AND RISK THEORIES
ANA-MARIA R ˘ADUCAN, LASZLO LAKATOS and GHEORGHIT¸ ˘A ZB ˘AGANU
We study a class of Lindley processes whose distribution can be computed. We call them computable Lindley processes and investigate their applications in queueing theory and ruin theory.
AMS 2000 Subject Classification: 60J15, 60K10, 60K25, 60G99.
Key words: queueing system, Lindley process, stochastic domination, ruin proba- bility, waiting time.
1. THE PROBLEM
I.AG/G/1queue is defined by two independent sequences (Xn)n≥1and (Yn)n≥1of i.i.d. random variables;Xnis theservice timeof thenth customer.
Forn≥2, Ynis the time between the arrival moment of the (n−1)th customer and the arrival moment of thenth customer whileY1 is the arrival moment of the first customer. Thus the nth customer arrives at tn=Y1+· · ·+Yn.Let FX, FY be the distributions of Xn and Yn. These two distributions are the parameters of the queue; we codify it as the queue hFX, FYi.
It is important to keep in mind that we shall denote by the same letter both the distribution and its distribution function: if F is a distribution – a probability measure on the real line – then F(x) means actually F((−∞, x]).
The right tail F((x,∞)) of F,will be denoted by F(x).
There are many characteristics of a queue worth to be studied. We shall focus on the waiting time of the nth customer: the time elapsed between joining the line and entering the service. This is a sequence of random variables (Wn)n≥0 constructed by the recurrence
(1.1) W1 = 0, Wn+1 = (Wn+Xn−Yn+1)+,
where x+ is the positive part of x: x+=xforx >0 andx+ = 0 forx≤0.
If we denote by Zn the random variablesXn−Yn+1,then (Zn)nwill be a sequence of i.i.d. random variables having the distribution F =FX∗F−Y . Replacing nby n−1, (to start the count from 0) recurrence (1.1) becomes (1.2) W0= 0, Wn+1 = (Wn+Zn)+.
REV. ROUMAINE MATH. PURES APPL.,53(2008),2–3, 239–266
Obviously,W= (Wn)n≥0 is a Markov chain with the transition kernel (1.3) Ux(B) :=P(Wn+1∈B |Wn=x) = (F∗δx)(+)(B), ∀B∈ R Borel set.
If Gis a distribution on the real lineR,we denote by G(+) the distribution (1.4) G(+)(B) =G(B∩(0,∞)) +δ0(B)G((−∞,0]).
Sometimes it is easier to understandG(+) by means of its right tail: we clearly have
(1.5) G(+)(x) =G(x) ∀x≥0, G(+)(x) = 1 if x <0.
LetGn be the distribution ofWn.Then we can write (1.2) as (1.6) G0 =δ0, Gn+1= (Gn∗F)(+).
Definition. A Markov chain (Wn)n given by recurrence (1.2) is called a Lindley process of parameter F. Thus, its distributions are G1 = F(+), G2 = (F(+)∗F)(+),G3= ((F(+)∗F)(+)∗F)(+), . . . .
II. A classic renewal risk model (the Sparre Andersen model)is a stochastic process of the form
(1.7) V(t) =SN(t)−ct,
whereS0= 0 andn≥1⇒Sn=X1+· · ·+Xn, N(t) = sup{k|tk ≤t},t0 = 0 and k≥1⇒tk=σ1+· · ·+σk.
Here, (Xn)n≥1 and (σn)n≥1 are two independent sequences of i.i.d. posi- tive random variables which have the following meaning: theXnare the values of the claims coming from insured persons to an insurer and the σn are the interarrival claim times. The nth claim has the sizeXnand comes at moment tn.The constant c is theintensity of the cash flow coming from the insured persons. Then V(t) is the loss of the insurer at timet: if V(t)≤0, it is OK, but if the initial capital of the insurer is u≥0 and it happens that V(t)> u, it is bad. The first moment t when V(t) > u stops the business: this is the ruin moment τ(u).
Of course, the ruin may only occur at claim arrival moments tn.But (1.8) V(tn) = (X1−cσ1) +· · ·+ (Xn−cσn) =Z1+· · ·+Zn,
where Zn=Xn−Yn andYn=cσn.
LetLn= max(0, S1, . . . , Sn), withS0 = 0 andn≥1⇒Sn=Z1+· · ·+ Zn.Then (Ln)n≥1 is not a Markov chain but, since
(1.9) Ln+1 = max(0, Z1+ max(0, Z2, Z2+Z3, . . . , Z2+· · ·+Zn+1))
= (L∗n+Z1)+,
it has the same distribution as a Markov chain given by the recurrence (1.10) L0= 0, Ln+1 = (Ln+Zn)+.
This is exactly recurrence (1.2). Thus, Ln has the same distribution as a Lindley process of parameterF,where F is the distribution ofZn.The right tailψn(u) :=P(Ln> u) ofLn is the probability that the ruin occurs beforen transactions.
It is well known (see, for instance [1]) that the necessary and sufficient condition thatLnhas a limit in distributionL∞is that EZn<0.Or, in terms of Xn and Yn, that EXn <EYn. The ratio EXEYn
n is denoted in queuing theory by ρ and it is called the traffic intensity. In the ruin theory the same ratio is denoted by 1+θ1 , whereθis the loading factor.
