Splitting the surplus

When two parties have a hundred pounds to divide and no way to create it except by agreeing, equilibrium analysis accepts every division from nothing to everything, which is not a prediction but a shrug.

The problem equilibrium cannot solve

Write the division as a game. Both parties simultaneously name a demand, $x$ for the first and $y$ for the second. If $x + y \leq 100$ each receives what they asked for; otherwise negotiations fail and both receive nothing.

Now hunt for Nash equilibria. Suppose the second party demands 40. The first party's best reply is 60: demanding more means the total exceeds 100 and both get nothing, demanding less means leaving money on the table. So $(60, 40)$ is an equilibrium. The identical argument works for $(50, 50)$ , for $(99, 1)$ , for $(1, 99)$ and for every other exactly exhausting pair. There is even an equilibrium at $(100, 100)$ , where both demand everything, both get nothing, and neither can improve alone, since a unilateral reduction still leaves the total above 100.

So the game has a continuum of equilibria, one for every split, plus a disastrous one. Refinement does not help: there are no unreached nodes, so subgame perfection has nothing to remove, and the folk theorem of the previous lesson says the same problem returns in worse form if the parties bargain repeatedly. The equilibrium concept is not merely failing to choose here. It is structurally incapable of choosing, because every split is a mutual best response and that is all the concept looks at.

John Nash, then twenty-one, published a different move in Econometrica in 1950. Stop deriving the split from individual optimisation. Instead state the properties a sensible answer ought to have, and see how many answers survive.

Writing the problem down

A bargaining problem is a pair $(S, d)$ . The feasible set $S$ is the set of utility pairs $(u_{1}, u_{2})$ the parties can achieve by agreement, taken to be convex, closed and bounded above. The disagreement point $d = (d_{1}, d_{2})$ is the utility pair they receive if they fail to agree.

Both objects need care. The feasible set is in utilities, not in pounds, which matters as soon as the parties are not equally happy about risk. And the disagreement point is not zero by convention; it is whatever each party actually gets from walking away. A union with a strike fund and a firm with six months of stock have a different disagreement point from a union with no savings and a firm with none, and it will turn out that this is the single most important number in the whole problem.

Convexity is worth a sentence, since it is doing real work. If two agreements are available, the parties can agree to flip a coin between them, and if payoffs are von Neumann-Morgenstern utilities, as the first lesson insisted, the value of that coin flip is the average of the two. So the feasible set automatically contains every point on the line between any two of its points.

Four properties

Nash asked for a rule $f$ that takes any bargaining problem $(S, d)$ and returns a single point of $S$ , subject to four requirements.

Efficiency. The chosen point is on the Pareto frontier: there is no other feasible point that is at least as good for both and strictly better for one. Leaving money unclaimed is not an answer.

Symmetry. If the problem is symmetric, meaning $d_{1} = d_{2}$ and $S$ is unchanged when the two coordinates are swapped, then the solution gives both parties the same utility. This says the rule uses nothing about the parties except what is in $S$ and $d$ , so it cannot favour the taller or the older or the one whose name comes first.

Invariance to affine transformations. If one party's utility scale is rescaled, $u_{1} \mapsto α u_{1} + β$ with $α > 0$ , the solution is the same agreement, described in the new units. This is the invariance from the first lesson, and it is what prevents a party from improving their share by writing their utilities in different units.

Independence of irrelevant alternatives. If the solution to a problem lies inside a smaller feasible set with the same disagreement point, then it is the solution to the smaller problem too. Removing options that were not going to be chosen changes nothing.

None of these sounds like it says much. Together they say everything.

One answer survives

Theorem (Nash, 1950). There is exactly one rule satisfying all four, and it selects the point of $S$ with $u_{i} \geq d_{i}$ that maximises the Nash product

(u_{1} - d_{1}) (u_{2} - d_{2})

The proof of uniqueness is a two-line trick worth knowing. Given any problem, use the affine invariance axiom to rescale both utilities so that the Nash product maximiser sits at $(1, 1)$ . One can then show the whole feasible set lies below the line $u_{1} + u_{2} = 2$ , since a point above it would allow a higher product. Enlarge the problem to the symmetric triangle under that line: by symmetry and efficiency, the rule must choose $(1, 1)$ there. By independence of irrelevant alternatives, since $(1, 1)$ is in the original smaller set, it must be chosen in the original problem too. Undo the rescaling and the theorem is proved.

