First-exit problems for integrated diffusion processes with
state-dependent jumps
Department of Mathematics and Industrial Engineering
Polytechnique Montréal
2500, chemin de Polytechnique, Montréal (Québec) H3T 1J4
CANADA
https://www.polymtl.ca/expertises/en/lefebvre-mario
Abstract: Let dX(t) = Y(t)dt, where Y(t)is a one-dimensional diffusion process. First-exit problems from
CR2are studied for the degenerate two-dimensional diffusion process (X(t), Y (t)) when the process leaves C
not later than after a random time having an exponential distribution. When Y(t)is a standard Brownian motion,
the Laplace transform of the moment-generating function Mof the first-exit time is computed explicitly, as well
as the Laplace transforms of the mean exit time mand the probability pof leaving Cthrough a given part of its
boundary. When Y(t)is a geometric Brownian motion, the functions M,mand pare obtained by making use
of the method of similarity solutions to solve the various partial differential equations, subject to the appropriate
boundary conditions.
Key-Words: Kolmogorov backward equation, Brownian motion, geometric Brownian motion, method of
similarity solutions.
1 Introduction
We consider degenerate two-dimensional diffusion
processes (X(t), Y (t)) defined by
dX(t) = Y(t)dt, (1)
dY(t) = f[Y(t)]dt+{v[Y(t)]}1/2 dB(t),(2)
where {B(t), t 0}is a standard Brownian motion
and the functions fand vare such that {Y(t), t 0}
is a diffusion process. The process {X(t), t 0}
(multiplied by 1) is known as an integrated diffusion
process, because we can write that
X(t) = X(0) Zt
0
Y(s)ds. (3)
Numerous papers have been written on first-exit
problems for integrated diffusion processes. For
the case when {Y(t), t 0}is a Wiener pro-
cess, see McKean [12], Goldman [2], Gor’kov [3],
Lachal [5] and Lefebvre [8]. Moreover, Lefebvre [7],
Makasu [11], Metzler [13], Caravelli et al. [1] and
Levy [10] considered the case of integrated geomet-
ric Brownian motions. Finally, Lefebvre [6] and
Hesse [4] studied problems for integrated Ornstein-
Uhlenbeck processes.
Next, let (X(0), Y (0)) = (x, y)CR2and
define the first-exit time from C
T(x, y) = inf{t > 0 : (X(t), Y (t)) /C}.(4)
Notice that if Y(t)is always positive in C, then X(t)
will be strictly decreasing with time. Therefore, this
type of process can serve as model in applications
where the variable of interest cannot increase, for in-
stance when X(t)represents the remaining lifetime
of a certain device.
Assume that α > 0. The function
M(x, y) := EheαT (x,y)i(5)
is the moment-generating function of the random
variable T(x, y). It satisfies the Kolmogorov back-
ward equation (see Lefebvre [9], for a generalization
of this result)
1
2v(y)Myy +f(y)Myy Mx=αM (6)
for (x, y)C, where Myy := 2
y2M, etc. Further-
more, since T(x, y) = 0 if (x, y) /C, we have the
boundary condition
M(x, y;α) = 1 for (x, y) /C. (7)
Now, suppose that at a random time τhav-
ing an exponential distribution with parameter λ, if
(X(t), Y (t)) is still inside Ca random quantity Zis
added to the value of Y(t)at that time instant, so that
MARIO LEFEBVRE
Received: April 25, 2022. Revised: October 25, 2022. Accepted: November 28, 2022. Published: December 31, 2022.
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the process (X(t), Y (t)) will exit the continuation re-
gion Cimmediately. We say that the process is killed
not later than at time τ.
We can prove the following proposition.
Proposition 1.1. The function M(x, y)satisfies the
partial differential equation (PDE)
αM(x, y) = 1
2v(y)Myy(x, y) + f(y)My(x, y)
y Mx(x, y) + λ[1 M(x, y)].(8)
The equation is valid for (x, y)Cand is subject to
the boundary condition (7).
We also have the following corollaries.
Corollary 1.1. If it exists, the function m(x, y) :=
E[T(x, y)] satisfies the PDE
1 = 1
2v(y)myy(x, y) + f(y)my(x, y)
y mx(x, y)λm(x, y)(9)
for (x, y)C. The boundary condition is
m(x, y) = 0 for (x, y) /C.(10)
Corollary 1.2. Let
p(x, y) := P[(X(T), Y (T)) C0],(11)
where C0is a subset of the boundary C of C. For
(x, y)C, the probability p(x, y)is a solution of the
PDE
0 = 1
2v(y)pyy(x, y) + f(y)py(x, y)
y px(x, y) + λ[p0p(x, y)],(12)
where p0:= P[(X(τ), Y (τ)) C0]. Furthermore,
the boundary condition is
p(x, y) = 1if (x, y)C0,
0if (x, y)D,(13)
where D := C \C0.
