2.1. Solving the equation for the Hawkes process

The result below follows from an algorithmic proof which will be given in Appendix A.1. The algorithmic construction can be used for simulations, which are shown in Figure 1.

Figure 1:

(a) Hawkes process with a positive reproduction function h. (b) Hawkes process with a general reproduction function h. The dots in the plane represent the atoms of the Poisson point process Q used for the construction. The atoms of the Hawkes processes are the green dots on the abscissa axis. The bold red curve corresponds to the intensity $\Lambda^h$ and the colored curves represent the partial cumulative contributions of the successive atoms of the Hawkes process. In (b), the bold blue curve corresponds to the intensity of the dominating Hawkes process with reproduction function $h^+$.

Proposition 2.1. Let Q be an $(\mathcal{F}_t)_{t\ge0}$-Poisson point process on $(0,+\infty)\times (0,+\infty)$ with unit intensity. Consider Equation (1.2), i.e.

\begin{equation*}\left\{\begin{aligned} &N^{h} = N^0+\int_{(0,+\infty)\times(0,+\infty)} \delta_u {\mathds{1}}_{\{\theta\leq \Lambda^{h}(u)\}}\,Q(du,d\theta)\,,\\&\Lambda^h(u) =\biggl(\lambda+\int_{(\!-\infty,u)} h(u-s)\,N^h(ds)\biggr)^+ \,, &&u >0\,,\end{aligned}\right.\end{equation*}

in which $h\,{:}\, (0,+\infty)\to \mathbb{R}$ is a signed measurable reproduction function, $\lambda>0$ an immigration rate, and $N^0$ an initial condition in $\mathcal{N}((\!-\infty,0])$ with law $\mathfrak{m}$. Consider the similar equation for $N^{h^+}$ in which h is replaced by $h^+$. Assume that

(2.1)\begin{equation} \|h^+\|_1 <1\end{equation}

and that the distribution $\mathfrak{m}$ of the initial condition $N^0$ satisfies

(2.2)\begin{equation} \forall t>0,\quad \int_0^t \mathbb{E}_{\mathfrak{m}} \bigg(\int_{(\!-\infty,0]}h^+(u-s)\,N^0(ds) \bigg)\ du < +\infty.\end{equation}

1. 1. Then there exists a pathwise unique strong solution $N^h$ of Equation (1.2), and this solution is a Hawkes process in the sense of Definition 1.1.

2. 2. The same holds for $N^{h^+}$, and moreover $N^h \le N^{h^+}$ a.s. (in the sense of measures).

The main novelty of this proposition is the coupling obtained in (2). Let us first note that the coupling is very strong since the comparison between $N^h$ and $N^{h^+ }$ holds in the sense of measures: each atom of $N^h$ is an atom of $N^{h^+ }$. Moreover, even though couplings are easily derived for Hawkes processes associated with nonnegative reproduction functions, it is not so when the reproductive functions are signed: if h and g are two signed functions such that $h\le g$, then it is not always possible to couple $N^h$ and $N^g$ in such a way that atoms of $N^h$ are atoms of $N^g$ as well. However, if h is signed and g is nonnegative, then our construction applies and $N^h\le N^g$; see Appendix A.1 for details. We present the result above using h and $h^+$ since $h^+$ is the least positive upper bound of h.

Remark 2.1. In order to prove the strong existence and pathwise uniqueness of the solution of Equation (1.2), we propose a proof based on an algorithmic construction similar to the Poisson embedding of [Reference Brémaud and Massoulié5], also referred to in [Reference Daley and Vere-Jones11] as thinning. Since this construction is rather classical, we postpone the proof to Appendix A.1. A similar result is also proved in these references using Picard iteration techniques, with the assumption (2.2) replaced by the stronger hypothesis that there exists $D_{\mathfrak{m}}>0$ such that

(2.3)\begin{equation}\forall t>0,\quad \mathbb{E}_{\mathfrak{m}} \bigg(\int_{(\!-\infty,0]}|h(t-s)| \,N^0(ds) \bigg) < D_{\mathfrak{m}}\,.\end{equation}

When h is nonnegative, the result can be deduced from the cluster representation of the self-exciting Hawkes process, since $N^h([0,t])$ is bounded above by the sum of the sizes of a Poisson number of subcritical Galton–Watson trees; see [Reference Hawkes and Oakes20], [Reference Reynaud-Bouret and Roy30].

Remark 2.2. Proposition 2.1 does not require that L(h) be finite. When $L(h)<\infty$, the assumption (2.2) can be rewritten as

(2.4)\begin{equation}\int_0^{L(h)} \mathbb{E}_\mathfrak{m} \bigg(\int_{(\!-L(h),0]}h^+(u-s) \,N^0(ds) \bigg)\ du < +\infty\,.\end{equation}

A sufficient condition for (2.4) to hold is that $\mathbb{E}_\mathfrak{m}(N^0(\!-L(h),0])<+\infty$. Indeed, using the Fubini–Tonelli theorem, the left-hand side of (2.4) can be bounded by $\|h^+\|_1 \,\mathbb{E}_\mathfrak{m}(N^0(\!-L(h),0])$. Therefore, the results of Proposition 2.1 hold under Assumption 1.1.

2.2. The cluster representation for nonnegative reproduction functions

In this subsection, we consider the case in which the reproduction function h is nonnegative. The intensity process of a corresponding Hawkes process can be written, for $t>0$, as

\begin{equation*}\Lambda^h(t) = \lambda + \int_{(\!-L(h),t)} h(t-u) \,N^h(du)\,.\end{equation*}

The first term can be interpreted as an immigration rate of ancestors. Let $(V_k)_{k\ge1}$ be the corresponding sequence of arrival times, forming a Poisson process of intensity $\lambda$.

The second term is the sum of all the contributions of the atoms of $N^h$ before time t and can be seen as self-excitation. If U is an atom of $N^h$, it contributes to the intensity by the addition of the function $t \mapsto h(t-U)$, hence generating new points regarded as its descendants or offspring. Each individual has a lifelength $L(h)=\sup(\textnormal{supp}(h))$, the number of its descendants follows a Poisson distribution with mean $\|h\|_1$, and the ages at which it gives birth to them have density $h/\|h\|_1$, all this independently. This induces a Galton–Watson process in continuous time; see [Reference Hawkes and Oakes20], [Reference Reynaud-Bouret and Roy30], and Figure 2.

Figure 2:

Cluster representation of a Hawkes process with positive reproduction function. The abscissas of the dots give its atoms. Offspring are colored according to their ancestor, and their ordinates correspond to their generation in this age-structured Galton–Watson tree.

To each ancestor arrival time $V_k$ we can associate a cluster of times composed of the times of birth of its descendants. The condition $\|h\|_1<1$ is a necessary and sufficient condition for the corresponding Galton–Watson process to be subcritical, which implies that the cluster sizes are finite a.s. More precisely, if we define $H_k$ by saying that $V_k+H_k$ is the largest time in the cluster associated with $V_k$, then the $(H_k)_{k\ge1}$ are i.i.d. random variables independent from the sequence $(V_k)_{k\ge1}$.

Reynaud-Bouret and Roy [Reference Reynaud-Bouret and Roy30] proved the following tail estimate for $H_1$.

Proposition 2.2. ([Reference Reynaud-Bouret and Roy30, Prop. 1.2]) Let us define

(2.5)\begin{equation}\gamma\triangleq\frac{\|h\|_1-\log(\|h\|_1)-1}{L(h)} >0\,.\end{equation}

Under Assumption 1.1, we have that

\begin{equation*}\forall x\ge0,\quad \mathbb{P}(H_1 >x)\le \exp(1-\|h\|_1)\,\exp (\!-\gamma x)\,,\end{equation*}

which provides a lower bound for the rate of decay of the cluster length.

When h is nonnegative, it is possible to associate to the Hawkes process an M/G/$\infty$ queue. For $A \ge L(h)$, we consider that the arrival times of ancestors $(V_k)_{k\ge1}$ correspond to the arrivals of customers in the queue and associate to the kth customer a service time $\widetilde{H}_k(A) \triangleq H_k+A$. We assume that the queue is empty at time 0, and then the number $Y_t$ of customers in the queue at time $t\ge0$ is given by

(2.6)\begin{equation}Y_t=\sum_{k\,{:}\,V_k\le t} {\mathds{1}}_{\{V_k+\widetilde{H}_k(A)>t\}}\,.\end{equation}

Let $\mathcal{T}_0=0$, and let the successive hitting times of 0 by the process $(Y_t)_{t\geq 0}$ be given by

(2.7)\begin{equation} \mathcal{T}_{k}=\inf\{t\geq \mathcal{T}_{k-1},\ Y_{t-}\not=0,\ Y_t=0\}, \quad \forall k\geq 1.\end{equation}

The time interval $[V_1,\mathcal{T}_1)$ is called the first busy period, and is the first time interval during which the queue is never empty. Note that the $\mathcal{T}_{k}$ are times at which the conditional intensity of the underlying Hawkes process has returned to $\lambda$ and there is no remaining influence of its previous atoms, since $\widetilde{H}_k(A) \triangleq H_k+A \ge H_k+L(h)$.

Thus the Hawkes process after $\mathcal{T}_{k}$ has the same law as the Hawkes process with initial condition the null point process $\emptyset \in \mathcal{N}((\!-A,0])$, translated by $\mathcal{T}_{k}$. This allows us to split the random measure $N^h$ into i.i.d. parts. We will prove all this in the next section.

We end this part by giving tail estimates for the $\mathcal{T}_{k}$, which depend on $\lambda$ and on $\gamma$ given in (2.5), which respectively control the exponential decays of $\mathbb{P}(V_1>x)$ and $\mathbb{P}(H_1>x)$.

Proof of Proposition 2.3. The proof follows from Proposition 2.2, from which we deduce that the service time $\widetilde{H}_1=H_1+A$ satisfies

(2.8)\begin{equation}\mathbb{P}(\widetilde{H}_1>x)=\mathbb{P}(H_1>x-A) \le \exp(\!-(x-A)\gamma +1-\|h\|_1) = O(\text{e}^{-\gamma x})\,.\end{equation}

We then conclude by applying Theorem A.1 to the queue $(Y_t)_{t\geq 0}$ defined by (2.6).

Theorem A.1 in the appendix establishes the decay rates for the tail distributions of $\mathcal{T}_1$ and of the length of the busy period $[V_1,\mathcal{T}_1)$. This result is of interest in itself, independently of the results for Hawkes processes considered here.

3.1. Definition of a strong Markov process

We suppose that Assumption 1.1 holds and consider the Hawkes process $N^h$ that is the solution of the corresponding Equation (1.2) constructed in Proposition 2.1. We recall that $L(h)<\infty$. Then, for any $t>0$ and $u\in (\!-\infty,-L(h)]$, we have $h(t-u)=0$, and thus

(3.1)\begin{equation} \Lambda^h(t) = \biggl(\lambda+\int_{(\!-\infty,t)}h(t-u)\,N^h(du)\biggr)^+= \biggl(\lambda+\int_{(\!-L(h),t)} h(t-u)\,N^h(du)\biggr)^+\,.\end{equation}

In particular, $N^h|_{(0,+\infty)}$ depends only on the restriction $N^0|_{(\!-L(h),0]}$ of the initial condition.

