Appendix A. Rate conservation

One can rewrite (6) as

\begin{equation*}\lambda p {\mathbb{E}} [Y]+ \alpha {\mathbb{E}}\big[ XY^2\big] =\lambda p {\mathbb{E}} [Y]+ \alpha {\mathbb{E}}[XY] {\mathbb{E}}\left[ \frac{\alpha XY}{\alpha {\mathbb{E}}[XY]} Y \right]=(\mu +\beta) {\mathbb{E}}[Y(Y-1)].\end{equation*}

Let I denote the stationary infection epoch point process. Its intensity is $a_i = \alpha {\mathbb{E}}[XY]$ , and by Papangelou’s theorem [Reference Baccelli and Brémaud1],

(53) \begin{equation}{\mathbb{E}}\left[ \frac{\alpha XY}{\alpha {\mathbb{E}}[XY]} Y \right] = {\mathbb{E}}^0_{I}[ Y],\end{equation}

with ${\mathbb{E}}^0_{I}$ denoting expectation with respect to the Palm probability with respect to I [Reference Baccelli and Brémaud1]. Hence the left-hand side of the last equality is $\lambda p {\mathbb{E}} [Y] + a_i {\mathbb{E}}^0_{I}[ Y]$ , in which we recognize half of the increase rate of ${\mathbb{E}}[Y(Y-1)]$ . Let $M_i$ denote the point process of recovery or departures of infected customers. The right-hand side of (53) can be rewritten as

$$(\mu +\beta) {\mathbb{E}}[Y] {\mathbb{E}}\left[ \frac{(\mu +\beta)Y}{(\mu +\beta) {\mathbb{E}}[Y]}\right. (Y-1) = d_i {\mathbb{E}}^0_{N} [Y-1],$$

with $d_i$ the intensity of $M_i$ and the same Palm probability notation as above. This is twice the decrease rate of $Y(Y-1)$ . Hence (6) is nothing else than the RCP for $\frac 1 2 Y(Y-1)$ .

The RCP for $Y^2$ in turn reads

(54) \begin{equation}\lambda p {\mathbb{E}} [2Y+1]+ \alpha {\mathbb{E}}[ XY(2Y+1)]=(\mu +\beta) {\mathbb{E}}[Y(2Y-1)].\end{equation}

Similarly, we can rewrite (7) as

\begin{equation*}\lambda q {\mathbb{E}} [X] +\beta {\mathbb{E}}[XY] =\alpha {\mathbb{E}}[ XY(X-1)] +\mu {\mathbb{E}}[X(X-1)]\end{equation*}

or equivalently

\begin{equation*}\lambda q {\mathbb{E}} [X] +\beta {\mathbb{E}}[XY] =\alpha {\mathbb{E}}[XY] {\mathbb{E}}\left[\frac{\alpha XY}{\alpha {\mathbb{E}}[XY]}(X-1)\right]+\mu {\mathbb{E}}[X] {\mathbb{E}}\left[\frac{\mu X}{\mu {\mathbb{E}}[X]}(X-1)\right].\end{equation*}

By the same arguments, the right-hand side is half the decrease rate of the second factorial moment of X due to infection and departures. The left-hand side is half the increase rate of the same quantity due to arrivals of susceptible customers and recoveries of infected customers. This equation is hence the RCP for $\frac 1 2 X(X-1)$ . The RCP for $X^2$ reads

(55) \begin{equation}\lambda q {\mathbb{E}} [2X+1] +\beta {\mathbb{E}}[Y (2X+1)] =\alpha {\mathbb{E}}[ XY(2X-1)] +\mu {\mathbb{E}}[X(2X-1)].\end{equation}

If we differentiate the PDE with respect to x and y (or y and x) and taking $x=y=1$ , we get

(56) \begin{equation}\lambda p {\mathbb{E}} [X] + \lambda q{\mathbb{E}} [Y] +\beta {\mathbb{E}} [Y(Y-X-1)]+ \alpha({\mathbb{E}}[ XY(X-Y-1])= 2\mu{\mathbb{E}} [XY].\end{equation}

It is easy to check that inserting the two relations from the last lemma into the equation above does not give anything new. It just confirms that $X+Y$ is Poisson- $\lambda/\mu$ . It is easy to check that this is in fact the RCP for XY.

Remark 7. Using the Palm interpretation given above, and PASTA, it is easy to check that (6) can be rewritten as

(57) \begin{equation}\lambda p {\mathbb{E}}_{A_i} [Y^-]+ a_i {\mathbb{E}}_{I}[ Y^-] =\mu_i {\mathbb{E}}_{D_i}[Y^+]+a_s {\mathbb{E}}_{R}[Y^+],\end{equation}

with $\mu_i=\mu{\mathbb{E}}[Y]$ the exogenous departure rate in this queue, $a_s=\beta{\mathbb{E}}[Y]$ the recovery rate, and $a_i=\alpha{\mathbb{E}}[XY]$ the infection rate. This is nothing but the classical property that, in the ‘infected queue’, the Palm expectation of the number of customers just before arrivals coincides with the Palm expectation of the number of customers just after departures. Similarly, (7) reads

(58) \begin{equation}\lambda q {\mathbb{E}}_{A_s} [X^-] +a_s {\mathbb{E}}_{R}[X^-] = a_i {\mathbb{E}}_{I}[ X^+] +\mu_s {\mathbb{E}}_{D_s}[X^+],\end{equation}

with similar notation.

