Optimality of Two-Parameter Strategies in Stochastic Control
Kazutoshi Yamazaki
Department of Mathematics, Kansai University
Simposio de Probabilidad y Procesos Estocásticos
November 16 – 20, 2015
Explicit Solutions in Stochastic Control
One Parameter Case
Optimal stopping – Mordecki (2002), Asmussen, Avram & Pistorius
(2004), Hilberink & Rogers (FS)
Singular control – w/o fixed costs – Avram, Palmowski, Pistorius
(2007), Loeffen (2008), Bayraktar, Kyprianou & Y. (2014).
Two Parameter Case
2 of 137
Impulse control – w/ fixed costs – Bensoussan, Liu & Sethi (2005),
Loeffen (2011), Bayraktar, Kyprianou & Y. (2013).
Games b/w two players – Egami, Leung & Y. (2013),
Hernandez-Hernandez & Y. (2015)
Two-sided singular control – Baurdoux & Y. (2015)
One Parameter Case
Suppose the NPV va (x) can be computed.
Smooth/Continuous fit (or first-order) condition to derive the
candidate threshold a∗ .
Typically one obtains a monotone function a 7→ Γ(a):
20
15
10
I(b)
5
0
-5
-10
-15
-20
-20
-15
-10
-5
0
b
Obtain the root a∗ s.t. Γ(a∗ ) = 0.
Then verify the optimality of va∗ .
3 of 137
5
10
15
20
Two Parameter Case
Suppose the NPV va,b (x) can be computed.
Two Smooth/Continuous fit (or two first-order) conditions to
derive the candidate thresholds a∗ and b ∗ .
Typically one obtains a function like (a, b) 7→ Γ(a, b) and the
conditions are equivalently to Γ(a∗ , b ∗ ) = ∂Γ(a∗ , b ∗ )/∂b = 0.
300
200
100
Gamma
0
-100
-200
-300
-400
-500
-3
-2
-1
0
1
2
3
b
Consider b 7→ Γ(a, b) and obtain the curve s.t. it starts at a∗ and
touches and is tangent to the x-axis at b ∗ .
Then verify the optimality of va∗ ,b∗ .
4 of 137
Part 1: One-parameter Strategies
5 of 137
Outline
Review of One-parameter strategies –
optimal stopping
singular control.
Regularity and smooth/continuous fit.
Lévy processes, scale functions, fluctuation theory
When do we need two parameters to describe optimal strategies –
6 of 137
Two-sided singular control
Stochastic games
Impulse control
Optimal Stopping
Want to maximize or minimize
i
hZ τ
e −qt f (Xτ )dt + e −qτ g (Xτ ) ,
v τ (x) := Ex
0
over all stopping times τ .
Examples include
7 of 137
American options
Change point detection
Sequential hypothesis testing
Secretary problems ...
When threshold strategy is expected to be optimal
Suppose we can compute for τb (the first time X up/down-crosses
b)
h Z τb
i
vb (x) := Ex
e −qt f (Xτb )dt + e −qτb g (Xτb ) .
0
Guess via Smooth/continuous fit
choose b such that vb is smooth/continuous at b
typically the condition is equivalent to the first order condition
If lucky, we obtain a monotone function:
20
15
10
I(b)
5
0
-5
-10
-15
-20
-20
-15
-10
-5
0
5
10
15
20
b
Its root becomes the optimal threshold level.
8 of 137
Regularity and Smooth/Continuous fit
A point b ∈ R is regular for an open or closed set B if
Pb {τ B = 0} = 1
where
τ B := inf{t > 0 : Xt ∈ B}.
Typically, we pursue a stopping region B ∗ (in the form (b ∗ , ∞) or
(−∞, b ∗ )) so that the value function vb is
9 of 137
smooth when b ∗ is regular for B ∗ ;
continuous when b ∗ is irregular for B ∗ .
Verification of v ∗ = vb∗
To show that our candidate τb∗ (v ∗ = vb∗ ) is indeed optimal, we need
smoothness - smooth enough to apply (a version of) Ito’s formula
already done by how b ∗ was chosen by smooth/continuous fit.
variational inequality
(i) v ∗ (x) ≥ g (x) for all x ∈ R,
(ii) (L − q)v ∗ (x) + f (x) = 0 for all x ∈
/ B ∗,
∗
(iii) (L − q)v (x) + f (x) < 0 for all x ∈ B ∗ .
where L is the infinitesimal generator.
some localizing arguments
v ∗ should not grow too rapidly in the tail – we want to take limits
inside expectation)
10 of 137
How Verification Works
1. Ito’s lemma + smoothness/continuity gives
Z t
e −qt v ∗ (Xt ) − v ∗ (X0 ) =
e −qs (L − q)v ∗ (Xs− )ds + Mt ,
0
where M is a local martingale.
2. With Tm := inf{t > 0 : |Xt | > m},
∗
x
h
v (x)≥ E e
−q(t∧τ ∧Tm ) ∗
Z
v (Xt∧τ ∧Tm ) +
Z
h
x
−q(t∧τ ∧Tm )
≥E e
g (Xt∧τ ∧Tm ) +
0
3. Finally take t, m ↑ ∞.
11 of 137
t∧τ ∧Tm
0
t∧τ ∧Tm
e −qs f (Xs )ds
e −qs f (Xs )ds
i
i
Singular Control
We want to maximize/minimize
Z
hZ ∞
e −qt f (Xt ± Utπ )dt ±
v π (x) := Ex
0
Examples include
Optimal dividends
Inventory models
12 of 137
[0,∞)
i
e −qt dUtπ .
Singular Control: Guessing
We pursue a reflection barrier b ∗ so that the process stays on
[b ∗ , ∞) or (−∞, b ∗ ].
Control region B ∗ is its complement.
We obtain b ∗ so that the expected NPV value functional vb∗ is
twice-differentiable when b ∗ is regular for B ∗ ;
differentiable when b ∗ is irregular for B ∗ .
If lucky, we obtain a monotone function:
20
15
10
I(b)
5
0
-5
-10
-15
-20
-20
-15
-10
-5
0
5
10
15
20
b
Its root becomes the optimal threshold level.
13 of 137
Singular Control: Verifying
To show that our candidate v ∗ = vb∗ is indeed optimal, we need
smoothness - smooth enough to apply (a version of) Ito’s formula
variational inequality
(i)
(ii)
(iii)
(iv)
already done by how b ∗ was chosen by the smoothness condition.
0
v ∗ (x) ≥ 1 for all x ∈ R\B ∗ ,
0
v ∗ (x) = 1 for all x ∈ B ∗ ,
(L − q)v ∗ (x) + f (x) = 0 for all x ∈ R\B ∗ ,
(L − q)v ∗ (x) + f (x) ≤ 0 for all x ∈ B ∗ .
where L is the infinitesimal generator.
some localizing arguments
v ∗ should not grow too rapidly in the tail – we want to take limits
inside expectation)
14 of 137
Example: part of Y. (2014)
Suppose we want to maximize
Z τ
−rt
−r τ
τ
e f (Xt )dt + e g (Xτ )1{τ <∞} ,
u (x) := Ex
0
where
P
ai x for some constants K ∈ R, b ≥ 0
g (x) = K − bx − N
i=1 ci e
and ci , ai > 0, 1 ≤ i ≤ N, N ≥ 0,
f (·) is continuous, piecewise differentiable, and increasing (and
some condition on its tail).
15 of 137
Spectrally Negative Lévy Processes
Let X be a spectrally negative Lévy process with a
Laplace exponent:
h
i
sX1
ψ(s) := log E e
1
= cs + σ 2 s 2 +
2
such that
R
(−∞,0) (1
Z
(e sz − 1 − sz1{−1<z<0} )ν(dz),
(−∞,0)
∧ z 2 )ν(dz) < ∞.
It
R has paths of bounded variation if and only if σ = 0 and
(−1,0) z ν(dz) < ∞.
We exclude the case X is a subordinator.
16 of 137
Scale Functions
Laplace exponent ψ(s) = log E e sX1 .
Fix any q > 0, there exists a function called the q-scale function
W (q) : R → [0, ∞),
which is zero on (−∞, 0), continuous and strictly increasing on
[0, ∞), and is characterized by the Laplace transform:
Z ∞
1
e −sx W (q) (x)dx =
,
s > Φ(q),
ψ(s) − q
0
where
Φ(q) := sup{λ ≥ 0 : ψ(λ) = q}.
Define
Z
(q)
Z
W
(x) := 1 + q
0
17 of 137
x
(q)
(y )dy ,
Z
(q)
Z
(x) :=
0
x
Z (q) (y )dy .
Spectrally Negative Case
Z τA
e −qt f (Xt )dt + e −qτA g (XτA )1{τA <∞}
uA (x) := Ex
0
q
= K Z (q) (x − A) −
W (q) (x − A)
Φ(q)
N
X
q − ψ(ai ) (q−ψ(ai ))
i ))
−
ci e ai x Za(q−ψ(a
(x − A) −
Wai
(x − A)
i
Φ(q) − ai
i=1
h (q)
ψ 0 (0+) (q)
ψ 0 (0+)
− b Z (x − A) + A −
Z (x − A) +
q
q
i
q − ψ 0 (0+)Φ(q) + qAΦ(q) (q)
W (x − A)
−
Φ(q)2
Z ∞
+ W (q) (x − A)
e −Φ(q)y f (y + A)dy
0
Z x
−
W (q) (x − y )f (y )dy .
18 of 137
A
Continuous/smooth fit
For SN Lévy process of bounded variation, A is irregular for
(−∞, A) – we apply continuous fit.
