Exit from an interval by a difference of two renewal processes
Integral transforms of the joint distribution of the first exit time from an interval,
 the value of the overshoot through a boundary, and the value of a linear component
 at the epoch of the exit are determined for the difference of two renewal processes.
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Інститут математики НАН України
2005
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| Цитувати: | Exit from an interval by a difference of two renewal processes / V. Kadankov // Theory of Stochastic Processes. — 2005. — Т. 11 (27), № 3-4. — С. 92–96. — Бібліогр.: 10 назв.— англ. |
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Digital Library of Periodicals of National Academy of Sciences of Ukraine| _version_ | 1860146229461123072 |
|---|---|
| author | Kadankov, V. |
| author_facet | Kadankov, V. |
| citation_txt | Exit from an interval by a difference of two renewal processes / V. Kadankov // Theory of Stochastic Processes. — 2005. — Т. 11 (27), № 3-4. — С. 92–96. — Бібліогр.: 10 назв.— англ. |
| collection | DSpace DC |
| description | Integral transforms of the joint distribution of the first exit time from an interval,
the value of the overshoot through a boundary, and the value of a linear component
at the epoch of the exit are determined for the difference of two renewal processes.
|
| first_indexed | 2025-12-07T17:49:58Z |
| format | Article |
| fulltext |
Theory of Stochastic Processes
Vol. 11 (27), no. 3–4, 2005, pp. 92–96
UDC 519.21
VICTOR KADANKOV
EXIT FROM AN INTERVAL BY A
DIFFERENCE OF TWO RENEWAL PROCESSES
Integral transforms of the joint distribution of the first exit time from an interval,
the value of the overshoot through a boundary, and the value of a linear component
at the epoch of the exit are determined for the difference of two renewal processes.
1. Main definitions
Let η, ξ, κ, δ ∈ (0,∞) be positive random variables. By F (x) = P [ η ≤ x ], G(y) =
P [ ξ ≤ y ] we denote the distribution functions of the random variables η, ξ. Introduce
the sequences {η, ηn}, {ξ, ξn}, n ∈ N, of independent identically distributed variables
and, for x, y ≥ 0, define the random sequences
η0(x) = 0, η1(x) = ηx, ηn+1(x) = ηx + η1 + · · · + ηn, n ∈ N,
ξ0(y) = 0, ξ1(y) = ξy , ξn+1(y) = ξy + ξ1 + · · · + ξn, n ∈ N,(1)
where ηx, ξy are the random variables given by their distribution functions
P [ ηx ≤ u ] =
F (x + u) − F (x)
1 − F (x)
, P [ ξy ≤ v ] =
G(y + v) − G(y)
1 − G(y)
, u, v ≥ 0.
For all x, y ≥ 0, we define the renewal processes generated by sequences (1):
αy(t) = max {n ∈ N ∪ 0 : ξn(y) ≤ t }, βx(t) = max {n ∈ N ∪ 0 : ηn(x) ≤ t }.
Introduce the monotone independent random walks generated by κ, δ:
κ0 = 0, κn = κ
′
1 + · · · + κ
′
n, δ0 = 0, δn = δ′1 + · · · + δ′n, n ∈ N.
Here, {κ, κ
′
n}, {δ, δ′n}, n ∈ N, are the sequences of independent identically distributed
random variables. Define the right-continuous step process
(2) {Dxy(t)}{t≥0} = καy(t) − δβx(t) ∈ R, x, y ≥ 0; Dxy(0) = 0.
This process takes positive jumps at the epochs ξn(y), n ∈ N, of the value κ
′
n. At the
epochs ηn(x) n ∈ N, there occur negative jumps of the value δ′n. Let {Dxy(t) }{t≥0} be
the difference of two renewal processes. Note that this process is not Markovian (apart
from the case where η, ξ are exponentially distributed). For all t ≥ 0, we introduce the
right-continuous linear components
(3) η +
x (t) =
{
t + x, 0 ≤ t < ηx,
t − ηβx(t)(x), t ≥ ηx
∈ R+, x ≥ 0,
(4) ξ +
y (t) =
{
t + y, 0 ≤ t < ξy,
t − ξαy(t)(y), t ≥ ξy
∈ R+, y ≥ 0.
2000 AMS Mathematics Subject Classification. Primary 60G40, 60K20.
Key words and phrases. Exit problem, difference of two renewal processes, successive iterations.
