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Stationarity at zero gives ${\pi }_0=q{\pi }_0+q{\pi }_1$, hence ${\pi }_1=(p/q){\pi }_0$. At one, ${\pi }_1=p{\pi }_0+q{\pi }_2$; substituting $p{\pi }_0=q{\pi }_1$ gives $q{\pi }_2=p{\pi }_1$. More generally the interior equations are ${\pi }_i=p{\pi }_{i-1}+q{\pi }_{i+1}$. Induction givesSince $p/q<1$, normalization of this geometric series givesThese probabilities satisfy detailed balance ${\pi }_{i}p={\pi }_{i+1}q$, including the boundary balance with the holding probability. Thus they do define a stationary distribution. Irreducibility then gives its uniqueness.

The baryon excess requires a mechanism that produces different net baryon numbers from an initially symmetric ensemble. In the usual local, CPT-invariant framework, the Sakharov conditions are baryon-number violation, C and CP violation, and departure from thermal equilibrium. Baryon conservation would preserve zero net baryon number; exact conjugation symmetries equate the conjugate production processes; equilibrium inverse reactions and detailed balance erase a would-be asymmetry. These necessary conditions do not determine its magnitude.^[ref]

For example, let equal numbers of a heavy particle $X$ and its antiparticle decay into channels $i$ with baryon numbers $B_i$. Their mean net yield per pair isIt can be nonzero even though CPT requires equal total widths: the branching fractions differ among channels carrying different baryon numbers. A nonzero decay-rate asymmetry requires interference with both a CP-odd weak phase and an absorptive phase. If the expansion rate overtakes the inverse-reaction rate, decays can occur out of equilibrium; later baryon-violating washout must then be weak enough for the asymmetry to survive.

In electroweak baryogenesis, CP-asymmetric transport near a first-order phase-transition boundary can bias baryon-violating sphaleron transitions. CP violation alone is insufficient: the transition and washout dynamics also matter. In leptogenesis, out-of-equilibrium decays of heavy Majorana neutrinos can generate lepton asymmetry, part of which sphalerons convert into baryon asymmetry while conserving $B-L$. The CKM phase demonstrates CP violation in the Standard Model but does not by itself supply a successful cosmological mechanism. The key quantity is a surviving net charge yield, not merely the existence of a CP-violating reaction.

A grand unified theory embeds the strong and electroweak gauge groups in a larger group, for example $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y\subset SU(5)$. quarks and leptons can belong to common representations, while heavy extra gauge or scalar bosons connect them. The resulting interactions can violate baryon number, unlike ordinary perturbative Standard Model reactions. Running gauge couplings and proton-decay searches test such ideas; unification is a theoretical possibility, not an established observation.^[ref]

baryogenesis must explain why a nearly matter-antimatter-symmetric early state evolves into a surviving baryon asymmetry. In the usual local CPT-invariant framework, the Sakharov conditions require baryon-number violation, C and CP violation, and departure from thermal equilibrium. baryon conservation would forbid creating a net excess; exact conjugation symmetries would pair baryon-producing processes with equal antibaryon production; equilibrium with no imposed asymmetry would erase a source term through detailed balance.

For heavy particles $X$ and $\bar{X}$, let decay channels carry different baryon numbers. CP-violating interference can make their branching fractions differ, even though CPT symmetry requires equal total widths. Defining the mean baryon number produced per decay as $ϵ$, an out-of-equilibrium abundance can generate $n_B/s$ of order $ϵ$ times its initial number per entropy, reduced by inefficient production and washout. A rough condition is that decays become effective after inverse production has ceased, with the relevant reaction rate compared with the Hubble parameter. Complex couplings alone do not guarantee a nonzero rate asymmetry: a relative absorptive phase is also needed.

Survival matters as much as production. Electroweak sphalerons violate $B+L$ while conserving baryon-minus-lepton number $B-L$, so a minimal GUT-generated asymmetry with $B-L=0$ can be washed out if sphalerons and the other relevant reactions equilibrate. Generating nonzero $B-L$, as in suitable heavy-Majorana-neutrino leptogenesis, can avoid this problem. A viable GUT baryogenesis model must generate an asymmetry and demonstrate that subsequent reactions do not erase it. The Sakharov conditions are necessary ingredients under these assumptions, not a numerical prediction of the observed abundance.

