contributions in the direction of finding its answer.
To clarify our contribution to the literature on integral equations and optimal stopping we would like to recall that nonlinear equations of Volterra type arise frequently in
problems of optimal stopping on finite time horizon. Such equations are indeed used to
compute the related optimal stopping boundaries (see [17] for a collection of examples)
and their study in this context goes back to McKean [13] and Van Moerbeke [21] among
others (see also [2], [10], [11] and [14] for the American put exercise boundary). The particular equations that we consider here (see Theorem 2.5 below) are more general than the
ones usually addressed in optimal stopping literature and in fact ours reduce to the more
“standard” ones in the special case of a target measure µ having density with respect to
the Lebesgue measure. In this sense our first contribution is then to establish existence of
solutions for such broader class of equations. Secondly, the question of uniqueness of the
solution for the type of equations normally found in optimal stopping theory was a long
standing problem. It found a positive answer in the paper by Peskir [16] where uniqueness was proved in the class of continuous functions. Here we prove that such uniqueness
holds for a more general class of integral equations and, more interestingly, we prove that
the solution is indeed unique in the class of right-continuous functions, thus extending
Peskir’s previous results (note that our result obviously holds for all the cases covered by
Integral equations for Rost’s barriers
3
[16]). We remark that although the line of proof is based on a 4-step procedure as in [16],
the greater generality of our setting requires new arguments to prove each one of the four
steps so that our proof could not be derived by the original one in [16].
Before concluding this introduction we would like to point out that our work is closely
related to a recent one by Gassiat, Mijatović and Oberhauser [8]. There the authors
consider another solution of the Skorokhod embedding problem, namely Root’s solution
[19], which is also given in terms of a barrier set in time-space. The boundary of Root’s
barrier is expressed as a function of space x 7→ r(x) rather than a function of time. Under the assumption of atom-less target measures µ (as in our case) [8] prove that the
boundary r( · ) solves a Volterra-type equation which is reminiscent of the ones obtained
here. Uniqueness is provided via arguments based on viscosity theory in the special cases
of laws µ’s which produce continuous boundaries r( · ). This additional requirement excludes for instance continuous distributions singular with respect to the Lebesgue measure
(e.g. Cantor distribution) which are instead covered in our paper.
It is important to notice that although Rost’s and Root’s solutions are conceptually
related there seems to be no easy way of obtaining one from the other. Therefore even if
equations found here and those in [8] may look similar we could not establish any rigorous
link. Rost’s barriers enjoy nicer regularity properties than Root’s ones (e.g. monotonicity
and left continuity) and this allowed us to prove uniqueness of solution for the integral
equations even in the case of continuous singular measures µ. For the same reason the
Volterra equations that we obtain here are relatively easy to compute numerically; indeed
we can run our algorithm forward in time starting from known initial values s± (0) of the
boundaries at time zero. The equations obtained in [8] are instead more complicated as
there are no clear boundary conditions on r( · ) that can be used in general to initialise
the algorithm. In fact the numerical evaluation of the boundaries in [8] is restricted to the
class of µ’s that produce symmetric, continuous, monotonic on [0, ∞], maps x 7→ r(x).
Finally the existence results in our paper and in [8] are derived in completely different
ways.
The rest of the paper is organised as follows. In Section 2 we provide some background
material on Rost’s barriers, recall the link to a suitable optimal stopping problem and
then state our main result in Theorem 2.5. Section 3 is entirely devoted to the proof of
such theorem.
2
Setting, background material and results
In the initial part of this section we introduce the notation and some background material
needed to properly set our problem. In particular we briefly recall the results obtained in
[4] and [6] as these constitute the starting point of our analysis.
1. Let (Ω, F, P) be a probability space, B := (Bt )t≥0 a one dimensional standard
Brownian motion and denote (Ft )t≥0 the natural filtration of B augmented with P-nullsets. Throughout the paper we will equivalently use the notations Ef (Btx ) and Ex f (Bt ),
f : R → R, Borel-measurable, to refer to expectations under the initial condition B0 = x.
We also use the notation X ∼ µ for a random variable X with law µ.
Let µ and ν be probability measures on R. We denote Fµ (x) := µ((−∞, x]) and
Fν (x) := ν((−∞, x]) the (right-continuous) cumulative distributions functions of µ and ν.
Throughout the paper we make the following set of standing assumptions unless otherwise
Integral equations for Rost’s barriers
4
specified:
(A) Let a+ := sup{x ∈ R : x ∈ supp ν} and a− := − inf{x ∈ R : x ∈ supp ν}, then
0 ≤ a± < +∞;
(B) µ([−a− , a+ ]) = 0;
(C) Let µ+ := sup{x ∈ R : x ∈ supp µ} and µ− := − inf{x ∈ R : x ∈ supp µ},
then µ± ≥ 0 possibly infinite; if µ+ < ∞ (resp. µ− < ∞) then µ({µ+ }) = 0
(resp. µ({−µ− }) = 0); if µ+ = +∞ (resp. µ− = +∞) then there exists x > 0 such
that Fµ ∈ C(x, +∞) (resp. Fµ ∈ C(−∞, −x)), i.e. µ is always continuous locally at
the endpoints of its support;
(D.1) There exist numbers b̂+ ≥ a+ and b̂− ≥ a− such that (−b̂− , b̂+ ] is the largest interval
containing 0 with µ((−b̂− , b̂+ ]) = 0;
(D.2) If b̂+ = a+ (resp. b̂− = a− ) then ν({a+ }) > 0 (resp. ν({−a− }) > 0);
(D.3) µ({±b̂± }) = 0 (i.e. µ is continuous locally at b̂± ).
The above set of assumptions covers a large class of probability measures. It should be
noted in particular that in the canonical example of ν(dx) = δ0 (x)dx those conditions
hold for any µ such that µ({0}) = 0 and (C) is true.