Apply it to money first, where it is almost too simple. With linear utilities, a pot of $M$ and a disagreement point $(d_{1}, d_{2})$ , the frontier is $u_{1} + u_{2} = M$ and the product $(u_{1} - d_{1}) (M - u_{1} - d_{2})$ is a downward parabola in $u_{1}$ with roots at $d_{1}$ and $M - d_{2}$ . Its peak is halfway between them:

u_{1} = d_{1} + \frac{M - d_{1} - d_{2}}{2}

So the Nash solution takes each party's outside option off the top and splits the remaining surplus down the middle. That is not an assumption; it is what four axioms about fairness and consistency force.

Example. Two firms can jointly earn £100,000 from a project. Without the deal, firm 1 has an alternative worth £30,000 and firm 2 has nothing. Both are risk neutral. What does the Nash solution give?

Maximise $(x - 30) (100 - x)$ in thousands. The roots are 30 and 100, so the peak is at $x = 65$ . Firm 1 takes £65,000, firm 2 takes £35,000, and each has gained £35,000 over its outside option. The £30,000 alternative was worth exactly £30,000 in the bargain, no more and no less, which is the clearest reason for a negotiator to invest in an alternative before negotiating rather than in rhetoric during.

Now you. Same £100,000, but firm 1's outside option is £20,000 and firm 2's is £5,000. Find the split.

Answer

Maximise $(x - 20) (95 - x)$ , whose roots are 20 and 95, so $x = 57.5$ . Firm 1 takes £57,500 and firm 2 takes £42,500, and each gains £37,500 over its own fallback. The surplus being divided is $100 - 20 - 5 = £ 75 {,} 000$ , split evenly, with the fallbacks returned on top.

When the frontier is not a straight line

Not every bargain trades pound for pound. If one party's pound costs the other two, because of tax, transport or a difference in what the asset is worth to each, the frontier tilts and the Nash solution moves with it.

Example. A frontier runs $x + 2 y = 120$ , so every pound given to party 2 costs party 1 two pounds. The disagreement point is $(0, 0)$ and both are risk neutral. Where does the Nash solution sit?

Maximise $x y$ subject to the constraint. Substituting $y = (120 - x) / 2$ gives the product $x (120 - x) / 2$ , a parabola with roots 0 and 120, peaking at $x = 60$ . So $x = 60$ and $y = 30$ , with a Nash product of 1800. Party 1 takes twice as much in nominal terms, and in utility terms the two are equally distant from disagreement given the exchange rate between them. The general pattern for a linear frontier through the origin is that each party takes half of what they could have taken alone.

Now you. The frontier is $x + 3 y = 90$ with disagreement at $(0, 0)$ . Find the Nash solution.

Answer

Maximise $x (90 - x) / 3$ , roots 0 and 90, peaking at $x = 45$ , so $y = 15$ and the Nash product is 675. Party 1 could have taken 90 alone and takes 45; party 2 could have taken 30 alone and takes 15. Both take half their maximum, which is what a linear frontier through the disagreement point always gives.

What risk aversion costs

Now the point where working in utilities rather than money stops being pedantry.

Two people split £100. The first has utility $u_{1} (x) = \sqrt{x}$ , so they are risk averse in the sense of the first lesson. The second is risk neutral, $u_{2} (y) = y$ . Disagreement pays both nothing. Maximise the Nash product $\sqrt{x} (100 - x)$ .

Taking logs turns the product into $\frac{1}{2} \ln x + \ln (100 - x)$ , and setting the derivative to zero gives $\frac{1}{2 x} = 1 / (100 - x)$ , so $100 - x = 2 x$ and $x = 33.33$ . The risk-averse party receives £33.33 and the risk-neutral one £66.67.

