Applied Probability Trust (17 March 2004)

A SECOND ORDER MARKOV MODULATED FLUID QUEUE WITH LINEAR SERVICE RATE LANDY RABEHASAINA, BRUNO SERICOLA,∗ IRISA-INRIA

Abstract We consider an infinite capacity second order fluid queue governed by a continuous time Markov chain and with linear service rate. The variability of the traffic is modeled by a Brownian motion and a local variance function modulated by the Markov chain and proportional to the fluid level in the queue. The behavior of this second order fluid flow model is described by a linear stochastic differential equation, satisfied by the transient queue level. We study the transient level’s convergence in distribution under weak assumptions and we obtain an expression of the stationary queue level. For the first order case, we give a simple expression of all its moments as well as of its Laplace transform. For the second order model we compute its two first moments. Keywords: Fluid queue; Markov chain; Second order model; Risk theory AMS 2000 Subject Classification: Primary 60K25; 60J27 Secondary 90B05

1. Introduction and motivations We consider in this paper a infinite capacity fluid queue of which level at time t is denoted by Q(t). Fluid arrives into this queue according to a nondecreasing process A(t), and leaves the queue at rate µt at time t. We consider a second order model. This means that the fluid level is not only represented by the incoming fluid and the service rate, but also by a variability factor, which is modeled by a local variance function σt and a Brownian motion Bt . In that case Q(t) satisfies the following stochastic differential equation dQ(t) = ∗

dA(t) − µt dt + σt dBt + dLt

Postal address: Campus de Beaulieu, 35042 RENNES, FRANCE

1

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L. Rabehasaina, B. Sericola

where Lt is a nondecreasing process, interfering only when Q(t) hits 0 and preventing it from being negative. In practise, we consider the case where dA(t) = λ(X(t), Q(t))dt, µt = µ(X(t), Q(t)) and σt = σ(X(t), Q(t)) where X(t) is an external stochastic process modulating the queue and also called the environment process. In that case, Q(t) is solution to the stochastic differential equation reflected at 0 dQ(t) = (λ(X(t), Q(t)) − µ(X(t), Q(t)))dt + σ(X(t), Q(t))dBt + dLt . This model was introduced in [8], and was studied already extensively with different hypotheses on X(t) and on the functions λ and µ, particularly when X(t) is a continuous time Markov chain: see e.g. Karandikar and Kulkarni [4] for a study where λ and µ do not depend on Q(t), or Chen, Hong and Trivedi [2] for a set of differential equations satisfied by the density of Q(t). Kella and Stadje [5] obtain the explicit distribution of the steady state distribution of Q(t) when X(t) is a two state Markov chain and σ = 0. Our main study is focused on the linear model. More precisely, the service rate at time t is µ(X(t))Q(t) (modulated linear release rate), where of course µ is assumed to be nonnegative. We will also suppose that σ(X(t), Q(t)) = σ(X(t))Q(t). This model has been studied by Asmussen and Kella [1], Kella and Whitt [7] and Kella and Stadje [6] (with a network background in the two latter cases), when A is a Poisson process or a Markov modulated L´evy process, and when σ = 0 (first order model). The authors identify in [6] a functional equation that the steady-state Laplace-Stieljes transform must satisfy. As pointed out by the authors, even though it is not clear how to solve this functional equation, it is used to compute the two first moments. We consider in the present paper dA(t) of the form λ(X(t))dt, where {X(t),t ∈ R} is a stationary continuous time Markov chain and we obtain explicit analytical results for all order moments as well as for the Laplace transform of the stationary distribution when σ = 0. When σ is a function non identical to 0 then it is much harder to get the distribution of the stationary regime: we however obtain its two first moments. It is not then difficult to see that Q(t) verifies the following stochastic differential equation dQ(t) = λ(X(t))dt − µ(X(t))Q(t)dt + σ(X(t))Q(t)dBt ,

(1.1)

fluid queue with linear service rate

3

the term in dLt having disappeared, 0 being then an impenetrable barrier for Q(t) (see e.g. [1]). Due to its linearity, this model has some application in finance. Let us see the connection between finance and the model described by (1.1) by considering an insurance risk model, similar to the one in [1]. More precisely, let us consider a Markov chain {I(t),t ∈ R} and an insurance risk process {R(t),t ∈ R} with interest rate ρ(I(t)) and volatility s(I(t)) at time t. Let J(t0 ) − J(t) =

Z

t0

ν(I(s))ds t

be the total claim in the period [t, t0 ], where ν is a non negative function. R(t) then satisfies the linear stochastic differential equation (with the notations traditionally used in finance) dR(t)

= −dJ(t) + R(t)[ρ(I(t))dt + s(I(t))dw(t)] = −ν(I(t))dt + ρ(I(t))R(t)dt + s(I(t))R(t)dw(t)

(1.2)

where {w(t),t ∈ R} is a Brownian motion. {I(t),t ∈ R} typically captures the market behavior: for example its state space is {−1, +1} where +1 represents an up-trend of the market, whereas −1 means a down-trend. The state space can be even larger, leading to a more precise description of the market trend (see [13]). Typically, the volatility is small during an up-trend period, as investors are cautious and move slowly, and it increases during a down-trend period, as investors get panicked and the market becomes more erratic (see again [13]). Let R∗ (t) := R(−t) be the reversed process of R (also called the dual process). From (1.2), R∗ satisfies dR∗ (t) = ν(I ∗ (t))dt − ρ(I ∗ (t))R∗ (t)dt + s(I ∗ (t))R∗ (t) ◦ dw∗ (t)

(1.3)

where I ∗ and w∗ are respectively the reversed Markov chain and Brownian motion defined by I ∗ (t) := I(−t) and w∗ (t) := w(−t). s(I ∗ (t))R∗ (t) ◦ dw∗ (t) is the Stratonovitch integral, linked to the usual Ito integral via the equality s(I ∗ (t))R∗ (t) ◦ dw∗ (t) = s(I ∗ (t))2 R∗ (t)dt + s(I ∗ (t))R∗ (t)dw∗ (t)

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L. Rabehasaina, B. Sericola

(see e.g. [12]). Thus (1.3) can be rewritten in the following way: dR∗ (t) = ν(I ∗ (t))dt − [ρ(I ∗ (t)) − s(I ∗ (t))2 ]R∗ (t)dt + s(I ∗ (t))R∗ (t)dw∗ (t).