Now we recall a particular result recently obtained by Cox and Peskir [4] in the context
of a more general work on Rost’s solutions of the Skorokhod embedding problem. Here
we give a different statement for notational convenience.
Theorem 2.1 (Cox & Peskir [4]). Let W ν := (Wtν )t≥0 be a standard Brownian motion
with initial distribution ν and let us only assume (A) and (B). There exists a unique couple
of left continuous, increasing functions s+ : R+ → R+ ∪{+∞} and s− : R+ → R+ ∪{+∞}
such that if we define
σ∗ := inf t > 0 : Wtν ∈
/ − s− (t), s+ (t)
(2.1)
then Wσν∗ ∼ µ.
Moreover, letting ∆s± (t) := s± (t+) − s± (t) ≥ 0, for any t ≥ 0 such that s± (t) < +∞
it also holds
∆s+ (t) > 0 ⇔ µ s+ (t), s+ (t+) = 0
(2.2)
∆s− (t) > 0 ⇔ µ − s− (t+), −s− (t) = 0.
(2.3)
Notice that the second part of the theorem asserts that jumps of the boundaries s±
may only occur in correspondence of intervals where µ charges no mass.
2. For 0 < T < +∞ and (t, x) ∈ [0, T ] × R we denote
Z x
G(x) :=2
Fν (z) − Fµ (z) dz
(2.4)
0
and introduce the following optimal stopping problem
V T (t, x) :=
sup
0≤τ ≤T −t
Ex G(Bτ )
(2.5)
Integral equations for Rost’s barriers
5
where the supremum is taken over all (Ft )-stopping times in [0, T − t]. Here the upper
index T in the definition of the value function is used to account for the time horizon
explicitly. As usual the continuation set CT and the stopping set DT of (2.5) are given by
CT := {(t, x) ∈ [0, T ] × R : V T (t, x) > G(x)}
(2.6)
DT := {(t, x) ∈ [0, T ] × R : V T (t, x) = G(x)}.
(2.7)
Throughout the paper we will often use the following notation: for a set A ⊂ [0, T ] × R
we denote A ∩ {t < T } := {(t, x) ∈ A : t < T }.
Problem (2.5) was addresses in [6] where it was shown that the geometry of the
continuation and stopping sets can be characterised in terms of the boundaries s± of
the reversed Rost’s barrier. Here we summarise those results in the following theorem
under the full set of assumptions (A)–(D.3)
Theorem 2.2 (De Angelis [6]). For each T > 0 it holds
CT = (t, x) ∈ [0, T ) × R : x ∈ − s− (T − t), s+ (T − t)
DT = (t, x) ∈ [0, T ) × R : x ∈ − ∞, −s− (T − t)] ∪ [s+ (T − t), +∞
and the smallest optimal stopping time of problem (2.5) is given by
τ∗T (t, x) = inf s ∈ [0, T − t] : (t + s, Bsx ) ∈ DT
(2.8)
(2.9)
(2.10)
for (t, x) ∈ [0, T ] × R. Moreover the functions s± fulfill the following additional properties:
s± (T − t) > a± for t ∈ [0, T ) and s± (0+) = b̂± .
A crucial observation follows from Theorem 2.1 (see also Corollary 4.3 of [6]) which
we summarise in the next
Corollary 2.3. If the measure µ is atom-less then the boundaries s+ and s− are strictly
monotone.
3. Now that we have introduced the necessary background material we can provide
the main result of this work, i.e. the characterisation of the boundaries s± as the unique
solution of suitable nonlinear integral equations of Volterra type. The proof of such result
will be then given in the next section.
For T > 0 it is convenient to introduce the function
U T (t, x) := V T (t, x) − G(x),
(t, x) ∈ [0, +∞) × R.
(2.11)
Applying Itô-Tanaka-Meyer formula to G inside the expectation of (2.5) we find an alternative representation for U T which will also be useful in the rest of the paper, i.e.
Z
T
U (t, x) = sup
Ex Lyτ (ν − µ)(dy)
(2.12)
0≤τ ≤T −t
R
where (Lys )s≥0 is the local time of B at a point y ∈ R. Let us also denote
(x−y)2
1
e− 2(s−t) ,
p(t, x, s, y) := p
2π(s − t)
for t < s, x, y ∈ R
(2.13)
Integral equations for Rost’s barriers
6
the Brownian motion transition density. As in Section 4 of [6] we introduce the (generalised) inverse of s± defined by


 inf{t ≥ 0 : −s− (t) < x}, x ≤ −s− (0)
0,
x ∈ (−s− (0), s+ (0))
ϕ(x) :=
(2.14)


inf{t ≥ 0 : s+ (t) > x},
x ≥ s+ (0)
Note that x ∈ (−s− (t), s+ (t)) if and only if ϕ(x) < t. Moreover ϕ is positive, decreasing left-continuous on R− and increasing right-continuous on R+ (hence upper semicontinuous).
Remark 2.4. In the special case of a law µ with no atoms we can actually invert s± (see
Corollary 2.3) and obtain the simpler definition

−1

 (−s− ) (x), x ≤ −s− (0)
0,
x ∈ (−s− (0), s+ (0))
ϕ(x) :=
(2.15)


(s+ )−1 (x),
x ≥ s+ (0).
The next is the paper’s main theorem.
Theorem 2.5. Assume that µ is atom-less. Then one has:
i) for any T > 0 the following equivalent representations of (2.11) hold
Z T Z s+ (T −u)
T
U (t, x) =
p(t, x, u, y) ν − µ (dy) ds
t
T
Z
U (t, x) =
(2.16)
−s− (T −u)
1{y : ϕ(y)<T −t} Ex LyT −t−ϕ(y) ν − µ (dy)
(2.17)
R
for (t, x) ∈ [0, T ] × R;
ii) the couple (s+ , s− ) is the unique couple of left-continuous, increasing, positive functions fulfilling (2.2) and (2.3) that solve the system
Z T Z s+ (T −u)
p(t, ±s± (T − t), u, y) ν − µ (dy) du = 0, t ∈ [0, T )
(2.18)
t
−s− (T −u)
with initial conditions s± (0) = b̂± . Equivalently we may express (2.18) in terms of
ϕ as
Z
1{y : ϕ(x)>ϕ(y)} Ex Lyϕ(x)−ϕ(y) ν − µ (dy) = 0, x ∈ R.
