Duration of stay inside an interval by the poisson process with a negative exponential component
Several two-boundary problems for the Poisson process with an exponential component are solved in the present article. The integral transforms of the joint distribution of the epoch of the first exit from the interval and the value of the overshoot through boundaries at the epoch of the exit are ob...
Gespeichert in:
| Datum: | 2006 |
|---|---|
| 1. Verfasser: | |
| Format: | Artikel |
| Sprache: | English |
| Veröffentlicht: |
Інститут математики НАН України
2006
|
| Online Zugang: | http://dspace.nbuv.gov.ua/handle/123456789/4441 |
| Tags: |
Tag hinzufügen
Keine Tags, Fügen Sie den ersten Tag hinzu!
|
| Назва журналу: | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| Zitieren: | Duration of stay inside an interval by the poisson process with a negative exponential component / T. Kadankova // Theory of Stochastic Processes. — 2006. — Т. 12 (28), № 1-2. — С. 55–67. — Бібліогр.: 11 назв.— англ. |
Institution
Digital Library of Periodicals of National Academy of Sciences of Ukraine| id |
irk-123456789-4441 |
|---|---|
| record_format |
dspace |
| spelling |
irk-123456789-44412009-11-11T12:00:34Z Duration of stay inside an interval by the poisson process with a negative exponential component Kadankova, T. Several two-boundary problems for the Poisson process with an exponential component are solved in the present article. The integral transforms of the joint distribution of the epoch of the first exit from the interval and the value of the overshoot through boundaries at the epoch of the exit are obtained. The Laplace transform of the total duration time of the process’s stay inside the interval is found. 2006 Article Duration of stay inside an interval by the poisson process with a negative exponential component / T. Kadankova // Theory of Stochastic Processes. — 2006. — Т. 12 (28), № 1-2. — С. 55–67. — Бібліогр.: 11 назв.— англ. 0321-3900 http://dspace.nbuv.gov.ua/handle/123456789/4441 519.21 en Інститут математики НАН України |
| institution |
Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| collection |
DSpace DC |
| language |
English |
| description |
Several two-boundary problems for the Poisson process with an exponential component
are solved in the present article. The integral transforms of the joint distribution
of the epoch of the first exit from the interval and the value of the overshoot through boundaries at the epoch of the exit are obtained. The Laplace transform of the total duration time of the process’s stay inside the interval is found. |
| format |
Article |
| author |
Kadankova, T. |
| spellingShingle |
Kadankova, T. Duration of stay inside an interval by the poisson process with a negative exponential component |
| author_facet |
Kadankova, T. |
| author_sort |
Kadankova, T. |
| title |
Duration of stay inside an interval by the poisson process with a negative exponential component |
| title_short |
Duration of stay inside an interval by the poisson process with a negative exponential component |
| title_full |
Duration of stay inside an interval by the poisson process with a negative exponential component |
| title_fullStr |
Duration of stay inside an interval by the poisson process with a negative exponential component |
| title_full_unstemmed |
Duration of stay inside an interval by the poisson process with a negative exponential component |
| title_sort |
duration of stay inside an interval by the poisson process with a negative exponential component |
| publisher |
Інститут математики НАН України |
| publishDate |
2006 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/4441 |
| citation_txt |
Duration of stay inside an interval by the poisson process with a negative exponential component / T. Kadankova // Theory of Stochastic Processes. — 2006. — Т. 12 (28), № 1-2. — С. 55–67. — Бібліогр.: 11 назв.— англ. |
| work_keys_str_mv |
AT kadankovat durationofstayinsideanintervalbythepoissonprocesswithanegativeexponentialcomponent |
| first_indexed |
2025-07-02T07:41:08Z |
| last_indexed |
2025-07-02T07:41:08Z |
| _version_ |
1836520118981492736 |
| fulltext |
Theory of Stochastic Processes
Vol. 12 (28), no. 1–2, 2006, pp. 55–67
UDC 519.21
TETYANA KADANKOVA
DURATION OF STAY INSIDE AN INTERVAL BY THE POISSON
PROCESS WITH A NEGATIVE EXPONENTIAL COMPONENT
Several two-boundary problems for the Poisson process with an exponential compo-
nent are solved in the present article. The integral transforms of the joint distribution
of the epoch of the first exit from the interval and the value of the overshoot through
boundaries at the epoch of the exit are obtained. The Laplace transform of the total
duration time of the process’s stay inside the interval is found.
1. Main definitions and auxiliary results
Let ξ(t) ∈ R, t ≥ 0 be a homogeneous process with independent increments [1] with
the cumulant
(1) k(p) =
1
2
p2σ2 − αp +
∫ ∞
−∞
(
e−px − 1 +
px
1 + x2
)
Π(dx), �p = 0.
Here and in the sequel, the paths of the process are supposed to be right-continuous
and ξ(0) = 0. Note that the introduced process is a space homogeneous and a strong
Markov process [2]. This property will be used continually while setting up equations.
