On extension of the limit theorem of renewal theory and its application
An extension of the limit theorem of renewal theory on the case of semi-Markov process with an atom of the stationary distribution of the embedded Markov chain is obtained in this paper. The obtained result is used for finding an analytic solution of a reliability problem for a system with protection...
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Цитувати: | On extension of the limit theorem of renewal theory and its application / A. Bondarenko // Theory of Stochastic Processes. — 2006. — Т. 12 (28), № 3-4. — С. 12–19. — Бібліогр.: 6 назв.— англ. |
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irk-123456789-44532009-11-12T12:00:40Z On extension of the limit theorem of renewal theory and its application Bondarenko, A. An extension of the limit theorem of renewal theory on the case of semi-Markov process with an atom of the stationary distribution of the embedded Markov chain is obtained in this paper. The obtained result is used for finding an analytic solution of a reliability problem for a system with protection. 2006 Article On extension of the limit theorem of renewal theory and its application / A. Bondarenko // Theory of Stochastic Processes. — 2006. — Т. 12 (28), № 3-4. — С. 12–19. — Бібліогр.: 6 назв.— англ. 0321-3900 http://dspace.nbuv.gov.ua/handle/123456789/4453 en Інститут математики НАН України |
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An extension of the limit theorem of renewal theory on the case of semi-Markov process with an atom of the stationary distribution of the embedded Markov chain is obtained in this paper. The obtained result is used for finding an analytic solution of a reliability problem for a system with protection. |
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Bondarenko, A. |
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Bondarenko, A. On extension of the limit theorem of renewal theory and its application |
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Bondarenko, A. |
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Bondarenko, A. |
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On extension of the limit theorem of renewal theory and its application |
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On extension of the limit theorem of renewal theory and its application |
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On extension of the limit theorem of renewal theory and its application |
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On extension of the limit theorem of renewal theory and its application |
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On extension of the limit theorem of renewal theory and its application |
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on extension of the limit theorem of renewal theory and its application |
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Інститут математики НАН України |
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2006 |
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http://dspace.nbuv.gov.ua/handle/123456789/4453 |
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On extension of the limit theorem of renewal theory and its application / A. Bondarenko // Theory of Stochastic Processes. — 2006. — Т. 12 (28), № 3-4. — С. 12–19. — Бібліогр.: 6 назв.— англ. |
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AT bondarenkoa onextensionofthelimittheoremofrenewaltheoryanditsapplication |
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2025-07-02T07:41:41Z |
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2025-07-02T07:41:41Z |
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Theory of Stochastic Processes
Vol. 12 (28), no. 3–4, 2006, pp. 12–19
ANNA BONDARENKO
ON EXTENSION OF THE LIMIT THEOREM OF
RENEWAL THEORY AND ITS APPLICATION
An extension of the limit theorem of renewal theory on the case of semi-
Markov process with an atom of the stationary distribution of the em-
bedded Markov chain is obtained in this paper. The obtained result is
used for finding an analytic solution of a reliability problem for a system
with protection.
1. Introduction
An extension of the limit theorem of renewal theory on the case of semi-
Markov process with an atom of the stationary distribution of the embedded
Markov chain is obtained in this paper. The result is illustrated by appli-
cation to the solution of reliability problem for system with protection.
Consider the renewal process ξ(t) which satisfies conditions:
B1. Distribution function of time between renewals is absolutely contin-
uous: Q(t) =
t∫
0
q(s)ds;
B2. ml =
∞∫
0
tlq(t)dt < ∞, l = 1, k + 2, k ≥ 1.
Let h(t) be the renewal density of the renewal process ξ(t).
In the paper by C. J. Stone (1965) and B. A. Sevastianov (1968) it is proved
that under conditions B1, B2
∞∫
0
tn|h∗(t)|dt < ∞, n = 0, k, where h∗(t) = h(t) − 1
m1
.
In the present paper we extend this result on the case of strongly regu-
lar semi-Markov process ξ(t) that satisfies conditions C1 – C3, stated be-
low, and the condition that the stationary distribution of embedded in ξ(t)
Markov chain has an atom of distribution.
Note that conditions B1, B2 are a particular case of conditions C1-C3 for
the renewal process.
2. Extension of the limit theorem of renewal theory
2000 Mathematics Subject Classification. Primary 60-XX, 60J35.
Key words and phrases. Strongly regular semi-Markov process, embedded Markov
chain, uniformly recurring, atom of stationary distribution.