(2.19)
R
3
Proof of the main theorem
In this section we provide the proof of Theorem 2.5. This requires to show first that the
value function V T of (2.5) is C 1 in the whole space.
1. It was already shown in [6], and it was one of the paper’s main technical results, that
the time-derivative VtT of the value function is indeed continuous on [0, T ) × R. However
Integral equations for Rost’s barriers
7
it was also mentioned there that when µ has atoms the continuity of VxT (t, ·) could break
down across the boundaries s± (let alone continuity of VxT (·, ·)). For this reason here we
focus only on continuous target measures µ.
As a first technical step we want to extend some continuity results regarding the
mappings (t, x) 7→ τ∗T (t, x) (see (2.10)) which were partly provided in Lemma 3.6 and
Corollaries 3.7 and 3.8 of [6]. Since the proof seems fairly standard we give it in appendix
for completeness.
Lemma 3.1. Fix T > 0, let (t, x) ∈ CT and h > 0 such that (t, x + h) ∈ CT and let τ∗T be
as in (2.10). Then it holds
lim τ∗T (t, x + h) = τ∗T (t, x),
P − a.s.
h→0
(3.1)
We also summarise below the results of Lemma 3.6 and Corollary 3.7 of [6] which will
be needed in the following Proposition.
Lemma 3.2. For any (t, x) ∈ ∂CT ∩ {t < T } and any sequence (tn , xn )n ⊂ CT such that
(tn , xn ) → (t, x) as n → ∞ it holds
lim τ∗T (tn , xn ) = 0,
n→0
P − a.s.
(3.2)
Finally we provide C 1 regularity of V T in the next
Proposition 3.3. Assume that µ is atom-less. Then for any T > 0 it holds V T ∈
C 1 ([0, T ) × R).
Proof. (a). As already mentioned one can check [6, Prop. 3.11] for VtT ∈ C([0, T ) × R).
(b). Here we prove that VxT ∈ C([0, T ) × R). The claim is obvious in the interior
of the continuation set CT where V T ∈ C 1,2 and in the interior of the stopping set DT
where V T = G. Then we must only consider points along the two boundaries. Let
(t, x) ∈ ∂CT ∩ {t < T } be arbitrary but fixed and let (tn , xn )n ⊂ CT be a sequence
of points converging to (t, x) as n → ∞. Fix an arbitrary n ∈ N and for simplicity
denote τn := τ∗T (tn , xn ) (see (2.10)) and τn,h := τ∗T (tn , xn + h) for h > 0 such that
(tn , xn + h) ∈ CT . From an application of the mean value theorem, noting that τn and τn,h
are optimal stopping times respectively for the initial conditions (tn , xn ) and (tn , xn + h)
it follows
i
V T (tn , xn + h) − V T (tn , xn )
1 h
≥ E G(Xτxnn +h ) − G(Xτxnn ) = EG0 (ξ1h )
(3.3)
h
h
i
V T (tn , xn + h) − V T (tn , xn )
1 h
n +h
n
≤ E G(Xτxn,h
) − G(Xτxn,h
) = EG0 (ξ2h )
(3.4)
h
h
n
n +h
where ξ1h ∈ (Xτxnn , Xτxnn +h ) and ξ2h ∈ (Xτxn,h
, Xτxn,h
), P-a.s. From continuity of the sample
xn +h
n +h
paths and Lemma 3.1 we get that Xτn
and Xτxn,h
converge to Xτxnn as h → 0, P-a.s.,
hence ξ1h , ξ2h → Xτxnn as h → 0 as well. The latter limits, dominated convergence (G is
Lipschitz) and (3.3) and (3.4) allow us to conclude
VxT (tn , xn )
V T (tn , xn + h) − V T (tn , xn )
= EG0 (Xτxnn ).
= lim
h→0
h
(3.5)
Integral equations for Rost’s barriers
8
Now we use continuity of sample paths and Lemma 3.2 to obtain Xτxnn → x as n → ∞.
Hence another application of dominated convergence and (3.5) give
lim VxT (tn , xn ) = G0 (x)
(3.6)
n→∞
and from arbitrariness of the point (t, x) and of the sequence (tn , xn ) we conclude that
VxT is continuous across the boundaries and hence everywhere in [0, T ) × R.
2. It now follows from the above Proposition and standard arguments based on strong
Markov property (see for instance [17, Sec. 7.1]) that V T ∈ C 1,2 in CT and if µ is continuous
then V T is globally C 1 in [0, T ) × R and it solves
T
VtT + 12 Vxx
(t, x) = 0,
(t, x) ∈ CT
(3.7)
V T (t, x) = G(x),
(t, x) ∈ ∂CT
(3.8)
VxT (t, x) = G0 (x),
(t, x) ∈ ∂CT ∩ {t < T }
(3.9)
VtT (t, x) = 0,
(t, x) ∈ ∂CT ∩ {t < T }
(3.10)
x ∈ R.
(3.11)
V T (T, x) = G(x),
Notice that CT is expressed in terms of the curves s± (T − ·) as in (2.8). For U T = V T − G
(see (2.11)), under the assumption that µ is continuous we have U T ∈ C 1 ([0, T ) × R) and
the above boundary value problem becomes
T
(t, x) = −(ν − µ)(dx),
(t, x) ∈ CT
(3.12)
UtT + 12 Uxx
U T (t, x) = 0,
UxT (t, x) = UtT (t, x) = 0,
U T (T, x) = 0,
(t, x) ∈ ∂CT
(3.13)
(t, x) ∈ ∂CT ∩ {t < T }
(3.14)
x∈R
(3.15)
where (3.12) holds in the sense of distributions.
