22
A Note on Impulse Control with Outside Jumps*

Makoto Goto

Faculty of Economics and BusinessHokkaido University

Seiya Kanagawa
GCA Corporation

1

Introduction

Reversible investment or capacity choice have been studied extensively as singular control or

impulse control problems. Especially, capacity choice is one of the biggest interest of compa‐
nies and to optimize their decision makings about capacity choice based on product prices orsomething that affect their production planning brings economic profits into them. However,capacity choice is usually represented by impulse control problem with outside jumps and any

effective methods that give solutions of the problem have not been found yet. If those methods
are found, more general and complicated capacity choice problems can be solved explicitly andthe theory of capacity choice or even stochastic control as a whole dramatically makes progress.
One of the most epoch‐making study of capacity choice in these days is Guo and Tomecek(2008). In Guo and Tomecek (2008), the theoretical connection between singular control andoptimal switching control in two‐regime case that was discussed in Vath and Pham (2007)
was established and they were successful in solving a multidimensional capacity choice problem
which is defined as a singular control by applying the connection. It means that they obtained
solutions of singular control problem by solving corresponding optimal switching problem and at
the same time, the conversion from the outside control in singular control to the inside control

in optimal switching control is observed. Moreover, they expanded their theory in Guo (2009).
This approach is more informative than other previous approaches because it does not require
any special forms of utility functions which are required by previous researches and, furthermore,
it allows non‐smooth utility functions. However, many capacity choice problems include not only
proportional cost but also fixed cost, so impulse control is more suitable for representing thosekind of problems. And it is difficult to solve the problems through usual approaches as reported

in Goto et al. (2006).
In this thesis, we consider a novel approach that gives solutions of impulse control prob‐

lem with outside jumps and that is inspired by Guo and Tomecek (2008). If we can connect
impulse control with optimal switching control, we can solve the problems through converting

outside jumps of impulse control into inside jumps of optimal switching control. In our case,controls are accompanied by impulse, so we should change the form of optimal switching control
to make it correspond to impulse control and we call the changed optimal switching control

transformed switching control. For solving the problem, we have to clear three things. First, it*This work was supported by the Research Institute for Mathematical Sciences, a Joint Usage/Research Center

located in Kyoto University.

23
is necessary that the theoretical connection between impulse control and transformed switchingexists. The transformed switching that we now consider is not derived from the impulse control
mathematically but assumed for convenience, so it might be possible that there is no connection
between them in practice. Second, we need to reveal whether transformed switching has math‐ematical explicit solutions or not. It means that the transformed switching has to meet HJB
equation but we currently do not know wether it meets the equation. Third, we have to verify

that transformed switching can be solved practically and analyze the parameters’ effects on thesolutions. However, sophisticated mathematical discussion is unavoidable to prove the first andsecond problems, so we establish that the purpose of present study is to obtain the solutions oftransformed switching and give examples of them under some different parameters.

2

Hypothesis for Correspondence between Impulse Control withTransformed Switching Control

2.1

ProblemLet (\Omega, F, F, P) be a filtered probability space, and assume a given bounded interval [a, b]\in(-\infty, \infty) . Consider the following problem.

V(x, y)= \sup_{\omega\ni W}J(x, y, \xi_{i}) ,

(1)

with

J(x, y, \xi_{\dot{i} ) :=E[\int_{0}^{\infty}e^{-\rho t}H(Y_{t})X_{t}^{x}dt- \sum_{i=1}^{\infty}e^{-\rho\tau_{i} C(\xi_{\dot{i} )]subject toC(\xi)=\{ begin{ar y}{l K_{1}+E_{1}\xi, \xi>0, 0, \xi=0,C(0)=K_{1}, K_{0}+E_{0}\xi, \xi<0, \end{ar y}

,

(2)

(3)

dX_{t}^{x}=\mu X_{t}dt+\sigma X_{t}dW_{t}, X_{0}=x>0,\omega=(\tau_{1}, \tau_{2}, \ldots;\xi_{1}, \xi_{2}, \ldots),

N(t)= \sup[i\geq 0 : \tau_{i}\leq t],

Yt:=y+ \sum_{\dot{i}=1}^{N(t)}\xi_{i}, y\in[a, b],

H:[a, b], H(y)=H(a)+ \int_{a}^{y}h(z)dz,

K_{1}, K_{0}, E_{1}>0.

where the market price X is modeled by a geometric Brownian motion; an impulse control \omega isrepresented by stopping time \tau and impulse \xi ; the last stopping time N(t) is defined by terminal
time t ; the capacity level Y is controlled process represented by \xi_{i} ; the resource extraction rate24
is modeled by a function H(Y);K_{1}>0 is the fixed cost per investment and K_{0}>0 is the fixed

cost per disinvestment; E_{1}>0 is the cost of capacity increase and E_{0} is the cost of capacity
reduction. The aim of the firm is to maximize

2.2

J by controlling Y.