We use the stationary weights of a reflected birth-death chain, also verifying why they decide the question. If a stationary distribution $\pi$ exists, its equation at zero is ${\pi }_0={\pi }_1{d}_1={\pi }_0{u}_0$. At $k\ge 1$, stationarity givesThus the edge flow is constant, and its value at the boundary is zero. Every stationary distribution must therefore satisfy detailed balance on each edge. Since irreducibility forces ${\pi }_0>0$, it must be proportional toConversely, if ${\sum }_k{w}_k<\infty$, normalizing these positive weights gives a probability distribution satisfying detailed balance, hence a stationary distribution. Existence of a stationary distribution for an irreducible Markov chain on a countable state space is equivalent to positive recurrence.

The ratio isFor $p<1/2$, the limit is less than one, so the ratio test gives summability. For $p\ge 1/2$, observe that $u_k=p+\theta q/(k+\theta )>p\ge 1/2$, whereas $d_{k+1}<q\le 1/2$. Thus $w_{k+1}/w_k>1$ for every $k$, so the weights cannot be summable. This also resolves $p=1/2$ without making an inconclusive appeal to the limiting ratio. Therefore

Let $I_n$ count immigration events during the first $n$ transitions, and let $A_n$ count upward steps during those transitions. Write $h(k)=\theta /(k+\theta )$ for the conditional immigration probability. If $J_m$ is the indicator that transition $m$ to $m+1$ is immigration, then $J_m-h(X_m)$ is a bounded martingale difference with respect to the history that records event types as well as population sizes. The strong law for martingales with bounded increments impliesSince $h$ is bounded, the countable-state Markov chain ergodic theorem givesApplying the same argument to upward-step indicators gives $A_n/n\to {\sum }_k{\pi }_k{u}_k$ almost surely. To evaluate this denominator, sum detailed balance over all neighboring edges:Since $u_k+d_k=1$ and ${\sum }_k{\pi }_k=1$, both sums equal $1/2$. The denominator limit is positive, so division is valid and the immigration fraction among upward steps isThis is the requested answer in terms of $\pi$. Substituting part (iii) additionally gives ${\sum }_k{\pi }_k\theta /(k+\theta )=(1-r)/2$, so the fraction simplifies to $1-r=(1-2p)/(1-p)$.

The Markov property says that, conditional on the present state, the next state is independent of the earlier history:Two states communicate if each is reachable from the other with positive probability in finitely many steps. The equivalence classes are communicating classes; a class is closed if no positive-probability transition leaves it.

The displayed probabilities require $0\le p\le 1/2$; negative values allowed by the printed one-sided inequality are not transition probabilities. For $p=0$, every state is an absorbing singleton, so all singleton classes are closed. For $0<p\le 1/2$, neighboring states communicate in both directions, so there is one closed irreducible class $S$ and no open classes.

For $m=3$, the detailed balance relations give $2{\pi }_0={\pi }_1={\pi }_2=2{\pi }_3$. Normalization yieldsFor $0<p<1/2$, positive holding probabilities make the finite irreducible chain aperiodic, so its transition probabilities converge to ${\pi }_j$ from every initial state. At $p=1/2$, every step changes parity and the period is two; convergence from a fixed initial state fails. A stationary initial mixture still has distribution $\pi$ at every time, but this does not give convergence of the transition probabilities asked for.

For the Metropolis–Hastings algorithm, choose a proposal transition $q(x,y)$ and initialize at a state with $\pi (x)>0$. From current $x$, propose $Y\sim q(x,\cdot )$ and accept it with probabilityOtherwise stay at $x$. The accepted off-diagonal transition satisfies detailed balance, since $\pi (x)q(x,y)\alpha (x,y)=\min \{\pi (x)q(x,y),\pi (y)q(y,x)\}$ is symmetric in $x,y$. Thus $\pi$ is stationary. For convergence from arbitrary starting states in the support, choose the proposal so the resulting chain is irreducible and aperiodic. Only ratios of target probabilities are needed, so an unknown normalizing constant cancels.