3. We can now prove Theorem (2.5).
Proof of Theorem 2.5. The proof is divided in two main blocks. Firstly we obtain the
integral representation (2.16) for U T which easily yields the integral equations for the
boundaries s± . Secondly we deal with the issue of uniqueness of the solution to the
system (2.18).
For the sake of this proof it is convenient to denote b± (t) := s± (T − t) for t ∈ [0, T ]
and use b± instead of s± . Similarly we drop the apex T and denote U = U T .
(a) Existence. The representation (2.16) for U T is obtained by an application of an
extension of Itô-Tanaka-Meyer formula to the time-space framework which is inspired by
[1]. However since the result of [1, Prop. 4] cannot be applied directly to our setting we
provide a full proof for completeness.
We start by observing that for each t ∈ [0, T ) the map x 7→ Ux (t, x) may be seen as the
cumulative distribution function of a signed measure on R with support on [−b− (t), b+ (t)].
In fact we set Uxx (t, (a, b]) := Ux (t, b) − Ux (t, a) for any interval (a, b) ⊂ R (recall that
Integral equations for Rost’s barriers
9
Fν is only right continuous) and we notice that due to (3.14) Uxx (t, {±b± (t)}) = 0 for all
t ∈ [0, T ), i.e. Uxx (t, dx) has no atoms at the boundary points b± (t). It then follows
Uxx (t, dx) = 1{x∈(−b− (t),b+ (t))} Vxx (t, x)dx − 2(ν − µ)(dx) , (t, x) ∈ [0, T ) × R. (3.16)
We continuously extend U to R2 by taking U (t, x) = U (0, x) for t < 0 and U (t, x) =
U (T, x) = 0 for t > T . Then we pick positive functions f, g ∈ Cc∞ (R) such that
Z
Z
f (x)dx =
g(x)dx = 1
R
R
and for n ∈ N we set fn (x) := nf (nx), gn (x) = ng(nx). We use standard mollification to
obtain a double indexed sequence of functions (Un,m )n,m∈N2 ⊂ Cc∞ (R2 ) defined by
Z Z
U (s, y)fn (x − y)gm (t − s) dy ds, (t, x) ∈ R2 .
(3.17)
Un,m (t, x) :=
R
R
To avoid technicalities related to the endpoints of the time domain fix (t, x) ∈ (ε, T −ε)×R,
ε > 0 arbitrary, then by applying classical Dynkin’s formula for each n and m we find the
expression
Z T −ε
h
i
t,x
t,x
∂
1 ∂2
Un,m (t, x) = E Un,m (T − ε, BT −ε ) −
U
+
U
(s,
B
)ds
.
(3.18)
s
∂ t n,m
2 ∂ x2 n,m
t
By the standard properties of mollifiers it is not
taking limits as n, m → ∞
hard to see that
t,x
and as ε → 0 one has Un,m (t, x) → U (t, x), E Un,m (T − ε, BT −ε )] → E U (T, BTt,x )] = 0.
It only remains to analyse the integral term. Note that when extending U to R2 we
have implicitly taken b± (t) = b± (0) for t < 0 and b± (t) = 0 for t > T . Note as well that
there is no loss of generality assuming that the index m in (3.18) is large enough to have
gm supported in (−ε, +ε). Then for (s, z) ∈ (t, T − ε) × R it holds
∂2
U
∂ x2 n,m
(s, z) =
Z Z
R
∂
U
∂ t n,m
(s, z) =
Z
b+ (v)
−b− (v)
Z b+ (v)
U (v, y)fn00 (z − y)gm (s − v) dy dv
(3.19)
0
U (v, y)fn (z − y)gm
(s − v) dy dv
(3.20)
−b− (v)
R
where the integrals in dv must only be taken for v ∈ (s − ε, s + ε) ⊂ (0, T ).
Use integration by parts twice, the smooth-fit (3.14) and (3.16) to obtain
∂2
U
∂ x2 n,m
(s, z) =
Z Z
R
=
fn (z − y)Uxx (v, dy) gm (s − v) dv
(3.21)
−b− (v)
Z Z
R
b+ (v)
b+ (v)
fn (z − y) Vxx (v, y)dy − 2(ν − µ)(dy) gm (s − v) dv
−b− (v)
and similarly
∂
U
∂ t n,m
Z Z
b+ (v)
Ut (v, y)fn (z − y)gm (s − v) dy dv.
(s, z) =
R
−b− (v)
(3.22)
Integral equations for Rost’s barriers
10
Since Ut = Vt and (3.7) holds, then adding up (3.21) and (3.22) (where now z = Bst,x ) the
above gives
h Z T −ε
i
t,x
∂
1 ∂2
E
U
+ 2 ∂ x2 Un,m (s, Bs )ds
(3.23)
∂ t n,m
t
h Z T −ε n Z Z b+ (v)
o i
=−E
fn (Bst,x − y)(ν − µ)(dy) gm (s − v) dv ds
t
R
T −εn Z
Z Z
=−
t
R
Z
−b− (v)
b+ (v)
o
fn (z − y)(ν − µ)(dy) gm (s − v) dv p(t, x, s, z)ds dz.
−b− (v)
R
Recall that supp gm ⊂ (−ε, +ε). For (s, z) ∈ (t, T − ε) × R we set
Z
b+ (v)
fn (z − y)(ν − µ)(dy),
Λn (v, z) :=
v ∈ (s − ε, s + ε)
(3.24)
−b− (v)
Z
Λn (v, z)gm (s − v)dv
Λn,m (s, z) :=
(3.25)
R
so that we rewrite (3.23) in the form
E
hZ
T −ε
∂
U
∂ t n,m
t
+
1 ∂2
U
2 ∂ x2 n,m
i
t,x
(s, Bs )ds = −
Z Z
T −ε
Λn,m (s, z)p(t, x, s, z)ds dz.
