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An open migration process is a continuous-time Markov chain of colony populations with external Poisson processes, state-dependent departure rates, and fixed probabilistic routing that eventually exits the system. Its traffic equations of an open migration process determine the total arrival rates, and its equilibrium has a product-form stationary distribution of an open migration process when the normalizing sums are finite.

Use finitely many colonies and let $n_i$ be the population of colony $i$. An open migration process has independent external Poisson processes of rates ${\nu }_i$, local total departure rates ${\varphi }_i(n_i)$, and fixed routing probabilities $p_{ij}$ to other colonies and $p_{i0}$ out of the system. Here ${\varphi }_i(0)=0$, $0<{\varphi }_i(k)<\infty$ for $k\ge 1$, $p_{ii}=0$ and $p_{i0}+{\sum }_j{p}_{ij}=1$. The continuous-time Markov chain has intensitiesRequire the routing matrix $P$ to be transient: every individual eventually exits. For finitely many colonies this is equivalent to every colony having a directed path to an exit. Restrict to colonies reachable from external arrivals, so their traffic rates are positive and the relevant chain is irreducible. The total population cannot exceed its initial value plus the number of external arrivals, so on any finite time interval it is bounded by a finite random number. There are only finitely many states below that bound and finite rates in each, proving nonexplosion even when the departure functions grow rapidly.

The traffic equations of an open migration process, in row-vector form, areTransience makes the inverse ${\sum }_{k\ge 0}{P}^k$ finite. DefineProvided every $Z_i$ is finite, the candidate product-form stationary distribution of an open migration process isThe normalizability condition is exactly $Z_i<\infty$ for every active colony. For example, constant total service ${\varphi }_i(k)={\mu }_i$ for $k>0$ requires ${\alpha }_i<{\mu }_i$ and gives a geometric distribution; independent individual departures ${\varphi }_i(k)=k{\mu }_i$ give a Poisson distribution of mean ${\alpha }_i/{\mu }_i$. In the irreducible case the normalizable invariant law is the unique equilibrium. If the weights cannot be normalized, no stationary probability law exists, by uniqueness up to scale of invariant measures for an irreducible recurrent chain.

We establish both stationarity and the time reversal of an open migration process assertion by adjacent-weight calculations. DefineThe identity $b_i(k){\varphi }_i(k)={\alpha }_i{b}_i(k-1)$ givesThus the proposed time reversal of an open migration process retains the local service functions and hasThe traffic equation for colony $i$ implies $p_{i0}^{\ast }+{\sum }_j{p}_{ij}^{\ast }=1$. Summing all traffic equations also gives ${\sum }_i{\nu }_i^{\ast }={\sum }_i{\nu }_i$. Therefore the total outgoing rate from every state is the same for $q^{\ast }$ and $q$:Consequently ${\sum }_{m\ne n}\pi (m)q(m,n)=\pi (n){\sum }_{m\ne n}q(n,m)$, proving global balance for a continuous-time Markov chain. The same weight recurrence gives ${\mathbb{E}}_{\pi }{\varphi }_i(n_i)={\alpha }_i$, so the stationary mean total jump rate is finite. The candidate is a stationary distribution, and the displayed intensities are indeed its stationary time reversal of a continuous-time Markov chain. Reverse routing is also open: reverse an original arrival-to-colony path to obtain a path to a colony with $p_{i0}^{\ast }>0$.

For the last request, let $S$ be all colonies reachable by routing from $k$, including $k$ itself. The no-return-path hypothesis gives $j\notin S$. By construction no routing edge leads from $S$ to its complement $C$. The populations in $C$ therefore evolve autonomously as an open migration process, with moves into $S$ treated as exits. Its stationary marginal is the corresponding product of the factors above. Retain the original destination as a mark on each such exit.

In this autonomous process, a marked exit $j\to k$ has intensity ${\varphi }_j(n_j)p_{jk}$. The same reversed-weight calculation makes its reverse a marked external immigration channel of constant rate ${\alpha }_j{p}_{jk}$. External immigration channels can be constructed as independent Poisson processes. Stationary reversal of a two-sided Poisson process preserves its law, so reversing this immigration channel givesThis Poisson migration stream without a return path argument is needed because a general internal migration stream in a network with feedback is not automatically Poisson. A zero routing probability gives the empty Poisson stream of rate zero.

