Maskin Monotonicity

This scene tests whether a desired rule can arise as a game's equilibrium. Two people choose among three options.

The test is called Maskin monotonicity. Suppose some option wins in one situation. In another situation, that option has fallen for nobody. If the rule then picks something else, it fails. No game can ever implement such a rule. A dictator rule passes the test. The pick-the-bottom rule gets a red violation.

For example, your class votes on what to play. This week ⚽ soccer wins. Next week nobody has soured on soccer. If the result suddenly flips to 📚 reading, the rule is fishy. Such a rule cannot come from any fair game.

f(θ)=x,  x never fell from θ to θ′  ⟹  f(θ′)=xf(\theta) = x,\; x \text{ never fell from } \theta \text{ to } \theta' \implies f(\theta') = x

Part of these learning paths

Examples & Exercises

Counterexample

At θ\theta agent 1 ranks c≻a≻bc \succ a \succ b and agent 2 ranks a≻b≻ca \succ b \succ c; perturbation 1 changes agent 2's ranking to a≻c≻ba \succ c \succ b. With fanti(θ)f_{\text{anti}}(\theta) defined as agent 2's bottom choice, verify step by step that this perturbation violates Maskin monotonicity.

fanti(θ)=bottom2(θ),θ:1:c≻a≻b,  2:a≻b≻cf_{\text{anti}}(\theta) = \text{bottom}_2(\theta), \qquad \theta: 1: c \succ a \succ b,\; 2: a \succ b \succ c
θ′:1:c≻a≻b,  2:a≻c≻b\theta': 1: c \succ a \succ b,\; 2: a \succ c \succ b
Verdict

The claim is false — the antiplurality rule fantif_{\text{anti}} violates Maskin monotonicity.

fantif_{\text{anti}}, perturbation 1 — the violation case.
Show proof

Counterexample and verification

  1. First compute the rule's value at both profiles. At θ\theta agent 2's bottom choice is cc, so x=fanti(θ)=cx = f_{\text{anti}}(\theta) = c. At θ′\theta' agent 2 ranks a≻c≻ba \succ c \succ b, so the bottom choice is bb: x′=fanti(θ′)=b≠cx' = f_{\text{anti}}(\theta') = b \neq c.

  2. Now check whether x=cx = c fell. Agent 1's ranking did not move, so nothing could fall there. For agent 2, check each pair: (c,a)(c, a) — cc was below aa at θ\theta and is still below at θ′\theta', not a fall. (c,b)(c, b) — cc was below bb at θ\theta but is above bb at θ′\theta': that is a rise, not a fall. A fall requires "above, then below", so cc did not fall for agent 2 either.

  3. So monotonicity's hypothesis ("xx fell for no agent") holds, yet the outcome changed from cc to bb — a Maskin monotonicity violation. The point: cc "rose" from bottom to middle for agent 2, and the rule dropped it anyway.

    x=c never fell, but fanti(θ′)=b≠c=fanti(θ)x = c \text{ never fell, but } f_{\text{anti}}(\theta') = b \neq c = f_{\text{anti}}(\theta)

Prove in general that the dictatorial rule fdict(θ)f_{\text{dict}}(\theta) = agent 1's top choice satisfies Maskin monotonicity (under strict preferences).

fdict(θ)=top1(θ)f_{\text{dict}}(\theta) = \text{top}_1(\theta)
fdictf_{\text{dict}}, perturbation 1 — monotonicity respected.
Show proof

  1. Let x=fdict(θ)x = f_{\text{dict}}(\theta): xx is agent 1's top choice at θ\theta, so x≻1yx \succ_1 y for every y≠xy \neq x. Assume monotonicity's hypothesis — moving to θ′\theta', xx fell for no agent.

  2. By the definition of "did not fall", every yy with x≻1yx \succ_1 y at θ\theta still satisfies x≻1′yx \succ_1' y at θ′\theta'. Since xx topped agent 1's ranking at θ\theta, this means x≻1′yx \succ_1' y for every y≠xy \neq x — xx is still agent 1's top choice at θ′\theta'.

  3. Hence fdict(θ′)=x=fdict(θ)f_{\text{dict}}(\theta') = x = f_{\text{dict}}(\theta). Since the conclusion follows for every (θ,θ′)(\theta, \theta') pair satisfying the hypothesis, the dictatorial rule is Maskin monotonic. The intuition: "if xx was never overtaken, xx is still first."

    x top1 at θ,  x never fell for 1  ⟹  x top1 at θ′x \text{ top}_1 \text{ at } \theta,\; x \text{ never fell for } 1 \implies x \text{ top}_1 \text{ at } \theta'

22. Implementation Theory