R
t
(3.26)
For fixed n and for any z ∈ R and s ∈ (t, T − ε) the map v 7→ Λn (v, z) is bounded
and continuous on (s − ε, s + ε) since µ is flat at possible jumps of b± (cf. (2.2), (2.3) of
Theorem 2.1 and recall that b± (·) = s± (T − ·)). Notice also that gm (s − · ) converges
weakly to δ(s − ·) as m → ∞ and hence for all z ∈ R and s ∈ (t, T − ε) we get
lim Λn,m (s, z) = Λn (s, z).
m→∞
(3.27)
For
fixed n
it holds kfn k ≤ Kn for suitably large Kn > 0 and hence it also holds
Λn,m (s, z) ≤ 2Kn . We keep n fixed and take limits as m → ∞, then dominated convergence and (3.27) give
Z Z
lim
m→∞
T −ε
Z Z
Λn (s, z)p(t, x, s, z)ds dz.
Λn,m (s, z)p(t, x, s, z)ds dz =
R
t
T −ε
R
(3.28)
t
In order to take limits as n → ∞ we analyse the the right-hand side of the expression
above. In particular we recall from (2.14) that
y ∈ (−b− (s), b+ (s)) ⇔ y ∈ (−s− (T − s), s+ (T − s)) ⇔ ϕ(y) < T − s.
(3.29)
Integral equations for Rost’s barriers
11
Hence Fubini’s theorem and the occupation formula for local time give
Z Z T −ε
Λn (s, z)p(t, x, s, z)ds dz
R t
h Z T −εZ b+ (s)
i
=Et,x
fn (Bs − y)(ν − µ)(dy) ds
t
Z
=
Et,x
hZ
=
−b− (s)
T −ε
i
1{s<T −ϕ(y)} fn (Bs − y)ds (ν − µ)(dy)
t
R
Z
=
(3.30)
Et,x
R
Z Z
R
hZ
(T −ε)∧([T −ϕ(y)]∨t)
i
fn (Bs − y)ds (ν − µ)(dy)
t
fn (z − y)Et,x Lz(T −ε)∧([T −ϕ(y)]∨t) dz (ν − µ)(dy)
R
where we recall that (Lzs )s≥0 is the local time of the Brownian motion at a point z ∈ R.
With no loss of generality we may assume that there exists a compact
K ⊂ R such
z
that supp fn ⊂ K for all n. For fixed t and x the quantity Et,x L(T −ε)∧([T −ϕ(y)]∨t) is
bounded uniformly with respect to all y, z ∈ R such that y − z ∈ K (use for example
Itô-Tanaka’s formula). Moreover, for fixed t, x and y the map z 7→ Et,x Lz(T −ε)∧([T −ϕ(y)]∨t)
is bounded and continuous for z ∈ R such that z − y ∈ K. Finally, for each y ∈ R one
has fn (· − y) → δ(· − y) weakly as n → ∞ and hence the above expression and dominated
convergence (with respect to the measure ν − µ) give
Z Z T −ε
Z
Λn (s, z)p(t, x, s, z)ds dz =
Et,x Ly(T −ε)∧([T −ϕ(y)]∨t) (ν − µ)(dy). (3.31)
lim
n→∞
R
t
R
Now, collecting (3.18), (3.26), taking limits as m → ∞ first, then as n → ∞ and
finally as ε → 0 and using (3.31) we obtain
Z
Z
y
U (t, x) = Et,x L[T −ϕ(y)]∨t (ν − µ)(dy) =
1{y:ϕ(y)<T −t} Et,x LyT −ϕ(y) (ν − µ)(dy)
R
R
(3.32)
Z
=
1{y:ϕ(y)<T −t} Ex LyT −t−ϕ(y) (ν − µ)(dy).
R
The equivalent expression (2.16) can be obtained by recalling the well known representation for the expectation of the Brownian local time
Z T −ϕ(y)
y
Ex LT −t−ϕ(y) =
p(t, x, s, y)ds.
(3.33)
t
Then by using once again (3.29) we obtain
Z Z T
U (t, x) =
1{s<T −ϕ(y)} p(t, x, s, y)ds (ν − µ)(dy)
R
t
Z Z T
=
1{y∈(−b− (s),b+ (s))} p(t, x, s, y)ds (ν − µ)(dy)
R
(3.34)
t
and hence (2.16) follows by a simple application of Fubini’s theorem and by putting
b± (t) = s± (T − t).
Integral equations for Rost’s barriers
12
The system (2.18) of integral equations is now easily obtained by taking x = ±s± (T −
t), t < T in (2.16) and observing that U (t, ±s± (T − t)) = 0.
(b) Uniqueness. We prove uniqueness of the solution to (2.18) (equivalently to (2.19))
via a contradiction argument. Let us assume that there exists a couple of functions
(r− , r+ ) with the following properties: r± : R+ → R+ ∪ {+∞}, strictly increasing, leftcontinuous, such that µ does not charge jumps of r± (see (2.2) and (2.3)), r± (T − t) > a±
for t ∈ [0, T ) and such that (r− , r+ ) solves (2.18) with terminal conditions r± (0+) = b̂± .
As in (2.14) we can define a generalised inverse of r± and we denote it by ψ : R → [0, ∞).
Again it is convenient to consider a time reversal and denote c± (t) := r± (T − t),
t ∈ [0, T ]. Motivated by (2.17) let us define a function
Z
c
U (t, x) :=
Ex Ly(T −t−ψ(y))+ (ν − µ)(dy), (t, x) ∈ [0, T ] × R
(3.35)
R
and notice that for ψ = ϕ we simply have U c = U (observe that 1{y:ψ(y)<T −t} LyT −t−ψ(y) =
Ly(T −t−ψ(y))+ ). By assumption (c− , c+ ) solves (2.18) and hence
U c (t, ±c± (t)) = 0,
t ∈ [0, T ), and U c (T, x) = 0, x ∈ R.
