2. Trajectorial moderate-deviation limit theorems

Assume as given a sequence of many-server queues with unlimited waiting room indexed by n, where n represents the number of servers. Service is performed on a first-come-first-served basis. If, upon a customer’s arrival, there are available servers, then the customer starts being served by one of the available servers, chosen arbitrarily. Otherwise, the customer joins the queue and awaits her turn to be served. When the service is complete, the customer leaves, relinquishing the server.

Let $Q_n(t)$ denote the number of customers present at time t. Of those customers, $Q_n(t)\wedge n $ customers are in service and $(Q_n(t)-n)^+$ customers are in the queue. The service times of the customers in the queue at time 0 and the service times of customers exogenously arriving after time 0 are denoted by $\eta_1,\eta_2,\ldots$ (in the order in which they enter service) and come from a sequence of independent and identically distributed (i.i.d.) positive unbounded random variables with continuous CDF F. It is thus assumed that

\[F(0)=0, F(x)<1, \text{ for all }x.\]

The mean service time $\mu^{-1}=\int_0^\infty x\,dF(x)$ is assumed to be finite. The residual service times of customers in service at time 0 are denoted by $\eta^{(0)}_1,\eta^{(0)}_2,\ldots$ and are assumed to be i.i.d. with CDF $F_0$ , which is the CDF of the delay in a stationary renewal process with inter-renewal CDF F. Thus,

(2.1) \begin{equation} F_0(x)=\mu\int_0^x(1-F(y))\,dy.\end{equation}

Let $A_n(t)$ denote the number of exogenous arrivals by time t, with $A_n(0)=0$ . It is assumed that the process $A_n(t)$ has unit jumps. The entities $Q_n(0)$ , $\{\eta^{(0)}_1,\eta^{(0)}_2,\ldots\}$ , $\{\eta_1,\eta_2,\ldots\}$ , and $A_n=(A_n(t),\,t\in\mathbb R_+)$ are assumed to be independent. All stochastic processes are assumed to have right-continuous paths with left-hand limits. Let ${{\hat A}_n}(t)$ denote the number of customers that enter service after time 0 and by time t, with ${{\hat A}_n}(0)=0$ . Since the random variables $\eta_i$ are continuous, the process $\hat A_n=(\hat A_n(t),t\in\mathbb R_+)$ has unit jumps almost surely. Balancing the arrivals and departures yields the equation

(2.2) \begin{equation} Q_n(t)=Q_n^{(0)}(t)+(Q_n(0)-n)^++A_n(t)-\int_0^t\int_0^t \mathbf{1}_{\{x+s\le t\}}\,d\,\sum_{i=1}^{{{\hat A}_n}(s)}\mathbf{1}_{\{\eta_i\le x\}},\end{equation}

where

(2.3) \begin{equation} Q_n^{(0)}(t)= \sum_{i=1}^{Q_n(0)\wedge n}\,\,\mathbf{1}_{\{\eta^{(0)}_i> t\}}\,,\end{equation}

which represents the number of customers present at time t out of those in service at time 0, and

\[ \int_0^t\int_0^t \mathbf{1}_{\{x+s\le t\}}\,d\,\sum_{i=1}^{{{\hat A}_n}(s)}\mathbf{1}_{\{\eta_i\le x\}}=\sum_{i=1}^{\hat A_n(t)}\,\mathbf{1}_{\{\eta_i+\hat\tau_{n,i}\le t\}}\,,\]

which represents the number of customers that enter service after time 0 and leave by time t, with $\hat\tau_{n,i}$ denoting the ith jump time of $\hat A_n$ , i.e., $ \hat\tau_{n,i}=\inf\{t:\,\hat A_n(t)\ge i\}.$ In addition, since each customer that is either in the queue at time 0 or has arrived exogenously by time t must either be in the queue at time t or have entered service by time t,

(2.4) \begin{equation} (Q_n(0)-n)^++A_n(t)=(Q_n(t)-n)^++{\hat A}_n(t).\end{equation}

For the existence and uniqueness of a solution to (2.2)–(2.4), the reader is referred to [Reference Puhalskii and Reed21].

Given $r_n\to\infty$ , as $n\to\infty$ , a sequence $\mathbb P_n$ of probability laws on the Borel $\sigma$ -algebra of a metric space M, and a $[0,\infty]$ -valued function I on M such that the sets $\{y\in M:\,I(y)\le \gamma\}$ are compact for all $\gamma\ge0$ , the sequence $\mathbb P_n$ is said to obey the LDP for rate $r_n$ with deviation function I, also referred to as a rate function, provided $ \lim_{n\to\infty}1/r_n\,\ln \mathbb P_n(W)=-\inf_{y\in W}I(y),$ for all Borel sets W such that the infima of I over the interior and the closure of W agree.

We now introduce the deviation function for the number-in-the-system process. For $T>0$ and $m\in\mathbb N$ , let $\mathbb D([0,T],\mathbb R^m)$ and $\mathbb D(\mathbb R_+,\mathbb R^m)$ represent the Skorokhod spaces of right-continuous $\mathbb R^m$ -valued functions with left-hand limits defined on [0, T] and $\mathbb R_+$ , respectively. These spaces are endowed with metrics rendering them complete separable metric spaces; see [Reference Ethier and Kurtz6, Reference Jacod and Shiryaev9] for more detail. Given $q=(q(t),t\in\mathbb R_+)\in\mathbb D(\mathbb R_+,\mathbb R)$ and $x_0\in\mathbb R$ , let

(2.5) \begin{equation} I_{x_0}^Q(q)=\frac{1}{2}\inf\bigg\{\int_0^1\dot w^0(x)^2\,dx+\int_0^\infty \dot w(t)^2\,dt+\int_0^\infty\int_0^1 \dot k(x,t)^2\,dx\,dt\bigg\}, \end{equation}

the infimum being taken over $w^0=(w^0(x),x\in[0,1])\in \mathbb D([0,1],\mathbb R)$ , $w=(w(t),t\in\mathbb R_+)\in\mathbb D(\mathbb R_+,\mathbb R)$ , and $k=((k(x,t),x\in[0,1]),t\in\mathbb R_+)\in\mathbb D(\mathbb R_+,\mathbb D([0,1],\mathbb R))$ such that $w^0(0)=w^0(1)=0$ , $w(0)=0$ , $k(x,0)=k(0,t)=k(1,t)=0$ ; $w^0$ , w, and k are absolutely continuous with respect to the Lebesgue measures on [0, 1], $\mathbb R_+$ , and $[0,1]\times \mathbb R_+$ , respectively; and, for all t,

(2.6) \begin{multline}q(t)=(1-F(t))x_0^+-(1-F_0(t))x_0^- -\beta F_0(t) +\int_0^tq(t-s)^+\,dF(s)+w^0(F_0(t))\\[5pt] +\int_0^t\bigl(1-F(t-s)\bigr)\sigma\,\dot w(s)\,ds+\int_{\mathbb R_+^2} \,\mathbf{1}_{\{x+s\le t\}}\,\,\dot k(F(x),\mu s)\,dF(x)\,\mu \,ds,\end{multline}

where $x_0^-=(\!-x_0)^+$ , and $\dot w^0(x)$ , $\dot w(t)$ and $\dot k(x,t)$ represent the respective Radon–Nikodym derivatives. If $w^0$ , w, and k as indicated do not exist, then $I_{x_0}^Q(q)=\infty$ . Note that $I_{x_0}^Q(q)=\infty$ unless $q(0)=x_0$ and q(t) is a continuous function, as the right-hand side of (2.6) is a continuous function of t. It is proved in Lemma 8 that if F is, in addition, absolutely continuous with respect to Lebesgue measure, then q(t) in (2.6) is absolutely continuous too. By Lemma B.1 in [Reference Puhalskii and Reed21], the equation (2.6) has a unique solution q(t) in the space of essentially locally bounded functions.

Let the process $X_n=(X_n(t),t\in\mathbb R_+)$ be defined by

(2.7) \begin{equation} X_n(t)=\frac{\sqrt{n}}{b_n}\bigg(\frac{Q_n(t)}{n}-1\bigg).\end{equation}

The next theorem verifies and refines Conjecture 1 in [Reference Puhalskii20]. Its proof is presented in Section 3.

Theorem 1. Suppose, in addition, that $A_n$ is a renewal process of rate $\lambda_n$ . Let $\rho_n=\lambda_n/(n\mu)$ , $\beta\in\mathbb R$ , $x_0\in\mathbb R$ , and $\sigma>0$ . Suppose that, as $n\to\infty$ ,

(2.8) \begin{equation} \frac{\sqrt{n}}{b_n}(1-\rho_n)\to\beta\end{equation}

and the sequence of random variables $\sqrt{n}/b_n\,(Q_n(0)/n-1)$ obeys the LDP in $\mathbb R$ for rate $b_n^2$ with deviation function $I_{x_0}(y)$ such that $I_{x_0}(x_0)=0$ and $I_{x_0}(y)=\infty$ , for $y\not=x_0$ . Suppose that the sequence of processes $\bigl((A_n(t)-\lambda_nt)/(b_n\sqrt{n}),t\in\mathbb R_+\bigr)$ obeys the LDP in $\mathbb D(\mathbb R_+,\mathbb R)$ for rate $b_n^2$ with deviation function $I^A(a)$ such that $I^A(a)=1/(2\sigma^2)\int_0^\infty\dot a(t)^2\,dt$ , provided $a=(a(t),t\in\mathbb R_+)$ is an absolutely continuous function with $a(0)=0$ , and $I^A(a)=\infty$ , otherwise. If, in addition,

(2.9) \begin{equation} b_n^6n^{1/b_n^2-1}\to0,\end{equation}

then the sequence $X_n$ obeys the LDP in $\mathbb D(\mathbb R_+,\mathbb R)$ for rate $b_n^2$ with deviation function $I_{x_0}^Q(q)$ .

As suggested by a referee, the next statement provides a version for the case of infinitely many servers. Consider a $GI/GI/\infty$ queue with renewal arrival process $A_n$ of rate $\lambda_n=n\lambda$ . All the assumptions and notation concerning the service times are the same as in Theorem 1. The arrival process, the initial number of customers, and the service times are independent. With $\overline Q_n(t)$ denoting the number of customers present at time t, the equations (2.2) and (2.3) are replaced with the respective equations

(2.10) \begin{equation} \overline{Q}_n(t)=\overline{Q}_n^{(0)}(t)+A_n(t)-\int_0^t\int_0^t \mathbf{1}_{\{x+s\le t\}}\,d\,\sum_{i=1}^{{A_n}(s)}\mathbf{1}_{\{\eta_i\le x\}}\end{equation}

and

(2.11) \begin{equation} \overline{Q}_n^{(0)}(t)= \sum_{i=1}^{\overline{Q}_n(0)}\,\,\mathbf{1}_{\{\eta^{(0)}_i> t\}}.\end{equation}

Given $q_0\in\mathbb R_+$ , let

(2.12) \begin{equation} \overline q(t)=q_0(1-F_0(t))+\lambda t-\lambda\int_0^t(t-s)\,dF(s)\end{equation}

and

(2.13) \begin{equation} \overline{X}_n(t)=\frac{\sqrt{n}}{b_n}\bigg(\frac{\overline Q_n(t)}{n}-\overline q(t)\bigg).\end{equation}

Theorem 2. Suppose that the sequence of processes $\bigl((A_n(t)-\lambda_nt)/(b_n\sqrt{n}),t\in\mathbb R_+\bigr)$ obeys the LDP in the hypotheses of Theorem 1. Given $x_0\in\mathbb R$ , suppose that the sequence $\overline{X}_n(0)$ obeys the LDP with deviation function $\overline I_{x_0}(y)$ such that $\overline I_{x_0}(x_0)=0$ and $\overline I_{x_0}(y)=\infty$ , for $y\not=x_0$ . If, in addition, (2.9) holds, then the sequence $\overline{X}_n$ obeys the LDP in $\mathbb D(\mathbb R_+,\mathbb R)$ for rate $b_n^2$ with deviation function $\overline I^Q_{q_0,x_0}(q)$ given by the right-hand side of (2.5), provided

\begin{multline*} q(t)=(1-F_0(t))x_0+\sqrt{q_0}\,w^0(F_0(t))+\int_0^t\bigl(1-F(t-s)\bigr)\sigma\,\dot w(s)\,ds\\[5pt] +\int_{\mathbb R_+^2} \,\mathbf{1}_{\{x+s\le t\}}\,\,\dot k(F(x),\lambda s)\,dF(x)\,\lambda \,ds,\end{multline*}

and equals $\infty$ otherwise.