Preliminary
Throughout the thesis, we definem<0<1<n to be the roots of \sigma^{2}x^{2}+(\mu-\sigma^{2})x-\rho=0,

so thatm, n= \frac{-(\mu-\sigma^{2})\pm\sqrt{(\mu-\sigma^{2})^{2}+4\sigma^{2}\rho} {2\sigma^{2} .

(4)\eta:=\frac{1}{\rho-\mu}=\frac{-mn}{(n-1)(1-m)\rho}=\frac{1}{\sigma^{2}(n-1)(1 -m)} .

(5)

We also observe the identity \rho=-\sigma^{2} mn and define the useful quantity \eta>0 :Next, let R(x, y) :=J(x, y, 0) be the no‐action expected payoff. Then,

R(x, y) := E[\int_{0}^{\infty}e^{-\rho t}H(y)X_{t}^{x}dt]=\eta H(y)x ,r(x, y) :=R_{y}(x, y)= E[\int_{0}^{\infty}e^{-\rho t}h(y)X_{t}^{x}dt]=\eta h(y) x .
2.3

(6)
(7)

Corresponding transformed switching and the value function

Optimal switching control corresponding with singular control in Guo and Tomecek (2008)is as follows.

v_{k}(x,z):=\kap a_{0}=k\sup_{\alpha\inB}E[\int_{0}^{\infty}e^{-\rhot} [h(z)X_{t}^{x}]I_{t}dt-\sum_{n=1}^{\infty}e^{-\rho\tau_{n} K_{\kap a_{n} ],(8)where \alpha=(\tau_{i}, \kappa_{n})_{n\geq 0} is an admissible switching control, B is the subset of admissible switching
controls \alpha=(\tau_{i}, \kappa_{n})_{n\geq 0} such that E[\sum_{n=1}^{\infty}e^{-\rho\tau_{n}}]<0 , and I_{t} is the regime indicator function
defined as follows.

I_{t}:= \sum_{n=0}^{\infty}\kap a_{n}1_{\{\tau_{n}<t\leq\tau_{n+1}\} , I_{0}= \kap a_{0}.

(9)Here, we consider the two‐regime switching between the stoping status and the operating status,

k\in[0,1] . Finally, we have the value function represented by the payoff of optimal switching
control:V(x, y)= \eta H(a)x+\int_{a}^{y}v_{1}(x, z)dz+\int_{y}^{b}v_{0}(x, z)dz

.

(10)

According to the theoretical connection between singular control and optimal switching control

established in Guo and Tomecek (2008), the value function represents that of singular control
problem.

As for impulse control with outside jumps, we assume that the representation of the valuefunction of impulse control is same as that of singular control. However, we define transformed
switching may corresponds with impulse control as follows.

v_{k}(x,z) := \kap a_{0}=k\sup_{\alpha\inB}E[\int_{0}^{\infty}e^{-\rhot} [h(z)X_{t}^{x}]I_{t}dt-\sum_{n=1}^{\infty}e^{-\rho\tau_{n} (K_{\kap a_{n} + E_{\kap a_{n} \xi_{n})]

(11)

25
Transformed switching deals with the switching cost of capacity increase or reduction in addition
to the fixed switching cost. Moreover, we consider that the two‐regime switching between two
different operating statuses in this case. Therefore, we know the value function of impulse control

problem is (10) and the payoff \nu_{0} and \nu_{1} are the solutions of transformed switching problem(11) are as follows:v_{0}(x, z)= \kap a_{0}=0\sup_{\alpha\in B}E[\int_{0}^{\infty}e^{-\rho t}h(z)X_ {t}^{x}I_{t}dt-\sum_{n=1}^{\infty}e^{-\rho\tau_{n} (K_{1}+E_{1}\xi_{n})]v_{1}(x, z)= \kap a_{0}=1\sup_{\alpha\in B}E[\int_{0}^{\infty}e^{-\rho t}(-h(z) X_{t}^{x})(1-I_{t})dt-\sum_{n=1}^{\infty}e^{-\rho\tau_{n} (K_{0}+E_{0}\xi_{n})]
,

3

(12)

.