(3.36)
Using that Ex Ly(T −t−ψ(y))+ = Ex B(T −t−ψ(y))+ − y − |x − y| it is not difficult to prove
that U c ∈ C([0, T ] × R). Now we want to show that the process Y c defined as
Z
c
c
Ys := U (t + s, Bs ) + Lys∧(T −t−ψ(y))+ (ν − µ)(dy), s ∈ [0, T − t]
(3.37)
R
is a Px -martingale for any x ∈ R and t ∈ [0, T ). By the definition (3.35) it follows
i
hZ
c
(3.38)
1{t+s<T −ψ(y)} EBs Ly(T −(t+s)−ψ(y)) (ν − µ)(dy)
Ex U (t + s, Bs ) = Ex
R
and by strong Markov property
h
i EBs Lyu = Ex Bs+u − y Fs − Bs − y = Ex Lys+u Fs − Lys − Ms ,
Px − a.s.
(3.39)
for all u ≥ 0 and with (Ms )s≥0 a Px -martingale. In particular for u = T − (t + s) − ψ(y),
by Fubini’s theorem and (3.39) we obtain
Z
c
Ex U (t + s, Bs ) =
1{s<T −t−ψ(y)} Ex Ly(T −t−ψ(y)) − Lys (ν − µ)(dy).
(3.40)
R
Now we add and subtract U c (t, x) from the right-hand side of the above expression, use
(3.35) and simplify the indicator variables to get
Ex U c (t + s, Bs ) =U c (t, x)
Z
+
1{s<T −t−ψ(y)} − 1{0<T −t−ψ(y)} Ex Ly(T −t−ψ(y)) (ν − µ)(dy)
R
Z
(3.41)
− 1{s<T −t−ψ(y)} Ex Lys (ν − µ)(dy)
R
Integral equations for Rost’s barriers
13
Z
=U (t, x) − 1{0<T −t−ψ(y)≤s} Ex Ly(T −t−ψ(y)) (ν − µ)(dy)
R
Z
− 1{s<T −t−ψ(y)} Ex Lys (ν − µ)(dy)
R
Z
c
=U (t, x) − Ex Lys∧(T −t−ψ(y))+ (ν − µ)(dy).
c
(3.42)
R
Therefore the process Y c is a continuous martingale as claimed and from now on we will
refer to this property as to the martingale property of U c . Further, we notice that in the
particular case of c± = b± we obtain what we will refer to as the martingale property of
U . Then for any (t, x) ∈ [0, T ] × R and any stopping time τ ∈ [0, T − t] it holds
Z
i
h
y
(3.43)
U (t, x) = Ex U (t + τ, Bτ ) + Lτ ∧(T −t−ϕ(y))+ (ν − µ)(dy) ,
R
Z
h
i
U c (t, x) = Ex U c (t + τ, Bτ ) + Lyτ ∧(T −t−ψ(y))+ (ν − µ)(dy) .
(3.44)
R
We now proceed in four steps inspired by Peskir [16]. The equations considered by
Peskir can be seen as a special case of (2.18) and (2.19) and here we obtain uniqueness
in a stronger sense, i.e. in the class of right-continuous functions rather than continuous
ones as in [16].
Figure 1:
(b.1). First we show that
U c (t, x) = 0,
for t ∈ [0, T ), x ∈ (−∞, −c− (t)] ∪ [c+ (t), ∞).
(3.45)
We fix t ∈ [0, T ) and with no loss of generality it is sufficient to consider x > c+ (t)
since the proof for x < −c− (t) follows from analogous arguments. We set τ := inf{s ∈
(0, T − t] : Bs ≤ c+ (t + s)} and note that since s 7→ Bs (ω) − c+ (t + s) is positive for
s < τ , right-continuous, with only upwards jumps, then Bs (ω) − c+ (t + s) must cross
zero in a continuous way and it must be Bτ = c+ (t + τ ), Px -a.s. Then (3.36) implies
Integral equations for Rost’s barriers
14
U c (t + τ, Bτ ) = 0, Px -a.s. and it follows from (3.41) that
Z
c
U (t, x) =
Ex Lyτ ∧(T −t−ψ(y))+ (ν − µ)(dy).
(3.46)
R
The support of y 7→ Lyτ ∧(T −t−ψ(y))+ is contained in [b̂+ , c+ (t)), Px -a.s. by definition of τ
and of the set {z ∈ R : ψ(z) < T − t} and by recalling that c+ (T ) = b̂+ . Since c+ is
strictly monotonic decreasing, for any y > 0 it is not hard to see that:
i) if τ (ω) ≤ T − t − ψ(y), then Bs (ω) > y for s ≤ τ (ω);
ii) if τ (ω) > T − t − ψ(y), then Bs (ω) > y for s < T − t − ψ(y);
hence Lyτ ∧(T −t−ψ(y))+ = 0, Px -a.s. and (3.45) follows (see also Figure 1). Since U is non
negative by definition (see (2.11)) then it also follows
0 = U c (t, x) ≤ U (t, x),
for t ∈ [0, T ), x ∈ (−∞, −c− (t)] ∪ [c+ (t), ∞).
(3.47)
(b.2). Next we prove that U c ≤ U on [0, T ] × R. Since U c (T, x) = U (T, x), x ∈ R and
(3.47) holds it only remains to consider t ∈ [0, T ) and x ∈ (−c− (t), c+ (t)). Fix any such
(t, x) and denote σ := inf{s ∈ [0, T − t) : Bs ≤ −c− (t + s) or Bs ≥ c+ (t + s)}. Then from
(2.12) and the martingale property of U c we obtain
Z
c
U (t, x) − U (t, x) ≥ Ex Lyσ − Lyσ∧(T −t−ψ(y))+ (ν − µ)(dy)
(3.48)
RZ
≥ − Ex Lyσ − Lyσ∧(T −t−ψ(y))+ µ(dy)
R
since u 7→ Lyu (ω) is increasing for each y ∈ R. We claim that
Lyσ = Lyσ∧(T −t−ψ(y))+ , Px -a.s. for y ∈ {z ∈ R : ψ(z) < T − t} = (−c− (t), c+ (t)).
