The goal of this paper is to establish positive recurrence of the model $M_n/G/1/\infty$ under certain assumptions. The Foster condition will be used for this. Intensity of service is assumed only partially at zero (as a lower left derivative value at zero of the “integrated intensity”); in addition, an integral type condition on the “integrated intensity” over intervals of some length is assumed.

For the recent history of the topic see [1] – [7], [14] and many more; see also [15,18]. One of the reasons – although not the only one – why various versions of this system are so popular may be explained by the fact of their intrinsic links to important topics of mathematical insurance theory, see [3].

In this paper we return to the less involved single–server system $M_n/GI/1/\infty$ where the intensity of arrivals may only depend on the number of customers in the system, with the goal of reviewing conditions of its positive recurrence. An importance of this property may be commented, for example, by the publications [1,12] where the investigation of the model assumes that it is in the “steady state”, which is the synonym of stationarity. As is well-known, positive recurrence along with some mild mixing or coupling properties guarantees the existence of a stationary regime of the system. One particular aspect of this issue is how to achieve bounds without assuming existence of intensity of service in $M/GI/1$ model and, more generally, in Erlang – Sevastyanov’s type systems. Certain results in this direction were recently established in [19] for a slightly different model. Still, in [19] it is essential that the absolute continuous part of the distribution function F (in our notation) were non-degenerate; in the present paper this is not required and the approach is different.

Note that in such a model certain close results may be obtained by the methods of regenerative processes if it is assumed that the same distribution function F (see below) has enough number of moments. However, our conditions and methods are different. The main (moderate) hope of the author is that possibly this approach may also be useful in studying ergodic properties of Erlang – Sevastyanov type models, as it happened with the earlier results and approaches based on the intensity of service as in [15,18] successfully applied in [16]. The present paper is some initial attempt toward the programme of developing tools that could help attack the problem outlined recently in [17].

The paper consists of the introduction in Section 1, of the setting and main result in Section 2, of two simple auxiliary lemmata in Section 3 and of the proof of the main result in Section 4, and two simple examples in Section 5 for the comparison of sufficient conditions of theorem 1 with conditions in terms of the intensity of service in the case if the latter does exist.

Recall the notation $\mathsf{\Lambda }={sup}_n{\lambda }_n$ and assumption (1).

0. The proof will be split into several steps. We shall consider the embedded Markov chain, namely, the process $X_t$ at times $t=0,1,\dots$, and it will be shown that this process hits some suitable compact around “zero state” $(0,0)$ in time which admits a finite expectation. From this property the main result will follow. The reader is warned that after this first hit the definition of the embedded Markov chain will change, as further times may become random and possibly non-integer, see step 4 of this proof.

The main goal is to establish the bound (7), from which the estimate (8) follows as a corollary.

1. Let us choose $\epsilon$ so that (see condition (4)). NB: We highlight that this value $\epsilon$ will be fixed for what follows and will not tend to zero.

Once $\epsilon$ is chosen, let (see (2) for the definition of $\tilde{F}\left(1\right)$) and let us choose $x\left(\epsilon \right)\ge 2$ such that Since $F\left(x\right)\uparrow 1$ as $x\uparrow \infty$, this is possible for any $\epsilon >0$. Let us introduce an auxiliary stopping time Denote Function L will serve as a Lyapunov function outside the compact set $K_{\epsilon }$.

First of all, we are going to estimate the first moment of $\tau$, namely, to prove that there exists $C>0$ such that Recall that and highlight that the definition of $\tau ={\tau }_{\epsilon }$ is quite different from that of ${\tau }_0$.

Let $X_0=X$. We have for $X_t=(n_t,x_t)\ne (0,0)$, where $M_t$ is a local martingale (see, e.g., [13]), and It is assumed that $J^1\left(0\right)=J^2\left(0\right)=J^3\left(0\right)=0$ by definition. The martingale $M_t$ is, in fact, a “normal”, that is, non-local one because due to the assumptions expectations of all terms in the integral version of (16) are finite. The following bound will be established: with some $C>0$, if $(n,x)\notin K_{\epsilon }$.