For SN Lévy process of unbounded variation, A is regular for
(−∞, A) – we apply smooth fit.
Both turn out to be equivalent to Λ(A) = 0 where
q
qA − ψ 0 (0+) q
K +b
+
Λ(A):= −
Φ(q)
Φ(q)2
Φ(q)
Z
N
x
X
q − ψ(ai )
+
ci e ai A
+
W (q) (x − y )f (y )dy .
Φ(q) − ai
A
i=1
Λ(A) is continuous and increasing.
If limA↓−∞ Λ(A) < 0 < limA↑∞ Λ(A), there exists a unique root
A∗ ∈ R such that Λ(A∗ ) = 0.
∗
Otherwise, let A∗ = −∞ if lim
A↓−∞ Λ(A) ≥ 0 and let A = ∞ if
19 oflim
137A↑∞ Λ(A) ≤ 0.
Value functions
Candidate Value function becomes
uA∗ (x)= KZ (q) (x − A∗ )
h
ψ 0 (0+) (q)
ψ 0 (0+) i
(q)
−b Z (x − A∗ ) + A∗ −
Z (x − A∗ ) +
q
q
Z
N
x
X
(q−ψ(ai ))
−
c i e ai x Z ai
(x − A∗ ) −
W (q) (x − y )f (y )dy .
i=1
A∗
Verification
(i) v ∗ (x) ≥ g (x) for all x ∈ R −→ easy
(ii) (L − q)v ∗ (x) + f (x) = 0 for all x ∈ (A∗ , ∞) −→ easy
(iii) (L − q)v ∗ (x) + f (x) < 0 for all x ∈ (−∞, A∗ ) −→ hard.
For (i), observe smooth/continuous fit ⇔ first order condition
w.r.t. A.
(q)
For (ii), use (L − q)Z (q) (x) = (L − q)(Z (x) + ψ 0 (0+)/q) = 0.
20 of 137
Optimal Dividend
A strategy π = {Lπt , t ≥ 0} is a nondecreasing and adapted process
starting at zero.
A controlled risk process is the difference:
Utπ := Xt − Lπt ,
t ≥ 0.
Time of ruin: σ π := inf {t > 0 : Utπ < 0}.
We want to maximize, for q > 0,
Z σπ
−qt
π
vπ (x) := Ex
e dLt ,
0
π + ∆X for all
over the set of all strategies Π satisfying ∆Lπt ≤ Ut−
t
π
t ≤ σ a.s.
We want to obtain the value function:
v (x) := sup vπ (x),
π∈Π
21 of 137
x ≥ 0.
Solution Procedures
We follow a classical approach “guess” and “verify”.
Guess that an optimal strategy is a barrier strategy (reflected Lévy
process) πa := {Lat ; t ≤ σa } in the form:
Lat := sup (Xs − a) ∨ 0,
0≤s≤t
Uta
:= Xt − Lat ,
with the corresponding ruin time σa := inf {t > 0 : Uta < 0}.
Choose the value of a using some smoothness condition.
Verify that
Z σa
−qt
a
va (x) := Ex
e dLt ≥ sup vπ (x).
0
22 of 137
π∈Π
SN case: Avram et al. (2004) etc
As in Avram, Palmowski & Pistorius (2004)

(q)
 W 0(x) ,
0 ≤ x ≤ a,
W (q) (a)
va (x) =
(q)
 (x − a) + W 0(a) , x > a,
W (q) (a)
a is regular for (a, ∞) – twice-differentiability should hold for a –
00
or W (q) (a) = 0.
It turned out that optimality is not guaranteed, but Loeffen (2009)
obtained a sufficient condition on the Lévy measure.
23 of 137
SP case: Bayraktar et al. (2013)
Unlike the SN case, the SP case is straightforward.
Lat := sup (Xs − a) ∨ 0
0≤s≤t
and Uta := Xt − Lat .
By an Application of Avram, Palmowski, Pistorius (2004),

 −k(a − x) + Z (q) (a−x) k(a), 0 ≤ x ≤ a,
Z (q) (a)
va (x) =
 k(a)
,
x ≥ a,
(q)
Z
(a)
where
Z
k(y ) :=
0
y
Z (q) (z)dz −
1
ψ 0 (0+)
Z (q) (y ) +
.
Φ(q)
q
X is of bounded var. → a is irregular for (a, ∞) → pursue C 1 .
X is of unbounded var. → a is regular for (a, ∞) → pursue C 2 .
24 of 137
SP case: Bayraktar et al. (2013)
We will denote our candidate barrier level by

 (q) −1 |ψ0 (0+)|
Z
> 0 if ψ 0 (0+) < 0,
q
a∗ =
0
if ψ 0 (0+) ≥ 0,
Rx
(q)
which is well-defined because Z (x) := 0 Z (q) (z)dz is monotone.
Theorem (Bayraktar, Kyprianou & Y. (Astin Bull., 2013))
We have
va∗ (x) := sup vπ (x),
x ≥ 0,
π∈Π
where
va∗ (x) =
25 of 137
(
(q)
−Z (a∗ − x) −
x,
ψ 0 (0+)
,
q
if ψ 0 (0+) > 0,
if ψ 0 (0+) ≤ 0.
Motivation to extend to the two-parameter case
Optimal Stopping: monotone function → v-shaped function.
One player → two players.
Proportional costs/rewards → proportional + fixed rewards.
If two thresholds characterize the optimal strategy
apply smooth/continuous fit at the two boundaries → this gives
two equations.
Solutions (if any) will be our candidate thresholds.
Extra work on verification.
26 of 137
References
[1] E. Bayraktar, A. E. Kyprianou and K. Yamazaki. On Optimal Dividends
in the Dual Model. ASTIN Bulletin, 43(3):359-372, 2013.
[2] K. Yamazaki. Contraction Options and Optimal Multiple-Stopping in
Spectrally Negative Levy Models. Applied Mathematics and
Optimization, 72(1):147-185, 2015.
27 of 137
Part 2: Inventory Control: from one-parameter
to two parameters.
28 of 137
Inventory Control: One-sided & No Fixed Cost Case
(Ω, F, P) hosting a spectrally negative Lévy process X .
An admissible inventory strategy π := {Utπ ; t ≥ 0} –
nondecreasing, right-continuous
and
hR
i adapted processes such that
π
−qt
π
U0− = 0 and Ex [0,∞) e dUt < ∞.
We want to minimize
hZ
π
v (x) := Ex
∞
e
−qt
f (Ytπ )dt
Z
+
[0,∞)
0
with the controlled process
Ytπ := Xt + Utπ .
29 of 137
i
e −qt C dUtπ ,
Assumptions on Inventory Cost
Assumption
1. f is a piecewise continuously differentiable function with f (0) = 0.
2. There exists a number a such that the function
f˜(x) := f (x) + Cqx,
x ∈ R,
is increasing on (a, ∞) and decreasing and convex on (−∞, a).
3. There exist a c0 > 0 and an x0 ≥ a such that f˜0 (x) ≥ c0 for x ≥ x0 .
30 of 137
Outline
One-sided & No Fixed Cost Case – one-parameter strategy is
optimal.
Two extensions:
One-sided & Fixed Cost Case – two-parameter strategy is optimal
(often called (s, S)-policy)
Two-sided & No Fixed Cost Case – two-parameter strategy is optimal.
31 of 137
Solutions for One-sided & No Fixed Cost Case
Focus on a set of barrier strategies Uta that reflect the process at a
so that the inventory stays at or above a:
Uta := sup (a − Xt 0 ) ∨ 0,
0≤t 0 ≤t
Yta := Xt + Uta .
The NPV corresponding to this strategy
Z
hZ ∞
va (x) := Ex
e −qt f (Yta )dt +
0
i
e −qt C dUta ,
[0,∞)
can be written in terms of the scale function if X is a spectrally
negative Lévy process.
32 of 137
Spectrally Negative Lévy Processes
Let X be a spectrally negative Lévy process with a
Laplace exponent:
h
i
sX1
ψ(s) := log E e
1
= cs + σ 2 s 2 +
2
such that
R
(−∞,0) (1
Z
(e sz − 1 − sz1{−1<z<0} )ν(dz),
(−∞,0)
∧ z 2 )ν(dz) < ∞.
It
R has paths of bounded variation if and only if σ = 0 and
(−1,0) z ν(dz) < ∞.
We exclude the case X is a subordinator.
33 of 137
Scale Functions
Laplace exponent ψ(s) = log E e sX1 .
Fix any q > 0, there exists a function called the q-scale function
W (q) : R → [0, ∞),
which is zero on (−∞, 0), continuous and strictly increasing on
[0, ∞), and is characterized by the Laplace transform:
Z ∞
1
e −sx W (q) (x)dx =
,
s > Φ(q),
ψ(s) − q
0
where
Φ(q) := sup{λ ≥ 0 : ψ(λ) = q}.
Define
Z
(q)
Z
(x) := 1 + q
0
34 of 137
x
W (q) (y )dy .
How we choose a
X is of bounded var. → a is irregular for (−∞, a) → pursue C 1 .
X is of unbounded var. → a is regular for (−∞, a) → pursue C 2 .
Define
Z
Ψ(s; h) :=
∞
e −Φ(q)y h(y + s)dy =
0
e −Φ(q)(y −s) h(y )dy .
s
There exists a unique number a0 such that Ψ(a0 ; f˜0 ) = 0,
Ψ(x; f˜0 ) < 0 if x < a0 and Ψ(x; f˜0 ) > 0 if x > a0 where we recall
f˜(x) := f (x) + Cqx,
∞
Z
This a0 satisfies the above condition.
