A Direct Method for Solving Optimal Switching Problems of
One-Dimensional Diffusions
Masahiko Egami
Abstract
In this paper, we propose a direct solution method for optimal switching problems of one-dimensional
diffusions. This method is free from conjectures about the form of the value function and switching
strategies, or does not require the proof of optimality through quasi-variational inequalities. The direct
method uses a general theory of optimal stopping problems for one-dimensional diffusions and charac-
terizes the value function as sets of the smallest linear majorants in their respective transformed spaces.
1 Introduction
Stochastic optimal switching problems (or starting and stopping problems) are important subjects both in
mathematics and economics. Since there are numerous articles about real options in the economic and
financial literature in recent years, the importance and applicability of control problems including optimal
switching problems cannot be exaggerated.
A typical optimal switching problem is described as follows: The controller monitors the price of natural
resources for optimizing (in some sense) the operation of an extraction facility. She can choose when to start
extracting this resource and when to temporarily stop doing so, based upon price fluctuations she observes.
The problem is concerned with finding an optimal switching policy and the corresponding value function. A
number of papers on this topic are well worth mentioning : Brennan and Schwarz (1985) in conjunction with
convenience yield in the energy market, Dixit (1989) for production facility problems, Brekke and Øksendal
(1994) for resource extraction problems, Yushkevich (2001) for positive recurrent countable Markov chain,
and Duckworth and Zervos (2001) for reversible investment problems. Hamdadène and Jeanblanc (2004)
analyze a general adapted process for finite time horizon using reflected stochastic backward differential
equations. Carmona and Ludkovski (2005) apply to energy tolling agreement in a finite time horizon using
Monte-Carlo regressions.
A basic analytical tool for solving switching problems is quasi-variational inequalities. This method is
indirect in the sense that one first conjectures the form of the value function and the switching policy and
next verifies the optimality of the candidate function by proving that the candidate satisfies the variational
inequalities. In finding the specific form of the candidate function, appropriate boundary conditions includ-
ing the smooth-fit principle are employed. This formation shall lead to a system of non-linear equations that
are often hard to solve and the existence of the solution to the system is also difficult to prove. Moreover,
http://arxiv.org/abs/0704.0991v1
this indirect solution method is specific to the underlying process and reward/cost structure of the prob-
lem. Hence a slight change in the original problem often causes a complete overhaul in the highly technical
solution procedures.
Our solution method is direct in the sense that we first show a new mathematical characterization of
the value functions and, based on the characterization, we shall directly find the value function and optimal
switching policy. Therefore, it is free from any guesswork and applicable to a larger set of problems (where
the underlying process is one-dimensional diffusions) than the conventional methods. Our approach here
is similar to Dayanik and Karatzas (2003) and Dayanik and Egami (2005) that propose direct methods of
solving optimal stopping problems and stochastic impulse control problems, respectively.
The paper is organized in the following way. In the next section, after we introduce our setup of one
dimensional optimal switching problems, in section 2.1, we characterize the optimal switching times as
exit times from certain intervals through sequential optimal stopping problems equivalent to the original
switching problem. In section 2.2, we shall provide a new characterization of the value function, which
leads to a direct solution method described in 2.3. We shall illustrate this method through examples in
section 3, one of which is a new optimal switching problem. Section 4 concludes with comments on an
extension to a further general problem.
2 Optimal Switching Problems
We consider the following optimal switching problems for one dimensional diffusions. Let (Ω,F ,P) be a
complete probability space with a standard Brownian motion W = {Wt; t ≥ 0}. Let Zt be the indicator
vector at time t, Zt ∈ {z1, z2, ..., zm} , Z where each vector zi = (a1, a2, ..., ak) with a is either 0 (closed)
or 1 (open), so that m = 2k. In this section, we consider the case of k = 1. That is, Zt takes either 0 or 1.
The admissible switching strategy is
w = (θ0, θ1, θ2, ..., θk, ...; ζ0, ζ1, ζ2, ..., ζk, ...)
with θ0 = 0 where where where 0 ≤ θ1 < θ2 < .... are an increasing sequence of Ft-stopping times and ζ1,
ζ2... are Fθi-measurable random variables representing the new value of Zt at the corresponding switching
times θi (in this section, ζi = 1 or 0). The state process at time t is denoted by (Xt)t≥0 with state space
I = (c, d) ⊆ R and X0 = x ∈ I , and with the following dynamics:
If ζ0 = 1 (starting in open state), we have, for m = 0, 1, 2, .....,
dXt =
dX1t = µ1(X
1)dt+ σ1(X
1)dWt, θ2m ≤ t < θ2m+1,
dX0t = µ0(X
0)dt+ σ0(X
0)dWt, θ2m+1 ≤ t < θ2m+2,
(2.1)
and if ζ0 = 0 (starting in closed state),
dXt =
dX0t = µ0(X
1)dt+ σ0(X
0)dWt, θ2m ≤ t < θ2m+1,
dX1t = µ1(X
1)dt+ σ1(X
1)dWt, θ2m+1 ≤ t < θ2m+2.
(2.2)
We assume that µi : R → R and σi : R → R are some Borel functions that ensure the existence and
uniqueness of the solution of (2.1) for i = 1 and (2.2) for i = 0.
Our performance measure, corresponding to starting state i = 0, 1, is
Jwi (x) = E
e−αsf(Xs)ds −
e−αθjH(Xθj− , ζj)
 (2.3)
where H : R×Z → R+ is the switching cost function and f : R → R is a continuous function that satisfies
e−αs|f(Xs)|ds
<∞. (2.4)
In this section, the cost functions are of the form:
H(Xθ−, ζ) =
H(Xθ−, 1) opening cost,
H(Xθ−, 0) closing cost.
The optimal switching problem is to optimize the performance measure for i = 0 (start in closed state) and
1 (start in open state). That is to find, for both i = 1 and i = 0,
vi(x) , sup
Jw(x) with X0 = x (2.5)
where W is the set of all the admissible strategies.
2.1 Characterization of switching times
For the remaining part of section 2, we assume that the state space X is I = (c, d) where both c and d are
natural boundaries of X. But our characterization of the value function does not rely on this assumption. In
fact, it is easily applied to other types of boundaries, for example, absorbing boundary.
The first task is to characterize the optimal switching times as exit times from intervals in R. For this
purpose, we define two functions g0 and g1 : R+ → R with
g1(x) , sup
Jw1 (x) and g0(x) , sup
Jw0 (x). (2.6)
where W0 , {w ∈ W : w = (θ0, ζ0, θ1 = +∞)}. In other words, g1(·) is the discounted expected revenue
by starting with ζ0 = 1 and making no switches. Similarly, g0(·) is the discounted expected revenue by
staring with ζ0 = 0 and making no switches.
We set w0 , g1 and y0 , g0. We consider the following simultaneous sequential optimal stopping
problems with wn : R+ → R and yn : R+ → R for n = 1, 2, ....:
wn(x) , sup
e−αsf(Xs)ds + e
−ατ (yn−1(Xτ )−H(Xτ−, 1− Zτ−))
, (2.7)
yn(x) , sup
e−αsf(Xs)ds+ e
−ατ (wn−1(Xτ )−H(Xτ−, 1− Zτ−))
, (2.8)
where S is a set of Ft stopping times. Note that for each n, the sequential problem 2.7 (resp. (2.8)) starts in
open (resp. closed) state.
On the other hand, we define n-time switching problems for ζ0 = 1:
q(n)(x) , sup
Jw1 (x), (2.9)
where
Wn , {w ∈W ;w = (θ1, θ2, ...θn+1; ζ1, ζ2, ...ζn); θn+1 = +∞}.
In other words, we start with ζ0 = 1 (open) and are allowed to make at most n switches. Similarly, we
define another n-time switching problems corresponding to ζ0 = 0:
p(n)(x) , sup
Jw0 (x). (2.10)
We investigate the relationship of these four problems:
Lemma 2.1. For any x ∈ R, wn(x) = q(n)(x) and yn(x) = p(n)(x).
Proof. We shall prove only the first assertion since the proof of the second is similar. We have set y0(x) =
g0(x). Now we consider w1 by using the strong Markov property of X:
w1(x) = sup
e−αsf(Xs)ds+ e
−ατ (g0(Xτ )−H(Xτ−, 0))
= sup
e−αsf(Xs)ds −
e−αsf(Xs)ds− e−ατ (g0(Xτ )−H(Xτ−, 0))
= sup
e−ατ (g0(Xτ )− g1(Xτ )−H(Xτ−, 0))
+ g1(x).
On the other hand,
q(1)(x) = sup
e−αsf(Xs)ds− e−αθ1H(Xθ1− , ζ1)
= sup
[∫ θ1
e−αsf(Xs)ds +
e−αsf(Xs)ds− e−αθ1H(Xθ1− , 0)
= sup
(g1(x)− e−αθ1g1(Xθ1))− e
−αθ1(g0(Xθ1)−H(Xθ1− , 0))
= sup
e−αθ1(g0(Xθ1)− g1(Xθ1)−H(Xθ1− , 0))
+ g1(x).
Since both τ and θ1 are Ft stopping times, we have w1(x) = q(1)(x) for all x ∈ R. Moreover, by the theory
of the optimal stopping (see Appendix A, especially Proposition A.4), τ and hence θ1 are characterized as
an exit time from an interval. Similarly, we can prove y1(x) = p
(1)(x). Now we consider q(2)(x) which is
the value if we start in open state and make at most 2 switches (open → close → open). For this purpose, we
consider the performance measure q̄(2) that starts in an open state and is allowed two switches: For arbitrary
switching times θ1, θ2 > θ1 ∈ S , we have
q̄(2)(x) , Ex
e−αsf(Xs)ds−
e−αθjH(Xθj− , ζj)
e−αsf(Xs)ds +
e−αsf(Xs)ds +
e−αsf(Xs)ds
− e−αθ1H(Xθ1−, 0)− e
−αθ2H(Xθ2−, 1)
g1(x)− Ex[e−αθ1g1(Xθ1)]
x[e−αθ1g0(Xθ1)− e−αθ2g0(Xθ2)]
+ Ex[e−αθ2g1(Xθ2)]
− Ex[e−αθ1H(Xθ1−, 0) + e−αθ2H(Xθ2−, 1)].
Hence we have the following multiple optimal stopping problems:
q̄(2)(x) = sup
(θ1,θ2)∈S2
e−αθ1
(g0 − g1)(Xθ1)−H(Xθ1−, 0)
+ e−αθ2
(g1 − g0)(Xθ2)−H(Xθ2−, 1)
+ g1(x)
where S2 , {(θ1, θ2); θ1 ∈ S; θ2 ∈ Sθ1} and Sσ = {τ ∈ S; τ ≥ σ} for every σ ∈ S . Let us denote
h1(x) , g1(x)− g0(x)−H(x, 0), h2(x) , g0(x)− g1(x)−H(x, 1),
V1(x) , sup
e−ατh1(Xτ )
and V2(x) , sup
e−ατ (h2(Xτ ) + V1(Xτ ))
We also define
Γ1 , {x ∈ I : V1(x) = h1(x)} and Γ2 , {x ∈ I : V2(x) = h2(x) + V1(x)}
with σn , inf{t ≥ 0 : Xt ∈ Γn}. By using Proposition 5.4. in Carmona and Dayanik (2003), we conclude
that θ1 = σ1 and θ2 = θ1 + σ2 ◦ s(θ1) is optimal strategy where s(·) is the shift operator. Hence we only
consider the maximization over the set of admissible strategy W ∗2 where
W ∗2 , {w ∈W2 : θ1, θ2 are exit imes from an interval in I},
and can use the relation θ2 − θ1 = θ ◦ s(θ1) with some exit time θ ∈ S .
q(2)(x) = sup
w∈W ∗2
e−αsf(Xs)ds−
e−αθjH(Xθj− , ζj)
= sup
w∈W ∗2
e−αsf(Xs)ds +
e−αsf(Xs)ds +
e−αsf(Xs)ds
− e−αθ1(H(Xθ1− , 0) + e−α(θ2−θ1)H(Xθ2− , 1))
= sup
w∈W ∗2
e−αsf(Xs)ds + e
[(∫ θ
e−αsf(Xs)ds− e−αθH(Xθ−, 1)
− e−αθ1H(Xθ1− , 0)
Now by using the result for p(1), we can conclude
q(2)(x) = sup
w∈W ∗2
[∫ θ1
e−αsf(Xs)ds + e
p(1)(Xθ1)−H(Xθ1− , 0)
= sup
[∫ θ1
e−αsf(Xs)ds+ e
y1(Xθ1)−H(Xθ1− , 0)
= w2(x)
Similarly, we can prove y2(x) = p
(2)(x) and we can continue this process inductively to conclude that
wn(x) = q
(n)(x) and yn(x) = p
(n)(x) for all x and n.
Lemma 2.2. For all x ∈ R, limn→∞ q(n)(x) = v1(x) and limn→∞ p(n)(x) = v0(x).
Proof. Let us define q(x) , limn→∞ q
(n)(x). Since Wn ⊂ W , q(n)(x) ≤ v1(x) and hence q(x) ≤ v1(x).
To show the reverse inequality, we define W+ to be a set of admissible strategies such that
W+ = {w ∈W : Jw1 (x) <∞ for all x ∈ R}.
Let us assume that v1(x) < +∞ and consider a strategy w+ ∈ W+ and another strategy wn that coincides
with w+ up to and including time θn and then takes no further interventions.
