Bargaining, Commitment, and Repeated Cooperation

The game has a unique subgame-perfect equilibrium. Agreement is immediate, and the proposer obtains $(1-{\delta }_j)/(1-{\delta }_1{\delta }_2)$, where $j$ is the responder: at the base values player 1 obtains $\frac{5}{7}$ when it proposes first and player 2 obtains $\frac{5}{14}$ when it does. In the $T$-period game the first proposer's equilibrium share $x_T$ satisfies $|x_T-\frac{5}{7}|\le ({\delta }_1{\delta }_2{)}^{\lfloor (T-1)/2\rfloor }$. With ${\delta }_i=1/(1+r_i\Delta )$, player 1's share as first proposer is $r_2(1+r_1\Delta )/(r_1+r_2+r_1{r}_2\Delta )$, which decreases to $r_2/(r_1+r_2)=\frac{9}{13}$ as $\Delta \to 0$: the division of the asymmetric Nash bargaining solution with weights $1/r_1:1/r_2$.

With commitment the proposer asks $1$, earns $\frac{2}{3}$, and agreement fails with probability $\frac{1}{3}$; the total surplus is $\frac{2}{3}$ against $\frac{5}{6}$ if every deal were made. With two offers and $0<\delta <1$, write $\hat{p}=(1-\delta )v_H+\delta v_L$.

• After a first offer $p_1\le v_L$ both types accept and the proposer earns $p_1$. After $v_L<p_1<\hat{p}$ the high type accepts at once and the low type buys at $v_L$ next; the proposer earns $\frac{2}{3}{p}_1+\frac{1}{6}\delta$.

• At the boundary $p_1=\hat{p}$ the high type's probability $\alpha$ of accepting at once can be anything in $[\frac{1}{2},1]$ (the last price is then $v_L$); the proposer earns $\frac{1}{2}\delta +\frac{2}{3}\alpha (1-\delta )$.

After $\hat{p}<p_1<1$ the high type accepts with probability $\frac{1}{2}$, which leaves the belief at $v_L/v_H=\frac{1}{2}$ after a rejection; the proposer then asks $1$ with probability $1-2(1-p_1)/\delta$, keeping the high type indifferent, and earns $\frac{1}{3}{p}_1+\frac{1}{3}\delta$.

At $p_1=1$, $\alpha$ can be anything in $[0,\frac{1}{2}]$ (last price $1$), and the proposer earns $\frac{2}{3}(\delta +\alpha (1-\delta ))$.

Above $1$ both types reject and the proposer earns $\frac{2}{3}\delta$.

• The proposer screens (at $\hat{p}$, with $\alpha =1$: $\frac{2}{3}-\frac{1}{6}\delta$) if $\delta <\frac{2}{3}$ and holds out (at $1$, with $\alpha =\frac{1}{2}$: $\frac{1}{3}+\frac{1}{3}\delta$) if $\delta >\frac{2}{3}$; for $0<\delta <1$, at the optimum the upper endpoint of each boundary family is forced, since otherwise a nearby price in the adjacent open interval would do better (at $\delta =1$ every $\alpha$ at $p_1=1$ earns the same $\frac{2}{3}$). At $\delta =\frac{2}{3}$ both first offers ($\frac{2}{3}$ and $1$) are optimal, earning $\frac{5}{9}$, with total surplus $\frac{7}{9}$ and $\frac{5}{9}$ respectively, and any mixture of the two is also an equilibrium. At $\delta =\frac{9}{10}$ the proposer asks $1$ twice and earns $\frac{19}{30}$; half the high types buy at once and half after a delay; the low type never trades; the total surplus is $\frac{19}{30}$, below the $\frac{2}{3}$ of the committed offer.

Mutual cooperation, enforced by permanent mutual defection after any defection, is a subgame-perfect equilibrium if and only if $\delta \ge (T-R)/(T-P)=\frac{1}{2}$. Every payoff pair that is achievable by some distribution of stage outcomes and gives each player strictly more than its minmax value $P=1$ is a subgame-perfect equilibrium payoff for all $\delta$ close enough to $1$.