35 of 137
x ∈ R.
The Case with No Fixed Costs (Cont’d)
Theorem
The barrier strategy La0 is optimal and the value function is given by
va0 (x) = −Cx −
C ψ 0 (0+) Z (q) (x − a0 ) ˜
+
f (a0 ) − ϕa0 (x; f˜).
q
q
where
Z
ϕs (x; h) :=
s
36 of 137
x
W (q) (x − y )h(y )dy ,
x ∈ R.
Part 2-1: Inventory control with fixed costs
37 of 137
Inventory Models with Fixed Costs
1. Uncontrolled surplus: Xt , t ≥ 0 defined on (Ω, F, F, P).
2. An (ordering) policy
n
o
X
π := Utπ =
uiπ ; t ≥ 0
i:Tiπ ≤t
in the form of an impulse control (T1π , u1π ; T2π , u2π ; · · · ) where
{Ti ; i ≥ 1} is an increasing sequence of F-stopping times and
ui > 0 is an FTi -measurable random variable for i ≥ 1.
3. Corresponding to every policy π, the (controlled) surplus process is
Ytπ := Xt + Utπ ,
t ≥ 0.
4. The problem is to minimize the total expected cost:
∞
hZ ∞
i
X
π
vπ (x) := Ex
e −qt f (Ytπ )dt +
e −qTi (K + Cuiπ ) .
0
38 of 137
i=1
The Model of X
Compound Poisson models
Brownian Motion models
Sum of Compound Poisson and Brownian Motion
Bensoussan, Liu & Sethi (SICON, 2005), Benkherouf & Bensoussan
(SICON, 2009).
Spectrally negative Lévy models (this talk) – a general Lévy process
with only positive jumps that is not a negative of a subordinator.
39 of 137
Assumptions
The assumptions are the same as those in Bensoussan, Liu & Sethi
(SICON, 2005), Benkherouf & Bensoussan (SICON, 2009).
Assumption
g (y ) := Cy + K , y > 0, for some unit cost of the item C ∈ R and
fixed ordering cost K > 0.
Assumption
1. f is a piecewise continuously differentiable function with f (0) = 0.
2. There exists a number a such that the function
f˜(x) := f (x) + Cqx,
x ∈ R,
is increasing on (a, ∞) and decreasing and convex on (−∞, a).
3. There exist a c0 > 0 and an x0 ≥ a such that f˜0 (x) ≥ c0 for x ≥ x0 .
40 of 137
The (s, S)-Policy
For −∞ < s < S < ∞, an (s, S)-policy, πs,S := Uts,S ; t ≥ 0 ,
brings the level of the surplus process Y s,S := X + U s,S up to S
whenever it goes below s, with the corresponding NPV:
hZ ∞
i
X
e −qt f (Yts,S )dt +
e −qt g (∆Uts,S )1{∆U s,S >0} .
vs,S (x) := Ex
0
t
0≤t<∞
We aim to prove that the (s ∗ , S ∗ )-policy is optimal for some
−∞ < s ∗ < S ∗ < ∞. Toward this end,
Write vs,S analytically using the scale function.
Choose the value of s ∗ and S ∗ using some smoothness condition.
Verify that the Vs ∗ ,S ∗ satisfies QVI (quasi-variational inequality)
41 of 137
Applications of Scale Functions
Let us define the first down- and up-crossing times, respectively, by
τa− := inf {t ≥ 0 : Xt < a}
τb+ := inf {t ≥ 0 : Xt > b} .
Then we have for any b > 0
h
i W (q) (x)
+
Ex e −qτb 1{τ + <τ − } = (q) ,
0
b
W (b)
i
h
−
W (q) (x)
Ex e −qτ0 1{τ + >τ − } = Z (q) (x) − Z (q) (b) (q) ,
0
b
W (b)
where
Z x
(q)
W (x) :=
W (q) (y )dy ,
0
Z
42 of 137
(q)
(x) := 1 + qW
(q)
(x).
Computation of vs,S (x)
1. We shall first rewrite
hZ ∞
i
X
vs,S (x) := Ex
e −qt f (Yts,S )dt +
e −qt g (∆Uts,S )1{∆U s,S >0} ,
t
0
0≤t<∞
in terms of the scale function.
2. By the strong Markov property, it must satisfy,
vs,S (x) = Ex
hZ
τs−
e −qt f (Xt )dt
i
0
h
+ Ex e
−qτs−
i
i
h
−
(C (S − Xτs− ) + K ) + Ex e −qτs vs,S (S).
3. Solving for x = S gives vs,S (S) once we know these expectations.
43 of 137
Computation of vs,S (x)
Instead, we write
ṽs,S (x) := vs,S (x) + Cx
= k(s, x) + 1 −
(q)
q
(q)
Θ (x − s) ṽs,S (S),
Φ(q)
(q)
x > s,
(x) := W (q) (x) − Φ(q)W (x) and
h Z τs−
i
h
i
−
k(s, x) := Ex
e −qt f (Xt )dt − C Ex e −qτs Xτs−
h 0 −i
+ K Ex e −qτs + Cx, x > s.
with Θ
Then,
ṽs,S := ṽs,S (S) =
44 of 137
Φ(q) k(s, S)
,
q Θ(q) (S − s)
S > s.
Computation of vs,S (x)
Define, for any measurable function h and s ∈ R,
Z ∞
Z ∞
e −Φ(q)y h(y + s)dy =
e −Φ(q)(y −s) h(y )dy ,
Ψ(s; h) :=
s
Z0 x
(q)
ϕs (x; h) :=
W (x − y )h(y )dy , x ∈ R,
s
(q)
G(s, x) := Φ(q)Ψ(s; f˜)W (x − s) + K − ϕs (x; f˜),
x > s.
For any x > s,
(q)
k(s, x) = Θ
45 of 137
(x − s) Ψ(s; f˜) −
q
Φ(q)
C ψ 0 (0+)
K+
+ G(s, x).
q
Choosing Candidates for (s, S)
To narrow down the candidates for (s, S), we shall choose these
values so that
the function vs,S is continuous/smooth enough and
its slope at S is the same as the proportional cost C .
Lemma
Suppose (s, S) are such that G(s, S) = H(s, S) = 0 where
H(s, x) :=
∂
G(s, x).
∂x
Then
1. vs,S is continuous (resp. differentiable) at s when X is of bounded
(resp. unbounded) variation,
0 (S) = 0 or equivalently v 0 (S) = C .
2. ṽs,S
s,S
46 of 137
Plots of G(s, S) and H(s, S)
150
120
100
100
80
H
G
60
50
40
20
0
0
-20
-50
-3
-2
-1
0
1
-40
-3
2
-2
-1
s,S
0
1
2
0
1
2
s,S
unbounded variation case
140
120
120
100
100
80
80
60
40
H
G
60
20
40
20
0
0
-20
-20
-40
-60
-3
-2
-1
0
1
2
-40
-3
-2
-1
s,S
s,S
bounded variation case
47 of 137
Existence of (s ∗ , S ∗ )
1. Recall
Z
∞
e
Ψ(s; h) :=
−Φ(q)y
Z
h(y + s)dy =
0
∞
e −Φ(q)(y −s) h(y )dy .
s
2. There exists a unique number a0 such that Ψ(a0 ; f˜0 ) = 0,
Ψ(x; f˜0 ) < 0 if x < a0 and Ψ(x; f˜0 ) > 0 if x > a0 .
3. There exists s ∗ < a0 and S ∗ > a0 such that
∗
s := sup s < a0 : inf G(s, S) = 0
and S ∗ ∈ arg inf G(s, S),
S≥s
holds with H(s ∗ , S ∗ ) = G(s ∗ , S ∗ ) = 0.
48 of 137
S≥s
Verification for Optimality
Proposition
1. (L − q)vs ∗ ,S ∗ (x) + f (x) = 0 for x > s ∗ ,
2. (L − q)vs ∗ ,S ∗ (x) + f (x) ≥ 0 for x < s ∗ .
Proposition
For every x ∈ R, we have vs ∗ ,S ∗ (x) ≤ K + inf u≥0 [Cu + vs ∗ ,S ∗ (x + u)].
Theorem
The (s ∗ , S ∗ )-policy is optimal and the value function is given by
vs ∗ ,S ∗ (x) =
49 of 137
Φ(q)
C ψ 0 (0+)
Ψ(s ∗ ; f˜)Z (q) (x − s ∗ ) − ϕs ∗ (x; f˜) −
− Cx,
q
q
x>
Meromorphic Lévy Processes
A class of the meromorphic Lévy process Kuznetsov, Kyprianou &
Pardo (AAP, 2012) admits the Lévy measure in the form:
ν(dz) =
∞
X
pj ηj e −ηj z 1{z>0} dz,
z ∈ R,
j=1
for some {pk , ηk ; k ≥ 1}. The equation ψ(·) = q has a countable
negative real-valued roots {−ξk,q ; k ≥ 1} that satisfy the
interlacing condition:
· · · < −ηk < −ξk,q < · · · < −η2 < −ξ2,q < −η1 < −ξ1,q < 0.
The scale function can be written as
∞
W (q) (x) =
X
e Φ(q)x
1
−
e −ξi,q x ,
0
0
ψ (Φ(q))
|ψ (−ξi,q )|
i=1
50 of 137
x ≥ 0.