An open migration process has finitely many colonies and state $n\in {\mathbb{N}}^J$. External arrivals to colony $j$ form independent Poisson processes with rates ${\nu }_j$. When $n_j=k>0$, individuals leave that colony at total rate $g_j(k)>0$, with $g_j(0)=0$. A departing individual moves to $k$ with probability $p_{jk}$ or exits with probability $p_{j0}=1-{\sum }_k{p}_{jk}$. The routing choices and all arrival and service randomness are independent. The routing matrix $P=(p_{jk})$ is substochastic and transient: every individual eventually exits, equivalently its spectral radius is less than one. Self-routing is allowed, but creates no state change.

The traffic equations of an open migration process, using row vectors, areThe inverse matrix exists because ${\sum }_{m\ge 0}{P}^m$ converges; its $(k,j)$ entry counts expected visits to $j$ by an individual entering at $k$. Thus ${\alpha }_j$ includes both external arrivals and internal migrations. Restrict first to the accessible colonies with ${\alpha }_j>0$.

SetIf each $Z_j<\infty$, the candidate product-form stationary distribution of an open migration process isHere is a global balance for a continuous-time Markov chain proof, which does not incorrectly assume that the routing is reversible. The weight ratio is $\pi (n+e_j)/\pi (n)={\alpha }_j/g_j(n_j+1)$. After dividing the incoming balance equation at $n$ by $\pi (n)$, external arrivals contribute ${\sum }_{j:n_j>0}{\nu }_j{g}_j(n_j)/{\alpha }_j$, exits contribute ${\sum }_j{\alpha }_j{p}_{j0}$, and migrations from $k$ to $j\ne k$ contribute ${\sum }_{j:n_j>0}{\sum }_{k\ne j}{g}_j(n_j){\alpha }_k{p}_{kj}/{\alpha }_j$. Combining the first and third contributions using the traffic equations of an open migration process yields ${\sum }_j{g}_j(n_j)(1-p_{jj})$. Summing those same traffic equations of an open migration process gives ${\sum }_j{\alpha }_j{p}_{j0}={\sum }_j{\nu }_j$. The incoming rate is therefore exactly the outgoing rateThis proves stationarity. With finite arrival rates, finite service rates in each state and transient routing, the process is nonexplosive: only finitely many individuals enter in a finite interval, and each makes finitely many routing transitions before exit, almost surely. On the accessible irreducible Markov chain, a finite normalizer gives the unique stationary distribution and positive recurrence. Colonies with zero effective arrivals are empty in this equilibrium. For a nonempty accessible colony, a divergent $Z_j$ prevents normalization of the equilibrium weights.

For the single-server specialization, $g_j(k)={\varphi }_j$ for $k>0$. Each colony has the stationary distribution of an M-M-1 queue:requiring ${\varphi }_j>{\alpha }_j$ on every colony with positive traffic. The service rates do not change the traffic equations of an open migration process.

For positive ${\alpha }_j$, write $B=F-{\sum }_j{\alpha }_j$ and $s_j={\varphi }_j-{\alpha }_j$. A stable allocation requires $B>0$ and $s_j>0$, with ${\sum }_j{s}_j=B$. The Cauchy-Schwarz inequality givesEquality holds precisely when $s_j$ is proportional to $\sqrt{{\alpha }_j}$. Hence the unique optimal service-capacity allocation in an open migration process and its mean population areThere is no stable allocation when $F\le {\sum }_j{\alpha }_j$ and some traffic is positive. If some ${\alpha }_j=0$, those empty colonies contribute no population; allocate the spare capacity to the positive-traffic colonies by the same rule. Allowing zero capacity at an unused colony attains this optimum. If strictly positive ${\varphi }_j$ are required even there, the displayed value is only an infimum, approached as their capacities tend to zero. If every ${\alpha }_j=0$, the equilibrium is empty and any admissible allocation minimizes its mean population at zero.