In the next section, we will compute the Laplace
transform of the functions M(x, y),m(x, y)and
p(x, y)when {Y(t), t 0}is a standard Brown-
ian motion and T(x, y)is the first time that the two-
dimensional process ((X(t), Y (t)) leaves a rectangle
located in the first quadrant. In Section 3, the set
Cwill be the region between two straight lines and
{Y(t), t 0}will be a particular geometric Brown-
ian motion. With the help of the method of similarity
solutions, we will be able to get the exact solutions to
the PDE’s (8), (9) and (12). We will end this paper
with a few remarks in Section 4.
2 Integrated Brownian motion
In this section, we assume that v(y)1and f(y)
0, so that {Y(t), t 0}is a standard Brownian mo-
tion, and we define
T1(x, y)=inf{t > 0 : X(t) = 0 or Y(t) = aor b},
(14)
where x > 0and 0a<y<b. Moreover, the
quantity Zadded to Y(t)at time τis
Z=bY(τ)with probability p1(0,1),
aY(τ)with probability 1p1.
(15)
Because Y(t)is always positive in Cand dX(t) =
Y(t)dt,X(t)will be strictly decreasing with time.
Moreover, Cis actually a finite rectangle located in
the first quadrant.
Solving the PDE (8) is not an easy task. We can
however obtain the Laplace transform of the function
M(x, y)explicitly. Indeed, let
L(y) := Z
0
eβx M(x, y)dx, (16)
where β > 0. Since M(x, y)(0,1) for (x, y)C
and M(0, y) = 1, we have
Z
0
eβx Mx(x, y)dx=eβx M(x, y)
0
+β L
=1 + β L. (17)
It follows, taking the Laplace transform of each side
of Eq. (8), that
αL(y) = 1
2L00(y)y[1 + β L(y)]
+λ[(1/β)L(y)].(18)
The above ordinary differential equation (ODE) is
subject to the boundary conditions
L(a) = L(b) = 1/β. (19)
Making use of the mathematical software program
Maple, we find that the general solution of Eq. (18)
can be written as follows:
L(y) = c1Ai(η) + c2Bi(η)
+22/3 π
β4/3 Ai(η)ZBi(η)(β y +λ)dy
Bi(η)ZAi(η)(β y +λ)dy,(20)
where Ai and Bi are Airy functions and
η:= 21/3
β2/3 (β y +λ+α).(21)
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Mario Lefebvre
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The arbitrary constants c1and c2can be determined
from the boundary conditions (19) for any choice of
aand b. The expression obtained is rather involved.
Therefore, inverting the Laplace transform to obtain
the function M(x, y)would be really difficult.
Next, let
L1(y) := Z
0
eβx m(x, y)dx. (22)
Proceeding as above, we find that Eq. (9) becomes the
ODE
1
β=1
2L00
1(y)(y β +λ)L1(y),(23)
whose general solution is
L1(y) = c1Ai(ξ) + c2Bi(ξ)(24)
+22/3 π
β4/3 Ai(ξ)ZBi(ξ)dyBi(ξ)ZAi(ξ)dy,
where
ξ:= 21/3
β2/3 (β y +α).(25)
The constants c1and c2must be chosen so that
L1(a) = L1(b) = 0.
Finally, let
p1(x, y) := P[X(T) = 0].(26)
Since p0=P[X(τ) = 0] = 0, we may write that the
function
L2(y) := Z
0
eβx p1(x, y)dx(27)
satisfies the ODE
0 = 1
2L00
2(y)(y β +λ)L2(y) + y. (28)
Indeed, we calculate
Z
0
eβx
xp1(x, y)dx=eβx p1
0
+β L2
= [0 p1(0, y)] + β L2
=1 + β L2.(29)
Maple gives us the following expression for the gen-
eral solution of Eq. (28):
L2(y) = c1Ai(ξ) + c2Bi(ξ)(30)
+22/3 π
β1/3 Ai(ξ)ZBi(ξ)dyBi(ξ)ZAi(ξ)dy.
Notice that the functions L1(y)and L2(y)are almost
the same. Moreover, the constants c1and c2are also
such that L2(a) = L2(b) = 0.
In the next section, we will be able to obtain the
exact expressions for the functions of interest.
3 Integrated geometric Brownian
motion
Suppose that the diffusion process {Y(t), t 0}is
defined by
dY(t) = k Y (t)dt+Y(t)dB(t).(31)
That is, {Y(t), t 0}is a geometric Brownian
motion with infinitesimal mean k y and infinitesimal
variance y2. This process can be expressed as the ex-
ponential of a Wiener process with infinitesimal pa-
rameters k1
2and 1. Therefore, if Y(0) >0, the
process will remain positive for any t > 0, and X(t)
will be strictly decreasing with t.