Numerical Examples
1. A spectrally negative Lévy process is said to be in the β-family if
1 2 2 $
z
ψ(z) = δ̂z + σ z +
B(α + , 1 − λ) − B(α, 1 − λ)
2
β
β
for some δ̂ ∈ R, α > 0, β > 0, $ ≥ 0, λ ∈ (0, 3)\{1, 2} and the
beta function B(x, y ) := Γ(x)Γ(y )/Γ(x + y ).
2. We suppose δ̂ = 0.1, λ = 1.5, α = 3, β = 1 and $ = 0.1. With
this specification, the process has jumps of infinite activity (and of
bounded variation), which is not covered in the framework of
Bensoussan, Liu & Sethi (SICON, 2005).
3. We consider σ = 0 and σ = 0.2 so as to study both the bounded
and unbounded variation cases.
4. We let q = 0.03 and for the surplus cost we consider the quadratic
case f (x) = x 2 , x ∈ R.
51 of 137
Sensitivity w.r.t. the Proportional Cost C
800
700
700
600
600
500
value function
value function
500
400
300
300
200
200
100
100
0
-5
400
0
5
x
unbounded variation case (σ > 0)
0
-5
0
5
x
bounded variation case (σ = 0)
Figure : The value functions for various values of the proportional cost C .
52 of 137
Sensitivity w.r.t. the Fixed Cost K
800
700
700
600
600
500
value function
value function
500
400
300
300
200
200
100
100
0
-5
400
0
5
x
unbounded variation case (σ > 0)
0
-5
0
bounded variation case (σ = 0)
Figure : The value functions for various values of the fixed cost K .
53 of 137
5
x
References
[1] K. Yamazaki. Inventory Control for Spectrally Positive Lévy Demand
Processes. arXiv:1303.5163, 2013.
54 of 137
Part 2-2: Inventory control with two-sided
control (joint with E. Baurdoux)
55 of 137
Model
(Ω, F, P) hosting a spectrally negative Lévy process X .
An admissible strategy π := {(Utπ , Dtπ ); t ≥ 0} – nondecreasing,
π = Dπ = 0
right-continuous
and adapted processes
such that U0−
0−
hR
i
and Ex [0,∞) e −qt (dUtπ + dDtπ ) < ∞.
We want to minimize
Z
hZ ∞
π
−qt
π
v (x) := Ex
e f (Yt )dt +
i
e −qt (CU dUtπ + CD dDtπ ) ,
[0,∞)
0
with the controlled process
Ytπ := Xt + Utπ − Dtπ ,
for the case X is a spectrally negative Lévy process.
56 of 137
Assumptions
1. The unit proportional costs CU and CD can be negative but must
satisfy
CU + CD > 0.
2. We assume that E[X1 ] = ψ 0 (0+) ∈ (−∞, ∞).
3. About f (common assumptions as in Bensoussan et al. (2005))
convex (can be generalized slightly);
grows (or decreases) at most polynomially;
There exists a number a ∈ R such that the function
f˜(x) := f (x) + CU qx,
x ∈ R,
is increasing on (a, ∞) and is decreasing on (−∞, a).
57 of 137
Spectrally Positive Lévy Processes
Let X be a spectrally positive Lévy process with Laplace exponent:
h
i
ψ(s) := log E e −sX1
1
= cs + σ 2 s 2 +
2
such that
R
(0,∞) (1
Z
(e −sz − 1 + sz1{0<z<1} )ν(dz),
(0,∞)
∧ z 2 )ν(dz) < ∞.
It
R has paths of bounded variation if and only if σ = 0 and
(0,1) z ν(dz) < ∞.
We exclude the trivial case in which X is a subordinator.
58 of 137
Scale Functions
Recall that X is a spectrally
positive
Lévy process with Laplace
exponent ψ(s) = log E e −sX1 .
Fix any q > 0, there exists a function called the q-scale function
W (q) : R → [0, ∞),
which is zero on (−∞, 0), continuous and strictly increasing on
[0, ∞), and is characterized by the Laplace transform:
Z ∞
1
e −sx W (q) (x)dx =
,
s > Φ(q),
ψ(s) − q
0
where
Φ(q) := sup{λ ≥ 0 : ψ(λ) = q}.
59 of 137
Scale Functions
Let us define the first down- and up-crossing times, respectively, by
τa− := inf {t ≥ 0 : Xt < a}
τb+ := inf {t ≥ 0 : Xt > b} .
Then we have for any b > 0
h
i W (q) (x)
−
Ex e −qτ0 1{τ + >τ − } = (q) ,
0
b
W (b)
h
i
+
W (q) (x)
Ex e −qτb 1{τ + <τ − } = Z (q) (x) − Z (q) (b) (q) ,
0
b
W (b)
where
(q)
Z (q) (x) := 1 + qW (x),
Z x
(q)
W (x) :=
W (q) (y )dy .
0
60 of 137
Double reflection strategies
A doubly reflected Lévy process given by
Yta,b := Xt + Uta,b − Dta,b ,
t ≥ 0, a < b,
which is reflected at two barriers a and b so as to stay on the
interval [a, b].
The corresponding NPV of costs becomes
Γ(a, b)
Z (q) (x − a) − CU R (q) (x − a)
qW (q) (b − a)
Z x
f (a)
(q)
+
−
W (x − y )f 0 (y )dy .
q
a
va,b (x) =
For x ≥ b, we have va,b (x) = va,b (b) + CD (x − b).
Here
ψ 0 (0+)
(q)
R (q) (y ) := Z (y ) +
, y ∈ R.
q
61 of 137
Selection of a and b
Define
Γ(a, b) := CD + CU Z (q) (b − a) + f (b)W (q) (0)+
Z b
0
f (y )W (q) (b − y )dy − W (q) (b − a)f (a),
a
∂
Γ(a, b−).
γ(a, b) :=
∂b
We see that the values of (a, b) such that Γ(a, b) and γ(a, b)
vanish simultaneously attain smoothness.
62 of 137
Selection of a and b (Cont’d)
Taking a derivative and then limits
0
va,b
(b−) = CD
0
and va,b
(a+) =
Γ(a, b)
W (q) (0) − CU .
− a)
W (q) (b
Taking another derivative and then limits
Γ(a, b)
0
W (q) ((b − a)−) − γ(a, b),
− a)
Γ(a, b)
0
00
W (q) (0+) − f˜0 (a+)W (q) (0).
va,b
(a+)= (q)
W (b − a)
00
va,b
(b−)=
W (q) (b
Recall also that W (q) (0) = 0 iff X is of unbounded variation.
(q)0 ((b − a)−) = γ(a, b) = 0, then v
If W Γ(a,b)
a,b is
(q) (b−a) W
differentiable (resp. twice-differentiable) at a when X is of bounded
(resp. unbounded) variation.
it is twice-differentiable at b.
63 of 137
Existence of a∗ and b ∗
Plots of b 7→ Γ(a, b) for five fixed values of a – the red one is what we
want.
300
200
100
Gamma
0
-100
-200
-300
-400
-500
-3
-2
-1
0
1
b
64 of 137
2
3
Candidate Thresholds
There exist a∗ and b ∗ such that Γ(a∗ , x) ≥ 0 for all x ∈ [a∗ , ∞) and
either Case 1 or Case 2 defined below holds.
Case 1 a∗ < b ∗ and
Γ(a∗ , b ∗ ) = 0.
Moreover, if γ(a∗ , b ∗ ) is continuous at u ∗ then we also
have that γ(a∗ , b ∗ ) = 0.
Case 2 a∗ ∈ R and b ∗ = ∞ and
Γ(a∗ , b)
= 0.
b→∞ W (q) (b − a∗ )
lim
65 of 137
Verification of optimality
With our choice of (a∗ , b ∗ ),
ψ 0 (0+)
f˜(a∗ )
+x +
Z (q) (x − a∗ )
va∗ ,b∗ (x) = −cR
q
q
Z x
W (q) (x − y )f˜0 (y )dy .
−
a
Verification of optimality requires
1. −CU ≤ va0 ∗ ,b∗ (x) ≤ CD for all x ∈ R.
2. (L − q)va∗ ,b∗ (x) + f (x) ≥ 0 for all x > b ∗ .
3. (L − q)va∗ ,b∗ (x) + f (x) = 0 for a∗ < x < b ∗ .
4. (L − q)va∗ ,b∗ (x) + f (x) ≥ 0 for x ≤ a∗ .
66 of 137
Verification of optimality (Cont’d)
Hard ones to prove are:
1. −CU ≤ va0∗ ,b∗ (x) for all x ∈ (a∗ , b ∗ ) – simple under the convexity of f ,
2. (L − q)va∗ ,b∗ (x) + f (x) ≥ 0 for all x > b ∗ – six page proof;
others hold even w/o the convexity of f .
We use the convexity of f and
va0 ∗ ,b∗ (x) = −Γ(a∗ , x) + CD ,
a∗ ≤ x ≤ b ∗ .
Plots of γ(a, x) = ∂Γ(a, x)/∂x:
150
50
40
100
30
50
gamma
gamma
20
0
10
0
-50
-10
-100
67 of 137
-3
-2
-1
0
1
b
2
3
-20
-3
-2
-1
0
1
b
2
3
Numerical results
The β-family introduced by Kuznetsov (AAP, 2010):
z
1 2 2 $
B(α + , 1 − λ) − B(α, 1 − λ)
ψ(z) = δ̂z + σ z +
2
β
β
for some δ̂ ∈ R, α > 0, β > 0, $ ≥ 0, λ ∈ (0, 3)\{1, 2} and the
beta function B(x, y ).
ψ is rational and hence can be inverted to obtain an analytical
form of the scale function.
