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Multiply the master equation by $z^n$ and sum over $n$. Shifting indices producesThe apparent negative power causes no problem at $z=0$ when $r_0=0$. Let $D=z{\partial }_z$, so $D{z}^n=n{z}^n$. If the rates are polynomials in $n$, replace their arguments by $D$ to obtain a finite-order partial differential equation:For arbitrary rates the weighted sums still give an exact generating-function evolution, but need not reduce to a finite-order differential operator. The probability generating function obeyswhich is the normalization condition. The evolution has ${\partial }_{t}F(1,t)=0$ when the rate sums converge, so an initially normalized distribution stays normalized. Its derivatives at $z=1$ also give the factorial moments, and expansion around $z=0$ recovers individual probabilities.

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.

Write the probability column vector as $\mathbf{p}=(p_0,p_A,p_B{)}^T$, with $0$ denoting the uncharged state. The arrows in the original figure give transitions $0\to A$ at rate $f_1$, $A\to B$ at rate $f_2$, $B\to 0$ at rate $f_3$, and $B\to A$ at rate $f_4$. Exclusion of double occupation means injection occurs only from state $0$. Thus the master equation isFor example, ${\dot{p}}_B=f_2{p}_A-(f_3+f_4)p_B$. Every allowed jump adds to one probability and subtracts from another, which verifies the zero column sums.

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 $f>0$ set $\tau =ft$ and $y(\tau )=p_0(t)$. The master equation becomesUsing $p_A=1-y-p_B$ and $p_B=(y^{\prime }+y)/2$ givesInitially the molecule occupies $B$, so $y(0)=0$ and $y^{\prime }(0)=2$. The characteristic roots are $-2,-3$; the constant particular solution is $1/3$. ConsequentlywhenceIt has the required initial slope $2f$, tends to the stationary uncharged probability $1/3$, and remains nonnegative. As a further check, $p_B=1/6-e^{-2ft}/2+4e^{-3ft}/3$ and $p_A=1/2-e^{-2ft}/2$ are nonnegative and sum with $p(t)$ to one. If $f=0$, no jump occurs and $p(t)=0$, also obtained directly from the boxed expression.

The one-step master equation for the biased random walk isExpand in a Taylor series and retain the first time derivative and the first two space derivatives. Since $w_{+}+w_{-}=1$ and $w_{-}-w_{+}=2ϵ$,Thus the usual continuum truncation gives the requested Fokker-Planck equation,The diffusion coefficient has units $\mathrm{lengt}{\mathrm{h}}^2/time$, while $C$ has units of velocity. The physical drift velocity is $v=-C=a(w_{+}-w_{-})/\tau$, because the standard advection term is $-v{\partial }_{x}P$. A rightward bias therefore means $ϵ<0$, $C<0$, and $v>0$.

The approximation must be stated carefully. The truncation is a controlled diffusion limit of a weakly biased random walk when $a\to 0$, $\tau \propto a^2$, and $ϵ\propto a$, keeping $D_0$ and $C$ finite. It is also the spatial diffusion approximation for a continuous-time jump process with exponentially distributed waiting times of mean $\tau$.

If the original fixed-step-time walk has a finite bias, its exact variance from (a) instead gives the diffusion coefficient of the Gaussian envelope about the drifting mean:One obtains the same correction by retaining $\frac{1}{2}{\tau }^2{\partial }_t^{2}P$ in the time expansion and using the leading advection equation to replace ${\partial }_t^{2}P$ by $v^2{\partial }_x^{2}P$. The result is $D_{\mathrm{fixed}}=D_0-v^2\tau /2$. In particular, a deterministic one-way walk has zero variance, as this expression requires. Thus $D_0$ is the usual intended first-time-derivative approximation, not an exact variance-growth coefficient at arbitrary bias. The later continuum estimates use that convention, with the fixed-time alternative identified where relevant.

For this radial random walk on a regular tree, the drawing is an infinite regular tree of degree $z$, not a planar lattice with closed loops. The radial coordinate is the number of edges in the unique path to the root. A non-root vertex has one neighbour inward and $z-1$ neighbours outward, so equal probabilities for its $z$ edges giveLet $r$ be radial distance measured in edge units. For $r>1$, probability arriving in shell $r$ comes from shell $r-1$ by an outward step or from shell $r+1$ by an inward step:This has precisely the same bulk master equation as the one-dimensional biased random walk, withThe outward bias is geometric: beyond the first shell, the number of vertices is $M_r=z(z-1{)}^{r-1}$, so there are more outward than inward choices even though no edge direction is dynamically preferred. $P(r,N)$ is the probability summed over an entire shell, not the probability at one specified vertex; no additional shell-size factor belongs in the boxed recurrence.

The equality concerns the evolution equation away from the root, not equality to the unrestricted full-line binomial distribution at all positions. The root imposes a different boundary rule.

Farther than a few edge lengths from the root, replace the radial coordinate $r$ by a physical coordinate $x=ar$ and smooth the master equation. It has the same constant-coefficient Fokker-Planck equation as in (b). Under the conventional first-time-derivative truncation,The radial drift $v=-C=a(z-2)/(z\tau )$ is outward. For $z=2$ the tree is the infinite line, its radial distance is an unbiased reflected walk, and $C=0$.

The root cannot absorb probability or allow negative radial distance. Write the equation as a continuity equation, using the Fokker-Planck probability currentThe appropriate continuum boundary condition is reflecting, or zero flux:It is not generally ${\partial }_{x}P(0,t)=0$: that simpler condition is correct only when $C=0$. With the finite-bias fixed-step Gaussian convention from (b), use $D=D_{\mathrm{fixed}}=2a^2(z-1)/(z^2\tau )$ instead; the drift sign and zero-current boundary condition are unchanged. A continuum boundary condition represents the microscopic root region after coarse-graining, rather than reproducing each discrete root probability at individual step times.