1 (x)− Jw1 (x) = Ex
e−αs(f(Xs)− f(Xs−θn))−
i≥n+1
e−αθiH(Xθi−, ζi)
 , (2.11)
which implies
|Jw+1 (x)− Jw1 (x)| ≤ Ex
e−αθn −
i≥n+1
e−αθiH(Xθi−, ζi)
As n→ +∞, the right hand side goes to zero by the dominated convergence theorem. Hence it is shown
v1(x) = sup
Jw1 (x) = sup
w∈∪nWn
Jw1 (x)
so that v1(x) ≤ q(x). Next we consider v1(x) = +∞. Then we have some m ∈ N such that wm(x) =
q(m)(x) = ∞. Hence q(n)(x) = ∞ for all n ≥ m. The second assertion is proved similarly.
We define an operator L : H → H where H is a set of Borel functions
Lu(x) , sup
e−αsf(Xs)ds+ e
−ατ (u(Xτ )−H(Xτ−, 1 − Zτ−))
Lemma 2.3. The function w(x) , limn→∞wn(x) is the smallest solution, that majorizes g1(x), of the
function equation w = Lw.
Proof. We renumber the sequence (w0, y1, w2, y3...) as (u0, u1, u2, u3....). Since un is monotone increas-
ing, the limit u(x) exists. We have un+1(x) = Lun(x) and apply the monotone convergence theorem
by taking n → ∞, we have u(x) = Lu(x). We assume that u′(x) satisfies u′ = Lu′ and majorizes
g1(x) = u0(x). Then u
′ = Lu′ ≥ Lu0 = u1. Let us assume, for induction argument that u′ ≥ un, then
u′ = Lu′ ≥ Lun = un+1.
Hence we have u′ ≥ un for all n, leading to u′ ≥ limn→∞ un = u. Now we take the subsequence in
(w0, y1, w2, y3....) to complete the proof.
Proposition 2.1. For each x ∈ R, limn→∞wn(x) = v1(x) and limn→∞ yn(x) = v0(x). Moreover, the
optimal switching times, θ∗i are exit times from an interval.
Proof. We can prove the first assertion by combining the first two lemmas above. Now we concentrate on
the sequence of wn(x). For each n, finding wn(x) by solving (2.7) is an optimal stopping problem. By
Proposition A.4, the optimal stopping times are characterized as an exit time of X from an interval for all
n. This is also true in the limit: Indeed, by Lemma 2.3, in the limit, the value function of optimal switching
problem v1(x) = w(x) satisfies w = Lw, implying that v1(x) is the solution of an optimal stopping
problem. Hence the optimal switching times are characterized as exit time from an interval.
2.2 Characterization of the value functions
We go back to the original problem (2.3) to characterize the value function of the optimal switching prob-
lems. By the exit time characterization of the optimal switching times, θ∗i are given by
θ∗i =
inf{t > θi−1;X1t ∈ Γ1}
inf{t > θi−1;X0t ∈ Γ0}
(2.12)
where Γ1 = R \C1 and Γ0 = R \C0. We define here Ci and Γi to be continuation and stopping region for
Xit , respectively. We can simplify the performance measure J
w considerably. For ζ0 = 1, we have
Jw1 (x) = E
e−αsf(Xs)ds −
e−αθjH(Xθj− , ζj)
e−αsf(Xs)ds+
e−αsf(Xs)ds
− e−αθ1
H(Xθ1−, 0) +
e−α(θi−θ1)H(Xθj− , ζj)
e−αsf(Xs)ds+ e
−αθ1E
e−αsf(Xs)ds−
e−αθjH(Xθj− , ζj)
− e−αθ1H(Xθ1−, 0)
We notice that in the time interval (0, θ1), the process X is not intervened. The inner expectation is just
Jw0 (Xθ1). Hence we further simplify
Jw1 (x) = E
e−αsf(Xs)ds+ e
−αθ1(Jw0 (Xθ1)−H(Xθ1−, 0))
−e−αθ1g1(Xθ1) + e−αθ1(Jw0 (Xθ1)−H(Xθ1−, 0))
+ g1(x)
−e−αθ1g1(Xθ1) + e−αθ1Jw1 (Xθ1)
+ g1(x).
The third equality is a critical observation. Finally, we define u1 , J1 − g1 and obtain
u1(x) = J
1 (x)− g1(x) = Ex
e−αθ1u1(Xθ1)
. (2.13)
Since the switching time θ1 is characterized as a hitting time of a certain point in the state space, we can
represent θ1 = τa , inf{t ≥ 0 : Xt = a} for some a ∈ R. Hence equation (2.13) is an optimal stopping
problem that maximizes
u1(x) = J
1 (x)− g1(x) = Ex
e−ατau1(Xτa)
. (2.14)
among all the τa ∈ S . When θ1 = 0 (i.e., x = Xθ1),
Jw1 (x) = E
x [−g1(x) + Jw0 (x)−H(x, 0)] + g1(x)
and hence
u1(x) = J
0 (x)−H(x, 0)− g1(x).
In other words, we make a switch from open to closed immediately by paying the switching cost. Similarly,
for ζ0 = 0, we can simplify the performance measure J
0 (·) to obtain
Jw0 (x) = E
−e−αθ1g0(Xθ1) + e
−αθ1Jw0 (Xθ1)
+ g0(x).
By defining u0 , J
0 − g0, we have
u0(x) = J
0 (x)− g0(x) = Ex
e−αθ1u0(Xθ1)
Again, by using the characterization of switching times, we replace θ1 with τb,
u0(x) = J
0 (x)− g0(x) = Ex
e−ατbu0(Xτb)
. (2.15)
In summary, we have
u1(x) =
u0(x) + g0(x)−H(x, 0) − g1(x), x ∈ Γ1,
x [e−ατau1(Xτa)] = E
x [e−ατa(u0(Xτa) + g0(Xτa)− g1(Xτa)−H(Xτa , 0))] , x ∈ C1,
(2.16)
u0(x) =
x [e−ατbu0(Xτb)] = E
x [e−ατb(u1(Xτb) + g1(Xτb)− g0(Xτb)−H(Xτb , 1))] , x ∈ C0,
u1(x) + g1(x)−H(x, 1) − g0(x), x ∈ Γ0.
(2.17)
Hence we should solve the following optimal stopping problems simultaneously:
v̄1(x) , supτ∈S E
x [e−ατ (u1(Xτ )]
v̄0(x) , supσ∈S E
x [e−ασ(u0(Xσ)]
(2.18)
Now we let the infinitesimal generators of X1 and X0 be A1 and A0, respectively. We consider (Ai −
α)v(x) = 0 for i = 0, 1. This ODE has two fundamental solutions, ψi(·) and ϕi(·). We set ψi(·) is
an increasing and ϕi(·) is a decreasing function. Note that ψi(c+) = 0, ϕi(c+) = ∞ and ψi(d−) =
∞, ϕi(d−) = 0. We define
Fi(x) ,
ψi(x)
ϕi(x)
and Gi(x) , −
ϕi(x)
ψi(x)
for i = 0, 1.
By referring to Dayanik and Karatzas (2003), we have the following representation
x[e−ατr1{τr<τl}] =
ψ(l)ϕ(x) − ψ(x)ϕ(l)
ψ(l)ϕ(r) − ψ(r)ϕ(l)
, Ex[e−ατr1{τl<τr}] =
ψ(x)ϕ(r) − ψ(r)ϕ(x)
ψ(l)ϕ(r) − ψ(r)ϕ(l)
for x ∈ [l, r] where τl , inf{t > 0;Xt = l} and τr , inf{t > 0;Xt = r}.
By defining
W1 = (u1/ψ1) ◦G−11 and W0 = (u0/ϕ0) ◦ F
the second equation in (2.16) and the first equation in (2.17) become
W1(G1(x)) =W1(G1(a))
G1(d)−G1(x)
G1(d) −G1(a)
+W1(G1(d))
G1(x)−G1(a)
G1(d) −G1(a)
x ∈ [a, d), (2.19)
W0(F0(x)) =W0(F0(c))
F0(b)− F0(x)
F0(b)− F0(c)
+W0(F0(b))
F0(x)− F0(c)
F0(b)− F0(c)
, x ∈ (c, b], (2.20)
respectively. We should understand that F0(c) , F0(c+) = ψ0(c+)/ϕ0(c+) = 0 and that G1(d) ,
G1(d−) = −ϕ1(d−)/ψ1(d−) = 0. In the next subsection, we shall explain W1(G1(d−)) and W0(F0(c+))
in details. Both W1 and W0 are a linear function in their respective transformed spaces. Hence under the
appropriate transformations, the two value functions are linear functions in the continuation region.
2.3 Direct Method for a Solution
We have established a mathematical characterization of the value functions of optimal switching problems.
We shall investigate, by using the characterization, a direct solution method that does not require the recur-
sive optimal stopping schemes described in section 2.1. Since the two optimal stopping problems (2.18)
have to be solved simultaneously, finding u0 in x ∈ C0, for example, requires that we find the smallest
F0-concave majorant of (u1(x) + g1(x)− g0(x)−H(x, 1))/ϕ0(x) as in (2.17) that involves u1.
There are two cases, depending on whether x ∈ C1 ∩C0 or x ∈ Γ1 ∩C0, as to what u1(·) represents.
In the region x ∈ Γ1 ∩C0, u1(·) that shows up in the equation of u0(x) is of the form u1(x) = u0(x) +
g0(x)−H(x, 1, 0) − g1(x). In this case, the “obstacle” that should be majorized is in the form
u1(x) + g1(x)− g0(x)−H(x, 1)
= (u0(x) + g0(x)−H(x, 0)− g1(x)) + g1(x)− g0(x)−H(x, 1)
= u0(x)−H(x, 0)−H(x, 1) < u0(x). (2.21)
This implies that in x ∈ Γ1∩C0, the u0(x) function always majorizes the obstacle. Similarly, in x ∈ Γ0∩C1,
the u1(x) function always majorizes the obstacle.
Next, we consider the region x ∈ C0 ∩ C1. The u0(·) term in (2.16) is represented, due to its linear
characterization, as
W0(F0(x)) = β0(F0(x)) + d0
with some β0 ∈ R and d0 ∈ R+ in the transformed space. (The nonnegativity of d0 will be shown.) In
the original space, it has the form of ϕ0(x)(β0F0(x) + d0). Hence by the transformation (u1/ψ1) ◦ G−1,
W1(G1(x)) is the smallest linear majorant of
K1(x) + ϕ0(x)(β0F0(x) + d0)
ψ1(x)
K1(x) + β0ψ0(x) + d0ϕ0(x)
ψ1(x)
on (G1(d−), G1(a∗)) where
K1(x) , g0(x)− g1(x)−H(x, 0). (2.22)
This linear function passes a point (G1(d−), ld) where G1(d−) = 0 and
ld = lim sup
(K1(x) + β0ψ0(x) + d0ϕ0(x))
ψ1(x)
Let us consider further the quantity ld ≥ 0. By noting
lim sup
(K1(x) + β0ψ0(x))
ψ1(x)
≤ lim sup
(K1(x) + β0ψ0(x) + d0ϕ0(x))
ψ1(x)
≤ lim sup
(K1(x) + β0ψ0(x))
ψ1(x)
+ lim sup
d0ϕ0(x)
ψ1(x)
and lim supx↑d
d0ϕ0(x)
ψ1(x)
= 0, we can redefine ld by
ld , lim sup
(K1(x) + β0ψ0(x))
ψ1(x)
(2.23)
to determine the finiteness of the value function of the optimal switching problem, v1(x), based upon Propo-
sition A.5-A.7. Let us concentrate on the case ld = 0.
Similar analysis applies to (2.17). u1(x) in (2.17) is represented as
W1(G1(x)) = β1G1(x) + d1
with some β1 ∈ R and d1 ∈ R+. Note that d1 = ld ≥ 0. In the original space, it has the form of
ψ1(x)(β1G1(x) + d1). Hence by the transformation (u0/ϕ0(x)) ◦ F−1, W0(F0(x)) is the smallest linear
majorant of
K0(x) + ψ1(x)(β1G1(x) + d1)
ϕ0(x)
K0(x)− β1ϕ1(x) + d1ψ1(x)
ϕ0(x)
on (F0(c+), F0(b
∗)) where
K0(x) , g1(x)− g0(x)−H(x, 1). (2.24)
This linear function passes a point (F0(c+), lc) where F0(c+) = 0 and
lc = lim sup
(K0(x)− β1ϕ1(x) + d1ψ1(x))+
ϕ0(x)
Hence we have lc = d0 ≥ 0. By the same argument as for ld, we can redefine
lc , lim sup
(K0(x)− β1ϕ1(x))+
ϕ0(x)
. (2.25)
Remark 2.1. (a) Evaluation of ld or lc does not require knowledge of β0 or β1, respectively unless the
orders of max(K1(x), ψ1(x)) and ψ0(x) are equal, for example. (For this event, see Proposition 2.4.)
Otherwise, we just compare the order of the positive leading terms of the numerator in (2.23) and
(2.25) with that of the denominator.
(b) A sufficient condition for ld = lc = 0: since we have
0 ≤ ld ≤ lim sup
(K1(x))
ψ1(x)
+ lim sup
(β0ψ0(x))
ψ1(x)
a sufficient condition for ld = 0 is
lim sup
(K1(x))
ψ1(x)
= 0 and lim sup
ψ0(x)
ψ1(x)
= 0. (2.26)
Similarly,
0 ≤ lc ≤ lim sup
(K0(x))
ϕ0(x)
+ lim sup
(−β1ϕ1(x))+
ϕ0(x)
Hence a sufficient condition for lc = 0 is
lim sup
(K0(x))
ϕ0(x)
= 0 and lim sup
ϕ1(x)
ϕ0(x)
= 0. (2.27)
Moreover, it is obvious β1 < 0 and β0 > 0 since the linear majorant passes the origin of each
transformed space. Recall a points in the interval (c, d) ∈ R+ will be transformed by G(·) to
(G(c), G(d−)) ∈ R−.
We summarize the case of lc = ld = 0:
Proposition 2.2. Suppose that ld = lc = 0, the quantities being defined by (2.23) and by (2.25), respectively.