68 of 137
Quadratic Case
Suppose the running cost function is f ≡ fQ where
fQ (x) := x 2 ,
x ∈ R.
300
150
200
100
100
50
gamma
Gamma
0
-100
0
-200
-300
-50
-400
-500
-3
-2
-1
0
1
b
Γ(a, ·)
69 of 137
2
3
-100
-3
-2
-1
0
1
b
γ(a, ·)
2
3
Quadratic Case (Cont’d)
80
200
70
150
60
value function
value function
50
40
30
100
50
20
0
10
0
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
x
with respect to CU
70 of 137
2
-50
-2
-1.5
-1
-0.5
0
0.5
1
1.5
x
with respect to CD
2
Linear Case
Suppose the running cost function is f ≡ fL where
fL (x) := |x|,
x ∈ R.
120
50
100
40
80
30
20
gamma
Gamma
60
40
10
20
0
0
-10
-20
-40
-3
-2
-1
0
1
b
Γ(a, ·)
71 of 137
2
3
-20
-3
-2
-1
0
1
b
γ(a, ·)
2
3
Linear Case (Cont’d)
300
180
160
250
140
120
150
value function
value function
200
100
100
80
60
40
50
20
0
0
-50
-8
-6
-4
-2
0
x
with respect to CU
72 of 137
2
-20
-3
-2
-1
0
1
x
with respect to CD
2
References
[1] E. J. Baurdoux and K. Yamazaki. Optimality of Doubly Reflected Lévy
Processes in Singular Control. Stochastic Processes and their
Applications, 125(7):2727-2751, 2015.
73 of 137
Part 3: Optimal Dividend Problem: from
one-parameter to two parameters.
74 of 137
Part 3-1: De Finetti’s Problem with Fixed
Costs
75 of 137
Outline
Solutions to the dual model
with review on fluctuation theory of (reflected) Lévy processes.
Extensions with transaction costs (impulse control):
vπ (x) := Ex
hZ
0
σπ
i
X
β1{∆Lπs >0} .
e −qt d Lπt −
0≤s<t
Spectrally negative model has been solved by Loeffen (IME, 2009)
Numerical Results.
76 of 137
The Classical Dual Model
A strategy π = {Lπt , t ≥ 0} is a nondecreasing, right-continuous
and adapted process starting at zero.
A controlled risk process is the difference:
Utπ := Xt − Lπt ,
t ≥ 0.
Time of ruin: σ π := inf {t > 0 : Utπ < 0}.
We want to maximize, for q > 0,
Z σπ
−qt
π
vπ (x) := Ex
e dLt ,
0
π + ∆X
over the set of all strategies Π satisfying Lπt − Lπt− ≤ Ut−
t
π
for all t ≤ σ a.s.
We want to obtain the value function:
v (x) := sup vπ (x),
π∈Π
77 of 137
x ≥ 0.
Solution Procedures
We follow a classical approach “guess” and “verify”.
Guess that an optimal strategy is a barrier strategy (reflected Lévy
process) πa := {Lat ; t ≤ σa } in the form:
Lat := sup (Xs − a) ∨ 0,
0≤s≤t
Uta
:= Xt − Lat ,
with the corresponding ruin time σa := inf {t > 0 : Uta < 0}.
Choose the value of a using some smoothness condition.
Verify that
Z σa
−qt
a
va (x) := Ex
e dLt ≥ sup vπ (x).
0
78 of 137
π∈Π
Main Results
Direct application of Avram, Palmowski and Pistorius (AAP, 2007).
Let µ := EX1 . We will denote our candidate barrier level by

 (q) −1 µ
Z
q > 0 if µ > 0,
a∗ =
0
if µ ≤ 0,
which is well-defined because Z
(q)
(x) :=
Rx
0
Z (q) (z)dz is monotone.
Theorem (Bayraktar, Kyprianou & Y. (Astin Bull., 2013))
We have
va∗ (x) := sup vπ (x),
x ≥ 0,
π∈Π
where
va∗ (x) =
79 of 137
(
(q)
−Z (a∗ − x) −
x,
ψ 0 (0+)
q
= −Z
(q)
(a∗ − x) + µq , if µ > 0,
if µ ≤ 0.
Extension with Transaction Costs
Consider an extension where we want to maximize
h Z σπ
i
X
vπ (x) := Ex
e −qt d Lπt −
β1{∆Lπs >0} ,
0
0≤s<t
for some fixed unit transaction cost β > 0.
A strategy π = {Lπt , t ≥ 0} is assumed to be a pure-jump,
nondecreasing, right-continuous and adapted process starting at
zero.
Spectrally negative model has been solved by Loeffen (IME, 2009)
80 of 137
The (c1 , c2 )-policy
For c2 > c1 ≥ 0, a (c1 , c2 )-policy, πc1 ,c2 := Lct 1 ,c2 ; t ≥ 0 , brings
the level of the controlled risk process U c1 ,c2 := X − Lc1 ,c2 down to
c1 whenever it reaches or exceeds c2 .
We aim to prove that a (c1∗ , c2∗ )-policy is optimal for some
c2∗ > c1∗ ≥ 0.
We shall express, in terms of the scale function,
h Z σc1 ,c2
i
X
vc1 ,c2 (x) := Ex
e −qt d Lct 1 ,c2 −
β1{∆Lsc1 ,c2 >0} ,
0
where σc1 ,c2
time.
81 of 137
0≤s<t
:= inf t > 0 : Utc1 ,c2 < 0 is the corresponding ruin
Computing vc1 ,c2
By the strong Markov property, it must satisfy for every
0 ≤ x ≤ c2 and 0 ≤ c1 < c2
h
i
+
vc1 ,c2 (x) = Ex e −qτc2 1{τc+ <τ − } (Xτc+ − c1 − β)
0
2
h
i 2
−qτc+2
+ Ex e
1{τc+ <τ − } v̄c1 ,c2 ,
0
2
where v̄c1 ,c2 := vc1 ,c2 (c1 ).
Solving for x = c1 , we have
h
i
+
Ec1 e −qτc2 1{τc+ <τ − } (Xτc+ − c1 − β)
2
h2 0 +
i
,
v̄c1 ,c2 =
−qτc2
1 − Ec1 e
1{τc+ <τ − }
2
0 ≤ c1 < c2 .
0
These can be rewritten in terms of the scale function.
82 of 137
Candidate Value function
Lemma (Bayraktar, Kyprianou & Y. (IME, 2014))
For 0 < x < c2 and 0 ≤ c1 < c2 ,
µ
+ γ(c1 , c2 )Z (q) (c2 − x)
q
W (q) (c2 − x)
− G (c1 , c2 )
,
W (q) (c2 )
vc1 ,c2 (x) = −Z
(q)
(c2 − x) +
where
γ(c1 , c2 ) := v̄c1 ,c2 + c2 − c1 − β −
G (c1 , c2 ) := γ(c1 , c2 )Z (q) (c2 ) − Z
For x ≥ c2 , vc1 ,c2 (x) = x − c1 − β + v̄c1 ,c2 .
83 of 137
µ
,
q
(q)
(c2 ) +
µ
.
q
Solution Procedures
Choose 0 ≤ c1∗ < c2∗ such that vc1 ,c2 (x) (or v̄c1 ,c2 − c1 ) is
maximized.
Examine the smoothness of vc1∗ ,c2∗ (x); it turns out that
at x = c2∗ , vc1∗ ,c2∗ (x) is C 0 (C 1 ) when X is of bounded (unbounded)
variation;
at x = c1∗ , vc01∗ ,c2∗ (c1∗ ) = 1 when c1∗ > 0 and vc01∗ ,c2∗ (c1∗ ) ≤ 1 when
c1∗ = 0.
Verify the optimality.
Show uniqueness of (c1∗ , c2∗ ).
84 of 137
Main Results
Theorem (Bayraktar, Kyprianou & Y. (IME, 2013))
There exist unique 0 ≤ c1∗ < c2∗ < ∞ such that G (c1∗ , c2∗ ) = 0 and
either
case 1 c1∗ > 0 with H(c1∗ , c2∗ ) = 0, or
case 2 c1∗ = 0 with H(c1∗ , c2∗ ) ≥ 0,
where
µ
(q)
G (c1 , c2 ):= γ(c1 , c2 )Z (q) (c2 ) − Z (c2 ) + ,
q
h
i
(q)
(q)
H(c1 , c2 ):= q γ(c1 , c2 )W (c2 − c1 ) − W (c2 − c1 ) .
The (c1∗ , c2∗ )-strategy is optimal and the value function is
vc1∗ ,c2∗ (x) = −Z
85 of 137
(q)
(c2∗ − x) +
µ
+ γ(c1∗ , c2∗ )Z (q) (c2∗ − x),
q
x ≥ 0.
Some properties
Proposition
Let v β denote the value function corresponding to the dividend
payment problem when the fixed transaction cost is β (defined as
above), and
v̂ the value function when there are no-transaction costs.
Then v β converges to v̂ uniformly as β ↓ 0.
Proposition
If µ := EX1 ≤ 0, we must have c1∗ = 0.
86 of 137
Numerical Results
Suppose
Xt − X0 = −dt + σBt +
Nt
X
Zn ,
0 ≤ t < ∞,
n=1
for some d ∈ R and σ ≥ 0. Here
B = {Bt ; t ≥ 0} is a standard Brownian motion,
N = {Nt ; t ≥ 0} is a Poisson process with arrival rate λ, and
Z = {Zn ; n = 1, 2, . . .} is an i.i.d. sequence of
phase-type-distributed r.v. with representation (m, α, T).