3. Large-deviation convergence and proofs of Theorems 1 and 2

It is convenient to recast Theorem 1 as a statement on LD convergence. Introduce

(3.1) \begin{equation} Y_n(t)=\frac{\sqrt{n}}{b_n}\,\bigl(\frac{A_n(t)}{n}-\mu t\bigr)\end{equation}

and let $Y_n=(Y_n(t),\,t\in\mathbb R_+)$ . For the statement and proof of the next theorem, Appendix A is recommended reading.

Theorem 3. Suppose that, as $n\to\infty$ , the sequence $X_n(0)$ LD converges in distribution in $\mathbb R$ at rate $b_n^2$ to an idempotent variable X(0), the sequence $Y_n$ LD converges in distribution in $\mathbb D(\mathbb R_+,\mathbb R)$ at rate $b_n^2$ to an idempotent process Y with continuous paths, and (2.9) holds. Then the sequence $X_n$ LD converges in distribution in $\mathbb D(\mathbb R_+,\mathbb R)$ at rate $b_n^2$ to the idempotent process $X=(X(t),t\in\mathbb R_+)$ that is the unique solution to the equation

(3.2) \begin{multline} X(t)=(1-F(t))X(0)^+- (1-F_0(t))X(0)^-+\int_0^tX(t-s)^+\,dF(s)+W^0(F_0(t))\\[5pt] +Y(t)-\int_0^t Y(t-s)\,dF(s)+\int_0^t\int_{0}^t \,\mathbf{1}_{\{x+s\le t\}}\,\,\dot K( F(x),\mu s)\,dF(x)\,\mu\,ds, \end{multline}

where $W^0=(W^0(x),x\in[0,1])$ is a Brownian bridge idempotent process and $K=(K(x,t),(x,t)\in [0,1]\times\mathbb R_+)$ is a Kiefer idempotent process, X(0), Y, $W^0$ , and K being independent.

Theorem 1 is obtained as a special case. Suppose $A_n$ is a renewal process of rate $\lambda_n$ , the condition (2.8) holds, and the sequence $((A_n(t)-\lambda_nt)/(b_n\sqrt{n})\,,t\in\mathbb R_+)$ LD converges in distribution in $\mathbb D(\mathbb R_+,\mathbb R)$ at rate $b_n^2$ to $\sigma W$ , where $\sigma>0$ and $W=(W(t),t\in\mathbb R_+)$ is a standard Wiener idempotent process. Then, in the statement of Theorem 3, $Y(t)=\sigma W(t)-\beta\mu t$ , so that the limit idempotent process X solves the equation

\begin{multline*} X(t)=(1-F(t))X(0)^+-(1-F_0(t))X(0)^- -\beta F_0(t)+\int_0^tX(t-s)^+\,dF(s)\\[5pt] +W^0(F_0(t))+\int_0^t\bigl(1-F(t-s)\bigr)\sigma\,\dot W(s)\,ds+\int_0^t\int_{0}^t \,\mathbf{1}_{\{x+s\le t\}}\,\,\dot K(F(x),\mu s)\,dF(x)\,\mu \,ds, \end{multline*}

with X(0), W, $W^0$ , and K being independent. The assertion of Theorem 1 follows on observing that $\exp(\!-I^Q(y))$ , with $y\in\mathbb D(\mathbb R_+,\mathbb R)$ , is the deviability density of the idempotent distribution of X. To see the latter, note that the mapping $(w^0,w,k)\to q$ , as specified by (2.6), is continuous when restricted to the set $\{(w^0,w,k):\Pi^{W^0,W,K}(w^0,w,k)\ge a\}$ , where $\Pi^{W^0,W,K}(w^0,w,k)=\Pi^{W^0}(w^0)\Pi^{W}(w)\Pi^K(k)$ and $a\in(0,1]$ , so that X is strictly Luzin on $\bigl(\mathbb D([0,1],\mathbb R)\times\mathbb D(\mathbb R_+,\mathbb R)\times\mathbb D(\mathbb R_+,\mathbb D(\mathbb R_+,\mathbb R)),\Pi^{W^0,W,K}\bigr)$ ; see Appendix A for the definition and properties of being strictly Luzin. Therefore,

\[ \Pi^X(q)=\Pi^{W^0,W,K}(X=q)=\text{sup}_{(w^0,w,k):\,\eqref{eq:1}\text{ holds}}\Pi^{W^0}(w^0)\Pi^W(w)\Pi^K(k).\]

It is noteworthy that the limit idempotent process in (3.2) is analogous to the limit stochastic process on p. 139 in [Reference Puhalskii and Reed21].

The proof of Theorem 3 relies on an analogue of the weak convergence of the sequential empirical process to the Kiefer process; see, e.g., [Reference Krichagina and Puhalskii12]. Let random variables $\zeta_i$ be independent and uniform on [0, 1]. Define the centred and normalised sequential empirical process by

(3.3) \begin{align} K_n(x,t)&=\frac{1}{b_n\sqrt{n}}\,\sum_{i=1}^{\lfloor nt \rfloor}\bigl(\,\mathbf{1}_{\{\zeta_i\le x\}}\,-x\bigr),\end{align}

and let

(3.4) \begin{align}B_n(x,t)&=\frac{1}{b_n\sqrt n}\sum_{i=1}^{\lfloor nt\rfloor}\bigg(\,\mathbf{1}_{\{\zeta_i\le x\}}\,-\int_0^{x\wedge \zeta_i}\frac{dy}{1-y}\bigg),\end{align}

where $x\in[0,1]$ and $t\in\mathbb R_+$ . It is a simple matter to check that

(3.5) \begin{equation}K_n(x,t)=-\int_0^x\frac{K_n(y,t)}{1-y}\,dy+B_n(x,t).\end{equation}

Let $K_n=\bigl((K_n(x,t),\,x\in[0,1]),t\in\mathbb R_+\bigr)$ and $B_n=\bigl((B_n(x,t),\,x\in[0,1]),t\in\mathbb R_+\bigr)$ . Both processes are considered as random elements of $\mathbb D(\mathbb R_+,\mathbb D([0,1],\mathbb R))$ . Let $B=((B(x,t),x\in[0,1]),t\in\mathbb R_+)$ represent a Brownian sheet idempotent process, which is the canonical coordinate process on $\mathbb D(\mathbb R_+,\mathbb D([0,1],\mathbb R))$ , endowed with deviability $\Pi$ . Let $K=((K(x,t),x\in[0,1]),t\in\mathbb R_+)$ be defined as the solution of the equation

(3.6) \begin{equation} K(x,t)=-\int_0^x\frac{K(y,t)}{1-y}\,dy+B(x,t).\end{equation}

It is a Kiefer idempotent process by Lemma 7.

Proof. The proof draws on the proof of Lemma 3.1 in [Reference Krichagina and Puhalskii12]; see also [Reference Jacod and Shiryaev9, Chapter IX, §4c]. Given $t\in\mathbb R_+$ , let $K_n(t)=(K_n(x,t),\,x\in[0,1])\in\mathbb D([0,1],\mathbb R)$ , $B_n(t)=(B_n(x,t),\,x\in[0,1])\in\mathbb D([0,1],\mathbb R)$ , $K(t)=(K(x,t),\,x\in[0,1])\in\mathbb D([0,1],\mathbb R)$ , and $B(t)=(B(x,t),\,x\in[0,1])\in\mathbb D([0,1],\mathbb R).$ We prove first that, for $0\le t_1<t_2<\ldots<t_k$ , the sequence of $\mathbb D([0,1],\mathbb R)^{2k}$ -valued stochastic processes $\bigl((K_n(t_1),B_n(t_1)), ((K_n(t_2),B_n(t_2)),\ldots,(K_n(t_k),B_n(t_k))\bigr)$ LD converges to the $\mathbb D([0,1],\mathbb R)^{2k}$ -valued idempotent process $\bigl((K(t_1),B(t_1)),(K(t_2),B(t_2)),\ldots,(K(t_k),B(t_k))\bigr)$ in $\mathbb D([0,1],\mathbb R)^{2k}$ , as $n\to\infty$ . Since both the stochastic processes $((K_n(t),B_n(t)),t\in\mathbb R_+)$ and the idempotent process $((K(t),B(t)),t\in\mathbb R_+)$ have independent increments in t (see Lemma 7), it suffices to prove convergence of one-dimensional distributions, so we work with $((K_n(x,t),B_n(x,t)), x\in[0,1])$ and $((K(x,t),B(x,t)),x\in[0,1])$ , holding t fixed. By (3.4) and [Reference Jacod and Shiryaev9, Chapter II, §3c], the stochastic process $(B_n(x,t),x\in[0,1])$ is a martingale with respect to the natural filtration with the measure of jumps

\[ \mu^{n,B}([0,x],\Gamma)=\,\mathbf{1}_{\{1/(b_n\sqrt{n})\in\Gamma\}}\,\sum_{i=1}^{\lfloor nt\rfloor}\,\mathbf{1}_{\{\zeta_i\le x\}}\,,\]

the predictable measure of jumps

\[ \nu^{n,B}([0,x],\Gamma)=\,\mathbf{1}_{\{1/(b_n\sqrt{n})\in\Gamma\}}\,\sum_{i=1}^{\lfloor nt\rfloor}\int_0^{x\wedge \zeta_i}\frac{dy}{1-y},\]

and the predictable quadratic variation process

(3.7) \begin{multline} \langle B_n\rangle (x,t)=\int_0^x\int_\mathbb R u^2\nu^{n,B}(dy,du)=\frac{1}{b_n^2n}\,\nu^{n,B}([0,x],\{1/(b_n\sqrt{n})\})\\[5pt] =\frac{1}{b_n^2n}\,\sum_{i=1}^{\lfloor nt\rfloor}\int_0^{x\wedge \zeta_i}\frac{dy}{1-y}=\frac{\lfloor nt\rfloor}{b_n^2n}\,x+\frac{1}{b_n\sqrt{n}}\,K_n(x,t)-\frac{1}{b_n\sqrt{n}}\,B_n(x,t),\end{multline}

where $\Gamma\subset\mathbb R\setminus\{0\}$ .

We show next that

(3.8) \begin{equation} \lim_{r\to\infty}\limsup_{n\to\infty}\mathbb P(\text{sup}_{x\in[0,1]}\lvert B_n(x,t)\rvert>r)^{1/b_n^2}=0.\end{equation}

Since the process

\begin{multline*} \exp\bigl( b_n^2 B_n(x,t)-\int_0^x\int_\mathbb R(e^{ b_n^2 u}-1- b_n^2u)\nu^{n,B}(dy,du)\bigr)\\[5pt] = \exp\bigg( b_n^2 B_n(x,t)-(e^{ b_n/\sqrt{n}}-1- \frac{b_n}{\sqrt{n}})\sum_{i=1}^{\lfloor nt\rfloor}\int_0^{x\wedge \zeta_i}\frac{dy}{1-y}\bigg)\end{multline*}

is a local martingale with respect to x (see, e.g., [Reference Puhalskii19, Lemma 4.1.1, p. 294]), for any stopping time $\tau$ ,

\[ \mathbb E\exp\bigg( b_n^2 B_n(\tau,t)-(e^{ b_n/\sqrt{n}}-1- \frac{b_n}{\sqrt{n}})\sum_{i=1}^{\lfloor n t\rfloor}\int_0^{\tau\wedge \zeta_i}\frac{dy}{1-y}\bigg)\le 1.\]

Lemma 3.2.6 on p. 282 in [Reference Puhalskii19] implies that, for $r>0$ and $\gamma>0$ ,

\begin{multline*} \!\!\!\!\mathbb P(\text{sup}_{x\in[0,1]}e^{ b_n^2 B_n(x,t)}\ge e^{ b_n^2 r})\le e^{ b_n^2(\gamma- r)}+\mathbb P\bigg(\!\!\exp\!\bigg(\!\bigg(\!e^{ b_n/\sqrt{n}}-1- \frac{b_n}{\sqrt{n}}\!\bigg)\!\sum_{i=1}^{\lfloor n t\rfloor}\int_0^{ \zeta_i}\frac{dy}{1-y}\!\bigg)\ge e^{ b_n^2\gamma}\!\bigg)\\[5pt] \le e^{ b_n^2(\gamma- r)}+ e^{- b_n^2\gamma}\mathbb E\bigg(\!\!\exp\bigg(\bigg(e^{ b_n/\sqrt{n}}-1- \frac{b_n}{\sqrt{n}}\bigg)\sum_{i=1}^{\lfloor n t\rfloor}\int_0^{ \zeta_i}\frac{dy}{1-y}\bigg)\bigg)\\[5pt] =e^{ b_n^2(\gamma- r)}+ e^{- b_n^2\gamma}\big(1-(e^{ b_n/\sqrt{n}}-1- b_n/\sqrt{n})\big)^{-\lfloor nt\rfloor},\end{multline*}

with the latter equality holding for all n large enough because $e^{ b_n/\sqrt{n}}-1- b_n/\sqrt{n}\to0$ . Hence, assuming that $e^{ b_n/\sqrt{n}}-1- b_n/\sqrt{n}\le1/2$ , we have

\[ \mathbb P(\text{sup}_{x\in[0,1]}e^{ b_n^2 B_n(x,t)}\ge e^{ b_n^2 r})^{1/b_n^2}\le e^{ \gamma- r}+e^{- \gamma}2^{\lfloor nt\rfloor/b_n^2}.\]

Since $n/b_n^2\to\infty$ , it follows that

(3.9) \begin{equation} \lim_{ r\to\infty}\limsup_{n\to\infty} \mathbb P(\text{sup}_{x\in[0,1]} B_n(x,t)\ge r)^{1/b_n^2}=0.\end{equation}

A similar convergence holds with $-B_n(x,t)$ substituted for $B_n(x,t)$ . The limit (3.8) has been proved.