The method we propose to solve two-boundary problems is based on the application of
integral transforms of the one-boundary functionals of the process which were determined
by B.A. Rogozin, E.A. Peĉerskii, A.A. Borovkov, V.M. Zolotarev and others. Let us
define the one-boundary functionals we require. For x ≥ 0, we introduce the random
variables
τ x = inf{t : ξ(t) > x}, T x = ξ(τx) − x, τx = inf{t : ξ(t) < −x }, Tx = −ξ(τx) − x
the first passage time of the level x and the value of the overshoot through this level at
the epoch of passage, the first passage time of the level −x and the value of the overshoot
this level at the epoch of the passage. The integral transforms of the joint distribution
of { τ x, T x }, { τx, T x } for s > 0, � p ≥ 0 satisfy the following equalities
E [ e−sτx
exp{−pT x} ] =
(
E e−pξ+(νs)
)−1
E [ e−p(ξ+(νs)−x); ξ+(νs) > x ],
(2) E [ e−sτx exp{−pTx} ] =
(
E e pξ−(νs)
)−1
E [ ep(ξ−(νs)+y); −ξ−(νs) > x ],
2000 AMS Mathematics Subject Classification. Primary 60G40, 60K20.
Key words and phrases. Poisson process with an exponential component, exit problem, summary
time, scale function.
This research has been partially supported by the Belgian Federal Science Policy Office, Interuniver-
sity Attraction Pole Programme P5/24.
55
56 TETYANA KADANKOVA
where ξ+(t) = supu≤t ξ(u), ξ−(t) = infu≤t ξ(u), νs is the exponential variable with
parameter s > 0 independent of the process, P[ νs > t ] = exp{−st}, and
E exp{−p ξ±(νs) } = exp
{∫ ∞
0
1
t
e−st E [ e−pξ(t) − 1; ± ξ(t) > 0 ] dt
}
, ±� p ≥ 0.
Formulae (2) were obtained by E.A. Peĉerskii and B.A. Rogozin in [3]. A simple proof of
these equalities is given in [4]. In this framework we solve several two-boundary problems
for the Poisson process with a negative exponential component. Let us give the rigorous
definition of the process.
Let η ∈ (0,∞) be a positive random variable, and γ be an exponential variable with
parameter λ > 0 : P [ γ > x ] = e−λx, x ≥ 0. Introduce a random variable ξ ∈ R by its
distribution function
F (x) = a exλ I{x≤0} + (a + (1 − a)P [ η < x ]) I{x>0}, a ∈ (0, 1), λ > 0.
Define a right-continuous step Poisson process ξ(t) ∈ R, t ≥ 0 with cumulant
(3) k(p) = c
∫ ∞
−∞
(e−xp − 1) dF (x) = a1
p
λ − p
+ a2(E e−pη − 1), c > 0, � p = 0,
where a1 = ac, a2 = (1−a)c. Note that the jumps of the process ξ(t), t ≥ 0, occur at the
time epochs that are exponentially distributed with parameter c. With the probability
1−a, there occur positive jumps with value distributed as η, and, with the probability a,
there occur negative jumps (whose value γ is exponentially distributed with parameter
λ). Here and in the sequel, we will call such a process as the Poisson process with a
negative exponential component.
The first term of (3) is the simplest case of a rational function, while the second is
nothing but the cumulant of a monotone Poisson process with positive jumps of value η.
It is well known fact (see, for instance, [5],[6]) that, in this case, the equation k(p)−s = 0,
s > 0, has a unique root c(s) ∈ (0, λ) in the semi-plane � p > 0. Moreover, for the integral
transforms of the random variables ξ+(νs), ξ−(νs), the following formulae hold:
(4) E e−pξ−(νs) =
c(s)
λ
λ − p
c(s) − p
, � p ≤ 0,
E e−pξ+(νs) =
sλ
c(s)
(p − c(s))R(p, s), � p ≥ 0,
where
(5) R(p, s) =
(
a1p + (p − λ)[s − a2(E e−pη − 1)]
)−1
, � p ≥ 0, p �= c(s)
It follows from (2) and (4) after some calculations that the integral transforms of the
joint distributions {τx, Tx}, {τx, T x} of the Poisson process with a negative exponential
component satisfy the equalities
(6) E[ e−sτx ; Tx ∈ du ] = (λ − c(s)) e−xc(s) e−λu du = E e−sτx P [ γ ∈ du ],∫ ∞
0
e−px E e−sτx−zξ(τx) dx =
1
p
(
1 − p + z − c(s)
z − c(s)
R(p + z, s)
R(z, s)
)
, � p > 0, � z ≥ 0.
The first equality of (6) yields that τx and Tx are independent. Moreover, for all x ≥ 0,
the value of the overshoot through the lower level Tx is exponentially distributed with
parameter λ. This fact serves as a characterizing feature of the Poisson process with a
negative exponential component.
DURATION OF STAY INSIDE AN INTERVAL 57
Observe that the function R(p, s) defined by (5) is analytic in the semi-plane � p >
c(s), and limp→∞ R(p, s) = 0. Therefore, it allows the representation in the form of an
absolutely convergent Laplace integral [7]
(7) R(p, s) =
∫ ∞
0
e−pxRx(s) dx, � p > c(s).
We call the function Rx(s), x ≥ 0, as the resolvent of the Poisson process with a negative
exponential component. We assume that Rx(s) = 0, for x < 0. Notice that R0(s) =
limp→∞ p R(p, s) = (c + s)−1, and equality (4) implies
P [ ξ−(νs) = 0 ] =
c(s)
λ
, P [ ξ+(νs) = 0 ] =
λ
c(s)
s
s + c
.