The value functions in the transformed space are the smallest linear majorants of
R1(·) ,
1 (·))
1 (·))
and R0(·) ,
0 (·))
0 (·))
where
r1(x) , g0(x)− g1(x) + β0ψ0(x)−H(x, 0)
r0(x) , g1(x)− g0(x)− β1ϕ1(x)−H(x, 1)
β0 > 0 and β1 < 0. (2.28)
Furthermore, Γ1 and Γ0 in (2.16) and (2.17) are given by
Γ1 , {x ∈ (c, d) :W1(G1(x)) = R1(G1(x))}, and Γ0 , {x ∈ (c, d) :W0(F0(x)) = R0(F0(x))}.
Corollary 2.1. If either of the boundary points c or d is absorbing, then (F0(c),W0(F0(c)) or (G1(d),W1(G1(d)))
is obtained directly. We can entirely omit the analysis of lc or ld. The characterization of the value function
(2.19) and (2.20) remains exactly the same.
Remark 2.2. An algorithm to find (a∗, b∗, β∗0 , β
1) can be described as follows:
1. Start with some β′1 ∈ R.
2. Calculate r0 and then R0 by the transformation R0(·) =
0 (·))
0 (·))
3. Find the linear majorant of R0 passing the origin of the transformed space. Call the slope of the linear
majorant, β0 and the point, F0(b), where R0 and the linear majorant meet .
4. Plug b and β0 in the equation for r1 and calculate R1 by the transformation R1(·) =
1 (·))
1 (·))
5. Find the linear majorant of R1 passing the origin of the transformed space. Call the slope of the linear
majorant, β1 and the point, G1(a), where R1 and the linear majorant meet.
6. Iterate step 1 to 5 until β1 = β
If both R1 and R0 are differentiable functions with their respective arguments, we can find (a
∗, b∗) analyti-
cally. Namely, we solve the following system for a and b:
dR0(y)
y=F0(b)
(F0(b)− F0(c)) = R0(F0(b))
dR1(y)
y=G1(a)
(G1(a)−G1(d)) = R1(G1(a))
(2.29)
where dR0(y)
y=F0(b∗)
= β∗0 and
dR1(y)
y=G1(a∗)
= β∗1 .
Once we find W1(·) and W0(·), then we convert to the original space and add back g1(x) and g0(x)
respectively so that v1(x) = ψ1(x)W1(G1(x)) + g1(x) and v0(x) = ϕ0(x)W0(F0(x)) + g0(x). Therefore,
by (2.16) and (2.17), the value functions v1(·) and v0(·) are given by:
Proposition 2.3. If the optimal continuation regions for both of the value functions are connected and if
lc = ld = 0, then the pair of the value functions v1(x) and v0(x) are represented as
v1(x) =
v̂0(x)−H(x, 0), x ≤ a∗,
v̂1(x) , ψ1(x)W1(G1(x)) + g1(x), a
∗ < x,
v0(x) =
v̂0(x) , ϕ0(x)W0(F0(x)) + g0(x) x < b
v̂1(x)−H(x, 1), b∗ ≤ x,
for some a∗, b∗ ∈ R with a∗ < b∗.
Proof. If the optimal continuation regions for both of the value functions are connected and if ld = lc = 0,
then the optimal intervention times (2.30) have the following form:
θ∗i =
inf{t > θi−1;Xt /∈ (a∗, d)}, Z = 1,
inf{t > θi−1;Xt /∈ (c, b∗)}, Z = 0.
(2.30)
Indeed, since we have lc = ld = 0, the linear majorants W1(·) and W0(·) pass the origins in their respective
transformed coordinates. Hence the continuation regions shall necessarily of the form of (2.30).
By our construction, both v1(x) and v0(x) are continuous in x ∈ R. Suppose we have a∗ > b∗. In this
case, by the form of the value functions, v0(b−)−H(b, 1, 0) = v1(b). Since the cost function H(·) > 0 and
continuous, it follows v0(b−) > v1(b). On the other hand, v0(b+) = v1(b)−H(b, 0, 1) implying v0(b+) <
v1(b). This contradicts the continuity of v0(x). Also, a
∗ = b∗ will lead to v1(x) = v1(x)−H(x, 1, 0) which
is impossible. Hence if the value functions exist, then we must necessarily have a∗ < b∗.
In relation to Proposition 2.3, we have the following observations:
Remark 2.3. (a) It is obvious that
v0(x) = v̂0(x) > v̂0(x)−H(x, 0) = v1(x), x ∈ (c, a∗),
v1(x) = v̂1(x) > v̂1(x)−H(x, 1) = v0(x), x ∈ (b∗, d).
(b) Since u1(x) is continuous in (c, d), the “obstacle” u1(x) + g1(x)− g0(x)−H(x, 1) to be majorized
by u0(x) on x ∈ C0 = (c, b∗) is also continuous, in particular at x = a∗. We proved that u0(x)
always majorizes the obstacle on (c, a∗). Hence F (a∗) ∈ {y : W0(y) > R0(y)} if there exists a
linear majorant of R0(y) in an interval of the form (F0(q), F0(d)) with some q ∈ (c, d): otherwise,
the continuity of u1(x) + g1(x) − g0(x) −H(x, 1) does not hold. Similarly, we have F (b∗) ∈ {y :
W1(y) > R1(y)} if there exists a linear majorant of R0(y) in an interval of the form (G1(c), G1(q)).
Finally, we summarize other cases than lc = ld = 0:
Proposition 2.4.
(a) If either ld = +∞ or lc = +∞, then v1(x) = v0(x) ≡ +∞.
(b) If both ld and lc are finite, then ld = lc = 0.
Proof. (a) The proof is immediate by invoking Proposition A.5. (b) When lc is finite, we know by Proposi-
tion A.5 that the value function v0(x) is finite. On x ∈ (c, a∗), u1(x)+g1(x)−g0(x)−H(x, 1) < u0(x) <
+∞ is finite (see (2.21)) and thereby
lc = lim sup
u1(x) + g1(x)− g0(x)−H(x, 1)
ϕ0(x)
The same argument for ld = 0.
Therefore, we can conclude that ld = 0 for the situation where the orders of max(K1(x), ψ1(x)) and ψ0(x)
are equal (⇒ ld is finite) as described in Remark 2.1 (a).
3 Examples
We recall some useful observations. If h(·) is twice-differentiable at x ∈ I and y , F (x), then we define
H(y) , h(F−1(y))/ϕ(F−1(y)) and we obtain H
(y) = m(x) and H
(y) = m
(x)/F
(x) with
m(x) =
(x), and H
(y)(A− α)h(x) ≥ 0, y = F (x) (3.1)
with strict inequality if H
(y) 6= 0. These identities are of practical use in identifying the concavities of
H(·) when it is hard to calculate its derivatives explicitly. Using these representations, we can modify (2.29)
F ′0(b)
(b)(F0(b)− F0(c)) = r0(b)ϕ0(b)
G′1(a)
(a)(G1(a)−G1(d)) = r1(a)ψ1(a)
(3.2)
Example 3.1. Brekke and Øksendal (1994): We first illustrate our solution method by using a resource
extraction problem solved by Brekke and Øksendal (1994). The price Pt at time t per unit of the resource
follows a geometric Brownian motion. Qt denotes the stock of remaining resources in the field that decays
exponentially. Hence we have
dPt = αPtdt+ βPtdWt and dQt = −λQtdt
where α, β, and λ > 0 (extraction rate) are constants. The objective of the problem is to find the optimal
switching times of resource extraction:
v(x) = sup
Jw(x) = sup
e−ρt(λPtQt −K)Ztdt−
e−ρθiH(Xθi−, Zθi)
where rho ∈ R+ is a discount factor with ρ > α, K ∈ R+ is the operating cost and H(x, 0) = C ∈ R+
and H(x, 1) = L ∈ R+ are constant closing and opening costs. Since P and Q always show up in the form
of PQ, we reduce the dimension by defining Xt = PtQt with the dynamics:
dXt = (α− λZt)Xtdt+ βXtdWt.
Solution: (1) We shall calculate all the necessary functions. For Zt = 1 (open state), we solve (A1 −
ρ)v(x) = 0 where A1 = (α − λ)xv′(x) + 12β
2x2v′′(x) to obtain ψ1(x) = x
ν+ and ϕ1(x) = x
ν− where
ν+,− = β
−α+ λ+ 1
(α− λ− 1
β2)2 + 2ρβ2
. Similarly, for Zt = 0 (closed state), we solve
(A0 − ρ)v(x) = 0 where A0 = αxv′(x) + 12β
2x2v′′(x) to obtain ψ0(x) = x
µ+ and ϕ0(x) = x
µ− where
µ+,− = β
−α+ 1
(α− 1
β2)2 + 2ρβ2
. Note that under the assumption ρ > α, we have
ν+, µ+ > 1 and ν−, ν− < 0.
By setting ∆1 =
(α− λ− 1
β2)2 + 2ρβ2 and ∆0 =
(α− 1
β2)2 + 2ρβ2, we have G1(x) =
−ϕ1(x)/ψ1(x) = −x−2∆1/β
and F0(x) = ψ0(x)/ϕ0(x) = x
2∆0/β
. It follows thatG−11 (y) = (−y)−β
2/2∆1
and F−10 (y) = y
β2/2∆0 . In this problem, we can calculate g1(x), g0(x) explicitly:
g1(x) = E
e−ρs(λXs −K)ds
ρ+ λ− α
and g(x) = 0. Lastly, K1(x) = g0(x) − g1(x) − H(x, 0) = −
ρ+λ−α
− C and K0(x) =
g1(x)− g0(x)−H(x, 1) = xρ+λ−α −
(2) The state space of X is (c, d) = (0,∞) and we evaluate lc and ld. Let us first note that ∆0−∆1+λ > 0.
Since limx↓0
ϕ1(x)
ϕ0(x)
= limx↓0 x
∆0−∆1+λ
β2 = 0 and limx↓0(K0(x))
+/ϕ0(x) = 0, we have lc = l0 = 0 by
(2.27). Similarly, by noting limx↑+∞
ψ0(x)
ψ1(x)
= limx↑+∞ x
−(∆0−∆1+λ)
β2 = 0 and limx↑+∞(K1(x))
+/ϕ0(x) =
0, we have ld = l+∞ = 0 by (2.26).
(3) To find the value functions together with continuation regions, we set
r1(x) = −
ρ+ λ− α
− C + β0ψ0(x) and r0(x) =
ρ+ λ− α
− L− β1ϕ1(x)
and make transformations R1(y) = r1(F
−1(y))/ψ1(F
−1(y)) and R0(y) = r0(F
−1(y))/ϕ0(F
−1(y)), re-
spectively. We examine the shape and behavior of the two functions R1(·) and R0(·) with an aid of (3.1).
By calculating (r0/ϕ0)
′(x) explicitly to examine the derivative of R0(y), we can find a critical point x = q,
at which R0(F (x)) attains a local minimum and from which R0(F (x)) is increasing monotonically on
(F0(q),∞). Moreover, we can confirm that limy→∞R′0(y) = limx→∞
(r0/ϕ0)
F ′0(x)
= 0, which shows that
there exists a finite linear majorant of R0(y). We define
p(x) = β1ωx
ν− − (ρ− α)
ρ+ λ− α
+ (K + ρL)
such that (A0 − ρ)r0(x) = p(x) where ω ,
β2ν−(ν− − 1)− αν−
(∆0 − ∆1 + λ)(∆0 +
∆1−λ) > 0. By the second identity in (3.1), the sign of the second derivative R′′0(y) is the same as the sign
of p(x). It is easy to see that p(x) has only one critical point. For any β1 < 0, the first term is dominant as
x → 0, so that limx↓0 p(x) < 0. As x gets larger, for |β1| sufficiently small, p(x) can take positive values,
providing two positive roots, say x = k1, k2 with k1 < k2. We also have limx→+∞ p(x) = −∞. In this
case, R0(y) is concave on (0, F (k1) ∪ (F (k2),+∞) and convex on (F (k1), F (k2)). Since we know that
R0(y) attains a local minimum at y = F (q), we have q < k2, and it implies that there is one and only on
tangency point of the linear majorant W (y) and R0(y) on (F (q),∞), so that the continuation region is of
the form (0, b∗).
¿From this analysis of the derivatives of R0(y), there is only one tangency point of the linear majorant
W0(y) and R0(y). (See Figure 3.1-(a)). A similar analysis shows that there is only one tangency point of
the linear majorant W1(y) and R1(y). (See Figure 3.1-(b)).
0.5 1 1.5 2 2.5 3 3.5
RHFH.L,WHFH.LL
-200 -150 -100 -50
RHGH.L,WHGH.LL
0.5 1 1.5 2 2.5
v0HxL
0.1 0.2 0.3 0.4 0.5 0.6 0.7
v1HxL
Figure 1: A numerical example of resource extraction problem. with parameters (α, β, λ, ρ,K, L,C) =
(0.01, 0.25, 0.01, 0.05, 0.4, 2, 2)(a) The smallest linear majorant W0(F0(x)) and R0(F0(x)) with b
∗ = 1.15042 and
= 10.8125. (b)The smallest linear majorantW1(G1(x)) andR1(G1(x)) with a
∗ = 0.18300 and β∗
= −0.695324.
(c) The value function v0(x). (d) The value function v1(x).
(4) By solving the system of equations (2.29), we can find (a∗, b∗, β∗0 , β
1). We transform back to the original
space to find
v̂1(x) = ψ1(x)W1(G1(x)) + g1(x) = ψ1(x)β
1G1(x) + g1(x)
= −β∗1ϕ1(x) + g1(x) = −β∗1xν− +
ρ+ λ− α
v̂0(x) = ϕ0(x)W0(F0(x)) + g0(x) = ϕ0(x)β
0F0(x) + g0(x) = β
0ψ0(x) + g0(x) = β
Hence the solution is
v1(x) =
µ+ − C, x ≤ a∗,
−β∗1xν− +
ρ+λ−α
, x > a∗,
v0(x) =
µ+ , x ≤ b∗,
−β∗1xν− +
ρ+λ−α
− L, x > b∗,
which agrees with Brekke and Økesendal (1994).