We next prove that, similarly,

(3.10) \begin{equation} \lim_{r\to\infty}\limsup_{n\to\infty}\mathbb P(\text{sup}_{x\in[0,1]}\lvert K_n(x,t)\rvert>r)^{1/b_n^2}=0.\end{equation}

Since, by (3.3), $(K_n(x,t),x\in[0,1])$ is distributed as $(\!-K_n(1-x,t),x\in[0,1])$ , it suffices to prove that

(3.11) \begin{align} \lim_{r\to\infty}\limsup_{n\to\infty}\mathbb P(\text{sup}_{x\in[0,1/2]}\lvert K_n(x,t)\rvert>r)^{1/b_n^2}=0.\end{align}

By (3.5) and the Gronwall–Bellman inequality,

(3.12) \begin{equation} \text{sup}_{x\in[0,1/2]}\lvert K_n(x,t)\rvert\le e^2\text{sup}_{x\in[0,1/2]}\lvert B_n(x,t)\rvert,\end{equation}

so that (3.11) follows from (3.9).

By (3.7), (3.8), and (3.10), for $x\in[0,1]$ ,

(3.13) \begin{equation} \lim_{n\to\infty}\mathbb P(\lvert b_n^2\langle B_n\rangle (x,t)-tx\rvert>\epsilon)^{1/b_n^2}=0.\end{equation}

If we extend it past $x=1$ by letting $B_n(x,t)=B_n(1,t)$ , the process $(B_n(x,t),x\in\mathbb R_+)$ is a square-integrable martingale with predictable quadratic variation process $(\langle B_n\rangle(x\wedge1,t), x\in\mathbb R_+)$ , so, by (3.13) and Theorem 5.4.4 on p. 423 in [Reference Puhalskii19], where one takes $\beta_\phi=b_n\sqrt{n}$ , $\alpha_\phi=n$ , and $r_\phi=b_n^2$ , the sequence of the extended processes $(B_n(x,t),x\in\mathbb R_+)$ LD converges in $\mathbb D(\mathbb R_+,\mathbb R)$ to the idempotent process $(B(x\wedge1,t),x\in\mathbb R_+)$ . By (3.5), (3.6), and the continuous mapping principle, for $0\le x_1\le\ldots\le x_l<1$ , the $((K_n(x_i,t),B_n(x_i,t)),i\in\{1,2,\ldots,l\})$ LD converge in $\mathbb R^{2l}$ to $((K(x_i,t),B(x_i,t)),i\in\{1,2,\ldots,l\})$ . Since $K_n(1,t)=0$ and $K(1,t)=0$ (see Appendix A), the latter convergence also holds if $x_l=1$ .

We now show that the sequence $(K_n(x,t),x\in[0,1])$ is $\mathbb C$ -exponentially tight of order $b_n^2$ in $\mathbb D([0,1],\mathbb R)$ . (The definition and basic properties of $\mathbb C$ -exponential tightness are reviewed in Appendix A.) By Theorem 8, (3.10) needs to be complemented with

(3.14) \begin{equation} \lim_{\delta\rightarrow0}\limsup_{n\to\infty}\text{sup}_{x\in[0,1]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n(x+y,t)-K_n(x,t)|\ge \eta)^{1/b_n^2}=0,\end{equation}

for arbitrary $\eta>0$ , where $K_n(x,t)=0$ when $x\ge1$ . We use an argument similar to that used in the proof of (3.10). Defining $\overline{K}_n(x,t)=-K_n(1-x,t)$ for $x\in[0,1]$ and $\overline{K}_n(x,t)=0$ for $x\ge 1$ , we have by (3.3) that

(3.15) \begin{align} &\text{sup}_{x\in[0,1]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n(x+y,t)-K_n(x,t)|\ge \eta)\nonumber\\[5pt] & \le \text{sup}_{x\in[0,1/2]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n( x+y,t)-K_n(x,t)|\ge \eta)\nonumber\\[5pt] &+\text{sup}_{x\in[1/2,1-\delta]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n(x+y,t)- K_n(x,t)|\ge \eta)\nonumber\\[5pt] &+\text{sup}_{x\in[1-\delta,1]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n(x+y,t)- K_n(x,t)|\ge \eta)\nonumber\\[5pt] &\le\text{sup}_{x\in[0,1/2]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n( x+y,t)-K_n(x,t)|\ge \eta)\nonumber\\[5pt] &+\text{sup}_{x\in[\delta,1/2]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n(1-x+y,t)- K_n(1-x,t)|\ge \eta) \\[5pt] &+\mathbb P(\text{sup}_{u\in[1- \delta,1]}|K_n(u,t)|\ge \eta/2)\le\text{sup}_{x\in[0,1/2]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n( x+y,t)-K_n(x,t)|\ge \eta)\nonumber\\[5pt] &+\text{sup}_{x\in[\delta,1/2]}\mathbb P(\text{sup}_{y\in[0, \delta]}|\overline K_n(x-y,t)-\overline K_n(x,t)|\ge \eta)\nonumber\\[5pt] &+\mathbb P(\text{sup}_{u\in[0, \delta]}|\overline K_n(u,t)|\ge \eta/2)\nonumber\\[5pt] &\le\text{sup}_{x\in[0,1/2]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n( x+y,t)-K_n(x,t)|\ge \eta)\nonumber\\[5pt] &+\text{sup}_{x\in[0,1/2]}\mathbb P(\text{sup}_{y\in[0, \delta]}|\overline K_n(x+y,t)-\overline K_n(x,t)|\ge \eta)\nonumber\\[5pt] & +\mathbb P(\text{sup}_{u\in[0, \delta]}|\overline K_n(u,t)|\ge \eta/2).\end{align}

Since the random variables $\zeta_i$ are independent and uniformly distributed on [0,1], $\overline{K}_n$ has the same finite-dimensional distributions as $K_n$ , so that

(3.16) \begin{multline} \text{sup}_{x\in[0,1]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n(x+y,t)-K_n(x,t)|\ge \eta)\\[5pt] \le2\text{sup}_{x\in[0,1/2]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n( x+y,t)-K_n(x,t)|\ge \eta)+\mathbb P(\text{sup}_{u\in[0, \delta]}| K_n(u,t)|\ge \eta/2).\end{multline}

Since $x+y\le 2/3$ when $x\in[0,1/2]$ and $y\in[0,\delta]$ provided $\delta$ is small enough, by (3.5), for $x\in[0,1/2]$ and $\delta$ small enough,

(3.17) \begin{align} & \text{sup}_{y\in[0,\delta]}\lvert K_n(x+y,t)-K_n(x,t)\rvert\le\nonumber\\[5pt]& 3\delta\text{sup}_{u\in[0,1]}\lvert K_n(u,t)\rvert+\text{sup}_{y\in[0,\delta]}\lvert B_n(x+y,t)-B_n(x,t)\rvert.\end{align}

Similarly,

(3.18) \begin{equation} \text{sup}_{u\in[0, \delta]}| K_n(u,t)|\le\frac{\delta}{1-\delta}\,\text{sup}_{u\in[0,1]}\lvert K_n(u,t)\rvert+\text{sup}_{u\in[0, \delta]}| B_n(u,t)|.\end{equation}

By (3.16), (3.17), (3.18), and the fact that $B_n(t)$ LD converges to B(t),

(3.19) \begin{align} &\limsup_{n\to\infty}\text{sup}_{x\in[0,1]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n(x+y,t)-K_n(x,t)|\ge \eta)^{1/b_n^2}\nonumber\\[5pt]&\le\limsup_{n\to\infty}\mathbb P(3\delta\text{sup}_{u\in[0,1]}\lvert K_n(u,t)\rvert\ge\eta/2)^{1/b_n^2}\nonumber\\[5pt] &+\limsup_{n\to\infty}\mathbb P(\frac{\delta}{1-\delta}\,\text{sup}_{u\in[0,1]}\lvert K_n(u,t)\rvert\ge\eta/4)^{1/b_n^2}\nonumber\\[5pt]&+\Pi(\text{sup}_{x\in[0,1/2]}\text{sup}_{y\in[0,\delta]}\lvert B(x+y,t)-B(x,t)\rvert\ge \eta/2)\nonumber\\[5pt] &+\Pi(\text{sup}_{u\in[0, \delta]}| B(u,t)|\ge\eta/4).\end{align}

By (3.10),

(3.20) \begin{align} &\lim_{\delta\to0}\limsup_{n\to\infty}\text{sup}_{x\in[0,1]}\mathbb P(\text{sup}_{y\in[0, \delta]}|K_n(x+y,t)-K_n(x,t)|\ge \eta)^{1/b_n^2} \nonumber\\[5pt] &\le\lim_{\delta\to0} \Pi(\text{sup}_{x\in[0,1/2]}\text{sup}_{y\in[0,\delta]}\lvert B(x+y,t)-B(x,t)\rvert\ge \eta/2)\nonumber\\[5pt]&+\lim_{\delta\to0}\Pi(\text{sup}_{u\in[0, \delta]}| B(u,t)|\ge\eta/4).\end{align}

The idempotent process $B=((B(x,t),x\in[0,1]),\,t\in\mathbb R_+)$ has trajectories from the space of continuous functions $\mathbb C(\mathbb R_+,\mathbb C([0,1],\mathbb R))$ ; see Appendix A. Since the collections of sets $\{b\in\mathbb C(\mathbb R_+,\mathbb C([0,1],\mathbb R)) :\,\text{sup}_{x\in[0,1/2]}\text{sup}_{y\in[0,\delta]}\lvert b(x+y,t)-b(x,t)\rvert\ge\eta/2\}$ and $\{b\in\mathbb C(\mathbb R_+,\mathbb C([0,1],\mathbb R)) :\,\text{sup}_{x\in[0,\delta]}\lvert b(x,t)\rvert\ge \eta/4\}$ are nested collections of closed sets as $\delta\downarrow0$ , the limit on the right of (3.20) is (see Appendix A)

\[ \Pi(\text{sup}_{x\in[0,1/2]}\text{sup}_{y\in[0,0]}\lvert B(x+y,t)-B(x,t)\rvert\ge \eta/2)+\Pi(\text{sup}_{u\in[0,0]}| B(u,t)|\ge\eta/4)=0,\]

which concludes the proof of (3.14).