(1.4)

Note then that, provided that ρ − s2 ≥ 0, (1.4) is exactly (1.1) with λ = ν, µ = ρ − s2 , s = σ, X = I ∗ , B = w∗ and Q = R∗ . Now the fundamental duality relation between R∗ and R is the following: let us set τ (x) := inf{t ≥ 0 | R(t) = 0} the ruin time of the process R, where R(0) = x, and let W be a random variable towards which R∗ (t) converges in distribution as t → +∞ (which, with the notations of the present paper, and because of the matching correspondence Q = R∗ in (1.4) and (1.1), amounts to Q(t) converging in distribution to W ), then the relation between τ (x) and W is given by (see [1] and [12]) P (τ (x) < ∞) = P (W > x). In other words, there is a close connection between the probability of ruin P (τ (x) < ∞) and the distribution of W . This paper is then dedicated to giving some characteristics of W , when the volatility s (= σ) is equal to 0 or when it is not identically equal to 0. Throughout this paper, we will suppose that {X(t),t ∈ R} is a stationary ergodic continuous time Markov chain on a finite state space S = {1, ..., N } with stationary distribution π = (π1 , ..., πN ). We will denote by Q = (qij )(i,j)∈S×S its generator matrix. {X(t),t ∈ R} and {Bt ,t ∈ R} will be taken independent. Let us remember that λ is nonnegative. µ is also non negative, and verifies µ(i) > 0 for an i ∈ S (the function µ is then not identically equal to 0). The paper is organized as follows. In Section 2 we show, under suitable assumptions, the convergence in distribution of the queue level. In Section 3, we give an expression for the first moment of the stationary distribution. In Section 4 we study the stationary regime when σ is the function identically equal to 0. We give explicitly the moments of all orders of the stationary distribution as well as of its Laplace transform. To finish, in Section 5 we compute the second moment of the stationary distribution when σ is not necessarily identically equal to 0, under some mild assumption relating µ and σ. We recall a lemma about exponential martingales (see [11]) that will be used throughout the paper:

fluid queue with linear service rate

5

Lemma 1.1. Let {Z(t), t ∈ R} and {w(t), t ∈ R} be respectively a stationary process and a brownian motion. Let us suppose that the two processes are independent. Then for all bounded function g and all u ∈ R, the process    Z t Z 1 t 2 g(Z(v))dw(v) − g(Z(v)) dv , t ≥ u exp 2 u u is a martingale given {Z(t), t ∈ R} adapted to the filtration (Ft )t∈R defined by Ft = σ(w(s) − w(u), u ≤ s ≤ t). Besides, for all t ≥ u,    Z t Z 1 t g(Z(v))dw(v) − g(Z(v))2 dv Z = 1. E exp 2 u u 2. Stationary regime of the queue Let us then denote by {Qyu (t), t ≥ u} the process solving (1.1) for t ≥ u and such that Qyu (u) = y. It is then standard (See e.g. [11]) that one has an explicit solution for Qyu (t): for t ≥ u we have

 Z t  Z t y exp − [µ(X(s)) + σ(X(s))2 /2]ds + σ(X(s))dBs u u  Z t  Z t Z t exp − [µ(X(v)) + σ(X(v))2 /2]dv + σ(X(v))dBv λ(X(s))ds.(2.1) +

Qyu (t)

=

u

s

s

Besides, the relation between Qyu (t) and Qyu (t0 ), t ≥ t0 ≥ u, is  Z t  Z t [µ(X(s)) + σ(X(s))2 /2]ds + σ(X(s))dBs Qyu (t) = Qyu (t0 ) exp − t0 t0  Z t  Z t Z t exp − [µ(X(v)) + σ(X(v))2 /2]dv + σ(X(v))dBv λ(X(s))ds.(2.2) + t0

s

s

Out of legibility, we will write, since y will be fixed throughout the paper, Q(t) := Qy0 (t). We begin by finding a stationary process W (t) solving (1.1). Theorem 2.1. Let us set  Z t  Z t Z t 2 exp − [µ(X(v)) + σ(X(v)) /2]dv + σ(X(v))dBv λ(X(s))ds. W (t) = −∞

s

s

(2.3) Then

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1. W (t) is finite for all t. 2. {W (t), t ∈ R} is a stationary process solving (1.1). 3. Q(t) converges in distribution to W (0), independently of the initial condition. Proof. Let us begin by showing the first point with t = 0 without loss of generality. The integral (2.3) lies in [0, +∞], and we only need to prove that it is finite almost surely. For each u ≤ t we have  Z 0 Z 0 Z exp − [µ(X(v)) + σ(X(v))2 /2]dv + E u

Z

0

= u

s

0 s

  σ(X(v))dBv λ(X(s))ds (2.4)

  Z 0 Z E exp − [µ(X(v)) + σ(X(v))2 /2]dv + Z

≤ sup λ(i) i∈S

0

u

s

s

s

s

s

s



 Z µ(X(v))dv E exp

s

s

0



0 s

σ(X(v))dBv

ds.



0

σ(X(v))dBv

   Z 0 Z 2 [µ(X(v)) + σ(X(v)) /2]dv + = E E exp − 0

  σ(X(v))dBv λ(X(s)) ds

  Z 0 Z 2 E exp − [µ(X(v)) + σ(X(v)) /2]dv +

Now,   Z 0 Z [µ(X(v)) + σ(X(v))2 /2]dv + E exp −

  Z = E exp −

0

s

0

  σ(X(v))dBv X

σ(X(v))dBv −

Z

0 s

  σ(X(v)) /2dv X . 2

By Lemma 1.1 with w = B, Z = X and g = σ, the process    Z t Z t 2 σ(X(v))dBv − σ(X(v)) /2dv , t ≥ s exp s

s

is a martingale given X, hence  Z 0 Z σ(X(v))dBv − E exp s

s

0

  σ(X(v))2 /2dv X = 1.

In order to show that (2.4) admits a finite limit as u → −∞, it is then sufficient to show that   Z 0  1 µ(X(v))dv > 0. α = lim inf ln E exp − s→−∞ s s

(2.5)

Since µ(X(v)) ≥ 0 we already know that α ≥ 0. Now if α was equal to 0 we R0 would have that s µ(X(v))dv = 0 almost surely for all s ≤ 0, which is impossible because of µ(i) > 0 for an i in S and the positive recurrence of X. This implies that

fluid queue with linear service rate

  Z E exp −

7



0

≤ K exp(αs) for some constant K. Thus (2.4) admits a

µ(X(v))dv

s

finite limit as u → −∞. This in particular implies that W (0) < +∞ almost surely. The second point is proved as in Proposition 3.3 of [9]. Let us first notice that, because X is stationary and B has stationary increments, Q0u (t) has the same distribution as Q0u−t (0) for t ≥ u. Those two random variable converge respectively to W (t) and W (0) as u → −∞. Thus W (t) and W (0) are identical in distribution. To see that {W (t), t ∈ R} solves (1.1), we write that, in view of (2.2), for t ≥ 0,  Z t  Z t 0 0 [µ(X(s)) + σ(X(s))2 /2]ds + σ(X(s))dBs Qu (t) = Qu (0) exp − 0 0  Z t  Z t Z t exp − [µ(X(v)) + σ(X(v))2 /2]dv + σ(X(v))dBv λ(X(s))ds + 0

0

0

and we let u → −∞. Finally, we prove that {Qy0 (t), t ≥ 0} converges in distribution to W (0) for all y ≥ 0 as t tends to infinity. We saw that Q0−t (0) has the same distribution as Q00 (t) for t ≥ 0, hence Q00 (t) converges in distribution to W (0). To see that Qy0 (t) converges in distribution to W (0) we write that, thanks to (2.1), for t ≥ 0,  Z t  Z t y 0 2 [µ(X(s)) + σ(X(s)) /2]ds + σ(X(s))dBs . Q0 (t) − Q0 (t) = y exp − 0