Example 3.2. Ornstein-Uhrenbeck process: We shall consider a new problem involving an Ornstein-
Uhrenbeck process. Consider a firm whose revenue solely depends on the price of one product. Due to its
cyclical nature of the prices, the firm does not want to have a large production facilty and decides to rent
additional production facility when the price is favorable. The revenue process to the firm is
dXt = δ(m −Xt − λZt)dt+ σdWt,
where λ = r/δ with r being a rent per unit of time. The firm’s objective is to maximize the incremental
revenue generated by renting the facility until the time τ0 when the price is at an intolerably low level.
Without loss of generality, we set τ0 = inf{t > 0 : Xt = 0}. We keep assuming constant operating cost K ,
opening cost, L and closing cost C . Now the value function is defined as
v(x) = sup
Jw(x) = sup
e−αt(Xt −K)Ztdt−
θi<τ0
e−αθiH(Xθi−, Zθi)
Solution: (1) We denote, by ψ̃(·) and ϕ̃(·), the functions of the fundamental solutions for the auxiliary
process Pt , (Xt −m+ λ)/σ, t ≥ 0, which satisfies dPt = −δPtdt+ dWt. For every x ∈ R,
ψ̃(x) = eδx
2/2D−α/δ(−x
2δ) and ϕ̃(x) = eδx
2/2D−α/δ(x
which leads to ψ1(x) = ψ̃((x −m + λ)/σ), ϕ1(x) = ϕ̃((x −m + λ)/σ), ψ0(x) = ψ̃((x −m)/σ), and
ϕ0(x) = ϕ̃((x −m)/σ) where Dν(·) is the parabolic cylinder function; (see Borodin and Salminen (2002,
Appendices 1.24 and 2.9) and Carmona and Dayanik (2003, Section 6.3)). By using the relation
Dν(z) = 2−ν/2e−z
2/4Hν(z/
2), z ∈ R (3.3)
in terms of the Hermite function Hν of degree ν and its integral representation
Hν(z) =
Γ(−ν)
2−2tzt−ν−1dt, Re(ν) < 0, (3.4)
(see for example, Lebedev(1972, pp 284, 290)). Since Ex[Xt] = e
−δtx + (1 − e−δt)(m − λ), we have
g0(x) = 0 and g1(x) =
x−(m−λ)
+ m−λ−K
(2) The state space of X is (c, d) = (0,+∞). Since the left boundary 0 is the absorbing, the linear majorant
passes (0, F0(0)). Since limx→+∞ ψ0(x)/ψ1(x) = 0, we have ld = 0.
(3) We formulate
r1(x) = −
x− (m− λ)
δ + α
m− λ−K
− C + β0ψ0(x)
r0(x) =
x− (m− λ)
δ + α
m− λ−K
− L− β1ϕ1(x)
and make transformations: R1(y) = r1(F
−1(y))/ψ1(F
−1(y)) and R0(y) = r0(F
−1(y))/ϕ0(F
−1(y)),
respectively. We examine the shape and behavior of the two functions R1(·) and R0(·) with an aid of (3.1).
First we check the sign of R′0(y) and find a critical point x = q, at which R0(F (x)) attains a local minimum
and from which R0(F (x)) is increasing monotonically on (F0(q),∞). It can be shown that R
0(+∞) = 0
by using (3.3) and (3.4) and the identity H′ν(z) = 2νHν−1(z), z ∈ R (see Lebedev (1972, p.289), for
example.) This shows that there must exist a (finite) linear majorant of R0(y) on (F (q),∞). To check
convexity of R0(y), we define
p(x) = −
ϕ′′1(x) + δ(m− x− λ)
δ + α
− β1ϕ′1(x)
− αr0(x)
such that (A0−α)r0(x) = p(x). We can show easily limx→+∞ p(x) = −∞ since ϕ1(+∞) = ϕ′1(+∞) =
ϕ′′1(+∞) = 0. Due to the monotonicity of ϕ1(x) and its derivatives, p(x) can have at most one critical point
and p(x) = 0 can have one or two positive roots depending on the value of β1. In either case, let us call the
largest positive root x = k2. We also have limx→+∞ p(x) = −∞. Since we know that R0(y) attains a local
minimum at y = F (q) and is increasing thereafter, we have q < k2. It follows that there is one and only on
tangency point of the linear majorant W (y) and R0(y) on (F (q),∞), so that the continuation region is of
the form (0, b∗). A similar analysis shows that there is only one tangency point of the linear majorant W1(y)
and R1(y).
(4) Solving (3.2), we we can find (a∗, b∗, β∗0 , β
1). We transform back to the original space to find
v̂1(x) = ψ1(x)W1(G1(x)) + g1(x) = ψ1(x)β
1G1(x) + g1(x) = −β∗1ϕ1(x) + g1(x)
= −β∗1e
δ(x−m+λ)2
2σ2 D−α/δ
(x−m+ λ)
x− (m− λ)
δ + α
v̂0(x) = ϕ0(x)W0(F0(x)) + g0(x) = ϕ0(x)β
0 (F0(x)− F0(0)) + g0(x)
= β∗0{ψ0(x)− F0(0)ϕ0(x)}+ g0(x)
= β∗0e
(x−m+λ)2
D−α/δ
x−m+ λ
− F (0)D−α/δ
Hence the solution is, using the above functions,
v1(x) =
v̂0(x)− C, x ≤ a∗,
v̂1(x), x > a
v0(x) =
v̂0(x), x ≤ b∗,
v̂1(x)− L, x > b∗.
See Figure 3.2 for a numerical example.
0.5 1 1.5 2 2.5 3
v0HxL
0.5 1 1.5 2 2.5 3
v1HxL
Figure 2: A numerical example of leasing production facility problem with parameters (m,α, σ, δ, λ,K, L,C) =
(5, 0.105, 0.35, 0.05, 4, 0.4, 0.2, 0.2): (a) The value function v0(x) with b
∗ = 1.66182 and β∗
= 144.313. (b)The
value function v1(x) with a
∗ = 0.781797 and β∗
= −2.16941.
4 Extensions and conclusions
4.1 An extension to the case of k ≥ 2
It is not difficult to extend to a general case of k ≥ 2 where more than one switching opportunities are
available. But we put a condition that z ∈ Z is of the form z = (a1, a2, ...., ak) where only one element of
this vector is 1 with the rest being zero, i.e., z = (0, 0, 0, ...., 1, 0, 0) for example.
We should introduce the switching operator M0 on h ∈ H,
M0h(u, z) = max
ζ∈Z\{z}
{h(u, ζ)−H(u, z; ζ)} . (4.1)
In words, this operator would calculate which production mode should be chosen by moving from the current
production mode z. Now the recursive optimal stopping (2.7) becomes
wn+1(x) , sup
e−αsf(Xs)ds+ e
−ατMwn(Xτ )
Accordingly, the optimization procedure will become two-stage. To illustrate this, we suppose k = 2 so that
i = 0, 1, and 2. By eliminating the integral in (4.1), we redefine the switching operator,
Mhz(x) , max
ζ∈Z\{z}
{hζ(x) + gζ(x)− gz(x)−H(x, z, ζ)} , (4.2)
where
gz(x) , sup
Jwz (x) = E
e−αsf(Xs)ds
Hence (2.13) will be modified to uz(x) = E
x[e−ατMuz(Xτ )]. It follows that our system of equations
(2.18) is now
v̄2(x) , supτ∈S E
x [e−ατMv̄2(Xτ )]
v̄1(x) , supτ∈S E
x [e−ατMv̄1(Xτ )]
v̄0(x) , supτ∈S E
x [e−ατMv̄0(Xτ )]
(4.3)
The first stage is optimal stopping problem. One possibility of switching production modes is (0 → 1, 1 →
2, 2 → 0). First, we fix this switching scheme, say c, and solve the system of equations (4.3) as three
optimal stopping problems. All the arguments in Section 2.3 hold. This first-stage optimization will give
(x∗0(c), x
1(c), x
2(c), β
0 (c), β
1 (c), β
2 (c)), where xi’s are switching boundaries, depending on this switching
scheme c.
Now we move to another switching scheme c′ and solve the system of optimal stopping problems until
we find the optimal scheme.
4.2 Conclusions
We have studied optimal switching problems for one-dimensional diffusions. We characterize the value
function as linear functions in their respective spaces, and provide a direct method to find the value functions
and the opening and switching boundaries at the same time. Using the techniques we developed here as
well as the ones in Dayanik and Karazas (2003) and Dayanik and Egami (2005), we solved two specific
problems, one of which involves a mean-reverting process. This problem might be hard to solve with just
the HJB equation and the related quasi-variational inequalities. Finally, an extension to more general cases
is suggested. We believe that this direct method and the new characterization will expand the coverage of
solvable problems in the financial engineering and economic analysis.
A Summary of Optimal Stopping Theory
Let (Ω,F ,P) be a complete probability space with a standard Brownian motion W = {Wt; t ≥ 0} and
consider the diffusion process X0 with state pace I ⊆ R and dynamics
dX0t = µ(X
t )dt+ σ(X
t )dWt (A.1)
for some Borel functions µ : I → R and σ : I → (0,∞). We emphasize here that X0 is an uncontrolled
process. We assume that I is an interval with endpoints −∞ ≤ a < b ≤ +∞, and that X0 is regular in
(a, b); in other words, X0 reaches y with positive probability starting at x for every x and y in (a, b). We
shall denote by F = {Ft} the natural filtration generated by X0.
Let α ≥ 0 be a real constant and h(·) a Borel function such that Ex[e−ατh(X0τ )] is well-defined for
every F-stopping time τ and x ∈ I . Let τy be the first hitting time of y ∈ I by X0, and let c ∈ I be a fixed
point of the state space. We set:
ψ(x) =
x[e−ατc1{τc<∞}], x ≤ c,
1/Ec[e−ατx1{τx<∞}], x > c,
ϕ(x) =
e−ατx1{τx<∞}
, x ≤ c,
x[e−ατc1{τc<∞}], x > c,
F (x) ,
, x ∈ I. (A.2)
Then F (·) is continuous and strictly increasing. It should be noted that ψ(·) and ϕ(·) consist of an increasing
and a decreasing solution of the second-order differential equation (A − α)u = 0 in I where A is the
infinitesimal generator of X0. They are linearly independent positive solutions and uniquely determined
up to multiplication. For the complete characterization of ψ(·) and ϕ(·) corresponding to various types of
boundary behavior, refer to Itô and McKean (1974).
Let F : [c, d] → R be a strictly increasing function. A real valued function u is called F -concave on
[c, d] if, for every a ≤ l < r ≤ b and x ∈ [l, r],
u(x) ≥ u(l)
F (r)− F (x)
F (r)− F (l)
+ u(r)
F (x)− F (l)
F (r)− F (l)
We denote by
V (x) , sup
x[e−ατh(X0τ )], x ∈ [c, d] (A.3)
the value function of the optimal stopping problem with the reward function h(·) where the supremum is
taken over the class S of all F-stopping times. Then we have the following results, the proofs of which we
refer to Dayanik and Karatzas (2003).
Proposition A.1. For a given function U : [c, d] → [0,+∞) the quotient U(·)/ϕ(·) is an F -concave function
if and only if U(·) is α-excessive, i.e.,
U(x) ≥ Ex[e−ατU(X0τ )],∀τ ∈ S,∀x ∈ [c, d]. (A.4)
Proposition A.2. The value function V (·) of (A.3) is the smallest nonnegative majorant of h(·) such that
V (·)/ϕ(·) is F -concave on [c, d].
Proposition A.3. Let W (·) be the smallest nonnegative concave majorant of H , (h/ϕ) ◦ F−1 on
[F (c), F (d)], where F−1(·) is the inverse of the strictly increasing function F (·) in (A.2). Then V (x) =
ϕ(x)W (F (x)) for every x ∈ [c, d].
Proposition A.4. Define
S , {x ∈ [c, d] : V (x) = h(x)}, and τ∗ , inf{t ≧ 0 : X0t ∈ S}. (A.5)
If h(·) is continuous on [c, d], then τ∗ is an optimal stopping rule.
When both boundaries are natural, we have the following results:
Proposition A.5. We have either V ≡ 0 in (c, d) or V (x) < +∞ for all (c, d). Moreover, V (x) < +∞ for
every x ∈ (c, d) if and only if
lc , lim sup
h+(x)
and ld , lim sup
h+(x)
(A.6)
are both finite.
In the finite case, furthermore,
Proposition A.6. The value function V (·) is continuous on (c, d). If h : (c, d) → R is continuous and
lc = ld = 0, then τ
∗ of (A.5) is an optimal stopping time.
Proposition A.7. Suppose that lc and ld are finite and one of them is strictly positive, and h(·) is continuous.
Define the continuation region C , (c, d) \ Γ. Then τ∗ of (A.5) is an optimal stopping time, if and only if
there is no r ∈ (c, d) such that (c, r) ⊂ C if lc > 0 and
there is no l ∈ (c, d) such that (l, d) ⊂ C if ld > 0.
References
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Preprint. Princeton University.
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Preprint. University of Michigan.
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one-dimensional diffusions. Preprint. www.math.lsa.umich.edu/∼ egami.
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investments. Preprint.
Itô, K. and McKean, Jr., H. P. (1974). Diffusions processes and their sample paths. Springer-Verlag, Berlin.
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Yushkevich, A. (2001). Optimal switching problem for countable Markov chains: average reward criterion.
Math. Meth. Oper. Res.53, 1–24.
