Why Zero Needs Its Own Equation
For a birth–death process, the states are usually nonnegative integers and transitions change the state by one. Immigration supplies arrivals even when the population is zero. But a death from zero is impossible: there is no state \(-1\). This small boundary fact changes the balance equation at zero. Applying an interior equation there as if the chain could move down from zero can produce incorrect stationary probabilities.
Let \((X_t)_{t\ge0}\) be a continuous-time Markov chain on \(S=\{0,1,2,\ldots\}\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). From state \(n\), let the chain move to \(n+1\) at rate \(\lambda_n\), and, for \(n\ge1\), to \(n-1\) at rate \(\mu_n\). Assume \(\lambda_n>0\) and \(\mu_n>0\) for the infinite irreducible model; a finite-capacity model may instead have \(\lambda_N=0\) at its upper boundary \(N\). Set \(\mu_0=0\) as notation for the absent downward rate, not as a transition to another state.
A probability vector \(\pi=(\pi_0,\pi_1,\ldots)\) is a candidate stationary distribution if its coordinates are nonnegative and sum to one. When the stationary forward equations apply, the derivative of the probability of state \(n\) is the total rate flowing into \(n\) minus the total rate flowing out. A stationary vector must make each such derivative zero. For unbounded rates, applying these equations requires the relevant forward-equation and integrability conditions; the algebra below concerns vectors for which the coordinate equations are valid.
The Boundary and Interior Balance Equations
At state zero, probability flows in from state one by a death, at rate \(\mu_1\), and flows out to state one by an arrival, at rate \(\lambda_0\). There is no incoming flow from \(-1\). Hence the boundary equation is
For an interior state \(n\ge1\), probability flows in from \(n-1\) by a birth and from \(n+1\) by a death. It flows out by a birth or death from \(n\). Therefore
These equations distinguish the boundary from the interior. They are not obtained by pretending that state \(-1\) exists. At a finite upper boundary \(N\), there is also no transition from \(N\) to \(N+1\); with \(\lambda_N=0\), the top equation becomes \(\pi_N\mu_N=\pi_{N-1}\lambda_{N-1}\).
Proof. Define the net upward probability flux across the edge between \(n\) and \(n+1\) by \(J_n=\pi_n\lambda_n-\pi_{n+1}\mu_{n+1}\). The boundary equation gives \(J_0=0\). For \(n\ge1\), subtracting the interior equation into this expression gives \(J_{n-1}-J_n=\pi_{n-1}\lambda_{n-1}+\pi_{n+1}\mu_{n+1}-\pi_n\mu_n-\pi_n\lambda_n=0\). Thus \(J_n=J_{n-1}\). Starting from \(J_0=0\), induction gives \(J_n=0\) on every existing edge. This proves the forward implication. Conversely, if every \(J_n=0\), then the boundary equation is \(J_0=0\), and for each interior state \(n\), \(J_{n-1}-J_n=0\) rearranges to its interior equation. At a finite upper boundary, the last edge has no outgoing birth from \(N\), and the same rearrangement gives the top equation. \(\square\)
This result explains why the usual neighboring-state detailed-balance ratios are valid here when the balance equations hold: in a one-dimensional birth–death chain, the boundary forces the flux on every edge to be zero. The boundary itself must still be written correctly. The theorem does not say that any arbitrary recurrence obtained by extending an interior formula is valid.
Worked Examples
Worked Example: A Finite-Capacity Chain
Consider states \(S=\{0,1,2\}\). Immigration occurs at rate \(2\) in states \(0\) and \(1\), there is no birth from state \(2\), and deaths occur at rate \(3\) from states \(1\) and \(2\). The generator is
Check the rows before using the generator: \(-2+2+0=0\), \(3-5+2=0\), and \(0+3-3=0\). The balance equations are the boundary equation \(2\pi_0=3\pi_1\), the state-one equation \(5\pi_1=2\pi_0+3\pi_2\), and the top-boundary equation \(3\pi_2=2\pi_1\).
The edge ratios give \(\pi_1=\tfrac23\pi_0\) and \(\pi_2=\tfrac23\pi_1=\tfrac49\pi_0\). Normalization yields \(\pi_0+\pi_1+\pi_2=\pi_0(1+\tfrac23+\tfrac49)=\tfrac{19}{9}\pi_0=1\). Therefore \(\pi_0=\tfrac9{19}\), \(\pi_1=\tfrac6{19}\), and \(\pi_2=\tfrac4{19}\).
Check: The boundary gives \(2\cdot\tfrac9{19}=\tfrac{18}{19}=3\cdot\tfrac6{19}\). The state-one equation gives \(5\cdot\tfrac6{19}=\tfrac{30}{19}\), while \(2\cdot\tfrac9{19}+3\cdot\tfrac4{19}=\tfrac{18}{19}+\tfrac{12}{19}=\tfrac{30}{19}\). The top equation gives \(3\cdot\tfrac4{19}=\tfrac{12}{19}=2\cdot\tfrac6{19}\). Also \((9+6+4)/19=1\). Thus \(\pi Q=0\); for this finite-state chain, \(\pi e^{tQ}=\pi\), as follows by expanding the matrix exponential and using \(\pi Q=0\).