Since the sequence of stochastic processes $((K_n(x,t),B_n(x,t)),x\in[0,1])$ LD converges to the idempotent process $((K(x,t),B(x,t)),x\in[0,1])$ in the sense of finite-dimensional distributions and is $\mathbb C$ -exponentially tight, the LD convergence holds in $\mathbb D([0,1],\mathbb R^2)$ ; see Theorem 7 in Appendix A. It has thus been proved that the sequence of stochastic processes $\bigl(((K_n(x,t_1),B_n(x,t_1)),\,x\in[0,1]),\ldots,((K_n(x,t_l),B_n(x,t_l)),\,x\in[0,1])\bigr)$ LD converges in $\mathbb D([0,1],\mathbb R^2)^l$ to the idempotent process $\bigl(((K(x,t_1),B(x,t_1)),\,x\in[0,1]),\ldots,((K(x,t_l),B(x,t_l)),\,x\in[0,1])\bigr)$ , for all $t_1\le t_2\le\ldots\le t_l$ . The proof of the lemma will be complete if the sequence $\bigl(((K_n(x,t),B_n(x,t)),x\in[0,1]),t\in\mathbb R_+\bigr)$ is shown to be $\mathbb C$ -exponentially tight of order $b_n^2$ in $\mathbb D(\mathbb R_+,\mathbb D([0,1],\mathbb R^2))$ . The definition of exponential tightness implies that it is sufficient to prove that each of the sequences $\{K_n,n\ge1\}$ and $\{B_n,n\ge1\}$ is $\mathbb C$ -exponentially tight of order $b_n^2$ in $\mathbb D(\mathbb R_+,\mathbb D([0,1],\mathbb R))$ . By (3.10) and Theorem 8, the $\mathbb C$ -exponential tightness of $\{K_n,n\ge1\}$ would follow if, for all $L>0$ and $\eta>0$ ,

(3.21) \begin{eqnarray} \lim_{\delta\rightarrow0}\limsup_{n\to\infty}\text{sup}_{s\in[0,L]}\mathbb P(\text{sup}_{t\in[0, \delta]}\text{sup}_{x\in[0,1]}|K_n(x,s+t)-K_n(x,s)|\ge \eta)^{1/b_n^2}=0. \qquad \end{eqnarray}

Since, in analogy with the reasoning in (3.15),

\begin{multline*} \mathbb P(\text{sup}_{t\in[0, \delta]}\text{sup}_{x\in[0,1]}|K_n(x,t+s)-K_n(x,s)|\ge \eta)\\[5pt] \le\mathbb P(\text{sup}_{t\in[0, \delta]}\text{sup}_{x\in[0,1/2]}|K_n(x,t+s)-K_n(x,s)|\ge \eta)\\[5pt] +\mathbb P(\text{sup}_{t\in[0, \delta]}\text{sup}_{x\in[1/2,1]}|K_n(x,t+s)-K_n(x,s)|\ge \eta)\\[5pt] =\mathbb P(\text{sup}_{t\in[0, \delta]}\text{sup}_{x\in[0,1/2]}|K_n(x,t+s)-K_n(x,s)|\ge \eta)\\[5pt] + \mathbb P(\text{sup}_{t\in[0, \delta]}\text{sup}_{x\in[0,1/2]}|\overline{K}_n(x,t+s)-\overline{K}_n(x,s)|\ge \eta)\\[5pt] =2\mathbb P(\text{sup}_{t\in[0, \delta]}\text{sup}_{x\in[0,1/2]}|K_n(x,t+s)-K_n(x,s)|\ge \eta),\end{multline*}

(3.21) is implied by

(3.22) \begin{eqnarray} &&\lim_{\delta\rightarrow0}\limsup_{n\to\infty}\text{sup}_{s\in[0,L]}\mathbb P(\text{sup}_{t\in[0, \delta]}\text{sup}_{x\in[0,1/2]}|K_n(x,t+s)-K_n(x,s)|\ge \eta)^{1/b_n^2}=0.\nonumber\\ && \end{eqnarray}

By (3.3), with x being held fixed, the process $(K_n(x,t+s)-K_n(x,s),t\in\mathbb R_+)$ is a locally square-integrable martingale, so $(\text{sup}_{x\in[0,1/2]}(K_n(x,t+s)-K_n(x,s)),t\in\mathbb R_+)$ , is a submartingale; hence by Doob’s inequality [Reference Liptser and Shiryaev14, Theorem 3.2, p. 60],

(3.23) \begin{align} &\mathbb P(\text{sup}_{t\in[0, \delta]}\text{sup}_{x\in[0,1/2]}|K_n(x,t+s)-K_n(x,s)|\ge \eta)\notag \\[5pt] & \qquad \le\frac{1}{\eta^{2b_n^2}}\,\mathbb E\text{sup}_{x\in[0,1/2]}(K_n(x,s+\delta)-K_n(x,s))^{2b_n^2}.\end{align}

As noted earlier, with t being held fixed, the process $(B_n(x,t),x\in[0,1])$ is a square-integrable martingale [Reference Jacod and Shiryaev9, Chapter II, §3c]. Equation (3.5) yields, by the Gronwall–Bellman inequality, as in (3.12),

(3.24) \begin{eqnarray} \text{sup}_{x\in[0,1/2]}|K_n(x,s+\delta)-K_n(x,s)|\le e^2\text{sup}_{x\in[0,1]}|B_n(x,s+\delta)-B_n(x,s)|.\end{eqnarray}

Since $(B_n(x,s+\delta)-B_n(x,s),x\in[0,1])$ is a square-integrable martingale, by another application of Doob’s inequality (see [Reference Jacod and Shiryaev9, Theorem I.1.43] and [Reference Liptser and Shiryayev13, Theorem I.9.2]), as well as by Jensen’s inequality,

\begin{eqnarray} \notag \mathbb E\text{sup}_{x\in[0,1]}(B_n(x,s+\delta)-B_n(x,s))^{2b_n^2}\le \Bigl(\frac{2b_n^2}{2b_n^2-1}\Bigr)^{2b_n^2}\mathbb E(B_n(1,s+\delta)-B_n(1,s))^{2b_n^2}.\end{eqnarray}

By (3.4), the fact that $1-\zeta_1$ and $\zeta_1$ have the same distribution, and the bound (5.6) in the proof of Theorem 19 in [Reference Petrov15, Chapter III, §5],

\begin{multline*} \mathbb E(B_n(1,s+\delta)-B_n(1,s))^{2b_n^2}\le(b_n\sqrt{n})^{-2b_n^2}\bigl((b_n^2+1)^{2b_n^2} (n\delta+1) \mathbb E(1+\ln\zeta_1)^{2b_n^2}\\[5pt] +2b_n^2(b_n^2+1)^{b_n^2}e^{b_n^2+1} (n\delta+1)^{b_n^2}\bigl(\mathbb E(1+\ln\zeta_1)^2\bigr)^{b_n^2}\bigr).\end{multline*}

(More specifically, the following bound is used. Suppose $X_1,\ldots,X_n$ are i.i.d. with $\mathbb EX_1=0$ . Then, provided $p\ge2$ and $r>p/2$ ,

\[ \mathbb E\lvert \sum_{i=1}^nX_i\rvert^p\le r^pn\mathbb E\lvert X_1\rvert^p+pr^{p/2}e^rn^{p/2}(\mathbb EX_1^2)^{p/2}.\]

See also [Reference Whittle25] for similar results.)

As $\zeta_1$ is uniform on [0,1], $\mathbb E(\ln\zeta_1)^{2b_n^2}=(2b_n^2)!$ , so that, with the use of Jensen’s inequality,

\begin{multline*}\limsup_{n\to\infty}(\mathbb E(B_n(1,s+\delta)-B_n(1,s))^{2b_n^2})^{1/b_n^2}\\[5pt] \le \limsup_{n\to\infty}(b_n\sqrt{n})^{-2}\Bigl(4n^{1/b_n^2}(b_n^2+1)^{2}\bigl((2b_n^2)!\bigr)^{1/b_n^2}+(b_n^2+1)e( n\delta+1)\mathbb E(1+\ln\zeta_1)^2\Bigr),\end{multline*}

which implies, via Stirling’s formula, on recalling that $b_n^6n^{1/b_n^2-1}\to0$ , that

\[\lim_{\delta\to0}\limsup_{n\to\infty}\text{sup}_{s\in[0,L]}\Bigl(\mathbb E(B_n(1,s+\delta)-B_n(1,s))^{2b_n^2}\Bigr)^{1/b_n^2}=0.\]

Recalling (3.23), (3.24), and (3.25) yields (3.22). The proof of the $\mathbb C$ -exponential tightness of $B_n$ is similar. (It is actually simpler.)

Going back to the set-up of Theorem 3, let

(3.25) \begin{align} H_n(t)&=Y_n(t)-\int_0^tY_n(t-s)\,dF(s), \end{align}

(3.26) \begin{align} X^{(0)}_n(t)&=\frac{\sqrt{n}}{b_n}\,\bigl(\frac{1}{n}\, Q_n^{(0)}(t)-(1-F_0(t))\bigr), \end{align}

(3.27) \begin{align} U_n(x,t)&=\frac{1}{b_n\sqrt{n}}\,\sum_{i=1}^{{{\hat A}_n}(t)}\bigl(\,\mathbf{1}_{\{\eta_i\le x\}}\,-F(x)\bigr),\end{align}

and

(3.28) \begin{align} \Theta_n(t)&=-\int_{\mathbb R_+^2} \mathbf{1}_{\{x+s\le t\}}\,dU_n(x,s).\end{align}

As, owing to (2.1),

(3.29) \begin{equation} 1-F_0(t)+\mu t-\mu\int_0^t(t-s)dF(s)=1,\end{equation}

(2.2), (2.4), and (2.7) imply that

(3.30) \begin{align} X_n(t)&=(1-F(t))X_n(0)^++ X_n^{(0)}(t)+\int_0^tX_n(t-s)^+\,dF(s)+H_n(t)+\Theta_n(t).\end{align}

The equation (3.2) is written in a similar way: introducing

(3.31) \begin{align} H(t)&=Y(t)-\int_0^t Y(t-s)\,dF(s),\end{align}

(3.32) \begin{align}X^{(0)}(t)&=W^0(F_0(t))-(1-F_0(t))X(0)^-,\end{align}

(3.33) \begin{align}U(x,t)&=K(F(x),\mu t),\end{align}

and

(3.34) \begin{align} \Theta(t)&=-\int_{\mathbb R_+^2} \mathbf{1}_{\{x+s\le t\}}\,dU(x,s)=-\int_{\mathbb R_+^2} \mathbf{1}_{\{x+s\le t\}}\,\dot K(F(x),\mu s)\,dF(x)\,\mu ds\end{align}

yields

(3.35) \begin{equation} X(t)=(1-F(t))X(0)^++X^{(0)}(t)+\int_0^tX(t-s)^+\,dF(s)+H(t)+\Theta(t).\end{equation}

Let $H=(H(t),\,t\in\mathbb R_+)$ , $H_n=(H_n(t),\,t\in\mathbb R_+)$ , $X^{(0)}=(X^{(0)}(t),\,t\in\mathbb R_+)$ , $ X_n^{(0)}=( X_n^{(0)}(t),\,t\in\mathbb R_+)$ , $U=((U(x,t),x\in\mathbb R_+),t\in\mathbb R_+)$ , $U_n=((U_n(x,t),x\in\mathbb R_+),t\in\mathbb R_+)$ , $\Theta=(\Theta(t),t\in\mathbb R_+)$ , and $\Theta_n=(\Theta_n(t),\,t\in\mathbb R_+)$ .

The groundwork needs to be laid first. Let

(3.36) \begin{equation} L_n(x,t)=\frac{1}{b_n\sqrt{n}}\,\sum_{i=1}^{{{\hat A}_n}(t)}\bigg(\mathbf{1}_{\{\eta_i\le x\}}\,-\int_0^{\eta_i\wedge x}\frac{dF(u)}{1-F(u)}\bigg).\end{equation}

Since the random variables $F(\eta_i)$ are i.i.d. and uniform on [0, 1], in view of (3.3), (3.4), and (3.27), it may be assumed that

(3.37) \begin{equation} L_n(x,t)=B_n(F(x),\frac{\hat A_n(t)}{n})\end{equation}

and that

(3.38) \begin{equation} U_n(x,t)=K_n\bigg(F(x),\frac{\hat A_n(t)}{n}\bigg).\end{equation}

By (3.5),

(3.39) \begin{equation} U_n(x,t)=-\int_0^x\frac{U_n(y,t )}{1-F(y)}\,dF(y)+L_n(x,t).\end{equation}

By (3.28),

(3.40) \begin{equation} \Theta_n(t)=J_n(t)-M_n(t),\end{equation}

with

(3.41) \begin{align} J_n(t)&= \int_0^t \frac{U_n(x,t-x)}{1-F(x)}\,dF(x)\end{align}

and

(3.42) \begin{align} M_{n}(t)&=\int_{\mathbb R_+^2} \,\mathbf{1}_{\{x+s\le t\}}\,\,dL_n(x,s).\end{align}

Lemma 2. Under the hypotheses,

(3.43) \begin{align} \lim_{r\to\infty}\limsup_{n\to\infty}\mathbb P(\text{sup}_{s\in[0,t]}\lvert U_n(s,t-s)\rvert>r)^{1/b_n^2}=0 \end{align}

and

(3.44) \begin{align} \lim_{r\to\infty}\limsup_{n\to\infty}\mathbb P(\text{sup}_{s\in[0,t]}\lvert M_n(s)\rvert>r)^{1/b_n^2}=0. \end{align}

Proof. Note that

(3.45) \begin{equation} \lim_{r\to\infty} \limsup_{n\to\infty}\mathbb P\bigg(\frac{A_n(t)}{n}>r\bigg)^{1/b_n^2}=0, \end{equation}

which is a consequence of the LD convergence at rate $b_n^2$ of the $Y_n$ to Y (see (3.1)). Similarly, the LD convergence of $X_n(0)$ to X(0) implies that

(3.46) \begin{equation} \lim_{r\to\infty} \limsup_{n\to\infty}\mathbb P\bigg(\frac{Q_n(0)}{n}>r\bigg)^{1/b_n^2}=0.\end{equation}

By (2.4), ${{\hat A}_n}(t)\le(Q_n(0)-n)^++A_n(t)$ , so that (3.45) and (3.46) imply that

(3.47) \begin{equation} \lim_{L\to\infty} \limsup_{n\to\infty}\mathbb P\bigg(\frac{\hat A_n(t)}{n}>L\bigg)^{1/b_n^2}=0. \end{equation}

By (3.38),

(3.48) \begin{equation} \mathbb P(\text{sup}_{s\in[0,t]}\lvert U_n(s,t-s)\rvert>r)\le\mathbb P\bigg(\frac{\hat A_n(t)}{n}>L\bigg)+\mathbb P(\text{sup}_{\substack{s\in [0,L],\\ x\in[0,1]}}\lvert K_n(x,s)\rvert>r).\end{equation}

By the LD convergence of $K_n$ to K in Lemma 1, and since the trajectories of K are continuous,

\[ \lim_{r\to\infty} \limsup_{n\to\infty}\mathbb P\bigl(\text{sup}_{\substack{s\in [0,L],\\ x\in[0,1]}}\lvert K_n(x,s)\rvert>r)^{1/b_n^2}=0.\]

Combined with (3.47) and (3.48), this proves (3.43).

