Statistics and Probability Letters 131 (2017) 78–86
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Statistics and Probability Letters
journal homepage: www.elsevier.com/locate/stapro
Piecewise linear process with renewal starting points
Nikita Ratanov
Universidad del Rosario, Cl. 12c, No. 4-69, Bogotá, Colombia
a r t i c l e i n f o a b s t r a c t
Article history: This paper concerns a Markovian piecewise linear process, based on a continuous-time
Received 9 June 2017 Markov chain with a finite state space. The process describes the movement of a particle
Received in revised form 19 August 2017 that takes a new linear trend starting from a new random point (with state-dependent
Accepted 21 August 2017 distribution) after each trend switch. The distribution of particle’s position is derived in a
Available online 30 August 2017 closed form. In some special cases the distributions of the level passage times are provided
MSC: explicitly.
primary 60J27 © 2017 Elsevier B.V. All rights reserved.
secondary 60J75
60K99
Keywords:
Continuous-time homogeneous Markov
chain
Piecewise linear process
Renewal process
First passage time
1. Introduction and main definitions
Let ε( = ε()t), t ≥ 0, be a continuous-time homogeneous Markov chain with the finite state space E. Let Φ(t) =
etΛ = Φij(t) i j E be the transition semi-group with the infinitesimal generator Λ. Consider the sequence of switching, ∈
times, 0 < T1 < T2 < · · · < Tn < · · · , T0 = 0, and denote by N(t) the number of switchings occurred up to time t,
N(t) = max{n | Tn ≤ t}.
We study the piecewise linear random process
X(t) = xN(t) + cε(TN(t))(t − TN(t)), t ≥ 0. (1.1)
Here ci, i ∈ E, are deterministic constants, and xn, n ≥ 0, are independent random variables with distributions determined
by the current state ε(Tn).
Process X(t), t ≥ 0, describes the location of a particle thatmoves linearly and takes a new starting point after each trend
switch. Process X resembleswell-studiedMarkovian growth-collapse processes, see e.g. Boxma et al. (2006). In contrastwith
(1.1) the growth-collapse processes presume a constant trendwith additive ormultiplicative down jumps. Suchmodels occur
in insurance mathematics and related fields, see Asmussen (2003, XIV-5) o∫r Rolski et al. (1999, Chapters 5 and 11), and inproduction/inventory models studied by Shanthikumar and Sumita (1983), among other∫s.
On the other hand, (1.1) is related to jump-telegraph process T(t) t:= 0 c
t
ε(u)du + 0 Yε(u−)dNu, which is a piecewise
linear process jumping from the current position, see e.g. Di Crescenzo et al. (2013) and López and Ratanov (2012); piecewise
linear processes with jumps are presented by Ratanov (2014). Jump-telegraph processes are widely applied in various fields,
including financial market modelling, see Kolesnik and Ratanov (2013).
E-mail address: nikita.ratanov@urosario.edu.co.
http://dx.doi.org/10.1016/j.spl.2017.08.010
0167-7152/© 2017 Elsevier B.V. All rights reserved.
N. Ratanov / Statistics and Probability Letters 131 (2017) 78–86 79
In contrast to jump-telegraph process, X(t), t ≥ 0, (1.1), completely updates starting points. Possible applications of this
type of processes could be in financial modelling and insurance mathematics.
The paper is structured as follows. In Section 2 we give explicit formulae for transition densities, expectations and limit
distributions of X(t) (as t → ∞). Section 3 is devoted to a detailed analysis of the level passage times in the case of the
two-state underlying Markov process ε.
2. Distribution
Define the transition probabilities p(t, dy | x) and p(t, dy) by entries
pi(t, dy | x) = P{X(t) ∈ dy | ε(0) = i, X(0) = x}, −∞ < x, y < ∞, (2.1)
and ∫
∞
pi(t, dy) = pi(t, dy | x)gi(dx), −∞ < y < ∞, i ∈ E, (2.2)
−∞
where gi(dx) is the distribution of the initial starting point at the initial state i = ε(0). By conditioning on the first switching
one can obtain the following integral e∑quati∫onst
p (t, dy x) e−λitδ (dy) λ e−λ τi | = x+cit + ij i pj(t − τ , dy)dτ , i ∈ E, (2.3)
j 0∈E, j̸=i ∑
where δa(dy) denotes δ-measure l∑ocalis∫ed at point a and λi = j E, j iλij. Further, by (2.2)∈ ̸=t
p (t, dy) e−λitg tci (dy) λ e−λ τi = i + ij i pj(t − τ , dy)dτ , i ∈ E. (2.4)
j 0∈E, j̸=i
Here gai∫(dy) is the displace∫ment of measure gi: for any integrable test-function φ
∞ ∞
φ(y)gai (dy) = φ(y + a)gi(dy).