The second formula of (4) yields
(8) R(p, s) =
c(s)
sλ
1
p − c(s)
E e−pξ+(νs), � p > c(s).
The functions
1
p − c(s)
=
∫ ∞
0
e−u(p−c(s))du, � p > c(s),
E e−pξ+(νs) =
∫ ∞
0
e−updP [ ξ+(νs) < u ], � p ≥ 0,
which enter the right-side of (8) are the Laplace transforms for � p > c(s). Therefore,
the original functions of the left- and right-hand sides of (8) coincide, and
(9) Rx(s) =
c(s)
sλ
∫ x
−0
ec(s)(x−u)dP [ ξ+(νs) < u ], x ≥ 0.
The latter formula is another representation for the resolvent of the Poisson process with
a negative exponential component. Note that the representation for the resolvent similar
to (9) for a semi-continuous process with independent increments was obtained by V.M.
Shurenkov and V.N. Suprun in [8,9]. Representation (9) implies that the function Rx(s),
x ≥ 0 is positive, monotone, continuous, and increasing, and Rx(s) < A(s) exp{xc(s)},
A(s) < ∞. Consequently,∫ ∞
0
Rx(s)e−αxdx < ∞, α > c(s).
Moreover, in the neighbourhood of any x ≥ 0, the function Rx(s) has bounded variation.
Hence, the inversion formula [7] is valid:
(10) Rx(s) =
1
2πi
∫ α+i∞
α−i∞
expR(p, s) dp, α > c(s).
The latter equality together with (7) determines the resolvent of the Poisson process with
a negative exponential component. Now we state the main results.
2 Exit from an interval
In this section, the integral transform of the joint distribution of the instant of the first
exit from an interval by the Poisson process with a negative exponential component and
the value of the overshoot through a boundary are determined. These integral transforms
will be obtained as a corollary from a general theorem for a homogeneous process with
independent increments.
58 TETYANA KADANKOVA
Let ξ(t) ∈ R, t ≥ 0, ξ(0) = 0 be a homogeneous process with independent increments
with cumulant (1), B > 0 is fixed. Introduce the random variable
χ(y) = inf{ t : y + ξ(t) /∈ [0, B] }, y ∈ [0, B]
the instant of the first exit by the process y + ξ(t) from the interval [0, B]. Introduce
events AB = { ξ(χ(y)) > B }— the exit from the interval by the process takes place
through the upper level; A 0 = { ξ(χ(y)) < 0 }— the exit from the interval by the process
takes place through the lower level. Denote, by
X(y) = (ξ(χ(y)) − B) IA B + (−ξ(χ(y))) IA 0 , P [ AB + A 0 ] = 1,
the value of the overshoot through a boundary at the epoch of the exit from the interval
by the given process. Here, IA = IA(ω) is an indicator of the event A. The following
theorem is true.
Theorem 1 [4]. Let ξ(t) ∈ R, t ≥ 0 be a homogeneous process with independent incre-
ments with cumulant (1), B > 0 is fixed, y ∈ [0, B], x = B − y, ξ(0) = 0, and
χ(y) = inf{ t : y + ξ(t) /∈ [0, B] }, X(y) = (ξ(χ(y)) − B) IA B + (−ξ(χ(y))) IA 0
the instant of the first exit by the process y + ξ(t) from the interval [0, B] and the value
of the overshoot through a boundary at the epoch of the first exit from the interval
by the given process. The Laplace transforms of the joint distribution of the variables
{χ(y), X(y) } for s > 0 satisfy the formulae
E [ e−sχ(y); X(y) ∈ du, AB ] = fs
+(x, du) +
∫ ∞
0
fs
+(x, dv)Ks
+(v, du),
(11) E [ e−sχ(y); X(y) ∈ du, A 0 ] = fs
−(y, du) +
∫ ∞
0
fs
−(y, dv)Ks
−(v, du),
where
fs
+(x, du) = E [ e−sτx
; T x ∈ du ] −
∫ ∞
0
E [ e−sτy ; Ty ∈ dv ]E [ e−sτv+B
; T v+B ∈ du ],
fs
−(y, du) = E [ e−sτy ; Ty ∈ du ] −
∫ ∞
0
E [ e−sτx
; T x ∈ dv ]E [ e−sτv+B ; Tv+B ∈ du ];
and Ks±(v, du) =
∑∞
n=1 K
(n)
± (v, du, s), v ≥ 0, is the series of successive iterations;
(12) K
(1)
± (v, du, s) = K±(v, du, s), K
(n+1)
± (v, du, s) =
∫ ∞
0
K
(n)
± (v, dl, s)K±(l, du, s)
are the successive iterations (n ∈ N) of the kernels K±(v, du, s) which are given by the
defining equalities
K+(v, du, s) =
∫ ∞
0
E [ e−sτv+B ; Tv+B ∈ dl ]E [ e−sτ l+B
; T l+B ∈ du ],
(13) K−(v, du, s) =
∫ ∞
0
E [ e−sτv+B
; T v+B ∈ dl ]E [ e−sτl+B ; Tl+B ∈ du ].
We apply now the formulae of the theorem in the case where the underlying process
is the Poisson process with an exponentially distributed negative component.