Corollary 4. The following relations hold:

(59) \begin{equation}\lambda \left(q +\frac \beta \mu\right)\left(1 +\frac \beta \alpha\right)+\left(\lambda q -\frac \beta \alpha(\mu+\alpha+\beta)\right) {\mathbb{E}} [X]-\mu {\mathbb{E}}[X^2]=\alpha {\mathbb{E}}\big[ X^2Y\big]\end{equation}

and

(60) \begin{eqnarray}& & \hspace{-1.6cm} \frac \lambda \mu\left(\lambda p +\frac {\mu +\beta} {\mu\alpha} (2\mu(\mu q+\beta) -\lambda \alpha)\right)-\left(\lambda p +\mu+\beta +\frac 2 \alpha (\mu+\beta)^2\right) {\mathbb{E}} [X] \nonumber \\& & +(\mu+\beta) {\mathbb{E}}[X^2] =-\alpha {\mathbb{E}}\big[ XY^2\big].\end{eqnarray}

Appendix B. Proofs of Lemmas 5, 6, and 7

(1) Monotonicity. Consider the typical busy cycle. Assume that the first customer arrives at an empty system at time 0 and let $T>0$ be the first time when the system becomes empty again. We have to show that the mean number of infected customers that depart from the system within the time interval [0, T] is an increasing function of p.

For that, we consider two models, with input probabilities of being infected p and $\widehat{p}>p$ . For the input probability p and for $0\le t \le T$ , we denote by N(t) the total number of customers at time $t>0$ , by $N_I(t)$ the number of infected customers, and by $N_S(t)$ the number of susceptible customers. Clearly, $N(t)=N_I(t)+N_S(t)$ . Let $\widehat{N}(t)$ , $\widehat{N}_I(t)$ , and $\widehat{N}_S(t)$ by the corresponding processes related to probability $\widehat{p}$ , with $\widehat{N}(t)=\widehat{N}_I(t)+\widehat{N}_S(t)$ for $t\ge 0$ .

We produce a coupling of the two models such that, for all $0\le t\le T$ , $N(t)=\widehat{N}(t)$ a.s., while $\widehat{N}_I(t)\ge N_I(t)$ and, therefore, $\widehat{N}_S(t)\le N_S(t)$ . Thus, for the total numbers of departures of infected customers within the first busy cycle, we get $D_I(T) \le \widehat{D}_I(T)$ a.s. Moreover, we will show that ${\mathbb{P}}(D_I(T) < \widehat{D}_I(T))>0$ .

Let $t_n$ , $n\ge 1$ , be exponential- $\lambda$ random variables; let $\sigma_{n,j}$ , $n,j\ge 1$ , be exponential- $\mu$ random variables; let $a_{n,i,j}$ be exponential- $\alpha$ random variables; let $b_{n,j}$ be exponential- $\beta$ random variables; and let $u_{n}$ have a uniform distribution in the interval (0, 1). We assume all these random variables to be mutually independent.

We introduce embedded epochs $T_0=0<T_1< T_2 < \ldots < T_{\psi}=T$ , where $\psi$ is a random natural number representing the total number of events (jumps) that occur in both systems. This is not restrictive, as some of these events will be fictitious in either system. Both processes $(N_S(t),N_I(t))$ and $(\widehat{N}_S(t),\widehat{N}_I(t))$ are piecewise constant and may make jumps only at time instants $T_n$ . We assume the processes to be right-continuous, and customers to be numbered in each group at any time, in a way that we will make precise in due time.

At time $T_0=0$ , we let

\begin{align*}N_S(0)={\mathbf I} (u_1>p) = 1 -N_I(0) \quad \mbox{and}\quad \widehat{N}_S(0)={\mathbf I} (u_1>\widehat{p}) = 1 -\widehat{N}_I(0)\end{align*}

and

\begin{align*}D_S(0)=D_I(0)=\widehat{D}_S(0)=\widehat{D}_I(0)=0.\end{align*}

Then we evaluate the values of the processes recursively: that is, we produce both time $T_{n+1}$ and their values at time $T_{n+1}$ given time $T_n$ , their values at time $T_n$ , and the values of $t_{n+1}$ , $u_{n+1}$ , $\{\sigma_{n+1,j}\}_{j\ge 1}$ , $\{a_{n+1,i,j}\}_{i,j\ge 1}$ , $\{b_{n+1,j}\}_{j\ge 1}$ .

Assume we have introduced both processes up to time $T_n$ and shown that $\widehat{N}_I(T_n)\ge N_I(T_n)$ and $\widehat{N}(T_n)=N(T_n)$ . Assume that $\sigma_{n+1,j}$ , $1\le j \le N(T_n)$ , are the remaining service times of all present customers, where customers numbered $1,\ldots,N_I(T_n)$ are infected in both systems, customers numbered $N_I(T_n)+1,\ldots,\widehat{N}_I(T_n)$ are susceptible in the first system and infected in the second system, and customers numbered $\widehat{N}_I(T_n)+1, \ldots,N(T_n)$ are susceptible in both systems. Assume that the random variables $b_{n+1,j}$ , for $j=N_I(T_n)+1,\ldots,\widehat{N}_I(T_n)$ , represent the recovery times for the corresponding infected customers in the second system, and that for $j=1,\ldots, N_I(T_n)$ they represent the recovery times for customers that are infected in both systems. Assume that if customer j is infected and customer i is susceptible at time $T_n$ , then $a_{n+1,j,i}$ is the instant when j may infect i.

Consider the random sets of integers $C_1(n) = \{j\,:\, 1\le j \le N_I(T_n)\}$ , $C_2(n) = \{j\,:\, N_I(T_n)< j \le \widehat{N}_I(T_n)\}$ , and $C_3(n) = \{j\,:\, \widehat{N}_I(T_n)< j \le N(T_n)\}$ . For $k=1,2,3$ , let

\begin{align*}\Sigma_{n+1,k} = \min_{j\in C_k(n)} \sigma_{n+1,j} \quad \mbox{and}\quad B_{n+1,k} = \min_{j\in C_k(n)} b_{n+1,j},\end{align*}

where $\min_{\emptyset} = \infty$ , by convention. Next, for $k,l=1,2,3$ , let

\begin{align*}A_{n+1,k,l} = \min_{j\in C_k(n), i\in C_l(n)} a_{n+1, j,i}.\end{align*}

Finally, let

(61) \begin{align}\theta_{n+1} = \min (t_{n+1}, \min_{1\le k \le 3} \Sigma_{n+1,k},\min_{1\le k \le 2} B_{n+1,k},\min\! (A_{n+1,2,3}, A_{n+1,1,3},A_{n+1,1,2})).\end{align}