In other words, the sequence (Gn)n of distributions has a proper limit G∞ iff ρ <1.We will always assume this condition in the sequel.
A last remark: the case where the distributionFY is exponential is spe- cial. In ruin theory this is the Cram`er-Lundberg model while in queuing theory is the “M/G/1 queue”. Most of the interesting results are proved under this hypothesis.
In short: if we are interested only in the waiting timesWnin a queue or in the ruin probabilities in a risk model, then we deal with the same mathematical object: a Lindley process of parameter F, whereF =FX ∗F−Y.We will also denote this Lindley process (maybe it is not very usual) by hFX, FYi and its trajectories by (Ln)n.The distribution of Ln will be denoted byGn; hence
G0 =δ0, Gn+1= (Gn∗F)(+).
Now, we state the problem. Suppose that we want to compare two Lindley processes: the first one has parameter F (and distributions Gn) and the second one has parameter F0 (and distributionsG0n).Write F ≺++F0 iff Gn≺stG0n ∀nand F ≺+F0 iffG∞≺st G0∞.Or, in terms of random variables, Z ≺++ Z0 ⇔ Ln ≺st L0n ∀n and Z ≺+ Z0 ⇔ L∞ ≺st L0∞. Do we have computational criteria to decide if F ≺++F0 orF ≺+F0?
We shall deal with particular types of Lindley process of the formhFX, FYi and hFX0, FY0i. In that case we write hFX, FYi ≺≺ hFX0, FY0i instead of FX∗F−Y ≺++FX0∗F−Y0 andhFX, FYi ≺ hFX0, FY0iinstead ofFX∗F−Y ≺+ FX0 ∗F−Y0.
In [4] was considered the particular case FX0 = Exp(a), FY = Exp(b), FX0 = Exp(a0), FY0 = Exp(b0).The parameters of the two Lindley processes are Fa,b=FX∗F−Y and Fa0,b0 =FX0∗F−Y0.In that situation, the result was Fa,b≺++Fa0,b0 ⇔a≥a0,ρ≤ρ0, where ρ= EXEY = ab,ρ0 = EXEY00 = ba00.In terms of queues,the first queue is better if the service time is smaller and the traffic intensity is smaller. In the same paper a counterexample was given of two Lindley processes hFX, FYi and hFX0, FY0i for which this is not true: ρ =ρ0, X0 ≺st X and, on the contrary, hFX, FYi ≺ hFX0, FY0i. The domination goes in the opposite way.
The result was generalized a bit in [5]: suppose that the pairshFX, FYi and hFX0, FY0i areconjugated, to mean that
(1.11) X, X0>0 (a.s.), FX∗F−Y =pFX+qF−Y, FX0∗F−Y0 =p0FX0+q0F−Y0
In this case a weaker result than that from [4] holds, namely:
(1.12)
X≺stX0 and ρ≤ρ0 ⇒ hFX, F−Yi ≺≺ hFX0, F−Y0i ⇒
⇒FX ≤ p
p0FX0 and ρ≤ρ0, where ρ= EXEY and ρ0 = EXEY00.
Remark 1.1. Property (1.11) means that the convolution of FX and F−Y is a mixture of these distributions, and the same property hold for the convolution ofFX0 andF−Y0. Notice that if this is indeed the case, thenpand p0 are uniquely determined by the expectations. It is easy to see that
p= ρ
ρ+ 1, p0= ρ0
ρ0+ 1, q= 1
ρ+ 1, q0 = 1 ρ0+ 1, where ρ= EXEY and ρ0 = EXEY00.
In this paper we generalize this result and give more examples of Lindley processes for which the first implication from (1.12) holds. In terms of ruin theory, we want to give a partial answer to the question: for what kind of risk processes it is still true that a process with smaller claims and greater loading is safer?
Remark 1.2. We shall use the following notation:
– Negbin(k, p) is the negative binomial distribution of parameterskand p: that is the kth convolution of Negbin(1, p) :=
0 1 2 3 · · · p pq pq2 pq3 · · ·
; – Negbin(k,−p) is the distribution of −X ifX ∼Negbin(k, p); it is the kth convolution of Negbin(1,−p) :=
0 −1 −2 −3 · · · p pq pq2 pq3 · · ·
; – Gamma(m, a) = Exp(a)∗m;
– Exp(−b) is the distribution with the distribution function bebx, x≤0.
It is the distribution of −X forX∼Exp(b);
– Geom(1, p) =
1 2 3 4 · · · p pq pq2 pq3 · · ·
= Negbin(1, p)∗δ1
– Geom(m, p) = Geom(1, p)∗m = Negbin(m, p)∗δm;
– Geom(m,−p) is the distribution of −X forX∼Geom(m, p);
– Poisson(−λ) is the distribution of−X forX∼Poisson(λ);
– Uniform(a, b) is the distribution with density 1(a,b)/(b−a).
