advanced30 min studyLesson 652 in Mathematics

Recurrence, transience and hitting times

Distinguish returning almost surely from returning quickly, and solve stopped-chain equations with the correct boundary condition.

On this page

What you’ll understand

  • Separate hitting times from strictly positive return times.
  • Relate recurrence to return probabilities and the Green sum.
  • Solve finite absorption probabilities and mean durations by first-step equations.

A chain may revisit a state forever, return only finitely many times, or return almost surely after a waiting time with infinite expectation. These are different behaviors even when every state communicates with every other. Hitting probabilities and expected hitting times provide a concrete way to distinguish them. They also require care about boundaries: an equation obtained by conditioning on the first step can admit extra solutions on an infinite state space. The probabilistically meaningful solution is selected by nonnegativity and minimality, not just by algebraic manipulation of a recurrence.

Returns are events, durations are random variables

The hitting time of A allows time zero, while the return time to x starts at time one. Thus a chain started in x has hitting time of x equal to zero regardless of its later behavior. State x is recurrent when its strictly positive return time is finite almost surely, and transient otherwise. A recurrent state is positive recurrent when that return time also has finite expectation; otherwise it is null recurrent. Recurrence type is shared within an irreducible communicating class, but a reducible chain can contain several different types.

The expected total number of visits to x, including the initial visit when starting there, is the Green sum of the diagonal transition probabilities. If the probability of ever returning is f less than one, successive returns and the strong Markov property give a geometric visit count with expectation one divided by one minus f. If f equals one, the sum diverges. Consequently recurrence is equivalent to divergence of that Green sum. This criterion tests return frequency over the entire future; observing no return during a finite simulation cannot establish transience.

τA=inf⁡{n≥0:Xn∈A},τx+=inf⁡{n≥1:Xn=x},G(x,x)=∑n≥0Pn(x,x)={(1−fx)−1fx<1,∞fx=1.\tau_A=\inf\{n\geq0:X_n\in A\},\qquad\tau_x^+=\inf\{n\geq1:X_n=x\},\qquad G(x,x)=\sum_{n\geq0}P^n(x,x)=\begin{cases}(1-f_x)^{-1}&f_x<1,\\\infty&f_x=1.\end{cases}
Here fₓ is the probability, starting at x, of a strictly positive return. Infinite Green sum characterizes recurrence, not finite mean return time.↔ Scroll wide equations horizontally.

First-step equations need their boundary data

For a target set A, the hitting probability h(x) equals one on A and is harmonic outside A: average h over the next-state distribution. The mean hitting time g is zero on A and satisfies a similar equation with an additional one for the transition just taken. These statements hold in the extended nonnegative sense when expectations can be infinite. In a finite chain that reaches A almost surely, the killed transition matrix on the complement has spectral radius less than one, and the associated finite linear system has a unique finite solution.

On a countable space, the first-step equations alone need not force uniqueness. The actual hitting probability is the smallest nonnegative solution with the stated boundary values, obtained by increasing finite-horizon hitting probabilities. The expected hitting time is likewise the minimal nonnegative solution of its duration equation. Arbitrary harmonic additions can produce algebraic solutions with no probabilistic interpretation. Before using an optional stopping identity, one must also justify its stopping and integrability assumptions; the existence of a formally constant martingale expectation at deterministic times is not enough.

h(x)=∑yP(x,y)h(y) (x∉A),h∣A=1,g(x)=1+∑yP(x,y)g(y) (x∉A),g∣A=0h(x)=\sum_yP(x,y)h(y)\ (x\notin A),\quad h|_A=1,\qquad g(x)=1+\sum_yP(x,y)g(y)\ (x\notin A),\quad g|_A=0
The first equation concerns a probability and the second an expected duration. Minimal nonnegative solutions select the intended quantities on infinite spaces.↔ Scroll wide equations horizontally.

Finite certainty does not extend to finite expectation

For simple symmetric random walk on the integers, the origin is recurrent but its mean positive return time is infinite. A finite interval provides a useful approximation: stop on reaching either endpoint and solve there, then enlarge the interval. A fair walk started at i in the interval from zero to N hits N before zero with probability i/N, while its expected time to either endpoint is i(N−i). This expected duration grows with the interval, so a collection of finite absorption calculations does not automatically supply a uniform mean bound.

For a biased walk with probability p strictly between one half and one of moving right and q=1−p of moving left, a start one unit to the right of zero hits zero with probability q/p. From one unit to the left it hits zero almost surely. Combining these two possibilities after the first transition shows that the return probability from zero is less than one. This example separates a directional escape mechanism from periodicity: both biased and unbiased nearest-neighbor walks have period two, but only the biased walk is transient.

An absorbing fair walk has two different answers

A fair nearest-neighbor walk on {0,1,2,3,4} stops at 0 or 4. Starting at 1, find the probability of absorption at 4 and the expected number of transitions until absorption.

  1. The harmonic hitting equation with h(0)=0,h(4)=1 has solution h(i)=i/4, since averaging adjacent values preserves this linear function.

  2. The duration equation with g(0)=g(4)=0 has solution g(i)=i(4−i). Its adjacent average is g(i)−1, which verifies the added-one equation.

  3. At i=1 these give h(1)=1/4 and g(1)=3. At the central state g(2)=4, confirming that mean duration is not the probability of a particular endpoint.

The result

The chance of reaching 4 first is 1/4, whereas the expected stopping duration is three transitions. Both are finite because absorption occurs almost surely in this finite interval.

Explore the model

Interactive lab

Interactive lab · markov

A marginal probability is not a hitting-time distribution

Pause and Reset, then set p=0.25,q=0. Predict P(state 1)=1−(0.75)ⁿ after n Steps; here state 1 is absorbing, so it also equals the probability of having hit 1 by step n. At n=2 predict 0.4375. For a fresh comparison Reset and set p=0.25,q=0.50. After two Steps predict 0.3125: return transitions now make occupancy different from ever having hit. The SVG displays exact marginal probabilities, not sampled hitting times.

A marginal probability is not a hitting-time distributionExact 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. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

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 forOccupancy equals a hitting probability for an absorbing target; allowing exits breaks that identification.

Check your understanding

A recurrent countable-state chain has infinite mean return time to a state. How is that state classified?

Try it yourself

On the integers, a nearest-neighbor walk moves right with probability 3/5 and left with probability 2/5. Using the one-sided hitting probabilities, compute its positive return probability from zero and classify the walk.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Condition on the first transition; a start at +1 and a start at −1 have different probabilities of reaching zero.

Compare with the solution

From +1, the probability of reaching zero is q/p=2/3; from −1 it is one. The return probability is therefore (3/5)(2/3)+(2/5)·1=4/5. It is strictly below one, so zero is transient, and irreducibility makes every state transient. The expected number of visits to zero including the initial visit is 1/(1−4/5)=5. This does not say that every path makes exactly five visits.

Take it with you

  • A return time excludes time zero, unlike a hitting time.
  • Almost-sure return can coexist with infinite mean return time.
  • Infinite-space first-step equations require the minimal nonnegative solution.

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