Therefore, $\theta_{n+1}$ is the next time instant when something happens in any of the systems—either the arrival of a new customer, the departure of a present customer, the recovery of a present customer, or the infection of one present customer by another one. Then

\begin{align*}T_{n+1} = T_n+\theta_{n+1}.\end{align*}

If $\theta_{n+1}=t_{n+1}$ , i.e. a new customer arrives, then

\begin{align*}D_S(T_{n+1})=D_S(T_n), D_I(T_{n+1})=D_I(T_n),\widehat{D}_S(T_{n+1})=\widehat{D}_S(T_n), \widehat{D}_I(T_{n+1})=\widehat{D}_I(T_n)\end{align*}

and $N(T_{n+1})=\widehat{N}(T_{n+1})=N(T_n)+1$ ,

If $\theta_{n+1}= \min_{1\le k\le 3} \Sigma_{n+1,k}$ , then one of the present customers departs. If $\Sigma_{n+1,1}$ is the smallest among the three, then there is a departure of a customer that is infected in both systems. Therefore, $D_I(T_{n+1})=D_I(T_n)+1$ , $\widehat{D}_I(T_{n+1})=\widehat{D}_I(T_n)+1$ , $D_S(T_{n+1})=D_S(T_n)$ , $\widehat{D}_S(T_{n+1})=\widehat{D}_S(T_n)$ , and $N_I(T_{n+1})=$ $N_I(T_n)-1$ , $\widehat{N}_I(T_{n+1})=\widehat{N}_I(T_n)-1$ , $N_S(T_{n+1})=N_S(T_n)$ , $\widehat{N}_S(T_{n+1})=\widehat{N}_S(T_n)$ . By the symmetry, if $\sigma_{n+1,3}$ is the smallest, then there is a departure of a customer that is susceptible in both systems. Therefore, $D_S(T_{n+1})=D_S(T_n)+1$ , $\widehat{D}_S(T_{n+1})=\widehat{D}_S(T_n)+1$ , $D_I(T_{n+1})=D_I(T_n)$ , $\widehat{D}_I(T_{n+1})=\widehat{D}_I(T_n)$ , and $N_S(T_{n+1})=N_S(T_n)-1$ , $\widehat{N}_S(T_{n+1})=\widehat{N}_S(T_n)-1$ , $N_I(T_{n+1})=N_I(T_n)$ , $\widehat{N}_I(T_{n+1})=\widehat{N}_I(T_n)$ .

Next, if $\Sigma_{n+1,2}$ is the smallest, then this means that the strict inequality $N_I(T_n)<\widehat{N}_I(T_n)$ holds and there is a departure of a customer that is susceptible in the first system and infected in the second. Then $D_I(T_{n+1})=D_I(T_n)$ , $\widehat{D}_I(T_{n+1})=\widehat{D}_I(T_n)+1$ , $D_S(T_{n+1})=D_S(T_n)+1$ , $\widehat{D}_S(T_{n+1})=\widehat{D}_S(T_n)$ , and $N_I(T_{n+1})=N_I(T_n)$ , $\widehat{N}_I(T_{n+1})=\widehat{N}_I(T_n)-1$ , $N_S(T_{n+1})=N_S(T_n)-1$ , $\widehat{N}_S(T_{n+1})=\widehat{N}_S(T_n)$ . One can see that in all three cases

(62) \begin{align}N_I(T_{n+1})\le \widehat{N}_I(T_{n+1})\quad \mbox{and}\quad D_I(T_{n+1})\le \widehat{D}_I(T_{n+1}).\end{align}

Similarly, if $\theta_{n+1} = \min_{k=1,2} B_{n+1,k}$ , then the numbers of departures do not change, and either (a) there is a simultaneous recovery of a customer that was infected in both systems, or (b) there was a customer that was susceptible in the first system and infected in the second, it has recovered in the second system, and nothing has changed in the first system. Again, one can see that the needed inequalities (62) continue to hold.

Finally, if $\theta_{n+1} = \min (A_{n+1,2,3}, A_{n+1,1,3},A_{n+1,1,2})$ , then the number of departures stays the same and there are again three scenarios. If $A_{n+1,2,3}$ is the smallest among the three, then the set $C_2(n)$ of customers that are susceptible in the first system and infected in the second system at time $T_n$ is non-empty, one of them has infected one of the susceptible customers in the second system, and nothing has changed in the first system. Therefore, the number of susceptible customers in the second system decreases and the number of susceptible customers in the first system stays the same. Thus, the needed inequalities (62) continue to hold. If $A_{n+1,1,2}$ is the smallest, then the set $C_2(n)$ is non-empty again, so we have strict inequality $N_S(T_n)>\widehat{N}_S(T_n)$ , and at time $T_{n+1}$ , one of customers from this set becomes infected in the first system, while nothing changes in the second system. Therefore,

\begin{align*}N_S(T_{n+1})=N_S(T_n) \le \widehat{N}_S(T_n)-1 =\widehat{N}_S(T_{n+1}),\end{align*}

as required. If $A_{n+1,1,3}$ is the smallest, then one of the customers that is susceptible at time $T_n$ in both systems becomes infected in both systems. So the required inequalities continue to hold.

This completes the proof of the fact that $D_I(t)\le \widehat{D}_I(t)$ at any time $0<t<T$ and, in particular, $D_I\equiv D_I(T) \le \widehat{D}_I\equiv \widehat{D}_I(T)$ a.s. It is left to show that the inequality may be strict with positive probability. However, this is almost obvious:

\begin{align*}\{D_I < \widehat{D}_I\} \supseteq \{D_I=0, \widehat{D}_I=1\} =\{\sigma_{1,1}<t_1\} \cap\{ p<u_1<\widehat{p}\}, \end{align*}

where the events on the right are independent and of positive probabilities.

(2) Concavity. We now introduce a slightly different coupling that allows us to consider simultaneously models with different parameters.