Recall the shift operator $S_t$ defined in (1.3) and (1.4). Note that if $t,s\geq 0$ then $S_{s+t}N^h=S_t S_s N^h=S_s S_tN^h$. Let $A<\infty$ be such that $A \ge L(h)$. Consider the $(\mathcal{F}_t)$-adapted process $X=(X_t)_{t\ge 0}$ defined by

(3.2)\begin{equation}X_t = (S_tN^h ) |_{(\!-A,0]}=N^h|_{(t-A,t]}(\cdot+t) \,,\end{equation}

i.e.,

\begin{equation*}\begin{array}{ccl}X_t \,{:}\, \mathcal{B}((\!-A,0]) &\quad \rightarrow\quad &\quad \mathbb{R}_+\\ B &\quad \mapsto\quad &\quad X_t(B) = N^h|_{(t-A,t]}(B+t). \end{array}\end{equation*}

The measure $X_t$ is the point process $N^h$ in the time window $(t-A,t]$, shifted back to $(\!-A,0]$. This is a function of $N^h|_{(\!-A,+\infty)}$. Using Equation (3.1) and the remark below it, we see that the law of $N^h|_{(\!-A,+\infty)}$ depends on the initial condition $N^0$ only through $N^0|_{(\!-A,0]}$. Therefore, with abuse of notation, when dealing with the process $(X_t)_{t \ge 0}$ we shall use the notation $\mathbb{P}_\mathfrak{m}$ and $\mathbb{E}_\mathfrak{m}$ even when $\mathfrak{m}$ is a law on $\mathcal{N}((\!-A,0])$, and $\mathbb{P}_{\nu}$ and $\mathbb{E}_{\nu}$ even when $\nu$ is an element of $\mathcal{N}((\!-A,0])$.

Note that X depends on A, and that we omit this in the notation.

Proof. This follows from the fact that $N^h$ is the unique solution of Equation (1.2). Indeed, let T be a stopping time. On $\{T<\infty\}$, by definition

\begin{equation*}X_{T+t} = (S_{T+t}N^h ) |_{(\!-A,0]} = (S_t S_T N^h ) |_{(\!-A,0]}\,.\end{equation*}

Using that $N^h$ satisfies Equation (1.2) driven by the process Q, we have

\begin{align*}S_T N^h & = S_T (N^h|_{(\!-\infty,T]}) +S_T (N^h|_{(T,+\infty)})\\[3pt] & = (S_T N^h)|_{(\!-\infty,0]} +\int_{(T,+\infty)\times (0,+\infty)} \delta_{u-T} {\mathds{1}}_{\{\theta\leq \Lambda^h(u)\}}Q(du,d\theta)\\[3pt] & = (S_T N^h)|_{(\!-\infty,0]} +\int_{(0,+\infty)\times (0,+\infty)} \delta_{v}{\mathds{1}}_{\{\theta\leq \widetilde{\Lambda}^h(v)\}} \ S_T Q(dv,d\theta),\end{align*}

where $S_T Q$ is the (randomly) shifted process with bivariate cumulative distribution function given by

(3.3)\begin{equation}S_TQ((0,t]\times (0,a]) = Q((T,T+t]\times (0,a])\,,\qquad t,a>0,\end{equation}

and where for $v>0$,

\begin{align*}\widetilde{\Lambda}^h(v)& = \Lambda^h(v+T)= \bigg(\lambda+\int_{(\!-\infty,v)}h(v-s)S_T N^h(ds)\bigg)^+.\end{align*}

This shows that $S_T N^h$ satisfies Equation (1.2) driven by $S_TQ$ with initial condition $(S_T N^h)|_{(\!-\infty,0]}$. Since $A\ge L(h)$, moreover $S_TN^h|_{(0,+\infty)}$ actually depends only on $(S_T N^h) |_{(\!-A,0]} \triangleq X_T$.

Let us now condition on $\{T<\infty\}$ and on $\mathcal{F}_T$. Since Q is an $(\mathcal{F}_t)_{t\ge0}$-Poisson point process with unit intensity, $S_TQ$ is an $(\mathcal{F}_{T+t})_{t\ge0}$-Poisson point process with unit intensity; see Lemma A.2 for this classic fact. In particular it is independent of the $\mathcal{F}_T$-measurable random variable $X_T$. Additionally, $X_T$ satisfies the assumption (2.2), which becomes in this case the following: for all $r>0$,

\begin{equation*}\int_0^r \int_{(\!-A,0]}h^+(u-s) (S_T N^h)(ds) \ du < +\infty \qquad \mathbb{P}_{\mathfrak{m}}\textnormal{-a.s.}\end{equation*}

We have indeed that

\begin{align*}\int_0^r \int_{(\!-A,0]} & h^+(u-s) (S_TN^h)(ds) du \nonumber \\[3pt] & = \int_0^r \int_{(\!-A+T,T]}h^+(T+u-s) N^h(ds) \ du \nonumber \\[3pt] &= \int_T^{T+r} \int_{(\!-A+T,T]}h^+(v-s) N^h(ds) \ dv \nonumber \\[3pt] & \le \int_T^{T+r} \int_{(\!-\infty,0]}h^+(v-s) N^0(ds) \ dv + \int_T^{T+r} \int_{(0,T]}h^+(v-s) N^h(ds) \ dv \nonumber \\[3pt] & \le \int_T^{T+r} \int_{(\!-\infty,0]}h^+(v-s) N^0(ds) \ dv + \|h^+\|_1 N^h(0,T] \nonumber\\[3pt] & < +\infty\qquad \mathbb{P}_{\mathfrak{m}}\textnormal{-a.s.},\end{align*}

since the distribution $\mathfrak{m}$ of $N^0$ satisfies (2.2), and since we have shown at the end of the proof of Proposition 2.1 that $\mathbb{E}_{\mathfrak{m}}(N^h(0,t]) < +\infty$ for all $t >0$.

Thus the assumptions of Proposition 2.1 are satisfied, which yields that $(X_{T+t} )_{t\ge0}$ is the pathwise unique, and hence weakly unique, strong solution of Equation (1.2) started at $X_T$ and driven by the $(\mathcal{F}_{T+t})_{t\ge0}$-Poisson point process $S_TQ$. Hence, it is a process started at $X_T$ which is an $(\mathcal{F}_{T+t})_{t\ge0}$-Markov process with same transition semi-group as $(X_t)_{t\geq 0}$. If we wish to be more specific, for every bounded Borel function F on $\mathbb{D}(\mathbb{R}_+,\mathcal{N}((\!-A,0]))$ we set

\begin{equation*}\Pi F(x) \triangleq \mathbb{E}_x(F((X_t)_{t\geq 0}))\end{equation*}

and note that existence and uniqueness in law for (1.2) yield that

\begin{equation*}\mathbb{E}_x(F((X_t)_{t\geq 0}) \,|\, T<\infty, \mathcal{F}_T) = \Pi F(X_T)\,.\end{equation*}

This is the strong Markov property we set out to prove.

3.2. Renewal of X at $\emptyset$

Using $(X_t)_{t\geq 0}$ and Proposition 3.1, we obtain that if T is a stopping time such that $N^h|_{(T-A,T]}=\emptyset$, then $N^h|_{(T,+\infty)}$ is independent of $N^h|_{(\!-\infty,T]}$ and behaves the same as $N^h$ started from $\emptyset$ and translated by T. Such renewal times lead to an interesting decomposition of $N^h$ which illuminates its dependence structure.

The successive hitting times of $\emptyset \in \mathcal{N}((\!-A,0])$ for the Markov process X are such renewal times. This subsection is devoted to the study of their properties. Recall that we have introduced in (1.7) the first hitting time of $\emptyset \in \mathcal{N}((\!-A,0])$ for X, given by

\begin{equation*}\tau \triangleq \inf\{t>0\,{:}\,X_{t-}\neq \emptyset, X_{t} =\emptyset\}=\inf\{t>0\,{:}\, N^h[t-A,t)\neq 0, N^h(t-A,t] =0\}\,.\end{equation*}

It depends on A, but this is omitted in the notation. It is natural to study whether $\tau$ is finite or not. When the reproduction function h is nonnegative, we introduce the queue $(Y_t)_{t \ge 0}$ defined by (2.6), and its return time to zero, $\mathcal{T}_1$, defined in (2.7). The following result will yield the finiteness of $\tau$.

Proof. We use the notation defined in Section 2.2. To begin with, we remark that $\tau >V_1$. First, let us consider t such that $V_1<t< \mathcal{T}_1$. By definition, there exists $i\ge1$ such that

\begin{equation*}V_i\le t\le V_i+\widetilde{H}_i(A) =V_i+H_i+A.\end{equation*}

Since the interval $[V_i, V_i+H_i]$ corresponds to the cluster of descendants of $V_i$, there exists a sequence of points of $N^h$ in $[V_i, V_i+H_i]$ which are distant by less than L(h) and thus by less than A. Therefore, if $t\in[V_i, V_i+H_i]$, then $N^h(t-A,t]>0$.

If $t\in[ V_i+H_i, V_i+H_i+A]$, then $N^h(t-A,t]>0$ as well, since $V_i+H_i \in N^h$ (it is the last birth time in the Galton–Watson tree stemming from $V_i$, by definition of $H_i$). Since this reasoning holds for any $t\le \mathcal{T}_1$, it follows that $\tau \ge\mathcal{T}_1$.

Conversely, for any $t \in [V_1,\tau)$, by definition of $\tau$, necessarily $N^h(t-A,t]>0$. Thus there exists an atom of $N^h$ in $(t-A,t]$, and from the cluster representation, there exists $i \ge 1$ such that this atom belongs to the cluster of $V_i$, hence to $[V_i,V_i+H_i]$. We easily deduce that

\begin{equation*}V_i\le t\le V_i+H_i+A\end{equation*}

and thus $Y_t\ge1$, for all $t \in [V_1,\tau)$. This proves that $\tau\le \mathcal{T}_1$ and concludes the proof.□

To extend the result concerning the finiteness of $\tau$ to the case where no assumption is made on the sign of h, we use the coupling between $N^h$ and $N^{h^+}$ stated in Proposition 2.1(2).

Proof. We use the coupling $(N^h,N^{h^+})$ of Proposition 2.1(2), which satisfies $N^h \le N^{h^+}$. If $\tau=+\infty$, since the immigration rate $\lambda$ is positive, for any $t\ge0$ we necessarily have $N^h(t-A,t]>0$ and thus $N^{h^+}(t-A,t]>0$, which implies that $\tau^+=+\infty$ also, a.s.