The state-zero stationary distribution equation is ${\pi }_0=q{\pi }_0+q{\pi }_1$, so ${\pi }_1=(p/q){\pi }_0$. The state-one equation is ${\pi }_1=p{\pi }_0+q{\pi }_2$, giving ${\pi }_2=(p/q){\pi }_1$. More generally, the stationary probability flow across the cut $\{0,\dots ,i\}$ is balanced, so $p{\pi }_i=q{\pi }_{i+1}$. This also follows inductively from ${\pi }_i=p{\pi }_{i-1}+q{\pi }_{i+1}$.

Hence ${\pi }_i=(p/q{)}^i{\pi }_0$. Since $p<q$, the geometric series is summable, and normalization givesThese probabilities satisfy detailed balance and therefore all the stationarity equations. The Markov chain is an irreducible Markov chain, so this normalized stationary distribution is unique.

Let $L$ be the number of index tables, including the zero table. If distinct feasible states $A,B$ differ by an index table, then $P(A,B)=1/L=P(B,A)$, using the sign-reversal bijection for the reverse move. If they do not differ by an index table, both probabilities are zero. Since $\pi$ is uniform,For equal states detailed balance is tautological. Thus the chain is reversible and its uniform distribution is stationary.

The left panel below shows paddles in a bath connected by an axle to an asymmetric toothed wheel, a spring-loaded pawl and a load. A naive argument lets random paddle kicks turn the wheel in one direction while the pawl blocks reverse motion, apparently extracting work from a single bath. The missing ingredient is the pawl's own thermal fluctuations. At the same temperature, it sometimes lifts and allows reverse slips. Correct forward and reverse transition rates obey detailed balance, so no sustained directed rotation can do work at single-bath thermal equilibrium. The device illustrates, rather than disproves, the Second law of thermodynamics.

A macroscopic shaken ratchet is driven far from thermal equilibrium: imposed shaking supplies organized mechanical energy, and friction or inelastic collisions dissipate it. Its kilogram-scale parts are not moved appreciably by molecular thermal fluctuations. Rectification requires broken spatial symmetry from the teeth or pawl and appropriate dynamical coupling: the shaking must sometimes disengage or drive the pawl over a tooth while contact and dissipation prevent equal reverse motion. The load must remain below the available mean driving torque. Isotropic, aperiodic shaking can satisfy these conditions because the asymmetric mechanism breaks directional symmetry; without the external energy supply, a working rectifier and dissipation, there is no guaranteed directed motion.

At $7000\mathrm{K}$, $k_{B}T=0.603\mathrm{eV}$ andThe $2p$ ground term has fine-structure levels $J=1/2$ and $J=3/2$, with energy-level degeneracy $g_1=2$ and $g_2=4$. Detailed balance gives $C_{21}=C_{12}(g_1/g_2)e^{\Delta E/(k_{B}T)}=7.17\times 10^{-14}{\mathrm{m}}^3{\mathrm{s}}^{-1}$. The critical density is thereforeWith collisions and spontaneous decay both retained, $N_2/N_1=n_e{C}_{12}/(A_{21}+n_e{C}_{21})$. A Boltzmann distribution, $N_2/N_1\to 2e^{-\Delta E/(k_{B}T)}\simeq 1.97$, requires $n_e\gg n_{\mathrm{crit}}$, not merely equality. For example $n_e\gtrsim 3\times 10^8{\mathrm{m}}^{-3}$ brings the ratio within about ten per cent of that limit.