	Introduction
	Optimal Switching Problems
	Characterization of switching times
	Characterization of the value functions
	Direct Method for a Solution
	Examples
	Extensions and conclusions
	An extension to the case of k2
	Conclusions
	Summary of Optimal Stopping Theory

  In this paper, we propose a direct solution method for optimal switching
problems of one-dimensional diffusions. This method is free from conjectures
about the form of the value function and switching strategies, or does not
require the proof of optimality through quasi-variational inequalities. The
direct method uses a general theory of optimal stopping problems for
one-dimensional diffusions and characterizes the value function as sets of the
smallest linear majorants in their respective transformed spaces.

Introduction
Stochastic optimal switching problems (or starting and stopping problems) are important subjects both in
mathematics and economics. Since there are numerous articles about real options in the economic and
financial literature in recent years, the importance and applicability of control problems including optimal
switching problems cannot be exaggerated.
A typical optimal switching problem is described as follows: The controller monitors the price of natural
resources for optimizing (in some sense) the operation of an extraction facility. She can choose when to start
extracting this resource and when to temporarily stop doing so, based upon price fluctuations she observes.
The problem is concerned with finding an optimal switching policy and the corresponding value function. A
number of papers on this topic are well worth mentioning : Brennan and Schwarz (1985) in conjunction with
convenience yield in the energy market, Dixit (1989) for production facility problems, Brekke and Øksendal
(1994) for resource extraction problems, Yushkevich (2001) for positive recurrent countable Markov chain,
and Duckworth and Zervos (2001) for reversible investment problems. Hamdadène and Jeanblanc (2004)
analyze a general adapted process for finite time horizon using reflected stochastic backward differential
equations. Carmona and Ludkovski (2005) apply to energy tolling agreement in a finite time horizon using
Monte-Carlo regressions.
A basic analytical tool for solving switching problems is quasi-variational inequalities. This method is
indirect in the sense that one first conjectures the form of the value function and the switching policy and
next verifies the optimality of the candidate function by proving that the candidate satisfies the variational
inequalities. In finding the specific form of the candidate function, appropriate boundary conditions includ-
ing the smooth-fit principle are employed. This formation shall lead to a system of non-linear equations that
are often hard to solve and the existence of the solution to the system is also difficult to prove. Moreover,
http://arxiv.org/abs/0704.0991v1
this indirect solution method is specific to the underlying process and reward/cost structure of the prob-
lem. Hence a slight change in the original problem often causes a complete overhaul in the highly technical
solution procedures.
Our solution method is direct in the sense that we first show a new mathematical characterization of
the value functions and, based on the characterization, we shall directly find the value function and optimal
switching policy. Therefore, it is free from any guesswork and applicable to a larger set of problems (where
the underlying process is one-dimensional diffusions) than the conventional methods. Our approach here
is similar to Dayanik and Karatzas (2003) and Dayanik and Egami (2005) that propose direct methods of
solving optimal stopping problems and stochastic impulse control problems, respectively.
The paper is organized in the following way. In the next section, after we introduce our setup of one
dimensional optimal switching problems, in section 2.1, we characterize the optimal switching times as
exit times from certain intervals through sequential optimal stopping problems equivalent to the original
switching problem. In section 2.2, we shall provide a new characterization of the value function, which
leads to a direct solution method described in 2.3. We shall illustrate this method through examples in
section 3, one of which is a new optimal switching problem. Section 4 concludes with comments on an
extension to a further general problem.
2 Optimal Switching Problems
We consider the following optimal switching problems for one dimensional diffusions. Let (Ω,F ,P) be a
complete probability space with a standard Brownian motion W = {Wt; t ≥ 0}. Let Zt be the indicator
vector at time t, Zt ∈ {z1, z2, ..., zm} , Z where each vector zi = (a1, a2, ..., ak) with a is either 0 (closed)
or 1 (open), so that m = 2k. In this section, we consider the case of k = 1. That is, Zt takes either 0 or 1.
The admissible switching strategy is
w = (θ0, θ1, θ2, ..., θk, ...; ζ0, ζ1, ζ2, ..., ζk, ...)
with θ0 = 0 where where where 0 ≤ θ1 < θ2 < .... are an increasing sequence of Ft-stopping times and ζ1,
ζ2... are Fθi-measurable random variables representing the new value of Zt at the corresponding switching
times θi (in this section, ζi = 1 or 0). The state process at time t is denoted by (Xt)t≥0 with state space
I = (c, d) ⊆ R and X0 = x ∈ I , and with the following dynamics:
If ζ0 = 1 (starting in open state), we have, for m = 0, 1, 2, .....,
dXt =
dX1t = µ1(X
1)dt+ σ1(X
1)dWt, θ2m ≤ t < θ2m+1,
dX0t = µ0(X
0)dt+ σ0(X
0)dWt, θ2m+1 ≤ t < θ2m+2,
(2.1)
and if ζ0 = 0 (starting in closed state),
dXt =
dX0t = µ0(X
1)dt+ σ0(X
0)dWt, θ2m ≤ t < θ2m+1,
dX1t = µ1(X
1)dt+ σ1(X
1)dWt, θ2m+1 ≤ t < θ2m+2.
(2.2)
We assume that µi : R → R and σi : R → R are some Borel functions that ensure the existence and
uniqueness of the solution of (2.1) for i = 1 and (2.2) for i = 0.
Our performance measure, corresponding to starting state i = 0, 1, is
Jwi (x) = E
e−αsf(Xs)ds −
e−αθjH(Xθj− , ζj)
 (2.3)
where H : R×Z → R+ is the switching cost function and f : R → R is a continuous function that satisfies
e−αs|f(Xs)|ds
<∞. (2.4)
In this section, the cost functions are of the form:
H(Xθ−, ζ) =
H(Xθ−, 1) opening cost,
H(Xθ−, 0) closing cost.
The optimal switching problem is to optimize the performance measure for i = 0 (start in closed state) and
1 (start in open state). That is to find, for both i = 1 and i = 0,
vi(x) , sup
Jw(x) with X0 = x (2.5)
where W is the set of all the admissible strategies.
2.1 Characterization of switching times
For the remaining part of section 2, we assume that the state space X is I = (c, d) where both c and d are
natural boundaries of X. But our characterization of the value function does not rely on this assumption. In
fact, it is easily applied to other types of boundaries, for example, absorbing boundary.
The first task is to characterize the optimal switching times as exit times from intervals in R. For this
purpose, we define two functions g0 and g1 : R+ → R with
g1(x) , sup
Jw1 (x) and g0(x) , sup
Jw0 (x). (2.6)
where W0 , {w ∈ W : w = (θ0, ζ0, θ1 = +∞)}. In other words, g1(·) is the discounted expected revenue
by starting with ζ0 = 1 and making no switches. Similarly, g0(·) is the discounted expected revenue by
staring with ζ0 = 0 and making no switches.
We set w0 , g1 and y0 , g0. We consider the following simultaneous sequential optimal stopping
problems with wn : R+ → R and yn : R+ → R for n = 1, 2, ....:
wn(x) , sup
e−αsf(Xs)ds + e
−ατ (yn−1(Xτ )−H(Xτ−, 1− Zτ−))
, (2.7)
yn(x) , sup
e−αsf(Xs)ds+ e
−ατ (wn−1(Xτ )−H(Xτ−, 1− Zτ−))
, (2.8)
where S is a set of Ft stopping times. Note that for each n, the sequential problem 2.7 (resp. (2.8)) starts in
open (resp. closed) state.
On the other hand, we define n-time switching problems for ζ0 = 1:
q(n)(x) , sup
Jw1 (x), (2.9)
where
Wn , {w ∈W ;w = (θ1, θ2, ...θn+1; ζ1, ζ2, ...ζn); θn+1 = +∞}.
In other words, we start with ζ0 = 1 (open) and are allowed to make at most n switches. Similarly, we
define another n-time switching problems corresponding to ζ0 = 0:
p(n)(x) , sup
Jw0 (x). (2.10)
We investigate the relationship of these four problems:
Lemma 2.1. For any x ∈ R, wn(x) = q(n)(x) and yn(x) = p(n)(x).
Proof. We shall prove only the first assertion since the proof of the second is similar. We have set y0(x) =
g0(x). Now we consider w1 by using the strong Markov property of X:
w1(x) = sup
e−αsf(Xs)ds+ e
−ατ (g0(Xτ )−H(Xτ−, 0))
= sup
e−αsf(Xs)ds −
e−αsf(Xs)ds− e−ατ (g0(Xτ )−H(Xτ−, 0))
= sup
e−ατ (g0(Xτ )− g1(Xτ )−H(Xτ−, 0))
+ g1(x).
On the other hand,
q(1)(x) = sup
e−αsf(Xs)ds− e−αθ1H(Xθ1− , ζ1)
= sup
[∫ θ1
e−αsf(Xs)ds +
e−αsf(Xs)ds− e−αθ1H(Xθ1− , 0)
= sup
(g1(x)− e−αθ1g1(Xθ1))− e
−αθ1(g0(Xθ1)−H(Xθ1− , 0))
= sup
e−αθ1(g0(Xθ1)− g1(Xθ1)−H(Xθ1− , 0))
+ g1(x).
Since both τ and θ1 are Ft stopping times, we have w1(x) = q(1)(x) for all x ∈ R. Moreover, by the theory
of the optimal stopping (see Appendix A, especially Proposition A.4), τ and hence θ1 are characterized as
an exit time from an interval. Similarly, we can prove y1(x) = p
(1)(x). Now we consider q(2)(x) which is
the value if we start in open state and make at most 2 switches (open → close → open). For this purpose, we
consider the performance measure q̄(2) that starts in an open state and is allowed two switches: For arbitrary
switching times θ1, θ2 > θ1 ∈ S , we have
q̄(2)(x) , Ex
e−αsf(Xs)ds−
e−αθjH(Xθj− , ζj)
e−αsf(Xs)ds +
e−αsf(Xs)ds +
e−αsf(Xs)ds
− e−αθ1H(Xθ1−, 0)− e
−αθ2H(Xθ2−, 1)
g1(x)− Ex[e−αθ1g1(Xθ1)]
x[e−αθ1g0(Xθ1)− e−αθ2g0(Xθ2)]
+ Ex[e−αθ2g1(Xθ2)]
− Ex[e−αθ1H(Xθ1−, 0) + e−αθ2H(Xθ2−, 1)].
Hence we have the following multiple optimal stopping problems:
q̄(2)(x) = sup
(θ1,θ2)∈S2
e−αθ1
(g0 − g1)(Xθ1)−H(Xθ1−, 0)
+ e−αθ2
(g1 − g0)(Xθ2)−H(Xθ2−, 1)
+ g1(x)
where S2 , {(θ1, θ2); θ1 ∈ S; θ2 ∈ Sθ1} and Sσ = {τ ∈ S; τ ≥ σ} for every σ ∈ S . Let us denote
h1(x) , g1(x)− g0(x)−H(x, 0), h2(x) , g0(x)− g1(x)−H(x, 1),
V1(x) , sup
e−ατh1(Xτ )
and V2(x) , sup
e−ατ (h2(Xτ ) + V1(Xτ ))
We also define
Γ1 , {x ∈ I : V1(x) = h1(x)} and Γ2 , {x ∈ I : V2(x) = h2(x) + V1(x)}
with σn , inf{t ≥ 0 : Xt ∈ Γn}. By using Proposition 5.4. in Carmona and Dayanik (2003), we conclude
that θ1 = σ1 and θ2 = θ1 + σ2 ◦ s(θ1) is optimal strategy where s(·) is the shift operator. Hence we only
consider the maximization over the set of admissible strategy W ∗2 where
W ∗2 , {w ∈W2 : θ1, θ2 are exit imes from an interval in I},
and can use the relation θ2 − θ1 = θ ◦ s(θ1) with some exit time θ ∈ S .
q(2)(x) = sup
w∈W ∗2
e−αsf(Xs)ds−
e−αθjH(Xθj− , ζj)
= sup
w∈W ∗2
e−αsf(Xs)ds +
e−αsf(Xs)ds +
e−αsf(Xs)ds
− e−αθ1(H(Xθ1− , 0) + e−α(θ2−θ1)H(Xθ2− , 1))
= sup
w∈W ∗2
e−αsf(Xs)ds + e
[(∫ θ
e−αsf(Xs)ds− e−αθH(Xθ−, 1)
− e−αθ1H(Xθ1− , 0)
Now by using the result for p(1), we can conclude
q(2)(x) = sup
w∈W ∗2
[∫ θ1
e−αsf(Xs)ds + e
p(1)(Xθ1)−H(Xθ1− , 0)
= sup
[∫ θ1
e−αsf(Xs)ds+ e
y1(Xθ1)−H(Xθ1− , 0)
= w2(x)
Similarly, we can prove y2(x) = p
(2)(x) and we can continue this process inductively to conclude that
wn(x) = q
(n)(x) and yn(x) = p
(n)(x) for all x and n.
Lemma 2.2. For all x ∈ R, limn→∞ q(n)(x) = v1(x) and limn→∞ p(n)(x) = v0(x).
Proof. Let us define q(x) , limn→∞ q
(n)(x). Since Wn ⊂ W , q(n)(x) ≤ v1(x) and hence q(x) ≤ v1(x).
To show the reverse inequality, we define W+ to be a set of admissible strategies such that
W+ = {w ∈W : Jw1 (x) <∞ for all x ∈ R}.
Let us assume that v1(x) < +∞ and consider a strategy w+ ∈ W+ and another strategy wn that coincides
with w+ up to and including time θn and then takes no further interventions.