DURATION OF STAY INSIDE AN INTERVAL 59
Corollary 1. Let ξ(t) ∈ R, t ≥ 0 be a Poisson process with a negative exponential
component with cumulant (3), B > 0, y ∈ [0, B], x = B − y, ξ(0) = 0, and
χ(y) = inf{ t : y + ξ(t) /∈ [0, B] }, X(y) = (ξ(χ(y)) − B) IA B + (−ξ(χ(y))) IA 0
the instant of the first exit from the interval and the value of the overshoot through one
of the boundaries. Then, for s > 0,
1) the integral transforms of the joint distribution {χ(y), X(y) } satisfy the equalities
(14)
E[e−sχ(y); X(y) ∈ du, A0] = (λ − c(s)) e−yc(s)−λu
(
1 − E[ e−sτx−c(s)ξ(τx) ]
)
K(s)−1du,
E[e−sχ(y); X(y) ∈ du, AB] = E[e−sτx
; T x ∈ du]−E [e−sχ(y); A0]E [e−sτγ+B
; T γ+B ∈ du],
where
K(s) = 1 − E exp{−sτB} E exp{−sτγ+B − c(s)T γ+B},
E exp{−sτγ+B − c(s)T γ+B} =
∫ ∞
0
λ e−λuE exp{−sτu+B − c(s)T u+B} du,
and, in particular,
E[ e−sχ(y); A 0] =
(
1 − c(s)
λ
)
e−yc(s)
(
1 − E[ e−sτx−c(s)ξ(τx) ]
)
K(s)−1,
(15) E[e−sχ(y); AB ] = E exp{−sτx} − E[e−sχ(y); A 0] E exp{−sτγ+B};
2) for the Laplace transforms of the random variable χ(y), the following representa-
tions hold:
E[e−sχ(y); X(y) ∈ du, A0] = e−λ(u+B) Rx(s)
R̂B(λ, s)
du, E[e−sχ(y); A0] =
1
λ
e−λB Rx(s)
R̂B(λ, s)
,
E[ e−sχ(y); AB] = 1 − Rx(s)
R̂B(λ, s)
[
1
λ
e−λB + sλ ŜB(λ, s)
]
+ sλSx(s),
(16)
∫ ∞
0
e−stP [χ(y) > t] dt = λ
Rx(s)
R̂B(λ, s)
ŜB(λ, s) − λSx(s),
where Rx(s), x ≥ 0, is the resolvent of the Poisson process with a negative exponential
component defined by (10),
Sx(s) =
∫ x
0
Ru(s) du, R̂B(λ, s) =
∫ ∞
B
e−λuRu(s) du, ŜB(λ, s) =
∫ ∞
B
e−λuSu(s) du.
Proof. We use the results of Theorem 1 for the Poisson process with a negative expo-
nential component which takes a simplified form in this case. Using equality (6) and
defining formulae (13) of the kernels K±(v, du, s) yields
K+(v, du, s) =
(
1 − c(s)
λ
)
e−c(s)(v+B) E [ e−sτγ+B
; T γ+B ∈ du ],
K−(v, du, s) = (λ − c(s)) e−c(s)B−λu E [ e−sτv+B−c(s)T v+B
] du,
60 TETYANA KADANKOVA
where γ is the exponentially distributed random variable with parameter λ. The latter
equalities imply the following form of the successive iterations K
(n)
± (v, du, s), n ∈ N, of
the kernels K±(v, du, s) :
K
(n)
− (v, du, s) =E [ e−sτv+B−c(s)T v+B
]E e−sτB (1 − K(s))n−1
λ e−λu du,
K
(n)
+ (v, du, s) =e−vc(s) E e−sτB (1 − K(s))n−1 E[ e−sτγ+B
; T γ+B ∈ du ].
The series Ks±(v, du) of successive iterations K
(n)
± (v, du, s) are, in this case, geometric
series, and their sums are given by
Ks
−(v, du) =
∞∑
n=1
K
(n)
− (v, du, s) = E[ e−sτv+B−c(s)T v+B
] E e−sτB K(s)−1λ e−λu du,
Ks
+(v, du) =
∞∑
n=1
K
(n)
+ (v, du, s) = e−vc(s) E e−sτB K(s)−1 E[ e−sτγ+B
; T γ+B ∈ du ].
Substituting the obtained expressions for Ks
±(v, du) into equalities (11) of Theorem 1,
we get formulae (14) of Corollary 1. Integrating (14) with respect to u ∈ R+ yields
equalities (15) of Corollary 1. Further, utilizing representation (10) for the resolvent
and formulae (6), we obtain the resolvent representations for the functions E exp{−sτx},
E exp{−sτ x − c(s)ξ(τx)} :
E exp{−sτx} = 1 − sλ
c(s)
Rx(s) + sλSx(s),
E exp{−sτ x − c(s)ξ(τx)} = 1 − e−xc(s) Rx(s) r(c(s), s),
where
Sx(s) =
∫ x
0
Ru(s) du, r(c(s), s) =
d
dp
R(p, s)−1
∣∣∣∣
p=c(s)
.
Substituting the resolvent representations into (15) yields formulae (16) of Corollary 1.
Note that, for a random walk with a geometrically distributed negative component, the
resolvent representations similar to (16) were obtained in [10].