An open migration process is a continuous-time Markov chain recording the populations $n=(n_1,\dots ,n_J)$ of finitely many colonies. Independent external Poisson processes bring individuals to colony $j$ at rate ${\nu }_j$. When that colony contains $k>0$ individuals, a departure opportunity occurs at total rate ${\varphi }_j(k)>0$, with ${\varphi }_j(0)=0$. The departing individual moves to colony $l$ with probability $p_{jl}$, or leaves the system with probability $p_{j0}=1-{\sum }_l{p}_{jl}$. Routing choices and the clocks are independent. A return to the same colony does not change the state.

The word open means that the routing matrix $P=(p_{jl})$ is transient: every individual eventually leaves, or equivalently the spectral radius of $P$ is less than one. The traffic equations of an open migration process, with row-vector convention, give the total arrival rates:These rates include migrations as well as external arrivals. On the active colonies assume ${\alpha }_j>0$ and the usual irreducible Markov chain condition. PutThe product form and its normalization condition areTo establish the product-form stationary distribution of an open migration process, first use its unnormalized weight $b(n)={\prod }_j{b}_j(n_j)$. Its ratios satisfy $b_j(k){\varphi }_j(k)={\alpha }_j{b}_j(k-1)$. The incoming external arrivals into $n$, divided by $b(n)$, contribute ${\sum }_j{\nu }_j{\varphi }_j(n_j)/{\alpha }_j$. Incoming exits contribute ${\sum }_i{\alpha }_i{p}_{i0}$. Incoming migrations from $i$ to $j\ne i$ contribute ${\alpha }_i{p}_{ij}{\varphi }_j(n_j)/{\alpha }_j$. Terms involving an empty destination are zero. Thus the full incoming rate divided by $b(n)$ isHere the traffic equations of an open migration process give the second equality, including ${\sum }_i{\alpha }_i{p}_{i0}={\sum }_j{\nu }_j$. The right side is precisely the total outgoing rate. This proves global balance for a continuous-time Markov chain; pairwise detailed balance is not required for general routing.

For finite rates at each finite population, the continuous-time Markov chain is nonexplosive Markov chain: its total population is bounded over a finite time interval by the initial population plus the finitely many external arrivals, so it visits a finite set on which all jump rates are bounded. Consequently finite $Z_j$ normalize the invariant weight to a stationary distribution. Under the stated irreducible Markov chain assumptions, this is the unique stationary distribution, and it is a positive recurrent Markov chain. Conversely, the unique invariant measure of an irreducible recurrent chain is proportional to $b$, so a positive recurrent Markov chain requires its normalization to be finite. A sufficient condition is ${\mathrm{liminf}}_{k\to \infty }{\varphi }_j(k)>{\alpha }_j$; the series criterion itself is the exact condition. Colonies with ${\alpha }_j=0$ have zero stationary population and can be removed from the active class.

For the single-server M/M/1 queue specialization, ${\varphi }_j(k)={\varphi }_j$ for $k>0$, each stationary colony has the geometric distribution of an M/M/1 queue:The traffic equations of an open migration process do not involve the service rates, so the ${\alpha }_j$ remain fixed during this service-capacity allocation in an open migration process. Write $A={\sum }_j{\alpha }_j$, $B=F-A$ and $b_j={\varphi }_j-{\alpha }_j$. A stationary allocation requires $B>0$, with $b_j>0$ and ${\sum }_j{b}_j=B$. The Cauchy-Schwarz inequality givesEquality holds exactly when $b_j$ is proportional to $\sqrt{{\alpha }_j}$. Hence, for positive arrival rates, the unique optimal service allocation and the minimum expected value of the total population areIf $F\le A$ and some arrival rate is positive, no stable service allocation exists. Zero-arrival colonies need no service in their stationary empty class: allowing zero service there gives the same formula restricted to active colonies. If strictly positive service is demanded even at inactive colonies, the displayed minimum is an infimum approached as that unused allocation tends to zero. If all arrivals vanish, the stationary mean population is zero for every allocation that drains the initial population.