From Local Balance to a Probability Distribution
On a finite state space, positive weights can always be normalized by dividing by their sum. On an infinite state space, that sum may diverge, so the same procedure may fail. For a birth-death process, detailed balance supplies a natural sequence of candidate weights. The key question is whether those weights have a finite total mass.
Let \((X_t)_{t\ge0}\) be a continuous-time Markov chain on the probability space \((\Omega,\mathcal{F},\mathbb{P})\), with state space \(S=\{0,1,2,\ldots\}\). From state \(n\), the process jumps to \(n+1\) at rate \(\lambda_n\), and, for \(n\ge1\), to \(n-1\) at rate \(\mu_n\). Set \(\mu_0=0\). Assume throughout that all \(\lambda_n\) and \(\mu_n\) for \(n\ge1\) are finite and strictly positive, and that the chain is nonexplosive: with probability one, it makes only finitely many jumps on every bounded time interval. This assumption ensures that its transition probabilities \(P_{ij}(t)=\mathbb{P}(X_t=j\mid X_0=i)\) sum to one over \(j\).
The generator \(Q=(q_{ij})\) has \(q_{n,n+1}=\lambda_n\), \(q_{n,n-1}=\mu_n\) for \(n\ge1\), \(q_{nn}=-(\lambda_n+\mu_n)\) for \(n\ge1\), and \(q_{00}=-\lambda_0\); all other entries are zero. Each row sums to zero. A probability distribution \(\pi=(\pi_n)_{n\ge0}\) is stationary if \(\sum_{i\ge0}\pi_iP_{ij}(t)=\pi_j\) for every \(j\) and every \(t\ge0\).
The denominator in the recursion is \(\mu_{n+1}\), the rate of a jump from \(n+1\) back to \(n\). The weights themselves need not sum to one; they are candidate relative masses. The following result states precisely when they yield a stationary probability distribution.
Proof. First suppose \(Z<\infty\) and set \(\pi_n=w_n/Z\). Then \(\sum_n\pi_n=1\), and the recursion gives \(\pi_nq_{n,n+1}=\pi_{n+1}q_{n+1,n}\). We show this local equality gives stationarity here, using nonexplosion.
Fix states \(i,j\) and a time \(t>0\). A path from \(i\) to \(j\) with exactly \(k\) jumps is described by a sequence \(i=i_0,i_1,\ldots,i_k=j\) of neighboring states and jump times \(0<s_1<\cdots<s_k<t\). Its probability density, with respect to the jump times, is the product of the jump rates along the path times the holding-time factor \[ \exp\!\left(-\sum_{\ell=0}^{k}q_{i_\ell}^{*}(s_{\ell+1}-s_\ell)\right), \] where \(s_0=0\), \(s_{k+1}=t\), and \(q_n^{*}=\lambda_n+\mu_n\) is the total rate out of state \(n\). For \(k=0\), the path is possible only if \(i=j\), and its probability is \(e^{-q_i^{*}t}\). Reversing a path reverses its state sequence and its holding intervals. The holding-time factor is unchanged, because the same states and interval lengths occur in reverse order. Also, multiplying the detailed-balance equalities along the path gives \[ \pi_{i_0}\prod_{\ell=0}^{k-1}q_{i_\ell i_{\ell+1}} = \pi_{i_k}\prod_{\ell=0}^{k-1}q_{i_{\ell+1}i_\ell}. \] Thus the probability mass, weighted by \(\pi_i\), of each path from \(i\) to \(j\) equals the mass of its reversed path, weighted by \(\pi_j\). Reversal also maps the jump-time simplex onto itself. Integrating and summing over paths and over \(k\) gives \(\pi_iP_{ij}(t)=\pi_jP_{ji}(t)\). Nonexplosion ensures that paths with finite numbers of jumps account for all transition probability. Therefore \[ \sum_{i\ge0}\pi_iP_{ij}(t) =\pi_j\sum_{i\ge0}P_{ji}(t) =\pi_j, \] where the last sum is one because the chain is nonexplosive. This proves stationarity and detailed balance.