Lemma 3.1 in [Reference Puhalskii and Reed21] implies that the process $M_n=(M_n(t),\,t\in\mathbb R_+)$ is a local martingale with respect to the filtration $\mathbf{G}_n$ defined as follows. For $t\in\mathbb R_+$ , let $\hat{\mathcal{G}}_n(t)$ denote the complete $\sigma$ -algebra generated by the random variables $\,\mathbf{1}_{\{\hat\tau_{n,i}\le s\}}\,\,\mathbf{1}_{\{ \eta_i\le x\}}$ , where $x+ s\le t$ and $i\in\mathbb N$ , and by the ${{\hat A}_n}(s)$ (or, equivalently, by the $\,\mathbf{1}_{\{\hat\tau_{n,i}\le s\}}\,$ for $i\in\mathbb N$ ), where $s\leq t$ . Define $\mathcal{G}_n(t)=\cap_{\epsilon>0}\hat{\mathcal{G}}_n(t+\epsilon)$ and $\mathbf{G}_n=(\mathcal{G}_n(t), t\in\mathbb R_+)$ . By (3.36) and (3.42),

(3.49) \begin{equation} M_n(t)=\frac{1}{b_n\sqrt{n}}\,\sum_{i=1}^{{{\hat A}_n}(t)}\bigg(\,\mathbf{1}_{\{\eta_i+\hat\tau_{n,i}\le t\}}\,-\int_0^{\eta_i\wedge (t-\hat\tau_{n,i})}\frac{dF(u)}{1-F(u)}\bigg).\end{equation}

Thus, the measure of jumps of $M_n$ is

(3.50) \begin{equation} \mu_n([0,t],\Gamma)=\,\mathbf{1}_{\{1/(b_n\sqrt{n})\in\Gamma\}}\,\sum_{i=1}^{{{\hat A}_n}(t)}\,\mathbf{1}_{\{\eta_i+\hat\tau_{n,i}\le t\}}\,\end{equation}

and the associated $\mathbf{G}_n$ -predictable measure of jumps is

(3.51) \begin{equation} \nu_n([0,t],\Gamma)=\,\mathbf{1}_{\{1/(b_n\sqrt{n})\in\Gamma\}}\,\sum_{i=1}^{{{\hat A}_n}(t)}\int_0^{\eta_i\wedge (t-\hat\tau_{n,i})}\frac{dF(u)}{1-F(u)}.\end{equation}

Note that it is a continuous process. (For $\hat A_n$ being predictable, see Lemma C.1 in [Reference Puhalskii and Reed21].) The associated stochastic cumulant is (see, e.g., [Reference Puhalskii19, p. 293])

(3.52) \begin{equation} G_n(\alpha,t)=\bigg(e^{\alpha/(b_n\sqrt{n})}-1-\frac{\alpha}{b_n\sqrt{n}}\bigg)\sum_{i=1}^{{{\hat A}_n}(t)}\int_0^{\eta_i\wedge (t-\hat\tau_{n,i})}\frac{dF(u)}{1-F(u)}.\end{equation}

By Lemma 4.1.1 on p. 294 in [Reference Puhalskii19], the process $(e^{\alpha M_n(t)-G_n(\alpha,t)},t\in\mathbb R_+)$ is a local martingale, so that $\mathbb E e^{\alpha M_n(\tau)-G_n(\alpha,\tau)}\le1$ , for arbitrary stopping time $\tau$ . Lemma 3.2.6 on p. 282 in [Reference Puhalskii19] implies that, for $\gamma>0$ ,

\begin{multline*} \mathbb P(\text{sup}_{s\in[0,t]}e^{\alpha b_n^2 M_n(s)}\ge e^{\alpha b_n^2 r})\le e^{\alpha b_n^2(\gamma- r)}+\mathbb P(e^{G_n(\alpha b_n^2,t)}\ge e^{\alpha b_n^2\gamma})\\[5pt] \le e^{\alpha b_n^2(\gamma- r)}+\mathbb P(e^{\hat G_n(\alpha b_n^2,t)}\ge e^{\alpha b_n^2\gamma}),\end{multline*}

where

(3.53) \begin{equation} \hat G_n(\alpha,t)=\bigg(e^{\alpha/(b_n\sqrt{n})}-1-\frac{\alpha}{b_n\sqrt{n}}\bigg)\sum_{i=1}^{{{\hat A}_n}(t)}\int_0^{t-\hat\tau_{n,i}}\frac{dF(u)}{1-F(u)}.\end{equation}

Hence, for $\alpha>0$ ,

(3.54) \begin{equation} \mathbb P(\text{sup}_{s\in[0,t]} M_n(s)\ge r)^{1/b_n^2}\le e^{\alpha(\gamma- r)}+\mathbb P(\hat G_n(\alpha b_n^2,t)\ge\alpha b_n^2\gamma)^{1/b_n^2}.\end{equation}

On writing

(3.55) \begin{multline} \hat G_n(\alpha b_n^2,t)=\bigg(e^{\alpha b_n/\sqrt{n}}-1-\frac{\alpha b_n}{\sqrt{n}}\bigg)\int_0^t\int_0^{t-s}\frac{dF(u)}{1-F(u)}\,d\hat A_n(s)\\[5pt] =\bigg(e^{\alpha b_n/\sqrt{n}}-1-\frac{\alpha b_n}{\sqrt{n}}\bigg)\int_0^t\hat A_n(t-u)\,\frac{dF(u)}{1-F(u)}\end{multline}

and noting that $(n/b_n^2) (e^{\alpha b_n/\sqrt{n}}-1-\alpha b_n/\sqrt{n})\to\alpha^2/2$ , one can see, thanks to (3.47), that

\[ \lim_{n\to\infty}\mathbb P(\hat G_n(\alpha b_n^2,t)\ge\alpha b_n^2\gamma)^{1/b_n^2}=0, \]

provided $\alpha$ is small enough, which proves that

\[ \lim_{r\to\infty}\limsup_{n\to\infty}\mathbb P(\text{sup}_{s\in[0,t]}M_n(s)>r)^{1/b_n^2}=0.\]

The argument for $\text{sup}_{s\in[0,t]}(\!-M_n(s))$ is similar. The convergence (3.44) has been proved.

Lemma 3. For arbitrary $\epsilon>0$ and $t>0$ ,

\[\lim_{n\to\infty}\mathbb P\bigg(\text{sup}_{s\in[0,t]}\lvert \frac{\hat A_n(s)}{n}-\mu s\rvert>\epsilon\bigg)^{1/b_n^2}=0.\]

Proof. By (2.4), (2.2), (3.27), and (3.28),

(3.56) \begin{multline} \frac{1}{n}\,Q_n(t)=\bigg(\frac{1}{n}\,Q_n(0)-1\bigg)^+(1-F(t))+\frac{1}{n}\,Q_n^{(0)}(t)+\frac{1}{n}\,A_n(t)-\frac{1}{n}\,\int_{0}^tA_n(t-s)\,dF(s)\\[5pt] +\frac{1}{n}\,\int_0^t(Q_n(t-s)-n)^+ \,dF(s)+\frac{b_n}{\sqrt{n}}\,\Theta_n(t).\end{multline}

By (3.40), (3.41), (3.43), and (3.44), on recalling that $F(t)<1$ , we have

(3.57) \begin{equation} \lim_{r\to\infty}\limsup_{n\to\infty}\mathbb P(\text{sup}_{s\in[0,t]}\lvert \Theta_n(s)\rvert>r)^{1/b_n^2}=0.\end{equation}

The LD convergence at rate $b_n^2$ of $Y_n$ to Y implies that, for $\epsilon>0$ ,

(3.58) \begin{equation} \lim_{n\to\infty}\mathbb P\bigg(\text{sup}_{s\in[0,t]}\lvert \frac{A_n(s)}{n}-\mu s\rvert>\epsilon\bigg)^{1/b_n^2}=0.\end{equation}

By (2.3), (3.46), and Lemma 1,

\[ \lim_{n\to\infty}\mathbb P\bigg(\text{sup}_{s\in[0,t]}\lvert \frac{1}{n}\,Q_n^{(0)}(s)-(1-F_0(s))\rvert>\epsilon\bigg)^{1/b_n^2}=0.\]

Recalling (3.29) implies that

\[ \lim_{n\to\infty}\mathbb P\bigg(\text{sup}_{s\in[0,t]}\lvert \frac{1}{n}\,Q_n^{(0)}(s)+\frac{1}{n}\,A_n(s)-\frac{1}{n}\,\int_{0}^sA_n(s-x)\,dF(x)-1\rvert>\epsilon\bigg)^{1/b_n^2}=0.\]

In addition, the LD convergence of $X_n(0)$ to X(0) implies that (3.46) can be strengthened as follows:

(3.59) \begin{equation} \limsup_{n\to\infty}\mathbb P\bigg(\lvert \frac{Q_n(0)}{n}-1\rvert>\epsilon\bigg)^{1/b_n^2}=0.\end{equation}

Hence, by (3.56) and (3.57),

\[ \frac{1}{n}\,Q_n(t)-1=\,\int_0^t\bigg(\frac{1}{n}\,Q_n(t-s)-1\bigg)^+ \,dF(s)+\theta_n(t),\]

where

\begin{multline*}\!\!\!\!\!\theta_n(t)= \bigg(\frac{1}{n}\,Q_n(0)-1\bigg)^+(1-F(t))+\frac{1}{n}\,Q_n^{(0)}(t)+\frac{1}{n}\,A_n(t)-\frac{1}{n}\,\int_{0}^tA_n(t-s)\,dF(s)+\frac{b_n}{\sqrt{n}}\,\Theta_n(t)\end{multline*}

and

\[\!\!\!\!\!\lim_{n\to\infty} \mathbb P(\text{sup}_{s\in[0,t]}\lvert \theta_n(s)\rvert>\epsilon)^{1/b_n^2}=0.\]

Lemma B.1 in [Reference Puhalskii and Reed21] implies that there exists a function $\rho$ , which depends only on the function F, such that

\[ \text{sup}_{s\in[0,t]}\lvert \frac{1}{n}\,Q_n(s)-1\rvert\le \rho(t)\text{sup}_{s\in[0,t]}\lvert \theta_n(s)\rvert.\]

Therefore,

(3.60) \begin{equation} \lim_{n\to\infty} \mathbb P\bigg(\text{sup}_{s\in[0,t]}\lvert \frac{1}{n}\,Q_n(s)-1\rvert>\epsilon\bigg)^{1/b_n^2}=0.\end{equation}

When combined with (2.4) and (3.58), this yields the assertion of the lemma.

Let $L_n=\bigl((L_n(x,t),x\in\mathbb R_+),t\in\mathbb R_+\bigr)$ and $L=\bigl((L(x,t),x\in\mathbb R_+),t\in\mathbb R_+\bigr)$ , with $L(x,t)=B(F(x),\mu t)$ .