0

  Z t  µ(X(v))dv . SimThen E(|Qy0 (t) − Q00 (t)|) = E(Qy0 (t) − Q00 (t)) = yE exp − 0

ilarly as in the first point, this expression tends to 0 as t → +∞ exponentially fast because of the positive recurrence of X. Thus Qy0 (t) − Q00 (t) tends to 0 in L1 (Ω) as t → +∞. The convergence in distribution of Qy0 (t) towards W (0) follows. Let us remark that W (0) can also be expressed in the following way:  Z s  Z s Z ∞ exp − [µ(X ∗ (v)) + σ(X ∗ (v))2 /2]dv + σ(X ∗ (v))dBv∗ λ(X ∗ (s))ds W (0) = 0

0

0

(2.6) where X ∗ and B ∗ are the reversed versions of the processes X and B, namely for all t, X ∗ (t) = X((−t)− ) and Bt∗ = B−t . Obviously, B ∗ is still a Brownian motion. Moreover, it is standard that X ∗ is a continuous time Markov chain of transition matrix ∗ ∗ )(i,j)∈S×S where qij = qji πj /πi . Q∗ = (qij

For clarity purpose, we will often use the notation W instead of W (0).

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3. First moment of the stationary distribution We are interested in this section in the first moment of the queue level in stationary regime. Note that in Section 4 we get the same expression by another method for the first order case (i.e. σ = 0). For some similar results concerning the first moment of the stationary distribution for first order models see also [1] and [6]. Actually the result obtained in this section is slightly more general than the computation of the first moment. Moreover the originality of this section is the martingale approach used to prove the result (as opposed to the method in [1] and [6], where moments are computed by using a Markov renewal equation). Besides, a similar approach to the one presented here, although much more subtle, will be used in the computation of the second moment in Section 5. Let us set, for t ≥ 0, Z t Z t ∗ ∗ 2 [µ(X (s)) + σ(X (s)) /2]ds + σ(X ∗ (s))dBs∗ , Yt = Y0 − 0

0

where Y0 is a bounded random variable. We then have the following result, which also yields the expression of E(W ): Theorem 3.1. For all h : S −→ R, −1

E(W h(X(0))) = πH (Dµ − Q∗ )

Λ1

(3.1)

where H = diag (h(1), ..., h(N )), Dµ = diag (µ(1), ..., µ(N )), Λ = diag (λ(1), ..., λ(N )) and 1 = (1, ..., 1)0 . In particular, if h(i) = 1 for all i ∈ S we get the expression of the first moment of W : E(W ) = π (Dµ − Q∗ )

−1

Λ1

(3.2)

Proof. Let us set g(y, i) = exp(y)f (i), y ∈ R, i ∈ S, where f will be conveniently chosen later on. By Ito’s formula, Z t Z t ∗ ∗ ∗ Ag(Ys , X (s))ds = σ(X ∗ (s)) ∂y g(Ys , X ∗ (s))dBs∗ g(Yt , X (t)) − g(Y0 , X (0)) − 0

where Ag(y, i) = =

0

  X 1 1 2 ∗ qij g(y, j) − µ(i) + σ(i) ∂y g(y, i) + σ(i)2 ∂y2 g(y, i) + 2 2 j X ∗ qij f (j). −µ(i) exp(y)f (i) + exp(y) j

fluid queue with linear service rate

9

Let us then set Mt

Z t = g(Yt , X ∗ (t)) − g(Y0 , X ∗ (0)) − Ag(Ys , X ∗ (s))ds 0 Z t σ(X ∗ (s)) ∂y g(Ys , X ∗ (s))dBs∗ . = 0

In general, {Mt , t ≥ 0} is a local martingale given X ∗ , adapted to Ft = σ(Bs∗ −B0∗ , 0 ≤ s ≤ t). In the present case, let us show that Mt is actually a real martingale. Since ∂y g(y, i) = exp(y)f (i), we have Z t σ(X ∗ (s)) ∂y g(Ys , X ∗ (s))dBs∗ Mt = 0 Z t = σ(X ∗ (s)) exp(Ys )f (X ∗ (s))dBs∗ 0

 Z s  Z s with exp(Ys ) = exp Y0− [µ(X ∗ (v)) + σ(X ∗ (v))2 /2]dv + σ(X ∗ (v))dBv∗ . Since µ, 0

0

f and Y0 are bounded, it is easy to verify that Mt is a martingale if  Z t  Z t σ(X ∗ (s))2 /2ds + σ(X ∗ (s))dBs∗ Nt := exp − 0

0

admits a finite moment of order 2. It is easy to check that Z t  σ(X ∗ (s))2 ds Pt Nt2 = exp 0

 Z t  Z t ∗ 2 ∗ ∗ where Pt := exp − 2σ(X (s)) ds + 2σ(X (s))dBs . From Lemma 1.1 with 0

0

Z = X ∗ , w = B ∗ and g = 2σ, Pt is a martingale given X ∗ adapted to Ft . Thus   Z t   2 2 ∗ ∗ 2 ∗ σ(X (s)) ds Pt |X E(Nt ) = E(E(Nt |X )) = E E exp 0    Z t ∗ 2 ∗ σ(X (s)) ds E (Pt |X ) = E exp 0   Z t σ(X ∗ (s))2 ds = E exp 0

which is finite (Remember that σ takes its values in the finite set {σ(1), ..., σ(N )}.) Mt is then a real martingale. Now with the expression of g one can write Ag in the following way: Ag(y, i) = exp(y)[(Q∗ − Dµ )f ](i) where we likewise used the notation f = (f (1), ..., f (N ))0 , and where for all column vector v = (v1 , ..., vN )0 , we set v(i) := vi .

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Now since (Mt )t≥0 is a martingale given X ∗ , we have that E(Mt |X ∗ (0)) = 0. Hence we get  Z t ∗ ∗ ∗ exp(Ys )[(Q − Dµ )f ](X (s))ds h(X (0)) E 0   Z t ∗ ∗ ∗ ∗ exp(Ys )[(Q − Dµ )f ](X (s))ds h(X (0)) X (0) =E E 0    Z t ∗ ∗ ∗ ∗ exp(Ys )[(Q − Dµ )f ](X (s))ds X (0) h(X (0)) =E E 0    Z t Ag(Ys , X ∗ (s))ds X ∗ (0) h(X ∗ (0)) =E E 0   = E − E(exp(Y0 )f (X ∗ (0))|X ∗ (0)) h(X ∗ (0)) + E(exp(Yt )f (X ∗ (t))|X ∗ (0)) h(X ∗ (0))   = E − E(exp(Y0 )f (X ∗ (0)) h(X ∗ (0))|X ∗ (0)) + E(exp(Yt )f (X ∗ (t)) h(X ∗ (0))|X ∗ (0))     (3.3) = −E exp(Y0 )f (X ∗ (0)) h(X ∗ (0)) + E exp(Yt )f (X ∗ (t)) h(X ∗ (0)) . The matrix Dµ is diagonal with at least one non-zero entry and, the matrix Q∗ is irreducible because Q is irreducible by hypothesis, so Q∗ − Dµ is invertible. We now take Y0 = 0 and f = (Q∗ − Dµ )−1 Λ1. Then (3.3) reads  Z s   Z s exp − [µ(X ∗ (v)) + σ(X ∗ (v))2 /2]dv + σ(X ∗ (v))dBv∗ λ(X ∗ (s))ds h(X ∗ (0)) 0 0 0  = −E [(Q∗ − Dµ )−1 Λ1](X ∗ (0)) h(X ∗ (0))   Z t [µ(X ∗ (s)) + σ(X ∗ (s))2 /2]ds +E exp − 0   Z t ∗ ∗ ∗ −1 ∗ ∗ σ(X (s))dBs [(Q − Dµ ) Λ1](X (t)) h(X (0)) . (3.4) + Z