92
EXIT FROM AN INTERVAL BY A DIFFERENCE 93
The process {η +
x (t)}{t≥0} increases linearly on the intervals [ηn(x), ηn+1(x)), n ∈ N ∪ 0,
and vanishes to zero at the epochs ηn(x), n ∈ N. The value of the process at the time
t0 ≥ ηx is the time elapsed since the last negative jump of the process {Dxy(t)}{t≥0} till
t0. The process {ξ +
x (t)}{t≥0} increases linearly on the interval [ξn(x), ξn+1(x)), n ∈ N∪0,
and vanishes to zero at the epochs ξn(x), n ∈ N. The value of the process at the time
t0 ≥ ξx is the time elapsed since the last positive jump of the process {Dxy(t)}{t≥0} till
t0.
Define a right-continuous Markov process
(5) Y t
xy =
{
Dxy(t), η +
x (t), ξ +
y (t)
}
{t≥0} ∈ R × R
2
+, Y 0
xy = {0, x, y}, x, y ≥ 0,
accompanying (governing) the process {Dxy(t)}{t≥0}. The so defined process is a Markov
process which is homogeneous in the first component [1]. If, at the instant t0, Y t0
xy =
{k, u, v}, then the future behaviour of the process {Y t
xy}{t≥t0} does not depend on the
value k of the first component, and the first negative jump of the process {Dxy(t)}{t≥t0}
of the value δ occurs after the random time ηu has elapsed, and the process {Dxy(t)}{t≥t0}
will take the first positive jump of the value κ after the time ξv has elapsed.
For k ∈ R+, we define
τk
xy = inf { t : Dxy(t) > k }, T k
xy = Dxy(τk
xy) − k, ηk
xy = η +
x (τk
xy)
which are the first passage time of the upper level k by the process {Dxy(t)}{t≥0}, the
value of the overshoot through the upper level k, and the value of the linear component
η +
x (·) at the passage instant. Note that the overleap of the upper level can take place
only at the epochs ξn(y), n ∈ N, and, in view of (4), ξ +
y (τk
xy) = 0.
Similarly, for k ∈ R+, we introduce the random variables
τxy
k = inf { t : Dxy(t) < −k }, T xy
k = −Dxy(τ
xy
k ) − k, ξxy
k = ξ +
y (τxy
k )
which are the first passage time of the lower level −k by the process {Dxy(t)}{t≥0}, the
value of the overshoot, and the value of the linear component ξ +
y (·) at the passage instant.
Observe that the overleap of the lower level occurs at the epochs ηn(x), n ∈ N, and, in
view of (3), η +
x (τxy
k ) = 0. Note, that the random variables τk
xy, τxy
k are the Markov times
of the process {Y t
xy}{t≥0}, because they take values from countable sets { ξn(y), n ∈ N},
{ ηn(x), n ∈ N}. The integral transforms of the joint distributions
(6)
{
τk
xy, T k
xy, ηk
xy
}
, {τxy
k , T xy
k , ξxy
k } , k, x, y ∈ R+,
have been obtained in [2] in terms of the solutions of integral equations. For particular
types of the process {Y t
xy}{t≥0}, the integral transforms of distribution (6) have been
studied in [3]–[6] and also in [8], [9]. We assume that the integral transforms of the
joint distributions of the boundary functionals (6) are known, and we will solve the two-
boundary problem for the process {Dxy(t)}{t≥0} in terms of these integral transforms.