Now, it is enough to prove that $\tau\leq \tau^+$ when both times are finite. In this case, since $N^{h^+}$ is locally finite a.s., $\tau^+-A$ is an atom of $N^{h^+}$ such that $N^{h^+}(\tau^+-A,\tau^+]=0$. This implies that $N^{h}(\tau^+-A,\tau^+]=0$. If $\tau^+-A $ is also an atom of $N^h$, then $\tau\leq \tau^+$.

Otherwise, we first prove that $N^h(\!-A,\tau^+-A)>0$. The result is obviously true if $N^0\not= \emptyset$. When $N^0=\emptyset$, the first atoms of $N^h$ and $N^{h^+}$ coincide because $\Lambda^h_0=\Lambda_0^{h^+}$, where these functions are as defined in (A.1). This first atom is necessarily before $\tau^+-A$, and hence $N^h(\!-A,\tau^+-A)>0$. The last atom U of $N^h$ before $\tau^+-A$ is thus well defined, and necessarily satisfies $N^h(U,U+A]=0$ and $N^h [U, U+A)\neq 0$, so that $\tau\leq U+A\leq \tau^+$. We have thus proved that $\tau\leq \tau^+$, $\mathbb{P}_{\mathfrak{m}}$-a.s., as needed.□

We now prove that the regeneration time $\tau$ admits an exponential moment which ensures that it is finite a.s. The results will rely on the coupling between $N^h$ and $N^{h^+}$ and on the results obtained in Section 2.1. Let us define

\begin{equation*}\gamma^+\triangleq\frac{\|h^+\|_1-\log(\|h^+\|_1)-1}{L(h^+)}>0\,.\end{equation*}

Proposition 3.3. Let Assumption 1.1 hold. Let $A<\infty$ be such that $A \ge L(h)$, and assume that $\mathbb{E}_\mathfrak{m}(N^0(\!-A,0])<+\infty$. Then $\tau$ given by (1.7) satisfies

\begin{equation*}\forall \alpha < \min(\lambda,{\gamma^+})\,,\quad \mathbb{E}_{\mathfrak{m}}(\text{e}^{\alpha \tau}) < +\infty\,.\end{equation*}

In particular $\tau$ is finite, $\mathbb{P}_{\mathfrak{m}}$-a.s., and $\mathbb{E}_{\mathfrak{m}}(\tau) < +\infty$.

Proof. By Proposition 3.2, it is sufficient to prove this for $\tau^+$. When $\mathfrak{m}$ is the Dirac measure at $\emptyset$, the result is a direct consequence of Lemma 3.1 and Proposition 2.3. We now turn to the case when $\mathfrak{m}$ is different from $\delta_{\emptyset}$. The proof is separated into three steps.□

Step 1: Analysis of the problem. To control $\tau^+$, we distinguish the points of $N^h$ coming from the initial condition from the points coming from ancestors that arrived after zero. We thus let $K= N^0((\!-A,0])$ denote the number of atoms of $N^0$, $(V_i^0)_{1\le i \le K}$ the atoms themselves, and $(\widetilde{H}_i^0(A))_{1\le i \le K}$ the durations such that $V_i^0 + \widetilde{H}_i^0(A)-A$ is the time of birth of the last descendant of $V_i^0$. Note that $V_i^0$ has no offspring before time 0, so that the reproduction function of $V_i^0$ is a truncation of h. We finally define the time when the influence of the past before 0 has vanished, given by

\begin{equation*}E=\max_{1 \le i \le K}\Big(V_i^0+\widetilde{H}_i^0(A)\Big),\end{equation*}

with the convention that $E=0$ if $K=0$. If $K>0$, since $V_i^0\in (\!-A,0]$ and $\widetilde{H}_i^0(A)\geq A$, we have $E>0$. Note that $\tau^+ \ge E$.

We now consider the sequence $(V_i)_{i \ge 1}$ of ancestors arriving after time 0 at rate $\lambda$. We recall that these can be viewed as customers arriving in an M/G/$\infty$ queue with service times given by $\widetilde{H}_1(A)$. In our case, the queue may not be empty at time 0, when $E>0$. In that case, the queue returns to 0 when all the customers that arrived before time 0 have left the system (which is the case at time E) and when all the busy periods containing the customers that arrived at times between 0 and E are over. The first hitting time of 0 for the queue is thus equal to

(3.4)\begin{equation} \tau^+ = \left\{ \begin{array}{ccl} E & \quad \mbox{ if } & \quad Y_E=0\,, \\[5pt] \inf\{t \ge E\,{:}\, Y_t=0 \} & \quad\mbox{ if } & \quad Y_E>0\,, \end{array}\right. \end{equation}

where $Y_t$ is as given in (2.6):

\begin{equation*}Y_t=\sum_{k\,{:}\,0\leq V_k\le t} {\mathds{1}}_{\{V_k+\widetilde{H}_k(A)>t\}}.\end{equation*}

Step 2: Exponential moments ofE. In (3.4), E depends only on $N^0$, and $(Y_t)_{t\geq 0}$ depends only on the arrivals and service times of customers entering the queue after time 0. A natural idea is then to condition with respect to E, and for this it is important to gather estimates on the moments of E. Since $V_i^0\leq 0$, we have that

\begin{equation*}0\leq E\leq \max_{1\leq i\leq K} \widetilde{H}^0_i(A).\end{equation*}

The truncation mentioned in Step 1 implies that the $\widetilde{H}^0_i(A)$ are stochastically dominated by independent random variables distributed as $\widetilde{H}_1$, which we denote by $\bar{H}^0_i(A)$. Thus, for $t>0$, using (2.8), we have

\begin{align*}\mathbb{P}_{\mathfrak{m}}(E>t)&\leq \mathbb{P}_{\mathfrak{m}}\Big(\max_{1\leq i\leq K} \bar{H}^0_i(A)>t\Big)\\&= 1-\mathbb{E}_{\mathfrak{m}}\Big(\big(1-\mathbb{P}(\widetilde{H}_1(A)> t)\big)^K\Big)\\&\leq 1 -\mathbb{E}_{\mathfrak{m}}\big((1-C \text{e}^{-{\gamma^+} t})^K\big)\,.\end{align*}

Thus there exists $t_0>0$ such that for any $t>t_0$,

(3.5)\begin{align}\mathbb{P}_{\mathfrak{m}}(E>t) \leq C \mathbb{E}_\mathfrak{m}(N^0(\!-A,0]) \text{e}^{-{\gamma^+}t}.\end{align}

As a corollary, we have for any $\beta \in (0,{\gamma^+})$ that

(3.6)\begin{equation}\mathbb{E}_\mathfrak{m}\big(\text{e}^{\beta E}\big)<+\infty\,.\end{equation}

Step 3: Estimate of the tail distribution of$\tau^+$. For $t>0$, we have

\begin{align*}\mathbb{P}_{\mathfrak{m}}(\tau^+>t)& = \mathbb{P}_{\mathfrak{m}}\big(\tau^+>t,\,E> t\big)+\mathbb{P}_{\mathfrak{m}}\big(\tau^+>t,\,E\leq t\big)\\& \leq\mathbb{P}_{\mathfrak{m}}(E> t)+\mathbb{E}_{\mathfrak{m}}\Big({\mathds{1}}_{\{E\leq t\}}\, \mathbb{P}_{\mathfrak{m}}\big(\tau^+>t \,|\, E\big)\Big).\end{align*}

The first term is controlled by (3.5). For the second term, we use Proposition A.2, which is a consequence of Theorem A.1. For this, let us introduce a constant $\kappa$ such that $\kappa <{\gamma^+}$ if ${\gamma^+} \leq \lambda$ and $\kappa=\lambda$ if $\lambda<{\gamma^+}$. We have

\begin{equation*}\mathbb{E}_{\mathfrak{m}}\Big({\mathds{1}}_{\{E\leq t\}} \, \mathbb{P}\big(\tau^+>t \,|\, E\big)\Big) \leq \mathbb{E}_{\mathfrak{m}}\big({\mathds{1}}_{\{E\leq t\}} \, \lambda C E\,\text{e}^{-\kappa(t-E)}\big)= \lambda C \text{e}^{-\kappa t} \mathbb{E}_\mathfrak{m}\big({\mathds{1}}_{\{E\leq t\}} \, E\,\text{e}^{\kappa E}\big).\end{equation*}

Since $\kappa<{\gamma^+}$, it is always possible to choose $\beta\in (\kappa,{\gamma^+})$ such that (3.6) holds, which implies that $\mathbb{E}_\mathfrak{m}\big({\mathds{1}}_{\{E\leq t\}} \,E\,\text{e}^{\kappa E}\big)$ can be bounded by a finite constant independent of t.

Gathering all the results, we have

\begin{align*}\mathbb{P}_{\mathfrak{m}}(\tau^+>t) & \leq C \mathbb{E}_\mathfrak{m}(N^0(\!-A,0]) \text{e}^{-{\gamma^+}t}+ \lambda C' \text{e}^{-\kappa t}=O\big(\text{e}^{-\kappa t}\big).\end{align*}

This yields that $\mathbb{E}_\mathfrak{m}(\text{e}^{\alpha \tau^+})<+\infty$ for any $\alpha<\kappa$, i.e. $\alpha<\min (\lambda,{\gamma^+})$.

Note that if Assumption 1.1 holds, then $\tau$ given by (1.7) satisfies $\mathbb{E}_\emptyset(\tau)<\infty$, and hence the null measure $\emptyset$ is a positive recurrent state for the strong Markov process $X=(X_t)_{t\ge 0}$.

Theorem 3.1. Let Assumption 1.1 hold. The strong Markov process $X=(X_t)_{t\ge 0}$ with values in $\mathcal{N}((\!-A,0])$ defined by (3.2) admits a unique invariant law $\pi_A$ defined as in (1.8); i.e., for every Borel nonnegative function f on $\mathcal{N}((\!-A,0])$,

\begin{equation*}\pi_A\,f = \frac1{\mathbb{E}_\emptyset(\tau)} \mathbb{E}_\emptyset\biggl(\int_0^{\tau} f(X_t) \,dt\biggr)\,.\end{equation*}

Moreover, $\pi_A\{\emptyset\} = 1/(\lambda \mathbb{E}_\emptyset(\tau))$.

Proof. These facts are classic in the presence of the positive recurrent state $\emptyset$, which is reachable from all states.□

The strong Markov property of X yields a sequence of regeneration times $(\tau_k)_{k\geq 0}$, which are the successive visits of X to the positive recurrent state $\emptyset$, defined as follows (the time $\tau_0$ has already been introduced in (1.10)):

\begin{align*}\tau_0 &= \inf\{t\ge 0\,{:}\, X_t =\emptyset\}\,&&\text{(first entrance time of $\emptyset$),}\\\tau_k &= \inf\{t>\tau_{k-1}\,{:}\, X_{t-}\neq \emptyset, X_{t} =\emptyset\}\,,\quad k\ge1\,&&\text{(successive return times at $\emptyset$).}\end{align*}

These provide a useful decomposition of the path of X into i.i.d. excursions.