Now suppose \(\nu\) is any stationary probability distribution. For each \(n\ge0\), consider the bounded indicator function of the finite set \(\{0,\ldots,n\}\). The generator applied to this function is zero except at states \(n\) and \(n+1\): its values there are \(-\lambda_n\) and \(\mu_{n+1}\), respectively. These are finite. Under a stationary distribution, the expected change of this indicator over a time interval is zero. Dividing by the interval length and taking its limit to zero yields \[ 0=-\nu_n\lambda_n+\nu_{n+1}\mu_{n+1}. \] This differentiation is valid because the generator applied to the indicator is bounded; only the two boundary states contribute. Hence \(\nu_{n+1}=\nu_n\lambda_n/\mu_{n+1}\). Recursion gives \(\nu_n=\nu_0w_n\) for every \(n\). If \(\nu_0=0\), the recursion forces every \(\nu_n=0\), contradicting that \(\nu\) is a probability distribution. Thus \(\nu_0>0\), and summing its masses gives \(1=\nu_0Z\). In particular \(Z<\infty\), \(\nu_0=1/Z\), and \(\nu_n=w_n/Z\). This also proves uniqueness. \(\square\)
Computing and Checking the Normalizer
Worked Example: Geometric Weights with Varying Rates
For \(n\ge0\), take \(\lambda_n=\frac{n+1}{2(n+2)}\), and for \(n\ge1\), take \(\mu_n=\frac{n}{n+1}\). These rates are positive and bounded. The generator row at 0 has entries \(q_{00}=-\frac14\), \(q_{01}=\frac14\), so its sum is \(0\). For \(n\ge1\), the entries are \(q_{n,n+1}=\frac{n+1}{2(n+2)}\), \(q_{n,n-1}=\frac{n}{n+1}\), and \(q_{nn}=-\frac{n+1}{2(n+2)}-\frac{n}{n+1}\); their sum is \(0\). Bounded rates ensure nonexplosion: the total rate is at most \(2\), so the number of jumps by time \(t\) is bounded in distribution by a Poisson count of rate \(2\).
For every \(n\ge0\), \(\mu_{n+1}=\frac{n+1}{n+2}\), so \[ \frac{w_{n+1}}{w_n} =\frac{(n+1)/(2(n+2))}{(n+1)/(n+2)} =\frac12. \] Starting from \(w_0=1\), this gives \(w_n=2^{-n}\). The finite geometric sum is \(\sum_{n=0}^{m}2^{-n}=2(1-2^{-(m+1)})\); letting \(m\) increase gives \(Z=2\). Therefore \(\pi_n=2^{-(n+1)}\).
Check: The masses sum to \(\sum_{n\ge0}2^{-(n+1)}=1\). At any edge, \(\pi_n\lambda_n=2^{-(n+1)}\frac{n+1}{2(n+2)}=2^{-(n+2)}\frac{n+1}{n+2}=\pi_{n+1}\mu_{n+1}\), as required.
Worked Example: Constant Weights Do Not Normalize
Take \(\lambda_n=\frac{n+1}{n+2}\) and \(\mu_n=\frac{n}{n+1}\) for \(n\ge1\), with \(\mu_0=0\). The generator row at 0 is \((-1/2,1/2,0,\ldots)\) and sums to zero. For \(n\ge1\), the only nonzero entries are \(q_{n,n+1}=\frac{n+1}{n+2}\), \(q_{n,n-1}=\frac{n}{n+1}\), and the negative of their sum on the diagonal, so every row sums to zero. The total rates are bounded by \(2\), which ensures nonexplosion.
Here \(\mu_{n+1}=\frac{n+1}{n+2}=\lambda_n\), so \(w_{n+1}/w_n=1\) and \(w_n=1\) for all \(n\). The partial sums are \(\sum_{n=0}^{m}w_n=m+1\), which increase without bound. Thus \(Z=\infty\), and the theorem shows there is no stationary probability distribution.