Its Laplace exponent is then
1
ψ(s) = ds + σ 2 s 2 + λ α(sI − T)−1 t − 1 .
2
87 of 137
Numerical Results
Suppose {−ξi,q ; i ∈ Iq } is the set of the roots of the equality
ψ(s) = q with negative real parts.
If these are assumed distinct, the scale function can be written
h
i
X
W (q) (x) =
Ci,q e Φ(q)x − e −ξi,q x , σ > 0,
i∈Iq
W (q) (x) =
X
i∈Iq
h
i 1
Ci,q e Φ(q)x − e −ξi,q x + e Φ(q)x ,
d
σ = 0.
Here {ξi,q ; i ∈ Iq } and {Ci,q ; i ∈ Iq } are possibly complex-valued.
We choose (m, α, T) to approximate the Weibull distribution with
density function with α = 2 and γ = 1 (obtained using the
EM-algorithm).
88 of 137
Value functions
20
25
20
15
15
value
value
10
10
5
5
0
-5
0
0
5
10
15
-5
20
0
5
x
10
15
20
15
20
x
c1∗ = 0
c1∗ > 0
without BM
20
25
20
15
15
value
value
10
10
5
5
0
-5
0
0
5
10
15
20
-5
0
x
10
x
c1∗ = 0
c1∗ > 0
with BM
89 of 137
5
20
20
18
18
16
16
14
14
12
12
value
value
Convergence as β ↓ 0
10
10
8
8
6
6
4
4
2
0
2
0
5
10
15
0
20
0
5
x
10
15
20
15
20
x
µ>0
µ<0
20
20
18
18
16
16
14
14
12
12
value
value
without BM
10
10
8
8
6
6
4
4
2
0
2
0
5
10
15
20
0
0
x
10
x
µ>0
µ<0
with BM
90 of 137
5
References
[1] E. Bayraktar, A. E. Kyprianou and K. Yamazaki. On Optimal Dividends
in the Dual Model. ASTIN Bulletin, 43(3):359-372, 2013.
[2] R. Loeffen. An optimal dividends problem with transaction costs for
spectrally negative Lvy processes. Insurance: Mathematics and
Economics, 45(1):41-48, 2009.
[3] E. Bayraktar, A. E. Kyprianou and K. Yamazaki. Optimal Dividends in
the Dual Model under Transaction Costs. Insurance: Mathematics and
Economics, 54:133-143, 2014.
91 of 137
Part 4:
Zero-sum Game between Two Players
92 of 137
Part 4-1:
Zero-sum Game between Two Stoppers
93 of 137
Outline
We study the game version of Leung & Y. (2013).
Solve analytically for a general exponential spectrally negative Lévy
process.
Both the protection buyer and seller can step-up/down the
premium/notional anytime before contract termination.
Reduction to the Optimal Stopping Game.
Existence of Nash equilibrium.
Strategies and the value of the game expressed in terms of the
scale function.
Numerical Examples.
Equilibrium Strategies.
Value of Games.
Fair Premium.
94 of 137
Model
Let X = {Xt ; t ≥ 0} be a Lévy process defined on a probability
space (Ω, F, P).
The value of the reference entity (a company stock or other assets)
is assumed to evolve according to an exponential Lévy process
St = e Xt ,
t ≥ 0.
Following the Black-Cox structural approach, the default event is
triggered by S crossing a lower level D.
Without loss of generality, we can take log D = 0 by shifting the
initial value X0 . Henceforth, we shall work with the default time:
θ := inf{ t ≥ 0 : Xt ≤ 0 }.
95 of 137
Vanilla CDS
From the buyer’s perspective, the expected discounted payoff is
given by
Z θ∧T
x
−rt
−r θ
C̄ (x; p, α, T ) := E −
e p dt + αe 1{θ≤T } ,
0
where T is the maturity and r > 0 is the positive constant risk-free
interest rate.
The fair premium is
p̄(x; α, T ) =
α r ζT (x)
1 − ζT (x) − e −rT Px {θ > T }
where
h
i
ζT (x) := Ex e −r θ 1{θ≤T } .
96 of 137
Callable CDS (Leung & Y., 2013)
We consider a CDS contract with an embedded option that permits
the protection buyer to change
premium from p to p̂,
default payment from α to α̂,
any time before θ once for a fee γ ≥ 0. The value for the buyer is
V (x; T , p, p̂, α, α̂, γ)
Z τ
Z
x
−rt
:= sup E −
e p dt −
τ ∈ST
0
θ∧T
e −rt p̂dt
τ
i
−e −r τ γ1{τ <θ∧T } + e −r θ 1{θ<T } (α̂1{τ <θ} + α1{τ =θ} ) ,
where ST := {τ ∈ F : τ ≤ θ ∧ T a.s. }.
97 of 137
Putable CDS (Leung & Y., 2013)
We consider a CDS contract with an embedded option that permits
the protection seller to change
premium from p to p̂,
default payment from α to α̂,
any time before θ once for a fee γ. The value for the seller is
U(x; T , p, p̂, α, α̂, γ)
Z τ
Z
x
−rt
:= sup E
e p dt +
τ ∈ST
0
θ∧T
e −rt p̂ dt
τ
i
−e −r τ γ1{τ <θ∧T } − e −r θ 1{θ<T } (α̂1{τ <θ} + α1{τ =θ} ) ,
where ST := {τ ∈ F : τ ≤ θ ∧ T a.s. }.
98 of 137
Default Swap Games
Define, for any buyer’s exercise time τ ∈ S and seller’s exercise time
σ ∈ S with S := {τ ∈ F : τ ≤ θ a.s. },
Z τ ∧σ
Z θ
x
−rt
V (x; σ, τ ) := E −
e p dt + 1{τ ∧σ<∞} −
e −rt p̂ dt
τ ∧σ
0
+ e −r θ (α̂1{τ ∧σ<θ} + α1{τ ∧σ=θ} )
+ 1{τ ∧σ<θ} e
−r (τ ∧σ)
−γb 1{τ ≤σ} + γs 1{τ ≥σ}
1. This is the value of the CDS from the buyer’s perspective given
(σ, τ ).
2. The seller’s value becomes −V (x; σ, τ ).
3. The buyer wants to maximize V whereas the seller wants to
minimize V .
99 of 137
.
Nash Equilibria
Our problem is to find Nash equilibria, which means the existence of a
saddle point (σ ∗ , τ ∗ ) such that V (x) = V (x; σ ∗ , τ ∗ ) where
V (x; σ ∗ , τ ) ≤ V (x; σ ∗ , τ ∗ ) ≤ V (x; σ, τ ∗ ),
100 of 137
τ, σ ∈ S.
Step-up/down Game (Cont’d)
1. Step-up Game: if p̂ > p and α̂ > α, then the buyer and the seller
are allowed to increase the coverage once from α to α̂ by paying
the fee γb or γs and a higher premium p̂ thereafter.
2. Step-down Game: when p̂ < p and α̂ < α, then the buyer and the
seller can reduce the coverage once from α to α̂ by paying the fee
γb or γs and a reduced premium p̂ thereafter.
3. Cancellation Game: as a special case of the step-down option with
p̂ = α̂ = 0, the resulting contract allows the buyer and the seller to
terminate the CDS at time τ . This special case corresponds to a
step-down game.
101 of 137
Solution Methods
Let p̃ := p − p̂ and α̃ := α − α̂.
The buyer’s position for the step-down (-up) game with
(p̃, α̃, γb , γs ) is equivalent to the seller’s position for the step-up
(-down) game with (−p̃, −α̃, γs , γb ). Hence we
focus on the step-down game.
We denote the threshold strategy for the buyer and the seller,
respectively, by
τB := inf {t ≥ 0 : Xt ∈
/ (0, B)} ,
σA := inf {t ≥ 0 : Xt ∈
/ (A, ∞)} ,
B > 0,
A > 0.
We pursue A∗ and B ∗ that attains a Nash equilibrium via
continuous/smooth fit.
102 of 137
Optimal Threshold Levels
Define for every B > A > 0
p̃
Ψ(A, B) :=
−
+ γs Z (r ) (B − A)
r
Z ∞
Z u∧B−A
+ (α̃ − γs )
Π(du)
dzW (r ) (B − z − A).
p̃
− γb
r
A
0
For every fixed A, {Ψ(A, B), B ≥ A} has at most one local
maximum.
103 of 137
Optimal Threshold Levels (Cont’d)
1
0.2
A=1.5908
A=1.61
A=1.6292
A=2.35
A=4.7
0.8
0.6
A=1.5908
A=1.61
A=1.6292
A=2.35
A=4.7
0.15
0.1
small ψ(A,B)
large Ψ(A,B)
0.4
0.2
0
-0.2
0.05
0
-0.05
-0.4
-0.1
-0.6
-0.15
-0.8
-1
0
2
4
6
8
-0.2
1
10
2
3
B
5
B
∂
∂B Ψ(A, B)
Ψ(A, B)
For a regular case, A∗ and B ∗ such that
Ψ(A∗ , B ∗ ) =
∂
Ψ(A∗ , B ∗ ) = 0
∂B
constitute a saddle point (σA∗ , τB ∗ ).
104 of 137
4
6
7
8
Main Results
Theorem
There always exist 0 ≤ A∗ < B ∗ ≤ ∞ that satisfy the following.
