Detailed balance and time reversal
Reverse stationary paths with the correct support and distinguish pairwise flow symmetry from stationarity alone.
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The question for today
What you’ll understand
- Construct the reversed kernel on the support of a stationary law.
- Verify detailed balance through symmetric stationary edge flows.
- Exhibit a stationary chain that is not reversible.
Stationarity keeps the distribution at each time fixed; reversibility additionally makes a stationary movie indistinguishable from the same movie played backward. The distinction is visible in probability currents. A directed cycle can carry a permanent clockwise current while keeping equal mass at every vertex. Detailed balance rules out that current by matching every edge flow with its reverse. Time reversal makes the statement precise and introduces an essential support condition: division by stationary mass is legitimate only where that mass is positive.
Reverse the stationary joint law
Let π be a stationary probability on a finite or countable space. The stationary probability of seeing x followed by y is π(x)P(x,y). Conditioning that same pair in the reverse direction defines P*(y,x) by dividing by π(y). On the support Sπ of π, all denominators are positive. Stationarity makes each reversed row sum to one, and the reversed chain has the same stationary law. States of zero stationary mass do not occur in a stationary movie, so this construction does not determine reverse transitions starting from those states.
The support is closed under positive-probability transitions: if π(x)>0 and P(x,y)>0, stationarity forces π(y)>0. Thus restricting to Sπ loses no stationary path. For an irreducible positive recurrent chain the support is the entire state space. For a reducible chain, one must say which stationary law is being reversed, because different mixtures of closed classes can have different supports. No aperiodicity is required for time reversal; a rotating deterministic cycle has a perfectly meaningful reverse rotation.
Detailed balance is stronger than stationarity
Detailed balance requires π(x)P(x,y)=π(y)P(y,x) for every pair of states in the entire state space. For a candidate probability law, checking only positive-mass pairs is insufficient unless its support is also closed. It says exactly that P*=P there, and summing over x proves stationarity. The converse fails because stationarity balances total inflow and outflow at a vertex, allowing nonzero currents around cycles. A doubly stochastic finite matrix has uniform stationary law, but it need not be symmetric and hence need not be reversible. Testing a single vertex balance equation cannot establish reversibility of an entire chain.
Reversibility extends from pairs to full finite paths by repeatedly applying the balance identity: the probability of a stationary path equals the probability of its reversed path under the same kernel. In operator language, P is self-adjoint for the inner product weighted by π, making its eigenvalues real in the finite case. A real spectrum alone is not a substitute for checking self-adjointness. In later lessons the weighted structure supplies the correct variances, Dirichlet form and spectral-gap formulas; treating a nonuniform reversible kernel as an ordinary symmetric matrix can give the wrong calculation.
Weighted edges offer a constructive method
Choose symmetric nonnegative edge weights c(x,y) with finite positive row totals c(x). The walk P(x,y)=c(x,y)/c(x) has invariant measure c(x). If the sum of all row totals is finite, normalizing gives a reversible stationary probability because each stationary edge flow is c(x,y) divided by that common total. On a finite undirected graph this gives degree-proportional mass, not generally uniform mass. Adding identical holding probability at every state preserves this stationary law and reversibility.
This construction separates geometry from normalization and from speed. Scaling every edge weight by the same factor leaves the discrete kernel unchanged, while adding holds changes transition timing. On an infinite graph, symmetric weights can produce an infinite invariant measure rather than a stationary probability. The reverse-kernel formula based on a probability law should then not be presented as a stationary probability movie unless normalization is possible. Reversible invariant measures remain useful analytically, but their infinite mass does not remove the recurrence questions from the previous lesson.
Work it through
A stationary clockwise current
On {0,1,2} modulo three, hold with probability 1/4 and move clockwise with probability 3/4. Find the stationary law, reversed kernel and stationary probability of the path 0,1,2.
Every column and row sums to one, so the uniform law π(x)=1/3 is stationary.
The reversed kernel holds with probability 1/4 and moves counterclockwise with probability 3/4. Forward flow from 0 to 1 is 1/4, while forward flow from 1 to 0 is zero, so detailed balance fails.
The stationary probability of 0,1,2 is (1/3)(3/4)²=3/16. Under the reversed kernel, the path 2,1,0 has exactly that same probability.
Uniform stationarity coexists with nonreversibility. Reversing the stationary movie reverses its directed current rather than leaving the kernel unchanged.
The living notebook
Explore the model
Interactive lab · markov
Every irreducible two-state chain balances its only edge
Pause and Reset, then set p=0.20,q=0.40. Predict π(0)p=π(1)q=2/15. Step once: P(state 1)=0.20 is a transient mass, so multiplying that number by q is not the stationary edge flow. For a fresh comparison Reset and set p=0.40,q=0.20; predict the stationary masses exchange and the common flow remains 2/15. This two-state model cannot exhibit a nonzero cycle current or a nonreversible irreducible example.
- P(state 1) at step 0
- 0%
- Stationary P(state 1)
- 60%
- Chain behavior
- Monotone convergence
Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forDetailed balance uses stationary masses; instantaneous probabilities from the instrument’s state-0 start are different quantities.
Pause & reason
Check your understanding
Your turn
Try it yourself
Give edge weights 2 on {0,1} and 1 on {1,2}. At each transition hold with probability 1/2 and otherwise choose a neighbor proportionally to its edge weight. Compute π and the stationary flows across both edges.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Normalize weighted degrees first; the holding factor multiplies each nonloop transition by one half.
Compare with the solution
The weighted degrees are 2,3,1, so π=(1/3,1/2,1/6). The nonloop transitions are P01=1/2,P10=1/3,P12=1/6,P21=1/2, with diagonal entries 1/2. Thus π0P01=π1P10=1/6, while π1P12=π2P21=1/12. These equalities verify reversibility and therefore stationarity. Uniform mass would fail because vertex 1 has a larger weighted degree.
Keep the idea
Take it with you
- Time reversal is defined on the positive support of the chosen stationary law.
- Detailed balance forbids stationary cycle currents and implies stationarity.
- Weighted undirected walks are reversible with degree-proportional mass.
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