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Put $\mu ={\mu }_1+{\mu }_2$. In this heterogeneous two-server queue, the total population alone is not a Markov process: when only one customer is present, its departure rate depends on which server is occupied. Use states $0,a,b,2,3,\dots$, where $a$ and $b$ distinguish the two single-customer possibilities, and $n\ge 2$ records the total population. The memoryless property of the exponential distributions gives these nonzero transition rates:and $q_{n,n+1}=\nu$ for $n\ge 2$, $q_{n,n-1}=\mu$ for $n\ge 3$. In particular, a departure from state $2$ leaves the other server occupied.

The detailed balance equations suggestFor example, both ${\pi }_a\nu ={\pi }_2{\mu }_2$ and ${\pi }_b\nu ={\pi }_2{\mu }_1$ hold; these are the two transitions that would be lost by using only population as the state. All other adjacent-state detailed balance equations hold as well. Since $\nu /\mu <1$, the geometric series converges, and normalization givesThus the stationary distribution exists, and the requested probability isWith equal server rates ${\mu }_1={\mu }_2=m$, this reduces to ${\nu }^2/[m(2m+\nu )]$, as for an M-M-s queue with two servers.

To prove the output claim, put the continuous-time Markov chain in its stationary distribution. The preceding detailed balance equations make it a reversible Markov chain: time reversal of a continuous-time Markov chain has the same transition rates. A forward departure becomes a reversed arrival. In every reversed state, the total rate of such arrivals is $\nu$: from $0$ the rates $\nu /2$ add, from $a$ or $b$ the upward rate is $\nu$, and from every $n\ge 2$ it is also $\nu$. The reversed process can therefore be constructed with an independent rate-$\nu$ Poisson process of arrival epochs, choosing the idle server by a fair coin when appropriate. Consequently reversed arrival counts on disjoint time intervals are independent Poisson random variables with the corresponding means. Reversing those intervals proves that the stationary departure stream is a rate-$\nu$ Poisson process, rather than merely a stream with mean rate $\nu$.

Stationarity matters here. If the queue starts empty, a departure by time $h$ requires both an arrival and a completed service time, so its probability is $O(h^2)$; a rate-$\nu$ Poisson process would have probability $\nu h+O(h^2)$. Thus the output assertion is the equilibrium assertion in the context of the preceding stationary calculation.

For a fixed routing loss network, let $C_j$ be finite integer resource capacities and $A_{jr}$ the integer resource requirement of one route-$r$ call. Assume each route uses a resource, independent Poisson processes of calls with rates ${\nu }_r$, independent exponential distributions of holding times with parameters ${\mu }_r$, and immediate rejection when any required resource is unavailable. There is no waiting. The continuous-time Markov chain of call counts has feasible setPut ${\rho }_r={\nu }_r/{\mu }_r$ andThe ratios of adjacent weights satisfy $\pi (n){\nu }_r=\pi (n+e_r){\mu }_r(n_r+1)$ whenever the arrival is feasible. Thus detailed balance proves the product-form stationary distribution of a loss network,An arriving route-$r$ call is admitted precisely when $An\le C-A_{\cdot r}$. By Poisson arrivals see time averages, its exact blocking probability isDefine $Z$ to be zero if a capacity is negative. This is an exact finite sum, not an independence approximation for resource blocking events.

For the triangle, denote the physical link capacities by $C_{12},C_{13},C_{23}$ and the numbers of calls between the pairs by $n_{12},n_{13},n_{23}$. Assume unit capacity per call on each traversed link, unrestricted instantaneous rearrangement of existing calls, no interruption or extra resource cost during rearrangement, and independent exponential distributions of holding times whose parameters depend on the endpoint pair but not on the chosen path. Arrivals for each pair are independent Poisson processes, and admission occurs whenever some routing of all calls is feasible. These assumptions make the count vector a Markov process even though individual paths can change.

The necessary cut constraints areEach counts calls that must cross the two links separating one node from the others. They are also sufficient. If every pair count is at most its direct capacity, route all calls directly. Otherwise at most one pair can exceed its direct capacity: two such excesses would contradict their pairwise cut constraint. Suppose it is $12$, and put $d=n_{12}-C_{12}>0$. Route $C_{12}$ of these calls directly and divert the remaining $d$ through node $3$. Route the other calls directly. The loads on links $13$ and $23$ are respectively $n_{13}+d\le C_{13}$ and $n_{23}+d\le C_{23}$ by the two cut constraints involving $n_{12}$. This is an integer routing of every call. The other cases follow by relabelling.