2 Exit of the process {Dxy(t)}{t≥0} from the interval
Let us fix B ≥ 0, k ∈ [0, B], r = B − k, Y 0
xy = {0, x, y}, x, y ≥ 0, and introduce the
random variable
χ = inf { t : Dxy(t) /∈ [−r, k] }
as the first exit time from the interval [−r, k] by the process {Dxy(t)}{t≥0}. This ran-
dom variable is the Markov time of the process {Y t
xy}{t≥0}, since it takes values from a
countable set { ξn(y), n ∈ N} ∪ { ηn(x), n ∈ N}. Note that the exit can take place either
through the upper boundary k, or through the lower boundary −r. We introduce the
events A k = {Dxy(χ) > k } when the process exits the interval by crossing the upper
94 VICTOR KADANKOV
boundary and Ar = {Dxy(χ) < −r } when the process exits the interval by crossing the
lower boundary. Define the random variable,
X = (Dxy(χ) − k) IA k + (−Dxy(χ) − r) IAr , L = η +
x (χ) IA k + ξ +
y (χ) IAr
which are the value of the overshoot through the interval [−r, k] by the process
{Dxy(t)}{t≥0}
at the epoch of the first exit and the value of the linear component at the exit epoch,
where IA = IA(ω) is the indicator function of the set A. Denote
fk
xy(du, dl, s) = E
[
exp{−sτk
xy}; T k
xy ∈ du, ηk
xy ∈ dl
]
, k, x, y ≥ 0,
fxy
k (du, dl, s) = E [ exp{−sτxy
k }; T xy
k ∈ du, ξxy
k ∈ dl] , k, x, y ≥ 0,
V +
s (du, dl) = E [ e−sχ; X ∈ du, L ∈ dl, A k ],
V −
s (du, dl) = E
[
e−sχ; X ∈ du, L ∈ dl, Ar
]
.
Theorem 1. Let B ≥ 0, k ∈ [0, B], r = B − k, Y 0
xy = {0, x, y}, x, y ≥ 0. Then the
Laplace transforms V ±
s (du, dl) of the joint distribution of {χ, X, L }, being the first exit
time from the interval [−r, k] by the process {Dxy(t)}{t≥0}, the value of overshoot through
the boundary, and the value of the linear component at the exit epoch, satisfy the equalities
V +
s (du, dl) = F+
xy(k, du, dl, s) +
∫∫
R
2
+
F+
xy(k, du1, dl1, s)K+
u1l1
(du, dl, s),
V −
s (du, dl) = F−
xy(k, du, dl, s) +
∫∫
R
2
+
F−
xy(r, du1, dl1, s)K−
u1l1
(du, dl, s),(7)
where
F+
xy(k, du, dl, s) = fk
xy(du, dl, s) −
∫∫
R
2
+
fxy
r (du1, dl1, s) fu1+B
0l1
(du, dl, s),
F−
xy(r, du, dl, s) = fxy
r (du, dl, s) −
∫∫
R
2
+
fk
xy(du1, dl1, s) f l10
u1+B(du, dl, s);
K±
u1l1
(du, dl, s) =
∞∑
n=1
K±
u1l1
(du, dl, s)∗(n) are the series of iterations K±
u1l1
(du, dl, s)∗(n);
K±
u1l1
(du, dl, s)∗(1) = K±
u1l1
(du, dl, s),
K±
u1l1
(du, dl, s)∗(n+1) =
∫∫
R
2
+
K±
u1l1
(du2, dl2, s)K±
u2l2
(du, dl, s)∗(n), n ∈ N,(8)
are the successive iterations of the kernels K±
u1l1
(du, dl, s) which are given by the formulae
K+
u1l1
(du, dl, s) =
∫∫
R
2
+
f l10
u1+B(du2, dl2, s) fu2+B
0l2
(du, dl, s),
K−
u1l1
(du, dl, s) =
∫∫
R
2
+
fu1+B
0l1
(du2, dl2, s) f l20
u2+B(du, dl, s).(9)
Proof. The integral transforms of the joint distribution of the first exit time and the
value of the overshoot through the boundaries have been determined for the processes
with independent increments and random walks in [7]. These integral transforms were
obtained in terms of the distributions of one-boundary functionals of the process and the
random walk. We used the idea (utilizing the one-boundary functional of the process)
and the method (solving a system of linear integral equations) for this problem in the
EXIT FROM AN INTERVAL BY A DIFFERENCE 95
case where the process is a difference of two renewal processes. Following this framework,
we derive a system of integral equations to determine the function V ±
s (du, dl) :
fk
xy(du, dl, s) = V +
s (du, dl) +
∫∫
R
2
+
V −
s (du1, dl1) fu1+B
0l1
(du, dl, s),
fxy
r (du, dl, s) = V −
s (du, dl) +
∫∫
R
2
+
V +
s (du1, dl1) f l10
u1+B(du, dl, s).(10)
To derive these equations, we applied the total probability law and also used the fact that
the process {Y t
xy}{t≥0} is homogeneous in the first component and τk
xy, τxy
k , and χ are
Markov times. To get the first equation of (10), observe that the first crossing through
the upper boundary k by the process {Dxy(t)}{t≥0} (the expression on the left-hand side
of the equation) occurs on a sample path which does not intersect the lower level −r (the
first term on the right-hand side) or on the path which does intersect the lower level −r
and then crosses the upper boundary k (the second term on the right-hand side). The
second equation is derived analogously. This system of integral equations is similar to a
system of linear equations with two variables. Substituting the expression for V −
s (du, dl)
from the second equation into the first one yields
V +
s (du, dl) = fk
xy(du, dl, s) −
∫∫
R
2
+
fxy
r (du1, dl1, s) fu1+B
0l1
(du, dl, s)+
+
∫∫
u1,l1∈R
2
+
∫∫
u2,l2∈R
2
+
V +
s (du2, dl2) f l20
u2+B(du1, dl1, s) fu1+B
0l1
(du, dl, s).