Proof. The above items follow classically from the strong Markov property of X. Let us first prove the finiteness of the return times $\tau_k$. For any $\mathfrak{m}$, from the definition of $\tau_0$ and $\tau$, we have that $\tau_0\leq \tau$, $\mathbb{P}_\mathfrak{m}$-a.s. Then $\mathbb{P}_\mathfrak{m}(\tau_0<+\infty)=1$ follows from Proposition 3.3. For $k\geq 1$, using the strong Markov property of X, we have for any $\mathfrak{m}$ that

\begin{align*}\mathbb{P}_\mathfrak{m}(\tau_k<+\infty)&= \mathbb{E}_\mathfrak{m} \big( {\mathds{1}}_{\{\tau_{k-1}<+\infty\}} \, \mathbb{P}_{X_{\tau_{k-1}}}(\tau<+\infty)\big)\\ &= \mathbb{E}_\mathfrak{m}\big({\mathds{1}}_{\{\tau_{k-1}<+\infty\}} \,\mathbb{P}_\emptyset(\tau<+\infty)\big)\\& = \mathbb{P}_\mathfrak{m}(\tau_{k-1}<+\infty)=\cdots = \mathbb{P}_{\mathfrak{m}}(\tau_0<+\infty)=1.\end{align*}

Let us now prove (2) and (3). It is sufficient to consider $(X_t)_{t\in [0,\tau_0)}$, $(X_{\tau_0+t})_{t\in [0,\tau_1-\tau_0)}$, and $(X_{\tau_1+t})_{t\in [0,\tau_2-\tau_1)}$. Let $F_0$, $F_1$, and $F_2$ be three measurable bounded real functions on $\mathbb{D}(\mathbb{R}_+,\mathcal{N}(\!-A,0])$. Then, using the strong Markov property successively at $\tau_1$ and $\tau_0$, we obtain

\begin{align*}&\mathbb{E}_{\mathfrak{m}} \Big(F_0\big((X_t)_{t\in [0,\tau_0)}\big)\, F_1\big((X_{\tau_0+t})_{t\in [0,\tau_1-\tau_0)}\big) \,F_2\big((X_{\tau_1+t})_{t\in [0,\tau_2-\tau_1)}\big)\Big)\\&\quad = \mathbb{E}_{\mathfrak{m}} \Big(F_0\big((X_t)_{t\in [0,\tau_0)}\big)\Big) \,\mathbb{E}_{\emptyset}\Big(F_1\big((X_{t})_{t\in [0,\tau)}\big)\Big) \,\mathbb{E}_\emptyset\Big( F_2\big((X_{t})_{t\in [0,\tau)}\big)\Big).\end{align*}

This concludes the proof.□

Proof of Theorem 1.2(1)

This classic proof assumes first that $f\ge0$. Then using (4.1) and (4.2),

\begin{equation*}\frac1{K_T}\sum_{k=1}^{K_T} I_k f\le \frac1{K_T}\int_0^T f(X_t)\,dt\le\frac1{K_T}\int_0^{\tau_0} f(X_t)\,dt + \frac1{K_T}\sum_{k=1}^{K_T+1} I_kf,\end{equation*}

and the strong law of large numbers applied to the i.i.d. $I_k f$ yields that

\begin{equation*}\frac1{K_T}\int_0^T f(X_t)\,dt \xrightarrow[T\to\infty]{\mathbb{P}_\mathfrak{m}-\text{a.s.}}\mathbb{E}_\emptyset\biggl(\int_0^{\tau} f(X_t) \,dt\biggr) \triangleq \mathbb{E}_\emptyset(\tau) \,\pi_A\,f.\end{equation*}

Choosing $f=1$ yields that

(4.3)\begin{equation}\frac{T}{K_T} \xrightarrow[T\to\infty]{\mathbb{P}_\mathfrak{m}-\textnormal{a.s.}} \mathbb{E}_\emptyset(\tau) <\infty,\end{equation}

and dividing the first limit by the second concludes the proof for $f\ge0$. The case of $\pi_A$-integrable signed f follows using the decomposition $f=f^+ - f^-$.

Proof of Theorem 1.2(2)

This follows from a general result in Thorisson [Reference Thorisson36, Th. 10.3.3, p. 351], which says that if the distribution of $\tau$ under $\mathbb{P}_\emptyset$ has a density with respect to the Lebesgue measure and if $\mathbb{E}_\emptyset(\tau)<+\infty$, then there exists a probability measure $\mathbb{Q}$ on $\mathbb{D}(\mathbb{R}_+,\mathcal{N}(\!-A,0])$ such that, for any initial law $\mathfrak{m}$,

\begin{equation*}\mathbb{P}_\mathfrak{m}\bigl((X_{t+u})_{u\ge0}\in \cdot \bigr)\xrightarrow[t\to\infty]{\textnormal{total variation}}\mathbb{Q}\,.\end{equation*}

Since $\pi_A$ is an invariant law, $ \mathbb{P}_{\pi_A}\bigl((X_{t+u})_{u\ge0}\in \cdot \bigr) = \mathbb{P}_{\pi_A}(X\in \cdot) $ for every $t\geq 0$. Hence, taking $\mathfrak{m}=\pi_A$ in the above convergence yields that $\mathbb{Q}=\mathbb{P}_{\pi_A}(X\in \cdot)$.

It remains to check the assumptions of the theorem above. Proposition 3.3 yields that $\mathbb{E}_\emptyset(\tau)<+\infty$. Moreover, under $\mathbb{P}_{\emptyset}$ we can rewrite $\tau$ as

\begin{equation*}\tau=U_1^h+\inf\big\{t>0\,{:}\, \ X_{(t+U_1^h)_-}\not= \emptyset \mbox{ and } X_{t+U_1^h}= \emptyset\big\}.\end{equation*}

Using the strong Markov property, we easily prove independence of the two terms in the right-hand side. Since $U_1^h$ has an exponential distribution under $\mathbb{P}_{\emptyset}$, $\tau$ has a density under $\mathbb{P}_\emptyset$.

Proof of Theorem 1.3

Let $\tilde{f} \triangleq f -\pi_A\,f$, so that $\frac1T \int_0^T \tilde{f} (X_t)\,dt = \frac1T \int_0^T f(X_t)\,dt - \pi_A\,f $. With the notation (4.1) and (4.2), we have the decomposition

(4.4)\begin{equation}\int_0^T \tilde{f}(X_t)\,dt= \int_0^{\tau_0} \tilde{f}(X_t)\,dt+ \sum_{k=1}^{K_T} I_k \tilde{f} + \int_{\tau_{K_T}}^T \tilde{f}(X_t)\,dt\,.\end{equation}

The $I_k \tilde{f}$ are i.i.d. and are distributed as $\int_{0}^{\tau} \tilde{f} (X_t)\,dt$ under $\mathbb{P}_\emptyset$, with expectation 0 and variance $\mathbb{E}_\emptyset(\tau) \sigma^2(\,f)$; see Theorem 3.2. Since f is locally bounded, so is $\tilde{f}$, and

\begin{equation*}\frac1{\sqrt{T}} \int_0^{\tau_0} \tilde{f}(X_t)\,dt \xrightarrow[T\to\infty]{\mathbb{P}_\mathfrak{m}-\text{a.s.}} 0\,.\end{equation*}

Now, let $\varepsilon >0$. For arbitrary $a>0$ and $0<u\le T$,

\begin{equation*}\mathbb{P}_\mathfrak{m}\biggl(\biggl|\int_{\tau_{K_T}}^T \tilde{f}(X_t)\,dt \biggr| > a\biggr)\le\mathbb{P}_\mathfrak{m}(T-\tau_{K_T}> u) +\mathbb{P}_\mathfrak{m}\biggl(\sup_{0\le s \le u}\biggl|\int_{T-s}^T \tilde{f}(X_t)\,dt \biggr| > a\biggr)\,.\end{equation*}

But

\begin{equation*}\mathbb{P}_\mathfrak{m}(T-\tau_{K_T}> u)= 1 - \mathbb{P}_\mathfrak{m}(\exists t \in [T-u,T]\,{:}\, X_{t-}\neq \emptyset, X_t=\emptyset),\end{equation*}

and Theorem 1.2(2) yields that

\begin{equation*}\lim_{T\to\infty}\mathbb{P}_\mathfrak{m}(T-\tau_{K_T}> u)= 1-\mathbb{P}_{\pi_A}(\exists t \in [0,u]\,{:}\, X_{t-}\neq \emptyset, X_t=\emptyset),\end{equation*}

so that there exists $u_0$ large enough such that

\begin{equation*}\lim_{T\to\infty} \mathbb{P}_\mathfrak{m}(T-\tau_{K_T}> u_0) <\frac{\varepsilon}{2}\,.\end{equation*}

Moreover, Theorem 1.2(2) yields that

\begin{equation*}\lim_{T\to\infty}\mathbb{P}_\mathfrak{m}\biggl(\sup_{0\le s \le u_0}\biggl|\int_{T-s}^T \tilde{f}(X_t)\,dt \biggr| > a\biggr)=\mathbb{P}_{\pi_A}\biggl(\sup_{0\le s \le u_0}\biggl|\int_0^s \tilde{f}(X_t)\,dt \biggr| > a\biggr);\end{equation*}

thus there exists $a_0$ large enough that

\begin{equation*}\lim_{T\to\infty}\mathbb{P}_\mathfrak{m}\biggl(\sup_{0\le s \le u_0}\biggl|\int_{T-s}^T \tilde{f}(X_t)\,dt \biggr| > a_0\biggr) <\frac{\varepsilon}{2},\end{equation*}

and hence

\begin{equation*}\limsup_{T\to\infty} \mathbb{P}_\mathfrak{m}\biggl(\biggl|\int_{\tau_{K_T}}^T \tilde{f}(X_t)\,dt \biggr| > a_0\biggr) < \varepsilon\,.\end{equation*}

This implies in particular that

\begin{equation*}\frac1{\sqrt{T}} \int_{\tau_{K_T}}^T \tilde{f}(X_t)\,dt \xrightarrow[T\to\infty]{\text{probab.}} 0\,.\end{equation*}

It now remains to treat the second term in the right-hand side of (4.4). The classic central limit theorem yields that

\begin{equation*}\frac1{\sqrt{T}} \sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} I_k\tilde{f}\xrightarrow[T\to\infty]{\textnormal{in law}}\frac1{\sqrt{\mathbb{E}_\emptyset(\tau)}} \mathcal{N}(0, \mathbb{E}_\emptyset(\tau)\sigma^2(\,f)) = \mathcal{N}(0, \sigma^2(\,f)),\end{equation*}

and we are left to control

\begin{equation*}\Delta_T \triangleq\frac1{\sqrt{T}}\sum_{k=1}^{K_T} I_k\tilde{f}- \frac1{\sqrt{T}}\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} I_k\tilde{f}\,.\end{equation*}

