Which gives you the result above when you multiply it by (1 − δ) to average it. I have not gone senile. The reason for keeping the multiplication by 0 above is just to let you see clearly how you would calculate it if the payoff from (C, D) was something else.) Deviation will not be profitable when 2 ≥ 3/(1 + δ), or whenever δ ≥ 1 /2. 2. The last realization was (C, D). If player 1 follows TFT, the resulting sequence of outcomes will be (D,C),(C,D),(D,C),..., to which the payoff (as we just found out above) is 3/(1 + δ). If player 1 deviates and cooperates, the sequence will be (C,C),(C,C),(C,C),..., to which the payoff is 2. So, deviating will not be profitable as long as 3/(1 + δ) ≥ 2, which means δ ≤ 1 /2. (We are already in hot water here: Only δ = 1 /2 will satisfy both this condition and the one above!) 3. The last realization was (D, C). If player 1 follows TFT, the resulting sequence of outcomes will be (C,D),(D,C),(C,D),.... Using the same method as before, we find that the payoff to this sequence is 3δ/(1+δ). If player 1 deviates, the sequence of outcomes will be (D,D),(D,D),(D,D),..., to which the payoff is 1. Deviation will not be profitable whenever 3δ/(1 + δ) ≥ 1, which holds for δ ≥ 1 /2. 4. The last realization was (D, D). If player 1 follows TFT, the resulting sequence of outcomes will be (D,D),(D,D),(D,D),..., to which the payoff is 1. If he deviates instead, the sequence will be (C,D),(D,C),(C,D),...,towhichthepayoffis 3δ/(1 + δ). Deviation will not be profitable if 1 ≥ 3δ/(1 + δ), which is true only for δ ≤ 1 /2. It turns out that for (TFT,TFT) to be subgame perfect, it must be the case that δ = 1 /2, a fairly hefty knife-edge requirement. In addition, it is an artifact of the way we specified the payoffs. For example, changing the payoff from (D, C) to 4 instead of 3 results in the payoff for the sequence of outcomes (D,C),(C,D),(D,C),... to be 4/(1 + δ). To prevent deviation in case (a), we now want 2 ≥ 4/(1 + δ), which is only true if δ ≥ 1, which is not possible. So, with this little change, the SPE disappears. For general parameters, (TFT,TFT) is not subgame perfect, except for special cases where it may be for some knife-edge value of δ. Therefore, TFT is not all that is so often cracked up to be! • (TFT,GRIM) We first check the optimality of player 1’s strategy and then the optimality of player 2’s strategy. 1. No deviations have occurred in the past. If player 1 follows TFT, the result is perpetual cooperation, and the payoff is 2. If player 1 deviates, the resulting sequence of outcomes is (D, C), (C, D), (D,D),.... The payoff is (1 − δ)[3 + ∞ t=2 δt ] = 3 − 3δ + δ2 . Deviation is not profitable whenever 1 − 3δ + δ2 ≤ 0, which is true for any δ ≥ (3 − √ 5)/2. Similarly, player 2’s payoff from following GRIM is 2. If player 2 deviates, the sequence is (C,D),(D,D),(D,D),....Thepayoffis(1−δ)[3 + ∞ t=1 δt ] = 3 − 2δ. This is not profitable for any δ ≥ 1 /2. 2. Player 2 has deviated in the previous period. If player 1 follows TFT, the sequence of outcomes is perpetual defection, from which the payoff is 1. If player 1 deviates, 18
the sequence is (C,D),(D,D),..., for which the payoff is δ
Change language