1 (x)− Jw1 (x) = Ex
e−αs(f(Xs)− f(Xs−θn))−
i≥n+1
e−αθiH(Xθi−, ζi)
 , (2.11)
which implies
|Jw+1 (x)− Jw1 (x)| ≤ Ex
e−αθn −
i≥n+1
e−αθiH(Xθi−, ζi)
As n→ +∞, the right hand side goes to zero by the dominated convergence theorem. Hence it is shown
v1(x) = sup
Jw1 (x) = sup
w∈∪nWn
Jw1 (x)
so that v1(x) ≤ q(x). Next we consider v1(x) = +∞. Then we have some m ∈ N such that wm(x) =
q(m)(x) = ∞. Hence q(n)(x) = ∞ for all n ≥ m. The second assertion is proved similarly.
We define an operator L : H → H where H is a set of Borel functions
Lu(x) , sup
e−αsf(Xs)ds+ e
−ατ (u(Xτ )−H(Xτ−, 1 − Zτ−))
Lemma 2.3. The function w(x) , limn→∞wn(x) is the smallest solution, that majorizes g1(x), of the
function equation w = Lw.
Proof. We renumber the sequence (w0, y1, w2, y3...) as (u0, u1, u2, u3....). Since un is monotone increas-
ing, the limit u(x) exists. We have un+1(x) = Lun(x) and apply the monotone convergence theorem
by taking n → ∞, we have u(x) = Lu(x). We assume that u′(x) satisfies u′ = Lu′ and majorizes
g1(x) = u0(x). Then u
′ = Lu′ ≥ Lu0 = u1. Let us assume, for induction argument that u′ ≥ un, then
u′ = Lu′ ≥ Lun = un+1.
Hence we have u′ ≥ un for all n, leading to u′ ≥ limn→∞ un = u. Now we take the subsequence in
(w0, y1, w2, y3....) to complete the proof.
Proposition 2.1. For each x ∈ R, limn→∞wn(x) = v1(x) and limn→∞ yn(x) = v0(x). Moreover, the
optimal switching times, θ∗i are exit times from an interval.
Proof. We can prove the first assertion by combining the first two lemmas above. Now we concentrate on
the sequence of wn(x). For each n, finding wn(x) by solving (2.7) is an optimal stopping problem. By
Proposition A.4, the optimal stopping times are characterized as an exit time of X from an interval for all
n. This is also true in the limit: Indeed, by Lemma 2.3, in the limit, the value function of optimal switching
problem v1(x) = w(x) satisfies w = Lw, implying that v1(x) is the solution of an optimal stopping
problem. Hence the optimal switching times are characterized as exit time from an interval.
2.2 Characterization of the value functions
We go back to the original problem (2.3) to characterize the value function of the optimal switching prob-
lems. By the exit time characterization of the optimal switching times, θ∗i are given by
θ∗i =
inf{t > θi−1;X1t ∈ Γ1}
inf{t > θi−1;X0t ∈ Γ0}
(2.12)
where Γ1 = R \C1 and Γ0 = R \C0. We define here Ci and Γi to be continuation and stopping region for
Xit , respectively. We can simplify the performance measure J
w considerably. For ζ0 = 1, we have
Jw1 (x) = E
e−αsf(Xs)ds −
e−αθjH(Xθj− , ζj)
e−αsf(Xs)ds+
e−αsf(Xs)ds
− e−αθ1
H(Xθ1−, 0) +
e−α(θi−θ1)H(Xθj− , ζj)
e−αsf(Xs)ds+ e
−αθ1E
e−αsf(Xs)ds−
e−αθjH(Xθj− , ζj)
− e−αθ1H(Xθ1−, 0)
We notice that in the time interval (0, θ1), the process X is not intervened. The inner expectation is just
Jw0 (Xθ1). Hence we further simplify
Jw1 (x) = E
e−αsf(Xs)ds+ e
−αθ1(Jw0 (Xθ1)−H(Xθ1−, 0))
−e−αθ1g1(Xθ1) + e−αθ1(Jw0 (Xθ1)−H(Xθ1−, 0))
+ g1(x)
−e−αθ1g1(Xθ1) + e−αθ1Jw1 (Xθ1)
+ g1(x).
The third equality is a critical observation. Finally, we define u1 , J1 − g1 and obtain
u1(x) = J
1 (x)− g1(x) = Ex
e−αθ1u1(Xθ1)
. (2.13)
Since the switching time θ1 is characterized as a hitting time of a certain point in the state space, we can
represent θ1 = τa , inf{t ≥ 0 : Xt = a} for some a ∈ R. Hence equation (2.13) is an optimal stopping
problem that maximizes
u1(x) = J
1 (x)− g1(x) = Ex
e−ατau1(Xτa)
. (2.14)
among all the τa ∈ S . When θ1 = 0 (i.e., x = Xθ1),
Jw1 (x) = E
x [−g1(x) + Jw0 (x)−H(x, 0)] + g1(x)
and hence
u1(x) = J
0 (x)−H(x, 0)− g1(x).
In other words, we make a switch from open to closed immediately by paying the switching cost. Similarly,
for ζ0 = 0, we can simplify the performance measure J
0 (·) to obtain
Jw0 (x) = E
−e−αθ1g0(Xθ1) + e
−αθ1Jw0 (Xθ1)
+ g0(x).
By defining u0 , J
0 − g0, we have
u0(x) = J
0 (x)− g0(x) = Ex
e−αθ1u0(Xθ1)
Again, by using the characterization of switching times, we replace θ1 with τb,
u0(x) = J
0 (x)− g0(x) = Ex
e−ατbu0(Xτb)
. (2.15)
In summary, we have
u1(x) =
u0(x) + g0(x)−H(x, 0) − g1(x), x ∈ Γ1,
x [e−ατau1(Xτa)] = E
x [e−ατa(u0(Xτa) + g0(Xτa)− g1(Xτa)−H(Xτa , 0))] , x ∈ C1,
(2.16)
u0(x) =
x [e−ατbu0(Xτb)] = E
x [e−ατb(u1(Xτb) + g1(Xτb)− g0(Xτb)−H(Xτb , 1))] , x ∈ C0,
u1(x) + g1(x)−H(x, 1) − g0(x), x ∈ Γ0.
(2.17)
Hence we should solve the following optimal stopping problems simultaneously:
v̄1(x) , supτ∈S E
x [e−ατ (u1(Xτ )]
v̄0(x) , supσ∈S E
x [e−ασ(u0(Xσ)]
(2.18)
Now we let the infinitesimal generators of X1 and X0 be A1 and A0, respectively. We consider (Ai −
α)v(x) = 0 for i = 0, 1. This ODE has two fundamental solutions, ψi(·) and ϕi(·). We set ψi(·) is
an increasing and ϕi(·) is a decreasing function. Note that ψi(c+) = 0, ϕi(c+) = ∞ and ψi(d−) =
∞, ϕi(d−) = 0. We define
Fi(x) ,
ψi(x)
ϕi(x)
and Gi(x) , −
ϕi(x)
ψi(x)
for i = 0, 1.
By referring to Dayanik and Karatzas (2003), we have the following representation
x[e−ατr1{τr<τl}] =
ψ(l)ϕ(x) − ψ(x)ϕ(l)
ψ(l)ϕ(r) − ψ(r)ϕ(l)
, Ex[e−ατr1{τl<τr}] =
ψ(x)ϕ(r) − ψ(r)ϕ(x)
ψ(l)ϕ(r) − ψ(r)ϕ(l)
for x ∈ [l, r] where τl , inf{t > 0;Xt = l} and τr , inf{t > 0;Xt = r}.
By defining
W1 = (u1/ψ1) ◦G−11 and W0 = (u0/ϕ0) ◦ F
the second equation in (2.16) and the first equation in (2.17) become
W1(G1(x)) =W1(G1(a))
G1(d)−G1(x)
G1(d) −G1(a)
+W1(G1(d))
G1(x)−G1(a)
G1(d) −G1(a)
x ∈ [a, d), (2.19)
W0(F0(x)) =W0(F0(c))
F0(b)− F0(x)
F0(b)− F0(c)
+W0(F0(b))
F0(x)− F0(c)
F0(b)− F0(c)
, x ∈ (c, b], (2.20)
respectively. We should understand that F0(c) , F0(c+) = ψ0(c+)/ϕ0(c+) = 0 and that G1(d) ,
G1(d−) = −ϕ1(d−)/ψ1(d−) = 0. In the next subsection, we shall explain W1(G1(d−)) and W0(F0(c+))
in details. Both W1 and W0 are a linear function in their respective transformed spaces. Hence under the
appropriate transformations, the two value functions are linear functions in the continuation region.
2.3 Direct Method for a Solution
We have established a mathematical characterization of the value functions of optimal switching problems.
We shall investigate, by using the characterization, a direct solution method that does not require the recur-
sive optimal stopping schemes described in section 2.1. Since the two optimal stopping problems (2.18)
have to be solved simultaneously, finding u0 in x ∈ C0, for example, requires that we find the smallest
F0-concave majorant of (u1(x) + g1(x)− g0(x)−H(x, 1))/ϕ0(x) as in (2.17) that involves u1.
There are two cases, depending on whether x ∈ C1 ∩C0 or x ∈ Γ1 ∩C0, as to what u1(·) represents.
In the region x ∈ Γ1 ∩C0, u1(·) that shows up in the equation of u0(x) is of the form u1(x) = u0(x) +
g0(x)−H(x, 1, 0) − g1(x). In this case, the “obstacle” that should be majorized is in the form
u1(x) + g1(x)− g0(x)−H(x, 1)
= (u0(x) + g0(x)−H(x, 0)− g1(x)) + g1(x)− g0(x)−H(x, 1)
= u0(x)−H(x, 0)−H(x, 1) < u0(x). (2.21)
This implies that in x ∈ Γ1∩C0, the u0(x) function always majorizes the obstacle. Similarly, in x ∈ Γ0∩C1,
the u1(x) function always majorizes the obstacle.
Next, we consider the region x ∈ C0 ∩ C1. The u0(·) term in (2.16) is represented, due to its linear
characterization, as
W0(F0(x)) = β0(F0(x)) + d0
with some β0 ∈ R and d0 ∈ R+ in the transformed space. (The nonnegativity of d0 will be shown.) In
the original space, it has the form of ϕ0(x)(β0F0(x) + d0). Hence by the transformation (u1/ψ1) ◦ G−1,
W1(G1(x)) is the smallest linear majorant of
K1(x) + ϕ0(x)(β0F0(x) + d0)
ψ1(x)
K1(x) + β0ψ0(x) + d0ϕ0(x)
ψ1(x)
on (G1(d−), G1(a∗)) where
K1(x) , g0(x)− g1(x)−H(x, 0). (2.22)
This linear function passes a point (G1(d−), ld) where G1(d−) = 0 and
ld = lim sup
(K1(x) + β0ψ0(x) + d0ϕ0(x))
ψ1(x)
Let us consider further the quantity ld ≥ 0. By noting
lim sup
(K1(x) + β0ψ0(x))
ψ1(x)
≤ lim sup
(K1(x) + β0ψ0(x) + d0ϕ0(x))
ψ1(x)
≤ lim sup
(K1(x) + β0ψ0(x))
ψ1(x)
+ lim sup
d0ϕ0(x)
ψ1(x)
and lim supx↑d
d0ϕ0(x)
ψ1(x)
= 0, we can redefine ld by
ld , lim sup
(K1(x) + β0ψ0(x))
ψ1(x)
(2.23)
to determine the finiteness of the value function of the optimal switching problem, v1(x), based upon Propo-
sition A.5-A.7. Let us concentrate on the case ld = 0.
Similar analysis applies to (2.17). u1(x) in (2.17) is represented as
W1(G1(x)) = β1G1(x) + d1
with some β1 ∈ R and d1 ∈ R+. Note that d1 = ld ≥ 0. In the original space, it has the form of
ψ1(x)(β1G1(x) + d1). Hence by the transformation (u0/ϕ0(x)) ◦ F−1, W0(F0(x)) is the smallest linear
majorant of
K0(x) + ψ1(x)(β1G1(x) + d1)
ϕ0(x)
K0(x)− β1ϕ1(x) + d1ψ1(x)
ϕ0(x)
on (F0(c+), F0(b
∗)) where
K0(x) , g1(x)− g0(x)−H(x, 1). (2.24)
This linear function passes a point (F0(c+), lc) where F0(c+) = 0 and
lc = lim sup
(K0(x)− β1ϕ1(x) + d1ψ1(x))+
ϕ0(x)
Hence we have lc = d0 ≥ 0. By the same argument as for ld, we can redefine
lc , lim sup
(K0(x)− β1ϕ1(x))+
ϕ0(x)
. (2.25)
Remark 2.1. (a) Evaluation of ld or lc does not require knowledge of β0 or β1, respectively unless the
orders of max(K1(x), ψ1(x)) and ψ0(x) are equal, for example. (For this event, see Proposition 2.4.)
Otherwise, we just compare the order of the positive leading terms of the numerator in (2.23) and
(2.25) with that of the denominator.
(b) A sufficient condition for ld = lc = 0: since we have
0 ≤ ld ≤ lim sup
(K1(x))
ψ1(x)
+ lim sup
(β0ψ0(x))
ψ1(x)
a sufficient condition for ld = 0 is
lim sup
(K1(x))
ψ1(x)
= 0 and lim sup
ψ0(x)
ψ1(x)
= 0. (2.26)
Similarly,
0 ≤ lc ≤ lim sup
(K0(x))
ϕ0(x)
+ lim sup
(−β1ϕ1(x))+
ϕ0(x)
Hence a sufficient condition for lc = 0 is
lim sup
(K0(x))
ϕ0(x)
= 0 and lim sup
ϕ1(x)
ϕ0(x)
= 0. (2.27)
Moreover, it is obvious β1 < 0 and β0 > 0 since the linear majorant passes the origin of each
transformed space. Recall a points in the interval (c, d) ∈ R+ will be transformed by G(·) to
(G(c), G(d−)) ∈ R−.