Let $\varepsilon>0$ and

\begin{equation*}v(T,\varepsilon) \triangleq \{\lfloor (1 -\varepsilon^3)T/\mathbb{E}_\emptyset(\tau)\rfloor, \dots,\lfloor (1 +\varepsilon^3)T/\mathbb{E}_\emptyset(\tau)\rfloor\}\,.\end{equation*}

Note that $(1-\varepsilon^3)T/\mathbb{E}_\emptyset(\tau)<T/\mathbb{E}_\emptyset(\tau)<(1+\varepsilon^3)T/\mathbb{E}_\emptyset(\tau)$ and hence that $\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor$ belongs to $v(T,\varepsilon) $. In view of (4.3), there exists $t_\varepsilon$ such that if $T\ge t_\varepsilon$, then

\begin{equation*}\mathbb{P}_\mathfrak{m}(K_T \in v(T,\varepsilon)) >1-\varepsilon\,.\end{equation*}

For $T\ge t_\varepsilon$ we thus have on $\{K_T \in v(T,\varepsilon)\}$ that

\begin{align*}|\Delta_T |& \leq \Biggl| \frac{1}{\sqrt{T}} \sum_{k=\lfloor (1 -\varepsilon^3)T/\mathbb{E}_\emptyset(\tau)\rfloor}^{K_T} I_k\tilde{f} \Biggr|+ \Biggl| \frac{1}{\sqrt{T}}\sum_{k=\lfloor (1 -\varepsilon^3)T/\mathbb{E}_\emptyset(\tau)\rfloor}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} I_k\tilde{f} \Biggr|\\[5pt] & \leq \frac{2}{\sqrt{T}}\max_{n\in v(T,\varepsilon)}\Biggl|\sum_{k=\lfloor (1 -\varepsilon^3)T/\mathbb{E}_\emptyset(\tau)\rfloor}^n I_k\tilde{f}\Biggr|\, .\end{align*}

Using now Kolmogorov’s maximal inequality [Reference Feller16, Sec. IX.7, p. 234], we obtain that

\begin{equation*}\mathbb{P}_\mathfrak{m}(|\Delta_T | \ge \varepsilon)\le \frac{\lfloor (1 +\varepsilon^3)T/\mathbb{E}_\emptyset(\tau)\rfloor - \lfloor (1 -\varepsilon^3)T/\mathbb{E}_\emptyset(\tau)\rfloor}{\varepsilon^2 T/4} \mathbb{E}_\emptyset(\tau) \sigma^2(\,f) \le 8\sigma^2(\,f) \varepsilon \,.\end{equation*}

Since $\varepsilon >0$ is arbitrary, we conclude that

\begin{equation*}\Biggl| \frac1{\sqrt{T}}\sum_{k=1}^{K_T} I_k\tilde{f}- \frac1{\sqrt{T}}\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} I_k\tilde{f}\Biggr| \xrightarrow[T\to\infty]{\text{probab.}} 0\,.\end{equation*}

These three convergence results and Slutsky’s theorem yield the desired convergence result.

Proof of Theorem 1.4

With the notation $\tilde{f} \triangleq f -\pi_A\,f$, so that $\frac1T \int_0^T \tilde{f} (X_t)\,dt = \frac1T \int_0^T f(X_t)\,dt - \pi_A\,f $, and (4.2), let us consider the decomposition

(4.5)\begin{align}\int_0^T \tilde{f} (X_t)\,dt=\int_0^{\tau_0} \tilde{f}(X_t)\,dt+ \sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} I_k \tilde{f}+ \int_{\tau_{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor}}^T \tilde{f}(X_t)\,dt \,.\end{align}

The $I_k \tilde{f}$ are i.i.d. and distributed as $\int_{0}^{\tau} \tilde{f} (X_t)\,dt$ under $\mathbb{P}_\emptyset$, with expectation 0 and variance $\mathbb{E}_\emptyset(\tau) \sigma^2(\,f)$; see Theorem 3.2. Since f takes its values in [a, b], we have

\begin{equation*}\biggl|\int_0^{\tau_0} \tilde{f} (X_t)\,dt\biggr|\le |b-a| \tau_0\end{equation*}

and

\begin{equation*}\biggl|\int_{\tau_{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor}}^T \tilde{f} (X_t)\,dt\biggr|\le |b-a||T-\tau_{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor}|\,.\end{equation*}

Now,

\begin{align*}T-\tau_{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor}&= - \tau_0 - \sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} (\tau_k-\tau_{k-1}) +T\\[5pt] &= - \tau_0 -\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} (\tau_k-\tau_{k-1}-\mathbb{E}_\emptyset(\tau)) + T -\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor \mathbb{E}_\emptyset(\tau);\end{align*}

here

\begin{equation*}0\le T-\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor \mathbb{E}_\emptyset(\tau) < \mathbb{E}_\emptyset(\tau),\end{equation*}

and the $\tau_k-\tau_{k-1}-\mathbb{E}_\emptyset(\tau)$ are i.i.d., have the same law as $\tau-\mathbb{E}_\emptyset(\tau) $ under $\mathbb{P}_\emptyset$, and have expectation 0 and variance $\mbox{Var}_\emptyset (\tau)$. Thus,

\begin{align*}&\mathbb{P}_\mathfrak{m}\biggl( \biggl|\frac1T \int_0^T f(X_t)\,dt - \pi_A\,f \biggr| \ge \varepsilon \biggr)\\[5pt] &\le \mathbb{P}_\mathfrak{m}\!\left( \left| \sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} I_k \tilde{f} \right| + |b-a| \left( 2\tau_0 +\left|\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} (\tau_k-\tau_{k-1}-\mathbb{E}_\emptyset(\tau))\right|+ \mathbb{E}_\emptyset(\tau)\right)\ge T\varepsilon\right)\!.\end{align*}

Now, using that for any $u \in [0,1)$

\begin{equation*}T\varepsilon -|b-a| \mathbb{E}_\emptyset(\tau) -2|b-a| \mathbb{E}_\mathfrak{m}(\tau_0)= 2 \frac{(1-u)T\varepsilon - |b-a| \mathbb{E}_\emptyset(\tau)}2 + uT\varepsilon -2|b-a| \mathbb{E}_\mathfrak{m}(\tau_0),\end{equation*}

we obtain that

(4.6)\begin{align}&\mathbb{P}_\mathfrak{m}\biggl( \biggl|\frac1T \int_0^T f(X_t)\,dt - \pi_A\,f \biggr| \ge \varepsilon \biggr)\notag\\[4pt] &\quad \le\mathbb{P}_\mathfrak{m}\!\left(\left| \sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} I_k \tilde{f} \right|\ge \frac{(1-u)T\varepsilon - |b-a| \mathbb{E}_\emptyset(\tau)}2\right)\notag\\[4pt] &\qquad +\mathbb{P}_\mathfrak{m}\!\left(\left|\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} (\tau_k-\tau_{k-1}-\mathbb{E}_\emptyset(\tau))\right|\ge \frac{(1-u)T\varepsilon- |b-a| \mathbb{E}_\emptyset(\tau)}{2|b-a|}\right)\notag\\[4pt] &\qquad +\mathbb{P}_\mathfrak{m}\!\left(\tau_0-\mathbb{E}_\mathfrak{m}(\tau_0)\ge \frac{uT\varepsilon - 2|b-a| \mathbb{E}_{\mathfrak{m}}(\tau_0)}{2|b-a|}\right)\!.\end{align}

We aim to apply Bernstein’s inequality [Reference Massart25, Cor. 2.10, p. 25; (2.17), (2.18), p. 24] to bound the three terms of the right-hand side. We recall that for the application of Bernstein’s inequality to random variables $X_1,\dots X_N$, there should exist constants c and v such that

\begin{equation*} \sum_{k=1}^N\mathbb{E}_\mathfrak{m}\!\left[X_k^2\right]\le v \quad \text{ and } \quad \sum_{k=1}^N\mathbb{E}_\mathfrak{m}\!\left[(X_k)_+^n\right]\le \frac{n!}{2}vc^{n-2} \quad \forall n\ge3.\end{equation*}

First,

\begin{equation*}\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} \mathbb{E}_\mathfrak{m}\bigl( ( I_k \tilde{f})^2 \bigr)= \Big\lfloor\frac{T}{\mathbb{E}_\emptyset(\tau)}\Big\rfloor \mathbb{E}_\emptyset(\tau) \sigma^2(\,f)\le T \sigma^2(\,f)\end{equation*}

and, for $n\ge3$,

\begin{multline*}\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} \mathbb{E}_\mathfrak{m}\bigl( (I_k \tilde{f})_\pm^n \bigr)= \Big\lfloor \frac{T}{\mathbb{E}_\emptyset(\tau)}\Big\rfloor \mathbb{E}_\mathfrak{m}\bigl( (I \tilde{f})_\pm^n\bigr)\\ \le\frac{n!}2 T \sigma^2(\,f)\biggl(\sup_{k\ge3}\biggl(\frac2{k!}\frac{\mathbb{E}_\mathfrak{m}\bigl( (I\tilde{f})_\pm^k \bigr)}{\mathbb{E}_\emptyset(\tau)\sigma^2(\,f)}\biggr)^{\frac1{k-2}}\biggr)^{n-2}\triangleq\frac{n!}2 T \sigma^2(\,f)(c^\pm(\,f))^{n-2}\,.\end{multline*}

Then,

\begin{equation*}\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} \mathbb{E}_\mathfrak{m}\bigl( (\tau_k-\tau_{k-1}-\mathbb{E}_\emptyset(\tau))^2 \bigr)= \Big\lfloor\frac{T}{\mathbb{E}_\emptyset(\tau)}\Big\rfloor \mbox{Var}_\emptyset (\tau)\le T \frac{\mbox{Var}_\emptyset (\tau)}{\mathbb{E}_\emptyset(\tau)}\end{equation*}

and, for $n\ge3$,

\begin{multline*}\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} \mathbb{E}_\mathfrak{m}\bigl( (\tau_k-\tau_{k-1}-\mathbb{E}_\emptyset(\tau))_\pm^n \bigr) =\Big\lfloor T/\mathbb{E}_\emptyset(\tau)\Big\rfloor \mathbb{E}_\emptyset\bigl( (\tau-\mathbb{E}_\emptyset(\tau))_\pm^n \bigr)\\\le\frac{n!}2 T \frac{\mbox{Var}_\emptyset (\tau)}{\mathbb{E}_\emptyset(\tau)}\biggl(\sup_{k\ge3}\biggl(\frac2{k!}\frac{\mathbb{E}_\emptyset\bigl( (\tau-\mathbb{E}_\emptyset(\tau))_\pm^k \bigr)}{\mbox{Var}_\emptyset (\tau)}\biggr)^{\frac1{k-2}}\biggr)^{n-2}\triangleq\frac{n!}2 T \frac{\mbox{Var}_\emptyset (\tau)}{\mathbb{E}_\emptyset(\tau)}(c^\pm(\tau))^{n-2}\,.\end{multline*}