−∞ −∞
Systems (2.3) and (2.4) can be solved explicit∑ly. Note that when all trends vanish, ci = 0, i ∈ E, the distribution of
X(t) = xN(t) is given by P{X(t) ∈ dx | ε(0) = i} = j EΦij(t)gj(dx). where Φij(t) are the entries of the transition semi-group∈
etΛ introduced in Section 1.
Theorem 2.1. The transition probabilities p∫(t, dy | x) and⎡p(t, dy), (2.1)–(2.2), hav⎤e the form∑ t ∑
p (t, dy x) e−λ t τ ci | = i δ −λ τ kx+tc ⎣ k ⎦i (dy) + Φij(t − τ ) λjke gk (dy) dτ , (2.5)
∫j∈E
0
∑ ⎡
k∈E k ̸=j
∑ ⎤t
pi(t, dy) = e−λitg
tci ⎣ −λkτ τ ck ⎦
i (dy) + Φij(t − τ ) λjke gk (dy) dτ , i ∈ E. (2.6)
j E 0∈ k∈E k ̸=j
Let ∫
∞
Mi(t) = E {Xi(t) | ε(0) = i, Xi(0) = x} gi(dx), i ∈ E.
−∞
Then,
∑∫ ⎡t ∑ ⎤
M −λ t −λ τi(t) = e i (mi + tci) + Φij(t − τ )⎣ λjke k (mk + τ ck)⎦ dτ , (2.7)
∫ j 0∈E k∈E, k̸=j
where m ∞i = xgi(dx), i ∈ E.−∞
Proof. By conditioning on the last switching time and using the time-reversal property, see e.g. Brémaud (1999), one can
derive (2.5): the first term corresponds to the case of no switchings and other summands describe the movement of the
particle, which starts at time 0 from the state i ∈ E and makes the last switching at time t − τ . Eq. (2.6) follows from (2.2).
Formula (2.7) follows from (2.6). □
80 N. Ratanov / Statistics and Probability Letters 131 (2017) 78–86
Example 2.1. Let ci = 1, λi = λ, i ∈ E, and the random points xn are identically distributed with distribution g . In this
case (2.6) becomes ∫ t
p(t, dy) e−λt t= g (dy) + λe−λτ gτ (dy)dτ . (2.8)
0
After applying Fubini’s theorem[ one ca∫n see that th]e distribution (2.8) of X(t) isy
p(t, dy) e−λt= g t (dy) λe−λy+ eλxg(dx) dy. (2.9)
y−t
In particular, if g(dy) = δ(dy), that is xn = 0 n ≥ 0, then by (2.9)
p(t, dy) e−λt= δt (dy) λe−λy+ 10<y<tdy. (2.10)
Example 2.2. Consider the two-state Markov chain with alternating switching intensities λ0, λ1, so called flip-flop process,
Brémaud, (1999). Let 2λ = λ0(+ λ1 and 2ν = λ0 −(λ1. The tra)n)sition semi-group is defined by
1 λ ( λ −2λt −2λt+(t) exp(t ) 1 0e λ0 1 − eΦ = Λ = )
2 λ 1 e−2λt λ λ e−2λt
,
λ 1 − 0 + 1
see e.g. Kolesnik and Ratanov (2013, 4.1.23).
In this case formula (2.6) becomes ∫ t
tΛdiagp(t, dy) e− gtc
diag [ ]
= (dy) + λ0λ1 A(t −τΛ τc− τ )e g (dy) dτ , (2.11)
0
where ( )
λ 0
Λdiag 0= 0 λ1
and
⎝⎛ ( ( )⎞λ−1 νsinh(λt) ) λ−11 cosh(λt) − sinh(λt)A(t) e−λt= λν ⎠ . (2.12)
λ−10 cosh(λt) + sinh(λt) λ
−1 sinh(λt)
λ
Note, that in the case λ0 = λ1 = λ and g0 = g1 = g formula (2.11) coincides with (2.9).