From now on, we do not enumerate customers. Instead, we consider a countable number ${\mathcal Z}$ of ‘locations’ for them that coincide with ‘servers’. More precisely, we assume that an arriving customer chooses an empty server at random and stays there until its departure from the system.

We continue to consider a single busy cycle on the time interval [0, T]. We again denote by $T_0<T_1<T_2<\ldots$ the time instants when the state of the systems may change. Hence, the state of the system Z(t) at time t is a collection of pairs $\{ (z,c_z(t)), z\in{\mathcal Z}\}$ , where z is a server and $c_z$ is its ‘color’. Here $c_z(t) ={\bf n}$ (where ‘n’ means ‘no color’) if server z is empty at time t, and $c_z(t)$ has one of several other colors otherwise. (The meanings of these colors will be explained in due time.) Let ${\mathcal N}(t) = \{z \,:\, c_z(t) \ne {\bf n} \}$ be the set of occupied servers, and let $N(t) = \sum_{z\in {\mathcal z}} {\mathbf I} (c_z(t)\ne{\bf n})$ denote its cardinality.

We introduce the following sequences of random variables:

• • $\{u_n\}_{n\ge 0}$ are uniformly distributed in (0, 1);

• • $\{t_n\}_{n\ge 1}$ are exponential- $\lambda$ ;

• • $\{\sigma_{n,z}\}_{n\ge 0, z\in{\mathcal Z}}$ are exponential- $\mu$ ;

• • $\{b_{n,z}\}_{n\ge 0, z\in {\mathcal Z}}$ are exponential- $\beta$ ;

• • $\{a_{n,z,v}\}_{n\ge 0, z,v\in {\mathcal Z}, z\ne v}$ are exponential- $\alpha$ .

We assume that all these random variables are mutually independent.

Let $T_0=0$ . Introduce recursively $\{\theta_k\}$ and $T_k=\sum_1^k \theta_j$ . Let ${\mathcal N}_k = {\mathcal N} (T_k+0)$ . For $n\ge 1$ , assuming that $T_{n-1}$ is defined, we let $\Sigma_n = \min_{z\in {\mathcal N}_{n-1}} \sigma_{n,z}$ , $B_n = \min_{z\in {\mathcal N}_{n-1}} b_{n,z}$ , $A_n = \min_{z,v\in {\mathcal N}_{n-1}, z\ne v} a_{n,z,v}$ . Then let

\begin{align*}\theta_n = \min \{ t_n, \Sigma_n,A_n,B_n\} \quad \mbox{and} \quad T_{n} = T_{n-1}+\theta_n.\end{align*}

Now we are ready to define a ‘three-color’ process. Let $p\ge 0$ , $q\ge 0$ , $r>0$ be numbers that sum up to 1, $p+r+q=1$ . We assume that each arriving customer gets the red color with probability p, the magenta color with probability r, and the green color with probability q. It occupies a server (that gets the same color as the customer).

We assume the process to be piecewise constant between time instants $T_n$ . We introduce coloring by induction.

At time $T_0=0$ , we color the arriving customer magenta if $u_0\le r$ , red if $r<u_0\le r+p$ , and green if $u_0>r+p$ , and we assume that the customer keeps its color within the time interval $(T_0,T_1]$ .

Assume that coloring is done up to time $T_n$ . Proceed with coloring for the time interval $(T_n,T_{n+1}]$ .

If $\theta_n=t_n$ , then we place an arriving customer at one of the empty servers and color the customer magenta if $u_n\le r$ , red if $r<u_n\le r+p$ , and green if $u_n>r+p$ .

If $\theta_n=\sigma_{n,z}$ for some $z\in {\mathcal N}_n$ , then customer z leaves the system and the corresponding server becomes idle (‘no color’).

If $\theta_n = \beta_{n,z}$ for some $z\in {\mathcal N}_n$ , then customer z becomes green, regardless of what color it had before.

If $\theta_n = A_n$ , then we have to consider three cases. At time $T_n$ , let ${\mathcal N}_{n,m}, \ {\mathcal N}_{n,r}, \ {\mathcal N}_{n,g}$ be the sets of magenta, red, and green customers, where clearly

\begin{align*}{\mathcal N}_{n} = {\mathcal N}_{n,m} \cup {\mathcal N}_{n,r} \cup {\mathcal N}_{n,g}.\end{align*}

Let $N_{n,m}$ , $N_{n,r}$ , and $N_{n,g}$ be the corresponding cardinalities. If $\theta_n = a_{n,z,v}$ for some $z\in {\mathcal N}_{n,g}$ , then (since green means susceptible) there is no new infection and all colors stay the same (one could say that this is a ‘false coloring’).

If $\theta_n = a_{n,z,v}$ for some $z\in {\mathcal N}_{n,r}$ , then customer v (and the corresponding server) becomes red, whatever color it had before.

Finally, if $\theta_n = a_{n,z,v}$ for some $z\in {\mathcal N}_{n,m}$ , then customer v (and the corresponding server) becomes magenta if it was either magenta or green before—or remains red if it was red before.

Thus, we have defined the three-color dynamics within the busy cycle.

As a result of the proposed coupling construction, one can observe the following.

1. 1. If we do not distinguish between (i.e. we ‘merge’) the green and magenta colors (and recolor them as ‘new green’), then we get the two-color susceptible–infected model considered earlier, with the probability of green arrivals $r+q$ and the probability of red arrivals p; call this model the $(p, r+q)$ model.

2. 2. If we do not distinguish between (i.e. we ‘merge’) the magenta and red colors (and recolor them as ‘new red’), then we get the two-color susceptible–infected model considered earlier, with the probability of green arrivals q and the probability of red arrivals $r+p$ ; call this model the $(r+p,q)$ model.

This means that, for $0\le t\le T$ , the random variable $N_m(t)$ represents the ‘excess’ of the number of ‘red’ customers in the $(r+p,q)$ model in comparison with the $(p,r+q)$ model.