t

E

0

Using an argument similar to the one used to show that (2.4) tends to a finite limit as u tends to −∞, it is easy to show that the last term in the righthandside of the equality (3.4) converges to 0 as t tends to infinity. Since the lefthandside of (3.4) tends to E(W h(X ∗ (0))) = E(W h(X(0))) as t tends to infinity, by letting t → +∞ in (3.4) we then get  E(W h(X(0))) = E [(Dµ − Q∗ )−1 Λ1](X ∗ (0)) h(X ∗ (0))  = E [(Dµ − Q∗ )−1 Λ1](X(0)) h(X(0)) which can be written in matrix form like (3.1).

fluid queue with linear service rate

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4. Case of a first order model, stationary regime We study in this section the stationary regime of the queue, still in the first order model framework. From Theorem 2.1, Q(t) converges in distribution to W . Let us set F (x) := lim F (t, x) t→∞

where F (t, x) is defined by F (t, x) = (F1 (t, x), ..., FN (t, x))0 where Fi (t, x) = P (Q(t) ≤ x|X ∗ (0) = i). Then F (x) = (P (W ≤ x|X ∗ (0) = i))i∈S . Letting t → ∞ in Theorem 3.1 of [10] we get the following result: Theorem 4.1. For every x ≥ 0, we have (Λ − Dµ x)F 0 (x) = Q∗ F (x).

(4.1)

We will suppose in this section that inf i∈S µ(i) > 0 (the service rate is then always positive when the queue level is positive), so that W is bounded by supi∈S λ(i)/ inf i∈S µ(i) (and thus all its moments exist). The aim of this section is to find an expression of all moments of the stationary distribution. Let us start by recalling the following lemma. Lemma 4.1. Let H be the cumulative distribution function of a non-negative random variable. For every r ≥ 1, if the r-th order moment exists, we have Z ∞ Z ∞ r x dH(x) = r xr−1 (1 − H(x))dx. 0

0

Proof. See for instance [3]. Using Lemma 4.1, one can compute the moments of W as well as its Laplace transform. Let us set vi (k) = E(W k | X(0) = i) and V (k) the column vector containing the vi (k). The expression of the V (k)’s is given by the following corollary. Corollary 4.1. For every k ≥ 1, we have V (k) = (Dµ − Q∗ /k)−1 ΛV (k − 1).

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Proof. Since Q∗ 1 = 0, relation (4.1) can be written as (Λ − Dµ x)F 0 (x) = −Q∗ (1 − F (x)). Multiplying both sides by xk−1 , for k ≥ 1, and after integration, we get Z ∞ Z ∞ Z ∞ k−1 0 k 0 ∗ x F (x)dx − Dµ x F (x)dx = −Q xk−1 (1 − F (x)).dx Λ 0

0

0

Using Lemma 4.1, we easily get Z Z ∞ k−1 0 x F (x)dx − Dµ Λ 0

0



−1 ∗ Q x F (x)dx = k k

0

Thus, ΛV (k − 1) − Dµ V (k) =

Z



xk F 0 (x)dx.

0

−1 ∗ Q V (k), k

that is, (Dµ − Q∗ /k)V (k) = ΛV (k − 1). The matrix Dµ is diagonal with at least one non-zero entry and, so for k ≥ 1, the matrix Dµ − Q∗ /k is invertible. This leads to V (k) = (Dµ − Q∗ /k)−1 ΛV (k − 1), which completes the proof. Since V (0) = 1, we have V (k) = (Dµ − Q∗ /k)−1 Λ(Dµ − Q∗ /(k − 1))−1 Λ · · · (Dµ − Q∗ /2)−1 Λ(Dµ − Q∗ )−1 Λ1. The kth stationary moment of the queue level is then given by E(W k ) = πV (k). This corollary shows that the moments of W can be easily evaluated recursively by solving linear systems. Let us denote by φ the Laplace transform of W , that is φ(θ) := E(eθW ), which is defined for every θ in R since W is bounded by supi∈S λ(i)/ inf i∈S µ(i).

fluid queue with linear service rate

13

Corollary 4.2. For every θ in R, we have φ(θ) = π

∞ X V (k)θk . k! k=0

Proof. This is done by using the inequality W ≤ supi∈S λ(i)/ inf i∈S µ(i) and seeing that, since φ(k) (0) = πV (k), we have for all N ! N N X X φ(k) (0)θk W k θk = E exp(θW ) − φ(θ) − k! k! k=0 k=0 ! ∞ X W k θk = E k! k=N +1 k k  ∞ X θ ≤ −→ 0 as N → ∞. sup λ(i)/ inf µ(i) i∈S k! i∈E k=N +1

This completes the proof.

5. Second moment of the stationary distribution for the second order model We study the case of the second order linear model. We suppose in addition that (A) ∀i ∈ S,

µ(i) ≥ 4σ(i)2 and ∃i ∈ S,

µ(i) > 4σ(i)2 .

2

Let us set zt = (Q00 (t)) for t ≥ 0. The process zt then converges in distribution to 2

W 2 . Moreover, Ito’s formula applied to (Q00 (t)) yields 2

2

2

d(Q00 (t)) = 2λ(X(t))Q00 (t)dt+[−2µ(X(t))+σ(X(t))2 ](Q00 (t)) dt+2σ(X(t))(Q00 (t)) dBt and thus zt satisfies the following stochastic differential equation: dzt = 2λ(X(t))Q00 (t)dt + [−2µ(X(t)) + σ(X(t))2 ]zt dt + 2σ(X(t))zt dBt

(5.1)

with the initial condition z0 = 0. In a way, zt is a second order linear model with input rate equal to 2λ(X(t))Q00 (t), release rate equal to 2µ(X(t)) − σ(X(t))2 (which is assumed negative thanks to the condition µ(i) ≥ 4σ(i)2 ) and local variance 2σ(X(t))zt . Motivated by this remark, we set, for any u ∈ R, the solution (Ztu )t≥u to the following stochastic differential equation:

14

L. Rabehasaina, B. Sericola

  dZ u t  Zu u

=

2λ(X(t))W (t)dt + [−2µ(X(t)) + σ(X(t))2 ]Ztu dt + 2σ(X(t))Ztu dBt

=

0. (5.2)

Note then that, except for the initial condition, Ztu satisfies the same stochastic differential equation as W (t)2 (which again can be easily verified by applying Ito’s formula to W (t)2 ). In fact, we have the following lemma: Lemma 5.1. Zt0 converges in distribution towards Z