Let $U(x)$ be the static potential and $\gamma$ the drag coefficient, so terminal Genetic drift under force $-U^{\prime }$ is $v=-U^{\prime }/\gamma$. The probability current is $J=-D{U}^{\prime }p/(k_{B}T)$ only after imposing the desired relation; instead start independently from $J=-D{p}^{\prime }-U^{\prime }p/\gamma$. At thermal equilibrium $p\propto e^{-U/(k_{B}T)}$ and detailed balance requires $J=0$. Substitution yieldsAt 298 K with $a=0.5\mu$m, this gives approximately $4.37\times 10^{-13}$ m${}^2$ s${}^{-1}$. The argument assumes an overdamped particle in an equilibrium thermal bath, rather than arbitrary driven diffusion.

The Shockley-Queisser calculation treats a single-junction solar cell with a single band gap, in steady state, with one collected electron-hole pair per absorbed above-gap photon, an ideal absorption edge, thermalized carriers and radiative recombination as the limiting recombination process. The source spectrum, cell temperature, angular acceptance and optical concentration must be specified. Below-gap photons are transmitted; energy above the gap is lost during carrier thermalization; the cell must emit photons by detailed balance, producing a recombination current and an entropy-related voltage loss. Delivering power at finite current also puts the operating voltage below the open-circuit voltage. Nonradiative recombination, resistance and imperfect collection worsen a real cell but are not required for the ideal limit.

Three ways to change those assumptions are tandem solar cells, which divide the spectrum among gaps and reduce both transmission and thermalization losses; hot-carrier solar cells, which aims to collect excess carrier energy before thermalization; and carrier multiplication or spectral conversion, which aims to use a high-energy photon for more than one excitation or combine otherwise unusable sub-gap photons. These concepts have additional practical losses and do not automatically outperform an optimized conventional device. The usual unconcentrated single-junction optimum is approximately one third under a terrestrial solar spectrum, rather than a universal efficiency independent of illumination.

For the geometrical estimate, take the stated horizontal equinox irradiance to set a constant direct-normal reference. The equinox solar-noon elevation is ${\alpha }_e=90^{\circ }-51.75^{\circ }=38.25^{\circ }$, so this reference is $I_n=1/\sin {\alpha }_e\simeq 1.62\mathrm{kW}{\mathrm{m}}^{-2}$. A south-facing panel with tilt $\beta =41.1^{\circ }$ has noon incidence factor $\cos (51.75^{\circ }-\delta -41.1^{\circ })$, where $\delta =\pm 23.44^{\circ }$. Relative to the $1\mathrm{kW}{\mathrm{m}}^{-2}$ plane-of-array rating,These are the results of that idealized geometrical normalization, not credible seasonal production forecasts. Its inferred direct-normal irradiance already exceeds the approximately $1.36\mathrm{kW}{\mathrm{m}}^{-2}$ solar irradiance outside the atmosphere, so the supplied horizontal benchmark cannot describe a purely direct noon beam literally. Real direct and diffuse fractions, atmospheric attenuation at low winter elevation, cell temperature and inverter clipping are unspecified. An alternative assumption of $1\mathrm{kW}{\mathrm{m}}^{-2}$ direct-normal irradiance gives approximately $4.88$ and $4.14$ kW. A physically unique maximum cannot be inferred without an irradiance model.

The stated annual output corresponds to $4100/5=820$ full-rated hours, an average power of $0.468$ kW and a capacity factor of $9.4\%$. Night, cloud, seasonal day length and other losses explain the large difference from an instantaneous sunlit estimate.

For a national scale estimate, take UK area $2.44\times 10^{11}{\mathrm{m}}^2$, population $6.6\times 10^7$, and annual mean solar input $100\mathrm{W}{\mathrm{m}}^{-2}$. Covering five percent with twenty-percent-efficient modules givesThis is comparable with an order-$100$ kWh daily per-person primary-energy budget, and much larger than electricity use alone, of order $10$-$15$ kWh per person per day. It is a gross area-and-flux estimate: access land, shading, seasonal storage, curtailment and conversion between primary energy and electrical end use all affect its interpretation.