Assume now that $q>0$ and consider the three-color $(r,\widehat{p},\widehat{q})$ -model with $\widehat{p}>p$ and $r+\widehat{p}+\widehat{q}=1$ . Let $\widehat{\mathcal N}_m(t)$ be the set of magenta customers at time t in this model and $\widehat{N}_m(t)$ its cardinality. Then straightforward induction arguments show that $\widehat{\mathcal N}_m(t) \subseteq {\mathcal N}_m(t)$ and, therefore, $\widehat{N}_m(t)\le N_m(t)$ a.s., for any $0\le t \le T$ . In turn, this implies that the total number of departures within the first cycle of magenta customers in the corresponding systems, call them $D_m(T)$ and $\widehat{D}_m(T)$ , satisfy $D_m(T) \ge \widehat{D}_m(T)$ a.s. Recall the notation $p_o=g(p)$ . Then

\begin{eqnarray*} & & g(p) = \frac{{\mathbb{E}} D_g(T)}{{{\mathbb{E}} D(T)}}, \qquad g(p+r) = \frac{{\mathbb{E}}(D_g(T)+D_m(T))}{{{\mathbb{E}} D(T)}}, \\ & & g(\widehat{p}) = \frac{{\mathbb{E}} \widehat{D}_g(T)}{{{\mathbb{E}} D(T)}}, \qquad g(\widehat{p}+r) = \frac{{\mathbb{E}}(\widehat{D}_g(T)+\widehat{D}_m(T))}{{{\mathbb{E}} D(T)}},\end{eqnarray*}

where, as before, ${{\mathbb{E}} D(T)}$ is the mean number of customers served in the first busy cycle (which is the same as the total number of departures within the cycle). Since

\begin{align*}g(p+r) - g(p)= \frac{{\mathbb{E}} D_m(T)}{{{\mathbb{E}} D(T)}}\ge \frac{{\mathbb{E}} \widehat{D}_m(T)}{{{\mathbb{E}} D(T)}}=g(\widehat{p}+r) - g(\widehat{p}),\end{align*}

we get the required concavity.

(3) Strict concavity. It is enough to consider the latter coupling construction and to justify that, for any $r>0$ and any $0\le p <\widehat{p} \le 1-r$ , ${\mathbb{P}} (D_m(T)>\widehat{D}_m(T))>0$ . One can see that this strict inequality holds on the following event, which has strictly positive probability:

(63) \begin{align}\{ u_0\le r, \ \ \theta_1=t_1, \ \ u_1\in (r+p, r+\widehat{p}),\theta_2 = a_{2,2,1}, \ \theta_3=\sigma_{3,0}, \ \ \theta_4 = \sigma_{4,1}\}.\end{align}

The latter means that the following sequence of events occurs:

• • The first customer gets the magenta color in both three-color systems.

• • Then the second customer appears and gets the green color in the first system and red in the second.

• • Then the second customer attempts to recolor the first one. This does not work in the first system, where we continue to have one green and one magenta; however, the attempt is successful in the second system, where we get two reds.

• • Finally, the first customer leaves the system and then the second leaves the system; this ends the busy cycle.

One can see that $D_m(T)=1>\widehat{D}_m(T)=0$ on the event (63).

(4) Differentiability. Let $h>0$ be small. For $p>0$ , introduce a four-color model with the colors green, violet, magenta, and red. Consider the second coupling construction introduced in the subsection ‘Concavity’ above. If we have an arrival at time $T_n$ , then the new customer is colored green if $u_n<p-h$ , violet if $p-h\le u_n < p$ , magenta if $p\le u_n < p+h$ , and red if $u_n \ge p+h$ . Let $A_v$ be the event that there is only one violet arrival and no magenta arrival in the (first) busy cycle, and $A_m$ that there is only one magenta arrival and no violet arrival. These events have equal probabilities that are of order $ch+o(h)$ , where $c={\mathbb{E}} \nu$ and $\nu$ is the mean number of arrivals within a busy cycle; there appear only (at most) three colors (green, violet, and red) on the event $A_v$ and, similarly, at most three colors (green, magenta, and red) on the event $A_m$ . Moreover, the dynamics of the process on the events $A_v$ and $A_m$ are identical, with the obvious swap of violet and magenta customers. Therefore, the mean number ${\mathbb{E}} (D_v(T)\ | A_v)$ of violet departures given the event $A_v$ coincides with the mean number ${\mathbb{E}} (D_m(T) \ | A_m)$ of magenta departures given the event $A_m$ . Furthermore, the event that there are two or more violet and/or magenta arrivals has probability $O(h^2)=o(h)$ . Since

\begin{align*}& g(p)-g(p-h) = \frac{{\mathbb{E}} D_v(T)}{{\mathbb{E}} D(T)} = ch \cdot \frac{{\mathbb{E}} (D_v(T)\ | A_v))}{{\mathbb{E}} D(T)} + o(h)\\ &=ch \cdot \frac{{\mathbb{E}} (D_m(T)\ | A_m))}{{\mathbb{E}} D(T)} + o(h) = \frac{{\mathbb{E}} D_m(T)}{{\mathbb{E}} D(T)} +o(h) =g(p+h)-g(p) + o(h),\end{align*}

we get

$$\frac{g(p)-g(p-h)}{h} = \frac{g(p+h)-g(p)}{h} + o(1).$$

By letting h tend to infinity, we obtain the desired differentiability of the function g at the point p.

Proof of Lemma 7. Let $0<\widehat{\lambda}<\lambda$ . By the splitting theorem for Poisson processes, a Poisson- $\widehat{\lambda}$ process may be obtained from a Poisson- $\lambda$ process through independent and identically distributed thinning of points with acceptance probability $r=\widehat{\lambda}/\lambda$ .

Consider the first busy cycle of length T for the infinite-server queue with input rate $\lambda$ ; call it ‘system 1’. Let $D\equiv D(T)$ be the number of customers that are served in/depart from system 1 within the time interval [0, T]. Then in ‘system 2’ with input rate $\widehat{\lambda}$ , the total number $\widehat{D}$ of departures within [0, T] has a conditional binomial distribution Bin(D, r); i.e., given any value $D=k$ , we have $\widehat{D} \sim Bin (k,r)$ . Then ${\mathbb{E}} \widehat{D} = r {\mathbb{E}} D$ .