0

−∞

 Z 0 Z exp − [2µ(X(v)) + σ(X(v))2 ]dt + s

s



0

2σ(X(v))dBv

2λ(X(s))W (s)ds. (5.3)

Besides, this random variable is equal to W 2 in distribution. Proof. First note that the solution to (5.2) is (see [11]) Ztu

Z

t

= u

 Z t  Z t 2 exp − [2µ(X(v)) + σ(X(v)) ]dv + 2σ(X(v))dBv 2λ(X(s))W (s)ds. s

s

By an argument similar to the one used in Theorem 2.1, we have then that Zt0 has the same distribution as Z0−t . By letting t → ∞, we then get that Zt0 converges in distribution to (5.3). Let us now prove that (5.3) has the same distribution as W 2 . Since W (t)2 satisfies dW (t)2 = 2λ(X(t))W (t)dt + [−2µ(X(t)) + σ(X(t))2 ]W (t)dt + 2σ(X(t))W (t)dBt , which is the same stochastic differential equation as (5.2) except for the initial condition at t = u, we have that for t = 0 and 0 ≥ u, similarly to (2.2), W Z

 Z 0 Z W (u) exp − [2µ(X(s)) + σ(X(s))2 ]ds + 2

=

u

u

u

s

2

u

2σ(X(s))dBs

s

= W (u) M (u)  Z 0  Z 0 exp − [2µ(X(v)) + σ(X(v))2 ]dv + 2σ(X(v))dBv 2λ(X(s))W (s)ds (5.4)

0

+



0

 Z 0  Z 0 exp − [2µ(X(v)) + σ(X(v))2 ]dv + 2σ(X(v))dBv 2λ(X(s))W (s)ds

0

+ Z

2

s

s

fluid queue with linear service rate

15

 Z 0 Z where M (u) := exp − [2µ(X(s)) + σ(X(s))2 ]ds + u

0, A > 0, u ≤ 0,

0

u

 2σ(X(s))dBs . For all x >

P (W (u)2 M (u) > x) =

P (W (u)2 M (u) > x, W (u)2 > A)

+

P (W (u)2 M (u) > x, W (u)2 ≤ A)



P (W (u)2 > A) + P (A M (u) > x)

=

P (W 2 > A) + P (A M (u) > x),

(5.5)

the equality P (W (u)2 > A) = P (W 2 > A) holding because of the stationarity of W (t). Now P (AM (u) > x) ≤ AE(M (u))/x. Using Lemma 1.1, one can verify that    Z 0 2 (2µ(X(s)) − σ(X(s)) )ds . E(M (u)) = E exp − u

Condition (A) in particular implies that 2µ(i) − σ(i)2 ≥ 0 for all i ∈ S, and 2µ(i) − σ(i)2 > 0 for an i ∈ S. Thus, using an argument similar to the one in Theorem 2.1, one can show that

  Z 0 2 (2µ(X(s)) − σ(X(s)) )ds E exp − 

u

converges to 0 exponentially fast as u → −∞. Hence  .   Z 0 (2µ(X(s)) − σ(X(s))2 )ds x −→ 0 P (A M (u) > x) ≤ AE exp − u

as u → −∞. Taking the limsup in (5.5), we then get lim sup P (W (u)2 M (u) > x) ≤ P (W 2 > A) + 0, u→−∞

∀A > 0.

This equality holds for all A > 0, thus by letting A → ∞ we have that lim P (W (u)2 M (u) > x) = 0

u→−∞

for all x. Thus W (u)2 M (u) converges to 0 in probability as u tends to −∞. There thus exists a sequence (uk )k∈N verifying limk→∞ uk = −∞ such that W (uk )2 M (uk ) converges to 0 as k → ∞. Letting k tend to +∞ in (5.4) where u is replaced by uk , we conclude that W 2 equals in distribution to (5.3).

16

Z

L. Rabehasaina, B. Sericola

Again by reversing time, we have that (5.3) is equal to  Z s  Z s exp − [2µ(X ∗ (v)) + σ(X ∗ (v))2 ]dt + 2σ(X ∗ (v))dBv∗ 2λ(X ∗ (s))W ∗ (s)ds.



0

0

0

(5.6) where W ∗ (s) := W (−s). By reversing time in (2.3), we get  Z r  Z ∞ Z r exp − [µ(X ∗ (h)) + σ(X ∗ (h))2 /2]dh + σ(X ∗ (h))dBh∗ λ(X ∗ (r))dr. W ∗ (s) = s

s

s

We likewise set

 Z s Z ∗ ∗ 2 [2µ(X (v)) + σ(X (v)) ]dv + M (s) := M (−s) = exp −

s



0

2σ(X



0

(v))dBv∗

Then (5.6) can be rewritten in the following way Z ∞ M ∗ (s)2λ(X ∗ (s))W ∗ (s)ds.

 .

(5.7)

0

We now show that W and M ∗ (t) admits a moment of order 2, for all t ≥ 0: Lemma 5.2. E(W 2 ) < +∞. Proof. Since W 2 is equal in distribution to (5.7), we have, using Fubini’s theorem, Z ∞ E(M ∗ (s)2λ(X ∗ (s))W ∗ (s))ds E(W 2 ) = Z0 ∞ E(E(M ∗ (s)2λ(X ∗ (s))W ∗ (s)|Bv∗ , v ∈ [0, s], X))ds. (5.8) = 0

 Z s Z ∗ ∗ 2 ∗ [2µ(X (v)) + σ(X (v)) ]dv + But M (s) = exp − 0

[0, s], and X ∗ measurable, and so is λ(X ∗ (s)). Thus

s

2σ(X 0



(v))dBv∗

 is Bv∗ , v ∈

E(M ∗ (s)2λ(X ∗ (s))W ∗ (s)|Bv∗ , v ∈ [0, s], X ∗ ) = M ∗ (s)2λ(X ∗ (s))E(W ∗ (s)|Bv∗ , v ∈ [0, s], X ∗ ). Let us remember that  Z r Z ∞ Z exp − [µ(X ∗ (h)) + σ(X ∗ (h))2 /2]dh + W ∗ (s) = s

s

s

r

(5.9)

 σ(X ∗ (h))dBh∗ λ(X ∗ (r))dr,

thus E(W ∗ (s)|Bv∗ , v ∈ [0, s], X ∗ )  Z ∞  Z r Z r ∗ ∗ 2 ∗ ∗ ∗ exp − [µ(X (h)) + σ(X (h)) /2]dh + σ(X (h))dBh λ(X (r))dr Bv∗ , =E s s  s ∗ v ∈ [0, s], X .