2. COMPUTABLE LINDLEY PROCESSES
The crucial fact in the approach in [4] and [5] was the possibility to derive an algebraic formula forGn of the form
(2.1) Gn= ΓQne,
where
– Γ is a infinitely dimensional row vector (Γ0,Γ1,Γ2, . . .) with compo- nents the distributions Γn=FX∗FX ∗ · · · ∗FX ( thenth convolution of FX);
by convention, Γ0=δ0;
–Q is an infinite dimensional column stochastic matrix; and –eis the column vector (1,0,0, . . .).
In the case considered in [4] and [5], the components of the matrix Q were
q(i, j) =
qj+1 ifi= 0
pqj−i+1 ifi≥1, j ≥i−1 0 ifj < i−1.
Here, p= ρ+1ρ = EXEX+EY andq = ρ+11 = EX+EYEY .Or, explicitly,
Q=
q q2 q3 q4 · · · p pq pq2 pq3 · · · 0 p pq pq2 · · · 0 0 p pq · · · 0 0 0 p · · ·
· · · ·
.
If π =
· · · −2 −1 0 1
· · · pq3 pq2 pq p
and we write pi instead of π({i}), then we have
Q=
r0 r−1 r−2 r−3 · · · p1 p0 p−1 p−2 · · · 0 p1 p0 p−1 · · · 0 0 p1 p0 · · · 0 0 0 p1 · · ·
· · · ·
,
whererk=π((−∞, k]).Now, we see thatthe transposed of Qis the transition matrix of a new Lindley process of parameter π. Indeed, if we consider a sequence (ξn)n of i.i.d. random variables with distributionπ and the Markov chain given by the recurrence Nt+1 = (Nt+ξt+1)+, starting from some N0
independent of the ξn, then
P(Nt+1=n|Nt=k) =P((Nt+ξt+1)+ =n|Nt=k)
=P((k+ξt+1)+=n|Nt=k) =P((k+ξt+1)+=n) (because of independence!). Hence
P(Nt+1= 0|Nt=k) =P(ξt+1≤ −k) =r−k if n= 0 and
P(Nt+1=n|Nt=k) =P(ξt+1=n−k) =π({n−k}) =pn−k if n≥1.
Therefore, P(Nt+1=n|Nt =k) =qn,k. Moreover,Qneis the distribu- tion of Nn ifN0 = 0.
We can restate this remark in terms of random variables as follows Let N = (Nn)n≥0 be a Lindley process of parameter π starting from 0. Let X = (Xn)n be a sequence of i.i.d. random variables with distribution FX. Assume that X is independent of N. Let S(0) = 0 and for n ≥ 1 let S(n) =X1+· · ·+Xn.Then Ln has the same distribution as S(Nn).
(It seems that what really matter is the distributionFX whileF−Y plays a secondary role.). This remark motivated us to give the following
Definition. Let L = (Ln)n be a Lindley process of parameter F.If the distribution Gn- of Ln- can be written as
(2.2) Gn=ΓQne,
where Γm=µ∗m=µ∗µ∗· · ·∗µ,(themth convolution ofµ), µis a distribution on (0,∞) andQ is the transposed of a transition matrix of a Lindley process on the set of nonnegative integers of parameter
π=
... k−3 k−2 k−1 k ... pk−3 pk−2 pk−1 pk
for some k≥1, then we say that Lis acomputable Lindley process with characteristics µ and π.
Remark 2.1. It is easy to see that the above definition is equivalent to the existence (possibly on another suitable probability space) of a sequence U = (Un)n≥1 of i.i.d. positive random variables with distribution µ and of a discrete Lindley process N= (Nn)n≥0 of parameter π, which is independent of U and starts from 0 such that
(2.3) Ln=D X
i≤Nn
Ui. The Lindley process Nn is given by the recurrence (2.4) N0= 0, Nn+1 = (Nn+ξn+1)+
and the distribution ofξn isπ. Next, themth column ofQis the distribution of (m+ξn+1)+,namely,
qi,m=P((m+ξn+1)+=i)) =
( P(ξn+1≤ −m) =r−m ifi= 0 P(ξn+1=i−m) =pi−m ifi≥0 Remark 2.2. It is very important that the support of π is included in (−∞, k]. We call such a Lindley process computable because if we can compute the convolutions Γmform≤kn, then we can also computeGn.This happens becauseQneis the first column ofQnand has at mostkn+1 nonzero entries. We do not need to keep in the memory of the computer all the entries of Q– which is, of course, impossible. The reader can check that the product Qneis the same as QnQn−1· · ·Q2Q1e, where the matrices Qm are obtained from Q as follows:
–Q1is the (k+ 1)×1 matrix obtained from Qkeeping the first column and the first k+ 1 rows;
–Q2is the (2k+1)×(k+ 1) matrix obtained fromQkeeping the first k+
1 columns and the first 2k+ 1 rows;
–· · ·
– Qm is the (mk+ 1)×(mk−k+ 1) matrix obtained fromQ keeping only the first mk−k+ 1 columns and the first mk+ 1 rows.
Moreover, asqi,m=pi−mwe do not need to keep in memory the matrices Qm but only the lastnk+ 1 entries ofπ. Fork= 9, we were able to compute (Gn)nforn≤500.
Remark 2.3. It is obvious that any Lindley process of parameter F =
· · · (m−3)h (m−2)h (m−1)h mh
· · · pm−3 pm−2 pm−1 pm
is computable. Here, Γn=δnh and Qis the same matrix withh= 1.