Choose $k$ distinct coordinates uniformly for the starting ones; each element of $S_k$ then has starting probability $1/\left(\genfrac{}{}{0ex}{}{n}{k}\right)$. At each iteration choose one current one and one current zero uniformly and swap them. This implements the normalized proposal distribution from (c)(i), remains in $S_k$, and needs just one precomputed matrix inverse.

mymcmc <- function(x, k, sigma) {
  n <- length(x)
  stopifnot(length(k) == 1L, is.finite(k), k == floor(k),
            k > 0L, k < n, is.matrix(sigma),
            all(dim(sigma) == c(n, n)))
  sigma_inv <- invsigma(sigma)
  u <- integer(n)
  u[sample.int(n, k, replace = FALSE)] <- 1L
  draws <- matrix(0, 100L, n)
  for (iteration in seq_len(100L)) {
    ones <- which(u == 1L)
    zeros <- which(u == 0L)
    i <- ones[sample.int(length(ones), 1L)]
    j <- zeros[sample.int(length(zeros), 1L)]
    v <- u
    v[i] <- 0L
    v[j] <- 1L
    alpha <- mylogapp(x, u, v, sigma_inv)
    if (runif(1L) < alpha) u <- v
    draws[iteration, ] <- u
  }
  colMeans(draws)
}

Sampling positions with sample.int and then indexing into ones or zeros also works when $k=1$ or $n-k=1$. In R, a call such as sample(ones, 1) would incorrectly sample from 1:ones when ones is a single numeric value greater than one.

The code records the state after each of exactly $100$ transitions, including repetitions after rejection. Its answer is the vector Monte Carlo estimatorThis estimates the posterior mean coordinate by coordinate; because each coordinate is binary, its entries estimate the posterior probabilities of inclusion. Every stored state has coordinate sum $k$, so the estimated means lie in $[0,1]$ and sum to $k$.

For neighbouring states, the symmetry of the proposal and the Metropolis–Hastings acceptance probability give detailed balance:Thus the posterior distribution is a stationary distribution. Any two states in $S_k$ can be connected by moving the ones that occupy the wrong positions, and the multivariate normal distribution gives strictly positive target masses. The resulting Markov chain is therefore an irreducible Markov chain, and its long-run state averages estimate the posterior mean.

The requested $100$ iterations are a finite approximation, without a convergence guarantee or discarded burn-in. A randomly chosen starting subset has the prior distribution, rather than generally the posterior distribution. No aperiodicity assumption is needed for the finite-state Markov chain ergodic theorem for averages; indeed a uniform two-state target with $n=2,k=1$ gives a period-two alternating chain. This does not invalidate either the code or the state-average estimator.

A product-Poisson stationary distribution is suggested by balancing the three reaction fluxes around the directed cycle of reaction complexes. If their mean counts are $a,b$, this complex balance condition iswhich gives $a=4$ and $b=1/2$. This suggests the joint probability mass functionIt is normalized because it is the product of two Poisson distributions. We verify it directly in the chemical master equation, rather than assuming a product-Poisson stationary distribution theorem. The three incoming terms satisfyWith negative-index values set to zero, these identities also hold on the axes. Their sum equals $(2+n/2+nm)\pi (n,m)$, precisely the outgoing probability flux. The stationary average total exit rate is $2+{\mathbb{E}}_{\pi }n/2+{\mathbb{E}}_{\pi }(nm)=6<\infty$, so these stationary balance equations do define an invariant distribution of a continuous-time Markov chain for the nonexplosive model verified below. This is complex balance; it does not assert pairwise detailed balance for the irreversible chemical reactions.

To identify this invariant distribution with the limiting probability distribution in the question, first check that the model is a nonexplosive continuous-time Markov chain. On any finite time interval, $A_1$ is bounded by its initial count plus the finite number of arrivals in a rate-$2$ Poisson process. Conditional on that bound, the rate of producing $A_2$ is bounded, so only finitely many such productions occur. The number of joint removals cannot exceed the available molecules. Thus only finitely many chemical reactions occur in finite time.