fluid queue with linear service rate

17

The increments of Bh∗ for h ≥ s are independent from Bv∗ , v ∈ [0, s], so  Z ∞  Z r Z r ∗ ∗ 2 ∗ ∗ ∗ exp − [µ(X (h)) + σ(X (h)) /2]dh + σ(X (h))dBh λ(X (r))dr Bv∗ , E s s  s ∗ v ∈ [0, s], X   Z ∞  Z r Z r exp − [µ(X ∗ (h)) + σ(X ∗ (h))2 /2]dh + σ(X ∗ (h))dBh∗ λ(X ∗ (r))dr X ∗ =E s s  Zs r   Z r Z ∞  ∗ ∗ 2 E exp − [µ(X (h)) + σ(X (h)) /2]dh + σ(X ∗ (h))dBh∗ λ(X ∗ (r)) X ∗ dr. = s

s

s





Using Lemma 1.1 with Z = X , w = B and g = σ, we have that     Z r Z r ∗ ∗ ∗ 2 ∗ ∗ ∗ [µ(X (h)) + σ(X (h)) /2]dh + σ(X (h))dBh λ(X (r)) X E exp − s   Z rs = exp − µ(X ∗ (h))dh λ(X ∗ (r))  s Z r   Z r ∗ ∗ 2 ∗ ∗ ×E exp − σ(X (h)) /2dh + σ(X (h))dBh X s   Z r s µ(X ∗ (h))dh λ(X ∗ (r)), = exp − s

hence E(W ∗ (s)|Bv∗ , v ∈ [0, s], X)  Z r  Z r Z ∞  ∗ ∗ 2 ∗ ∗ E exp − [µ(X (h)) + σ(X (h)) /2]dh + σ(X (h))dBh = s s  s ∗ ∗ ×λ(X (r)) X dr  Z r  Z ∞ exp − µ(X ∗ (h))dh λ(X ∗ (r))dr. (5.10) = s

s

And thus from (5.8), (5.9) and (5.10) Z ∞  Z 2 E M ∗ (s)2λ(X ∗ (s)) E(W ) = 0

∞ s

 Z exp −

r









µ(X (h))dh λ(X (r))dr ds.

s

(5.11) Now, by reconditioning with respect to X ∗ we get  Z r    Z ∞ exp − µ(X ∗ (h))dh λ(X ∗ (r))dr E M ∗ (s)2λ(X ∗ (s)) s s Z r     Z ∞ ∗ ∗ exp − µ(X ∗ (h))dh λ(X ∗ (r))dr X ∗ = E E M (s)2λ(X (s)) s  sZ r    Z ∞ ∗ ∗ exp − µ(X (h))dh λ(X ∗ (r))dr E (M ∗ (s)| X ∗ . = E 2λ(X (s)) s

s

18

L. Rabehasaina, B. Sericola

We may write M ∗ (s) in the following way:  Z s  Z s ∗ ∗ ∗ 2 ∗ ∗ [2µ(X (v)) + σ(X (v)) ]dv + 2σ(X (v))dBv M (s) = exp −  0  Z0 s [2µ(X ∗ (h)) − σ(X ∗ (h))2 ]dh = exp − 0  Z s  Z s ∗ 2 ∗ ∗ 2σ(X (v)) dv + 2σ(X (v))dBv × exp − 0

0

Using again Lemma 1.1 with Z = X ∗ , w = B ∗ and g = 2σ, we then get easily that  Z s  ∗ ∗ ∗ ∗ 2 [2µ(X (h)) − σ(X (h)) ]dh . E (M (s)| X ) = exp − 0

Let us now notice that, since µ ≥ 4σ 2 yields that µ ≥ σ 2 ,  Z r  Z ∞ exp − µ(X ∗ (h))dh λ(X ∗ (r))dr E (M ∗ (s)| X ∗ ) s  Zs r   Z s  Z ∞ exp − µ(X ∗ (h))dh λ(X ∗ (r))dr exp − [2µ(X ∗ (h)) − σ(X ∗ (h))2 ]dh = s  Zs r   Z0 s  Z ∞ ∗ ∗ exp − µ(X (h))dh λ(X (r))dr exp − [µ(X ∗ (h)) − σ(X ∗ (h))2 ]dh = s 0  Z0 r  Z ∞ ∗ ∗ exp − µ(X (h))dh λ(X (r))dr, ≤ s

0

Thus we have from (5.11) Z ∞  Z 2 E 2λ(X ∗ (s)) E(W ) ≤ 0



2 Z  2 sup λ(i)

=

2 Z  2 sup λ(i)

i∈S

s

 Z exp −

Z





Z

0

s









µ(X (h))dh λ(X (r))dr ds

 Z exp −

r

  Z E exp −

r



s



r

0

E

0

i∈S









µ(X (h))dh dr ds

0

µ(X ∗ (h))dh

 drds.

(5.12)

0

  Z Let us now remember from Theorem 2.1 the estimate E exp −

r

0



µ(X (h))dh

 ≤

O(exp(−αr)) for the constant α defined by (2.5). It follows that  Z r  Z ∞  E exp − µ(X ∗ (h))dh dr ≤ O(exp(−αs)). Z

∞Z

Hence 0

is finite.

s

s

  Z ∞ E exp −

0

r



µ(X (h))dh

 drds is finite. Thus from (5.12), E(W 2 )

0

Lemma 5.3. E(M ∗ (t)2 ) tends to 0 exponentially fast as t → +∞.

fluid queue with linear service rate

19

Proof. For all t ≥ 0,  Z t  Z t ∗ 2 ∗ ∗ 2 ∗ ∗ M (t) = exp − [4µ(X (v)) + 2σ(X (v)) ]dv + 4σ(X (v))dBv 0 0   Z t Z t Z t = exp − [4µ(X ∗ (v)) − 6σ(X ∗ (v))2 ]dv − 8σ(X ∗ (v))2 dv + 4σ(X ∗ (v))dBv∗ . 0

0

0

Thus by the same martingale argument as in the proof of Theorem 2.1, we get that   Z t  [4µ(X ∗ (v)) − 6σ(X ∗ (v))2 ]dv . E(M ∗ (t)2 ) = E exp − 0

Condition (A) implies that 4µ(i) − 6σ(i)2 ≤ 0 for all i ∈ S, and that 4µ(i) − 6σ(i)2 < 0 for an i. Thus, again using the same argument as in Theorem 2.1, we deduce that E(M ∗ (t)2 ) tends to 0 exponentially fast as t → +∞. We may now state the expression of the second moment of W , which is the main result of this section: Theorem 5.1. Let us denote D2µ−σ2 := diag (2µ(1) − σ(1)2 , ..., 2µ(N ) − σ(N )2 ), Dµ−2σ2 := diag (µ(1) − 2σ(1)2 , ..., µ(N ) − 2σ(N )2 ). Then E(W 2 )

= 2π(D2µ−σ2 − Q∗ )−1 (Q∗ − Dµ−4σ2 )−1 Λ2 1 − 2π(Q∗ − Dµ−4σ2 )−1 Λ(Dµ − Q∗ )−1 Λ1.