We shall give some examples of computable Lindley processes. May be, there are smarter methods of proof, ours are computational.
All the examples we know fit into the pattern described below.
Theorem. Let µbe a probability distribution on (0,∞)and let Γn=µ∗n, Γ0=δ0.Suppose that
(i) FX = Γm for some m≥1;
(ii)F−Y is a distribution on (−∞,0] such that
(2.5) (Γn∗F−Y)(+)=q0,nΓ0+q1,nΓ1+· · ·+qn,nΓn.
Let Q.,n be the column vector (q0,n, q1,n, . . . , qn,n,0,0,0, . . .) and Q the column stochastic matrix obtained by putting together the columns starting with
Q.,m, i.e.,
(2.6) Q= (Q.,m;Q.,m+1;Q.,m+2;. . .).
Then the distribution of Ln is
Gn=ΓQne,
where Γis the row vector (Γ0,Γ1,Γ2, . . .)and ethe column vector (1,0,0, . . .).
Proof. For n= 1 the claim is true sinceG1 =ΓQe.Suppose it holds for n, and let us check it forn+ 1.We then have
Gn=αn,0Γ0+αn,1Γ1+· · ·+αn,nmΓnm⇒Gn+1= (Gn∗Γm∗F−Y)(+)=
= ((αn,0Γm+αn,1Γm+1+· · ·+αn,nmΓnm+m)∗F−Y)(+)=
= (αn,0Γm∗F−Y +αn,1Γm+1∗F−Y +· · ·+αn,nmΓnm+m∗F−Y)(+)=
=αn,0(Γm∗F−Y)(+)+αn,1(Γm+1∗F−Y)(+)+· · ·+αn,nm(Γnm+m∗F−Y)(+) (since (µ+ν)(+)=µ(+)+ν(+) for any measures µ, ν)
=αn,0ΓQ.,m+αn,1ΓQ.,m+1+· · ·+αn,nmΓQ.,nm+m =
= Γ(αn,0Q.,m+αn,1Q.,m+1+· · ·+αn,nmQ.,nm+m) =
= Γ(Q.,m;Q.,m+1;Q.,m+2;. . .)·(αn,0, αn,1, . . . , αn,nm,0,0, . . .)0 (here ν0 is the transposed of the vectorν!).
=ΓQQn−1e(by induction assumption) =ΓQn+1e.
To use this result, we give a combinatorial lemma.
Lemma 2.1. Let(A,+,∆) be an algebra over a commutative field K. Let a, b∈A. Suppose that
(2.7) ab=ba=pa+qb,
wherep, q∈K. Then
(2.8) anbk=
n
X
m=1
pk,n−mam+
k
X
i=1
qn,k−ibi,
where pk,m=
k+m−1 k−1
pkqm and qn,i=
n+i−1 n−1
piqn.
Proof. Induction. Notice that p1,m = pqm, q1,i = qpi, pk,0 = pk and qn,0 =qn.Forn=k= 1,(2.8) becomesab=p1,0a+q1,0b=pa+qband this is our very assumption (2.7). Next,a2b= (pa+qb)a=pa2+qab=pa2+pqa+q2b
⇒ a3b = (pa2+pqa+q2b)a= pa3+pqa2+q2(pa+qb); by recurrence, it is clear that (2.8) holds for k= 1 and forn= 1,that is,
(2.9) anb=
n
X
m=1
pqn−mam+qnb, abk=pka+
k
X
i=1
qpk−ibi.
It follows that (2.8) holds for n= 1 or for k= 1. Suppose it holds for a pair (n, k). We shall check it for (n+ 1, k) and (n, k+ 1), and this will end the proof. We have
an+1bk =a(anbk) =a n
X
m=1
pk,n−mam+
k
X
i=1
qn,k−ibi
=
=
n
X
m=1
pk,n−mam+1+
k
X
i=1
qn,k−iabi =
=
n+1
X
m=2
pk,n−m+1am+
k
X
i=1
qn,k−i(pia+
i
X
s=1
qpi−sbs) =
=
n+1
X
m=2
pk,n−m+1am+a
k
X
i=1
qn,k−ipi+
k
X
i=1 i
X
s=1
qn,k−iqpi−sbs. Now,
qn,k−i =
n+k−i−1 n−1
pk−iqn⇒
⇒
k
X
i=1
qn,k−ipi=
k
X
i=1
n+k−i−1 n−1
pkqn =
n+k−1 n
pkqn=pk,n hence
n+1
X
m=2
pk,n−m+1am+a
k
X
i=1
qn,k−ipi =
n+1
X
m=1
pk,n−m+1am. Next, interchanging the summation order we get
k
X
i=1 i
X
s=1
qn,k−iqpi−sbs =
i
X
s=1 k
X
i=s
qn,k−iqpi−sbs=
=
i
X
s=1 k
X
i=s
n+k−i−1 n−1
pk−iqnqpi−sbs=
=
k
X
s=1
k
X
i=s
n+k−i−1 n−1
qn+1pk−sbs=
i
X
s=1
n+k−s n
qn+1,k−sbs, hence (2.8) holds in this case, too.