Let
T2(x, y) := inf{t > 0 : X(t)/Y(t) = k1or k2},
(32)
where 0< k1< x/y < k2. Thus, the set Cis the
region between two straight lines going through the
origin.
Next, assume that
Z=(1
k1X(τ)Y(τ)with probability p1,
1
k2X(τ)Y(τ)with probability 1p1,
(33)
so that the process (X(t), Y (t)) will indeed leave the
continuation region Cnot later than at time τ.
The PDE given in (8) becomes
αM =1
2y2Myy +k y Myy Mx+λ[1 M(x, y)].
(34)
It is subject to the boundary conditions M(x, y)=1
if x/y=k1or k2.
Now, based on Eq. (34) and the boundary con-
ditions, we will look for a solution of the form
M(x, y) = N(r), where r:= x/yis called a sim-
ilarity variable. We find that Eq. (8) is transformed
into the ODE
1
2r2N00 +(rk r 1)N0(α+λ)N+λ= 0.(35)
The boundary conditions become N(k1) = N(k2) =
1.
The general solution of Eq. (35) is
N(r)= rkζ+1
2e2/rc1M1 + ζ
2+k, 1 + ζ, 2
r
+c2U1 + ζ
2+k, 1 + ζ, 2
r+λ
α+λ,(36)
where
ζ:= p4(k2k) + 8(α+λ)+1,(37)
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M(·,·,·)and U(·,·,·)are Kummer functions, and the
constants c1and c2are determined from the boundary
conditions N(k1) = N(k2) = 1. In the special case
when k= 1, the solution can be expressed in terms
of the Bessel functions Iν(·)and Kν(·)as follows:
N(r) = c1[(γ+ 1)r+ 2]Iγ
2(1/r)+2Iγ
2+1(1/r)
+c2[(γ+ 1)r+ 2]Kγ
2(1/r)2Kγ
2+1(1/r)
×e1/r
r+λ
α+λ,(38)
where
γ:= p8(α+λ)+1.(39)
Next, we assume that the function m2(x, y) :=
E[T2(x, y)] can be written as N1(r). Proceeding as
above, we find that the function N1(r)satisfies the
ODE
1 = 1
2r2N00
1(r)+(rk r1)N0
1(r)(α+λ)N1(r),
(40)
subject to N1(k1) = N1(k2)=0. The general solu-
tion of Eq. (40) is
N1(r) = c1M1 + ζ0
2+k, 1 + ζ0,2
r
+c2U1 + ζ0
2+k, 1 + ζ0,2
r
×rkζ0+1
2e2/r+1
λ,(41)
where
ζ0:= p4(k2k)+8λ+ 1.(42)
This solution can also be expressed in terms of Bessel
functions when k= 1.
Finally, we define
p2(x, y) = P[X(T2)/Y(T2) = k1].(43)
We assume that p2(x, y) = N2(r). To obtain the
function N2(r), we must find the solution of the ODE
0 = 1
2r2N00
2(r)+(rk r 1)N0
2(r)λN2(r)+λp1
(44)
that is such that N2(k1) = 1 and N2(k2) = 0. We
find that the general solution of the above equation is
N2(r) = c1M1 + ζ0
2+k, 1 + ζ0,2
r
+c2U1 + ζ0
2+k, 1 + ζ0,2
r
×rkζ0+1
2e2/r+p1.(45)
As in the previous cases, the function N2(r)can be
expressed in terms of Bessel functions when k= 1.
4 Conclusion
First-exit problems for diffusion processes have ap-
plications in many areas. To solve such problems,
one usually needs to find the solution of a differential
equation that satisfies certain boundary conditions. In
two or more dimensions, the equation to be solved is
a PDE. Therefore, the problem is difficult.
In this paper, we have added the constraint that
the two-dimensional process leaves the continuation
region at the latest at a random time that follows an
exponential distribution, which further increases the
difficulty of the problems considered.
In Section 2, we treated the case where the diffu-
sion process {Y(t), t 0}is a standard Brownian
motion and the continuation region Cis a rectangle
located in the first quadrant. We have succeeded in
obtaining the Laplace transform of the three functions
of interest.
In Section 3, {Y(t), t 0}was a geometric Brow-
nian motion and the set Cwas the region between two
straight lines that intersect at the origin. In this case,
we obtained, using the method of similarity solutions,
the exact expressions for the functions of interest.
As a continuation of this work, we could try to
solve this type of problem when the jumps are contin-
uous rather than discrete random variables. We could
also try to solve stochastic optimal control problems
defined in terms of the processes studied.
Acknowledgements
This research was supported by the Natural Sci-
ences and Engineering Research Council of Canada.
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