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Let $K(t,\cdot )$ be the Markov kernel of the Metropolis–Hastings algorithm. Away from the possibility of remaining at $t$, its transition density isMultiplication by the target density givesThis is detailed balance for every pair of distinct states.

Writefor the rejection probability, soFor every measurable $B\subseteq \Theta$, detailed balance and the Tonelli theorem giveHence $\mu$ is an invariant distribution of the chain.

A distribution $\pi$ is an invariant distribution whenfor every state $j$. The pair $(\pi ,P)$ satisfies detailed balance whenfor every $i,j$. Summing this identity over $i$ givesso detailed balance implies invariance.

For an irreducible positive recurrent Markov chain, Kac's lemma relates the invariant mass to the mean recurrence time:

The posterior isA random-walk Metropolis–Hastings algorithm proposes ${\vartheta }^{\prime }$ from a symmetric density about the current $\vartheta =(\mu ,\theta )$ and accepts with probabilityFor distinct states, multiplying the transition density by the target density giveswhich is symmetric and proves detailed balance. Run multiple dispersed chains, tune proposals during warm-up, inspect traces, effective sample sizes and convergence diagnostics, then estimate the period mean by averaging $2\pi /{\omega }_m$ over retained draws.

In equilibrium the chain is reversible by detailed balance. Reverse time. Forward departures become reverse accepted arrivals; reversibility identifies the reversed chain with the original one, whose attempted arrivals form a Poisson process. The accepted points are exactly those occurring while the state is below $N$. The standard quasi-reversibility argument therefore makes the equilibrium departure process Poisson, with rate equal to the mean departure rate

The chain is an irreducible birth-death process. The detailed balance equations areso ${\pi }_{i+1}={\pi }_i/2$. Normalization gives the stationary distributionAn irreducible countable-state Markov chain that possesses a stationary probability distribution is positive recurrent. Hence $X$ is positive recurrent.

Expand according to the hitting time and use detailed balance to reverse each finite path:The event in the final probability is $\{T_A^{+}>n\}$. The tail-sum formula for a nonnegative integer-valued random variable givesTherefore

Detailed balance says that each equilibrium transition is balanced by its time reverse:With the Boltzmann distribution $P_{\mathrm{eq}}\propto e^{-\beta H}$ and $\Delta H=H_2-H_1$, this requiresParts a and b instead give $\exp [-(2\zeta /{\sigma }^2)\Delta H]$. Equality for every pair of endpoints yields the fluctuation-dissipation relation for a Langevin particle

At stationarity,Thus every edge current vanishes and the two-state system satisfies detailed balance. A network containing only one undirected edge cannot sustain a stationary cycle current.

Write $d_G(x)$ for the degree of $x$ in $G$ and $d_A(x)$ for the number of its neighbours lying in $A$. Each trial succeeds with probability
$d_A(x)/d_G(x)$. The stated geometric-sum fact shows that the total holding time at $x$ is exponential with rate $d_A(x)/d_G(x)$. Conditional on success, each neighbour in $A$ is chosen uniformly. Hence

LetFor every edge $xy$ of $A$,Thus detailed balance holds, so $\pi$ is invariant. Connectedness of $A$ makes the chain irreducible and this invariant distribution unique.

Yes. The three nontrivial detailed balance identities areandThus the chain is a reversible Markov chain.

For $f,g\in L^2(\pi )$, detailed balance says that the measure $\pi (dx)K(x,dy)$ is invariant under exchanging $x$ and $y$. ThereforeThis is precisely the defining identity for a self-adjoint operator.

For $A+B\rightleftharpoons C+D$, the forward and reverse rates areAt thermochemical equilibrium, $r_f=r_r$ by detailed balance, whileafter the appropriate standard-concentration factors are included. The standard reaction Gibbs free energy isConsequentlyFor other stoichiometries, the same argument uses the corresponding activity product and its standard-state factors.

Subtracting the two quadratic actions givesThe functional chain rule identifies the last integral as $F_2-F_1$, so

Microscopic time-reversal invariance implies detailed balance. The equilibrium probability density of a configuration with free energy $F$ obeysThereforeComparison for arbitrary endpoint free energies yields the Model A fluctuation-dissipation relation

Take flat priors on $\alpha ,\beta ,\mu$ and scale priors $\pi ({\sigma }^2,{\tau }^2)\propto ({\sigma }^2{\tau }^2{)}^{-1}$ on the positive variances. The posterior distribution is thenA random-walk Metropolis–Hastings algorithm can update $(\alpha ,\beta ,\mu ,\log {\sigma }^2,\log {\tau }^2)$ with a symmetric proposal and accept a proposed state $z^{\prime }$ from $z$ with probability $1\wedge \pi (z^{\prime }\mid \mathcal{D})/\pi (z\mid \mathcal{D})$, including the Jacobian if the target is represented in transformed coordinates. Its transition kernel satisfieswhich is detailed balance; hence the posterior is stationary.

If distinct states $X,Y$ communicate in one step, they differ by a unique removed vertex and a unique inserted vertex. ConsequentlyThus the transition matrix is symmetric, so detailed balance holds for the uniform distribution on $S$. Irreducibility makes this invariant distribution unique.

Let $\pi$ denote the posterior and let the proposal density satisfy $q({\theta }^{\prime }\mid \theta )=q(\theta \mid {\theta }^{\prime })$. For distinct states, the Metropolis transition density isConsequentlyThe rejection mass on the diagonal also satisfies detailed balance, so the posterior is invariant.

For a finite irreducible Markov chain with at least two states satisfying detailed balance, its Dirichlet form of a Markov chain obeysExpand $f-\pi (f)$ in an orthonormal eigenbasis. The denominator is ${\sum }_{j\ge 2}{a}_j^2$, and the numerator is ${\sum }_{j\ge 2}(1-{\lambda }_j)a_j^2$. Their ratio is at least $1-{\lambda }_2$ and equality holds for the second eigenfunction. Thus the optimal Poincare inequality for a reversible Markov chain constant is $1/\gamma$, for the ordinary rather than absolute spectral gap.

In equilibrium, a queue is quasireversible when its present state is independent of subsequent arrival times and preceding departure times, for each customer class. For an M-M-s queue, detailed balance and time reversal of a continuous-time Markov chain show that past departures become future arrivals generated independently of the present state. This backward independence is useful in tandem calculations where downstream occupancy is determined by earlier departures.

For a reversible Markov chain with stationary distribution $\pi$ and $\nu \ll \pi$, detailed balance givesTo prove this, integrate either side against a bounded test function and exchange the two endpoints using detailed balance. If $\pi =h/Z$, the output probability density function is $q_t(x)=h(x){\mathbb{E}}_{B\sim K^t(x,\cdot )}[\nu (B)/h(B)]$. A reverse simulation can therefore estimate this density ratio without evaluating a transition probability density function.