Remark 1. The formulae of Corollary 1 can be also obtained after solving the following
system of integral equations:
E [ e−sτx
; T x ∈ du ] =E [ e−sχ(y); X(y) ∈ du, AB ]
+
∫ ∞
0
E [ e−sχ(y); X(y) ∈ dv, A 0 ] E [ e−sτv+B
; T v+B ∈ du ],
E [ e−sτy ; Ty ∈ du ] =E [ e−sχ(y); X(y) ∈ du, A 0 ]
+
∫ ∞
0
E [ e−sχ(y); X(y) ∈ dv, AB ] E [ e−sτv+B ; Tv+B ∈ du ].(17)
It takes a simple form for the Poisson process with a negative exponential component.
This system was solved in [4] for a process with the general cumulant (1).
Remark 2. We have obtained the integral transform of the joint distribution {χ(y), X(y)}
for the process with independent increments with the general cumulant (1) and, in partic-
ular, for a Poisson process with exponential component. The value of the overshoot X(y)
has a probabilistic interpretation, since it is expressed in terms of one-boundary func-
tionals T x, Tx. This fact allows us to effectively apply the distribution of {χ(y), X(y)}
to solve other exit problems.
DURATION OF STAY INSIDE AN INTERVAL 61
3 Duration of a stay inside the interval by the Poisson
process with a negative exponential component
In this section, we obtain the Laplace transform of the total duration of a stay of the
process inside the interval. Let ξ(t) ∈ R, t ≥ 0, ξ(0) = 0, be the Poisson process with a
negative exponential component. Denote, by
σy(t) =
∫ t
0
I { y + ξ(u) ∈ [0, B] } du, y ∈ R,
the total duration of a stay of the process y + ξ(·) inside [0, B] up to the moment t. We
will determine
C s
a (y) = E exp{−aσy(νs) }, y ∈ R, a ≥ 0,
the Laplace transform of the total duration of a stay of the process y + ξ(·) inside
the interval [0, B] on the exponential time interval [0, νs]. We require several auxiliary
functions to solve the stated problem. Let y ≥ 0, ξ(0) = 0, and let
τy = inf { t : y + ξ(t) < 0 } and σy = σy(τy)
be, respectively, the instant of the first exit from the upper semi-plane by the process
y + ξ(·) and the total duration of a stay of the process inside [0, B] on the time interval
[0, τy]. On the event { τy = ∞}, we set, by definition, σy = ∞. The following statement
is true.
Lemma 1. Let ξ(t) ∈ R, t ≥ 0, ξ(0) = 0, be the Poisson process with a negative
exponential component with cumulant (3). Then, for the integral transform
Ds
a(y) = E exp{−sτy − aσy }, y ≥ 0, a ≥ 0,
of the joint distribution { τy, σy }, the following equality holds for s > 0:
(18) Ds
a(y) =
[
V s
a (B − y) − a RB−y(s + a) e−c(s)(B−y)
]
V s
a (B)−1E e−sτy , y ≥ 0.
Here, Rx(s) = 0 for x < 0, and the continuous function V s
a (x), x ∈ R, is given by the
equality
(19) V s
a (x) = 1 + a(λ − c(s))
∫ x
0
e−uc(s)Ru(s + a) du, x ≥ 0, V s
a (x) = 1, x < 0.
Proof. For the functions Ds
a(y), y ≥ 0, according to the total probability law and the
strong Markov property of the underlying process, we can write the system
Ds
a(y) = E [e−(s+a)χ(y); A0 ] +
∫ ∞
0
E [e−(s+a)χ(y); X(y) ∈ dv, AB ] E [e−sτv ; Tv > B ]
+
∫ ∞
0
E [e−(s+a)χ(y); X(y) ∈ dv, AB ]
∫ B
0
E [e−sτv ; Tv ∈ du ] Ds
a(B − u), y ∈ [0, B],
Ds
a(y) = E [e−sτy−B ; Ty−B > B ] +
∫ B
0
E [e−sτy−B ; Ty−B ∈ du ] Ds
a(B − u), y > B.
The first equation of this system represents the fact that the total duration of a stay of
the process y+ξ(·), y ∈ [0, B], inside [0, B] on the time interval [0, τy] is realized either on
the path which does not intersect the upper level (the first term on the right-hand side
of the equation), or on the path which does intersect the upper level and then overleaps
the interval [0, B] (the second term on the right-hand side of the equation), or on the
path which intersects the upper level and then returns to the interval [0, B] (the third
term on the right-hand side of the equation). The second equation is written analogously.
62 TETYANA KADANKOVA
Now, using the first equality of (6) and equalities (16) of Corollary 1, we get, after some
simplifications,
D s
a (y) =
1
λ
e−λB RB−y(s + a)
R̂B(λ, s + a)
+ (λ − c(s))Es+a
y (c(s))
(
1
λ
e−λB + D̃ s
a (λ)
)
, y ∈ [0, B],
D s
a (y) =
(
1 − c(s)
λ
)
e−c(s)(y−B)e−λB + (λ − c(s)) e−c(s)(y−B)D̃ s
a (λ), y > B,
(20)
where
D̃ s
a (λ) =
∫ B
0
e−λuDs
a(B − u) du,
Es+a
y (c(s)) = E [ e−(s+a)χ(y)−c(s)X(y); AB ].