In the two-state case formula (2.7) f∫or expectations can be simplified also. LetM = (M0, M1). By (2.11) we havet
diag diag
M(t) −tΛ= e (m + tc) + λ0λ −τΛ1 A(t − τ )e (m + τc)dτ , (2.13)
0
where c = (c , c )′0 1 andm ′= (m0, m1) . Formula (2.13) can be expressed explicitly, but this is tedious. If λ0 = λ1 = λ > 0,
this formula looks simpler:
t ( ) 2 t ( )1 e−λ 1 e− λ− −
M(t) m c1 ∆c − λ∆m= + c + m c , (2.14)λ 0 2λ λ∆ − ∆
where ∆c = c0 − c1, ∆m = m0 − m1.
To prove (2.14) first note that∫in this case formula (2.13) becomest
M(t) e−λt= (m + tc) λ2 e−λτ+ A(t − τ )(m + τc)dτ ,
0
where ( )
1 1 e−2λt −2λtA(t) − 1 + e= ,
2λ 1 + e−2λt 1 e−2λt−
such that ( ) ( )
1 0 1 dA(t)A(0) and e−2λt 1 −1= 1 0 = .λ dt −1 1
Therefore, ( ) ∫ ( )
dM(t)
e−λt c0 − λ(∆m
t
+ t∆c) 2e−2λt eλτ ∆m + τ∆c= + λ dτ .
dt c1 + λ(∆m + t∆c) 0 −∆m − τ∆c
N. Ratanov / Statistics and Probability Letters 131 (2017) 78–86 81
By integrating we have ( ) ( )
dM(t)
e−λtc e−2λt ∆m (e−λt e−2λt ) ∆c= (− λ − −dt ) −(∆m ) −∆c
e−λt c1 e−2λt ∆c − λ∆m= c + ,0 λ∆m − ∆c
which gives (2.14).
From Theorem 2.1 one can obtain the limiting distribution, as t → ∞.
Let ε = ε(t) be an ergodic regular homogeneous Markov chain, Brémaud, (1999), that is
lim Φij(t) = πj, i, j ∈ E. (2.15)
t→∞
Theorem 2.2. Pro∑cess X(t) converges in distribution, as t → ∞. The limit distribution is
p̄(dy) = πjλjkψk(λk, dy), (2.16)
j,k∈∫E, k̸=j
where (s ) ∞ e−sτ gτ cψ , kk · = 0 k (·)dτ is the time-Laplace transform of distribution g
tck
k (·).
The limit of expec∑tations is∑given(by )c λ
lim Mi(t)
k jk
= πj mk + , i ∈ E. (2.17)
t→∞ λ λ
j∈E k∈E k k, k̸=j
Proof. Applying (2.15)∑to (2.6)∑by Lebes∫gue’s dominated convergence theorem we obtain∞
lim p (t, dy) π λ −λ τ τ cki = j jk e k gk (dy)dτ , (2.18)t→∞
j∈E k 0∈E, k̸=j
which gives (2.16).
By (2.7) ∑ ∫ ∞
lim M (t) π λ e−λ τi = j jk k (mk + τ ck)dτ ,
t→∞
j,k∈E, j 0̸=k
which gives (2.17). □
Formula (2.16) l∑ooks in detail as ∫ ∫πjλ [ y ∞ ]p̄(dy) jk= e−λky/ck eλkz/ckg (dz)1 λ z/c
c k {ck>0}
+ k k∑e gk(dz)1{ck<0} dy| |j,k∈E, k k −∞ y̸=j (2.19)λjk
+ πj gk(dy)1{ck=0}.λ
j,k∈E k j k, ̸=
To prov∫e this, no∫te that with any test-fun∫ction[∫φ applied to the integra]l term of (2.18) we have
∞ ∞ ∞ ∞
(y) e−λkτ gτ cφ kk (dy)dτ = φ(y + c
−λ τ
kτ )e k dτ gk(dy).
−∞ 0 −∞ 0
By applyi∫ng again Fubini’s[th∫eorem we obtain]
1 ∞ y
φ(y)e−λky/ck eλkz/ckg
c k
(dz) dy, if ck > 0;
k −∫∞ [ −∫∞ ]
1 ∞ ∞
φ(y) e−λky/ck eλkz/ckgk(dz) dy, if cc k
< 0;
| k| −∞ y
and ∫
1 ∞
φ(y)gk(dy), if ck = 0.