(3.49)
The latter is trivial if σ(ω) ≤ T − t − ψ(y). Consider instead σ(ω) > T − t − ψ(y) and with
no loss of generality assume y > 0; then strict monotonicity of c+ implies that Bs (ω) < y
for s ∈ (T − t − ψ(y), σ(ω)] (see also Figure 1) and the claim holds. The case of y > 0
follows from symmetric arguments relative to c− (·).
From (3.49) and observing that y 7→ Lyσ is supported in (−c− (t), c+ (t)), Px -a.s., we
conclude that the right hand side of (3.48) equals zero and U c ≤ U .
(b.3). Now we aim at proving that c± (t) ≤ b± (t) for t ∈ [0, T ). Let us assume that
there exists t ∈ [0, T ) such that c+ (t) > b+ (t) and take x > c+ (t). We denote τ 0 := inf{s ∈
[0, T − t) : Bs < b+ (t + s)} and from the same arguments as those used in (b.1) above we
have Bτ 0 = b+ (t + τ 0 ) and U (t + τ 0 , Bτ 0 ) = 0, Px -a.s. Since x > c+ (t) > b+ (t) we have also
U c (t, x) = U (t, x) = 0 and from (b.2) above it follows 0 = U (t + τ 0 , Bτ 0 ) ≥ U c (t + τ 0 , Bτ 0 ),
Px -a.s. The martingale property of U c and U now give (see (3.43) and (3.44))
Z
Ex Lyτ 0 ∧(T −t−ψ(y))+ − Lyτ 0 ∧(T −t−ϕ(y))+ (ν − µ)(dy) ≥ 0.
R
Integral equations for Rost’s barriers
15
Due to monotonicity of b+ , an argument similar to the one that allowed us to show that
the right-hand side of (3.46) vanishes, here implies Lyτ 0 ∧(T −t−ϕ(y))+ = 0, Px -a.s. Therefore
from the last inequality it follows
Z
Z
y
0≤
Ex Lτ 0 ∧(T −t−ψ(y))+ (ν − µ)(dy) = − Ex Lyτ 0 ∧(T −t−ψ(y))+ µ(dy) ≤ 0,
(3.50)
R
where we have also used that y ∈ supp ν ⇒
for s ≤ τ 0 (recall that b+ (·) > a+ in [0, T )).
Now (3.50) implies that
R
y
Lτ 0 ∧(T −t−ψ(y))+
= 0, Px -a.s. since x + Bs ≥ a+
for µ-a.e. y ∈ R, Lyτ 0 ∧(T −t−ψ(y))+ = 0, Px -a.s.
(3.51)
Since y 7→ Lyτ 0 ∧(T −t−ψ(y))+ is supported in [b̂+ , c+ (t)), Px -a.s. and x > c+ (t), then (3.51)
also implies that
τ 0 ∧ (T − t − ψ(y))+ ≤ ζy , Px -a.s., for µ-a.e. y ∈ R
(3.52)
with ζy := inf{s ≥ 0 : Bs = y}. Next we show that (3.52) leads to a contradiction.
Let I ⊂ (b+ (t), c+ (t)) be an open interval. For any y ∈ I, monotonicity of b+ implies
that τ 0 ≥ ζy , Px -a.s. and therefore it follows from (3.52) that for µ-a.e. y ∈ I it must be
(T − t − ψ(y))+ ≤ ζy , Px -a.s. However, for any fixed y ∈ I, by right-continuity of c+ there
must exist ε > 0 such that c+ (t + s) > y for s ∈ [0, ε] and equivalently ψ(y) < T − (t + ε).
For such y and ε this leads to (T − t − ψ(y))+ ≤ ζy ⇒ ζy > ε, Px -a.s., which is clearly
impossible since Px (ζy ≤ ε) > 0. Therefore c+ (t) ≤ b+ (t) for all t ∈ [0, T ).
To prove that c− (t) ≤ b− (t), t ∈ [0, T ) we assume that there exists t ∈ [0, T ) such
that c− (t) > b− (t); then we pick x < −c− (t) and follow arguments as above to show a
contradiction.
(b.4). At this point we prove c± = b± . Again it is enough to give the full argument
only for the upper boundaries as the case of the lower ones is similar.
We assume that there exists t ∈ [0, T ) such that c+ (t) < b+ (t), then by definition
of b+ there must exists δ > 0 and x ∈ (c+ (t), b+ (t)) such that U (t, x) ≥ δ. We denote
σ 0 := inf{s ∈ [0, T − t) : Bs ≤ −b− (t + s) or Bs ≥ b+ (t + s)} and we recall that σ 0 is
optimal for U (t, x) (i.e. σ 0 = τ∗ as in (2.10)), so that (2.12) gives
Z
U (t, x) =
Ex Lyσ0 (ν − µ)(dy).
(3.53)
R
Since c+ ≤ b+ (from (b.3) above) it follows that U c (t, x) = 0 and U c (t + σ 0 , Bσ0 ) = 0,
Px -a.s. by (3.45). Then subtracting (3.44) from (3.53) gives
Z
(3.54)
δ ≤ Ex Lyσ0 − Lyσ0 ∧(T −t−ψ(y))+ (ν − µ)(dy)
R
Z
= Ex 1{σ0 ≥(T −t−ψ(y))+ } Lyσ0 − Ly(T −t−ψ(y))+ (ν − µ)(dy).
R
We observe that for y ∈ supp ν one has ψ(y) = 0 (recall c± (t) > a± , t ∈ [0, T )) and then
for any such y and Px -a.e. ω ∈ {σ 0 ≥ T − t − ψ(y)} we have σ 0 (ω) = T − t − ψ(y). The
latter implies
Z
δ ≤ − Ex 1{σ0 ≥(T −t−ψ(y))+ } Lyσ0 − Ly(T −t−ψ(y))+ µ(dy) ≤ 0
(3.55)
R
which leads to a contradiction.