12
ON EXTENSION OF THE LIMIT THEOREM 13
Let semi-Markov process ξ(t) be given by the Markov renewal process
{ξn, τn; n � 0}, where ξn is an embedded Markov chain, τn are moments
of renewal. Let the following conditions holds:
C1. The Markov chain ξn, n ≥ 0, embedded in the ξ(t) is uniformly recur-
ring;
C2. sup
x∈X
Ml(x, B) < ∞, B ∈ B, l = 1, k + 2, k ≥ 1, where Ml(x, B) =
∞∫
0
tlQ(dt, x, B);
C3.The semi-Markov kernel of the process ξ(t) is absolutely continuous in t:
Q(t, x, B) =
t∫
0
q(s, x, B)ds, t ≥ 0, x ∈ X, B∈B.
Suppose that there exists a point x0 ∈ X such that ρ({x0}) > 0, in
other words, there exists at least one point from the set of states in which
stationary distribution ρ of the embedded Markov chain has an atom of
distribution.
Denote by x0F (t, x, B) the distribution function of the time of the first
attainment of the set of states B by process ξ(t) from the initial state x and
with forbidenness to get to x0:
x0F (t, x, B)= P{τn � t, ξn ∈ B, n � 1, ξν �∈ B, ξν �= x0, ν = 1, n − 1/ξ0 = x}
Denote by h(t, x, B) and hc(B) the density of the Markov renewal function
of the process ξ(t) and the stationary density of the Markov renewal function
correspondingly. Let x0Mn(x, B)
def
=
∞∫
0
tnx0F (dt, x, B).
Theorem 1. Let a strongly regular semi-Markov process ξ(t) satisfies con-
ditions C1 – C3 and let the stationary distribution of the embedded in ξ(t)
Markov chain has an atom of distribution at the point x0. Then there exist
Hn(x, B) =
∞∫
0
tnh∗(t, x, B)dt, n = 0, k, x ∈ X, B∈B,
where h∗(t, x, B) = h(t, x, B)−hc(B), and the following relations hold true:
(1)
Hn(x, x0) = x0Mn(x, x0) − hc(x0)
n + 1
x0Mn+1(x, x0)+
+
n∑
r=0
Cr
n x0Mn−r(x, x0)Hr(x0, x0),
(2)
Hn(x, B) = x0Mn(x, B) − hc(x0)
n + 1
x0Mn+1(x0, B)+
+
n∑
r=0
Cr
n Hn−r(x, x0)x0Mr(x0, B).
14 ANNA BONDARENKO
Proof. Let ξx0(t) be the sparse semi-Markov process with the state set
{x, x0}, where x = ξ(0) (see, for example, paper by V. S. Korolyuk, A. A. To-
musyak and A. F. Turbin (1979)). It is a general renewal process with the
initial state x and moments of renewal which coincides with moments of
hit of the process ξ(t) the state x0. Then from definition of x0F (t, y, x0)
it follows that for y = x it is the distribution function of time until the
first renewal of ξx0(t), and for y = x0 it is the distribution function of time
between any other renewals of ξx0(t).
Denote by hx0(t, y, x0), where y = x or x0, the renewal density of the
process ξx0(t), which, by definition of this process, coincides with h(t, y, x0),
the density of the Markov renewal function of the process ξ(t), were y = x
or x0, {x0}∈B.
From the book by A. N. Korlat, V. N. Kuznetsov, M. M. Novikov and
A. F. Turbin (1991) it follows, that under conditions of our theorem we get
x0Ml(y, x0) < ∞, l = 1, k + 2, y = x, x0,
and the distribution function of ξx0(t) is absolutely continuous:
x0F (t, y, x0) =
t∫
0
x0f(s, y, x0)ds, y = x, x0.
So the renewal process ξx0(t) satisfies conditions B1, B2. Therefore, as it
was proved by C. J. Stone (1965) and B. A. Sevastianov (1968)
(3)
∞∫
0
tn|h∗(t, y, x0)|dt < ∞, n = 0, k, y = x, x0.
Thus by Lebesgue theorem we may pass to the limit under the integral sign
in the Laplace transform:
(4) Hn(y, x0) =
∞∫
0
tnh∗(t, y, x0)dt = lim
p→0
∞∫
0
e−pth∗(t, y, x0)dt, y = x, x0
By applying the formula of total probability and taking into consideration
the first jump of the process ξx0(t) we will have the equation
h(t, x, x0) = x0f(t, x, x0) +
t∫
0
x0f(s, x, x0)h(t − s, x0, x0)ds.
Note that tn = (t − s + s)n =
n∑
r=0
Cr
nsn−r(t − s)r. So if multiply the last
equation by tn, n = 1, k, after not complicated algebraic transformation we
will get
tnh∗(t, x, x0) = tnx0f(t, x, x0) − hc(x0)t
n(1 − x0F (t, x, x0))+
ON EXTENSION OF THE LIMIT THEOREM 15
+
n∑
r=0
Cr
n
t∫
0
sn−r
x0f(s, x, x0)(t − s)rh∗(t − s, x0, x0)ds.