Proof. Let

(3.61) \begin{equation} \tilde X^{(0)}_n(x,t)=\frac{1}{b_n\sqrt{n}}\,\sum_{i=1}^{\lfloor nt\rfloor}\,(\,\mathbf{1}_{\{\eta^{(0)}_i> x\}}\,-(1-F_0(x))),\end{equation}

$\tilde X^{(0)}_n(t)=(\tilde X^{(0)}_n(x,t),x\in\mathbb R_+)$ and $\tilde X^{(0)}_n=(\tilde X^{(0)}_n(t),t\in\mathbb R_+)$ . By the hypotheses of Theorem 3, $X_n(0)$ LD converges to X(0), by Lemma 1 and because $F_0$ is strictly increasing, and $\tilde X^{(0)}_n$ LD converges to $((\tilde K(F_0(x),t),x\in\mathbb R_+), t\in\mathbb R_+)$ , where $\tilde K$ represents a Kiefer idempotent process that is independent of (X(0),K, B). Also $(K_n,B_n)$ LD converges to (K, B). By independence assumptions, these convergences hold jointly; cf. Appendix A. Since $Q_n(0)/n\to 1$ and $\hat A_n(t)/n\to\mu t$ super-exponentially in probability by (3.59) and (3.58), respectively, ‘Slutsky’s theorem’ (Lemma 6) yields joint LD convergence of $(X_n(0),\tilde X^{(0)}_n,K_n,B_n,Q_n(0)/n,(\hat A_n(t)/n,t\in \mathbb R_+))$ to $(X(0),((\tilde K(F_0(x),t),x\in\mathbb R_+), t\in\mathbb R_+),K,B,1,(\mu t,t\in\mathbb R_+))$ . In addition, by (2.3), (3.26), and (3.61),

\[ X^{(0)}_n(t)=\tilde X^{(0)}_n\bigg(t,\frac{Q_n(0)}{n}\wedge 1\bigg)-(1-F_0(t))X_n(0)^-.\]

In order to deduce the LD convergence of $(X_n(0), X_n^{(0)},U_n,L_n)$ to $(X(0), X^{(0)},U,L)$ , it remains to recall (3.32), (3.33), (3.37), and (3.38); note that, by Lemma 7, $(\tilde K(t,1),t\in[0,1])$ is a Brownian bridge idempotent process; and apply the continuous mapping principle, the associated composition mappings being continuous at continuous limits. (See [Reference Whitt24] for more background on continuous functions in the Skorokhod space context.)

Proof. Let

(3.62) \begin{equation}J(t)=\int_0^t \frac{U(x,t-x)}{1-F(x)}\,dF(x).\end{equation}

By Lemma 4, (3.41), (3.62), and the continuous mapping principle, $J_n=(J_n(t), t\in\mathbb R_+)$ LD converges to $J=J(t),t\in\mathbb R_+)$ , so the sequence $J_n$ is $\mathbb C$ -exponentially tight. By (3.40), it remains to check that the sequence $M_n$ is $\mathbb C$ -exponentially tight, which, according to Theorem 8, is implied by the following convergences:

\begin{align} \lim_{r\to\infty}\limsup_{n\to\infty}\mathbb P(\text{sup}_{s\in[0,t]}\lvert M_n(s)\rvert>r)^{1/b_n^2}=0\notag\end{align}

and

(3.63) \begin{align} \lim_{\delta\to0}\limsup_{n\to\infty}\text{sup}_{s\in[0,t]}\mathbb P(\text{sup}_{s'\in[0,\delta]}\lvert M_n(s+s')-M_n(s)\rvert>\epsilon)^{1/b_n^2}=0,\end{align}

where $t>0$ and $\epsilon>0$ . The former convergence has already been proved; see (3.44). The proof of (3.63) proceeds along similar lines. Since, with $\alpha\in\mathbb R$ , the process $ (\exp\bigl(\alpha( M_n(s+s')-M_n(s))-(G_n(\alpha,s+s')-G_n(\alpha,s))\bigr),s'\in\mathbb R_+)$ is a local martingale, so that, for arbitrary stopping time $\tau$ ,

$$\mathbb E e^{\alpha( M_n(s+\tau)-M_n(s))-(G_n(\alpha,s+\tau)-G_n(\alpha,s))}\le1,$$

by Lemma 3.2.6 on p. 282 in [Reference Puhalskii19], for arbitrary $\gamma>0$ , in analogy with (3.54), for $b_n\ge1$ , we have

\begin{multline*} \mathbb P(\text{sup}_{s'\in[0,\delta]}( M_n(s+s')-M_n(s))\\[5pt]\ge \epsilon)^{1/b_n^2}\le e^{\alpha(\gamma- \epsilon)}+\mathbb P(\hat G_n(\alpha b_n^2,s+\delta)-\hat G_n(\alpha b_n^2,s)\ge\alpha b_n^2\gamma)^{1/b_n^2}.\end{multline*}

By (3.55) and Lemma 3,

\[\frac{1}{b_n^2}\, \hat G_n(\alpha b_n^2,t)\to\frac{\alpha^2}{2}\,\mu\int_0^t( t-u)\frac{dF(u)}{1-F(u)}\]

super-exponentially in probability at rate $b_n^2$ . The fact that the latter super-exponential convergence in probability is locally uniform in t, as the limit is a monotonic continuous function starting at 0, implies that, for $\delta$ small enough, depending on $\alpha$ ,

\[ \limsup_{n\to\infty}\text{sup}_{s\in[0,t]} \mathbb P(\text{sup}_{s'\in[0,\delta]}( M_n(s'+s)-M_n(s))\ge \epsilon)^{1/b_n^2}\le e^{\alpha(\gamma- \epsilon)}.\]

Now, one chooses $\gamma<\epsilon$ and sends $\alpha\to\infty$ . A similar convergence holds with $-M_n(s')$ substituted for $M_n(s')$ . The limit (3.63) has been proved.

Proof. Let

(3.64) \begin{equation} \Theta^{(l)}_n(t)=-\int_{\mathbb R_+^2}I_{t}^{(l)}(x,s)dU_n(x,s),\end{equation}

where

\[ I_{t}^{(l)}(x,s)=\sum_{i=1}^\infty\,\mathbf{1}_{\{s\in(s^{(l)}_{i-1},s^{(l)}_i]\}}\,\,\mathbf{1}_{\{x\in[0,t-s^{(l)}_{i-1}]\}}\,,\]

for some $0=s^{(l)}_0<s^{(l)}_1<\ldots$ such that $s^{(l)}_i\to\infty$ , as $i\to\infty$ , and $\text{sup}_{i\ge1}(s^{(l)}_i-s^{(l)}_{i-1})\to0$ , as $l\to\infty$ . Evidently,

\[\Theta^{(l)}_n(t)=-\sum_{i=1}^\infty(U_n(t-s^{(l)}_{i-1},s^{(l)}_i)-U_n(t-s^{(l)}_{i-1}, s^{(l)}_{i-1}))\,\mathbf{1}_{\{s^{(l)}_{i-1}\le t\}}.\]

Similarly, let

\begin{multline*} \Theta^{(l)}(t)=-\int_{\mathbb R_+^2}I_{t}^{(l)}(x,s)\,dU(x,s)=\sum_{i=1}^\infty(U(t-s^{(l)}_{i-1},s^{(l)}_i)-U(t-s^{(l)}_{i-1},s^{(l)}_{i-1}))\,\mathbf{1}_{\{s^{(l)}_{i-1}\le t\}}.\end{multline*}

By the LD convergence of $Y_n$ to Y in the hypotheses of Theorem 3, Lemma 4, (3.25), (3.31), (3.41), (3.62), and the continuous mapping principle, the sequence $(X_{n}(0), X_n^{(0)},H_{n},J_{n},L_{n},U_{n})$ LD converges to $(X(0), X^{(0)},H,J,L,U)$ . Hence, the sequence $(X_{n}(0), X_n^{(0)},H_{n},J_{n},L_{n},U_{n},\Theta_{n}^{(l)})$ LD converges to $(X(0), X^{(0)},H,J,L,U,\Theta^{(l)})$ . In addition, by Lemma 5, the sequence $\Theta_n$ is $\mathbb C$ -exponentially tight. Since the idempotent processes $X^{(0)},H,J,L,U$ are seen to have continuous trajectories, the sequence $( X_n^{(0)},H_n,J_n,L_n,U_n,\Theta_n)$ is $\mathbb C$ -exponentially tight. That a limit point of $(X_n(0),X_n^{(0)},H_n,J_n,L_n,U_n,\Theta_n)$ has the same idempotent distribution as $(X(0), X^{(0)},H,J,L,U,\Theta)$ would follow from the LD convergence of finite-dimensional distributions of $(X_n(0), X_n^{(0)},H_n,J_n,L_n,U_n,\Theta_{n})$ to finite-dimensional distributions of $(X(0), X^{(0)},H,J,L,U,\Theta)$ . Owing to (3.33),

\[ \Theta^{(l)}(t)=-\int_{\mathbb R_+^2}I_{t}^{(l)}(x,s)\dot K(F(x),\mu s)\,dF(x)\,\mu ds,\]

which implies, by (3.34) and the Cauchy–Schwarz inequality, that $\Theta^{(l)}\to \Theta$ locally uniformly, as $l\to\infty$ . Since the sequence $(X_n(0), X_n^{(0)},H_n,J_n,L_n,U_n)$ LD converges to $(X(0), X^{(0)},H,J,L,U)$ , to prove the finite-dimensional LD convergence it suffices to prove that

(3.65) \begin{equation} \lim_{l\to\infty}\limsup_{n\to\infty}\mathbb P(\lvert \Theta_n^{(l)}(t)-\Theta_n(t)\rvert>\epsilon)^{1/b_n^2}=0.\end{equation}

Let

\begin{multline*} \hat \Theta_n^{(l)}(s,t)=\frac{1}{b_n\sqrt{n}}\sum_{i=1}^{\hat A_n(s)}\sum_{j=1}^\infty\,\mathbf{1}_{\{s^{(l)}_{j-1}\le t\}}\,\,\mathbf{1}_{\{\hat\tau_{n,i}\in(s^{(l)}_{j-1},s^{(l)}_j]\}}\,\bigl(\,\mathbf{1}_{\{\eta_i\in( t-\hat\tau_{n,i},t-s^{(l)}_{j-1}]\}}\,\\[5pt] -(F(t-s^{(l)}_{j-1})-F(t-\hat\tau_{n,i}))\bigr).\end{multline*}

By (3.28) and (3.64),

(3.66) \begin{equation} \Theta_n(t)-\Theta_n^{(l)}(t)=\hat \Theta_n^{(l)}(t,t).\end{equation}

Let $\mathcal{F}_{n}(s)$ represent the complete $\sigma$ -algebra generated by the random variables $\hat\tau_{n,j}\wedge \hat\tau_{n,\hat A_n(s)+1}$ and $\eta_{j\wedge \hat A_n(s)}$ , where $j\in \mathbb N$ . By Part 4 of Lemma C.1 in [Reference Puhalskii and Reed21], with t held fixed, the process $(\hat\Theta_n^{(l)}(s,t),s\in\mathbb R_+)$ is an $\mathbb F_n$ -locally square-integrable martingale. Its measure of jumps is

\begin{multline*} \mu_n^{(l)}([0,s],\Gamma)=\sum_{i=1}^{\hat A_n(s)}\sum_{j=1}^\infty\,\mathbf{1}_{\{s^{(l)}_{j-1}\le t\}}\,\,\mathbf{1}_{\{\hat\tau_{n,i}\in(s^{(l)}_{j-1},s^{(l)}_j]\}}\,\bigl(\,\mathbf{1}_{\{(1-(F(t-s^{(l)}_{j-1})-F(t-\hat\tau_{n,i}))/(b_n\sqrt{n})\in\Gamma\}}\,\\[5pt] \,\mathbf{1}_{\{\eta_i\in( t-\hat\tau_{n,i},t-s^{(l)}_{j-1}]\}}\,+\,\mathbf{1}_{\{(-(F(t-s^{(l)}_{j-1})-F(t-\hat\tau_{n,i}))/(b_n\sqrt{n})\in\Gamma\}}\,\,\mathbf{1}_{\{\eta_i\not\in( t-\hat\tau_{n,i},t-s^{(l)}_{j-1}]\}}\,\bigr),\end{multline*}

where $\Gamma\subset \mathbb R\setminus\{0\}$ . Accordingly, the $\mathbb F_n$ -predictable measure of jumps is

\begin{multline*} \nu_n^{(l)}([0,s],\Gamma)=\sum_{i=1}^{\hat A_n(s)}\sum_{j=1}^\infty\,\mathbf{1}_{\{s^{(l)}_{j-1}\le t\}}\,\,\mathbf{1}_{\{\hat\tau_{n,i}\in(s^{(l)}_{j-1},s^{(l)}_j]\}}\,\bigl(\,\mathbf{1}_{\{(1-(F(t-s^{(l)}_{j-1})-F(t-\hat\tau_{n,i}))/(b_n\sqrt{n})\in\Gamma\}}\,\\[5pt] (F(t-s^{(l)}_{j-1})-F( t-\hat\tau_{n,i}))+\,\mathbf{1}_{\{-(F(t-s^{(l)}_{j-1})-F(t-\hat\tau_{n,i}))/(b_n\sqrt{n})\in\Gamma\}}\,(1-(F(t-s^{(l)}_{j-1})-F( t-\hat\tau_{n,i})))\bigr).\end{multline*}

For $\alpha\in\mathbb R$ , as on p. 214 in [Reference Puhalskii19], define the stochastic cumulant