Integral equations for Rost’s barriers
16
4. We conclude this section with some diagrams (see Figures 2 and 3) relative to
boundaries obtained by numerical evaluation of (2.18). We rely upon a simple algorithm
based on a discretisation of the time integral in (2.18) and a (forward) recursive scheme
initialised by the known values s± (0) = b̂± (for details one may refer to Remark 4.2 in
[7]). The accuracy of the resulting boundaries depends on the size h of the intervals used
to discretise the time integral. In all the examples we consider for simplicity a Brownian
motion started from zero at time zero, i.e. ν(dx) = δ0 (x)dx.
(Left) Fµ0 (x) =
0.5 1[0,2] (x), h = 5 ×
(dashed line) and
= 0.5 1[0,0.4] (x) + 0.5 1[0.6,2.2] (x), h = 10−4
(solid line). (Right) Fµ0 (x) = 0.5 1[0,0.4] (x) + 0.5 1[0.6,2.2] (x), h = 4 × 10−3 (solid line),
h = 2 × 10−3 (dashed line), h = 2 × 10−4 (fine dashed line); as expected the sharpness of
the jump increases as h ↓ 0.
Figure 2: Boundary of Rost’s barriers for uniform-like distributions.
10−3
Fµ0 (x)
Figure 3: Boundary of Rost’s barriers: (Left) µ = N (m, σ 2 ) with mean m = 1 and variance
σ 2 = 1 (h = 10−3 ). (Right) µ = Exp(λ), h = 2 × 10−3 , λ = 1.5 (solid line), λ = 2.5 (fine
dashed line), λ = 3.5 (dashed line).
Integral equations for Rost’s barriers
4
17
Appendix
Proof of Lemma 3.1. For simplicity it is convenient to denote b± (t) := s± (T − t) for
t ∈ [0, T ] and use b± instead of s± . Similarly we simplify the notation for the stopping
times by setting τ := τ∗T (t, x) and τh := τ∗T (t, x + h). Fix an arbitrary ω ∈ Ω, then there
are two possibilities: either τ (ω) < T − t or τ (ω) = T − t.
i) We start by considering τ (ω) < T − t and again we have two subcases: either
Xτ hits the upper boundary or it hits the lower boundary. Assume first that Xτx (ω) ≥
b+ (t + τ (ω)), then by monotonicity of sample paths with respect to the initial point
Xτx+h (ω) ≥ b+ (t + τ (ω)) for all h > 0; hence τh (ω) ≤ τ (ω). From the same argument it
is easily verified that τh (ω) is increasing as h ↓ 0 with τh (ω) ↑ τ0 (ω) ≤ τ (ω) as h ↓ 0.
Moreover one has Xτx+h
(ω) ≥ b+ (t + τh (ω)) for all h > 0 and taking limits as h → 0 we get
h
x
Xτ0 (ω) ≥ b+ (t + τ0 (ω) −) where b+ (s−) denotes the left-limit of b+ . Since the boundary is
monotonic decreasing we conclude Xτx0 (ω) ≥ b+ (t + τ0 (ω)) and hence τ0 (ω) ≥ τ (ω) which
also implies τh (ω) → τ (ω) as h → 0.
Now we consider the case when Xτ hits the lower boundary, i.e. Xτx (ω) ≤ −b− (t+τ (ω)).
In this setting we have
d(ω) := inf b+ (t + s) − Xsx (ω) > 0 ⇒ inf b+ (t + s) − Xsx+h (ω) > 0
(4.1)
0≤s≤τ
0≤s≤τ
X·x+h (ω)
X·x (ω)
as well, we then have τh (ω) ≥ τ (ω) for all
>
for all h < d(ω). Since
h < d(ω). It then follows that in the limit one has
τ 0 (ω) := lim sup τh (ω) ≥ τ (ω).
(4.2)
h→0
Let us assume τ 0 (ω) > τ (ω), then there exists εω > 0 such that τ 0 (ω) ≥ τ (ω) + 3εω and a
subsequence (τhk (ω))k with hk → 0 as k → ∞, such that τhk (ω) > τ 0 (ω)−2εω ≥ τ (ω)+εω
for all k. Since this bound holds for all the elements of the subsequence and X x+hk is
decreasing in k then it must also hold
x+h
x
δω := inf
inf0
Xs k + b− (t + s) (ω) ≥
inf0
Xs + b− (t + s) (ω) > 0
k
0≤s≤τ −εω
0≤s≤τ −εω
(4.3)
with δω independent of hk . Now taking limits as k → ∞ and hk → 0 we reach the
following contradiction
x
δω ≤
inf0
Xs + b− (t + s) (ω) ≤ inf Xsx + b− (t + s) = 0,
(4.4)
0≤s≤τ −εω
0≤s≤τ
where in the second inequality we have used that τ 0 (ω) − εω > τ (ω). Therefore it must
be τ 0 (ω) = τ (ω) and also in this case τh (ω) → τ (ω) as h → 0.
ii) It only remains to consider the case τ (ω) = T −t. If b̂+ = −b̂− = 0 (see Assumption
D.1) then b± (T ) = 0 and ω ∈ {τ = T − t} ⊂ {XTx −t = 0} where the latter is a P-null
set. Then we consider the non trivial case b̂+ > −b̂− in which τ (ω) = T − t implies
Xτx (ω) ∈ (−b̂− , b̂+ ) and
x
Xs + b− (t + s) > 0.
(4.5)
d(ω) := inf
b+ (t + s) − Xsx ∧ inf
0≤s≤T −t
0≤s≤T −t
The latter easily implies that τh (ω) = T − t for all h < d(ω) and τh (ω) → τ (ω) as
h → 0.
Acknowledgments: The first named author was funded by EPSRC grant EP/K00557X/1.
Integral equations for Rost’s barriers
18
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