Then applying the Laplace transform to the last equation taking into con-
sideration (4)and passing to the limit as the parameter of the Laplace trans-
form p tends to 0 we will have (1).
By applying the formula of total probability and taking into consideration
the last until time t jump of the process ξ(t) in state x0 we will get
(5) h(t, x, B) = x0f(t, x, B) +
t∫
0
h(s, x, x0) x0f(t − s, x0, B)ds.
It follows from the results by A. N. Korlat, V. N. Kuznetsov, M. M. Novikov
and A. F. Turbin (1991) that
(6) hc(B) =
∞∫
0
hc(x0) x0f(t, x0, B)dt.
Since tn = (t − s + s)n =
n∑
r=0
Cr
nsn−r(t − s)r, from (5), (6) we get
tnh∗(t, x, B) = tn x0f(t, x, B) − hc(x0)t
n(x0F (∞, x0, B) − x0F (t, x0, B))+
(7) +
n∑
r=0
Cr
n
t∫
0
sn−rh∗(s, x, x0)(t − s)r
x0f(t − s, x0, B)ds.
Applying the Laplace transform to the last equation taking into consid-
eration (4) and passing to the limit as p tends to 0 we will have
lim
p→0
∞∫
0
e−pttnh∗(t, x, B)dt = x0Mn(x, B) − hc(x0)
n + 1
x0Mn+1(x0, B)+
+
n∑
r=0
Cr
n Hn−r(x, x0) x0Mr(x0, B).
So, to prove (2) we should to prove the existence of
∞∫
0
tnh∗(t, x, B)dt, n =
0, k. From (7) it follows that
|tnh∗(t, x, B)| ≤ tn x0f(t, x, B) + hc(x0)t
n(x0F (∞, x0, B) − x0F (t, x0, B))+
+
n∑
r=0
Cr
n
t∫
0
|sn−rh∗(s, x, x0)|(t − s)r
x0f(t − s, x0, B)ds.
16 ANNA BONDARENKO
Applying the Laplace transform to the right side of the last unequality and
then taking the parameter of the Laplace transform p equal to 0 we will get
∞∫
0
|tnh∗(t, x, B)|dt ≤ x0Mn(x, B) − hc(x0)
n + 1
x0Mn+1(x0, B) +
+
n∑
r=0
Cr
n
∞∫
0
tn|h∗(t, x, x0)|dt x0Mr(x0, B).
From the last unequality and (3) we have that
∞∫
0
tn|h∗(t, x, B)|dt < ∞. �
2. Reliability problem for system with protection
Consider a system with protection which consist of two independent el-
ements. Functioning of the first element is described by a renewal process
with the density of distribution function of the time of renewal f(t). Func-
tioning of the second element (system of protection) is described by an al-
ternating renewal process. This process models the alternating sequence of
periods of work (state 1) and periods of repair (state 2) of system of protec-
tion, with densities of distribution function g1(t) and g2(t) correspondingly.
Let f(t) be the density of the uniform distribution in interval [0, 1], let
g1(t) =
μn
1
Γ(n)
tn−1e−μ1t, t ≥ 0, n ∈ N, μ1 > 0 (Erlang distribution), and let
g2(t) = μ2e
−μ2t, t ≥ 0, μ2 > 0 (exponential distribution). If the moment of
renewal of the first element occurs in period of repair of the second element,
then the system faults.
Our problem is to find the average of distribution of time until the first
system fault (M), under condition that at starting time t = 0 both elements
start to work.
� � � �
� � �
f
g1 g1
g2
We will suppose that after a fault the system keep on functioning. Let us
describe functioning of this system by a semi-Markov process η(t) ( general
renewal process) with two states 0 and F . Let at starting time t = 0 the
process be in state 0 and stay there until the first system fault. In the
moment of system fault the process η(t) transfers to state F and stay there
until the next fault.
ON EXTENSION OF THE LIMIT THEOREM 17
� � �
0 F F
Let h(t, x, F ) be the renewal density of the process η(t), let hc be the
stationary renewal density of the process η(t). The process η(t) satisfies
conditions of Theorem 1. For this reason from equation (1) for n = 0 we
have:
(8)
∞∫
0
(h(t, 0, F ) − hc)dt = 1 +
∞∫
0
(h(t, F, F ) − hc)dt − hcM
Let’s calculate integrals that appear in equation (8). Let h1(t) be the den-
sity of renewal of the first element, let h1c
def
= lim
t→∞
h1(t), let Πi(t) be the
probability of being of the second element in state of repair at time t, under
condition that at starting time t = 0 it was in state i, i = 1, 2, and let
Πc
def
= lim
t→∞
Π1(t). Then because of independence of elements we find
h(t, 0, F ) = h1(t) Π1(t), h(t, F, F ) = h1(t) Π2(t), hc = h1c Πc.
