◄ WORLD V · SONNY 5DART 205 · a helldive at the net

THE HITTING TIME Kac: the mean return time is one over its share

How long until the chain first reaches a target state? First-passage times solve a linear system, just like absorption times. And Kac's lemma ties it back to the stationary distribution with startling economy: the mean time to return to a state i is exactly 1/πᵢ. The rarer the state — the smaller its long-run share — the longer the wait; a state visited a sixth of the time is returned to every six steps.

THE TECHNIQUE first-passage = linear solve; mean return to i = 1/πᵢ (Kac)

Compute first-passage times by relaxation, then check Kac's lemma directly: the mean return time to each state equals one over its stationary probability, so the rare state waits proportionally longer: live demo







  
HISTORY & CREDIT Mark Kac, 1947

  
“A frequently-visited state and a rare one take similar times to return to.” — no; the relationship is exact and inverse. Kac's lemma: mean return time = 1/(stationary probability). Halve a state's long-run share and you exactly double its expected return time. cited

  
1890 · Poincare recurrence — a measure-preserving system almost surely returns arbitrarily close to its start (the Acta Mathematica three-body memoir); but how long is left open.
1947 · Mark Kac — his recurrence lemma answers it: the mean return time to a set is one over the measure of that set.
now · first-passage and return times drive queueing, reliability, and the analysis of every MCMC sampler.

  
Kac closes the loop with the stationary distribution: π is not just where the chain lives long-run — its reciprocal is literally how often, in steps, each state comes back around. Kac 1947






  
RECOMMEND FOR I-13 return time = 1/π verified, computed

  
On the canonical compiler, first-passage times for the weather chain, fed into the return-time formula, land exactly on 1/π — the common state returns in 1.2 steps, the rare one in 6:

  
$ i13 run kac.i13   # weather chain, pi=[5/6,1/6]
first passage 1->0 = 2         first passage 0->1 = 9.999999999999995
mean return to state 0 = 1.2               (1/pi0 = 1 / (5/6) = 1.2)
mean return to state 1 = 5.999999999999997 (1/pi1 = 1 / (1/6) = 6)

  
Recommend: the hitting time is LIT for I-13 — verified first-passage 1→0 = 2 and 0→1 = 10 by relaxation, and Kac's lemma exactly: mean return to the common state (share 5/6) is 1.2 steps, to the rare state (share 1/6) is 6 — each the reciprocal of its stationary probability. The rare state waits exactly inversely to how often it is seen.