Why Detailed Balance Implies Stationarity

From Pairwise Balance to Stationarity

Stationarity is a statement about the total probability entering each state after one transition. Detailed balance gives a stronger, pairwise statement: for every pair of states, the probability mass flowing from the first to the second equals the mass flowing in the reverse direction. Summing those pairwise equalities over all possible incoming states gives stationarity. The argument is short, but its hypotheses and the direction of the sum matter, especially when the state space is countable.

Let \((X_n)_{n\ge0}\) be a discrete-time Markov chain on a finite or countable state space \(S\), defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with transition matrix \(P=(p_{ij})\). Each \(p_{ij}=\mathbb{P}(X_{n+1}=j\mid X_n=i)\) is nonnegative, and each row sums to \(1\). Write \(\mu_n\) for the row-vector distribution of \(X_n\). The transition rule gives \(\mu_{n+1}=\mu_nP\). A probability distribution \(\pi\) is stationary if \(\pi P=\pi\), so that starting the chain with distribution \(\pi\) leaves its distribution unchanged after one step.

For a countable state space, the key operation is summing over all \(i\in S\). The terms are nonnegative, so the sum is well-defined; the row-stochastic condition then evaluates the resulting sum. No finite-state assumption is needed.

Proof. Fix any state \(j\in S\). The probability mass at \(j\) after one transition from initial distribution \(\pi\) is \((\pi P)(j)=\sum_{i\in S}\pi(i)p_{ij}\). By detailed balance, each summand equals \(\pi(j)p_{ji}\), so \[ (\pi P)(j)=\sum_{i\in S}\pi(j)p_{ji} =\pi(j)\sum_{i\in S}p_{ji} =\pi(j). \] The last sum is the sum of row \(j\) of \(P\), which equals \(1\). Thus \((\pi P)(j)=\pi(j)\) for every \(j\), proving \(\pi P=\pi\). For countable \(S\), the sums contain only nonnegative terms, and the row sum is \(1\), so the same calculation is valid. \(\square\)

This proof uses the incoming transitions to a fixed state \(j\): it sums over the possible states \(i\) the chain could have occupied before arriving at \(j\). After applying detailed balance, that sum becomes the total of row \(j\), not column \(j\). Confusing those two sums can invalidate the argument.

Using Reversible Weights

Often the candidate distribution is not given in normalized form. Instead, one finds nonnegative weights \(w(i)\) satisfying the balance equations and then checks whether their total is finite. The following result states exactly when normalization works.

Proof. Since \(Z\) is positive and finite, \(\pi(i)=w(i)/Z\) is nonnegative and \(\sum_i\pi(i)=\sum_iw(i)/Z=1\). Dividing the weight balance equation by \(Z\) gives \(\pi(i)p_{ij}=\pi(j)p_{ji}\) for every pair \(i,j\). The theorem therefore gives \(\pi P=\pi\). \(\square\)

In a finite chain, any nonzero collection of finite nonnegative weights has a finite positive sum and can be normalized. On an infinite state space, finiteness of the sum is an additional requirement; balance equations alone do not guarantee a probability distribution.

Why Countable Sums Cause No Extra Problem

The theorem’s proof applies to countably many states because, for a fixed \(j\), the incoming terms \(\pi(i)p_{ij}\) are nonnegative. Detailed balance replaces each of them by \(\pi(j)p_{ji}\); summing then uses the row of transitions out of \(j\), whose total is \(1\). The sum is therefore finite and equals \(\pi(j)\). The argument does not require summing a conditionally convergent series or rearranging signed terms.

A Common Mistake: Checking the Wrong Distribution

A proposed distribution must balance the total incoming probability at every state. Checking only that its entries sum to \(1\), or checking only one incoming transition, is not enough. For example, consider \[ P=\begin{bmatrix}\tfrac12&\tfrac12\\[2pt]\tfrac14&\tfrac34\end{bmatrix}. \] Both rows sum to \(1\). A mistaken calculation might claim that \((\tfrac12,\tfrac12)\) is stationary because it is normalized and the chain has only two states. But the mass entering state \(1\) is \(\tfrac12\cdot\tfrac12+\tfrac12\cdot\tfrac14=\tfrac14+\tfrac18=\tfrac38\), not \(\tfrac12\). In fact, detailed balance also fails: \(\tfrac12\cdot\tfrac12=\tfrac14\), whereas \(\tfrac12\cdot\tfrac14=\tfrac18\).

Solving the balance equation gives \(\pi(1)\cdot\tfrac12=\pi(2)\cdot\tfrac14\), hence \(\pi(2)=2\pi(1)\). Together with \(\pi(1)+\pi(2)=1\), this gives \(\pi=(\tfrac13,\tfrac23)\). Substitution confirms the answer: the first incoming mass is \(\tfrac13\cdot\tfrac12+\tfrac23\cdot\tfrac14=\tfrac16+\tfrac16=\tfrac13\), and the second is \(\tfrac13\cdot\tfrac12+\tfrac23\cdot\tfrac34=\tfrac16+\tfrac12=\tfrac23\).

Detailed balance is a sufficient condition for stationarity: its pairwise equalities imply the required total incoming equalities by summation. The proof should always make that summation explicit, use the row-stochastic condition for the correct row, and verify that the proposed \(\pi\) is a probability distribution.