The fact that (2.8) holds for the pair (n, k+ 1) can be proved in the same manner.
Corollary 2.2. Let µ and ν be two conjugated distributions, to mean that
(2.10) µ∗ν =pµ+qν for somep, q≥0, p+q= 1.
Then
(2.11) µ∗n∗ν∗k=
n
X
m=1
pk,n−mµ∗m+
k
X
i=1
qn,k−iν∗i, where
pk,m=
k+m−1 k−1
pkqm= Negbin(k, p)({m}) and
qn,i=
n+i−1 n−1
piqn= Negbin(n, q)({i}).
Proof. This is an obvious particular case of Lemma 2.1.
Remark 2.4. Notice that since µ∗n∗ν∗k is a distribution, the equation below should also holds
(2.12)
n
X
m=1
pk,n−m+
k
X
i=1
qn,k−i=
= Negbin(k, p)({0,1, . . . , n−1})+Negbin(n, q)({0,1, . . . , k−1}) = 1.
This result gives us a formula for the tail of the negative binomial dis- tribution which, maybe, is well known. Not to us. Anyway, as a consequence we have
Corollary 2.3. Let µ and ν be two conjugated distributions, to mean (2.10). Suppose that µ((0,∞)) =ν((−∞,0]) = 1.Then
(2.13) (µ∗n∗ν∗k)(+)=rk,nΓ0+
n
X
m=1
pk,n−mΓm,
where Γm = µ∗m (with the convention Γ0 = δ0) and rk,n is the tail of Negbin(k, p) given by
(2.14) rk,n= Negbin(k, p)([n,∞)) = Negbin(n, q)({0,1, . . . , k−1}) (the last equality follows from (2.12)!)
Proof. We have
(µ∗n∗ν∗k)(+)= n
X
m=1
pk,n−mµ∗m+
k
X
i=1
qn,k−iν∗i
(+)
=
=
n
X
m=1
pk,n−m(µ∗m)(+)+
k
X
i=1
qn,k−i(ν∗i)(+)
(since (µ+ν)(+)=µ(+)+ν(+)) for any measuresµ, ν
=
n
X
m=1
pk,n−mµ∗m+
k
X
i=1
qn,k−iδ0=
(because µ((0,∞)) = 1⇒µ=µ+ andν((−∞,0]) = 1⇒ν(+)=δ0)
=rk,nΓ0+
n
X
m=1
pk,n−mΓm.
Now, we can give our first example of a computable Lindley process.
Proposition 2.4. Let µ and ν be two distributions. Suppose that (i)µ((0,∞)) =ν((−∞,0]) = 1;
(ii)FX =µ∗m, F−Y =ν∗k, m, k ≥1;
(iii) µand ν are conjugated, to mean (2.10).
Consider the Lindley process L= (Ln)n≥0 of parameter F =FX ∗F−Y. Then L is computable. Precisely, Ln
=D P
i≤Nn
Ui where U = (Ui)i≥1 is a se- quence of i.i.d. random variables with distribution µ and (Nn)n a Lindley process with parameter
(2.15) π =
· · · m−3 m−2 m−1 m
· · · pk,3 pk,2 pk,1 pk,0
, where pk,n = Negbin(k, p)({n}) =
k+n−1 k−1
pkqn. In terms of (2.2), Γn=µ∗n and
(2.16) Q=
rk,m rk,m+1 rk,m+2 rk,m+3 rk,m+4 · · · pk,m−1 pk,m pk,m+1 pk,m+2 pk,m+3 · · · pk,m−2 pk,m−1 pk,m pk,m+1 pk,m+2 · · ·
· · · · pk,1 pk,2 pk,3 pk,4 pk,5 · · · pk,0 pk,1 pk,2 pk,3 pk,4 · · · 0 pk,0 pk,1 pk,2 pk,3 · · · 0 0 pk,0 pk,1 pk,2 · · ·
· · · ·
,
where rk,n =pk,n+pk,n+1+pk,n+2+· · ·.
Proof. This is an immediate consequence of our theorem above.
Corollary 2.5. A Lindley process L = (Ln)n≥0 of parameter F = FX ∗F−Y is computable if
(i)FX = Gamma(m, a) and FY = Gamma(k, b).Here, Γ from(2.2) has the components Γn= Gamma(n, a) and Q is the matrix (2.16)with p= a+bb ,
q = a+ba .The limit distribution G∞ =G∞(m, a;k, b) does exist iff ma < kb ⇔ bm < ak.
(ii)FX = Geom(m, α)and FY = Negbin(k, β).Here,Γfrom(2.2)has the components Γn= Geom(n, α) and Q is the matrix (2.16) with p = α+β−αββ , q = α+β−αβα−αβ .The limit distribution G∞ =G∞(m, α;k, β) does exist iff mβ <
kα(1−β).
Proof. (i). We know [4] that if µ= Exp(a), ν = Exp(−b),thenµ∗ν =
b
a+bµ+a+ba ν.AsFX =µ∗m and F−Y =ν∗k, we can apply Proposition 2.4. It is well known that the limit distribution does exist iff EXn<EYn⇔ ma < kb.
