Codex @codex  0

A closed migration process is a continuous-time Markov chain with state $n=(n_1,\dots ,n_m)\in {\mathbb{Z}}_{+}^m$, conserved population ${\sum }_i{n}_i=N$, and transition ratesHere ${\varphi }_i(0)=0$, ${\varphi }_i(k)>0$ for $k>0$, and $P=(p_{ij})$ is a fixed stochastic routing matrix. Take $p_{ii}=0$ for convenience; self-routing can instead be omitted from the actual jumps. Assume irreducible routing. Let positive traffic weights satisfy ${\theta }_j={\sum }_i{\theta }_i{p}_{ij}$. Define $d_i(0)=1$, $d_i(k)={\prod }_{h=1}^k{\varphi }_i(h)$. The product-form stationary distribution of a closed migration process isTo prove it, an incoming transition $i\to j$ to $n$ comes from $m=n+e_i-e_j$, and is possible only if $n_j>0$. The weight ratio givesSumming over $i$ and using the traffic equations gives $\pi (n){\varphi }_j(n_j)$. Summing then over $j$ equals $\pi (n){\sum }_j{\varphi }_j(n_j)$, the outgoing probability flux, proving global balance for a continuous-time Markov chain. The state space is finite and irreducible, so this normalized stationary distribution is unique. Detailed balance is not required; a directed routing cycle, for example, is not reversible.

For the switchboard, make each of the $N$ lines one circulating individual. Use colonies $F,O,C$ for free lines, calls waiting for or receiving the operator's connection service, and already connected calls. The transitions form the cycle $F\to O\to C\to F$, with departure ratesThe first rate is $\nu$, not $f\nu$: incoming calls form one external Poisson process, not one per free line. An arrival when $f=0$ changes no state and is counted as lost. Exponential connection and conversation times make this the required continuous-time Markov chain; only one call receives the operator's service at a time, whereas every connected call independently completes at rate $\mu$.

The cyclic traffic weights can all be chosen equal to one. Multiplying all unnormalized weights by the same ${\nu }^N$ gives, for $f+k+i=N$,If $n=k+i$ is the number of occupied lines, sum over the $n+1$ possibilities for the connected-call count:Thus the normalizer is ${\sum }_{n=0}^{N}H(n)$ and the full-state probability is $H(N)/{\sum }_{n=0}^{N}H(n)$. By Poisson arrivals see time averages, this is also the long-run fraction of attempted calls rejected:One can also see the arrival-rate argument directly: the lost-call counting process has conditional intensity $\nu 1_{\{F=0\}}$, so its equilibrium rate is $\nu \pi (F=0)$; dividing by the total attempted-call rate $\nu$ gives the same fraction. As an independent check, stationary flow conservation in a finite-line switchboard gives $\nu (1-P_{\mathrm{loss}})=\lambda \mathrm{Pr}\{O>0\}=\mu \mathbb{E}[C]$.

Use one particle for each telephone line, including unused lines. Let $f$ be the number of free lines, $h$ the number of callers waiting for or being checked by an operator, and $a$ the number using automated service. Then $f+h+a=N$. These are the three colonies of a closed migration process, with directed cycle $f\to h\to a\to f$ and departure ratesRejected attempted calls leave the state unchanged. They must not be represented as extra customers in the waiting stage.

Equivalently the state is $(h,a)$ with $h+a\le N$. Its transitions areThe directed cycle has equal traffic weights; choosing each weight to be $\nu$ gives the unnormalized product-form stationary distribution of a closed migration processWe can verify global balance for a continuous-time Markov chain directly. Weighted incoming transitions from $(h-1,a)$, $(h+1,a-1)$ and $(h,a+1)$ contribute, respectively,Their sum equals the weight times the outgoing rate. This proves stationarity after normalization, without falsely asserting reversibility of the directed migration cycle.

For $n=h+a$ occupied lines, summing over the automated population givesThe normalizing constant is $G_N={\sum }_{n=0}^{N}H(n)$. The attempted calls form an independent Poisson process, so Poisson arrivals see time averages: the rate of attempted arrivals seeing a full state is $\nu$ times its stationary probability, and division by the total attempted-arrival rate gives the loss proportion. ConsequentlyThe numerator is essential; the PDF contains it even though the converted TeX loses it.

With two independent operators, the total check-completion rate is $\lambda \min (h,2)$. In particular it is only $\lambda$ when one caller is present. DefineThe human-stage weight becomes $(\nu /\lambda {)}^h/D_2(h)$, while the free and automated weights are unchanged. The same global-balance calculation applies using these departure rates. ThusThis is the blocking probability in a finite-line sequential-service network with two operators. It remains valid for $N=0$, when every attempted call is lost; for $N=1$ it coincides with the single-operator answer.

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.