Note that it is now sufficient to determine the function D̃ s
a (λ). Then, by means of (20),
we will derive the function D s
a (y), y ≥ 0. The function D̃ s
a (λ) can be obtained from the
first equation of system (20). To do this, we first make auxiliary calculations. Denote
T s
x (z) = E exp{−sτx − zT x }, x ≥ 0, � z ≥ 0.
Utilizing the second equality of (6) and the defining resolvent (10) yields
T s+a
x (c(s)) = exc(s) + a
λ − c(s)
c(s) − c(s + a)
Rx(s + a)
+ a(λ − c(s))
∫ x
0
e−c(s)(u−x)Ru(s + a) du, x ≥ 0.
Using the latter equality and relations (14) and (16) of Corollary 1, we find
Es+a
y (c(s)) = exc(s)V s
a (B − y)
− e−B(λ−c(s))
λ − c(s)
RB−y(s + a)
R̂B(λ, s + a)
V s
a (B) − aRB−y(s + a), y ∈ [0, B].(21)
Multiplying this equality by e−λ(B−y) and integrating with respect to y ∈ [0, B], we get
(22) (λ− c(s))
∫ B
0
e−λ(B−y)Es+a
y (c(s)) dy = 1−V s
a (B)
[
1 +
ŘB(λ, s + a)
R̂B(λ, s + a)
]
e−B(λ−c(s)),
where ŘB(λ, s + a) =
∫ B
0
e−λuRu(s + a) du. Further, multiplying the first equation of
(20) by e−λ(B−y) and integrating it with respect to y ∈ [0, B], we get
D̃ s
a (λ) =
1
λ
e−λB ŘB(λ, s + a)
R̂B(λ, s + a)
+
+
(
1
λ
e−λB + D̃ s
a (λ)
) (
1 − V s
a (B)
[
1 +
ŘB(λ, s + a)
R̂B(λ, s + a)
]
e−B(λ−c(s))
)
which is a linear equation for the function D̃ s
a (λ). It yields
D̃ s
a (λ) +
1
λ
e−λB =
1
λ
e−c(s)B V s
a (B)−1.
Substituting this expression for the function D̃ s
a (λ) into equalities (20) and taking into
account the fact that Rx(s) = 0 and V s
a (x) = 1 for x < 0, we get the equality of the
lemma.
DURATION OF STAY INSIDE AN INTERVAL 63
Now, for y ≥ 0, we determine the function
Q s
a (y) = E [ e−aσy(νs); τy > νs ], y ≥ 0, a ≥ 0,
being the Laplace transform of the total duration of a stay of the process y + ξ(·) inside
[0, B] on the time interval [0, νs] on the event {τy > νs}. The following statement is true.
Lemma 2. Let ξ(t) ∈ R, t ≥ 0, ξ(0) = 0 be a Poisson process with a negative exponential
component with cumulant (3). Then the function Q s
a (y), y ≥ 0, satisfies, for s > 0, the
equality
(23) Q s
a (y) = v s
a (B − y) − aRB−y(s + a) − v s
a (B)D s
a (y), y ≥ 0,
where Rx(s) = 0 for x < 0, the function D s
a (y), y ≥ 0, is given by (18), and the
continuous function v s
a (x), x ∈ R, is given by the formula
(24) v s
a (x) = 1 + aλ
∫ x
0
Ru(s + a) du, x ≥ 0, v s
a (x) = 1, x < 0.
Proof. In accordance with the total probability law and due to the fact that χ(y), τy are
the Markov times, the following system of equations for the function Q s
a (y), y ≥ 0, is
valid:
Qs
a(y) =
s
s + a
(
1 − E e−(s+a)χ(y)
)
+
∫ ∞
0
E [e−(s+a)χ(y); X(y) ∈ dv, AB ] (1 − E e−sτv)
+
∫ ∞
0
E [e−(s+a)χ(y); X(y) ∈ dv, AB ]
∫ B
0
E [e−sτv ; Tv ∈ du] Qs
a(B − u), y ∈ [0, B],
Qs
a(y) = 1 − E e−sτy−B +
∫ B
0
E [e−sτy−B ; Ty−B ∈ du] Qs
a(B − u), y > B.
Let us interpret the obtained equations. For the first equation, we have that the duration
of a stay inside [0, B] on the event {τy > νs} can be realized on the following disjunctive
events: 1) sample paths of the process do not leave the interval [0, B] (the first term), 2)
the paths intersect the upper level B and do not return to the interval (the second term),
3) the paths leave the interval [0, B] through the upper level B and then return to the
interval [0, B] (the third term of equation). The second equation is set up analogously.
After taking into account the first equality of (6) and equalities (16), it follows from the
latter system that
Q s
a (y) = 1 − 1
λ
e−λB RB−y(s + a)
R̂B(λ, s + a)
− aλ
RB−y(s + a)
R̂B(λ, s + a)
ŜB(λ, s + a)
+ aλSB−y(s + a) + (λ − c(s))Es+a
y (c(s))
(
Q̃ s
a (λ) − 1
λ
)
, y ∈ [0, B],
Q s
a (y) = 1 −
(
1 − c(s)
λ
)
e−c(s)(y−B) + (λ − c(s)) e−c(s)(y−B)Q̃ s
a (λ), y > B,
(25)
where the function Es+a
y (c(s)) = E [ e−(s+a)χ(y)−c(s)X(y); AB ] is determined by (21),
and
Q̃ s
a (λ) =
∫ B
0
e−λuQs
a(B − u) du,
ŜB(λ, s + a) =
∫ ∞
B
e−λuSu(s + a) du.