(13)

Solutions of Transformed Switching Control
We assume that \nu_{0} and \nu_{1} are the unique solutions with linear growth condition to thefollowing system of variational inequalities.

\min\{-\mathcal{L}v_{0}(x, z)-h(z)x, v_{0}(x, z)-v_{1}(x, z+\xi)+K_{1}+E_{1} \xi\}=0 ,(14)\min\{-\mathcal{L}v_{1}(x, z)-h(z)x, v_{1}(x, z)-v_{0}(x, z+\xi)+K_{0}+E_{0} \xi\}=0 ,(15)where \mathcal{L} is the generator of the diffusion X^{x} , killed at rate \rho , given by \mathcal{L}u(x, z)=\sigma^{2}u_{xx}(x, z)+\mu u_{x}(x, z)-pu(x, z) . Based on Vath and Pham (2007), we solve \nu_{0} and \nu_{1} in the following twocases.

3.1

Case 1:

E_{0}\geq 0

For each z\in(a, b) , we describe the switching regions as d(z) and u(z) which take values(0, \infty].
Firstly, for each z\in(a, b) such that h(z)=0 , it is never optimal to switch since E_{0}\geq
0, E_{1}>0, K_{0}>0, K_{1}>0, \xi_{n}>0 in \nu_{0} and \xi_{n}<0 in \nu_{1} , and so we have d(z)=\infty=u(z) . For

this case, v_{0}(x, z)=0=v_{1}(x, z) .

Secondly, for each z\in(a, b) such that h(z)>0 , it is optimal to switch from regime 0 to

regime 1 when X_{t}^{x}\in[u(z), \infty ), and to switch from regime 1 to regime 0 when X_{t}^{x}\in(0, d(z) ],where 0<d(z)<u(z)<\infty . Furthermore, we havev_{0}(x, z)=\{\begin{ar ay}{l } A(z)x^{n}+\eta h(z)x, x<u(z) , B(z)x^{m}+\eta h(z+\xi_{1})x-K_{1}-E_{1}\xi_{1}, x\geq u(z) , \end{ar ay}v_{1}(x, z)=\{\begin{ar ay}{l } A(z)x^{n}+\eta h(z+\xi_{0})x-K_{0}-E_{0}\xi_{0}, x<d(z) , B(z)x^{m}+\eta h(z)x, x\geq d(z) . \end{ar ay}

(16)

(17)

where \xi_{1}\geq 0 is accompanied by investment and \xi_{0}<0 is accompanied by disinvestment. The

value matching condition, the smooth pasting condition and the condition of a differentiation

26
with respect to an impulse lead to the following boundary conditions:\{ begin{ar y}{l A(z)u^{n}(z)+\etah(z)u =B(z)u^{m}(z)+\etah(z+\xi)u(z-K_{1}-E_{1}\xi_{1}, nA(z)u^{n-1}(z)+\etah(z)=mB(z)u^{m-1}(z)+\etah(z+\xi_{1}), \eta\frc{\partilh(z+\xi_{1}) \partil\xi_{1}u(z)-E_{1}=0, \end{ar y}\{begin{ar y}{l A(z)d^{n}(z)+\etah(z+\xi_{0})d(z-K_{0}-E_{0}\xi_{0}=B(z)d^{m}(z)+\etah(z)d , nA(z)d^{n-1}(z)+\etah(z+\xi_{0})=mB(z)d^{m-1}(z)+\etah(z), \etafrc{\partilh(z+\xi_{0}) \partil\xi_{0}d(z)-E_{0}=. \end{ar y}

(18)

(19)

Here, we assume h(z)=z^{\gamma} ( \gamma is an arbitrary constant value), so given d(z), u(z), A(z), B(z), \xi_{0}and \xi_{1} are solved from (16),\{beginary}lA(z)=fct1-{mh+\xi_}(z)uKE1{^n-\frac}m)=et(1{hz+\xi0-)dK_}E{^n(z\fracm-1),B=et{}n(hz+\xi_1)-uKE{}^m(z\fracn-1)=et{}(hz+\xi_0-)dKE{}^m(z1\fracn),K_+Exi{}u^-(z0\_)dn=eta1-frc{}m(hz+\xi_)u^n 0d{1-}h(z)+^n,K_{1E}\xiu-m(z)+0_{d^}=\eta1-frcn)(hz+xi_{u^m}\0d1-()hz{+^m},0=\etag(zxi_{1)m-}uE,0=\etag(z+xi_{)^m-1}dE0.\enary(20)