(ii). It was proved in [5] that if µ = Geom(1, α), ν = Negbin(1,−β), then µ∗ν = α+β−αββ µ+α+β−αβα−αβ ν. The same proof as before: FX =µ∗m and F−Y =ν∗k.As EXn= mα and EYn= k(1−β)β , G∞does exist iff mα < k(1−β)β . In other words, we arrived at the conclusion that if (Xn)n are i.i.d.
distributed as Gamma(m, a),(Yn)nare i.i.d. distributed as Gamma(k, b),Zn= Xn−Yn, S0 = 0, Sn =Z1+Z2+· · ·+Zn for n≥ 1, then we can compute the distribution of Ln = max(S0, S1, . . . , Sn). It can be represented as Ln =
P
j≤Nn
ξj, whereξj ∼Exp(a) and (Nn)nis another Lindley process of parameter Negbin(k,−p)∗δm with p= a+bb .What does happen if Yn is constant? This is our next example.
Proposition 2.6. Suppose that Xn ∼ Gamma(m, λ) and Yn = b = constant. Then L is a computable Lindley process. Precisely, Ln =D P
i≤Nn
Ui, where U = (Ui)i≥1 is a sequence of i.i.d. random variables distributed as Exp(λ) and (Nn)n is a Lindley process independent of U with parameter (2.17) π=
· · · m−3 m−2 m−1 m
· · · pbλ,3 pbλ,2 pbλ,1 pbλ,0
=δm∗Poisson(−bλ), where pa,n = Poisson(a)({n}) = an!ne−a.The limit distribution G∞ does exist iff bλ > m.
Proof. In order to apply our theorem, we notice that FX = Γm, where Γn = Gamma(n, λ) = Exp(λ)∗n. We have to compute (Γn∗F−Y)(+) = (Γn∗ δ−b)(+)in order to check (2.5). LetU ∼Exp(λ) andξ =U−b.The distribution of ξisF := Γ1∗δ−b.The density ofξ isf(x) =λe−λ(b+x)1(−b,∞)(x).We write F as a sum of two measures, the first concentrated on (0,∞) and the second on (−∞,0).Precisely, write F =µ+ν withµ, ν absolutely continuous such that dµ(x) = f(x)1(0,∞)(x)dx and dν(x) = f(x)1(−b,0)(x)dx. Thus, µ = e−λbΓ1
and supp(ν) = [−b,0].We claim that
(2.18) Γn∗ν=pbλ,nΓ1+pbλ,n−1Γ2+· · ·+pbλ,1Γn+νn,
whereνn is a measure concentrated on [−b,0].If this holds, then we can write Γn∗δ−b = Γn−1∗(Γ1∗δ−b) = Γn−1∗(µ+ν) =
= Γn−1∗(e−λbΓ1+ν) =pbλ,0Γn+ Γn−1∗ν=
=pbλ,0Γn+pbλ,1Γn−1+· · ·+pbλ,n−1Γ1+νn, hence
(2.19) (Γn∗F−Y)(+)=rbλ,nΓ0+pbλ,n−1Γ1+· · ·+pbλ,1Γn−1+pbλ,0Γn, where rλb,n = Poisson(λb)([n,∞)). By (2.6) the matrix Q from (2.2) would have the columns (Q.,m;Q.,m+1;. . .), where the transposed of Q.,n is (rλb,n, pλb,n−1, . . . , pλb,1, pλb,0,0,0, . . .).So, in order to complete the proof we have to check (2.18). We compute the density of Γn∗ν :
fΓn∗ν(x) = Z
fν(x−y)fΓn(y)dy
(fν is the density of ν and fΓn(x) = λ(n−1)!nxn−1e−λx1(0,∞)(x) is the density of Γn= Gamma(n, λ))
= Z
λe−λ(x−y+b)1(−b,0)(x−y)λnyn−1
(n−1)!e−λy1(0,∞)(y)dy=
=
b+x
Z
max(x,0)
λe−λ(x+b)λnyn−1
(n−1)!dy=λe−λ(x+b)
λ(b+x)
Z
max(λx,0)
yn−1 (n−1)!dy=
=λn+1e−λ(x+b)(b+x)n−(max(x,0))n
n! .
Thus, if x ∈ [−b,0] then fΓn∗ν (x) = λn+1e−λ(x+b) (b+x)n! n. This is the density of νn.
Ifx >0 then
fΓn∗ν(x) =λn+1e−λ(x+b)(b+x)n−xn
n! =λn+1e−λb
n−1
P
k=0
n!
k!(n−k)!xkbn−k
n! e−λx=
=
n−1
X
k=0
λk+1xke−λx k!
(λb)n−ke−λb (n−k)! =
n−1
X
k=0
fΓk+1(x)pλb,n−k⇒
⇒Γn∗ν =
n−1
P
k=0
pλb,n−kΓk+1+νn. A slight generalization is
Corollary 2.7. Suppose that Xn∼Gamma(m, λ) and Yn∼
b1 b2 · · · bk β1 β2 · · · βk
with 0 < b1 < · · · < bk. Then L is a computable Lindley process. Precisely, Ln
=D P
i≤Nn
Ui where U = (Ui)i≥1 is a sequence of i.i.d. random variables distributed as Exp(λ) and (Nn)n is a Lindley process independent of U with parameter
(2.20) π =
k
X
i=1
βiπi with πi =
· · · m−3 m−2 m−1 m
· · · pbiλ,3 pbiλ,2 pbiλ,1 pbiλ,0
, where pa,n= Poisson(a)({n}) = an!ne−a.