Let $D_I$ be the total number of infected departures from system 1 and $\widehat{D}_I$ from system 2, within the time interval [0, T]. Clearly, $ D_I\ge \widehat{D}_I$ a.s.

We have $G(h,\lambda) =\frac{{\mathbb{E}} D_I}{{\mathbb{E}}D}$ and $G(h,\widehat{\lambda}) =\frac{{\mathbb{E}} \widehat{D}_I}{r{\mathbb{E}}D}$ .

For small h, given D, the total number of arrivals of infected customers within [0, T] to system 1 is 0 with probability $1-hD +o(h)$ , 1 with probability $hD+o(h)$ , and more than 1 with probability o(h). Let $\kappa_i=1$ if the ith arrival is infected and $\kappa_i=0$ otherwise (here ${\mathbb{P}} (\kappa_i=1)=h$ ). Let $\zeta_i=1$ if the ith arrival to system 1 is selected for system 2 and $\zeta_i=0$ otherwise (here ${\mathbb{P}}(\zeta_i=1)=r$ ). Then

$${\mathbb{E}} D_I = \sum_k {\mathbb{P}} (D=k)\sum_{i=1}^k {\mathbb{P}} (A_{k,i}){\mathbb{E}}\left(D_I \ | \ D=k, A_{k,i} \right)= h C_{\lambda} +o(h),$$

where $C_{\lambda} = \sum_k {\mathbb{P}} (D=k)\sum_{i=1}^k{\mathbb{E}}\left(D_I \ | \ D=k, A_{k,i} \right)$ and, for $1\le i \le k$ , $A_{k,i }= \{ \kappa_i=1, \kappa_j=0 \ \mbox{for all} \ 1\le j \le k, \ j\neq i\}.$ Next,

\begin{align*} {\mathbb{E}} \widehat{D}_I &= \sum_k {\mathbb{P}} (D=k) {\mathbb{E}}\left( \sum_{i=1}^k\widehat{D}_I {\mathbf I} (A_{k,i}){\mathbf I}(\zeta_i=1) \ | \ D=k\right) + o(h)\\&\le\sum_k {\mathbb{P}} (D=k) {\mathbb{E}}\left( \sum_{i=1}^kD_I {\mathbf I} (A_{k,i}){\mathbf I}(\zeta_i=1) \ | \ D=k\right) + o(h)= r {\mathbb{E}} D_I + o(h),\end{align*}

and we get the needed monotonicity. In fact, there is strict inequality in the second line above, since the event $\{ D=2, \kappa_1=1, \kappa_2=0, \zeta_1=0, \zeta_2=1\}$ has positive probability and, assuming the occurrence in system 1 of the sequence of events that (1) customer 1 infects customer 2, (2) customer 1 leaves, and (3) customer 2 leaves, we get $D_I=2$ in the first system while $\widehat{D}_I=0$ in the second system.

Appendix C. Monotonicity in $\alpha$ and $\beta$

The construction of the coupling for the SIS reactor is as follows. The two processes are piecewise constant and may change their states (jump) only at embedded epochs $T_0=0<T_1<T_2<\ldots$ . Let $X_{i,n}=X_i(T_n+0)$ and $Y_{i,n}=Y_i(T_n+0)$ , for $i=1,2$ and $n=0, 1,2,\ldots$ . Furthermore, let $N_n=X_{1,n}+Y_{2,n}$ . At each time $T_n+0$ , we set an exponential clock of rate $\lambda$ , $N_n$ clocks of rate $\mu$ , $N_n$ clocks of rate $\beta$ , and $N_n(N_n-1)$ clocks of rate $\alpha_2$ . We equip all customers, $i=1,\ldots,N_n$ , with a $\mu$ –clock and a $\beta$ -clock, and all pairs of customers, say (i, j), with an $\alpha$ -clock. All clocks are mutually independent.

Let $T_{n+1}$ be the time when the first of these clocks rings. If this is the $\lambda$ -clock, then a new customer arrives and it (simultaneously) becomes either I, with probability p, or S, with probability $q=1-p$ . If it is the ith $\mu$ -clock, then the ith customer leaves both systems simultaneously. If it is the ith $\beta$ -clock, then the ith customer becomes S in both systems, regardless of its earlier state. And if it is the (i, j)th $\alpha$ -clock, then, in the second system (that with infection parameter $\alpha_2$ ), the jth customer becomes I if the ith customer is I, regardless of the history, and does not change its state if the ith was S. In the first system (with infection parameter $\alpha_1$ ), if the ith customer is I, then the jth customer becomes I with probability $\alpha_1/\alpha_2$ , and keeps its earlier state (no jump) in all other cases.

With this coupling, direct induction arguments provide the claimed monotonicity.

A similar coupling construction holds for the closed system with the following simplifications and modifications: here $N(t)\equiv N$ , for any t; there is no $\lambda$ -clock; the ringing of a $\mu$ -clock means that the corresponding customer moves at random to any of N stations, and we assume that in both systems, customers move to the same stations; when an $\alpha$ -clock rings that corresponds to the pair of customers (i, j), the jth customer becomes I if i and j are located at the same station.

Appendix D. The thermodynamic limits as nonlinear Markov processes

In the transient thermodynamic limit of SIS, at time t, a station has a Poisson arrival process of infected (resp. susceptible) customers with an intensity equal to $\mu {\mathbb{E}} [Y(t)]$ (resp. $\mu {\mathbb{E}}[X(t)]$ ), where Y(t) (resp. X(t)) is the the number of infected (resp. susceptible) customers in the station at time t. The aim of this section is to show that this can be described in terms of a nonlinear Markov dynamics.