Lastly, for each z\in(a, b) such that h(z)<0 , it is optimal to switch from regime 1 toregime 0 when X_{t}^{x}\in(0, d(z) ] and it is never optimal to switch from regime 0 to regime 1, so wehave d(z)<\infty and u(z)=\infty . Moreover, y_{0} and \nu_{1} are given byv_{0}(x, z)=\eta h(z)x ,(21)v_{1}(x,z)=\{ begin{ar ay}{l} \etah(z+\xi)x-K_{0}-E_{0}\xi, x\leqd(z), B(z)x^{m}+\etah(z)x, x>d(z), \end{ar ay}(22)

where \xi<0 . We obtain the boundary condition as follows.

\{begin{ar y}{l B(z)d ^{m}+\etah(z)d =\etah(z+\xi)d(z-K_{0}-E_{0}\xi, mB(z)d^{m-1}(z)+\etah(z)=\etah(z+\xi), \etafrc{\partilh(z+\xi)}{\partil\xi}d(z)-E_{0}=. \end{ar y}

(23)

27
Therefore, assuming h(z)=z^{\gamma}-\alpha ( \gamma is an arbitrary constant value and \alpha is a positive number)
for the existence of z\in(a, b) such that h<(0), B(z), d(z) and \xi are given by\{beginary}{lB(z)=\frac{et[(z+\xi)^{gam}-z^{\gam}]d(z)-K_{0E}\xi{d(z)^m},d(z)=\frac{m}-1\frac{K_0}+E \xi{eta[(z+\xi)^{gam}-z^{\gam}],0=\gam(z+\xi)^{gam-1}\frac{K_0}/E +\xi}{(z )^\gam}-z^{\gam}-\frac{m1} .\end{ary}3.2

Case 2:(24)

E_{0}<0First, for each z\in(a, b) such that h(z)=0 , it is never optimal to switch since E_{0}<0.
Therefore, d(z)=\infty=u(z) and v_{0}(x, z)=0=v_{1}(x, z) in this case.

Then, for each z\in(a, b) such that h(z)>0 , it is optimal to switch from regime 0 to regime

1 when X_{t}^{x}\in[u(z), \infty ) and it is never optimal to switch from regime 1 to regime 0 . That is,d(z)=\infty and u(z)<\infty and we havev_{0}(x,z)=\{ begin{ar ay}{l} A(z)x^{n}+\etah(z)x, x<u(z), h(z+\xi)x-K_{1}-E_{1}\xi, x\gequ(z), \end{ar ay}v_{1}(x, z)=\eta h(z)x .

(25)
(26)

According to the conditions, the boundary condition is as follows.

\{begin{ar y}{l A(z)u ^{n}+\etah(z)u =\etah(z+\xi)u(z-K_{1}-E_{1}\xi, nA(z)u^{n-1}(z)+\etah(z)=\etah(z+\xi), \etafrc{\partilh(z+\xi)}{\partil\xi}u(z)-E_{1}=0. \end{ar y}(27)

Here, h(z)=z^{\gamma} ( \gamma is an arbitrary constant value) is assumed. That is,\{beginary}{lA(z)=\frac{et[(z+\xi)^{gam}-z^{\gam}]u(z)-K_{1E}\xi{u(z)^n},u(z)=\frac{n}-1\frac{K_1}+E \xi{eta[(z+\xi)^{gam}-z^{\gam}],0=\gam(z+\xi)^{gam-1}\frac{K_1}/E +\xi}{(z )^\gam}-z^{\gam}-\frac{n1}.\end{ary}(28)Finally, for each z\in(a, b) such that h(z)>0 , it is optimal to switch from regime 1 toregime 0 when X_{t}^{x}\in(0, d(z) ] and it is never optimal to switch from regime 0 to regime 1, so
we have d(z)<\infty and u(z)=\infty . Therefore, we obtain clearly same conclusion as Case 1 witheach z\in(a, b) such that h(z)<0 in this case.
In short, we have 4 patterns of transformed switching, no switching, downside switching(only to switch from regime 1 to regime 0 ), upside switching(only to switch from regime 0 to
regime 1), bilateral switching (to switch between regime 0 and regime 1). They are summarized
as follows.