(3.67) \begin{multline} G^{(l)}_n(\alpha,s)=\int_0^s\int_\mathbb R(e^{\alpha x}-1-\alpha x)\nu^{(l)}_n(ds',dx)\\[5pt] =\sum_{i=1}^{\hat A_n(s)}\sum_{j=1}^\infty\,\mathbf{1}_{\{s^{(l)}_{j-1}\le t\}}\,\,\mathbf{1}_{\{\hat\tau_{n,i}\in(s^{(l)}_{j-1},s^{(l)}_j]\}}\,\bigl((e^{\alpha (1-(F(t-s^{(l)}_{j-1})-F(t-\hat\tau_{n,i})))/(b_n\sqrt{n})}\\[5pt] -1-\frac{\alpha}{b_n\sqrt{n}}(1-(F(t-s^{(l)}_{j-1})-F(t-\hat\tau_{n,i}))))(F(t-s^{(l)}_{j-1})-F(t-\hat\tau_{n,i}))\\[5pt] +(e^{-\alpha (F(t-s^{(l)}_{j-1})-F(t-\hat\tau_{n,i}))/( b_n\sqrt{n})}-1\\[5pt] +\frac{\alpha}{b_n\sqrt{n}}(F(t-s^{(l)}_{j-1})-F(t-\hat\tau_{n,i})))(1-(F(t-s^{(l)}_{j-1})-F(t-\hat\tau_{n,i})))\bigr).\end{multline}

The associated stochastic exponential is defined by

(3.68) \begin{equation} \mathcal{E}^{(l)}_n(\alpha,s)=e^{G^{(l)}_n(\alpha,s)}\prod_{0<s'\le s}(1+\Delta G^{(l)}_n(\alpha,s'))e^{-\Delta G^{(l)}_n(\alpha,s')},\end{equation}

where $\Delta G^{(l)}_n(s')$ represents the jump of $G^{(l)}_n(s'')$ with respect to s” evaluated at $s''=s'$ and the product is taken over the jumps. By Lemma 4.1.1 on p. 294 in [Reference Puhalskii19], the process $ \bigl( e^{\alpha b_n^2 \hat \Theta_n^{(l)}(s,t))}\mathcal{E}_n(\alpha b_n^2,s)^{-1},s\in\mathbb R_+\bigr)$ is a well-defined local martingale, so that, for any stopping time $\tau$ ,

\[ \mathbb Ee^{\alpha b_n^2 \hat \Theta_n^{(l)}(\tau,t))}\mathcal{E}^{(l)}_n(\alpha b_n^2,\tau)^{-1}\le1. \]

Lemma 3.2.6 on p. 282 in [Reference Puhalskii19] and (3.68) imply that, for $\alpha>0$ and $\gamma>0$ ,

(3.69) \begin{align} \mathbb P(\text{sup}_{s\in[0,t]} \hat \Theta_n^{(l)}(s,t)\ge \epsilon)\le e^{\alpha b_n^2(\gamma-\epsilon)}+\mathbb P (\mathcal{E}^{(l)}_n(\alpha b_n^2,t)\ge e^{\alpha\notag b_n^2\gamma})\\[5pt] \le e^{\alpha b_n^2(\gamma-\epsilon)}+\mathbb P (G^{(l)}_n(\alpha b_n^2,t)\ge \alpha b_n^2\gamma).\end{align}

By (3.67),

\begin{multline*}G^{(l)}_n(\alpha b_n^2,t)\le\hat A_n(t)\bigg(\text{sup}_{\lvert y\rvert\le1}\bigg(e^{\alpha b_ny/\sqrt{n}}-1-\alpha\frac{b_ny}{\sqrt{n}}\bigg)\,\text{sup}_j\big(F(t-s^{(l)}_{j-1})-F(t-s^{(l)}_j)\big)\\[5pt] +\text{sup}_j\text{sup}_{y\in[F(t-s^{(l)}_{j}),F(t-s^{(l)}_{j-1})]}\bigg(e^{-\alpha b_n y/\sqrt{n}}-1+\alpha\frac{b_n}{\sqrt{n}}y\bigg)\bigg). \end{multline*}

As $n/b_n^2(e^{\alpha b_n y/\sqrt{n}}-1-\alpha b_ny/\sqrt{n})\to\alpha^2y^2/2$ uniformly on bounded intervals, $\hat A_n(t)/n\to \mu t$ super-exponentially as $n\to\infty$ , and $\text{sup}_j(s^l_j-s^l_{j-1})\to0$ as $l\to\infty$ , it follows that

\[ \lim_{l\to\infty}\limsup_{n \to\infty}\mathbb P (G^{(l)}_n(\alpha b_n^2,t)\ge \alpha b_n^2\gamma)^{1/b_n^2}=0,\]

which implies, thanks to (3.69), that

\[\limsup_{l\to\infty} \limsup_{n\to\infty} \mathbb P(\text{sup}_{s\in[0,t]}\hat \Theta_n^{(l)}(s,t)\ge \epsilon)^{1/b_n^2}\le e^{\alpha (\gamma-\epsilon)}.\]

Picking $\gamma<\epsilon$ and sending $\alpha$ to $\infty$ shows that the latter left-hand side equals zero. A similar argument proves the convergence

\[\lim_{l\to\infty} \limsup_{n\to\infty} \mathbb P(\text{sup}_{s\in[0,t]}(\!-\hat \Theta_n^{(l)}(s,t))\ge \epsilon)^{1/b_n^2}=0.\]

Recalling (3.66) yields the convergence (3.65).

Theorem 4 has thus been proved. In order to obtain the assertion of Theorem 3, note that $-K$ has the same idempotent distribution as K and invoke the continuous mapping principle in (3.30), which applies by Lemma B.2 in [Reference Puhalskii and Reed21].

The proof of Theorem 2 proceeds along similar lines. First we prove an analogue of Theorem 3 to the effect that if the random variables $\overline X_n(0)$ LD converge to an idempotent variable $\overline X(0)$ , then the processes $\overline X_n$ LD converge to the idempotent process $\overline X$ that solves the following analogue of (3.2):

(3.70) \begin{multline} \overline X(t)=(1-F_0(t))\overline X(0)+\sqrt{q_0}\,W^0(F_0(t))+\int_0^t\bigl(1-F(t-s)\bigr)\sigma\,\dot W(s)\,ds\\[5pt] +\int_{\mathbb R_+^2} \,\mathbf{1}_{\{x+s\le t\}}\,\,\dot K( F(x),\lambda s)\,dF(x)\,\lambda\,ds.\end{multline}

The proof is a simpler version of the proof of Theorem 3. Essentially, one replaces $\hat A_n$ with $A_n$ and $\mu$ with $\lambda$ . A key element of the proof of Theorem 3 is the property, asserted in the statement of Lemma 3, that $\hat A_n(t)/n\to \mu t$ super-exponentially in probability. This property takes some effort to establish. Its counterpart for the infinite-server queue is that $ A_n(t)/n\to \lambda t$ super-exponentially in probability; this is a direct consequence of the hypotheses.

In some more detail, since, in analogy with (2.2),

\[ \overline Q_n(t)=Q_n^{(0)}(t)+A_n(t)-\int_0^t\int_0^t \mathbf{1}_{\{x+s\le t\}}\,d\,\sum_{i=1}^{{{ A}_n}(s)}\mathbf{1}_{\{\eta_i\le x\}},\]

grouping terms appropriately and recalling (2.12) yields

\begin{multline*} \!\!\!\!\!\frac{1}{n}\,\overline Q_n(t)-\overline q(t)=\bigg(\!\frac{1}{n}\,\overline{Q}_n(0)-q_0\!\bigg)(1-F_0(t))+\frac{1}{n}\,A_n(t)-\lambda t-\!\int_0^t\!\!\bigg(\frac{1}{n}\,A_n(t-s)-\lambda(t-s)\!\!\bigg) dF(s)\\[5pt] +\frac{1}{n}\,\sum_{i=1}^{\overline{Q}_n(0)}\,(\,\mathbf{1}_{\{\eta^{(0)}_i> t\}}\,-(1-F_0(t)))-\int_{\mathbb R_+^2}\,\mathbf{1}_{\{x+s\le t\}}\,d\sum_{i=1}^{A_n(s)}\bigl(\,\mathbf{1}_{\{\eta_i\le x\}}\,-F(x)).\end{multline*}

On introducing

\begin{align*}\overline Y_n(t)&=\frac{\sqrt{n}}{b_n}\,\bigg(\frac{A_n(t)}{n}-\lambda t\bigg),\\[5pt] \overline X_n^{(0)}(t)&=\frac{1}{b_n\sqrt{n}}\,\sum_{i=1}^{\overline{Q}_n(0)}\,(\,\mathbf{1}_{\{\eta^{(0)}_i> t\}}\,-(1-F_0(t))),\\[5pt] \overline H_n(t)&=\overline Y_n(t)-\int_0^t\overline Y_n(t-s)\,dF(s),\\[5pt] \overline U_n(x,t)&=\frac{1}{b_n\sqrt{n}}\,\sum_{i=1}^{{{ A}_n}(t)}\bigl(\,\mathbf{1}_{\{\eta_i\le x\}}\,-F(x)\bigr),\end{align*}

and

\begin{align*} \overline\Theta_n(t)&=-\int_{\mathbb R_+^2} \mathbf{1}_{\{x+s\le t\}}\,d\overline U_n(x,s),\end{align*}

we obtain the following analogue of (3.30):

(3.71) \begin{equation} \overline X_n(t)=(1-F_0(t))\overline X_n(0)+\overline X_n^{(0)}(t)+\overline H_n(t)+\overline \Theta_n(t).\end{equation}

The hypotheses imply that $\overline Y_n$ LD converges to $\sigma W$ , with the notation of the proof of Theorem 3 being reused. Since, in analogy with (3.38), $ \overline U_n(x,t)=K_n(F(x), A_n(t)/n)$ and $A_n(t)/n$ converges to $\lambda t$ super-exponentially in probability, the process $\overline U_n$ LD converges to $\overline U$ , where $\overline U(x,t)=K(F(x),\lambda t).$ Furthermore, similarly to Lemma 4, it is proved that $(\overline X_n(0),\overline X_n^{(0)},\overline U_n,\overline L_n)$ LD converges to $(\overline X(0),\overline X^{(0)},\overline U,\overline L)$ , where $\overline X^{(0)}(t)=\sqrt{q_0}W^0(F_0(t))$ and $\overline L(x,t)=B(F(x),\lambda t)$ . Put together, these properties imply the analogue of Theorem 4: that $(\overline X_n(0),\overline X_n^{(0)},\overline H_n,\overline \Theta_n)$ LD converges to $(\overline X(0),\overline X^{(0)},\overline H,\overline \Theta)$ , where

\begin{align*} \overline H(t)=\sigma W(t)-\sigma\int_0^t W(t-s)\,dF(s)=\sigma W(t)-\sigma\int_0^t(1- F(t-s))\,\dot W(s)\,ds\end{align*}

and

\begin{align*}\overline \Theta(t)=-\int_{\mathbb R_+^2} \mathbf{1}_{\{x+s\le t\}}\,d\overline U(x,s)=-\int_{\mathbb R_+^2} \mathbf{1}_{\{x+s\le t\}}\,\dot K(F(x),\lambda s)\,dF(x)\,\lambda ds. \end{align*}

An application of the continuous mapping principle to (3.71) concludes the proof.

4. Evaluating the deviation functions

This section is concerned with solving for $ I^Q_{x_0}(q)$ and $\overline I^Q_{q_0,x_0}(q)$ .

Theorem 5. Suppose that the CDF F is an absolutely continuous function and $I_{x_0}^Q(q)<\infty$ . Then q is absolutely continuous, $(\dot q(t)-\int_0^t\dot q(s)\,\mathbf{1}_{\{q(s)>0\}}\,F'(t-s)\,ds,t\in\mathbb R_+)\in\mathbb L_2(\mathbb R_+)$ , the infimum in (2.5) is attained uniquely, and

\[ I_{x_0}^Q(q)=\frac{1}{2}\,\int_0^\infty\hat{p}(t)\bigl(\dot q(t)-\int_0^t\dot q(s)\,\mathbf{1}_{\{q(s)>0\}}\,F'(t-s)\,ds+(\beta-x_0^-) F_0'(t)\bigr)\,dt,\]

where $\hat{p}(t)$ represents the unique $ \mathbb L_2(\mathbb R_+)$ -solution p(t) of the Fredholm equation of the second kind

(4.1) \begin{multline} (\mu+\sigma^2)p(t)= \dot q(t)-\int_0^t\dot q(s)\,\mathbf{1}_{\{q(s)>0\}}\,F'(t-s)\,ds+(\beta-x_0^-) F_0'(t)\\[5pt] +\sigma^2\int_0^\infty F'(\lvert s-t\rvert)p(s)\,ds-\sigma^2\int_0^\infty\int_0^{s\wedge t}F'(s-r)\, F'(t-r)\,dr\,p(s)\,ds,\end{multline}

with $\dot q$ , $F_0'$ , and F’ denoting derivatives.