λk −∞
Example 2.2 (continued). Note that in the case of Example 2.2 the limiting distribution (2.19) can be simplified to
λ λ
p̄(dy) 0 1= [ψ0(λ0, dy) + ψ1(λ1, dy)] , (2.20)2λ
82 N. Ratanov / Statistics and Probability Letters 131 (2017) 78–86
and formula (2.17) becomes [ ]
λ0λ1 m0 m1 c clim M0(t) = lim M1(t)
0 1
= + + + .
t→∞ t→∞ 2λ λ0 λ1 λ2 20 λ1
3. First passage time
Let T xi = T
x
i (y) = inf{t > 0 : Xi(t) ≥ y}, be the first passage time through the fixed level y starting with the initial point
x, Xi(0) = x < y; ε(0) = i ∈ E. To avoid overshooting we assume that the distributions of xn are supported in (−∞, y).
From viewpoint of possible applications, this means that the supporting level y is visible for a moving particle.
If all velocities are nonpositive, cj ≤ 0, j ∈ E, then T xj (y) = ∞, a.s. Note that in the case of a positive velocity ci the
distribution of T xi (y) has an atom: if ci > 0, then P
x
{Ti (y) = (y − x)/ci} = exp (−λi(y − x)/ci).
Let fi(x, y, t) be the density function of T xi (y),
fi(x, y, t) := P{T xi (y) ∈ dt}/dt, t > 0, i ∈ E.
Conditioning on the first switching we∑obt∫aint [∫ ]∧ξi ∞
fi(x, y, t) −λ ξ −λ= e i iδ(t − ξi) + λije iτ fj(z, y, t)gj(dz) dτ
j∑E j i ∫0 −∞∈ , ̸=ci>0 t [∫ ]∞ (3.1)
−λ
+ λije i§τ fj(z, y, t)gj(dz) dτ , i ∈ E,
j E j 0 −∞∈ , ̸=i
ci≤0
where δ(·) is Dirac’s δ-function and ξi = (y − x)/ci.
In what follows we study in detail the ‘‘flip-flop’’ case of the two-state Markov process ε, ε(t) ∈ E = {0, 1}.
3.1. Positive velocities
Let both trends be positive, c0 ≥ c1 > 0, and the alternating distributions g0, g1 of renewal starting points satisfy the
condition gi((−∞, y)) = 1, i ∈ {0, 1}.
Theorem 3.1. The first passage times T x0 and T
x
1 have proper distributions,
P{T x0 (y) <
x
∞} = P{T1 (y) < ∞} = 1, x < y. (3.2)
Let the renewal starting points be exponentially distributed over the half-line (−∞, y),
gi(dz) = ai exp(−ai(y − z))1{z<y}dz, ai > 0,
and µi = λi + ciai, i ∈ {0, 1}. Then, th[e density functions f0 and f1 are given by ]
∗ ∗
fi(x, y, t) e−λiξ= iδ(t − ξi) + λi A1−iu(t ∧ ξ , λ s∗i i − 0)e
−s0t + B1−iu(t ξ , λ s∗)e−s t∧ i i − 1 1 , i ∈ {0, 1}. (3.3)
Here
√ √
y − x µ0 + µ1 − D µ + µ + D
ξ ∗i := , s0 = , s
∗ 0 1
1 = , (3.4)ci 2 2
D (µ µ )2= 0 − 1 + 4λ0λ1; constants Ai and Bi are given by
µ0µ1 − λ0λ (µ λ )s∗ λ λ µ ∗1 − 0 − 0 0 0 1 − 0µ1 + (µ0 − λA B 0
)s1
0 = √ , 0 = √ , (3.5)
D D
µ0µ1 − λ0λ1 − (µ1 − λ1)s∗0 λ0λ1 − µ0µ1 + (µ1 − λ )s
∗
A B 11 = √ , 11 = √ , (3.6)
D D
and ∫ ⎧⎨1 e−βξξ −
u(ξ, β) −βτ , β ̸= 0,:= e dτ = β (3.7)
0 ⎩ ξ, β = 0.