As it is well known ( see, for example, D. Koks and V. Smeet (1967))
Π̃1(p) =
g̃1(1 − g̃2(p))
p(1 − g̃1(p)g̃2(p))
, Π̃2(p) =
1 − g̃2(p)
p(1 − g̃1(p)g̃2(p))
, h̃1(p) =
f̃(p)
1 − f̃(p)
,
where symbol ∼ denotes the Laplace transform.
As g̃1(p) =
μn
1
(μ1 + p)n
, g̃2(p) =
μ2
(μ2 + p)
, f̃(p) =
1 − e−p
p
, then
(9) Π̃1(p) =
(μ1)
n
(μ1 + p)n(p + μ2) − μn
1μ2
, Πc =
μ1
μ1 + nμ2
,
(10) Π̃2(p) =
(μ1 + p)n
(μ1 + p)n(p + μ2) − μn
1μ2
, h̃1(p) =
1 − e−p
(p − 1 + e−p)
, h1c = 2.
From Theorem 1 follows existence of
∞∫
0
(h1(t)− h1c)dt and
∞∫
0
(Πi(t)−Πc)dt,
i = 1, 2 in our case. Thus from (9), (10) it follows that there exists δ > 0
such that {p ∈ C : Rep ≥ −δ} is singularity-free domain for the functions
h̃1(p) − h1c
p
and Π̃i(p) − Πc
p
, i = 1, 2. Thus we may apply theorem about
composition of original functions and residue theorem (see, for example V.
Martynenko (1965)), according to which
∞∫
0
(h1(t)−h1c)(Πi(t)−Πc)dt =
∑
Res(Π̃i(p)−1
p
Πc)(h̃1(−p)+
1
p
h1c), i = 1, 2
18 ANNA BONDARENKO
where residues are calculated at singular points of function (Π̃i(p) − 1
p
Πc).
Consequently from (8) it follows that
(11) M =
1
h1cΠc
+
1
Πc
∞∫
0
(Π2(t) − Π1(t))dt+
+
1
h1cΠc
∑
Res(Π̃2(p) − 1
p
Πc)(h̃1(−p) +
1
p
h1c)−
− 1
h1cΠc
∑
Res(Π̃1(p) − 1
p
Πc)(h̃1(−p) +
1
p
h1c),
where residues are calculated at singular points of functions (Π̃2(p) − 1
p
Πc)
and (Π̃1(p)− 1
p
Πc) correspondingly. From (9), (10) it follows that functions
(Π̃2(p)− 1
p
Πc) and (Π̃1(p)− 1
p
Πc) have the same singular points, which are
nonzero roots of the equation (μ1 +p)n(p+μ2)−μn
1μ2 = 0. Noting that 0 is
the root of this equation, but it is not the singular point for this functions.
Thus considering (11), (9), (10), we get
M =
(μ1 + nμ2)
2μ1
+
n
μ1
+
(μ1 + nμ2)
2μ1
∑
Res
(2 − 2ep + p + pep)
(p + 1 − ep)p
×
× (μ1 + p)n − (μ1)
n
(μ1 + p)n(p + μ2) − μn
1μ2
where residues are calculated at points pi, i = 1, n, which are nonzero roots
of the equation (μ1 + p)n(p + μ2) − μn
1μ2 = 0.
In particular case, where μ1 = μ2, we have
M =
(n + 1)
2
+
n
μ1
+
(n + 1)
2
n∑
k=1
(2 − 2epk + pk + pke
pk)[(μ1 + pk)
n − (μ1)
n]
(n + 1)(pk + 1 − epk)pk(μ1 + pk)n
,
pk = μ1cos
2kπ
(n + 1)
+ isin
2kπ
(n + 1)
, k = 1, n.
Conclusion
Due to Theorem 1, proved in this paper, we succeeded in finding an analytic
solution of reliability problem for system with protection. Other methods
result in more complicated system of equations for which analytic solution
are not known.
ON EXTENSION OF THE LIMIT THEOREM 19
References
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models of renewal systems and queueing systems, Shtiintsa, Kishinev, (1991).
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Dumka”. (1979), 62 – 69.
4. Martynenko V. S., Operational calculus, Publisher of Kyiv University, Kyiv,
(1965).
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Department of Probability Theory and Mathematical Statistics, Kyiv
National Taras Shevchenko University, Kyiv, Ukraine
E-mail address: gannucia@ukr.net
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