This is best explained when looking first at a discrete-time version of the problem. Time is slotted with increments of duration h. At time $k=0$ , choose an initial condition in the infinite-dimensional thermodynamic limit. The latter consists of a product-form distribution of customers over stations, with a fixed joint distribution ${\mathcal P}_0$ of infected and susceptible customers in any given station, such that the sum of the two coordinates is Poisson- $\eta$ . The number of infected (resp. susceptible) arrivals in time slot 1 is Poisson with parameter $h\mu {\mathbb{E}} [Y_0]$ (resp. $h\mu {\mathbb{E}} [X_0]$ ), where the expectations are with respect to ${\mathcal P}_0$ . In addition, each of the $Y_0$ infected customers tosses a three-face die with respective weights

$$1-\exp\!({-}h(\beta+\mu)),\quad \frac{\beta}{\beta+\mu}\exp\!({-}h(\beta+\mu)),\quad \frac{\mu}{\beta+\mu}\exp\!({-}h(\beta+\mu)).$$

If the outcome is 1, the customer stays and keeps its SIS state. If the outcome is 2, it stays and changes its SIS state. If the outcome is 3, it leaves. Similarly, conditionally on $Y_0$ , each of the $X_0$ infected customers tosses a three-face die with respective weights

$$1-\exp\!({-}h(\alpha Y_0+\mu)),\quad \frac {\alpha Y_0} {\alpha Y_0+\mu}\exp\!({-}h(\alpha Y_0+\mu)),\quad \frac{\mu}{\alpha Y_0+\mu}\exp\!({-}h(\alpha Y_0+\mu)).$$

If the outcome is 1, the customer stays and keeps its SIS state. If the outcome is 2, it stays and changes its SIS state. If the outcome is 3, it leaves. All these define a state $(X_1,Y_1)$ . Let ${\mathcal P}_1$ be the distribution of $(X_1,Y_1)$ .

The construction for all k is then obtained by induction. Assume the triple $(X_k,Y_k, {\mathcal P}_k)$ is well defined. By applying the same dynamics to this triple, one gets at the same time a state $(X_{k+1},Y_{k+1})$ and a distribution ${\mathcal P}_{k+1}$ .

The key observations are then the following: (a) once the sequence of distributions $\{{\mathcal P}_k\}$ is determined, one can then see the evolution of $\{(X_k,Y_k)\}$ as that of a standard discrete-time discrete-space nonlinear Markov chain. (b) A stationary distribution is a distribution ${\mathcal P}$ such that if ${\mathcal P}_0={\mathcal P}$ , then ${\mathcal P}_1={\mathcal P}$ . (c) By tightness arguments, this can be extended to the continuous-time case by letting $h\to 0$ .

Appendix E. Proof of Theorem 2

Below, $p^*$ is the maximum root of the fixed-point equation $p=g(p)$ for the plain SIS reactor. This means that if $g^{\prime}(0)\le 1$ , then $p^*=0$ , whereas if $g^{\prime}(0)>1$ , then $p^*$ is positive and there are two solutions to the fixed-point equation.

Consider an M/M/ $\infty$ system in the stationary regime on the time horizon $[0,\infty)$ , with input rate $\lambda$ and service rate $\mu$ . The number of customers in the system at any time t, N(t), has a Poisson distribution with parameter $\eta = \lambda/\mu$ .

Monotonicity. Consider the SIS reactor with a Poisson input point process with a time-varying intensity like that discussed in Remark 8. Consider two variants of customer ‘coloring’ at time 0 and of input rates. In the first variant, there are initially Y(0) infected customers and the input point process of infected customers has an intensity equal to $\lambda p(.)$ , while in the second variant there are initially $\widehat Y(0)$ infected customers and the intensity of infected customers is $\lambda \widehat p(.)$ . Here p(t) and $\widehat p(t)$ are two given deterministic functions.

The next result follows from the monotonicity property established in Appendix B—more precisely, see Remark 8.

In what follows, ‘by simple monotonicity arguments’ means ‘by applying Lemma 11 ^′.

The upper process. We introduce an SIS model with a varying fraction of infected customers in the input process and instantaneous extra infections along the lines described in Remark 8.

Recall that we consider a total population dynamics which is that of an M/M/ $\infty$ system in the stationary regime on the time horizon $[0,\infty)$ , with input rate $\lambda$ and service rate $\mu$ .

Consider now the following infection/recovery mechanism. We introduce recursively deterministic times $T^{(0)}=0<T^{(1)}<T^{(2)}<\ldots$ and probabilities $p^{(1)}=1>p^{(2)}> \ldots$ such that

1. (i) $p^{(n)}\downarrow p^*$ as $n\to\infty$ ;

2. (ii) for any $n=1,2,\ldots$ , each customer that arrives in the system within the time interval $(T^{(n-1)},T^{(n)})$ is infected with probability $p^{(n)}$ and susceptible otherwise, independently of everything else;

3. (iii) for any $n=1,2,\ldots$ , at time $T^{(n-1)}$ , all customers present in the system become infected instantaneously.

In more detail, with $n=1,2,\ldots$ , we introduce recursively processes $(\widehat{X}^{(n)}(t),\widehat{Y}^{(n)}(t))$ , for $t\ge T^{(n-1)}$ , and then let

(64) \begin{align}\big(\widehat{X}(t),\widehat{Y}(t)\big) =\big(\widehat{X}^{(n)}(t),\widehat{Y}^{(n)}(t)\big) \quad \text{for} \ t\in \big[T^{(n-1)}, T^{(n)}\big).\end{align}

First we assume that, at time $T^{(0)}=0$ , all customers that are present in the system are infected, and that, starting from time 0, all arriving customers are infected too (i.e., each is infected with probability $p^{(1)}=1$ ). We have a time-homogeneous and irreducible Markov jump process on a countable state space, which is clearly aperiodic and positive recurrent. Therefore, there exists a unique stationary distribution which is the limit in law obtained for any initial state, and the speed of convergence to this stationary distribution is exponential. In the stationary regime, the output rate of infected customers is $p_o^{(1)}$ , which is strictly less than $p^{(1)}=1$ . In particular, if one denotes by $\big(\widehat{X}^{(1)}(t),\widehat{Y}^{(1)}(t)\big)$ the state process in this dynamics, one has ${\mathbb{E}} \widehat{Y}^{(1)}(t) \to p^{(1)}_o \eta$ exponentially fast, as $t\to\infty$ .