N. Ratanov / Statistics and Probability Letters 131 (2017) 78–86 83
Proof.⎪⎧For the two-state case, by (3.1) t∫he following coupled equations hold: for t > 0⎪⎪⎨ t∧ξ0f0(x, y, t) e−λ= 0ξ0δ(t − ξ0) λ e−λ τ+ ∫ 0
0 h1(y, t − τ )dτ ,
⎩⎪⎪⎪
0 (3.8)
t∧ξ1
f1(x, y, t) = e−λ1ξ1δ(t ξ ) λ e−λ1τ∫− 1 + 1 h0(y, t − τ )dτ ,0
where hi(y, t) = [Gifi] (y
∞
, t) := fi(z, y, t)gi(dz), i ∈ E = {0, 1}. Let−∞
αi(y, s) = [Lt→shi] (y, s) = [Lt→s (Gifi)] (y, s), s ≥ 0,
be the time-Laplace transform of hi, i ∈ {0, 1}.
By a{pplying to system (3.8) the time-Laplace transformationLt→s we have
[Lt→sf0] (x, y, s) = exp(−(λ0 + s)ξ0) + λ0u(ξ0, λ0 + s)α1(y, s),
(3.9)
[Lt→sf1] (x, y, s) = exp(−(λ1 + s)ξ1) + λ1u(ξ1, λ1 + s)α0(y, s),
since
Lt→sδ(t − ξ ) = exp(−sξ )
and, by Fubi∫ni’s theorem,t ξ ∫ ∫∧ ∞ t∧ξ
L e−∫λτh(y, t τ∫)dτ e−st −λτt→s − = dt e h(y, t − τ )dτ0 0 0ξ ∞
e−λτdτ e−st= h(y, t − τ )dt = u(ξ, λ + s)α(y, s).
0 τ
Here function u = u(ξ, β) is defined by (3.7).
Note that Lt→s is commutative with G0 and G1. Applying operator G0 to the first equation of (3.9) and operator G1 to the
second⎧⎪one, we obtain the linear algebraic system,⎪⎨ λ [ ]α0(y, s) 0= ĝ0 (β0(s)) + 1 − ĝ0 (β0(s)) α1(y, s),⎪⎪ λ0 + s⎩ λ [ ]α1(y 1, s) = ĝ1 (β1(s)) + 1 − ĝ1 (βs 1(s)) α0(y, s),λ1 +
where (s) ∫λ0+s λ +sβ0 = c , β 10 1(s) = c and1
∞
ĝ (β) e−β(y−x)i = gi(dx), β > 0, i ∈ {0, 1}. (3.10)
−∞
The integral in (3.10) converges by the condition, gi([y, ∞)) = 0.
The⎧⎪solution of the algebr{aic system is given by⎨⎪⎪ λ [ ] }α0(y, s) (Q (s))−1= ĝ0 (β0(s) 0) + 1 − ĝ0 (β0(s)) ĝ1 (β1(s)) ,
⎪⎪⎪
λ0 + s
⎩ { [ ]} (3.11)α (y, s) −1 λ11 = (Q (s)) (ĝ1 (β1(s)() + ))( ĝ0 (β0((s)) 1)−) ĝs 1 (β1(s)) ,λ1 +
where Q (s) 1 λ0λ1 1 λ +s λ +s= − ( s)( − ĝ
0 1 − ĝ 1 , s > 0.
λ0+ λ1+s) 0 c0 1 c1
By (3.11) it is easy to see that α0(y, 0) = α1(y, 0) = 1. Therefore, due to (3.9) the first passage times T x0 (y) and T
x
1 (y) have
proper distributions P{T x0 (y) <
x
∞} = P{T1 (y) < ∞} = 1, ∀y, y > x.
Functions α0(y, s) and α1(y, s) can be obtained explicitly whenever the renewal starting points are exponentially
distributed over half-line (−∞, y),
gi(dz) = ai exp(−ai(y − z))1{z<y}dz, i ∈ {0, 1},
where a0, a1 > 0 are constants, that is ĝi(β) = ai/(ai + β). In this case we have
β λ
1 i
+ s
− ĝi (β) | λi+s = | λi+s = , i ∈ {0, 1},β= c β=i ai + β ci µi + s
where µ0 = a0c0 + λ0, µ1 = a1c1 + λ1. Therefore
λ λ
Q (s) 1 0 1
q(s)
= − = , s > 0,
(µ0 + s)(µ1 + s) (µ0 + s)(µ1 + s)
84 N. Ratanov / Statistics and Probability Letters 131 (2017) 78–86
Fig. 1. Density functions f0(0, 2, t) (absolutely continuous parts) with c0 = 1, c1 = 0.5: for (a) a0 = a1 = 1; λ0 = λ1 = 1, 0.5, 0.2 (from top to bottom);
for (b) a0 = a1 = 1, 2, 5, respectively, λ0 = λ1 = 1, 0.5, 0.2 (from top to bottom).
