Trigger Strategy¶

Definition¶

In a repeated game, a trigger strategy for player $i$ prescribes a target action $a_i^c$ (the cooperative phase) as long as the history shows no deviation from the target profile $a^c$ since the cooperative phase began, and switches to a punishment phase once a deviation has occurred. In the punishment phase the players follow a specified punishment path, typically a stage-game Nash equilibrium $a^*$ played either forever (the grim trigger strategy) or for $k$ periods, after which play returns to $a^c$ .

Interpretation¶

A trigger strategy turns the future into a hostage: a player who grabs a one-period gain loses the difference between cooperative and punishment payoffs in later periods. Deviation is unprofitable when the discounted future loss exceeds the immediate gain, which requires enough patience $\delta$ and a punishment that is severe or long enough.

Per-period payoffs of player i in the Prisoner's Dilemma: cooperating gives 3 in every period, while deviating at t gives 5 once, then 1 for two punishment periods, then 3 again.

In the figure, $k=2$ and the payoffs are those of the example below. Deviating at $t$ raises the payoff from $3$ to $5$ , a gain of $2$ , but the two punishment periods pay $1$ instead of $3$ , losses worth $2\delta$ and $2\delta^2$ in period- $t$ terms. The deviation is unprofitable when $2\delta+2\delta^2\ge2$ $2$ , which is the condition $\delta+\delta^2\ge1$ of the example.

Two separate conditions must hold for a subgame-perfect equilibrium. On the cooperative path, nobody should gain by deviating. In the punishment phase, carrying out the punishment must itself be optimal; a threat to play something that is not a best response once the trigger has been pulled is not credible. An absorbing stage-game Nash punishment takes care of the second condition. For finite punishment the continuation rule also matters. In the example below, every deviation restarts all $k$ punishment periods; a stage Nash punishment with payoffs no higher than the target then makes punishment-phase deviations unprofitable.

A trigger strategy reacts to deviations, not to bad luck: it relies on actions being observed. When punishment payoffs are below the target, shorter punishments reduce the future loss and can require a higher $\delta$ ; a sufficiently short punishment may fail for every $\delta<1$ $1$ .

Key Results¶

With a finite punishment that restarts after any deviation, a stage-game Nash punishment and $v_i(a^c)\ge v_i(a^*)$ for every player are sufficient to rule out punishment-phase deviations. Being a stage Nash equilibrium is not necessary when a deviation also worsens the continuation payoff.
With $k$ -period punishment by $a^*$ , a one-period deviation from $a^c$ is unprofitable if and only if $\big(v_i(a^c)-v_i(a^*)\big)(\delta+\delta^2+\dots+\delta^k)\ge \max_{a_i\ne a_i^c}v_i(a_i,a_{-i}^c)-v_i(a^c)$ for every $i$ .
When $v_i(a^c)\ge v_i(a^*)$ $v_{i} (a^{*})$ , the left side is nondecreasing in $k$ and $\delta$ ; letting $k\to\infty$ gives the grim trigger condition.
With a finite horizon and a unique stage equilibrium, no subgame-perfect equilibrium, in trigger strategies or otherwise, plays anything but that equilibrium.

Examples¶

Short versus long punishment. Prisoner's Dilemma with $v(C,C)=(3,3)$ $(3, 3)$ , $v(D,C)=(5,0)$ $(5, 0)$ , $v(D,D)=(1,1)$ $(1, 1)$ : the gain from defecting is $5-3=2$ and the per-period loss during punishment is $3-1=2$ $2$ . The condition becomes

\delta+\delta^2+\dots+\delta^k\ge1.

$k=1$ $1$ : $\delta\ge1$ $1$ , impossible for $\delta<1$ $1$ ; one period of punishment never deters.
$k=2$ $2$ : $\delta^2+\delta-1\ge0$ $0$ , i.e. $\delta\ge(\sqrt{5}-1)/2\approx0.618$ $0.618$ .
$k=\infty$ (grim trigger): $\delta/(1-\delta)\ge1$ $1$ , i.e. $\delta\ge1/2$ $1/2$ .

A deviation during punishment yields $0$ instead of $1$ and restarts the punishment, so it never pays. For the symmetric two-action model in the interactive figure, write $R=v_i(C,C)$ $v_{i} (C, C)$ , $P=v_i(D,D)$ and $S=v_i(C,D)=0$ $0$ , with $R>P$ $P$ . When $j$ punishment periods remain, obeying gives $V_j=P/(1-\delta)+\delta^j(R-P)/(1-\delta)$ $δ)$ , whereas deviating once gives $S+\delta V_k$ $δ V_{k}$ . The smallest compliance margin occurs at $j=k$ $k$ , so all punishment states are optimal exactly when $P-S+\delta^k(R-P)\ge0$ $0$ . This check is separate from the cooperative-path inequality; for an absorbing grim punishment it reduces to $P\ge S$ $S$ .

Collusion. Firms in a repeated Cournot duopoly can produce the joint-monopoly quantities and revert to Cournot quantities after any deviation; the grim version is worked out on the grim trigger page.

Prerequisites¶

Sources¶

Game Theory · Week 4

Aliases: trigger strategies, punishment strategy Keywords: punishment phase, cooperative phase, finite punishment, k-period punishment, forgiving trigger, tacit collusion, Prisoner's Dilemma Symbols: delta, infinity, square root

Last updated: 27 October 2026 11:52