The limit distribution G∞ does exist iff λ(β1b1+· · ·+βkbk)> m.
Proof.We know from (2.19) that
(Γn∗δ−b)(+)=rbλ,nΓ0+pbλ,n−1Γ1+· · ·+pbλ,1Γn−1+pbλ,0Γn. This means that
(Γn∗F−Y)(+)=
k
X
i=1
βiΓn∗δ−bi=
k
X
i=1
βi(rλbi,nΓ0+pλbi,n−1Γ1+· · ·+pλbi,0Γn).
Applying our theorem completes the proof.
Remark 2.5. What does happen if we interchange X and Y? Precisely, what if Xn = a is constant while the Yn are distributed as Gamma(m, λ)?
Well, in this casehFX, FYiisnota computable Lindley process. Our theorem cannot be applied sinceFX =µ∗mfor some distributionµand positive integer m. Then µcan only be a Dirac measureµ=δh and in this case (2.5) cannot hold. But the reader can verify that an equation of the form (2.2) is manifestly excluded. However, in the particular case where m = 1 (thus Yn ∼ Exp(λ)) we are in the framework of the classical Cram`er-Lundberg model (see, for instance, [8] or [1]). Even if we still are not able tocomputethe distributions Gn,we know their limit distribution G∞: the Hinˇcin-Pollaczek formula (see, for instance [2] or [1]) yields
(2.21) G∞=
∞
X
n=0
(1−ρ)ρn(FX)∗nI , where ρ = EXEYn
n = λa < 1, and (FX)I is the integrated tail distribution de- fined by
(2.22) (FX)I(x) =
Rx
0 FX(y)dy R∞
0 FX(y)dy.
Since in our case it happens that (FX)I = Uniform(0, a), we can compute G∞(x) with reasonable precision (using the well-known formula for convolu- tions of uniform distributions:
Uniform(0, a)∗n(x) = xn−Cn1(x−a)n++Cn2(x−2a)n++Cn3(x−3a)n++· · ·
n!an ).
3. COMPARISON OF LINDLEY PROCESSES
Let L = (Ln)n≥0 and L0 = (L0n)n≥0 be two Lindley processes. Assume that the first one has parameter F while the second one has parameter F0 and, moreover, that the expectations of F and F0 are negative. In this case the limits L∞ and L0∞ exist in distribution. We denote by “F ≺++ F0” the fact that Ln ≺st L0n ∀n ≥ 1 and by “F ≺+ F0” the fact that L∞ ≺st L0∞. We may call the first domination the strong one and the second the weak one. ThusL isstrongly dominated byL0 iffLn≺stL0n∀n≥1 andweakly dominated byL0 iff L∞≺stL0∞.
In this very general situation, the only assertion that we can made is the trivial remark below.
Proposition 3.1. F ≺stF0 ⇒F ≺++F0 ⇒F ≺+F0.
Proof. Obvious. If Gn and G0n are the distributions of Ln and L0n, G∞
and G0∞ are the distributions of L∞ and L0∞, then we see thatF ≺st F0 ⇒ F(+) ≺st F(+)0 ⇔ G1 ≺st G01 ⇒ G1 ∗ F ≺st G01 ∗ F0 ⇒ (G1 ∗F)(+) ≺st (G01∗F0)(+)⇒G2 ≺stG02 ⇒ · · ·.
AsGn⇒G∞,andG0n⇒G0∞,the second implication is obvious, too.
Suppose that, moreover, both Land L0 are computable andGn and G0n are the distributions of Ln and L0n. According to (2.2), we have Gn =ΓQne and G0n=Γ0Q0ne, where Γm =µ∗m, Γ0m =µ0∗m, µand µ0 are distributions on (0,∞) andQ,Q0 are column stochastic matrices given by the distributionsπ and π0.Now, we can say more, even in a slightly more general context.
Definitions. 1. Let ∆= (∆n)n≥0 be a sequence of distributions on the real line.We say that ∆is monotone if ∆n≺st∆n+1∀n≥0.
2. Let Q be a column-stochastic infinitely dimensional matrix. We say that Q is monotone if Q.,n ≺st Q.,n+1, where Q.,n is the nth column of Q, considered as a probability distribution on the set of non-negative integers.
Lemma 3.2. Let ∆,∆0 be sequence of distributions on the real line and Q,Q0 infinitely dimensional column-stochastic matrices. Denote by ∆≺st∆ the fact that ∆n ≺st ∆0n ∀n ≥0 and by Q ≺st Q the fact that Q.,n ≺st Q0.,n
∀n≥0. Then
(i) if ∆ and Q are monotone then ∆Q is monotone, too;
(ii) if ∆,∆0and Q,Q0 are monotone and ∆≺st∆0,Q≺st Q0,then ∆Q
≺st∆0Q0.