Consequently the rearrangeable triangular loss network is equivalent, at the level of call counts and blocking, to a fixed routing loss network with three virtual node resources:The columns correspond to $12,13,23$, and a call uses its two endpoint resources. Its normalizer isWith $e_j$ now indexing virtual resources, the three exact answers areThe preceding detailed balance proof applies because departures occur at rates ${\mu }_{ab}{n}_{ab}$ regardless of the current routing. Without unrestricted rearrangement, feasibility can depend on the existing path assignments and this equivalence need not hold.

Use the local convention for the martingale problem, which is appropriate for arbitrary measurable coefficients. A continuous adapted process $X$ solves $\mathbf{M}(a,b)$ if the coefficient integrals are locally finite and, for every $f\in C_c^{\infty }(\mathbb{R})$,is a local martingale. Here locally finite means ${\int }_0^T(|a(X_s)|+|b(X_s)|)ds<\infty$ almost surely for each $T<\infty$. Suitable localization upgrades the identities to true martingale identities. With bounded coefficients and bounded test derivatives, this distinction disappears on finite horizons.

A Markov diffusion generated by $L$ is a time-homogeneous Markov process with continuous paths whose infinitesimal generator on the stated test domain is $L$. Its transition operators satisfy the generator limit, and Dynkin formula gives precisely the preceding martingale problem identities. Some terminology uses L-diffusion for the test-function identities alone; the Markov property is an additional assertion unless uniqueness supplies it.

Suppose now $a={\sigma }^2>0$. Use smooth compactly supported functions agreeing with $x$ and $x^2$ on $[-R,R]$, and stop before exiting this interval. Removing these stops shows thatis a continuous local martingale. Consequently $X$ is a continuous semimartingale. The martingale problem applied to the second cutoff function says thatis a local martingale. The Itô formula for $X^2$ also says thatis a local martingale. Subtracting, the continuous finite-variation process $[N{]}_t-{\int }_0^{t}a(X_s)ds$ is a local martingale and starts at zero. The theorem that a continuous finite-variation local martingale is constant givesDefineThe integrand is predictable and is stochastically integrable, since its integral against the bracket is exactly $t$. Thus $B$ is a continuous local martingale with $B_0=0$ and $[B{]}_t=t$. The Lévy characterization of Brownian motion states that these conditions make $B$ a Brownian motion relative to the given filtration. Associativity of stochastic integrals recovers $N=\int \sigma (X)dB$, and thereforeThe original process is a weak solution of a stochastic differential equation, with its driving Brownian motion constructed on the original space.

Conversely, the Itô formula applied to any weak solution with locally integrable coefficients gives the martingale problem identities. A Markov diffusion with generator $L$ gives them by Dynkin formula. A well-posed martingale problem, with the usual measurable path-space setup, identifies a unique family of diffusion laws and yields the Strong Markov property. Without uniqueness, the test-function identities alone do not assert that every chosen solution is Markov. This specifies both the equivalence with weak equations and the role of uniqueness in identifying a diffusion.

Assume the innovations are independent of the previous history and have finite exponential moments at every real argument, as required by the finite-valued affine definition. Conditioning on $X_{t-1}=x$ givesHence the autoregressive process of order one is an affine process, withHere and in the remaining parts, an affine process uses a time-homogeneous Markov process: the one-step transform is the same at every time. The displayed definition at time $1$ alone would not determine later transitions of an arbitrary time-inhomogeneous Markov process.

A steady state does not by itself give every fluctuating observable an entropy production symmetry. If the dimensionless total entropy production ${\Sigma }_t$ has the same-law finite relation and a large deviation principle with rate function $I$ for ${\Sigma }_t/t$, then, on its common domain,The first expression denotes leading exponential scaling; the second follows by taking the scaled logarithm of the probability ratio. A finite-state, autonomous Markov process with time-even states, positive bidirectional rates and a positive stationary distribution supplies a standard setting for a finite total-entropy production relation. The stochastic system entropy is then bounded, so its endpoint contribution does not change the long-time rate function of bath entropy production. ^[ref]

Unbounded state spaces can have endpoint fluctuations on the scale relevant to a large deviation principle. Dropping system entropy, changing the measured observable to heat, or changing the reverse preparation therefore requires a fresh argument. The exact finite integral fluctuation theorem, a same-law finite relation, and an asymptotic rate function symmetry are distinct claims.

The operator that averages a function over the next-state conditional law of a time-homogeneous Markov process. For a random walk with increments of law $\nu$ independent of the past including the initial state, $(Ph)(s)=\int h(s+y)\nu (dy)$. This additive operator preserves convexity when the integrals exist.

At each site, a rate-one clock rings and the site copies the state of a randomly chosen neighbor. The model is a Markov process on Ising spin configurations, interpreted through its local update infinitesimal generator. All constant configurations are absorbing state. Tracing copying arrows backwards gives the voter-model duality with coalescing random walks.