1. No switching:h(z)=0v_{0}(x, z)=v_{1}(x, z)=0, d(z)=u(z)=\infty.

28
2. Downside switching: h(z)=z^{\gamma}-\alpha<0, \xi<0\{ v_{1}(x,z)=\{\begin{aray}{l} \etah(z+\xi)x-K_{0}-E_{0}\xi,x\leqd(z) B(z)x^{m}+\etah(z)x, >d(z) \end{aray}\{
v_{0}(x, z)=\eta h(z)x

B(z)= \frac{\eta[(z+\xi)^{\gamma}-z^{\gamma}]d(z)-K_{0}-E_{0}\xi}{d(z)^{m} ,d(z)= \frac{m}{m-1}\frac{K_{0}+E_{0}\xi}{\eta[(z+\xi)^{\gamma}-z^{\gamma}]},0= \gamma(z+\xi)_{(z+\xi)^{\gamma}-z^{\gamma}}^{\gamma-1^{K_{0}/E_{0}+\xi} - \frac{m-1}{m}.

3. Upside switching: h(z)=z^{\gamma}>0, \xi>0, E_{0}<0\{\begin{ar ay}{l} v_{0}(x, z)=[Case] v_{1}(x, z)=\eta h(z)x. \end{ar ay}\{beginary}{lA(z)=\frac{et[(z+\xi)^{gam}-z^{\gam}]u(z)-K_{1}E \xi{u(z)^n},u(z)=\frac{n}-1\frac{K_1}+E \xi{[(z+)^\gam}-z^{\gam}],0=\gam(z+\xi)_{ ^\gam}-z^{\gam}^{\gam-1^{K_}/E1+\xi}-frac{n1}.\end{ary}4. Bilateral switching: h(z)=z^{\gamma}>0, \xi_{1}>0, \xi_{0}<0, E_{0}\geq 0\{_v{1}(x,z)=}^{v_0}(x,z)=}\{

A(z)x^{n}+\eta h(z)x,x<u(z),

B(z)x^{m}+\eta h(z+\xi_{1})x-K_{1}-E_{1}\xi_{1},x\geq u(z)A(z)x^{n}+\eta h(z+\xi_{0})x-K_{0}-E_{0}\xi_{0},x<d(z)

B(z)x^{m}+\eta h(z)x,x\geq d(z)\{beginary}lA(z)=fc1-mh+\xi_{u(z)K}E1^n-\frac{m)=et(1}hz+\xi_{0-)dKE}^n(z\frac{m-1),B=et}n(hz+\xi_{1)-uKE}^m(z\frac{n-1)=et}(hz+\xi_{0)d-KE}^m(z1\frac{n),edy

,

,

,

(K_{1}+E_{1}\xi_{1})u^{-n}(z)+(K_{0}+E_{0}\xi_{0})d^{-n}(z)= \eta(1-\frac{1}{m})(h(z+\xi_{1})u^{1-n}(z)+h(z+\xi_{0})d^{1-n}(z)-h(z)(u^{1- n}(z)+d^{1-n}(z))),(K_{1}+E_{1}\xi_{1})u^{-m}(z)+(K_{0}+E_{0}\xi_{0})d^{-m}(z)= \eta(1-\frac{1}{n})(h(z+\xi_{1})u^{1-m}(z)+h(z+\xi_{0})d^{1-m}(z)-h(z)(u^{1- m}(z)+d^{1-m}(z)))

0=\eta\gamma(z+\xi_{1})^{\gamma-1}u(z)-E_{1},0=\eta\gamma(z+\xi_{0})^{\gamma-1}d(z)-E_{0}.

,

29
4

Illustrations of Solutions

4.1

Constraint for calculation

In all cases, it is clear that A(z)>0 and B(z)>0 since A(z) and B(z) represent theoption values for waiting to switch from current regime to another regime. We also clearly haved(z)>0 and u(z)>0 because X_{t}^{x}\in(0, \infty) . In addition, \xi>0 in upside switching, \xi<0 indownside switching, \xi_{1}>0 and \xi_{0}<0 in bilateral switching.

As for \gamma , first, 0<\gamma<(n-1)/n for the convergence in upside switching since E_{1}>0

requires \gamma>0 and taking the limit of (28) proves \gamma<(n-1)/n . Then, in downside switching,
it is possible that \gamma\geq 0 and \gamma<0 since E_{0}<0 requires \gamma<0 and E_{0}\geq 0 requires \gamma\geq 0.
Lastly, \gamma>0 because E_{1}>0 requires \gamma>0.

