Solution
Asymmetric War of Attrition
Show the problem again
Two firms contest a prize. Firm A values it at 6 and Firm B values it at 12. Time is continuous. Each firm pays 1 per hour for as long as both remain in the contest. The first firm to quit loses (it has paid only its waiting costs), and the other firm receives its prize value at that moment, net of the waiting costs it has paid. Each firm chooses a quitting time, simultaneously and in advance. Consider the Nash equilibrium in which both firms use continuous quitting-time distributions with positive density on all of [0, ∞) (so neither quits immediately with positive probability). What is the probability that Firm A, the one with the lower valuation, wins? Give a reduced fraction.
Worked solution
The answer is 2/3. Suppose Firm j quits at time t with hazard rate h_j (its chance of quitting in the next dt, given it is still in, is h_j·dt). Firm i, still in at time t, compares quitting now with waiting dt more: waiting costs 1·dt and wins its value v_i with probability h_j·dt. Being willing to quit at any time requires h_j·v_i = 1, so h_j = 1/v_i. Hence Firm B quits at rate 1/6 (to keep A indifferent) and Firm A quits at rate 1/12 (to keep B indifferent): the firm with the higher value quits faster. Both quitting times are then independent exponentials, and A wins exactly when B quits first, which has probability (1/6) / (1/6 + 1/12) = 2/3. The expected contest length is 1/(1/6 + 1/12) = 4 hours.
Source: War of attrition after Maynard Smith (1974); asymmetric-value equilibrium as in Hendricks, Weiss and Wilson (1988). Statement written for AxiomIQ.