Firstly, we have,

In order to evaluate $J^3$ let us introduce a sequence of stopping times. Let To evaluate $J^3$, using the identity $1(t<\gamma )1(n_t>0)=1(t<\gamma )$ which holds provided that $n>0$, let us estimate, Let us estimate the term $F^1$. We have (recall that $\gamma \le 1$ by definition), Here the complementary part of the integral ${\mathsf{E}}_{n,x}1(\gamma <1/2){\int }_0^{\gamma }(1+x+t)dH(x+t)\ge 0$ was just dropped.

Further, it will be shown that For this aim, let us introduce by induction the sequence of stopping times, Note that the component $x_t$ might only have finitely many jumps down on any finite interval. The times of jumps on the interval $[0,1]$ are exactly the times ${\gamma }^n<1$, and possibly the last jump down on this interval may or may not occur at $t=1$. In any case, clearly, So, we have, where for each outcome this series is almost surely a finite sum. On each interval $[{\gamma }^n,{\gamma }^{n+1}]$ we may write down This is by virtue of assumption (5) and because at each stopping time ${\gamma }^n$ which is less than 1 we have $x_{{\gamma }^n}=0$. If ${\gamma }^n\ge 1$, then both sides in the latter inequality equal zero, so that the inequality still holds true. Therefore, taking a sum over n, we obtain (21), just without $1({\inf }_{0\le t\le 1}{n}_t>0)$ in the left hand side. This miltiplier $1({\inf }_{0\le t\le 1}{n}_t>0)$ in the right hand side guarantees that its presence in the left hand side of (21) still leads to a valid inequality, which means that the bound (21) is justified.

It follows from (20) and (21) that Due to lemma 1, if $n>M$ then (see (12)) (NB: In fact, at least n jobs should be completed, the first one on $[x,x+1]$; however, we prefer to have a bound independent of x. In any case, this does not change the scheme of the proof.) Likewise, if $x>x\left(\epsilon \right)$, then due to the choise of $x\left(\epsilon \right)$, see (13).

Recall that $\epsilon$ was chosen so that $r\epsilon <(r/2)-(1+\mathsf{\Lambda })$, see (11). Hence, Denote Thus, for any $(n,x)\notin K_{\epsilon }$ we obtain, The bound (17) follows with a constant C which may be evaluated via $r,\mathsf{\Lambda },\epsilon$.

2. So, with the choice of $M=M_{\epsilon }$ and $x\left(\epsilon \right)$ according to (12) and (13), respectively, we may write, The event $((n,x)\notin K_{\epsilon })$ may be equivalently expressed as $(\tau >0)$. Hence, the latter inequality may also be rewritten in the form suitable for the induction:

Similarly, $((n_1,x_1)\notin K_{\epsilon })$ may be equivalently expressed as $(\tau >1)$. So, we get Hence, by taking expectations we obtain,

Similarly, by induction (in what follows the notation $n_k=n_t{|}_{t=k}$ is used), Due to the elementary bound $1(k-1<\tau )\ge 1(k<\tau )$, this implies, Summing up and dropping the negative term in the right hand side, we obtain for any $N>0$. So, by the monotone convergence theorem, By virtue of the well-known relation for the expectation of $\tau$ the bound (23) implies the following inequality with $X_0=(n,x)$, In particular, this bound signifies that 3. Now, once the bound for the expected value of $\tau$ is established, we are ready to explain the details how to get a bound for ${\mathsf{E}}_{n,x}{\tau }_0$. The rest of the proof is devoted to this implication, with the last sentences related to the corollary about the invariant measure and convergence to it.

At $\tau$, the process $X_k$ attains the set $(X:X\in K_{\epsilon })$, while $X_0\notin K_{\epsilon }$; hence, both By definition, random variable $\tau$ is the first integer k where simultaneously Therefore, at $k-1$ we have either $n_k>M$, or $x_k>x\left(\epsilon \right)$, or both. If there are no completed jobs on $[k-1,k]$, then $n_t$ may only increase, or, at least, stay equal on this interval, while $x_t$ certainly increases. Therefore, $\tau =k$ may only be achieved by at least one completed job; it would mean a jump down by one of the n-component and simultaneously a jump down to zero of the x-component. Then at k we obtain $x_k\le 1$, which certainly makes it less than $x\left(\epsilon \right)$, irrespectively of whether or not there were other arrivals or completed jobs on $[k-1,k]$ (recall that in addition to inequality (13) we assumed that $x\left(\epsilon \right)\ge 2$).