The continuous-time Markov chain is also an irreducible Markov chain. From a Markov chain state with both counts positive, repeated joint removals reduce both. On the $A_2=0$ axis, producing one $A_2$ and then removing a pair reduces $A_1$ by one. On the $A_1=0$ axis, adding one $A_1$ and then removing a pair reduces $A_2$ by one. Each prescribed finite sequence has positive probability, so every Markov chain state reaches $(0,0)$. Conversely, arrivals followed by productions reach any $(n,m)$ with $n>0$; to reach $(0,m)$, reach $(1,m+1)$ and remove one pair. The continuous-time Markov chain is therefore positive recurrent, and the continuous-time Markov chain convergence theorem gives $\varphi =\pi$ from the empty initial condition.

The two Poisson factors happen to have cancelling numerical coefficients at $(2,2)$:

Treat $p_n(t)$ as the probability of the complete coarse-grained state $n$ at time $t$, with $p_n\ge 0$ and ${\sum }_n{p}_n=1$. In the column-vector convention of the master equation, $W_{nm}$ for $n\ne m$ is the transition rate from $m$ to $n$. The short-time transition law isNonnegative transition probabilities and conservation of total probability requireFor a finite state space these conditions define a continuous-time Markov chain. The Markov property means that, conditional on the present state, no further history affects the transition probabilities. A time-independent $W$ additionally assumes time-homogeneous rates; the waiting time in a state is exponential, and multiple jumps in an infinitesimal interval contribute beyond first order. A reduced description that omits relevant slow variables can retain memory and need not obey this closed equation. Quantum coherences are likewise absent from this classical population model. For infinite state spaces one also needs conditions preventing infinitely many jumps in finite time. Neither equal forward/backward rates nor detailed balance is required merely to conserve probability.

Here $p_n(t)$ is the probability of occupying state $n$, and $W_{nm}$ for $n\ne m$ is the transition rate from $m$ to $n$. More precisely, over $dt>0$ that transition has probability $W_{nm}dt+o(dt)$. This is a time-homogeneous continuous-time Markov chain: the present state contains all information needed for the next transition, the rates do not depend on time, and finite rates make multiple jumps have probability $o(dt)$. In a physical rate description, untracked coherent dynamics or memory must be negligible on the timescale being modeled.

The master equation separates incoming and outgoing terms:Columns, rather than rows, sum to zero with this indexing. These conditions preserve normalization and nonnegative probabilities. For infinitely many states one also needs a well-defined nonexplosive process. Detailed balance is an additional equilibrium condition, not a general requirement on a master equation.

Set the time derivatives in the master equation to zero. The first and third rows give $f_1{p}_0=f_3{p}_B$ and $f_2{p}_A=(f_3+f_4)p_B$. Normalization then gives, with $\mathcal{N}=f_2{f}_3+f_1(f_3+f_4)+f_1{f}_2$,These expressions assume positive rates, or a limiting choice for which $\mathcal{N}\ne 0$ and the stationary distribution is unique. If rates vanish so that multiple closed classes appear, the initial condition can matter.

The word “equilibrium” in this part must mean a stationary distribution. With positive $f_1,f_2,f_3$ the one-way contact transitions support a nonzero stationary electric current, so the process is not in thermodynamic equilibrium and does not satisfy detailed balance. In particular, $0\to A$ carries positive probability flow while the direct reverse rate is zero.

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.

A Poisson Metropolis chain samples a Poisson distribution by proposing an upward or downward unit step with equal probability and reflecting a downward proposal at zero into a self-proposal. Its upward and downward transition probabilities are the displayed $u_n,d_n$, with $d_0=0$, and the diagonal entry is $1-u_n-d_n$. For ${\pi }_n=e^{-\lambda }{\lambda }^n/n!$, adjacent stationary fluxes equal $\frac{1}{2}\min ({\pi }_n,{\pi }_{n+1})$. Hence detailed balance holds. Positive adjacent probabilities imply an irreducible Markov chain; $P(0,0)\ge 1/2$ implies aperiodicity. The normalized invariant Poisson distribution implies positive recurrence, so the countable-state Markov chain convergence theorem applies. The boundary proposal probability is essential: forcing every proposal at zero to go upward would require a different acceptance ratio.