Solutions to the example questions on the Repeated games page.
(a) Interpretation. [2]
\(\delta = 5/6\) is the probability that the game continues after each round (the die not showing a 1). The expected number of rounds is \(\dfrac{1}{1 - \delta} = \dfrac{1}{1/6} = 6\).
(b) Discounted payoffs. [6]
Hold the opponent's Grudger fixed and compare the focal player's two options. If the player conforms, the opponent is never triggered, so both play High in every round and the player earns 3 each round. If the player deviates to Low in the first round, they earn 5 that round against the opponent's High, but this trips the opponent's punishment, so both play Low for ever after, earning 1 each round. Hence
(c) Cooperation at \(\delta = 5/6\). [4]
Since \(18 > 10\), no player gains by deviating, so mutual cooperation is sustained.
(d) Threshold. [5]
Our die gives \(\delta = 5/6\), well above \(\tfrac{1}{2}\), so it made cooperation easy to sustain: the future cast a long enough shadow.
(e) A forgiving trigger. [7]
Now a one-off Low is punished by only two rounds of mutual Low before both return to High. A player who conforms still earns \(3\) every round, so \(U_{\text{conform}} = \dfrac{3}{1 - \delta}\). A player who deviates to Low in the first round earns \(5\) at once, then \(1\) in each of the two punishment rounds, and \(3\) for ever after:
Writing \(U_{\text{conform}} = 3 + 3\delta + 3\delta^{2} + \dfrac{3\delta^{3}}{1 - \delta}\), the tails cancel and
Cooperation is sustained when this is non-negative, that is when \(\delta^{2} + \delta - 1 \ge 0\), i.e. \(\delta \ge \dfrac{\sqrt{5} - 1}{2} \approx 0.618\). At \(\delta = 5/6\),
so our die still sustains cooperation. The threshold has risen from \(\tfrac{1}{2}\) under Grudger to \(\tfrac{\sqrt{5} - 1}{2} \approx 0.618\): shortening the punishment from a permanent grudge to two rounds weakens the deterrent, so the players must be more patient to keep cooperating.
(a) Definitions. (Bookwork.) [4]
(b)(i) Discounted payoffs. [3]
Hold the opponent's Grudger fixed. Conforming keeps both on mutual cooperation at 3 a round; defecting once earns 5 and then trips the opponent's punishment, after which both defect at 1 a round. Hence
(b)(ii) Smallest \(\delta\). [3]
(b)(iii) Subgame perfection. [5]
Grudger is subgame perfect when no player can gain by deviating in any subgame. There are two kinds of subgame. On the cooperative path, the condition is exactly the no-deviation inequality of part (ii), so it holds for \(\delta \ge \tfrac{1}{2}\). In the punishment phase, both players defect for ever, which is the stage Nash equilibrium repeated; a one-off deviation to cooperate earns \(0\) now and returns to mutual defection, so it cannot help. Hence for \(\delta \ge \tfrac{1}{2}\) Grudger is a subgame perfect equilibrium.
(c) Folk Theorem. (Bookwork.) [3]
Let \((u_1^{*}, u_2^{*})\) be a pair of Nash equilibrium payoffs of the stage game. For every individually rational pair \((v_1, v_2)\) there exists \(\bar{\delta}\) such that for all \(1 > \delta > \bar{\delta} > 0\) there is a subgame perfect Nash equilibrium with payoffs \((v_1, v_2)\).
(d) Individually rational payoff. [3]
The individually rational payoffs are the average payoff pairs that exceed the stage-game Nash equilibrium payoffs of both players. The stage game's unique Nash equilibrium is mutual defection, paying \(1\) to each player, so a pair is individually rational when each player's average payoff exceeds this stage Nash value of \(1\).
(e) Interpretation. [4]
Mutual cooperation gives each player \(3 > 1\), so the pair \((3, 3)\) exceeds the stage Nash payoff of \(1\) and is individually rational. By the Folk Theorem it can therefore be sustained in a subgame perfect equilibrium when players are patient enough, which matches the threshold \(\delta \ge \tfrac{1}{2}\) found above.
(a) Definition. (Bookwork.) [2]
A strategy profile is a subgame perfect equilibrium of a repeated game if, after every history of play, the strategies restricted to the subgame that follows form a Nash equilibrium of that subgame. Equivalently, no player can gain by deviating in any subgame, not only on the equilibrium path.
(b)(i) Backward induction. [4]
We argue by backward induction. In the last round \(N\) the players face a single play of the stage game, whose unique Nash equilibrium is mutual defection, so any subgame perfect equilibrium defects in round \(N\). Working back, suppose every round after \(k\) defects no matter what is played in round \(k\). Then the continuation payoff is fixed, so a player at round \(k\) maximises only the current stage payoff, where defection strictly dominates; hence both defect in round \(k\) as well. By induction the unique subgame perfect equilibrium defects in every round.
(b)(ii) Why cooperation cannot be sustained. [4]
By part (b)(i) the unique subgame perfect equilibrium plays mutual defection in every round after every history, so its path contains no cooperation. A cooperative agreement cannot be propped up by the threat of future punishment because no such threat is credible: any punishment would have to be carried out in a later subgame, yet every later subgame has mutual defection as its only subgame perfect behaviour, fixed by the last-round argument. A player's continuation payoff is therefore the same whether or not they cooperated, so it can neither reward cooperation nor penalise defection. With the future fixed, defection strictly dominates in the current stage, and cooperation unravels from the final round back to the first.