Lastly, $\mathbb{E}_\mathfrak{m}\bigl((\tau_0-\mathbb{E}_\mathfrak{m}(\tau_0))^2\bigr) = \mbox{Var}_\mathfrak{m}(\tau_0)$ and, for $n\ge3$,

\begin{align*}&\mathbb{E}_\mathfrak{m}\bigl((\tau_0-\mathbb{E}_\mathfrak{m}(\tau_0))_+^n\bigr)\\[4pt] &\quad \le \frac{n!}2 \mbox{Var}_\mathfrak{m}(\tau_0)\biggl(\sup_{k\ge3}\biggl(\frac2{k!}\frac{\mathbb{E}_\mathfrak{m}\bigl((\tau_0-\mathbb{E}_\mathfrak{m}(\tau_0))_+^k\bigr)}{\mbox{Var}_\mathfrak{m}(\tau_0)}\biggr)^{\frac1{k-2}}\biggr)^{n-2}\triangleq\frac{n!}2 \mbox{Var}_\mathfrak{m}(\tau_0)(c^+(\tau_0))^{n-2}\,.\end{align*}

Applying [Reference Massart25, Cor. 2.10, p. 25; (2.17), (2.18), p. 24] to the right-hand side of (4.6) yields that

\begin{align*}&\mathbb{P}_\mathfrak{m}\biggl( \biggl|\frac1T \int_0^T f(X_t)\,dt - \pi_A\,f \biggr| \ge \varepsilon \biggr)\notag\\[4pt] &\quad \le\exp\left(\!-\frac{((1-u)T\varepsilon - |b-a| \mathbb{E}_\emptyset(\tau))^2}{8 T \sigma^2(\,f) + 4 c^+(\,f)((1-u)T\varepsilon - |b-a| \mathbb{E}_\emptyset(\tau)) }\right)\notag\\[4pt] &\qquad +\exp\left(\!-\frac{((1-u)T\varepsilon - |b-a| \mathbb{E}_\emptyset(\tau))^2}{8 T \sigma^2(\,f) + 4 c^-(\,f)((1-u)T\varepsilon - |b-a|\mathbb{E}_\emptyset(\tau)) }\right)\notag\\[4pt] &\qquad +\exp\left(\!-\frac{((1-u)T\varepsilon - |b-a| \mathbb{E}_\emptyset(\tau))^2}{8T |b-a|^2\frac{\mbox{Var}_\emptyset (\tau)}{\mathbb{E}_\emptyset(\tau)}+ 4 |b-a| c^+(\tau)((1-u)T\varepsilon - |b-a| \mathbb{E}_\emptyset(\tau)) }\right)\notag\\[4pt] &\qquad +\exp\left(\!-\frac{((1-u)T\varepsilon - |b-a| \mathbb{E}_\emptyset(\tau))^2}{8T |b-a|^2\frac{\mbox{Var}_\emptyset (\tau)}{\mathbb{E}_\emptyset(\tau)}+ 4 |b-a| c^-(\tau)((1-u)T\varepsilon - |b-a| \mathbb{E}_\emptyset(\tau)) }\right)\notag\\[4pt] &\qquad +\exp\left(\!-\frac{(uT\varepsilon - 2|b-a| \mathbb{E}_\mathfrak{m}(\tau_0))^2}{8|b-a|^2\mbox{Var}_\mathfrak{m} (\tau_0) + 4 |b-a| c^+(\tau_0)(uT\varepsilon - 2|b-a| \mathbb{E}_\mathfrak{m}(\tau_0)) }\right)\!,\end{align*}

which is (1.11).

Proof of Corollary 1.1

Under $\mathbb{P}_\emptyset$, we have $\tau_0=0$, and thus (4.6) reads as follows:

(4.7)\begin{align}\mathbb{P}_\emptyset\biggl( \biggl|\frac1T \int_0^T f(X_t)\,dt - \pi_A\,f \biggr| \ge \varepsilon \biggr)&\le\mathbb{P}_\emptyset\!\left(\left| \sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} I_k \tilde{f} \right|\ge \frac{T\varepsilon - |b-a| \mathbb{E}_\emptyset(\tau)}2\right)\\[5pt] &+\mathbb{P}_\emptyset\!\left(\left|\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor}\! (\tau_k{-}\tau_{k-1}{-}\mathbb{E}_\emptyset(\tau))\right|{\ge} \frac{T\varepsilon- |b-a| \mathbb{E}_\emptyset(\tau)}{2|b-a|}\!\right)\!.\nonumber\end{align}

As in the proof of Theorem 1.4, we apply Bernstein’s inequality for each of the terms in the right-hand side. However, in order to simplify the obtained bound, we change the upper bounds of the moments of $I_k\tilde{f}$ and $\tau_k-\tau_{k-1}-\mathbb{E}_\emptyset(\tau)$. Namely, we use the fact that for all $n\ge1$,

\begin{equation*}\mathbb{E}_\emptyset(\tau^n)\le \frac{n!}{\alpha^n}\mathbb{E}_\emptyset(e^{\alpha\tau}) \quad \text{and} \quad \mathbb{E}_\emptyset(|\tau - \mathbb{E}_\emptyset(\tau)|^n)\le \frac{n!}{\alpha^n}\mathbb{E}_\emptyset(e^{\alpha\tau}) e^{\alpha \mathbb{E}_\emptyset(\tau)}.\end{equation*}

Since $\tau$ is a nonnegative random variable, we have $e^{\alpha \mathbb{E}_\emptyset(\tau)} \ge 1$, and in the sequel it will be more convenient to use the following upper bound: for all $n\ge1$,

\begin{equation*}\mathbb{E}_\emptyset(\tau^n)\le \frac{n!}{\alpha^n}\mathbb{E}_\emptyset(e^{\alpha\tau}) e^{\alpha \mathbb{E}_\emptyset(\tau)}.\end{equation*}

Then

\begin{equation*}\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} \mathbb{E}_\emptyset\bigl( ( I_k \tilde{f})^2 \bigr)\le \Big\lfloor\frac{T}{\mathbb{E}_\emptyset(\tau)}\Big\rfloor \mathbb{E}_\emptyset(\tau^2) (b-a)^2\le \frac{2(b-a)^2}{\alpha^2}\Big\lfloor\frac{T}{\mathbb{E}_\emptyset(\tau)}\Big\rfloor \mathbb{E}_\emptyset(e^{\alpha\tau}) e^{\alpha \mathbb{E}_\emptyset(\tau)}\,,\end{equation*}

and, for $n\ge3$,

\begin{equation*}\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} \mathbb{E}_\emptyset\bigl( |I_k \tilde{f})|^n \bigr) \le \frac{n!}{2} \left(\Big\lfloor\frac{T}{\mathbb{E}_\emptyset(\tau)}\Big\rfloor |b-a|^2 \frac{2}{\alpha^2}\mathbb{E}_\emptyset(e^{\alpha\tau})e^{\alpha \mathbb{E}_\emptyset(\tau)}\right) \ \Big(\frac{|b-a|}{\alpha}\Big)^{n-2}\,.\end{equation*}

Setting

\begin{equation*}v=\frac{2(b-a)^2}{\alpha^2}\Big\lfloor\frac{T}{\mathbb{E}_\emptyset(\tau)}\Big\rfloor \mathbb{E}_\emptyset(e^{\alpha\tau}) e^{\alpha \mathbb{E}_\emptyset(\tau)} \quad \text{and}\quad c= \frac{|b-a|}{\alpha},\end{equation*}

and applying Bernstein’s inequality, we obtain that

\begin{equation*} \mathbb{P}_\emptyset\!\left(\left| \sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} I_k \tilde{f} \right|\ge \frac{T\varepsilon - |b-a| \mathbb{E}_\emptyset(\tau)}2\right) \le 2 \exp\left(\!-\frac{\Bigl( T\varepsilon-|b-a|\mathbb{E}_\emptyset(\tau) \Bigr)^2}{4 \left(2v + (T\varepsilon-|b-a|\mathbb{E}_\emptyset(\tau))c \right)} \right)\!.\end{equation*}

Also,

\begin{equation*}\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} \mathbb{E}_\emptyset\bigl( (\tau_k-\tau_{k-1}-\mathbb{E}_\emptyset(\tau))^2 \bigr)\le \frac{2}{\alpha^2} \Big\lfloor \frac{T}{\mathbb{E}_\emptyset(\tau)}\Big\rfloor \mathbb{E}_\emptyset(e^{\alpha\tau})e^{\alpha \mathbb{E}_\emptyset(\tau)} \,,\end{equation*}

and, for $n\ge3$,

\begin{equation*}\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} \mathbb{E}_\emptyset\bigl( |\tau_k-\tau_{k-1}-\mathbb{E}_\emptyset(\tau)|^n \bigr)\le \frac{n!}{2}\left(\Big\lfloor\frac{T}{\mathbb{E}_\emptyset(\tau)}\Big\rfloor \frac{2}{\alpha^2}\mathbb{E}_\emptyset(e^{\alpha\tau})e^{\alpha \mathbb{E}_\emptyset(\tau)} \right) \frac{1}{\alpha^{n-2}}\,.\end{equation*}

Applying Bernstein’s inequality again, we obtain that

\begin{align*} &\mathbb{P}_\emptyset\!\left(\left|\sum_{k=1}^{\lfloor T/\mathbb{E}_\emptyset(\tau)\rfloor} (\tau_k-\tau_{k-1}-\mathbb{E}_\emptyset(\tau))\right|\ge \frac{T\varepsilon- |b-a| \mathbb{E}_\emptyset(\tau)}{2|b-a|}\right)\\&\qquad \le2 \exp\left(\!-\frac{\Bigl( T\varepsilon-|b-a|\mathbb{E}_\emptyset(\tau) \Bigr)^2}{4 \left(2v + (T\varepsilon-|b-a|\mathbb{E}_\emptyset(\tau))c \right)} \right)\!.\end{align*}

The inequality (4.7) gives that

\begin{align*}\mathbb{P}_\emptyset\biggl( \biggl|\frac1T \int_0^T f(X_t)\,dt - \pi_A\,f \biggr| \ge \varepsilon \biggr)&\le 4 \exp\left(\!-\frac{\Bigl( T\varepsilon-|b-a|\mathbb{E}_\emptyset(\tau) \Bigr)^2}{4 \left(2v + (T\varepsilon-|b-a|\mathbb{E}_\emptyset(\tau))c \right)} \right)\!.\end{align*}

To prove the second part of Corollary 1.1 we have to solve

(4.8)\begin{equation}\eta=4 \exp\left(\!-\frac{\Bigl( T\varepsilon-|b-a|\mathbb{E}_\emptyset(\tau) \Bigr)^2}{4 \left(2v + (T\varepsilon-|b-a|\mathbb{E}_\emptyset(\tau))c \right)} \right)\end{equation}

by expressing $\varepsilon$ as function of $\eta$, for any $\eta\in (0,1)$.