Let $\delta \in (0, 1)$ be arbitrarily small. We choose time $T^{(1)}$ as

(65) \begin{align}T^{(1)} = \min \{t\ge T^{(0)}\,:\, {\mathbb{E}} Y(u)/\eta \le p_o^{(1)}+ \delta \big(p^{(1)}-p^{(1)}_o\big)/2, \ \text{for all} \ u\ge t\}.\end{align}

Then let $p^{(2)}=p^{(1)}_o+\delta \big(p^{(1)}-p^{(1)}_o\big)/2$ . By (64), this completes the description of the dynamics of the process $\big(\widehat{X}(t), \widehat{Y}(t)\big)$ within the first time slot $[T^{(0)},T^{(1)})$ . Note that, by monotonicity and convexity of the function $p_o = g(p)$ , we get $p^{(1)}_0>p^*$ .

Assume we have introduced the processes $\big(\widehat{X}^{(i)}(t),\widehat{Y}^{(i)}(t)\big)$ on the time intervals $[T^{(i-1)},\infty)$ , for $i=1,\ldots,n-1$ , and therefore have defined the process $\big(\widehat{X}(t), \widehat{Y}(t)\big)$ on the interval $[0,T^{(n-1)})$ via (64). At the beginning of the nth time slot (at time $T^{(n-1)}$ ), we change all customers present in the system to infected and assume that, starting from this time, each arriving customer is either infected with probability $p^{(n)}>p^*$ , or susceptible otherwise. Then, starting from time $T^{(n-1)}$ , our system again behaves as an irreducible time-homogeneous Markov jump process whose distribution converges exponentially fast to its unique stationary distribution with output fraction of infected customers $p^{(n)}_o $ , which is strictly bigger than $p^*$ (thanks again to monotonicity and convexity of the function g). We denote this process by $(\widehat{X}^{(n)}(t),\widehat{Y}^{(n)}(t))$ , $t\ge T^{(n-1)}$ . Then we let

(66) \begin{align}T^{(n)} = \min \{t\ge T^{(n-1)}\,:\, {\mathbb{E}} \widehat{Y}^{(n)}(u)/\eta \le p_o^{(n)}+\delta \big(p^{(n)}-p^{(n)}_o\big)\cdot 2^{-n}, \forall\ u\ge t\}\end{align}

and $p^{(n+1)}=p^{(n)}_o+\delta \big(p^{(n)}-p^{(n)}_o\big)\cdot 2^{-n}$ . This completes the description of the process $(\widehat{X}(t),\widehat{Y}(t))$ within the nth time slot $[T^{(n-1)},T^{(n)})$ . Again, thanks to monotonicity and convexity of the function $p_o = g(p)$ , we get $p^{(n)}_o> p^{(n+1)}_o>p^*$ .

Thus, we have defined the process $(\widehat{X}(t),\widehat{Y}(t))$ for all $t\ge 0$ . It is left to explain the convergence (i). We know that there is no solution to the equation $p_0=g(p)$ bigger than $p^*$ , and the monotone sequences $p^{(n)}$ and $p^{(n)}_o$ converge to the same limit which is a solution to this fixed-point equation. The fact that this is the maximal solution proves the result.

Comparison of the upper process and the thermodynamic limit process.

As shown in Appendix D, under the propagation of chaos ansatz, the dynamics of any station in the SIS TL can be seen as a nonlinear Markov jump process (X(t), Y(t)) with a certain fraction $p^*(t)$ of infected customers at time t (the fraction such that at any time t, $\lambda p^*(t)=\mu {\mathbb{E}} Y(t)$ ).

By construction, ${X}(t)+{Y}(t)=N(t)$ for all $t\ge 0$ , where N(t) is the Poisson process describing the dynamics of the stationary M/M/ $\infty$ system and also $N(t)= \widehat{X}(t)+\widehat{Y}(t)$ a.s. (where the latter are the state variables in the upper system described above).

The next corollary follows from simple monotonicity arguments. Before stating it, we recall the following result on the fixed point of the SIS reactor. Assume $g^{\prime}(0)>0$ , which implies that $p^*>0$ . Any SIS reactor is a time-homogeneous Markov process $(\widetilde X(t),\widetilde Y(t))$ with a countable state space, which is irreducible, positive recurrent, and ergodic. In particular, for $p=p^*$ , for all initial conditions $(\widetilde X(0),\widetilde Y(0))$ , this Markov process converges to a limiting stationary Markov process with $\mu {\mathbb{E}} \widetilde Y= p^*\lambda$ . Denote this last stationary process by $\{(\widetilde X^*(t),\widetilde Y^*(t))\}_t$ . This last process can be used to build a specific example $(X^*(t), Y^*(t),{\mathcal P}^*_t)$ , $t\ge 0$ , of the nonlinear Markov chain of Appendix D, where the initial distribution ${\mathcal P}^*_0$ is that of $(\widetilde X^*(0),\widetilde Y^*(0))$ . This distribution is invariant for this nonlinear dynamics. By this, we mean that it makes the Markov chain homogeneous and stationary, with an input rate of infected customers which is constant, equal to $\lambda p^*$ , and equal to the mean output rate of infected customers.

Consider the nonlinear Markov process Y(t) defined in Appendix D with any initial distribution such that ${Y}(0) \ge Y^*(0)$ a.s., with $Y^*(.)$ defined above. Then $\widehat{Y}(t)\ge {Y}(t)\ge Y^*(t)$ a.s. and ${\mathbb{P}} (\widehat{Y}(t)= {Y}(t) = Y^*(t)) \to 1$ .

The only thing to comment upon is the last statement. It follows from the facts that all Y-variables are integer-valued and that the monotone convergence of integer-valued random variables is necessarily also a coupling (or total variation) convergence.