Changing the order of integration in the third term of the right-hand side of this equation,
we obtain a linear integral equation for the function V +
s (du, dl):
(11) V +
s (du, dl) = F+
xy(k, du, dl, s) +
∫∫
R
2
+
V +
s (du1, dl1)K+
u1l1
(du, dl, s).
Here,
K+
u1l1
(du, dl, s) =
∫∫
R
2
+
f l10
u1+B(du2, dl2, s) fu2+B
0l2
(du, dl, s)
is the kernel of this equation. Observe that the random variable τx0
k takes values from
the set {ηn(x), n ∈ N}. Therefore, for s > 0,
fx0
k (du, dl, s) ≤ E exp{−sτx0
k } ≤ fx(s) ≤ f∗(s),
where fx(s) = E exp{−sηx} and f∗(s) = supx≥0 fx(s). The analogous estimation is valid
for the function fk
0y(du, dl, s) :
fk
0y(du, dl, s) ≤ E exp{−sτk
0y} ≤ gy(s) ≤ g∗(s),
where gy(s) = E exp{−sξy}, g∗(s) = supy≥0 gy(s). Then, for s > s0 > 0, the kernel
K+
u1l1
(du, dl, s) satisfies the estimation
K+
u1l1
(du, dl, s) ≤ f∗(s) g∗(s) ≤ λ = f∗(s0) g∗(s0) < 1, s0 > 0.
Utilizing formula (8) defining the successive iterations of this kernel by the method of
mathematical induction, we get, for s > s0 > 0,
K+
u1l1
(du, dl, s)∗(n+1) < λn+1, n ∈ N.
Then the series of successive iterations converges uniformly with respect to u1, l1, u, l ∈
R+, s > s0, and
K+
u1l1
(du, dl, s) < λ(1 − λ)−1.
96 VICTOR KADANKOV
Therefore, we can apply the method of successive iterations [10] to solve the linear integral
equation (11). This yields the first equality of the theorem. The second equality of the
theorem can be proved analogously.
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E-mail : kadankov@voliacable.com
|
| id | nasplib_isofts_kiev_ua-123456789-4429 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 0321-3900 |
| language | English |
| last_indexed | 2025-12-07T17:49:58Z |
| publishDate | 2005 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Kadankov, V. 2009-11-09T15:34:47Z 2009-11-09T15:34:47Z 2005 Exit from an interval by a difference of two renewal processes / V. Kadankov // Theory of Stochastic Processes. — 2005. — Т. 11 (27), № 3-4. — С. 92–96. — Бібліогр.: 10 назв.— англ. 0321-3900 https://nasplib.isofts.kiev.ua/handle/123456789/4429 519.21 Integral transforms of the joint distribution of the first exit time from an interval,
 the value of the overshoot through a boundary, and the value of a linear component
 at the epoch of the exit are determined for the difference of two renewal processes. en Інститут математики НАН України Exit from an interval by a difference of two renewal processes Article published earlier |
| spellingShingle | Exit from an interval by a difference of two renewal processes Kadankov, V. |
| title | Exit from an interval by a difference of two renewal processes |
| title_full | Exit from an interval by a difference of two renewal processes |
| title_fullStr | Exit from an interval by a difference of two renewal processes |
| title_full_unstemmed | Exit from an interval by a difference of two renewal processes |
| title_short | Exit from an interval by a difference of two renewal processes |
| title_sort | exit from an interval by a difference of two renewal processes |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/4429 |
| work_keys_str_mv | AT kadankovv exitfromanintervalbyadifferenceoftworenewalprocesses |