Let us define the following decreasing bijection from $\mathbb{R}_+$ into $\mathbb{R}_-$:

\begin{equation*}\varphi(x)=-\frac{x^2}{4(2v+cx)}\,.\end{equation*}

The solution of (4.8) is then $\varepsilon_\eta=(|b-a|\mathbb{E}_\emptyset(\tau)+x_0)/T$, where $x_0$ is the unique positive solution of

\begin{equation*}\varphi(x)=\log\Big(\frac{\eta}{4}\Big)\quad \Leftrightarrow \quad x^2+4c\log\Big(\frac{\eta}{4}\Big) x + 8v\log\Big(\frac{\eta}{4}\Big)=0\,.\end{equation*}

Computing the roots of this second-order polynomial, we can show that there always exist one negative and one positive root as soon as $\eta<4$. More precisely,

\begin{equation*}x_0=-2c\log\Big(\frac{\eta}{4}\Big)+\sqrt{4c^2\log^2\Big(\frac{\eta}{4}\Big)-8 v \log\Big(\frac{\eta}{4}\Big)}\,,\end{equation*}

which concludes the proof.

A.2 Extension to the more general setting of Remark 1.2

As noted in Remark 1.2, the results of this article can be extended to a more general setting. A critical point for this extension is to construct a coupling of the Hawkes process $N^{h,\phi}$ with a Hawkes process $N^g$ satisfying Definition 1.1 for a nonnegative function g, in such a way that $N^{h,\phi} \le N^g$ (thinning). Then $N^g=\emptyset$ implies that $N^{h,\phi} =\emptyset$, and in particular this allows us to derive exponential bounds on the renewal time $\tau$ of $N^{h,\phi}$.

Proposition A.1. Assume that $N^{h,\phi}$ is a Hawkes process with conditional intensity $\Lambda^{h,\phi}$ defined in (1.13), and that the functions $\phi$ and h have the property that there exist $\lambda$ and a in $[0,\infty)$ such that for all $x\in\mathbb{R}$,

\begin{equation*}\phi(x)\le \lambda + a x^+\,\quad \text{and}\quad \,a\int h^+ <1.\end{equation*}

Let us define $g=a h^+$. Then there exists a coupling of $N^{h,\phi}$ with a Hawkes process $N^g$ in the sense of Definition 1.1 such that a.s. $N^{h,\phi}\le N^{g}$.

Scheme of the proof. As in the previous case, a key point is to establish an upper bound for the intensity $\Lambda^{h,\phi}$ on given time intervals. We have

\begin{align*}\phi\left(\int_{(\!-\infty,t)} h(t-u) N^{h,\phi} (du)\right)&\le \lambda + a \left(\int_{(\!-\infty,t)} h(t-u) N^{h,\phi} (du)\right)^+&&\text{(by assumption)}\\&\le \lambda + \int_{(\!-\infty,t)} g(t-u) N^{h,\phi} (du)&&\text{(since $g = a h^+$)}\\&\le \lambda + \int_{(\!-\infty,t)} g(t-u) N^g (du)&&\text{(thinning),}\end{align*}

and thus it is possible at each point U of $N^{g}$ to either include it into $N^{h,\phi}$ with probability

\begin{equation*}\frac{\phi\left(\int_{(\!-\infty,U)} h(U-u) N^{h,\phi} (du)\right)}{\lambda + \int_{(\!-\infty,U)} g(U-u) N^{g} (du)} \le1\end{equation*}

or else to reject it, independently of the rest. Then the conditional intensity of $N^{h,\phi}$ is given by

\begin{equation*}\frac{\phi\left(\int_{(\!-\infty,U)} h(t-u) N^{h,\phi} (du)\right)}{\lambda + \int_{(\!-\infty,t)} g(t-u) N^{g} (du)}\left(\lambda + \int_{(\!-\infty,t)} g(t-u) N^{g} (du)\right)= \phi\left(\int_{(\!-\infty,t)} h(t-u) N^{h,\phi} (du)\right)\!.\end{equation*}

A.3. Return time for M/G/$\infty$ queues

We now state a general result for the tail behavior of the time of return to zero $\mathcal{T}_1$ of an M/G/$\infty$ queue with a service time admitting exponential moments. All queues in this section start empty.

We recall that an M/G/$\infty$ queue has a Poisson process of customer arrivals with i.i.d. service times with a general distribution, and each customer starts its service immediately at arrival and leaves the system at its completion. For the Hawkes process with nonnegative reproduction function, we consider the ancestors to be customers (arriving as a Poisson process of intensity $\lambda$) with service times distributed as $\widetilde{H}_1(A) \triangleq H_1+A$, where $H_1$ is a cluster length (see Section 2.2), and then the queue empties exactly at the hitting times of $\emptyset$ by the auxiliary Markov process.

This result is of interest in itself, independently of the Hawkes process interpretation. Its proof is based on the computation of the Laplace transform $\mathbb{E}(\text{e}^{-s\mathcal{T}_1})$ on the half-plane $\{s\in\mathbb{C}\,{:}\, \Re(s) > 0\}$ by Takács [Reference Takács34, Reference Takács35]. We analytically extend this Laplace transform to $\{s\in\mathbb{C}\,{:}\, \Re(s) > s_c\}$ for an appropriate $s_c<0$, which yields exponential moments.

Theorem A.1. Consider an M/G/$\infty$ queue with arrival rate $\lambda >0$ and generic service duration H satisfying for some $\gamma >0$ that, for $t\ge0$,

\begin{equation*}\mathbb{P}(H>t) \triangleq1-G(t) = O(\text{e}^{-\gamma t})\,.\end{equation*}

Let $V_1$ denote the arrival time of the first customer, $\mathcal{T}_1$ the subsequent time of return of the queue to zero, and $B=\mathcal{T}_1-V_1$ the corresponding busy period.

1. 1. If $\beta <\gamma$ then $\mathbb{E}(\text{e}^{\beta B}) < \infty$. In particular $\mathbb{P}(B \ge t) = O(\text{e}^{-\beta t})$.

2. 2. If $\lambda <\gamma$, then $\mathbb{P}(\mathcal{T}_1 \ge t) = O(\text{e}^{-\lambda t})$. If $\gamma \le \lambda$, then $\mathbb{P}(\mathcal{T}_1 \ge t) = O(\text{e}^{-\alpha t})$ for $\alpha<\gamma$.

Proof. We have $\mathcal{T}_1=V_1+B$, and the strong Markov property of the Poisson process yields that $V_1$ and B are independent. Since $V_1$ is exponential of parameter $\lambda$, we need mainly to study B. Takács has proved in [Reference Takács34, Eq. (37)] (see also [Reference Takács35, Th. 1, p. 210]) that the Laplace transform of $\mathcal{T}_1$ satisfies

(A.6)\begin{equation}\mathbb{E}(\text{e}^{-s \mathcal{T}_1}) = 1-\frac1{\lambda + s} \frac1{\int_0^\infty \text{e}^{-st - \lambda \int_0^t[1-G(u)]\,\text{d} u}\,\text{d}t}\,,\qquad s\in\mathbb{C}\,,\;\Re(s) > 0\,.\end{equation}

Since the Laplace transform of $V_1$ is $\frac{\lambda}{\lambda + s}$, the Laplace transform of B satisfies

(A.7)\begin{equation}\mathbb{E}(\text{e}^{-s B}) =\frac{\lambda + s}{\lambda} -\frac1{\lambda} \frac1{\int_0^\infty \text{e}^{-st - \lambda \int_0^t[1-G(u)]\,\text{d} u}\,\text{d}t}\,,\qquad s\in\mathbb{C}\,,\;\Re(s) > 0\,.\end{equation}

There is an apparent singularity in the right-hand sides of (A.6) and of (A.7), since the integral term increases to infinity as s decreases to 0. This is normal, since these formulas remain valid for heavy-tailed service. Moreover, (A.6) is proved in [Reference Takács34] and [Reference Takács35] using the Laplace transform of a measure with infinite mass. We shall remove this apparent singularity and compute the abscissa of convergence of the Laplace transform in the left-hand side of (A.7).

The main point to prove is that the abscissa of convergence $\sigma_c$ of the Laplace transform in the left-hand side of (A.7) satisfies $\sigma_c \le -\gamma$. In order to remove the apparent singularity in the right-hand side of (A.7), we use integration by parts: on the half-line $\{s\in\mathbb{R}\,{:}\, s> 0\}$,

(A.8)\begin{align}\int_0^\infty \text{e}^{-st - \lambda \int_0^t[1-G(u)]\,\text{d} u}\,\text{d}t&= \left[\frac{\text{e}^{-st}}{-s} \text{e}^{- \lambda \int_0^t[1-G(u)]\,\text{d} u} \right]_{t=0}^\infty\notag\\[4pt] &\qquad - \int_0^\infty \frac{\text{e}^{-st}}{-s} (\!-\lambda[1-G(t)])\, \text{e}^{- \lambda \int_0^t[1-G(u)]\,\text{d} u}\,\text{d}t\notag\\[4pt] &= \frac1s - \frac{\lambda}s \int_0^\infty [1-G(t)]\, \text{e}^{-st- \lambda \int_0^t[1-G(u)]\,\text{d} u}\,\text{d}t\,.\end{align}

After inspection of the integral on the right-hand side, since $1-G(t) = O(\text{e}^{-\gamma t})$ and

\begin{equation*}\lambda \int_0^\infty [1-G(t)]\, \text{e}^{- \lambda \int_0^t[1-G(u)]\,\text{d} u}\,\text{d}t= \left[-\text{e}^{- \lambda \int_0^t[1-G(u)]\,\text{d} u} \right]_{t=0}^\infty = 1 - \text{e}^{- \lambda \mathbb{E}(H)} < 1,\end{equation*}

we are able to define a constant $\theta <0$ and an analytic function f by setting

(A.9)\begin{align}\theta &= \inf \left\{s \le 0\,{:}\, \lambda \int_0^\infty [1-G(t)] \,\text{e}^{-st- \lambda \int_0^t[1-G(u)]\,\text{d} u}\,\text{d}t <1\right\} \vee(\!- \gamma) \,,\notag\\[4pt] f(s) &=\frac{\lambda + s}{\lambda} - \frac{s}{\lambda}\frac1{1 - \lambda \int_0^\infty [1-G(t)]\, \text{e}^{-st- \lambda \int_0^t[1-G(u)]\,\text{d} u}\,\text{d}t}\,,\qquad s\in\mathbb{C}\,,\; \Re(s) > \theta \,.\nonumber\\\end{align}

The Laplace transform in the left-hand side of (A.7) has an abscissa of convergence $\sigma_c\le 0$ and is analytic in the half-plane $\{s\in\mathbb{C}\,{:}\, \Re(s) > \sigma_c\}$; see Widder [Reference Widder37, Th. 5a, p. 57]. Both this Laplace transform and f are analytic in the domain $\{s\in\mathbb{C}\,{:}\, \Re(s) > \max(\theta, \sigma_c)\}$, and since these two analytic functions coincide there on the half-line $\{s\in\mathbb{R}\,{:}\, s> 0\}$, they must coincide in the whole domain (see Rudin [Reference Rudin33, Th. 10.18, p. 208]), so that

\begin{equation*}\mathbb{E}(\text{e}^{-s B})=f(s) \,,\qquad s\in\mathbb{C} \,,\;\Re(s) > \max(\theta, \sigma_c)\,.\end{equation*}

This Laplace transform must have an analytic singularity at $s=\sigma_c$ (see Widder [Reference Widder37, Th. 5b, p. 58]), and since f is analytic in $\{s\in\mathbb{C}\,{:}\, \Re(s) > \theta\}$, necessarily $\sigma_c \le \theta$.

