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322 CHAPTER 10. TARGET-ZONE MODELSthe event that it loses a dollar on a particular intervention. 8 In theÞrst round, the probability that f hits ¯f is 1 2 .Thatis,P(Gc )= 1.By2implication, P(G) =1− P(G c )= 1 . It follows that before the Þrst2round starts, the probability that reserves eventually get driven to zeroisPr(L) = 1 2 Pr(L|G)+1 2 Pr(L|Gc ). (10.47)(10.47) true before the Þrst round and is true for any round as long asthe authorities still have at least one dollar in reserves.Let p j be the conditional probability that reserves eventually become0 given that the current level of reserves is j-dollars. For any j ≥ 1,(10.47) can be expressed as the difference equationp j = 1 2 p j+1 + 1 2 p j−1, (10.48)with p 0 =1. 9 Backward substitution gives p 2 =2p 1 − 1, p 3 =3p 1 − 2,p k = kp 1 − (k − 1),...,orequivalently,fork ≥ 2,p k =1− k(1 − p 1 ). (10.49)Since p k is a probability, it cannot exceed 1. Upon rearrangement yougetp 1 =1+ p kk − 1 → 1, as k →∞. (10.50)kbut if p 1 = 1, the recursion in (10.49) says that for any j ≥ 1, p j =1.Translation? It is a sure thing that any Þnite amount of reserves willeventually be exhausted.10.6 Imperfect Target-Zone CredibilityThediscreteinterventionruleismorerealisticthantheinÞnitesimalmarginal intervention rule. But if reserves run out with probability 1,8 G is the event that f hits f, andG c is the event that f hits ¯f.9 Clearly, p 0 = 1 since if j = 0, reserves have been exhausted. If j = 1, thereis a probability of 1 2that reserves are exhausted on the next intervention and aprobability of 1 that the central bank gains a dollar and survives to play again2at which time there will be a probability of p 2 that reserves will eventually beexhausted. That is, for j = 1, p 1 = 1 2 p 0 + 1 2 p 2. Continuingoninthisway,youget(10.48).