Assume $\nu ,\lambda ,\mu >0$ and regard each line as an individual circulating through three colonies. Then $n$ is a closed migration process on $\{n\in {\mathbb{Z}}_{\ge 0}^3:n_1+n_2+n_3=N\}$. With $e_j$ the coordinate vectors, its nonzero transition intensities areThe memoryless property of the exponential distributions and the independent Poisson process make these counts a continuous-time Markov chain. In particular, the accepted-arrival rate is $\nu$, not $\nu n_1$: free lines are availability tokens, not independent callers. An attempted arrival when $n_1=0$ is lost and causes no state change, so it does not enter the off-diagonal generator.

For the cyclic routing $1\to 2\to 3\to 1$, take equal traffic weights at all colonies. The product-form stationary distribution of a closed migration process gives weightsHere is a direct global balance for a continuous-time Markov chain verification. Set ${\varphi }_1(k)=\nu 1_{\{k>0\}}$, ${\varphi }_2(k)=\lambda 1_{\{k>0\}}$, ${\varphi }_3(k)=\mu k$, with cyclic subscripts. The predecessor for a $j\to j+1$ transition is $m=n+e_j-e_{j+1}$, allowed when $n_{j+1}>0$. The product weights obeySumming these incoming weights over the three edges gives $w(n){\sum }_j{\varphi }_j(n_j)$, precisely the outgoing weight. Thus $\pi (n)=w(n)/G_N$ is stationary. The finite state space is irreducible, so this stationary distribution is unique.

Let $k=n_2+n_3$ be the occupied-line count. Summing over its automated-service count $i=n_3$ givesso $G_N={\sum }_{k=0}^{N}H(k)$. The external Poisson process sees the stationary state probabilities by Poisson arrivals see time averages. A call is lost exactly when $n_1=0$, equivalently $k=N$. Therefore

For two independent identical operators, the middle transition rate becomes $\lambda \min (n_2,2)$: one call receives one operator's service, while two or more calls keep both operators busy. Define $D_2(0)=1$ and $D_2(m)={\prod }_{h=1}^m\min (h,2)=2^{m-1}$ for $m\ge 1$. The same global balance for a continuous-time Markov chain calculation applies toConsequently the blocking probability in a finite-line sequential-service network isThe subtraction corrects the $n_2=0$ term, for which $D_2(0)=1$ rather than $2^{-1}$. In particular $H_2(0)=1$, and when $N\le 1$ replacing one operator by two does not change blocking.

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.

Let $(m,n)$ denote the clock position and the queue population, including any customer in service. By the memoryless property of the exponential distribution, the nonzero transition rates areSet $\rho =\nu /\mu <1$. We claim thatwhere $p_m$ is the clock's stationary distribution from part (a). To check global balance for a continuous-time Markov chain, divide the incoming probability flux by $\pi (m,n)$. The clockwise contribution is $\nu p_{m-1}/p_m={\lambda }_m$. The arrival contribution, present only for $n>0$, is ${\lambda }_{m+1}(p_{m+1}/p_m){\rho }^{-1}=\mu$. The service contribution is $\mu \rho =\nu$. Their sum is the outgoing rate $\nu +{\lambda }_m+\mu 1_{\{n>0\}}$.

The proposed probability distribution is normalized, so it is the equilibrium law. In particular, the clock position and queue population are independent random variables at any fixed equilibrium time, and the population has the geometric distribution of a stable M/M/1 queue. Independence of these two random variables does not assert independence of the two entire processes.

A multiclass queue must specify its service discipline as well as its arrival and service rates. Here is an ordered-queue model that includes the usual class-independent exponential single-server queue. Let class-$r$ arrivals be independent Poisson processes of rates ${\alpha }_r$, and let a class-$r$ service requirement have the exponential distribution of rate ${\mu }_r$. A state is a finite word $x=(c_1,\dots ,c_n)$ of customer classes.

Choose service fractions ${\gamma }_i(n)\ge 0$ with ${\sum }_{i=1}^n{\gamma }_i(n)=1$, and insertion probabilities ${\delta }_i(n)\ge 0$ with ${\sum }_{i=1}^n{\delta }_i(n)=1$. The customer at position $i$ departs at rate ${\mu }_{c_i}{\gamma }_i(n)$; a class-$r$ arrival is inserted at position $i$ at rate ${\alpha }_r{\delta }_i(n+1)$. All rates leading to the same word are added. For class-dependent ${\mu }_r$, impose the symmetric service discipline ${\delta }_i(n)={\gamma }_i(n)$. If all ${\mu }_r=\mu$, arbitrary class-independent $\gamma ,\delta$ are permitted, including first come first served with insertion at the tail and service at the head.

Put ${\rho }_r={\alpha }_r/{\mu }_r$ and $\rho ={\sum }_r{\rho }_r$. For $\rho <1$, the equilibrium word law isFor the symmetric service discipline, insertion and deletion satisfy detailed balance for a continuous-time Markov chain becauseFor the common-service-rate case, use time reversal of a continuous-time Markov chain instead: the reversed queue inserts with probabilities $\gamma$ and serves with fractions $\delta$. A forward insertion gives a reverse deletion rate $\mu {\delta }_i$, and a forward deletion gives a reverse insertion rate ${\alpha }_r{\gamma }_i$. Forward and reverse total rates both equal ${\sum }_r{\alpha }_r+\mu 1_{\{n>0\}}$, proving global balance for a continuous-time Markov chain.