Worked Example: The False Equation at Zero
Use the same finite-capacity chain to see what goes wrong if the interior departure rate is incorrectly applied at zero. The true death rate there is \(\mu_0=0\). An erroneous extension might give zero an outgoing rate \(2+3\), as though a death from zero were possible. If one also sets the nonexistent incoming probability \(\pi_{-1}\) to zero, this false equation becomes \(5\pi_0=3\pi_1\).
That equation gives \(\pi_1=\tfrac53\pi_0\), whereas the actual boundary equation gives \(\pi_1=\tfrac23\pi_0\). The false ratio is not a small adjustment: it assigns the boundary a nonexistent downward departure. For a direct check, the correct vector from the preceding example has \(5\pi_0=\tfrac{45}{19}\), but \(3\pi_1=\tfrac{18}{19}\), so it fails the false equation. It does satisfy the actual equation \(2\pi_0=3\pi_1=\tfrac{18}{19}\).
The source of the error is treating \(\mu_0\) as if it were the positive interior death rate \(3\). The notation \(\mu_0=0\) records that the transition is absent; the boundary departure rate is only \(\lambda_0=2\). Nor should one add a probability \(\pi_{-1}\): \(-1\notin S\), so it is not a state with a probability coordinate.
Worked Example: Immigration with Linear Death Rates
Consider the immigration–death process on \(S=\{0,1,2,\ldots\}\), with constant immigration rate \(\lambda_n=2\) and death rate \(\mu_n=n\) for \(n\ge1\). The generator row at zero has entries \((-2,2)\), whose sum is zero. For \(n\ge1\), its only nonzero entries are \(q_{n,n-1}=n\), \(q_{n,n}=-(n+2)\), and \(q_{n,n+1}=2\), whose sum is \(n-(n+2)+2=0\).
The boundary equation is \(2\pi_0=\pi_1\). On the edge between \(n\) and \(n+1\), the zero-flux relation is \(2\pi_n=(n+1)\pi_{n+1}\). Starting with \(\pi_0\), this gives \(\pi_n=\pi_0\,2^n/n!\). The weights sum to \(\pi_0 e^2\), so normalization gives \(\pi_0=e^{-2}\), hence \(\pi_n=e^{-2}2^n/n!\).
Check: At zero, \(2\pi_0=2e^{-2}\) and \(\pi_1=e^{-2}2^1/1!=2e^{-2}\). At state one, the outgoing flow is \((2+1)\pi_1=6e^{-2}\); the incoming flow is \(2\pi_0+2\pi_2=2e^{-2}+2(e^{-2}2^2/2!)=6e^{-2}\). At state two, the outgoing flow is \((2+2)\pi_2=8e^{-2}\); the incoming flow is \(2\pi_1+3\pi_3=4e^{-2}+3(e^{-2}2^3/3!)=8e^{-2}\). The boundary and the first two interior equations agree with the formula, and the weights normalize because \(e^{-2}\sum_{n\ge0}2^n/n!=1\).
How to Check a Boundary Calculation
When constructing balance equations, begin with the actual transitions out of each state, rather than substituting into an interior formula. At zero, list only the transition to one. At a finite upper boundary, list only the transition to the state immediately below. Then write incoming flow from the existing neighbors and equate incoming and outgoing rates. This procedure makes the state space visible at every step and prevents a fictitious transition from entering the calculation.
A useful independent check is to compute the flux on each existing edge: \(\pi_n\lambda_n-\pi_{n+1}\mu_{n+1}\). The boundary equation forces the first flux to be zero, and the interior equations force successive fluxes to agree. If a proposed set of weights gives nonzero flux on an edge, it cannot satisfy these balance equations. Finally, normalization is a separate requirement: nonnegative weights that solve the equations are a probability distribution only if their sum is finite and can be scaled to one.
Common mistake. Extending \(\pi_n(\lambda_n+\mu_n)=\pi_{n-1}\lambda_{n-1}+\pi_{n+1}\mu_{n+1}\) to \(n=0\) silently imports the positive interior death rate into the boundary. For the finite example, this yields \(5\pi_0=3\pi_1\), hence \(\pi_1=\tfrac53\pi_0\). The correction is to write the boundary flows directly: \(\pi_0\lambda_0=\pi_1\mu_1\), so \(2\pi_0=3\pi_1\) and \(\pi_1=\tfrac23\pi_0\). The rate \(\mu_0\) is zero, and there is no state \(-1\).
Check Your Understanding
For each question, use only transitions that exist in the stated state space.
- A chain has \(\lambda_0=4\) and \(\mu_1=2\). Write the balance equation at zero and solve for \(\pi_1/\pi_0\), assuming \(\pi_0>0\).
- For an interior state \(n\ge1\), write the incoming and outgoing terms in its balance equation.
- In a finite-capacity chain with top state \(N\), what is the balance equation at \(N\) when \(\lambda_N=0\)?
- Define the net flux \(J_n\) across edge \(n,n+1\). Explain how the boundary and interior equations imply \(J_n=0\).
- Why is setting \(\pi_{-1}=0\) not a valid substitute for deriving the equation at zero from the transitions?