Check: The local equalities do hold: \(w_n\lambda_n=\frac{n+1}{n+2}=w_{n+1}\mu_{n+1}\). But dividing these weights by \(Z\) is impossible, because no finite constant can make the infinitely many unit weights sum to one. Local balance alone is not enough.
Worked Example: Summable Factorial Weights
Let \(\lambda_n=\frac{1}{n+2}\) for \(n\ge0\), and \(\mu_n=1\) for \(n\ge1\), with \(\mu_0=0\). At 0 the generator row has entries \(-\frac12,\frac12\), summing to zero. For \(n\ge1\), its off-diagonal entries are \(\frac{1}{n+2}\) and \(1\), and its diagonal entry is \(-(\frac{1}{n+2}+1)\), so that row also sums to zero. All total rates are at most \(2\), hence the chain is nonexplosive.
The recursion gives \(w_{n+1}=w_n/(n+2)\). Thus \(w_0=1\), \(w_1=1/2\), \(w_2=1/6\), and in general \(w_n=1/(n+1)!\). Consequently \[ Z=\sum_{n=0}^{\infty}\frac{1}{(n+1)!}=e-1, \qquad \pi_n=\frac{1}{(e-1)(n+1)!}. \] The series is finite, for example by comparison with the convergent exponential series.
Check: The normalization is \(\sum_n\pi_n=(e-1)/(e-1)=1\). Also, \(\pi_n\lambda_n=\frac{1}{(e-1)(n+2)!}=\pi_{n+1}\mu_{n+1}\) for every \(n\ge0\).
What Normalization Does—and Does Not—Say
The finite-sum test answers whether these detailed-balance weights can be made into a stationary probability distribution. It does not, by itself, assert that the distribution of \(X_t\) from every initial state converges to \(\pi\) as \(t\) tends to infinity. Stationarity means that starting with distribution \(\pi\) preserves that distribution at every time; convergence from other initial distributions is a separate question.
Nonexplosion is also a genuine condition in the sufficiency argument. Detailed balance compares the masses of finite paths and gives symmetry of the transition probabilities when all paths are accounted for. If explosion is possible, a process may leave the state space after infinitely many jumps in finite time, and its transition probabilities need not sum to one. The earlier path-reversal argument then cannot conclude invariance. The examples above have bounded total rates, which directly rules out this issue; the general criterion here explicitly assumes nonexplosion.
Common mistake: using the wrong death rate. For the first example, the correct ratio at \(n=1\) is \(w_2/w_1=\lambda_1/\mu_2=(1/3)/(2/3)=1/2\). Using \(\lambda_1/\mu_1\) instead gives \((1/3)/(1/2)=2/3\), which is wrong: \(\mu_1\) is the rate from state 1 to state 0, whereas the denominator needed to balance the edge from 1 to 2 is the rate \(\mu_2\) from 2 to 1.
For a new model, first write the rate ratio for each edge, then generate the weights from \(w_0=1\), and only then test whether their series converges. If it does, divide each weight by the same finite sum and verify both normalization and the adjacent-edge balance equations.
Check Your Understanding
Use the recursion and the summability criterion to answer the following.
- If \(w_{n+1}/w_n=1/3\) for every \(n\ge0\), compute \(Z\) and the normalized probabilities.
- If \(w_n=1/(n+1)\), explain why these weights cannot be normalized to a stationary probability distribution.
- For rates \(\lambda_0=1/4\), \(\lambda_1=1/3\), and \(\mu_1=1/2\), \(\mu_2=2/3\), compute \(w_1\) and \(w_2\).
- Why does the detailed-balance equation for the edge between \(n\) and \(n+1\) involve \(\mu_{n+1}\), not \(\mu_n\)?
- In the theorem, what role does nonexplosion play in proving that normalized detailed-balance weights are stationary?