4.2

Examples of numerical calculations

4.2.1

Downside switching

In this thesis, we use R programming codes to solve the problem. First of all, we set theparameter values such that \rho=0.1, \mu=0.01, \sigma=0.2, \gamma=0.3, K_{0}=1 and E_{0}=1 . Next,

we calculate B(z), d(z) and \xi with those parameters by the Newton method. Consequently, we
obtain the graph as in Fig. 1.

D^{-1}(z)=z+\xi>0 is the new z level after we add an impulse\xi to the current z level. This graph illustrates the switching region (0, d(z) ] and the continuous
region (d(z), \infty) , and when X_{t}^{x} reaches the switching region, the current capacity level jumps tothe new level D^{-1}(z) . As for the option values, B(z)>0 for each z\geq 2 (we start to calculate
from z=2 ) as in Fig. 2. So downside switching is satisfied with the all constraints under thesesettings.

Moreover, we also calculate a E_{0}<0 and \gamma<0 case and obtain the graphs as in Fig. 3(we set E_{0}=-1 and \gamma=-0.3 in this case). However, in this case, we have B(z)<0 for eachz\geq 2 as in Fig. 4. It proves that E_{0}<0 and \gamma<0 don’t match downside switching.

4.2.2

Upside switchingFirstly, we set the parameter values such that \rho=0.1, \mu=0.01, \sigma=0.2, \gamma=0.3, K_{1}=1and E_{1}=1 . Secondly, we determine
the graph as in Fig. 5.

A(z), u(z) and \xi based on the parameters. Finally, we haveU^{-1}(z)=z+\xi>0 is the newzlevel after we add an impulse \xi tothe current z level. In this figure, the switching region and the continuous region are describedas [u(z), \infty) and [0, u(z) ), respectively, and when X_{t}^{x} reaches the switching region, the current
capacity level jumps to the new level U^{-1}(z) . Furthermore, A(z)>0 for each z\geq 0 (calculation
starts from z=0 ) as in Fig. 6. So upside switching is satisfied with the all constraints underthese settings.
4.2.3

Bilateral switchingFirst, we set the parameter values such that \rho=0.1, \mu=0.01, \sigma=0.2, \gamma=0.3, K_{1}=1, K_{0}=1, E_{1}=2 and E_{0}=1 .

Then, we calculate A(z), B(z), u(z), d(z), \xi_{1} and \xi_{0} . Lastly,

30
the graph as in Fig. 7 is given below. We have the switching region [0, d(z) ) and [u(z), \infty )and the continuous region (d(z), u(z)) in bilateral switching, so when X_{t}^{x} reaches the downside
and upside switching regions, the current capacity level jumps to the new levelU^{-1}(z) , respectively.

D^{-1}(z) andMoreover, for each z\geq 5 (we start to calculate from z=5 ),A(z)>0 andB(z)>0 as in Figs. 8 and 9. So bilateral switching is satisfied with the all constraints underthese settings.
In bilateral switching, it is a critical problem that the calculation occasionally diverge at

higher z level. For the problem, we find rather high value of \sigma or \rho suppress the divergence ofthe calculation.

4.3

Sensitivity analysis

4.3.1

Downside switching

We analyze the parameters’ effects to the solutions. Firstly, as for downside switching,
we establish the standard condition such that\rho=0.1, \mu=0.01, \sigma=0.2, \gamma=0.3, K_{0}=1 andE_{0}=1 and the solutions under the condition are as in Fig.

10. Next, we see the following

cases under the different parameters. Fig. 11 is given by \gamma=0.4 , and the downside thresholdd(z) and the jumping destinations D(z) decrease at the same capacity level compared with thestandard condition. Fig. 12 is given by \sigma=0.3 , and d(z) decrease and D(z) is scarcely affected
at the same capacity level. Fig. 13 is given by \rho=0.2 , and d(z) and D(z) increase at the same
capacity level. Fig. 14 is given by \mu=0.04 , and d(z) and D(z) increase at the same capacity
level. We summarize the above findings from Figs. 11‐14 as follows: the increase of \gamma, \sigma, \mu andthe reduction of \rho reduce the downside threshold d(z) at the same capacity level; the increaseof \gamma, \mu and the reduction of \rho reduce the jumping destinations D(z) and the values of \sigma scarcely
effectD(z) .