1. If A∗ > 0, Nash equilibrium exists with saddle point (σA∗ , τB ∗ ).
2. If A∗ = 0,
2.1 if ν = 0, Nash equilibrium exists with saddle point (θ, τB ∗ );
2.2 if ν > 0, then the alternative equilibrium
v (x; σ0+ , τ ) ≤ v0+,B ∗ (x) ≤ v (x; σ, τB ∗ ),
σ, τ ∈ S,
holds and the value function satisfies v0+,B ∗ (x) = limε↓0 v (x; σε , τB ∗ )
where
v (x; σ0+ , τ ) := Ex e −r τ h(Xτ ) − (α̃ − γs )1{Xτ =0} 1{τ <∞} , τ ∈ S,
h
i
v0+,B ∗ (x) := Ex e −r τB ∗ h(XτB ∗ ) − (α̃ − γs )1{XτB ∗ =0} 1{τB ∗ <∞} .
105 of 137
References
[1] T. Leung and K. Yamazaki. American Step-Up and Step-Down Credit
Default Swaps under Levy Models. Quantitative Finance,
13(1):137-157, 2013.
[2] M. Egami, T. Leung and K. Yamazaki. Default Swap Games Driven by
Spectrally Negative Lévy Processes. Stochastic Processes and their
Applications, 123(2):347-384, 2013.
106 of 137
Part 4-2:
Zero-sum Game between Two
Controller-Stopper
107 of 137
The game between controller and stopper
There are two players: controller and stopper.
Given a stochastic process X , the controller chooses an adaptive
and nondecreasing process ξ; the resulting controlled process
becomes
Utξ := Xt + ξt ,
t ≥ 0.
The stopper chooses a stopping time τ at which the game is
terminated.
Given a pair of strategies (ξ, τ ), the payoff of the controller and
the stopper is −J(x; ξ, τ ) and J(x; ξ, τ ), respectively, where
J(x; ξ, τ ) :=
"Z
Ex
0
108 of 137
τ
e −qt h(Utξ )dt +
Z
[0,τ ]
#
e −qt dξt + e −qτ g (Uτξ )1{τ <∞} .
Saddle point (Nash equilibrium)
The controller minimizes J(x; ξ, τ ) over ξ while the stopper
maximizes it over τ .
The problem is to show the existence of a saddle point, or
equivalently a Nash equilibrium (ξ ∗ , τ ∗ ), such that
J(x; ξ ∗ , τ ) ≤ J(x; ξ ∗ , τ ∗ ) ≤ J(x; ξ, τ ∗ ) for all τ and ξ.
If it exists, J(x; ξ ∗ , τ ∗ ) is the value function of the game.
Hernández-Hernández, Simon, Zervos (2015) considers the
diffusion case.
109 of 137
Objective
For X , we consider (1) spectrally negative Lévy processes and (2)
spectrally positive Lévy processes
Recall
J(x; ξ, τ ) :=
"Z
Ex
τ
e
−qt
h(Utξ )dt
0
#
Z
+
e
−qt
dξt + e
−qτ
g (Uτξ )1{τ <∞}
[0,τ ]
We show under a suitable condition that a saddle point is given by
(ξ a , τa,b ) for some a < b, where we define
ξta := sup (a − Xt 0 ) ∨ 0,
t ≥ 0,
0≤t 0 ≤t
a
τa,b := inf{t ≥ 0 : Utξ > b}.
a
The controlled process U ξ = Xt + ξ a is a reflected Lévy process.
110 of 137
.
This talk
Consider the case X is
a spectrally negative Lévy process and
a spectrally positive Lévy process
and derive a saddle point.
Review of (reflected) spectrally negative/positve Lévy processes
and fluctuation theories.
Solution methods
Write expected net present values using the scale function.
Guessing of saddle point via smooth/continuous fit.
Verification of optimality.
Numerical results.
111 of 137
The problem revisited
X = {Xt ; t ≥ 0} is a spectrally negative Lévy process.
The controller chooses a process ξ ∈ Ξ, where
Ξ := the set of nondecreasing and right-continuous
adapted processes with ξ0− := 0.
The corresponding controlled process becomes
Utξ := Xt + ξt ,
t ≥ 0.
The stopper chooses the time τ ∈ Υ, where
Υ := the set of stopping times.
112 of 137
The problem revisited (Cont’d)
The controller minimizes and the stopper maximizes the common
criterion
J(x; ξ, τ )
"Z
:= Ex
0
τ
e −qt h(Utξ )dt +
Z
#
e −qt dξt + e −qτ g (Uτξ )1{τ <∞} .
[0,τ ]
The problem is to show the existence of a saddle point, or
equivalently a Nash equilibrium (ξ ∗ , τ ∗ ) ∈ Ξ × Υ, such that
J(x; ξ ∗ , τ ) ≤ J(x; ξ ∗ , τ ∗ ) ≤ J(x; ξ, τ ∗ ),
for any ξ ∈ Ξ and any stopping time τ ∈ Υ.
113 of 137
Assumptions
Assumption (on g )
We assume g (x) = C + Kx, x ∈ R, for some C ∈ R and K > −1.
Assumption (on h)
Define
h̃(x) := h(x) + qx,
x ∈ R.
We assume that h (and h̃) is continuously differentiable with an
absolutely continuous first derivative and
1. h̃0 (x) ≤ 0 for all x ∈ R;
2. h̃0 ≤ −c1 on [a1 , ∞) for some a1 ∈ R and c1 > 0;
3. h̃0 ≤ −c2 on (∞, a2 ] for some a2 ∈ R and c2 > 0.
114 of 137
Solution Procedures
We follow a classical approach “guess” and “verify”.
Guess that a saddle point is given by (ξ a , τa,b ) for some a < b,
where we define
ξta := sup (a − Xt 0 ) ∨ 0,
t ≥ 0,
0≤t 0 ≤t
a
τa,b := inf{t ≥ 0 : Utξ > b}.
Choose the value of a and b using some smoothness condition.
Verify the optimality.
115 of 137
Sample Paths
ξta := sup (a − Xt 0 ) ∨ 0,
t ≥ 0,
0≤t 0 ≤t
a
6
20
4
15
2
10
ξt
Ut
τa,b := inf{t ≥ 0 : Utξ > b}.
0
-2
5
0
5
10
t
116 of 137
Utξ
15
20
0
0
5
10
t
a
ξa
15
20
Results due to Fluctuation/Excursion Theory
For any a ≤ x ≤ b, we can write
h Z τa,b
i Z (q) (x − a)
a
ϕa (b; h) − ϕa (x; h),
Ex
e −qt h(Utξ )dt = (q)
Z (b − a)
0
h Z τa,b
i
l(b − a)
Ex
e −qt dξta = −l(x − a) + Z (q) (x − a) (q)
,
Z (b − a)
0
Ex [e −qτa,b ] =
Z (q) (x − a)
,
Z (q) (b − a)
where
x
Z
W (q) (x − y )h(y )dy ,
ϕa (x; h) :=
a, x ∈ R,
a
l(x) := Z
Z
(q)
Z (x) :=
117 of 137
(q)
0
(x) +
µ Z (q) (x)
−
,
q
Φ(q)
x
Z (q) (y )dy .
x ∈ R,
Rewrite via Scale Functions
a
Summing up these, for any a ≤ x ≤ b, and because Uτξa,b = b a.s.
(due to no positive jumps),
Ja,b (x) := J(x; ξ a , τa,b )
=
Z (q) (x − a)
[ϕa (b; h) + l(b − a) + g (b)] − ϕa (x; h) − l(x − a).
Z (q) (b − a)
For x < a, Ja,b (x) = a − x + Ja,b (a).
For x > b, Ja,b (x) = g (x).
118 of 137
Smoothness Condition
Define for b ≥ a,
Λ(a, b) := ϕa (b; h) + Z
λ(a, b) :=
(q)
(b − a) +
h(a) (q)
µ
+ g (b) −
Z (b − a),
q
q
∂
Λ(a, b) = ϕa (b; h0 ) + Z (q) (b − a) + K .
∂b
Proposition
Suppose a < b are such that Λ(a, b) = λ(a, b) = 0. Then
1. Ja,b is differentiable (resp. twice-differentiable) at a when X is of
bounded (resp. unbounded) variation;
2. Ja,b is differentiable at b for all cases.
119 of 137
Existence of (a∗ , b ∗ )
Proposition
There exist a∗ < b ∗ such that
1. Λ(a∗ , b ∗ ) = λ(a∗ , b ∗ ) = 0,
2. λ(a∗ , x) ≥ 0 for x ∈ (a∗ , b ∗ ) and λ(a∗ , x) ≤ 0 for x ∈ (b ∗ , ∞),
3. Λ(a∗ , x) ≤ 0 for all x ≥ a∗ .
120 of 137
Plots of Λ(a, b) and λ(a, b)
5
2
0
0
-2
-5
λ(a,b)
Λ(a,b)
-4
-10
-6
-8
-15
-10
-20
-12
-25
-6
-5
-4
-3
-2
-1
b
Λ(a, b)
121 of 137
0
1
2
-14
-6
-5
-4
-3
-2
-1
b
λ(a, b)
0
1
2
Verification for optimality
Proposition
The function Ja∗ ,b∗ is convex such that, for x ∈ R,
−1 ≤ Ja0 ∗ ,b∗ (x) ≤ K = g 0 (x) and
Ja∗ ,b∗ (x) ≥ g (x).
Proposition
The following relations are satisfied.
1. (L − q)Ja∗ ,b∗ (x) + h(x) ≤ 0 for x ∈ (b ∗ , ∞),
2. (L − q)Ja∗ ,b∗ (x) + h(x) = 0 for x ∈ (a∗ , b ∗ ),
3. (L − q)Ja∗ ,b∗ (x) + h(x) ≥ 0 for x ∈ (−∞, a∗ ).