64 TETYANA KADANKOVA
The function Q̃ s
a (λ) can be determined straightforwardly from the first equation of (25),
since we have already given the auxiliary calculations in the proof of the previous lemma.
Multiplying the first equation of (25) by e−λ(B−y) and integrating it with respect to
y ∈ [0, B], we get
Q̃ s
a (λ) − 1
λ
= − 1
λ
e−λB
[
1 +
RB−y(s + a)
R̂B(λ, s + a)
]
v s
a (B) + (λ − c(s)) Ẽs+a(c(s))
(
Q̃ s
a (λ) − 1
λ
)
(26)
which is a linear equation for the function Q̃ s
a (λ). The function
Ẽs+a(c(s)) =
∫ B
0
exp{−λ(B − y)}Es+a
y (c(s)) dy
is determined by (22), and the function v s
a (x), x ∈ R, is defined by equality (24) of
Lemma 2. To obtain this equation, we have also used the obvious identities
λŜB(λ, s + a) = R̂B(λ, s + a) + SB(s + a) e−λB,
λ
∫ B
0
e−λuSu(s + a) du =
∫ B
0
e−λuRu(s + a) du − SB(s + a) e−λB.
Utilizing equality (22), we find
Q̃ s
a (λ) − 1
λ
= − v s
a (B)
V s
a (B)
1
λ
e−c(s)B
from (26). Substituting the latter expression for the function Q̃ s
a (λ) into (25), and
taking into account the fact that v s
a (x) = V s
a (x) = 1 for x < 0, we obtain the equality
of Lemma 2.
Thus, the auxiliary functions D s
a (y), Q s
a (y), y ≥ 0, are now determined by equalities
(18) and (23). Hence, we can now obtain the integral transforms of the distribution of
the duration of a stay inside the interval by the process.
Theorem 2. Let ξ(t) ∈ R, t ≥ 0, ξ(0) = 0 be the Poisson process with a negative
exponential component with cumulant (3), B > 0, a ≥ 0, and let
σy(t) =
∫ t
0
I { y + ξ(u) ∈ [0, B] } du,
C s
a (y) = s
∫ ∞
0
e−stE exp{−aσy(t) } dt, y ∈ R,
be the total duration of a stay of the process y + ξ(·) inside the interval [0, B] on the
time interval [0, t] and the integral transform of the distribution of σy(t). Then, for the
function C s
a (y), y ∈ R, the following equalities hold for s > 0:
C s
a (y) = v s
a (B − y) − a RB−y(s + a) + D s
a (y)C∗(B), y ≥ 0,
C s
a (−y) = 1 − E e−sτy
+
∫ ∞
0
E [ e−sτy
; T y ∈ du ] C s
a (u), y > 0,(27)
where
C∗(B) =
aλ
c(s)
(
v s
a (B) − V s
a (B)ec(s)B
)
V s
a (B)
r(c(s), s) + a(λ − c(s))
∫ B
0
(
V s
a (x) − a e−xc(s)Rx(s + a)
)
dx
,
DURATION OF STAY INSIDE AN INTERVAL 65
v s
a (x) = 1 + aλ
∫ x
0
Ru(s + a) du, x ≥ 0, v s
a (x) = 1, x < 0,
V s
a (x) = 1 + a(λ − c(s))
∫ x
0
e−uc(s)Ru(s + a) du, x ≥ 0, V s
a (x) = 1, x < 0.
Proof. Since the negative jumps of the underlying process are exponentially distributed,
the random variables τx, Tx are independent (see the first formula of (6)), and, for all
x ≥ 0, the value of the overshoot through the lower level Tx is exponentially distributed
with parameter λ. Hence,
E [ exp{−sτy − aσy}; Tx ∈ du ] = D s
a (y)λ e−λu du.
Then, according to the total probability law and the strong Markov property of the
process for the function C s
a (y), y ≥ 0, we can write the system of equations
C s
a (y) = Q s
a (y) + D s
a (y) C̃ s
a (λ), y ≥ 0,
C̃ s
a (λ) = 1 − m s
γ +
∫ ∞
0
m s
γ (dy)C s
a (y),(28)
where
C̃ s
a (λ) =
∫ ∞
0
λ e−λxC s
a (−x) dx
is the unknown function which will be determined later on, and
m s
γ =
∫ ∞
0
λ e−λxE e−sτx
dx, m s
γ (dy) =
∫ ∞
0
λ e−λxE [ e−sτx
; T x ∈ dy ] dx.