Now, given $n\le M$ and $x\le x\left(\epsilon \right)$, by virtue of lemma 2, for any $T>0$ we have with any $x\le 1$, where Here T is any positive integer and Note that, of course, for non-integer values of $T>0$ there is a likewise bound, but it looks a bit more involved, and using integers T values suffices for the proof. Recall that it was assumed in (2) that $\tilde{F}\left(1\right)>0$, and it follows from the first line of (2) that $\tilde{F}\left(1\right)<1$. Inequality (25) implies that NB: Here the standard notations for homogeneous Markov processes are used, which means that $X_T$ after stopping time τ is, actually, the value $X_{\tau +T}$.

Note that for any $T>0$ the event $(n\left(X_{\tau +T}\right)=0)$ implies that Hence, we conclude, and, therefore, for any x, 4. Consider now the process X started at time $\tau$ from state $(n_{\tau },x_{\tau })$ with $x_{\tau }\le 1$ and $n_{\tau }\le M$.

Let $T:=\lceil x\left(\epsilon \right)\rceil$ and let us stop the process either at $\tau +T$, or at whatever happens earlier. In other words, consider the stopping time The event $({\chi }^1=\tau +T)$ implies that the process $L\left(X_t\right)$ does not exceed the level $M+2+T$ on the interval $[\tau ,\tau +T]$. On the other hand, according to the arguments of step 3 we have, Let and further, let us define two sequences of stopping times by induction, Note that all integers here in expressions like ${\chi }^k$ and ${\tau }^k$ stand for upper indices, not for power functions. Let us highlight that stopping time ${\tau }^{k+1}$ equals ${\chi }^k$ plus some integer, but ${\chi }^k$ may or may be not integer itself. All these stopping times are finite with probability one, and, moreover, Also, almost surely NB: If ${\inf }_n{\lambda }_n=0$, then, in general, we may not claim that ${lim}_{k\to \infty }{\tau }^k=\infty$, but it is not necessary for our aims.

Using strong Markov property at time ${\chi }^i$ (see [9]), we obtain by induction, 5. Denote Also by induction it follows from (24) and from the elementary bound by definition that there exist a constant C such that Using the representation and due to (30), we estimate, Now we are going to estimate this sum by some geometric type series in a combination with bounds (31) and (32). Some little issue is that we are not able to use Hölder’s, or Cauchy – Buniakowskii – Schwarz’ inequality here because we only possess a first moment bound for ${\tau }^k$, while higher moments are not available. This minor obstacle is resolved in the next step of the proof by the following arguments using conditional expectation with respect to suitable sigma-algebras.

6. We have, Let us investigate the last term. Since each random variable $1({\chi }^j\le {\tau }_0)$ is ${\mathcal{F}}_{{\chi }^k}$-measurable for any $j\le k$, and because, due to the strong Markov property and the bound (31), Moreover, by virtue of the inequality (15) and since by definition all $d^k\le T$, we have for $k\ge 1$, with some finite constant C.

Let us inspect the previous term. Using that ${\delta }^{k-1}$ and all random variables $1({\chi }^j<{\tau }_0)$ are ${\mathcal{F}}_{{\tau }^k}$-measurable for any $j\le k-1$, we obtain by induction,

Also by induction we find that a similar upper bound with the multiplier ${(1-p\left(T\right))}^k$ holds for each term in the sum in the right hand side of (33), for $2\le i<k$, and $k\ge 3$. Indeed, using the identity ${\displaystyle 1({\chi }^k<{\tau }_0)=\prod _{j=1}^{k}1({\chi }^j<{\tau }_0)}$ we have for $2\le i<k$, It was used that the random variable $({\delta }^{i-1}+d^i){\prod }_{j=1}^{k-1}1({\chi }^j<{\tau }_0)$ is ${\mathcal{F}}_{{\tau }^k}$-measurable.