Let $K=(p-q)(R-P)-(1-p)(T-R)$. If $K\le 0$, no profile in the class sustains cooperation. If $K>0$, cooperation is sustained if and only if $\delta \ge {\delta }_{\min }=(T-R)/(T-R+K)$; the best profile uses the smallest incentive-compatible punishment probability, ${\pi }^{\ast }=(1-\delta )(T-R)/(\delta K)$, and gives each player

whatever $\delta \ge {\delta }_{\min }$. At the base values $K=\frac{3}{5}$, ${\delta }_{\min }=\frac{10}{13}$, ${\pi }^{\ast }=\frac{10}{27}$ and $V^{\ast }=\frac{5}{2}$, against $3$ under perfect monitoring; cooperation of this kind is impossible once $q\ge \frac{4}{5}$.

In Model B4, consider perfect public equilibria in which both players use the same strategy after every public history, randomising independently, with public lotteries allowed. For every $\delta \in (0,1)$ and every probability $h$ of a good signal after mutual defection, no such equilibrium gives a player more than $V^{\ast }=\frac{5}{2}$. For $\delta \ge \frac{10}{13}$ the trigger profile of Proposition 3.9 (p. 133) attains it, so the optimality of $V^{\ast }$ extends from the trigger class to the whole strongly symmetric class.

The agreed shares are $(4,4)$ at the start and $(\frac{7}{4},\frac{17}{4})$ of what remains at the second stage. The unique schedule that pays out the agreed shares and leaves each party, at the second stage, exactly the solution of the remaining game — Petrosyan's imputation distribution procedure — pays $(\frac{9}{4},-\frac{1}{4})$ in the first stage and $(\frac{7}{4},\frac{17}{4})$ in the second: player 2 must pay $\frac{1}{4}$ at the start. The naive schedule that splits each stage's payoff equally gives player 2 $3$ in the second stage, less than the $\frac{7}{2}$ it can get by leaving. If player 2's second-stage disagreement value is $\frac{5}{2}$ instead, the time-consistent payments are $(\frac{7}{4},\frac{1}{4})$ and $(\frac{9}{4},\frac{15}{4})$, all nonnegative. Participation alone does not force the advance payment: with the original outside options the nonnegative schedule $(\frac{7}{4},\frac{1}{4})$ then $(\frac{9}{4},\frac{15}{4})$ pays the totals $(4,4)$ and leaves both parties above their second-stage outside options ($\frac{9}{4}>1$, $\frac{15}{4}>\frac{7}{2}$). What it violates is time consistency: its second-stage payment is not the recomputed Nash solution $(\frac{7}{4},\frac{17}{4})$.

$u(p)=\min \{4p,4(1-p),|4p-2|\}$ is piecewise linear with kinks at $\frac{1}{4},\frac{1}{2},\frac{3}{4}$, with $u(0)=u(\frac{1}{2})=u(1)=0$ and $u(\frac{1}{4})=u(\frac{3}{4})=1$; hence $\mathrm{cav}u(\frac{1}{2})=1>u(\frac{1}{2})=0$. Player 1 guarantees a long-run average of $1$ by a one-time signal that moves the uninformed player's belief to $\frac{1}{4}$ or $\frac{3}{4}$, each with probability $\frac{1}{2}$, and by then playing optimally in the corresponding average game; ignoring the information gives $0$, and so does revealing the state, after which the game at belief $0$ or $1$ is worth $0$. In a single round, using the information fully is worth $2$: the informed player's unique optimal plan is the top row in state 1 and the bottom row in state 2, and the minimiser may use any mixture of its three columns. Over $T$ rounds the splitting strategy loses at most a first-round amount, so its average falls short of $1$ by $O(1/T)$, not by zero. By the Aumann–Maschler theorem the value of the infinitely repeated game is $\mathrm{cav}u(\frac{1}{2})=1$.