Fig. 15 is given by the downside fixed cost K_{0}=2.4 , and we recognize that K_{0} that isover around 2.4 leads to a reversal between d(z) and D(z) . Moreover, the bigger K_{0} becomes,
the more expanded the reversal is as in Fig. 16, which is given by K_{0}=100 , so it is supposedthat the reversal is not dissolved at any z level in this case. However, it is not revealed why thethreshold of the reversal is around 2.4 and what affects it. Fig. 17 is given by E_{0}=10 , and it
proves that the downside proportional cost E_{0} becomes bigger, d(z) and D(z) increase with thesame magnification of E_{0} at the same capacity level.
4.3.2

Upside switchingFirst, we set the standard condition such that \rho=0.1, \mu=0.01, \sigma=0.2, \gamma=0.3, K_{1}=1and E_{1}=1 and the solutions under the condition are as in Fig. 18. Next, we see the following
cases under the different parameters and \gamma=0.4, \sigma=0.3, \rho=0.2, \mu=0.04, K_{1}=1000 and
E_{1}=10 . Fig. 19 is given by \gamma=0.4 , and the upside threshold u(z) and thejumping destinations

U(z) decrease at the same capacity level compared with the standard condition. Fig. 20 is givenby \sigma=0.3 , and u(z) increase and U(z) is scarcely affected at the same capacity level. Fig. 21
is given by \rho=0.2 , and u(z) and U(z) increase at the same capacity level. Fig. 22 is given by31
31\mu=0.04 , and u(z) increase and U(z) decrease at the same capacity level. We summarize theabove findings from Figs. 19‐22 as follows: the increase of \gamma, \mu and the reduction of \sigma, \rho reduce
the upside threshold u(z) at the same capacity level; the increase of \gamma and the reduction of \rho, \mureduce the jumping destinations U(z) and the values of \sigma scarcely effect U(z) .
Fig. 23 is given by the upside fixed cost K_{1}=1000 , and threshold u(z) and the amount of
impulse \xi also increase at the same capacity level. Fig. 24 is given by the upside proportional
cost E_{1}=10 , and it proves that u(z) and U(z) increase with the same magnification of E_{1} atthe same capacity level.

5

Conclusion
The main contribution of this thesis is giving a note on a new method to solve impulse

control with outside jumps. It is required to clear three things in order to solve the impulse

control problem with outside jumps in terms of transformed switching that we consider in thisthesis. Firstly, we have to prove that the theoretical connection between impulse control andtransformed switching really exists.

Secondly, we need to reveal that transformed switchingmeets HJB equation. Finally, it is necessary to verify that transformed switching can be solved
practically and how the parameters effect on the solutions. In this thesis, we only study thethird problem due to some mathematical difficulties of the first and second problems.
We assume transformed switching that is considered to corresponds to impulse control
problem and solve it. We are successful in obtaining the solutions, but the solutions of impulse\xi are not given as analytical solutions in all cases and neither are the solutions of the thresholds

d(z) and u(z) in bilateral switching. So we calculate the solutions with some different parametersand get the examples of the solutions using R programing. Consequently, although something
to be revealed still remain such as the convergence condition of the calculation, it is shown that
we can solve transformed switching by numerical calculations under some conditions and how
the effects are brought into the solutions by each parameter.
However, we also find the problems of the solutions of transformed switching. First, it is
not allowed to set \gamma<0 in all cases. It means that we have to consider that the utility function
H is convex and it may contradicts usual diminishing utilities.

that

H is concave, so the problem may be our original.

In Guo (2009), the authors say

Second, we can not consider the bigger

K_{0} in downside switching. This method can not deal with the problem with the bigger downsideproportional cost. Third, the bigger K_{0} is not allowed unless we consider the higher capacity

level in bilateral switching. Only in the z range in which the reversal between d(z) and D(z)is dissolved the solutions are valid. Lastly, especially in bilateral switching, strict parametersetting is required to converge the calculation because the divergence of the calculation is easily
observed. Due to these problems, it is possible that the solutions are useless. In addition, as theabove statement, we still leave the two more essential steps to complete our method. If theseassumptions are not valid, this method may be useless for solving impulse control problem withoutside jumps. Thus, there are a lot of things to be studied in order to prove our method’s

effectiveness, but our attempt must give some useful knowledge to the theory of capacity choice.

32
References

Mansur, E. T. (2013): Prices versus quantities: Environmental regulation and imperfect com‐
petition, Journal of Regulatory Economics, 44, 80‐102.

Guo, X and Tomecek, P. (2008): Connections between singular control and optimal switching,
SIAM Journal of Control and optimization, 47, 421‐443.