We summarize the case of lc = ld = 0:
Proposition 2.2. Suppose that ld = lc = 0, the quantities being defined by (2.23) and by (2.25), respectively.
The value functions in the transformed space are the smallest linear majorants of
R1(·) ,
1 (·))
1 (·))
and R0(·) ,
0 (·))
0 (·))
where
r1(x) , g0(x)− g1(x) + β0ψ0(x)−H(x, 0)
r0(x) , g1(x)− g0(x)− β1ϕ1(x)−H(x, 1)
β0 > 0 and β1 < 0. (2.28)
Furthermore, Γ1 and Γ0 in (2.16) and (2.17) are given by
Γ1 , {x ∈ (c, d) :W1(G1(x)) = R1(G1(x))}, and Γ0 , {x ∈ (c, d) :W0(F0(x)) = R0(F0(x))}.
Corollary 2.1. If either of the boundary points c or d is absorbing, then (F0(c),W0(F0(c)) or (G1(d),W1(G1(d)))
is obtained directly. We can entirely omit the analysis of lc or ld. The characterization of the value function
(2.19) and (2.20) remains exactly the same.
Remark 2.2. An algorithm to find (a∗, b∗, β∗0 , β
1) can be described as follows:
1. Start with some β′1 ∈ R.
2. Calculate r0 and then R0 by the transformation R0(·) =
0 (·))
0 (·))
3. Find the linear majorant of R0 passing the origin of the transformed space. Call the slope of the linear
majorant, β0 and the point, F0(b), where R0 and the linear majorant meet .
4. Plug b and β0 in the equation for r1 and calculate R1 by the transformation R1(·) =
1 (·))
1 (·))
5. Find the linear majorant of R1 passing the origin of the transformed space. Call the slope of the linear
majorant, β1 and the point, G1(a), where R1 and the linear majorant meet.
6. Iterate step 1 to 5 until β1 = β
If both R1 and R0 are differentiable functions with their respective arguments, we can find (a
∗, b∗) analyti-
cally. Namely, we solve the following system for a and b:
dR0(y)
y=F0(b)
(F0(b)− F0(c)) = R0(F0(b))
dR1(y)
y=G1(a)
(G1(a)−G1(d)) = R1(G1(a))
(2.29)
where dR0(y)
y=F0(b∗)
= β∗0 and
dR1(y)
y=G1(a∗)
= β∗1 .
Once we find W1(·) and W0(·), then we convert to the original space and add back g1(x) and g0(x)
respectively so that v1(x) = ψ1(x)W1(G1(x)) + g1(x) and v0(x) = ϕ0(x)W0(F0(x)) + g0(x). Therefore,
by (2.16) and (2.17), the value functions v1(·) and v0(·) are given by:
Proposition 2.3. If the optimal continuation regions for both of the value functions are connected and if
lc = ld = 0, then the pair of the value functions v1(x) and v0(x) are represented as
v1(x) =
v̂0(x)−H(x, 0), x ≤ a∗,
v̂1(x) , ψ1(x)W1(G1(x)) + g1(x), a
∗ < x,
v0(x) =
v̂0(x) , ϕ0(x)W0(F0(x)) + g0(x) x < b
v̂1(x)−H(x, 1), b∗ ≤ x,
for some a∗, b∗ ∈ R with a∗ < b∗.
Proof. If the optimal continuation regions for both of the value functions are connected and if ld = lc = 0,
then the optimal intervention times (2.30) have the following form:
θ∗i =
inf{t > θi−1;Xt /∈ (a∗, d)}, Z = 1,
inf{t > θi−1;Xt /∈ (c, b∗)}, Z = 0.
(2.30)
Indeed, since we have lc = ld = 0, the linear majorants W1(·) and W0(·) pass the origins in their respective
transformed coordinates. Hence the continuation regions shall necessarily of the form of (2.30).
By our construction, both v1(x) and v0(x) are continuous in x ∈ R. Suppose we have a∗ > b∗. In this
case, by the form of the value functions, v0(b−)−H(b, 1, 0) = v1(b). Since the cost function H(·) > 0 and
continuous, it follows v0(b−) > v1(b). On the other hand, v0(b+) = v1(b)−H(b, 0, 1) implying v0(b+) <
v1(b). This contradicts the continuity of v0(x). Also, a
∗ = b∗ will lead to v1(x) = v1(x)−H(x, 1, 0) which
is impossible. Hence if the value functions exist, then we must necessarily have a∗ < b∗.
In relation to Proposition 2.3, we have the following observations:
Remark 2.3. (a) It is obvious that
v0(x) = v̂0(x) > v̂0(x)−H(x, 0) = v1(x), x ∈ (c, a∗),
v1(x) = v̂1(x) > v̂1(x)−H(x, 1) = v0(x), x ∈ (b∗, d).
(b) Since u1(x) is continuous in (c, d), the “obstacle” u1(x) + g1(x)− g0(x)−H(x, 1) to be majorized
by u0(x) on x ∈ C0 = (c, b∗) is also continuous, in particular at x = a∗. We proved that u0(x)
always majorizes the obstacle on (c, a∗). Hence F (a∗) ∈ {y : W0(y) > R0(y)} if there exists a
linear majorant of R0(y) in an interval of the form (F0(q), F0(d)) with some q ∈ (c, d): otherwise,
the continuity of u1(x) + g1(x) − g0(x) −H(x, 1) does not hold. Similarly, we have F (b∗) ∈ {y :
W1(y) > R1(y)} if there exists a linear majorant of R0(y) in an interval of the form (G1(c), G1(q)).
Finally, we summarize other cases than lc = ld = 0:
Proposition 2.4.
(a) If either ld = +∞ or lc = +∞, then v1(x) = v0(x) ≡ +∞.
(b) If both ld and lc are finite, then ld = lc = 0.
Proof. (a) The proof is immediate by invoking Proposition A.5. (b) When lc is finite, we know by Proposi-
tion A.5 that the value function v0(x) is finite. On x ∈ (c, a∗), u1(x)+g1(x)−g0(x)−H(x, 1) < u0(x) <
+∞ is finite (see (2.21)) and thereby
lc = lim sup
u1(x) + g1(x)− g0(x)−H(x, 1)
ϕ0(x)
The same argument for ld = 0.
Therefore, we can conclude that ld = 0 for the situation where the orders of max(K1(x), ψ1(x)) and ψ0(x)
are equal (⇒ ld is finite) as described in Remark 2.1 (a).
3 Examples
We recall some useful observations. If h(·) is twice-differentiable at x ∈ I and y , F (x), then we define
H(y) , h(F−1(y))/ϕ(F−1(y)) and we obtain H
(y) = m(x) and H
(y) = m
(x)/F
(x) with
m(x) =
(x), and H
(y)(A− α)h(x) ≥ 0, y = F (x) (3.1)
with strict inequality if H
(y) 6= 0. These identities are of practical use in identifying the concavities of
H(·) when it is hard to calculate its derivatives explicitly. Using these representations, we can modify (2.29)
F ′0(b)
(b)(F0(b)− F0(c)) = r0(b)ϕ0(b)
G′1(a)
(a)(G1(a)−G1(d)) = r1(a)ψ1(a)
(3.2)
Example 3.1. Brekke and Øksendal (1994): We first illustrate our solution method by using a resource
extraction problem solved by Brekke and Øksendal (1994). The price Pt at time t per unit of the resource
follows a geometric Brownian motion. Qt denotes the stock of remaining resources in the field that decays
exponentially. Hence we have
dPt = αPtdt+ βPtdWt and dQt = −λQtdt
where α, β, and λ > 0 (extraction rate) are constants. The objective of the problem is to find the optimal
switching times of resource extraction:
v(x) = sup
Jw(x) = sup
e−ρt(λPtQt −K)Ztdt−
e−ρθiH(Xθi−, Zθi)
where rho ∈ R+ is a discount factor with ρ > α, K ∈ R+ is the operating cost and H(x, 0) = C ∈ R+
and H(x, 1) = L ∈ R+ are constant closing and opening costs. Since P and Q always show up in the form
of PQ, we reduce the dimension by defining Xt = PtQt with the dynamics:
dXt = (α− λZt)Xtdt+ βXtdWt.
Solution: (1) We shall calculate all the necessary functions. For Zt = 1 (open state), we solve (A1 −
ρ)v(x) = 0 where A1 = (α − λ)xv′(x) + 12β
2x2v′′(x) to obtain ψ1(x) = x
ν+ and ϕ1(x) = x
ν− where
ν+,− = β
−α+ λ+ 1
(α− λ− 1
β2)2 + 2ρβ2
. Similarly, for Zt = 0 (closed state), we solve
(A0 − ρ)v(x) = 0 where A0 = αxv′(x) + 12β
2x2v′′(x) to obtain ψ0(x) = x
µ+ and ϕ0(x) = x
µ− where
µ+,− = β
−α+ 1
(α− 1
β2)2 + 2ρβ2
. Note that under the assumption ρ > α, we have
ν+, µ+ > 1 and ν−, ν− < 0.
By setting ∆1 =
(α− λ− 1
β2)2 + 2ρβ2 and ∆0 =
(α− 1
β2)2 + 2ρβ2, we have G1(x) =
−ϕ1(x)/ψ1(x) = −x−2∆1/β
and F0(x) = ψ0(x)/ϕ0(x) = x
2∆0/β
. It follows thatG−11 (y) = (−y)−β
2/2∆1
and F−10 (y) = y
β2/2∆0 . In this problem, we can calculate g1(x), g0(x) explicitly:
g1(x) = E
e−ρs(λXs −K)ds
ρ+ λ− α
and g(x) = 0. Lastly, K1(x) = g0(x) − g1(x) − H(x, 0) = −
ρ+λ−α
− C and K0(x) =
g1(x)− g0(x)−H(x, 1) = xρ+λ−α −
(2) The state space of X is (c, d) = (0,∞) and we evaluate lc and ld. Let us first note that ∆0−∆1+λ > 0.
Since limx↓0
ϕ1(x)
ϕ0(x)
= limx↓0 x
∆0−∆1+λ
β2 = 0 and limx↓0(K0(x))
+/ϕ0(x) = 0, we have lc = l0 = 0 by
(2.27). Similarly, by noting limx↑+∞
ψ0(x)
ψ1(x)
= limx↑+∞ x
−(∆0−∆1+λ)
β2 = 0 and limx↑+∞(K1(x))
+/ϕ0(x) =
0, we have ld = l+∞ = 0 by (2.26).
(3) To find the value functions together with continuation regions, we set
r1(x) = −
ρ+ λ− α
− C + β0ψ0(x) and r0(x) =
ρ+ λ− α
− L− β1ϕ1(x)
and make transformations R1(y) = r1(F
−1(y))/ψ1(F
−1(y)) and R0(y) = r0(F
−1(y))/ϕ0(F
−1(y)), re-
spectively. We examine the shape and behavior of the two functions R1(·) and R0(·) with an aid of (3.1).
By calculating (r0/ϕ0)
′(x) explicitly to examine the derivative of R0(y), we can find a critical point x = q,
at which R0(F (x)) attains a local minimum and from which R0(F (x)) is increasing monotonically on
(F0(q),∞). Moreover, we can confirm that limy→∞R′0(y) = limx→∞
(r0/ϕ0)
F ′0(x)
= 0, which shows that
there exists a finite linear majorant of R0(y). We define
p(x) = β1ωx
ν− − (ρ− α)
ρ+ λ− α
+ (K + ρL)
such that (A0 − ρ)r0(x) = p(x) where ω ,
β2ν−(ν− − 1)− αν−
(∆0 − ∆1 + λ)(∆0 +
∆1−λ) > 0. By the second identity in (3.1), the sign of the second derivative R′′0(y) is the same as the sign
of p(x). It is easy to see that p(x) has only one critical point. For any β1 < 0, the first term is dominant as
x → 0, so that limx↓0 p(x) < 0. As x gets larger, for |β1| sufficiently small, p(x) can take positive values,
providing two positive roots, say x = k1, k2 with k1 < k2. We also have limx→+∞ p(x) = −∞. In this
case, R0(y) is concave on (0, F (k1) ∪ (F (k2),+∞) and convex on (F (k1), F (k2)). Since we know that
R0(y) attains a local minimum at y = F (q), we have q < k2, and it implies that there is one and only on
tangency point of the linear majorant W (y) and R0(y) on (F (q),∞), so that the continuation region is of
the form (0, b∗).
¿From this analysis of the derivatives of R0(y), there is only one tangency point of the linear majorant
W0(y) and R0(y). (See Figure 3.1-(a)). A similar analysis shows that there is only one tangency point of
the linear majorant W1(y) and R1(y). (See Figure 3.1-(b)).
0.5 1 1.5 2 2.5 3 3.5
RHFH.L,WHFH.LL
-200 -150 -100 -50
RHGH.L,WHGH.LL
0.5 1 1.5 2 2.5
v0HxL
0.1 0.2 0.3 0.4 0.5 0.6 0.7
v1HxL
Figure 1: A numerical example of resource extraction problem. with parameters (α, β, λ, ρ,K, L,C) =
(0.01, 0.25, 0.01, 0.05, 0.4, 2, 2)(a) The smallest linear majorant W0(F0(x)) and R0(F0(x)) with b
∗ = 1.15042 and
= 10.8125. (b)The smallest linear majorantW1(G1(x)) andR1(G1(x)) with a
∗ = 0.18300 and β∗
= −0.695324.
(c) The value function v0(x). (d) The value function v1(x).