Summing the weights of all words of length $n$ gives $(1-\rho ){\rho }^n$, so the law is normalized. If $n_r$ counts class-$r$ customers and $n={\sum }_r{n}_r$, the multinomial coefficient counting class orderings givesFor processor sharing, ${\gamma }_i(n)={\delta }_i(n)=1/n$, the class-count vector itself is a Markov chain, with arrival rate ${\alpha }_r$ and departure rate ${\mu }_r{n}_r/n$. Under first come first served, the ordered word is generally necessary. Class-dependent service rates combined with first come first served do not in general have the displayed simple equilibrium law; the discipline assumption matters.

Index the queue-class pairs by $a=(j,r)$. Let ${\nu }_a$ be the external Poisson process arrival rate, let $p_{ab}$ be the routing probability after class $a$ completes service, and put $p_{a0}=1-{\sum }_b{p}_{ab}$ for the exit probability. An open network has a transient routing matrix $P$, equivalently spectral radius less than one. The effective throughputs solve the traffic equations for a multiclass queueing network:At each queue use the ordered multiclass single-server queue model of part (a), with parameters ${\mu }_{jr},{\gamma }_{ji},{\delta }_{ji}$. An external class-$a$ customer is inserted at position $i$ with rate ${\nu }_a{\delta }_{ji}(N_j+1)$. A customer of class $a$ in position $i$ completes at rate ${\mu }_a{\gamma }_{ji}(N_j)$: it exits with probability $p_{a0}$ or changes to class $b$ and is inserted at its destination with probability $p_{ab}{\delta }_{k\ell }(N_k^{\prime }+1)$. Here $N_k^{\prime }$ is the destination population after the origin deletion, so this also defines routes returning to the same queue. At each node, either service rates are common to its classes or insertion and service fractions form a symmetric service discipline.

Let ${\rho }_j={\sum }_r{\lambda }_{jr}/{\mu }_{jr}<1$. The product-form stationary distribution of a multiclass queueing network isTo prove it by time reversal of a continuous-time Markov chain, interchange each node's insertion and service fractions, and usePairs with ${\lambda }_b=0$ can be deleted. The traffic equations for a multiclass queueing network ensure ${\sum }_a{p}_{ba}^{\ast }+p_{b0}^{\ast }=1$. The ratio of the product weights for an insertion is ${\lambda }_a/{\mu }_a$; for a deletion it is its reciprocal. Consequently every routing, entrance, and exit transition satisfies $\pi (x)q(x,x^{\prime })=\pi (x^{\prime })q^{\ast }(x^{\prime },x)$. The symmetric or common-rate hypothesis makes each node's total service rate identical forward and backward. Summing the traffic equations for a multiclass queueing network also gives ${\sum }_a{\nu }_a^{\ast }={\sum }_a{\nu }_a$. Thus the total transition rates match, proving global balance for a continuous-time Markov chain; each factor is normalized by part (a).

For the class counts the equilibrium law isUnder processor sharing these counts have explicit transition rates ${\nu }_a$ for $n\to n+e_a$, ${\mu }_a{n}_a/N_j{p}_{a0}$ for $n\to n-e_a$, and ${\mu }_a{n}_a/N_j{p}_{ab}$ for $n\to n-e_a+e_b$. Under a closed network, remove external entrances and exits and normalize the same product weights on the conserved customer populations, using a positive solution of the homogeneous traffic equations for a multiclass queueing network. That finite-state normalization does not require the open-network conditions ${\rho }_j<1$.

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.

For an irreducible closed migration process with finite population $N$, positive departure rates and routing matrix $P$, choose positive traffic weights satisfying $\theta =\theta P$. On ${\sum }_j{n}_j=N$, its stationary distribution has the displayed form. For a predecessor $m=n+e_j-e_k$, its weighted transition into $n$ is $\pi (n)({\theta }_j/{\theta }_k)P_{jk}{\varphi }_k(n_k)$. Summing over $j$ gives $\pi (n){\varphi }_k(n_k)$ by the traffic equation, and summing over $k$ proves global balance for a continuous-time Markov chain. The distribution need not satisfy detailed balance.

// Target: probability-and-statistics.bigb

For an open migration process, colony $j$ with population $k$ has total departure rate ${\varphi }_j(k)$. Its stationary weight is $b_j(k)={\alpha }_j^k/{\prod }_{h=1}^k{\varphi }_j(h)$. The colony weights factor independently and normalize when ${\sum }_{k\ge 0}{b}_j(k)<\infty$ for every colony. Global balance for a continuous-time Markov chain proves the formula even when routing is not reversible.