(c) Why backward induction does not apply to the infinite game. [3]
Backward induction relies on a final subgame to anchor the recursion: it pins down play in the last round, then in the round before, and so on. A finitely repeated game has such a final round \(N\), which is why the argument of part (b) succeeds. An infinitely repeated game has no final round, so the recursion has no base case and cannot begin; there is no last stage from which to reason backwards. This removes the unravelling argument, but does not by itself show that cooperation can be sustained. It only leaves room for strategies such as Grudger, whose threat of indefinite future punishment can deter a deviation when players are patient enough.
(d) Recomputed threshold with \(T = 4\). [4]
Cooperation is sustained when conforming is at least as good as a one-shot deviation against Grudger, that is \(U_{\text{conform}} \ge U_{\text{deviate}}\). With temptation \(T = 4\),
(e) Comment. [3]
Lowering the temptation from \(5\) to \(4\) lowers the threshold from \(\tfrac{1}{2}\) to \(\tfrac{1}{3}\). A smaller temptation makes a one-shot defection less attractive relative to staying on the cooperative path, so cooperation survives for a wider range of discount factors. The dependence is monotone: the threshold rises with the temptation payoff \(T\), as part (f) makes precise.
(f) General threshold. [5]
With \(R = 3\), \(P = 1\), \(S = 0\) and general temptation \(T\), the no-deviation condition against Grudger is
Multiplying through by \(1 - \delta > 0\) preserves the inequality and gives \(3 \ge T(1 - \delta) + \delta = T - \delta(T - 1)\), hence \(\delta(T - 1) \ge T - 3\). Since \(T > 1\) we have \(T - 1 > 0\), so dividing by \(T - 1\) keeps the direction:
Writing \(\delta^{*} = 1 - \dfrac{2}{T - 1}\) makes the behaviour transparent: as \(T\) grows, \(\dfrac{2}{T - 1}\) shrinks, so \(\delta^{*}\) increases towards \(1\). The greater the temptation, the more patient the players must be to sustain cooperation. For \(T \le 3\) we have \(\delta^{*} \le 0\) and cooperation is sustained for every \(\delta \in (0, 1)\).
(a) Conform and deviate payoffs. [4]
Conforming yields mutual cooperation for ever:
Defecting once earns the temptation payoff now, then \(T\) rounds of mutual defection, then a return to cooperation:
(b) The sustaining condition. [8]
Cooperation is sustained when \(V_{\text{C}} \ge V_{\text{D}}\). Multiplying through by \(1 - \delta\),
Expanding the right-hand side,
Rearranging gives \(4\delta \ge 2 + 2\,\delta^{\,T+1}\), that is
(c) Dependence on the punishment length. [6]
Let \(\varphi_T(\delta) = 2\delta - \delta^{\,T+1}\); the threshold \(\delta^{*}(T)\) solves \(\varphi_T(\delta) = 1\). For fixed \(\delta \in (0, 1)\) the term \(\delta^{\,T+1}\) decreases as \(T\) increases, so \(\varphi_T(\delta)\) increases with \(T\). A larger \(T\) therefore satisfies the inequality at a smaller \(\delta\), so \(\delta^{*}(T)\) is decreasing in \(T\). As \(T \to \infty\), \(\delta^{\,T+1} \to 0\) and the condition becomes \(2\delta \ge 1\), giving \(\delta^{*} \to \tfrac{1}{2}\). This is exactly the grim-trigger threshold, in which the punishment lasts for ever.
(d) Subgame perfection. [7]
By the one-shot deviation principle it suffices to rule out a profitable single deviation in each type of subgame, assuming conformity thereafter.
On the cooperative path, a one-shot defection gives the payoff \(V_{\text{D}}\) of part (a) against the continuation \(V_{\text{C}}\); the calculation of part (b) shows \(V_{\text{C}} \ge V_{\text{D}}\) exactly when \(2\delta - \delta^{\,T+1} \ge 1\), so no such deviation pays above the threshold.
During the punishment phase both players are due to defect. A one-shot deviation to cooperate earns \(S = 0\) this round instead of \(P = 1\), and leaves the remaining count and all future play unchanged, since the strategy ignores behaviour within the punishment phase. The deviation loses \(1 - 0 = 1\) now and gains nothing later, so it is never profitable.
No profitable one-shot deviation exists in any subgame, so the forgiving trigger is subgame perfect for \(\delta \ge \delta^{*}(T)\).
Marking exercise 1 (Question 2(b)(i) and (ii)).
The deviator's payoff stream is wrong: after the deviation both players defect for ever, and mutual defection pays \(P = 1\) per round, not nothing. The deviating payoff is
as in the solution above. The comparison then reads \(3 \ge 5(1 - \delta) + \delta\), giving the threshold \(\delta \ge \tfrac{1}{2}\), not \(\tfrac{2}{5}\). Dropping the punishment stream always understates the deviator's payoff and so understates the patience cooperation needs; the final \(\tfrac{2}{5}\) is confidently wrong. A fair mark is [1] of [3] for (i), where the conforming payoff is correct, and [1] of [3] for (ii), where the method of comparing the two streams is right but both the stream and the threshold are wrong.
Marking exercise 2 (Question 2(c) and (e)).
The statement drops both hypotheses of the Folk Theorem.
The closing flourish about real life inherits both errors. A fair mark is [1] of [3] for (c) and [1] of [4] for (e).