(4) By solving the system of equations (2.29), we can find (a∗, b∗, β∗0 , β
1). We transform back to the original
space to find
v̂1(x) = ψ1(x)W1(G1(x)) + g1(x) = ψ1(x)β
1G1(x) + g1(x)
= −β∗1ϕ1(x) + g1(x) = −β∗1xν− +
ρ+ λ− α
v̂0(x) = ϕ0(x)W0(F0(x)) + g0(x) = ϕ0(x)β
0F0(x) + g0(x) = β
0ψ0(x) + g0(x) = β
Hence the solution is
v1(x) =
µ+ − C, x ≤ a∗,
−β∗1xν− +
ρ+λ−α
, x > a∗,
v0(x) =
µ+ , x ≤ b∗,
−β∗1xν− +
ρ+λ−α
− L, x > b∗,
which agrees with Brekke and Økesendal (1994).
Example 3.2. Ornstein-Uhrenbeck process: We shall consider a new problem involving an Ornstein-
Uhrenbeck process. Consider a firm whose revenue solely depends on the price of one product. Due to its
cyclical nature of the prices, the firm does not want to have a large production facilty and decides to rent
additional production facility when the price is favorable. The revenue process to the firm is
dXt = δ(m −Xt − λZt)dt+ σdWt,
where λ = r/δ with r being a rent per unit of time. The firm’s objective is to maximize the incremental
revenue generated by renting the facility until the time τ0 when the price is at an intolerably low level.
Without loss of generality, we set τ0 = inf{t > 0 : Xt = 0}. We keep assuming constant operating cost K ,
opening cost, L and closing cost C . Now the value function is defined as
v(x) = sup
Jw(x) = sup
e−αt(Xt −K)Ztdt−
θi<τ0
e−αθiH(Xθi−, Zθi)
Solution: (1) We denote, by ψ̃(·) and ϕ̃(·), the functions of the fundamental solutions for the auxiliary
process Pt , (Xt −m+ λ)/σ, t ≥ 0, which satisfies dPt = −δPtdt+ dWt. For every x ∈ R,
ψ̃(x) = eδx
2/2D−α/δ(−x
2δ) and ϕ̃(x) = eδx
2/2D−α/δ(x
which leads to ψ1(x) = ψ̃((x −m + λ)/σ), ϕ1(x) = ϕ̃((x −m + λ)/σ), ψ0(x) = ψ̃((x −m)/σ), and
ϕ0(x) = ϕ̃((x −m)/σ) where Dν(·) is the parabolic cylinder function; (see Borodin and Salminen (2002,
Appendices 1.24 and 2.9) and Carmona and Dayanik (2003, Section 6.3)). By using the relation
Dν(z) = 2−ν/2e−z
2/4Hν(z/
2), z ∈ R (3.3)
in terms of the Hermite function Hν of degree ν and its integral representation
Hν(z) =
Γ(−ν)
2−2tzt−ν−1dt, Re(ν) < 0, (3.4)
(see for example, Lebedev(1972, pp 284, 290)). Since Ex[Xt] = e
−δtx + (1 − e−δt)(m − λ), we have
g0(x) = 0 and g1(x) =
x−(m−λ)
+ m−λ−K
(2) The state space of X is (c, d) = (0,+∞). Since the left boundary 0 is the absorbing, the linear majorant
passes (0, F0(0)). Since limx→+∞ ψ0(x)/ψ1(x) = 0, we have ld = 0.
(3) We formulate
r1(x) = −
x− (m− λ)
δ + α
m− λ−K
− C + β0ψ0(x)
r0(x) =
x− (m− λ)
δ + α
m− λ−K
− L− β1ϕ1(x)
and make transformations: R1(y) = r1(F
−1(y))/ψ1(F
−1(y)) and R0(y) = r0(F
−1(y))/ϕ0(F
−1(y)),
respectively. We examine the shape and behavior of the two functions R1(·) and R0(·) with an aid of (3.1).
First we check the sign of R′0(y) and find a critical point x = q, at which R0(F (x)) attains a local minimum
and from which R0(F (x)) is increasing monotonically on (F0(q),∞). It can be shown that R
0(+∞) = 0
by using (3.3) and (3.4) and the identity H′ν(z) = 2νHν−1(z), z ∈ R (see Lebedev (1972, p.289), for
example.) This shows that there must exist a (finite) linear majorant of R0(y) on (F (q),∞). To check
convexity of R0(y), we define
p(x) = −
ϕ′′1(x) + δ(m− x− λ)
δ + α
− β1ϕ′1(x)
− αr0(x)
such that (A0−α)r0(x) = p(x). We can show easily limx→+∞ p(x) = −∞ since ϕ1(+∞) = ϕ′1(+∞) =
ϕ′′1(+∞) = 0. Due to the monotonicity of ϕ1(x) and its derivatives, p(x) can have at most one critical point
and p(x) = 0 can have one or two positive roots depending on the value of β1. In either case, let us call the
largest positive root x = k2. We also have limx→+∞ p(x) = −∞. Since we know that R0(y) attains a local
minimum at y = F (q) and is increasing thereafter, we have q < k2. It follows that there is one and only on
tangency point of the linear majorant W (y) and R0(y) on (F (q),∞), so that the continuation region is of
the form (0, b∗). A similar analysis shows that there is only one tangency point of the linear majorant W1(y)
and R1(y).
(4) Solving (3.2), we we can find (a∗, b∗, β∗0 , β
1). We transform back to the original space to find
v̂1(x) = ψ1(x)W1(G1(x)) + g1(x) = ψ1(x)β
1G1(x) + g1(x) = −β∗1ϕ1(x) + g1(x)
= −β∗1e
δ(x−m+λ)2
2σ2 D−α/δ
(x−m+ λ)
x− (m− λ)
δ + α
v̂0(x) = ϕ0(x)W0(F0(x)) + g0(x) = ϕ0(x)β
0 (F0(x)− F0(0)) + g0(x)
= β∗0{ψ0(x)− F0(0)ϕ0(x)}+ g0(x)
= β∗0e
(x−m+λ)2
D−α/δ
x−m+ λ
− F (0)D−α/δ
Hence the solution is, using the above functions,
v1(x) =
v̂0(x)− C, x ≤ a∗,
v̂1(x), x > a
v0(x) =
v̂0(x), x ≤ b∗,
v̂1(x)− L, x > b∗.
See Figure 3.2 for a numerical example.
0.5 1 1.5 2 2.5 3
v0HxL
0.5 1 1.5 2 2.5 3
v1HxL
Figure 2: A numerical example of leasing production facility problem with parameters (m,α, σ, δ, λ,K, L,C) =
(5, 0.105, 0.35, 0.05, 4, 0.4, 0.2, 0.2): (a) The value function v0(x) with b
∗ = 1.66182 and β∗
= 144.313. (b)The
value function v1(x) with a
∗ = 0.781797 and β∗
= −2.16941.
4 Extensions and conclusions
4.1 An extension to the case of k ≥ 2
It is not difficult to extend to a general case of k ≥ 2 where more than one switching opportunities are
available. But we put a condition that z ∈ Z is of the form z = (a1, a2, ...., ak) where only one element of
this vector is 1 with the rest being zero, i.e., z = (0, 0, 0, ...., 1, 0, 0) for example.
We should introduce the switching operator M0 on h ∈ H,
M0h(u, z) = max
ζ∈Z\{z}
{h(u, ζ)−H(u, z; ζ)} . (4.1)
In words, this operator would calculate which production mode should be chosen by moving from the current
production mode z. Now the recursive optimal stopping (2.7) becomes
wn+1(x) , sup
e−αsf(Xs)ds+ e
−ατMwn(Xτ )
Accordingly, the optimization procedure will become two-stage. To illustrate this, we suppose k = 2 so that
i = 0, 1, and 2. By eliminating the integral in (4.1), we redefine the switching operator,
Mhz(x) , max
ζ∈Z\{z}
{hζ(x) + gζ(x)− gz(x)−H(x, z, ζ)} , (4.2)
where
gz(x) , sup
Jwz (x) = E
e−αsf(Xs)ds
Hence (2.13) will be modified to uz(x) = E
x[e−ατMuz(Xτ )]. It follows that our system of equations
(2.18) is now
v̄2(x) , supτ∈S E
x [e−ατMv̄2(Xτ )]
v̄1(x) , supτ∈S E
x [e−ατMv̄1(Xτ )]
v̄0(x) , supτ∈S E
x [e−ατMv̄0(Xτ )]
(4.3)
The first stage is optimal stopping problem. One possibility of switching production modes is (0 → 1, 1 →
2, 2 → 0). First, we fix this switching scheme, say c, and solve the system of equations (4.3) as three
optimal stopping problems. All the arguments in Section 2.3 hold. This first-stage optimization will give
(x∗0(c), x
1(c), x
2(c), β
0 (c), β
1 (c), β
2 (c)), where xi’s are switching boundaries, depending on this switching
scheme c.
Now we move to another switching scheme c′ and solve the system of optimal stopping problems until
we find the optimal scheme.
4.2 Conclusions
We have studied optimal switching problems for one-dimensional diffusions. We characterize the value
function as linear functions in their respective spaces, and provide a direct method to find the value functions
and the opening and switching boundaries at the same time. Using the techniques we developed here as
well as the ones in Dayanik and Karazas (2003) and Dayanik and Egami (2005), we solved two specific
problems, one of which involves a mean-reverting process. This problem might be hard to solve with just
the HJB equation and the related quasi-variational inequalities. Finally, an extension to more general cases
is suggested. We believe that this direct method and the new characterization will expand the coverage of
solvable problems in the financial engineering and economic analysis.
A Summary of Optimal Stopping Theory
Let (Ω,F ,P) be a complete probability space with a standard Brownian motion W = {Wt; t ≥ 0} and
consider the diffusion process X0 with state pace I ⊆ R and dynamics
dX0t = µ(X
t )dt+ σ(X
t )dWt (A.1)
for some Borel functions µ : I → R and σ : I → (0,∞). We emphasize here that X0 is an uncontrolled
process. We assume that I is an interval with endpoints −∞ ≤ a < b ≤ +∞, and that X0 is regular in
(a, b); in other words, X0 reaches y with positive probability starting at x for every x and y in (a, b). We
shall denote by F = {Ft} the natural filtration generated by X0.
Let α ≥ 0 be a real constant and h(·) a Borel function such that Ex[e−ατh(X0τ )] is well-defined for
every F-stopping time τ and x ∈ I . Let τy be the first hitting time of y ∈ I by X0, and let c ∈ I be a fixed
point of the state space. We set:
ψ(x) =
x[e−ατc1{τc<∞}], x ≤ c,
1/Ec[e−ατx1{τx<∞}], x > c,
ϕ(x) =
e−ατx1{τx<∞}
, x ≤ c,
x[e−ατc1{τc<∞}], x > c,
F (x) ,
, x ∈ I. (A.2)
Then F (·) is continuous and strictly increasing. It should be noted that ψ(·) and ϕ(·) consist of an increasing
and a decreasing solution of the second-order differential equation (A − α)u = 0 in I where A is the
infinitesimal generator of X0. They are linearly independent positive solutions and uniquely determined
up to multiplication. For the complete characterization of ψ(·) and ϕ(·) corresponding to various types of
boundary behavior, refer to Itô and McKean (1974).
Let F : [c, d] → R be a strictly increasing function. A real valued function u is called F -concave on
[c, d] if, for every a ≤ l < r ≤ b and x ∈ [l, r],
u(x) ≥ u(l)
F (r)− F (x)
F (r)− F (l)
+ u(r)
F (x)− F (l)
F (r)− F (l)
We denote by
V (x) , sup
x[e−ατh(X0τ )], x ∈ [c, d] (A.3)
the value function of the optimal stopping problem with the reward function h(·) where the supremum is
taken over the class S of all F-stopping times. Then we have the following results, the proofs of which we
refer to Dayanik and Karatzas (2003).
Proposition A.1. For a given function U : [c, d] → [0,+∞) the quotient U(·)/ϕ(·) is an F -concave function
if and only if U(·) is α-excessive, i.e.,
U(x) ≥ Ex[e−ατU(X0τ )],∀τ ∈ S,∀x ∈ [c, d]. (A.4)
Proposition A.2. The value function V (·) of (A.3) is the smallest nonnegative majorant of h(·) such that
V (·)/ϕ(·) is F -concave on [c, d].
Proposition A.3. Let W (·) be the smallest nonnegative concave majorant of H , (h/ϕ) ◦ F−1 on
[F (c), F (d)], where F−1(·) is the inverse of the strictly increasing function F (·) in (A.2). Then V (x) =
ϕ(x)W (F (x)) for every x ∈ [c, d].
Proposition A.4. Define
S , {x ∈ [c, d] : V (x) = h(x)}, and τ∗ , inf{t ≧ 0 : X0t ∈ S}. (A.5)
If h(·) is continuous on [c, d], then τ∗ is an optimal stopping rule.
When both boundaries are natural, we have the following results:
Proposition A.5. We have either V ≡ 0 in (c, d) or V (x) < +∞ for all (c, d). Moreover, V (x) < +∞ for
every x ∈ (c, d) if and only if
lc , lim sup
h+(x)
and ld , lim sup
h+(x)
(A.6)
are both finite.
In the finite case, furthermore,
Proposition A.6. The value function V (·) is continuous on (c, d). If h : (c, d) → R is continuous and
lc = ld = 0, then τ
∗ of (A.5) is an optimal stopping time.
Proposition A.7. Suppose that lc and ld are finite and one of them is strictly positive, and h(·) is continuous.
Define the continuation region C , (c, d) \ Γ. Then τ∗ of (A.5) is an optimal stopping time, if and only if
there is no r ∈ (c, d) such that (c, r) ⊂ C if lc > 0 and
there is no l ∈ (c, d) such that (l, d) ⊂ C if ld > 0.
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Math. Meth. Oper. Res.53, 1–24.
	Introduction
	Optimal Switching Problems
	Characterization of switching times
	Characterization of the value functions
	Direct Method for a Solution
	Examples
	Extensions and conclusions
	An extension to the case of k2
	Conclusions
	Summary of Optimal Stopping Theory