and (3.⎪⎧11) becomes⎨ a0c0s + µ1a0c0 + λ0a c(s) 1 1⎪α0 = ,⎩ q(s)a1c1s + µ0a1c1 + λ1a0c (3.12)0α1(s) = ,q(s)
where q(s) = (µ0 + s)(µ1 + s) − λ0λ1, s > 0.
Let s∗0 and s
∗
1 be defined by q(s)
∗ ∗ ∗ ∗
= (s + s0)(s + s1). It is easy to see that, s0, s1 > 0, and
Ai Bi
αi(s) = + , i ∈ {0, 1},s ∗ ∗+ s0 s + s1
where s∗0, s
∗
1 are given by (3.4) and coefficients Ai, Bi, i ∈ {0, 1}, are defined by (3.5)–(3.6).
Making the inverse Laplace transformation we obtain functions h0 and h1,
∗ ∗ ∗ ∗
h0(t) A e−s0t= 0 + B e−s1t , h (t) A e−s0t B e−s1t0 1 = 1 + 1 , t > 0. (3.13)
By (3.8) and (3.13) the density functions of T x0 (y) and T
x
1 (y) are given by (3.3) (see Fig. 1). □
Remark 3.1. Formulae (3.8) give the solution of the problem for arbitrary distributions g0 and g1: the density functions f0
and f1 are expressed by means of the inverse Laplace transforms h0 and h1 of α0 and α1, (3.11).
Example 3.1. Let c0 = c1 = 1, λ0 = λ1 = λ and a0 = a1 = a (see Example 2.1). Then, by (3.3) we have
f −λ(y−x)0 = f1 = e δ(t − (y − x)) + λa · u(t (y x), λ a)e−at∧ − − .
Remark 3.2. Themean of the first passage time T x(y) for aMarkovian growth-collapse process has been studied in detail, see
Löpker and Stadje (2011). Meanwhile, formula (3.3) presents the explicit representation of the distribution (in the special
case of two-state processes with exponentially distributed jumps).
3.2. Velocities of opposite signs
Let the trends be of opposite signs, c = c0 > 0 ≥ c1, and distributions g0, g1 satisfy the condition gi((−∞, y)) = 1, i ∈
{0, 1}.
Since c1 ≤ 0, the distributions of T x0 (y) and T
x
1 (y), x < y, do not depend on g1 and c1. Further, T
x
1 (y) does not depend on
the starting point x.
Let f0(x, y, t) and f1(y, t) be the density functions of distributions of T x0 (y) and T
x
1 (y) respectively.
Theorem 3.2. The first passage times T x0 and T
x
1 have proper distributions, (3.2).
Let the renewed starting point of the upwards movement be exponentially distributed, g0(dx) ≡ g(dx) = a exp(−a(y −
x))1{x<y}dx.
N. Ratanov / Statistics and Probability Letters 131 (2017) 78–86 85
Then, the density functions f0 = f0(x, y, t)[and f1 = f (y, t) ≡ f1(t) of T x0 (y) and T x1 (y) are g]iven by the following:acλ λ ∗ ∗
f (x y t) e−λ0ξ (t ) 0 1, , δ ξ u(t ξ, λ s∗)e−s0t u(t ξ, λ s∗ −s t0 = − + √ ∧ 0 − 0 − ∧ 0 − 1)e 1 (3.14)D
and
acλ [ ]∗ ∗
f (t) 1 e−s0t e−s1t1 = √ − , t > 0, (3.15)
D
where ξ = (y − x)/c and
1 √ 1 √
s∗0 = (λ
∗
0 + λ1 + ac − D), s1 = (λ
∗
0 + λ1 + ac + D), 0 < s0 < s
∗
1;2 2 (3.16)
D = (λ 2 20 + λ1 + ac) − 4acλ1 = (λ1 − ac) + λ0(λ0 + 2λ1 + 2ac)
and function u = u(ξ, β) is defined by (3.7).