Proof. (i) We have (∆Q)n= ∆0Q0,n+ ∆1Q1,n+ ∆2Q2,n+· · · . The tail of this distribution is
(3.1) (∆Q)n(x) = ∆0(x)Q0,n+ ∆1(x)Q1,n+ ∆2(x)Q2,n+· · · ,
where the ∆k(x) are the tails of the distributions ∆k.As we supposed that∆ is monotone, the sequences (∆k(x))k is non-decreasing, by the very definition of stochastic domination. On the other hand, the tail of the distribution (∆Q)n+1 is
(3.2) (∆Q)n+1(x) = ∆0(x)Q0,n+1+ ∆1(x)Q1,n+1+ ∆2(x)Q2,n+1+· · · . But we know thatQ.,n≺stQ.,n+1.This means that
∞
P
k=0
akQk,n ≤
∞
P
k=0
akQk,n+1 for any non-decreasing bounded sequence (ak)k≥0 while (∆k(x))k≥0 is exactly such kind of sequence.
ii) The tail of (∆Q)n computed at some real x is
∞
P
k=0
∆k(x)Qk,n while the tail of (∆0Q0)n computed at some real x is
∞
P
k=0
∆0k(x)Q0k,n. As ∆≺st ∆0, we have
∞
X
k=0
∆k(x)Qk,n≤
∞
X
k=0
∆0k(x)Qk,n ≤
∞
X
k=0
∆0k(x)Q0k,n. The last inequality holds since Q≺st Q0.
Here is a somewhat more general result that we need for computable Lindley processes.
Proposition 3.3. Suppose that (Ln)n≥0 and (L0n)n≥0 are two sequences of random variables. Let Gn and G0n be their distributions. Suppose that Gn=ΓQneand G0n=Γ0Q0neand
(i)Γ and Γ0 are monotone sequences of distributions on the real line;
(ii)Q and Q0 are monotone column-stochastic infinite dimensional ma- trices;
(iii) Γ≺stΓ0; (iv)Q≺stQ0; Then Gn ≺stG0n ∀n.
Proof. Let ∆n=ΓQn and ∆0n =Γ0(Q0)n. According to Lemma 3.2 (i),
∆nand ∆0nare monotone for every n.We can check by induction that ∆n≺st
∆0n ∀n≥0. For n= 0 this is true due to our assumption (iii). Suppose now
that ∆n≺st∆0n and let us prove that ∆n+1 ≺st∆0n+1.But ∆n+1= ∆nQ and
∆0n+1= ∆0nQ0.The claim follows now using Lemma 3.2 (ii).
Corollary 3.4. Let L and L0 be computable Lindley processes with characteristics µ, π and µ0, π0.Suppose that µ≺stµ0 and π ≺st π0.Then L is strongly dominated by L0.
Proof. LetGn, G0n be the distributions ofLn andL0n.ThenGn=ΓQne andG0n=Γ0Q0ne,Γn=µ∗n, Γ0n= (µ0)∗nwhile the column-stochastic matrices Q and Q0 are linked to the distributions π and π0 as follows: if ξ and ξ0 are random variables such that ξ∼π andξ0 ∼π0, then Q.,n is the distribution of (ξ+n)+and Q0.,nis the distribution of (ξ0+n)+.We are under the conditions of the previous proposition: µ ≺st µ0 ⇒ µ∗n ≺st (µ0)∗n ⇔ Γn ≺st Γ0n ∀n and π ≺st π0can be written in terms of random variables as ξ ≺stξ0, thus making obvious the relations (ξ +n)+ ≺st (ξ0+n)+ ∀n ≥ 0 (i.e., Q ≺st Q0 !) and (ξ+n)+ ≺st (ξ +n+ 1)+,(ξ0+n)+ ≺st (ξ0+n+ 1)+ ∀n (meaning that Q and Q0 are monotone). The fact that Γ and Γ0 are monotone is a simple consequence of the fact that µ and µ0 are distributions on (0,∞).So, we can apply Proposition 3.3. to conclude that L is strongly dominated by L0.
Now, we want to apply this result to our examples from Corollary 2.5 and Proposition 2.6.
Let L and L0 be two computable Lindley processes with characteristics µ, π and µ0, π0. We shall investigate the various implications between the fol- lowing assertions:
A.µ≺stµ0, π ≺st π0;
B.Gn ≺stG0n∀n(i.e. F ≺++F0);
C.FX ≺stFX0 andρ≤ρ0 (ρ and ρ0 are the traffic intensities);
D.G1 ≺stG01;
E.G∞ ≺stG0∞ (i.e.,F ≺F0).
We begin with processes from Corollary 2.5. The assumption from (i) are:
FX = Gamma(m, a), FY = Gamma(k, b), FX0 = Gamma(m0, a0), FY0 = Gamma(k0, b0) while the ones from (ii) are
FX = Geom(m, α), FY = Negbin(k, β), FX0 = Geom(m0, α0), FY0 = Negbin(k0, β0).
For short, in the first caseL is of type (m, k,Gamma) andL0 is of type (m0, k0,Gamma) while in the second case L is of type (m, k,Geom) and L0 is of type (m0, k0,Geom).
In the first case
(3.3) ρ= EX EY = mb
ka, ρ0 = EX0
EY0 = m0b0
k0a0, p= b
a+b, p0 = b0 a0+b0