Substituting the right-hand side of the first equation into the second equation of the
system implies
C̃ s
a (λ) = 1 − m s
γ +
∫ ∞
0
m s
γ (dy)Q s
a (y) + C̃ s
a (λ)
∫ ∞
0
m s
γ (dy)D s
a (y)
which is a linear equation for the function C̃ s
a (λ). Using expressions (18) and (23) for
the functions D s
a (y) and Q s
a (y), we find from the latter equation that
C̃ s
a (λ) = v s
a (B) +
aλ
1 − ∫ ∞
0
m s
γ (dy)D s
a (y)
×
×
(∫ B
0
m s
γ (dy)
∫ B−y
0
Ru(s + a) du − 1
λ
∫ B
0
m s
γ (dy)RB−y(s + a) − SB(s + a)
)
,
(29)
where SB(s + a) =
∫ B
0
Ru(s + a) du. Substituting the latter expression for C̃ s
a (λ) into
the first equation of system (28) yields
C s
a (y) = v s
a (B − y) − a RB−y(s + a) + D s
a (y)C∗(B), y ≥ 0
66 TETYANA KADANKOVA
which is the first equality of Theorem 2. Now we need to determine the constant C∗(B) =
C̃ s
a (λ) − v s
a (B). To do this, we require the following formulae∫ B
0
m s
γ (dy)RB−y(s + a) = λR̂B(λ, s + a)eλB − λR(λ, s)V s
a (B)ec(s)B ,
∫ B
0
m s
γ (dy)
∫ B−y
0
Ru(s + a) du = SB(s + a)+
R̂B(λ, s + a) eλB +
R(λ, s)
c(s)
(
(λ − c(s))v s
a (B) − λV s
a (B)ec(s)B
)
,
∫ B
0
m s
γ (dy)e−yc(s)
∫ B−y
0
e−uc(s)Ru(s + a) du =
λR(λ, s)
λ − c(s)
(
eB(λ−c(s)) − 1
)
+
λ
λ − c(s)
(∫ B
0
e−uc(s)Ru(s + a) du − ŘB(λ, s + a) e(λ−c(s))B
)
− λR(λ, s)
∫ B
0
V s
a (x) dx,
(30)
where
ŘB(λ, s + a) =
∫ B
0
e−λuRu(s + a) du, R̂B(λ, s + a) =
∫ ∞
B
e−λuRu(s + a) du.
The integrals which enter the left-hand sides of the latter formulae (30) are the con-
volutions of known functions and are easy to be calculated by utilizing the following
equalities:∫ ∞
0
e−yzm s
γ (dy) =
λ
λ − z
(
1 − λ − c(s)
z − c(s)
R(λ, s)
R(z, s)
)
, R(z, s)−1 = R(z, s+a)−1+a(λ−z),
The first equality follows from (6) when p = λ− z, and the second equality follows from
the definition of the function R(z, s) (5). Substituting formulae (30) into (29), we obtain
C∗(B) =
aλ
c(s)
(
v s
a (B) − V s
a (B)ec(s)B
)
V s
a (B)
r(c(s), s) + a(λ − c(s))
∫ B
0
(
V s
a (x) − a e−xc(s)Rx(s + a)
)
dx
,
where r(c(s), s) = d
dpR(p, s)−1
∣∣∣
p=c(s)
. The second equality of the theorem follows from
the total probability law and the fact that τy is a Markov time.
Remark 3. For a semi-continuous process with independent increments the integral trans-
forms of the distribution of the duration of a stay inside the interval have been obtained
in [11]. For a symmetric Wiener process, the distribution of the duration of stay inside
the interval was obtained in [11]. This distribution is the limit distribution of the distri-
bution of the duration of a stay of homogeneous processes and a random walk (after the
re-scaling of time and space). Such a limit theorem has been proved in the same paper
for a semi-continuous process with independent increments.
Bibliography
1. A.V. Skorokhod, Random Processes with Independent Increments, Nauka, Moscow, 1964.
2. I.I. Gikhman and A.V. Skorokhod, The Theory of Stochastic Processes, vol. 2, Nauka, Moscow,
1973.
3. E.A. Peĉerskii and B.A. Rogozin, Transformations of joint distributions of random variables
connected with fluctuations of a process with independent increments, Theor. Prob. and its
Appl. 14 (1969), no. 3, 431–444.
4. V.F. Kadankov and T.V. Kadankova, On the distribution of the first exit time from an interval
and the value of overshoot through the boundaries for processes with independent increments
and random walks, Ukr. Math. J. 57 (2005), no. 10, 1359–1384.
DURATION OF STAY INSIDE AN INTERVAL 67
5. A.A. Borovkov, Stochastic Processes in Queueing Theory, Nauka, Moscow, 1972.
6. N.S. Bratiichuk and D.V. Gusak, Boundary Problems for Processes with Independent Incre-
ments, Naukova Dumka, Kiev, 1990.
7. V.A. Ditkin and A.P. Prudnikov, Operational Calculus, Vysshaya Shkola, Moscow, 1966.
8. V.M. Shurenkov and V.N. Suprun, On the resolvent of a process with independent increments
terminating at the moment when it hits the negative real semiaxis., Studies in the Theory
of Stochastic process, Institute of Mathematics of the Academy of Sciences of UkrSSR, Kiev
(1975), 170–174.
9. V.N. Suprun, Ruin problem and the resolvent of a terminating process with independent incre-
ments., Ukr. Math. J. 27 (2005), no. 1, 53–61.
10. T.V. Kadankova, Two-boundary problems for random walk with negative jumps which have
geometrical distribution, Theor. Prob. and Math. Statist. 68 (2003), 60–71.
11. V.F. Kadankov and T.V. Kadankova, On the distribution of duration of stay in an interval of
the semi-continuous process with independent increments., Random Oper. and Stoch. Equ. 12
(2004), no. 4, 365–388.
E-mail : tetyana.kadankova@uhasselt.be
|