Invariant measures and positive recurrence
Normalize balance measures only when their total mass is finite and connect stationary probabilities to return cycles.
On this page
The question for today
What you’ll understand
- Distinguish invariant measures from stationary probability distributions.
- State the positive-recurrence criterion for an irreducible countable chain.
- Normalize a birth-and-death balance measure and compute a mean return time.
The equation νP=ν is a balance condition, but its solutions need not be probability laws. On an infinite state space, a perfectly valid invariant measure can have infinite total mass and therefore cannot describe an equilibrium distribution. Positive recurrence is the condition that turns return cycles into a normalized stationary law. This lesson keeps invariance, normalization and convergence separate: aperiodicity is unnecessary for the existence of a stationary probability, while it matters for convergence of the distribution from an individual starting state.
Balance does not perform normalization
An invariant measure assigns a nonnegative finite weight to each state and satisfies incoming balance at every state. Its total mass may be infinite. A stationary distribution is an invariant measure whose weights sum to one. If the chain starts with that law, its distribution stays the same at every time, even though individual paths continue moving. Multiplying an invariant measure by a positive constant preserves invariance, so uniqueness of an unnormalized invariant measure is normally understood up to a scalar under the relevant recurrence hypotheses.
For a finite irreducible chain, a stationary probability exists, is unique and is positive on every state. For an irreducible countable chain, existence of a stationary probability is equivalent to positive recurrence. In that case it is unique and assigns positive mass everywhere. An irreducible recurrent chain also possesses a nonzero statewise-finite invariant measure unique up to scaling, but in the null recurrent case its total mass is infinite. Transient chains may have invariant measures too, so merely writing down one does not certify recurrence or equilibrium.
A stationary mass is an inverse mean cycle length
For an irreducible positive recurrent countable chain, the stationary probability of x equals the reciprocal of the expected positive return time to x. A cycle beginning at x and ending immediately before its next return contains one visit to x. More generally, the expected number of visits to y during that cycle, divided by the expected cycle length, produces the stationary mass of y. This regenerative interpretation explains both normalization and the obstruction in null recurrence: an infinite mean cycle cannot yield a probability by this finite denominator.
The return formula does not require aperiodicity. A deterministic cycle of three states has stationary mass one third at each state and mean return length three, yet its distribution from one state never approaches that stationary law along all times. For an irreducible, positive recurrent, aperiodic countable chain, the distribution from each fixed initial state does converge in total variation to the stationary law. This is pointwise in the initial state; it need not imply a finite worst-case mixing time when the state space is infinite and starts arbitrarily far away remain far from equilibrium.
Birth-and-death weights expose the infinite tail
For a nearest-neighbor chain on the nonnegative integers, balancing adjacent flows gives a recursive construction of invariant weights. If the upward probability is p and the downward probability away from zero is q, both strictly positive, adjacent weights have ratio p/q, with any boundary holding chosen to make rows sum to one. A ratio below one gives a summable geometric tail. A ratio of one gives constant weights, and a ratio above one gives growing weights. This simple tail calculation can reveal whether a candidate stationary probability exists before solving a large linear system.
Failure of summability distinguishes absence of equilibrium from a missing normalization constant, but it does not by itself distinguish null recurrence from transience. For the reflecting nearest-neighbor chain with equal upward and downward probabilities one half, recurrence follows from the fair-walk hitting calculation and the invariant counting measure has infinite mass, so the chain is null recurrent. With outward drift it is transient. A truncated version of either chain has a stationary law because it is finite; those normalized finite laws can send their mass toward the truncation boundary instead of converging to an infinite-space stationary probability.
Work it through
A reflecting chain with a summable tail
Let P(0,1)=1/3,P(0,0)=2/3. For n≥1, let P(n,n+1)=1/3 and P(n,n−1)=2/3. Find its stationary law and mean positive return time to zero.
Adjacent balance gives ν(n+1)/ν(n)=(1/3)/(2/3)=1/2, so choose ν(n)=2⁻ⁿ.
The geometric sum of these weights is two, hence π(n)=2⁻ⁿ⁻¹. Adjacent flows balance and the self-loop at zero supplies the remaining boundary balance.
The chain is irreducible and the invariant probability proves positive recurrence. Its mass at zero is 1/2, so the mean positive return time is two transitions, including possible one-step returns through the boundary self-loop.
The unique stationary distribution is π(n)=2⁻ⁿ⁻¹, and E₀τ₀⁺=2. The boundary holding also makes this chain aperiodic.
The living notebook
Explore the model
Interactive lab · markov
Stationary mass and mean return length
Pause and Reset, then set p=0.20,q=0.40. Predict stationary P(state 0)=2/3 and P(state 1)=1/3, so their mean positive return times are 3/2 and 3. Step once to see transient P(state 1)=0.20 rather than 1/3. For a fresh comparison Reset and set p=1,q=1; predict stationary masses 1/2 but alternating transient probabilities. The two-state instrument cannot represent an infinite unsummable invariant measure.
- 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 forStationary probabilities determine finite mean return lengths even when periodicity prevents convergence from state 0.
Pause & reason
Check your understanding
Your turn
Try it yourself
In the reflecting chain above, replace the upward probability by 2/3 and downward probability by 1/3, keeping the remaining boundary probability as a hold at zero. Can adjacent balance be normalized? Compare with equal probabilities 1/2.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Set the weight at zero to one and inspect the ratio and sum of successive weights.
Compare with the solution
For outward drift, adjacent weights are ν(n)=2ⁿ, whose total mass diverges, so they cannot be normalized. The chain is irreducible but transient, as the probability of hitting zero from n is (1/2)ⁿ. With equal probabilities the weights are constant and also unsummable. That chain is recurrent by the symmetric-walk hitting argument, hence null recurrent rather than transient. Neither chain has a stationary probability, but their recurrence classifications differ.
Keep the idea
Take it with you
- Invariance is a linear balance condition; probability also requires total mass one.
- Positive recurrence is the countable irreducible criterion for a stationary probability.
- Aperiodicity governs convergence, not existence of the stationary law.
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