(5.13)

Proof. As in Theorem 3.1, we write, using Ito’s formula, Z t A0 g(W ∗ (s), M ∗ (s), X ∗ (s))ds g(W ∗ (t), M ∗ (t), X ∗ (t)) − g(W ∗ (0), M ∗ (0), X ∗ (0)) − 0 Z t σ(X ∗ (s))W ∗ (s)∂xw g(W ∗ (s), M ∗ (s), X ∗ (s))dBs∗ = 0 Z t + 2σ(X ∗ (s))M ∗ (s)∂xm g(W ∗ (s), M ∗ (s), X ∗ (s))dBs∗ . (5.14) 0

where A0 g(xw , xm , i) = − + +

[(µ(i) + σ(i)2 )xw − λ(i)]∂xw g(xw , xm , i) [2µ(i) − σ(i)2 ]xm ∂xm g(xw , xm , i) 1 σ(i)2 xw 2∂x2w g(xw , xm , i) + 2σ(i)2 xm 2∂x2m g(xw , xm , i) 2 X ∗ qij g(xw , xm , j). 2σ(i)2 xm xw ∂xm ∂xw g(xw , xm , i) + j

20

L. Rabehasaina, B. Sericola

(5.14) is a local martingale given X ∗ . We now take g(xw , xm , i) = xw xm f (i) where f D

will again be chosen later on. By Lemmas 5.2 and 5.3, W ∗ (t) = W and M ∗ (t) are in L2 (Ω), thus (5.14) is actually a real martingale. Taking the expectation in (5.14) with the expression of g and A0 , we then get, after some easy calculation,

Z

t

E 0

[(µ(X ∗ (s)) + σ(X ∗ (s))2 )W ∗ (s) − λ(X ∗ (s))]M ∗ (s)f (X ∗ (s))

−[2µ(X ∗ (s)) − σ(X ∗ (s))2 ]M ∗ (s)W ∗ (s)f (X ∗ (s))  X ∗ ∗ ∗ qX +2σ(X ∗ (s))2 M ∗ (s)W ∗ (s)f (X ∗ (s)) + ∗ (s),j W (s)M (s)f (j)ds j

= E(W ∗ (0)M ∗ (0)f (X ∗ (0))) − E(W ∗ (t)M ∗ (t)f (X ∗ (t)))

which can be rewritten, with the same notations as in Theorem 3.1,

Z E 0

t











W (s)M (s)[(Q − Dµ−4σ2 )f ](X (s))ds − E

Z

t









λ(X (s))M (s)f (X (s))ds 0

= −E(W ∗ (0)M ∗ (0)f (X ∗ (0))) + E(W ∗ (t)M ∗ (t)f (X ∗ (t))).

(5.15)

We are now goingZ to study each term of this equality.  t ∗ ∗ ∗ λ(X (s))M (s)f (X (s))ds is equal to The expression E 0

Z

t









λ(X (s))M (s)f (X (s))ds  Z s Z t  Z s exp − [2µ(X ∗ (v)) + σ(X ∗ (v))2 ]dv + 2σ(X ∗ (v))dBv∗ =E 0 0 0  ∗ ∗ ×λ(X (s))f (X (s))ds  Z s  Z s Z t  ∗ ∗ 2 ∗ ∗ = E exp − [2µ(X (v)) + σ(X (v)) ]dv + 2σ(X (v))dBv 0 0 0  ∗ ∗ ×λ(X (s))f (X (s)) ds. E

0

(5.16)

fluid queue with linear service rate

21

But, using again a conditioning argument,     Z s Z s E exp − [2µ(X ∗ (v)) + σ(X ∗ (v))2 ]dv + 2σ(X ∗ (v))dBv∗ λ(X ∗ (s))f (X ∗ (s)) 0   0 Z s  Z s [2µ(X ∗ (v)) + σ(X ∗ (v))2 ]dv + 2σ(X ∗ (v))dBv∗ = E E exp − 0 0  ∗ ∗ ∗ ×λ(X (s))f (X (s)) X      Z s Z s ∗ ∗ ∗ 2 ∗ ∗ [2µ(X (v)) + σ(X (v)) ]dv + 2σ(X (v))dBv X = E E exp − 0 0  ×λ(X ∗ (s))f (X ∗ (s))       Z s [2µ(X ∗ (v)) − σ(X ∗ (v))2 ]dv X ∗ λ(X ∗ (s))f (X ∗ (s)) = E E exp −     Z0 s  ∗ ∗ 2 ∗ ∗ [2µ(X (v)) − σ(X (v)) ]dv λ(X (s))f (X (s)) X ∗ = E E exp −     Z s 0 [2µ(X ∗ (v)) − σ(X ∗ (v))2 ]dv λ(X ∗ (s))f (X ∗ (s)) . (5.17) = E exp − 0

Thus from (5.16) and (5.17) Z

t









λ(X (s))M (s)f (X (s))ds   Z s   ∗ ∗ 2 ∗ ∗ E exp − [2µ(X (v)) − σ(X (v)) ]dv λ(X (s))f (X (s)) ds = 0 0  Z s   Z t ∗ ∗ 2 ∗ ∗ exp − [2µ(X (v)) − σ(X (v)) ]dv λ(X (s))f (X (s))ds . =E E

Z

0 t

0

0

The limit of this expression as t tends to +∞ is the first moment of the steady regime of the model with release rate 2µ(X(v)) − σ(X(v))2 , variance function 2σ(X(v)) and input rate λ(X(s))f (X(s)). Thus Formula (3.2) of Theorem 3.1 yields Z lim E

t→∞

0

t

λ(X ∗ (s))M ∗ (s)f (X ∗ (s))ds



= π(D2µ−σ2 − Q∗ )−1 F Λ1

where F = diag (f (1), ..., f (N )). Note that D2µ−σ2 −Q∗ is invertible because Condition (A) implies 2µ(i) − σ(i)2 ≥ 0 for all i ∈ S, and 2µ(i) − σ(i)2 > 0 for an i ∈ S. Let us consider the term E(W ∗ (t)M ∗ (t)f (X ∗ (t))) in (5.15) and let us show that it tends to 0 as t tends to +∞. Since f is bounded, we just need to show that

22

L. Rabehasaina, B. Sericola

E(W ∗ (t)M ∗ (t)) tends to 0. But  Z t  Z t [2µ(X ∗ (v)) + σ(X ∗ (v))2 ]dv + 2σ(X ∗ (v))dBv∗ M ∗ (t)W ∗ (t) = exp − 0 0  Z s  Z s Z ∞ exp − [µ(X ∗ (v)) + σ(X ∗ (v))2 /2]dv + σ(X ∗ (v))dBv∗ λ(X ∗ (s))ds × t t t   Z t Z t [µ(X ∗ (v)) + σ(X ∗ (v))2 /2]dv + σ(X ∗ (v))dBv∗ = exp − 0 0  Z s  Z s Z ∞ ∗ ∗ 2 exp − [µ(X (v)) + σ(X (v)) /2]dv + σ(X ∗ (v))dBv∗ λ(X ∗ (s))ds × t 0 0   Z t Z t [µ(X ∗ (v)) + σ(X ∗ (v))2 /2]dv + σ(X ∗ (v))dBv∗ ≤ exp − 0 0  Z s  Z s Z ∞ ∗ ∗ 2 exp − [µ(X (v)) + σ(X (v)) /2]dv + σ(X ∗ (v))dBv∗ λ(X ∗ (s))ds × 0 0 0   Z t Z t ∗ ∗ 2 ∗ ∗ [µ(X (v)) + σ(X (v)) /2]dv + σ(X (v))dBv W ∗ (0). = exp − 0

0

Then by the Cauchy-Schwartz inequality: E(M ∗ (t)W ∗ (t))   Z t 1/2 Z t ∗ ∗ 2 ∗ ∗ E(W ∗ (0)2 )1/2 ≤ E exp − [2µ(X (v)) + σ(X (v)) ]dv + 2σ(X (v))dBv 0



1/2

= E(M (t))

0

2 1/2

E(W )

.