Since $\theta<0$, by monotone convergence we have

\begin{equation*}\lim_{s\to\theta^+}f(s) =\frac{\lambda + \theta}{\lambda} - \frac{\theta}{\lambda}\frac1{1 - \lambda \int_0^\infty [1-G(t)]\, \text{e}^{-\theta t- \lambda \int_0^t[1-G(u)]\,\text{d} u}\,\text{d}t}= \mathbb{E}(\text{e}^{- \theta B}) \in [1,\infty]\,,\end{equation*}

which implies that

\begin{equation*} \lambda \int_0^\infty [1-G(t)]\, \text{e}^{-\theta t- \lambda \int_0^t[1-G(u)]\,\text{d} u}\,\text{d}t <1, \end{equation*}

and thus that $\theta = -\gamma$.

We conclude that $\sigma_c \le -\gamma$. Thus, if $\beta <\gamma$, then $\mathbb{E}(\text{e}^{\beta B}) < \infty$, and $\mathbb{P}(B \ge t) = O(\text{e}^{-\beta t})$ using the Markov inequality. Moreover, if $\mathbb{P}(B \ge t) = O(\text{e}^{-\alpha t})$ then

\begin{align*}\mathbb{P}(\mathcal{T}_1 \ge t) = \mathbb{P}(B+V_1 \ge t) &= \text{e}^{-\lambda t}+ \lambda \int_0^t \text{e}^{-\lambda u}\mathbb{P}(B \ge t-u)\,\text{d}u\\[4pt] &\le \text{e}^{-\lambda t} + C \int_0^t \text{e}^{-\lambda u -\alpha (t-u)} \,\text{d}u\,;\end{align*}

hence, if $\lambda <\gamma$, then choosing $\lambda < \alpha < \gamma$ yields that

\begin{equation*}\mathbb{P}(\mathcal{T}_1 \ge t) \le \text{e}^{-\lambda t}+C \text{e}^{-\lambda t}\int_0^t \text{e}^{ -(\alpha-\lambda) (t-u)} \,\text{d}u\le [1 + C/(\alpha-\lambda)]\text{e}^{-\lambda t} ,\end{equation*}

and if $\alpha <\gamma \le \lambda $, then

\begin{equation*}\mathbb{P}(\mathcal{T}_1 \ge t) \le \text{e}^{-\lambda t} + C\text{e}^{-\alpha t}\int_0^t \text{e}^{-(\lambda -\alpha)u} \,\text{d}u\le \Bigl[1 + \frac{C}{\lambda-\alpha}\Bigr]\text{e}^{-\alpha t} \,.\end{equation*}

We now provide a corollary to the previous result.

Proposition A.2. Consider an M/G/$\infty$ queue with arrival rate $\lambda >0$ and generic service duration H satisfying for some $\gamma >0$ that

\begin{equation*}\mathbb{P}(H>t) = O(\text{e}^{-\gamma t})\,.\end{equation*}

Let $Y_t$ denote the number of customers at time $t\ge0$, and for each $E\ge 0$ let

(A.10)\begin{equation}\tau_E=\inf\{ t\geq E\,{:}\, Y_t=0\}\end{equation}

be the first hitting time of zero after E. If $\lambda<\gamma$ then let $\alpha=\lambda$, and if $\gamma\leq \lambda$ then let $0<\alpha<\gamma$. Then there exists a constant $C<\infty$ such that

\begin{equation*}\mathbb{P}(\tau_E \ge t)\leq \lambda C E\ e^{-\alpha(t-E)}\,,\quad \forall t\geq E\,.\end{equation*}

Proof. The successive return times to zero $(\mathcal{T}_k)_{k\geq 0}$ of the process $(Y_t)_{t\geq 0}$ have been defined in (2.7). The events $\{\mathcal{T}_{k-1}\leq E,\ \mathcal{T}_k>E\}$ for $k\geq 1$ define a partition of $\Omega$, and for $t>E$,

\begin{align*}\mathbb{P}(\tau_E \ge t)& = \sum_{k=1}^{+\infty} \mathbb{P}\big(\tau_E \ge t,\,\mathcal{T}_{k-1}\leq E,\,\mathcal{T}_k>E\big)\\& = \sum_{k=1}^{+\infty}\mathbb{P}\big( \mathcal{T}_{k-1}\leq E,\,\mathcal{T}_k\ge t\big)\\& = \sum_{k=1}^{+\infty}\mathbb{E}\Big( {\mathds{1}}_{\{\mathcal{T}_{k-1}\leq E\}} \mathbb{P}\big(\mathcal{T}_k\ge t \,|\,\mathcal{F}_{\mathcal{T}_{k-1}}\big)\Big)\\&\leq \sum_{k=1}^{+\infty}\mathbb{E}\Big( {\mathds{1}}_{\{\mathcal{T}_{k-1}\leq E\}} \mathbb{P}\big(\mathcal{T}_k-\mathcal{T}_{k-1} \ge t-E \,|\,\mathcal{F}_{\mathcal{T}_{k-1}}\big)\Big),\end{align*}

so that, since $\mathcal{T}_k-\mathcal{T}_{k-1} $ is independent of $\mathcal{F}_{\mathcal{T}_{k-1}}$ and distributed as $\mathcal{T}_1$,

\begin{equation*}\mathbb{P}(\tau_E \ge t)\le \sum_{k=1}^{+\infty}\mathbb{E}\Big( {\mathds{1}}_{\{\mathcal{T}_{k-1}\leq E\}} \Big) \mathbb{P}\big(\mathcal{T}_1 \ge t-E\big)= \mathbb{P}\big(\mathcal{T}_1 \ge t-E\big) \mathbb{E}\Bigg(\sum_{k=1}^{+\infty} {\mathds{1}}_{\{\mathcal{T}_{k-1}\leq E\}} \Bigg)\,.\end{equation*}

By Theorem A.1, under the assumptions there exists a constant C such that

\begin{equation*}\mathbb{P}\big(\mathcal{T}_1 \ge t-E\big)\leq C \text{e}^{\alpha(t-E)}\,.\end{equation*}

Moreover, $\sum_{k=1}^{+\infty} {\mathds{1}}_{\{\mathcal{T}_{k-1}\leq E\}}$ is the number of returns to zero before time E. It is bounded by the number of arrivals between times 0 and E, which follows a Poisson law of parameter and expectation $\lambda E$. This leads to the stated inequality.□

A.4. Strong Markov property for homogeneous Poisson point processes

In this appendix, we prove a strong Markov property for homogeneous Poisson point processes on the line. This classic result is stated in [Reference Robert32, Prop. 1.18, p. 18] in the case when the filtration is the canonical filtration generated by the Poisson point process. Here, the filtration $(\mathcal{F}_t)_{t\ge 0}$ may contain additional information, for example coming from configurations on $\mathbb{R}_-$.

Proof. It is enough to prove that, for any stopping time T and any $h,a>0$, conditionally on $T<\infty$ the random variable $Q((T,T+h]\times (0,a])$ is $\mathcal{F}_{T+h}$-measurable, independent of $\mathcal{F}_T$, and Poisson of parameter ha. Indeed, in order to prove the strong Markov property at a given stopping time T, it is enough to apply the above to the stopping times $T+t$ for $t>0$ in order to see that $S_TQ$ satisfies that for every $t,h,a>0$, the random variable $Q((t,t+h]\times (0,a])$ is $\mathcal{F}_{t+h}$-measurable, independent of $\mathcal{F}_t$, and Poisson of parameter ha.

We first prove this for an arbitrary stopping time T with finite values belonging to an increasing deterministic sequence $(t_n)_{n\ge1}$. For each B in $\mathcal{F}_T$ and $k\ge0$, we have

\begin{align*}&\mathbb{P}(B\cap\{T<\infty\}\cap \{Q((T,T+h]\times (0,a])=k\})\\&\quad =\sum_{n\ge1} \mathbb{P}(B\cap\{T=t_n\}\cap\{Q((t_n,t_n+h]\times (0,a])=k\}),\end{align*}

in which, by definition of $\mathcal{F}_T$ and since $\mathcal{F}_{t_{n-1}}\subset \mathcal{F}_{t_n}$,

\begin{equation*}B\cap\{T=t_n\} = (B\cap \{T\le t_n\}) - (B\cap \{T\le t_{n-1}\}) \in \mathcal{F}_{t_n}\,.\end{equation*}

The $(\mathcal{F}_t)_{t\ge0}$-Poisson point process property then yields that

\begin{equation*}\mathbb{P}(B\cap\{T=t_n\}\cap\{Q((t_n,t_n+h]\times (0,a])=k\}) = \mathbb{P}(B\cap\{T=t_n\})\,\text{e}^{-ha}\frac{(ha)^k}{k!},\end{equation*}

and summation of the series yields that

\begin{equation*}\mathbb{P}(B\cap\{T<\infty\}\cap \{Q((T,T+h]\times (0,a])=k\}) =\mathbb{P}(B\cap\{T<\infty\})\,\text{e}^{-ha}\frac{(ha)^k}{k!}\,.\end{equation*}

Hence $Q((T,T+h]\times (0,a])$ is independent of $\mathcal{F}_T$ and Poisson of parameter ha. Moreover, for $k\ge0$, similarly

\begin{align*} &\{T<\infty,\, Q((T,T+h]\times (0,a])=k\} \cap \{T+h\le t\} \\ &\quad = \bigcup_{n\ge1} \{T=t_n,\, Q((t_n,t_n+h]\times (0,a])=k\} \cap \{t_n+h\le t\} \subset\mathcal{F}_t,\end{align*}

and hence $Q((T,T+h]\times (0,a])$ is $\mathcal{F}_{T+h}$-measurable.

In order to extend this to a general stopping time T, we approximate T by the discrete stopping times

\begin{equation*}T_n = \sum_{k=1}^{+\infty} \frac{k}{2^n} {\mathds{1}}_{\{\frac{k-1}{2^n} < T \le \frac{k}{2^n}\}}\,,\qquad n\ge1\,.\end{equation*}

The nondecreasing sequence $(T_n)$ satisfies $T_n \ge T$ a.s. As n goes to infinity, the right continuity of $t\mapsto Q((0,t]\times(0,a])$ and of $(\mathcal{F}_t)_{t\ge0}$ allows us to conclude.