Guo, X and Tomecek, P. (2009): A class of singular control problems and the smooth fit principle,
SIAM Journal of Control and optimization, 47, 3076‐3099.

Vath, V. and Pham, H. (2007): Explicit solution to an optimal switching problem in the two‐
regime case, SIAM Journal of Control and optimization, 46, 839‐876.

Cadenillas, A. (2000): Classical and impulse stochastic control of the exchange rate using interest
rates and reserves, Mathematical Finance, 10, 141‐156.

Goto, M. (2012): Real options theory and capacity choice problem: Stochastic control approach,Operations Research, 57, 560‐565 (in Japanese).
Goto, M., Takashima, R. and Tsujimura, M. (2006): Optimal Capacity Expansion and Contrac‐
tion with Fixed and Quadratic Adjustment Costs, Proceedings of the Sapporo Symposium on
Financial Engineering and Its Applications, 7‐20.

Goto, M. (2013): On the Impulse Control Problem with Outside Jumps, 17th Annual Interna‐
tional Real Options Conference, Tokyo, 2013‐7.

Faculty of Economics and BusinessHokkaido University, Sapporo 060‐0809, JapanE‐mail address: goto@econ.hokudai.ac.jp

GCA Corporation, Tokyo 100‐6230, Japan
GCA\Re X_{z}^{A}7\pm\Leftrightarrow
ftJll RrJffi

33

N

0

2

4

6

s

10

12

14

X_{t}^{x}
Figure 1: The downside thresholds and the jumping destinations from the current z level (E_{0}\geq

0)

.

n

N

B(z)

Figure 2: The option values in downside switching (E_{0}\geq 0) .

34

N

0 200 400 600 800 1000 1200 1400

X_{t}^{x}
Figure 3: The downside thresholds and the jumping destinations from the current z level (E_{0}<

0)

.

N

B(z)

Figure 4: The option values in downside switching (E_{0}<0) .

35

0

N

0

20

40

60

80

100

120

140

X_{t}^{X}
Figure 5: The upside thresholds and the jumping destinations from the current z level.

0

N

0.5

10

15

20

25

30

A(z)

Figure 6: The option values in upside switching.

36

N

0

20

40

60

80

100

X_{t}^{x}

Figure 7: The bilateral thresholds and the jumping destinations from the current z level.

0

N

0.2

04

0.6

0.8

10

A(z)

Figure 8: The upside option values in bilateral switching.

37

N

0

1000

2000

3000

4000

B(z)

Figure 9: The downside option values in bilateral switching.

N

0

2

4

6

s

10

12

14

X_{t}^{X}
Figure 10: The downside solutions under the standard condition.

38

N

0

2

4

6

s

10

12

14

X_{t}^{X}
Figure 11: The downside solutions under \gamma=0.4.

N

0

2

4

6

s

10

12

X_{t}^{X}
Figure 12: The downside solutions under \sigma=0.3.

14

39

N

0

2

4

6

s

10

12

14

X_{t}^{X}
Figure 13: The downside solutions under \rho=0.2.

N

0

2

4

6

s

10

12

X_{t}^{X}
Figure 14: The downside solutions under \mu=0.04.

14

40

N

0

2

4

6

s

10

12

14

X_{t}^{X}
Figure 15: The downside solutions under K_{0}=2.4.

N

0

2

4

6

s

10

12

X_{t}^{X}
Figure 16: The downside solutions under K_{0}=100.

14

41
41

N

0

20

40

60

80

100

120

140

X_{t}^{X}
Figure 17: The downside solutions under E_{0}=10.

N

0

20

40

60

80

100

120

140

X_{t}^{X}

Figure 18: The upside solutions under the standard condition.

42

N

0

20

40

60

80

100

120

140

X_{t}^{x}

Figure 19: The upside solutions under \gamma=0.4.

N

0

20

40

60

80

100

120

X_{t}^{X}

Figure 20: The upside solutions under \sigma=0.3.

140

43

N

0

20

40

60

80

100

120

140

X_{t}^{x}

Figure 21: The upside solutions under \rho=0.2.

N

0

20

40

60

80

100

120

X_{t}^{X}

Figure 22: The upside solutions under \mu=0.04.

140

44

N

0

20

40

60

80

100

120

140

X_{t}^{x}

Figure 23: The downside solutions under K_{1}=1000.

N

0

200

400

600

800

1000

1200

X_{t}^{X}

Figure 24: The downside solutions under E_{1}=10.

1400