122 of 137
Main Results
Theorem
Define v (x) := Ja∗ ,b∗ (x) as the value of the functional associated
with the strategies
∗
ξta := sup (a∗ − Xt 0 ) ∨ 0,
t≥0
0≤t 0 ≤t
τa∗ ,b∗ := inf{t ≥ 0 : Utξ
a∗
> b ∗ }.
Then, we have
1. v (x) ≤ J(x; ξ, τbξ∗ ) for any ξ ∈ Ξ, with τbξ∗ := inf{t ≥ 0 : Utξ > b ∗ };
∗
2. J(x; ξ a , τ ) ≤ v (x) for any τ ∈ Υ.
∗
In other words, the pair (ξ a , τa∗ ,b∗ ) is a saddle point and v (·) is
the value function of the game.
123 of 137
14
14
12
12
10
10
8
8
6
6
value
value
Numerical Examples
4
2
4
2
0
0
-2
-2
-4
-4
-6
-12
-10
-8
-6
-4
-2
0
-6
-12
2
-10
-8
-6
x
-4
-2
0
2
x
β = −2, 0, 2, 4
α = 0.5, 1, 2, 10
12
12
10
10
8
8
6
value
value
6
4
2
4
2
0
0
-2
-2
-4
-6
-12
-10
-8
-6
-4
-2
0
x
C = −2, 0, 2, 4
124 of 137
2
-4
-12
-10
-8
-6
-4
-2
0
2
x
K = 0.5, 1.0, 1.5, 2.0
Spectrally Positive Lévy Case
Assumption (on g )
We assume g (x) = C + Kx, x ∈ R, for some C ∈ R and K > −1.
Assumption (on h)
We assume h is affine:
h(x) = −αx + β,
x ∈ R.
Moreover, the function
ĥ(x) := h(x) − Kqx − [Cq + K µ̂],
is decreasing. In other words,
α̂ := α + Kq > 0.
125 of 137
x ∈ R,
Sample Paths
6
3.5
3
4
2.5
ξt
Ut
2
2
1.5
1
0
0.5
-2
0
5
10
t
Utξ
126 of 137
15
20
0
0
5
10
t
a
ξa
15
20
Fluctuation/Excursion Theory
For every a ≤ x ≤ b,
h Z τa,b
i W (q) (b − x)W (q) (0)
a
h(a)
Ex
e −qt h(Utξ )dt =
W (q)0 (b − a)
0
"
#
Z b
(q)0 (y − a)
W
+
h(y ) W (q) (b − x) (q)0
− W (q) (y − x) dy ,
W (b − a)
a
h Z τa,b
i
W (q) (b − x)
,
Ex
e −qt dξta = (q)0
W (b − a)
0
Ex [e −qτa,b ] = Z (q) (b − x) − qW (q) (b − a)
a
Ex [e −qτa,b (Uτξa,b − b)] = −Z
+ µ̂W
127 of 137
(q)
(b − x) +
(q)
W (q) (b − x)
,
W (q)0 (b − a)
(b − x)
Z (q) (b − a) − µ̂W (q) (b − a) (q)
W (b − x).
W (q)0 (b − a)
Rewrite via Scale Functions
We have
Ja,b (x) =



W (q) (b−a)
Γ(a, b)
0
W (q) (b−a)
(q)
W (b−x)
Γ(a, b)
0
W (q) (b−a)
+ Γ(a, b) − x + b, x ≤ a,
+ Γ(x, b) − x + b, a < x.
where we define
Γ(x, b) := 1 + W
(q)
Z
b
(0)h(x) +
0
h(y )W (q) (y − x)dy
x
(q)
− [(C + bK )q + K µ̂]W (b − x) + KZ (q) (b − x),
Z b
Γ(x, b) := x − b −
h(y )W (q) (y − x)dy
x
+ (C + bK )Z (q) (b − x) − K Z
µ̂ := −EX1 .
128 of 137
(q)
(b − x) + K µ̂W
(q)
(b − x),
Smoothness Condition
Define
γ(a, b) :=
∂
0
Γ(a, b) = −W (q) (b − a)ĥ(b) − α̂W (q) (b − a).
∂a
Proposition
0
1. If (a, b) are such that Γ(a, b)/W (q) (b − a) = 0, then Ja,b is
continuous (resp. differentiable) at b when X is of bounded (resp.
unbounded) variation.
2. If in addition γ(a, b) = 0, then Ja,b is twice-differentiable at a.
129 of 137
Existence of (a∗ , b ∗ )
Lemma
When q ≥ α, by setting b ∗ = b̄ and a∗ = ā(b ∗ ) = −∞, we have
0
0
Γ(a∗ , b ∗ )/W (q) (b ∗ − a∗ ) := lima↓−∞ Γ(a, b ∗ )/W (q) (b ∗ − a) = 0 and
Γ(·, b ∗ ) ≥ 0 and γ(·, b ∗ ) ≥ 0 on (−∞, b ∗ ).
Lemma (unbounded variation case)
Suppose q < α and X is of unbounded variation. Then, there exist
−∞ < a∗ < b ∗ such that b ∗ > b and Γ(a∗ , b ∗ ) = γ(a∗ , b ∗ ) = 0.
Lemma (bounded variation case)
Suppose q < α and X is of bounded variation.
(i) If (q + ν(0, ∞))(1 + K ) > α̂, there exist a∗ < b ∗ such that
b < b ∗ < B and Γ(a∗ , b ∗ ) = γ(a∗ , b ∗ ) = 0.
(ii) Otherwise, if we set a∗ = b ∗ = B, then Γ(a∗ , b ∗ ) = 0 and
Γ(a, b ∗ ) ≥ 0 and γ(a, b ∗ ) ≤ 0 on (−∞, a∗ ).
130 of 137
20
20
15
15
10
10
5
5
γ(a, b)
Γ(a,b)
Unbounded Variation Case
0
0
-5
-5
-10
-10
-15
-20
-12
-15
-10
-8
-6
-4
a
Γ(a, b)
131 of 137
-2
0
2
-20
-16
-14
-12
-10
-8
-6
a
γ(a, b)
-4
-2
0
2
20
20
15
15
10
10
5
5
γ(a, b)
Γ(a,b)
Bounded Variation Case with a∗ 6= b ∗
0
0
-5
-5
-10
-10
-15
-20
-12
-15
-10
-8
-6
-4
a
Γ(a, b)
132 of 137
-2
0
2
-20
-14
-12
-10
-8
-6
-4
a
γ(a, b)
-2
0
2
20
20
15
15
10
10
5
5
γ(a, b)
Γ(a,b)
Bounded Variation Case with a∗ = b ∗
0
0
-5
-5
-10
-10
-15
-20
-3
-15
-2.5
-2
-1.5
-1
-0.5
a
Γ(a, b)
133 of 137
0
0.5
1
-20
-3
-2.5
-2
-1.5
-1
-0.5
a
γ(a, b)
0
0.5
1
Verification for optimality
Proposition
The function Ja∗ ,b∗ is convex such that, for x ∈ R,
−1 ≤ Ja0 ∗ ,b∗ (x) ≤ K = g 0 (x) and
Ja∗ ,b∗ (x) ≥ g (x).
Proposition
The following relations are satisfied.
1. (L − q)Ja∗ ,b∗ (x) + h(x) ≤ 0 for x ∈ (b ∗ , ∞),
2. (L − q)Ja∗ ,b∗ (x) + h(x) = 0 for x ∈ (a∗ , b ∗ ),
3. (L − q)Ja∗ ,b∗ (x) + h(x) ≥ 0 for x ∈ (−∞, a∗ ).
134 of 137
Main Results
Theorem
Define v (x) := Ja∗ ,b∗ (x) as the value of the functional associated
with the strategies
∗
ξta := sup (a∗ − Xt 0 ) ∨ 0,
t≥0
0≤t 0 ≤t
τa∗ ,b∗ := inf{t ≥ 0 : Utξ
a∗
> b ∗ }.
Then, we have
1. v (x) ≤ J(x; ξ, τbξ∗ ) for any ξ ∈ Ξ, with τbξ∗ := inf{t ≥ 0 : Utξ > b ∗ };
∗
2. J(x; ξ a , τ ) ≤ v (x) for any τ ∈ Υ.
∗
In other words, the pair (ξ a , τa∗ ,b∗ ) is a saddle point and v (·) is
the value function of the game.
135 of 137
Numerical Examples
14
14
12
12
10
10
8
8
value
value
6
4
6
4
2
2
0
0
-2
-2
-4
-6
-12
-10
-8
-6
-4
-2
0
2
-4
-12
4
-10
-8
-6
-4
x
-2
0
2
4
x
β = −2, 0, 2, 4
α = 0.5, 1, 2, 10
12
20
10
15
8
10
6
value
value
5
4
0
2
-5
0
-10
-2
-4
-12
-10
-8
-6
-4
-2
0
2
x
C = −2, 0, 2, 4
136 of 137
4
-15
-12
-10
-8
-6
-4
-2
0
2
4
6
x
K = 0.5, 1.0, 1.5, 2.0
References
[1] D. Hernández-Hernández and K. Yamazaki. Games of Singular Control
and Stopping Driven by Spectrally One-sided Lévy Processes.
Stochastic Processes and their Applications, 125(1), 1-38, 2015.
[2] D. Hernández-Hernández and R.S. Simon and M. Zervos. Zero-sum
game between a singular stochastic controller and a discretionary
stopper, Ann. Appl. Prob., 25(1), 46-80, 2015.
137 of 137