By definitions (3.16) it turns out that λ0 < s∗1 always holds. Further, the equality λ0 = s
∗
0 holds, only if
acλ
λ = 10 ac .+λ1
Proof.⎪⎧System (3.1) becomes⎪⎨ ∫ t∧ξ⎪f0(x, y, t) e∫
−λ0ξ= δ(t −λ τ− ξ ) + λ0e 0 f1(y, t − τ )dτ ,
⎪⎩ t 0 (3.17)f1(y, t) = λ1e−λ1τh(y, t − τ )d∫τ , t > 0.0
Here ξ = (y − x)/c, c = c0, and h(y, t)
∞
= f0(z, y, t)g(dz), where g = g0 corresponds the distribution of the starting−∞
point of the upward motion.
Pass{ing to the time-Laplace transform, we get[Lt→sf0] (x, y, s) = exp(−(λ0 + s)ξ ) + λ0u(ξ, λ0 + s) [Lt→sf1] (y, s),
λ
[ f ] (y s) 1 (y s) (3.18)Lt→s 1 , = α , ,
λ1 + s
where ∫ [∫ ]
∞ ∞
α(y, s) = [Lt→sh] (y, s) e−st= f0(x, y, t)g(dx) dt
0 −∞
is the time-Laplace transform of h and function u = u(ξ, β) is defined by (3.7).
Similarly to the proof of Theorem 3.1 one can obtain
λ
(y s) ĝ(( s) c) 0
( ) λ
1 ĝ(( s) c) 1α , =∫ λ0 + / + − λ0 + / α(y, s),λ0 + s λ1 + s
where ĝ(β) ∞ e−β(y−x)= g(dx). Therefore
−∞
ĝ((λ0(+ s)/c)α(y, s) = ) . (3.19)
1 λ− 0λ1(λ0+s)(λ1+s) 1 − ĝ((λ0 + s)/c)
∫ ( )If the (distribution of) the renewed starting point is exponential, g(dx) = a exp(−a(y − x))1{x<y}dx, that is ĝ λ0+sc =y exp −(λ0 + s) y−x g(dx) acc = ac s , then (3.19) becomes−∞ +λ0+
ac(λ1 + s)
α = ,
q(s)
where q(s) = (ac λ 2 ∗ ∗ ∗ ∗+ 0 + s)(λ1 + s) − λ0λ1 = s + (λ0 + λ1 + ac)s + acλ1 = (s + s0)(s + s1) with s0 and s1 given by (3.16).
Hence, [ ]
ac λ ∗ ∗1 − s
α 0
s1 − λ1
= √ + ,
D s + s∗0 s
∗
+ s1
and the inverse[Laplace transform h can be expr]essed byac ∗ ∗
h ≡ √ (λ s∗)e−s1 − 0 0
t
− (λ s∗1 − 1)e
−s1t .
D
With this in hand, due to the second equation of (3.17) we easily obtain (3.15). Then, by the first equation of (3.17) one can
obtain (3.14). See Fig. 2. □
86 N. Ratanov / Statistics and Probability Letters 131 (2017) 78–86
Fig. 2. Density functions f0(0, 2, t) (absolutely continuous parts) with c = c0 > 0 > c1: for (a) with c = 1, a = 1, λ0 = 1 and λ1 = 2, 1; λ1 =
1, 2; λ1 = 0.5 3; for (b) with λ0 = λ1 = 1 and c = 1, a = 1 1; c = 2, a = 0.5 2; c = 5, a = 0.2 3.
Remark 3.3. The distribution f̄ = f̄ (t) of the times between consequent passages through the fixed level y satisfies the
equation ∫ t
f̄ (t) = λ e−λ τ0 0 f1(t − τ )dτ .
0
By (3.15) ∫
λ λ t [ ]
f̄ (t) 0 1
ac
e−λ0τ e−s
∗(t τ ) s∗0 − e− 1(t−τ )= √ [ − dτD 0λ0λ ]1ac ∗ ∗u(t, λ s∗)e−s0t u(t, λ s∗)e−s t= √ 0 − 0 − 0 − 1 .D 1
4. Conclusion
Distributions of a piecewise linear process which starts after each tendency switching from a new random point are
completely described. In the case of two-state process the distributions of the level passage time are presented explicitly.
Applications of these processes will be presented elsewhere later.
Acknowledgements
I would like to thank three anonymous referees for valuable remarks that considerably improved the primary version of
the paper.
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