How Long Does the Rain Last?

Question

Weather follows a two-state Markov chain: a rainy day is followed by another rainy day with probability 0.6, and a sunny day stays sunny with probability 0.9. It is raining today. What is the expected total length of this rainy spell, counting today?

Accepted Answer

Each morning of a rainy spell, rain continues with probability 0.6 and the spell ends with probability 0.4, independent of its history (Markov property). So the length L is geometric on 1, 2, dots with P(L = k) = 0.6^k-1(0.4) and mathbbE[L] = frac10.4 = 2.5 text days. General rule worth memorizing: expected sojourn time in a Markov state with stay-probability p is frac11-p. Cross-check via stationarity: sunny spells average frac10.1 = 10 days, so the long-run rainy fraction should be frac2.52.5 + 10 = 0.2 by renewal-reward. Solving the stationary equations directly: pi_R cdot 0.4 = pi_S cdot 0.1 gives pi_R = 0.2, the two computations agree, a consistency check worth volunteering. Trading relevance: two-state chains are the minimal regime model (calm/volatile, trend/chop), and desks quote expected regime duration as exactly frac11 - p_textstay. The memorylessness is the punchline people miss: being three days into a volatile regime tells you nothing about how much longer it lasts, under this model the expected remaining duration is a fresh 2.5 days every morning. Hazard rates, not elapsed time, drive the forecast; "this has gone on so long it must end soon" is the gambler's fallacy

How Long Does the Rain Last?

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