Now by Lemma 5.3, E(M ∗ (t)) tends to 0 as t tends to infinity. Thus the expression E(W ∗ (t)M ∗ (t)f (X ∗ (t))) tends to 0 as t tends to infinity. Let us now focus on E(W ∗ (0)M ∗ (0)f (X ∗ (0))) = E(W f (X(0)) in (5.15). Applying Formula (3.1) from Theorem 3.1 with h = f yields E(W ∗ (0)M ∗ (0)f (X ∗ (0)))

= E(W f (X(0))) −1

= πF (Dµ − Q∗ )

Λ1.

In view of this analysis, we may then let t → +∞ in (5.15) and get  Z ∞ ∗ ∗ ∗ ∗ W (s)M (s)[(Q − Dµ−4σ2 )f ](X (s))ds − π(D2µ−σ2 − Q∗ )−1 F Λ1 E 0

−1

= −πF (Dµ − Q∗ )

Λ1.

(5.18)

We now take F = 2(Q∗ − Dµ−4σ2 )−1 Λ. Thus f = 2(Q∗ − Dµ−4σ2 )−1 Λ1. In that case, with this expression for f , and since W 2 equals to (5.7) in distribution, we have  Z ∞ W ∗ (s)M ∗ (s)[(Q∗ − Dµ−4σ2 )f ](X ∗ (s))ds = E(W 2 ). E 0

fluid queue with linear service rate

23

Combining this equality with (5.18) yields (5.13).

References [1] Asmussen, S. and Kella, O. (1996). Rate modulation in dams and ruin problem. Journal of Applied Probability 33(2), 523–535. [2] Chen, D., Hong, Y. and Trivedi, K.S. (2002). Second-order stochastic fluid models with fluiddependent flow rates. Performance evaluation 49, 341–358. [3] Feller, W. (1957). An introduction to probability theory and its applications. Vol. 1, Wiley series in probability and mathematical statistics. [4] Karandikar, R.L. and Kulkarni, V.G. (1995). Second-Order Fluid Flow Models: Reflected Brownian Motion in a Random Environment. Operations Research 43(1), 77–88. [5] Kella, O. and Stadje, W. (2002). Exact results for a fluid model with state-dependent flow rates. Probability in the Engineering and Informational Sciences 16(4), 389–402. [6] Kella, O. and Stadje, W. (2002). Markov modulated linear fluid networks with Markov additive input. Journal of Applied Probability 39(2), 413–420. [7] Kella, O. and Whitt, W. (1999). Linear Stochastic Fluid Networks. Journal of Applied Probability 36(1), 244–260. [8] Kulkarni, V.G. (1997). Fluid models for single buffer systems. Frontiers In Queuing, Models and Applications In Science and Engineering, J. H. Dshalalow, editor, Probability and Stochastic Series, CRC Press, 321–338. [9] Rabehasaina, L. and Sericola, B. (2003). Stability of second order fluid flow models in a stationary ergodic environment. Annals of Applied Probability, 13(4), 1449–1473. [10] Rabehasaina, L. and Sericola, B. (2003). Transient Analysis of a Markov Modulated Fluid Queue with Linear Service Rate. 10th Conference in Analytical and Stochastic Modelling Techniques and Applications ASMTA’03 234–239. [11] Revuz, D. and Yor, M. (1999). Continuous Martingales and Brownian Motion, Springer. [12] Sigman, K. and Ryan, R. (2000). Continuous-time stochastic recursions and duality. Advances in Applied Probability, 32, 426–445. [13] Yao, D.D., Zhang, Q. and Zhou, X.Y. (2003). A regime-switching model for European option pricing.

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discontinuities. Vorticity-velocity scheme To deal with the advective term, we use the fol- lowing semidiscrete central scheme [13, 14]:. ∂tξi,j = −. Hx i+ 1. 2 ,j (t) − Hx i− 1. 2 ,j (t). ∆x. −. Hy i,j+ 1. 2(t) − Hy i,j− 1. 2. (t).

Bayesian Variable Order Markov Models
ference on Artificial Intelligence and Statistics (AISTATS). 2010, Chia Laguna .... over the set of active experts M(x1:t), we obtain the marginal probability of the ...

Linear and Quadratic Splines vs Circular-Queue Vector ...
2010 Electronics, Robotics and Automotive Mechanics Conference. Modeling Motion Prediction ... novel approaches for motion prediction ofobjecls in a limited sequence of .... degree of the polynomial increases that causes the oscillation of ...

Linear and Quadratic Splines vs Circular-Queue Vector ...
kinematics experiments either with constant velocity and accelerated motion. or with parabolic and circular ... which use the envelope and grid matrix for moving object description and tracing, the time for trajectory .... positive {negative} directi

Photochromic 1-benzofurylfulgides with modulated ... - Arkivoc
©ARKAT-USA, Inc. with the maxima at 320 and 370 nm corresponding to the S0→S3 and S0→S2 transitions of the ring-closed isomers of 1a-c (Table 1, Fig. 2).

Patient Queue - GitHub
If you are using a Notebook computer with Firefox follow these instructions: .... Claim: this option is selected by a clinician in order to start an exam on a patient. 9 ...

Realization Theory of Stochastic Jump-Markov Linear ...
JMLSs is the formulation and solution of a stochastic realization problem for a ... In turn, the solution ...... Theoretical Computer Science, 138:101–112, 1995.

Search game for a moving target with dynamically generated ... - Irisa
Jul 16, 2009 - agement and data fusion issues[6]. .... practice, we will take R = R(4) or R = R(A,$). As- .... but the conditionaI marginals are sufficient in practice.

Search game for a moving target with dynamically generated ... - Irisa
Jul 16, 2009 - egy is determined and then the target strategy. We ... big. An extensive definition is not possible. However, for algorithmic reason, it will be ...

A Simplified Lattice Structure of First-Order Linear ...
flexibility in filter design which leads to better image coding per- formance. However, the number of free parameters is often not the actual design freedom since ...

A Simplified Lattice Structure of First-Order Linear ...
order) and synthesis ones are equal to or longer than 2M was pre- sented. Its lattice structure of the analysis bank is. E(z) =diag(A1,A2)W. [. IM/2z. −1 −JF. 0M/2.

Second-Order Consensus of Multiple Agents with ...
Jan 15, 2009 - hicles (AUVs), automated highway systems, and so on. First-order consensus ... from E. *Corresponding author, E-mail: [email protected] ...

Subsystems of Second Order Arithmetic
Feb 7, 2006 - A preliminary version of this book was written with a software pack- .... Countable Well Orderings; Analytic Sets . ...... comparison map, 183.