Proof. Writing

(4.2) \begin{equation} \int_{\mathbb R_+^2} \,\mathbf{1}_{\{x+s\le t\}}\,\,\dot k(F(x),\mu s)\,dF(x)\,\mu\,ds= \int_0^{ t}\int_0^{F(t-s)}\dot k(x,\mu s)\,dx\,\mu\,ds,\end{equation}

we see that the equation (2.6) is of the form

\[ q(t)=f(t)+\int_0^tq(t-s)^+\,dF(s), t\in\mathbb R_+,\]

with the functions f(t) and F(t) being absolutely continuous. The function q(t) is absolutely continuous by Lemma 8. In addition, (4.2) implies that, almost everywhere (a.e.),

(4.3) \begin{equation} \frac{d}{dt}\,\int_{\mathbb R_+^2} \,\mathbf{1}_{\{x+s\le t\}}\,\,{\dot k}(F(x),\mu s)\,dF(x)\,\mu\,ds=\int_0^t{\dot k}(F(s),\mu (t-s))F'(s)\,\mu ds.\end{equation}

The infimum in (2.5) is attained uniquely by coercitivity and strict convexity of the function being minimised; cf. [Reference Ekeland and Temam5, Proposition II.1.2]. Differentiation in (2.6) with the account of (4.3) implies that, a.e.,

\begin{multline*}\dot w^0(F_0(t))F_0'(t)+\sigma\,\dot w(t)-\int_0^tF'(t-s)\sigma\,\dot w(s)\,ds+\int_0^t\dot k(F(s),\mu (t-s))F'(s)\,\mu\,ds\\[5pt] -\bigg( \dot q(t)-\int_0^t\dot q(s)\,\mathbf{1}_{\{q(s)>0\}}\,F'(t-s)\,ds+(\beta-x_0^-) F_0'(t)\bigg)=0.\end{multline*}

Introduce the map

\begin{multline*} \Phi:\, (\dot w^0,\dot w,\dot k)\to\bigg(\dot w^0(F_0(t))F_0'(t)+\sigma\,\dot w(t)-\int_0^tF'(t-s)\sigma\,\dot w(s)\,ds\\[5pt] +\int_0^t\dot k(F(s),\mu (t-s))F'(s)\,\mu\,ds,t\in\mathbb R_+\bigg).\end{multline*}

Since $F_0'(t)$ is bounded by (2.1), $\Phi$ maps $V=\mathbb L_2([0,1])\times\mathbb L_2(\mathbb R_+)\times\mathbb L_2( [0,1]\times\mathbb R_+)$ to $\mathbb L_2(\mathbb R_+)$ . For instance, using the fact that $\int_0^\infty F'(s)\,ds=1$ , we have

\[ \int_0^\infty\bigl(\int_0^tF'(t-s)\dot w(s)\,ds\bigr)^2\,dt\le \int_0^\infty\int_0^tF'(t-s)\dot w(s)^2\,ds\,dt=\int_0^\infty\dot w(s)^2\,ds<\infty\]

and

\begin{multline*} \int_0^\infty\bigl(\int_0^t\dot k(F(s),\mu (t-s))F'(s)\,\mu ds\bigr)^2\,dt\le \int_0^\infty\int_0^t\dot k(F(s),\mu (t-s))^2F'(s)\,\mu^2ds\,dt\\[5pt] =\mu^2\int_0^\infty\int_0^1\dot k( x,t)^2dx\,dt<\infty.\end{multline*}

The method of Lagrange multipliers, more specifically, [Reference Ekeland and Temam5, Proposition III.5.2] with $Y=\mathbb L_2(\mathbb R_+)$ and the set of nonnegative functions as the cone $\mathcal{C}$ , yields

(4.4) \begin{multline} I_{x_0}^Q(q)=\text{sup}_{p\in \mathbb L_2(\mathbb R_+)}\;\inf_{\substack{(\dot w^0,\dot w,\dot k)\in\mathbb L_2([0,1])\\[5pt] \times\mathbb L_2( \mathbb R_+)\times\mathbb L_2( [0,1]\times\mathbb R_+)}}\Bigl(\frac{1}{2}\int_0^1\dot w^0(x)^2\,dx+\frac{1}{2}\int_0^\infty \dot w(t)^2\,dt\\[5pt] +\frac{1}{2}\int_0^\infty\int_0^1 \dot k(x,t)^2\,dx\,dt+\int_0^\infty p(t)\bigl(\dot q(t)+F'(t)x_0^++(\beta-x_0^-) F_0'(t)\\[5pt] -\int_0^t\dot q(s)\,\mathbf{1}_{\{q(s)>0\}}\,F'(t-s)\,ds -\dot w^0(F_0(t))F_0'(t)-\sigma\,\dot w(t)+\int_0^tF'(t-s)\sigma\,\dot w(s)\,ds\\[5pt] -\int_0^t\dot k(F(s),\mu (t-s))F'(s)\,\mu\,ds\bigr)\,dt\Bigr).\end{multline}

Minimising in (4.4) yields, with $(\dot{\hat{ w}}^{0}(t),\dot{\hat w}(t), \dot{\hat{k}}(x,t))$ , being optimal,

\begin{align*} \dot{\hat w}^0(x)-p(F_0^{-1}(x))=0,\\[5pt] \dot{\hat w}(t)-\sigma p(t)+\sigma\int_0^\infty p(t+s)F'(s)\,ds=0,\\[5pt] \dot{\hat k}(x,t)-p(\frac{t}{\mu}+F^{-1}(x))=0.\end{align*}

(For the latter, note that

\begin{multline*} \int_0^\infty p(t)\int_0^t\dot k(F(s),\mu (t-s))F'(s)\,\mu ds\,dt=\int_0^\infty\int_s^\infty p(t)\dot k(F(s),\mu (t-s))F'(s)\,\mu\,dt\, ds\\[5pt] =\int_0^\infty\int_0^\infty p(t+s)\dot k(F(s),\mu t)F'(s)\,\mu\,dt\,ds=\int_0^\infty\int_0^1 p(\frac{t}{\mu}+F^{-1}(x))\dot k( x,t)\,dx\,dt.)\end{multline*}

Hence,

\begin{align*} I_{x_0}^Q(q)&=\text{sup}_{p\in \mathbb L_2(\mathbb R_+)}\bigg(\int_0^\infty p(t)\bigl(\dot q(t)-\int_0^t\dot q(s)\,\mathbf{1}_{\{q(s)>0\}}\,F'(t-s)\,ds+(\beta-x_0^-) F_0'(t)\bigr)\,dt\\[5pt] &\quad-\frac{1}{2}\bigg(\int_0^\infty p(s)^2F_0'(s)\,ds+\int_0^\infty(\sigma p(t)-\sigma\int_0^\infty p(t+s)F'(s)\,ds)^2\,dt\\[5pt]&\quad+\mu\int_0^\infty\int_0^\infty p(t+s)^2\,F'(s)ds\,dt\bigg)\bigg).\end{align*}

Noting that

(4.5) \begin{equation} \int_0^\infty p(s)^2F_0'(s)\,ds+\mu\int_0^\infty\int_0^\infty p(t+s)^2\,F'(s)ds\,dt=\mu\int_0^\infty p(s)^2\,ds\end{equation}

yields

(4.6) \begin{multline} I_{x_0}^Q(q)=\text{sup}_{p\in\mathbb L_2(\mathbb R_+)}\bigg(\int_0^\infty p(t)\bigg(\dot q(t)-\int_0^t\dot q(s)\,\mathbf{1}_{\{q(s)>0\}}\,F'(t-s)\,ds+(\beta-x_0^-) F_0'(t)\bigg)\,dt\\[5pt] -\frac{1}{2}\bigg(\mu\int_0^\infty p(s)^2\,ds+\int_0^\infty(\sigma p(t)-\sigma\int_0^\infty p(t+s)F'(s)\,ds)^2\,dt\bigg)\bigg).\end{multline}

The existence and uniqueness of a maximiser in (4.6) follows from Proposition II.1.2 in [Reference Ekeland and Temam5] because the expression in the supremum tends to $-\infty$ as $\lVert p\rVert_{\mathbb L_2(\mathbb R_+)}\to\infty$ . Varying p in (4.6) implies (4.1). As the maximiser in (4.6) is unique, so is an $\mathbb L_2(\mathbb R_+)$ -solution of the Fredholm equation.

It is noteworthy that the integral operator on $\mathbb L_2(\mathbb R_+)$ with kernel

\[ \tilde K(t,s)=F'(\lvert s-t\rvert)-\int_0^{s\wedge t}F'(s-r)\,F'(t-r)\,dr\]

is not generally either Hilbert–Schmidt or compact, so the existence and uniqueness of $\hat p(t)$ is not a direct consequence of the general theory.

The numerical solution of Fredholm equations, such as (4.1), is discussed at quite some length in the literature; see, e.g., [Reference Atkinson3] and references therein. For instance, the collocation method with a basis of ‘hat’ functions may be tried: for $i\in\mathbb N$ and $n\in\mathbb N$ , let $t_i=i/n$ and $\ell_i(t)=(1-\lvert t-t_i\rvert)\,\mathbf{1}_{\{t_{i-1}\le t\le t_i\}}$ , with $t_0=0$ . Then an approximate solution is

\[ p_n(t)=\sum_{i=1}^{n^2}p_n(t_i)\ell_i(t),\]

where the $p_n(t_i)$ , $i=1,\ldots,n^2$ , satisfy the linear system

\begin{multline*} (\mu+\sigma^2)p_n(t_i)-\sigma^2\sum_{j=1}^{n^2}p_n(t_j)\int_0^{n^2} \tilde K(t_i,s)\ell_j(s)\,ds\\[5pt] = \dot q(t_i)-\int_0^{t_i}\dot q(s)\,\mathbf{1}_{\{q(s)>0\}}\,F'(t_i-s)\,ds+(\beta-x_0^-) F_0'(t_i).\end{multline*}

For more background, see [Reference Atkinson3].

The evaluation of $\overline I^Q_{q_0,x_0}$ is carried out similarly:

\begin{multline*} \overline I^Q_{q_0,x_0}(q)=\text{sup}_{p\in \mathbb L_2(\mathbb R_+)}\; \inf_{ \substack{(\dot w^0,\dot w,\dot k)\in \mathbb L_2([0,1])\\[5pt] \times\mathbb L_2( \mathbb R_+)\times \mathbb L_2([0,1]\times\mathbb R_+)}}\Bigl(\frac{1}{2}\int_0^1\dot w^0(x)^2\,dx+\frac{1}{2}\int_0^\infty \dot w(t)^2\,dt\\[5pt] +\frac{1}{2}\int_0^\infty\int_0^1 \dot k(x,t)^2\,dx\,dt+\int_0^\infty p(t)\bigl(\dot q(t) +x_0 F_0'(t) -\sqrt{q_0}\dot w^0(F_0(t))F_0'(t) -\sigma\,\dot w(t)\\[5pt] +\int_0^tF'(t-s)\sigma\,\dot w(s)\,ds -\int_0^t\dot k(F(s),\lambda (t-s))F'(s)\,\lambda\, ds\bigr)\,dt\Bigr).\end{multline*}

The infimum is attained at $(\dot{\overline{w}}^0(t),\dot{\overline{w}}(t),\dot{\overline k}(x,t))$ such that

\begin{align*} \dot{\overline{w}}^0(x)-\sqrt{q_0} p(F_0^{-1}(x))=0,\\[5pt] \dot{\overline{w}}(t)-\sigma p(t)+\sigma\int_0^\infty p(t+s) F'(s)\,ds=0,\\[5pt] \dot{\overline k}(x,t)- p(\frac{t}{\lambda}+F^{-1}(x))=0.\end{align*}

Consequently, taking into account (2.1) and (4.5), we have

\begin{multline*} \overline I^Q_{q_0,x_0}(q)=\text{sup}_{p\in \mathbb L_2(\mathbb R_+)}\bigl(\int_0^\infty p(t)(\dot q(t) +x_0 F_0'(t))\,dt-\frac{1}{2}\bigl(q_0\mu\int_0^\infty p(s)^2\,(1-F(s))\,ds\\[5pt] +\lambda\int_0^\infty p(s)^2F(s)\,ds+\int_0^\infty(\sigma p(t)-\sigma\int_0^\infty p(t+s)F'(s)\,ds)^2\,dt\bigr)\bigr)\\[5pt] =\frac{1}{2}\int_0^\infty\overline p(t)(\dot q(t) +x_0 F_0'(t))\,dt,\end{multline*}

with $\overline p(t)$ being the $\mathbb L_2(\mathbb R_+)$ -solution p(t) to the Fredholm equation of the second kind

\[ (q_0\mu(1-F(t))+\lambda F(t)+\sigma^2)p(t)=\dot q(t)+x_0F^{\prime}_0(t)+\sigma^2\int_0^\infty \tilde K(t,s)p(s)\,ds.\]