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For a nonexplosive continuous-time Markov chain on the nonnegative integers, suppose zero is absorbing and each positive state has positive rate to zero. On every finite positive set the conditional chance of absorption at the next jump has a positive minimum. By the Strong Markov property, the chance of surviving $r$ departures from that set is bounded by $(1-\delta {)}^r$. Without absorption only finitely many such departures occur, and finite positive holding rates prohibit staying forever at a positive state. Thus eventually the path stays outside each finite set: it escapes to infinity. This conclusion needs no upward skip-free assumption.

Only routes with $n_r>0$ need a per-flow rate: absent flows contribute no utility and have no completion events. Interpret the proportional-change sum and the logarithmic objective on these active routes. With positive capacities and finitely many nonempty routes, a strictly positive feasible active-rate vector exists. Define its weighted logarithmic utility by $U(x)={\sum }_{r:n_r>0}{n}_r\log x_r$, with value $-\infty$ if an active rate is zero.

For positive feasible $x,y$, the elementary concavity inequality $\log u\le u-1$ givesThe last sum is nonpositive by proportional fairness. Feasible competitors with a zero active rate have utility $-\infty$, so they cannot improve it either. Thus a proportionally fair allocation maximizes weighted logarithmic utility. Conversely, if $x$ maximizes this utility, the feasible line segment $x+\theta (y-x)$ has nonpositive right derivative at $\theta =0$, giving precisely ${\sum }_r{n}_r(y_r-x_r)/x_r\le 0$. This also covers a competitor with zero coordinates, since $x+\theta (y-x)$ stays positive for sufficiently small $\theta$. The strictly concave function $U$ makes the active-rate optimizer unique.

For the linear flow network, put $M={\sum }_{i=1}^I{n}_i$ and $N=n_0+M$. If $n_i>0$ and resource $i$ is not saturated, its local per-flow rate could be increased without affecting any other resource. This would strictly increase weighted logarithmic utility. HenceIf $n_0>0$ and $M>0$, substitute $x_i=(1-n_0{x}_0)/n_i$ on all active local routes into the utility. The terms depending on $x_0$ areIts derivative vanishes when $1/x_0=M/(1-n_0{x}_0)$, so $x_0=1/N$. Its second derivative is negative, giving the unique maximizer. If $M=0<n_0$, utility increases up to the capacity endpoint $x_0=1/n_0=1/N$. If $n_0=0$, each active local route uses its entire local resource. Altogether the proportionally fair allocation on a linear flow network isAbsent-route rates may be set to zero, and departures are zero at the empty state $N=0$. Notice that these are per-flow rates: the aggregate through service is $n_0/N$, while each active local route gets aggregate service $M/N$.

In the document-transfer flow-level network model, let $e_r$ be the unit vector for route $r$. Independent Poisson processes of flow arrivals give transitions $n\to n+e_r$ at rate ${\nu }_r$. An active document on route $r$ receives service at rate $x_r(n)$ and has an exponential distribution of size with parameter ${\mu }_r$. By the memoryless property, its instantaneous completion rate is ${\mu }_r{x}_r(n)$. Summing over the active documents gives the continuous-time Markov chain intensitiesThus through-route departures have rate ${\mu }_0{n}_0/N$, and active local departures have rate ${\mu }_{i}M/N$. All other off-diagonal intensities vanish. The diagonal intensity is the negative of the sum of the outgoing intensities. Each route's aggregate service is at most one, so the total jump rate is bounded by ${\sum }_r({\nu }_r+{\mu }_r)$; the process is a nonexplosive continuous-time Markov chain.

Set ${\rho }_r={\nu }_r/{\mu }_r$, assume ${\mu }_r>0$, and define the balance function of a flow-level network byThe adjacent-state ratios of this binomial coefficient areThey equal the aggregate service rates just obtained. Therefore, with $w(n)=H(n){\prod }_r{\rho }_r^{n_r}$,These are the detailed balance for a continuous-time Markov chain equations on every adjacent pair. Summing incoming and outgoing pairs proves stationarity whenever the weights can be normalized. Hence the stationary law of a linear flow network isThe coefficient here is a binomial coefficient, not merely the total number of flows. Both it and the denominator in $x_0=1/N$ are visible in the original PDF but missing from the converted TeX.

We can also determine exactly when the normalizing constant is finite. For $0\le {\rho }_0<1$, summing first over $n_0$ with local counts fixed and applying the negative binomial series givesThe remaining sums are independent geometric series. Thusprovided ${\rho }_0+{\rho }_i<1$ for every $i$. These inequalities are also necessary: if ${\rho }_0\ge 1$, the states with every local count zero already give a divergent sum; otherwise any local ratio at least one makes its geometric series diverge. Thus the stationary formula requires strict offered load below capacity at every resource. With positive arrival rates, the process is irreducible, and these normalizable detailed balance weights give its unique stationary distribution. Outside these inequalities, the displayed weights are not a probability distribution. Zero arrival rates can be handled on the reachable class by setting the corresponding population to zero.

Fix $M$. Each time the chain is in $\{1,\dots ,M\}$ immediately before a jump, its conditional chance to jump to zero is at least the $\delta$ from part (b). Count all these opportunities, including returns to the same state after an excursion; the $T_m$ in part (b) need not count those returns. By the Strong Markov property at successive such visits, the probability of surviving $r$ opportunities without absorption is at most $(1-\delta {)}^r$. Hence infinitely many such opportunities without hitting zero have probability zero.

A positive state cannot be occupied forever: its finite positive holding rate gives an almost surely finite holding time. For a nonexplosive continuous-time Markov chain, visits at arbitrarily late times to the finite set therefore require infinitely many of the counted opportunities. On the event of no absorption, the chain consequently eventually stays above $M$, almost surely. Intersect this probability-one assertion over integers $M$. The result is $X_t\to \infty$. If zero is ever reached it remains occupied forever, and for initial state zero this is immediate. Thus $\boxed{{\mathbb{P}}_i(E\cup F)=1}$ for every initial state.

This uses the usual assumption implicit in an $I$-valued chain defined for all $t\ge 0$: no explosion with loss to a cemetery state. The rate conditions alone do not ensure it. For example rates $q_{i,i+1}=i^3$, $q_{i0}=1$ allow an infinite upward path with probability ${\prod }_i{i}^3/(i^3+1)>0$ and a finite total holding time, since ${\sum }_{i}1/(i^3+1)<\infty$. The minimal chain then explodes; with a cemetery state the printed dichotomy would require an additional explosion alternative.