The general version is worth having. If $u_{1} (x) = x^{α}$ with $0 < α \leq 1$ against a risk-neutral opponent, the same calculation gives

x = \frac{100 α}{1 + α}

At $α = 1$ the parties are equally risk neutral and split evenly. At $α = 0.5$ the share is £33.33, at $α = 0.25$ it is £20. The more sharply diminishing your marginal utility, the less you get.

The mechanism is not that risk aversion makes you a worse negotiator by temperament. It is that a bargaining problem is implicitly a problem about the risk of disagreement, and a party who suffers more from a bad outcome is willing to concede more to avoid it. That is a real and testable prediction about who does badly in negotiations, and it is also the sharpest warning against putting money in the payoff cells: had both parties' payoffs been written in pounds, the split would have come out at fifty each and the prediction would have been wrong.

Example. The same £100 with $u_{1} (x) = \sqrt{x}$ , but now the risk-averse party has an outside option worth £16 while the other has nothing. What is the split?

The disagreement point in utilities is $(\sqrt{16}, 0) = (4, 0)$ , so maximise $(\sqrt{x} - 4) (100 - x)$ . The derivative of the log gives $\frac{1}{2 \sqrt{x} (\sqrt{x} - 4)} = \frac{1}{100 - x}$ , so $100 - x = 2 x - 8 \sqrt{x}$ . Writing $s = \sqrt{x}$ gives $3 s^{2} - 8 s - 100 = 0$ , with positive root $s = (8 + \sqrt{64 + 1200}) / 6 = (8 + 35.55) / 6 = 7.259$ , so $x = 52.7$ . The outside option was worth £16 and it bought £19.4 of extra share, because it also removed part of the risk that made the party concede.

Now you. Back to no outside options, but with $u_{1} (x) = x^{0.25}$ against a risk-neutral partner. What does the risk-averse party get?

Answer

$x = 100 (0.25) / 1.25 = £ 20$ . Compared with the £33.33 at $α = 0.5$ and the £50 at $α = 1$ , the pattern is monotone: every increase in risk aversion transfers share to the other side. Nothing about the two parties differs except the curvature of one utility function.

The axiom that is doing the damage

Three of the four axioms are close to unarguable. The fourth, independence of irrelevant alternatives, is where the objections live, and the objection is easy to state with numbers.

Two parties split £100 with linear utilities and no outside options, so the Nash solution is £50 each. Now suppose a regulator caps party 2's share at £60. The cap removes only agreements that were not going to be chosen, so by independence of irrelevant alternatives the solution stays at £50 each. And yet something real has changed: party 1 can still imagine receiving the whole £100 while party 2's best case has fallen to £60, so their positions are no longer symmetric in any intuitive sense.

Ehud Kalai and Meir Smorodinsky proposed in 1975 dropping independence and requiring monotonicity instead: if the frontier expands so that one party's best possible outcome rises with the other's held fixed, that party should not do worse. Their solution equalises the ratio of each party's gain to their maximum possible gain. In the capped example the ideal point is $(100, 60)$ , so the solution sits where $x / 100 = y / 60$ on the frontier $x + y = 100$ . Writing both as $t$ times their ideal gives $160 t = 100$ , so $t = 0.625$ and the split is £62.50 to party 1 and £37.50 to party 2. The cap on party 2's upside has cost them £12.50 in a bargain they were never going to push that far.

Which is right is not settled and probably has no answer in the abstract, because they are answers to different questions. Nash's rule is the one that is consistent under removing options; Kalai and Smorodinsky's is the one that responds to how much each party could in principle have got. Experimental subjects, offered problems where the two disagree, do not systematically match either.

One further honesty is owed. Nash's theorem selects a point but describes no process. Nobody in it makes an offer, refuses one, or walks out. It is a statement about what a reasonable arbitrator would rule, and it is silent about two people in a room with conflicting interests and time to waste. Nash was aware of the gap and proposed closing it by building non-cooperative models whose equilibria reproduce the axiomatic answer, a research programme now called the Nash program.

Where this leaves us

Four modest requirements pick out one split from a continuum that equilibrium could not narrow at all. The split hands each party their outside option and divides what is left in half, shifting against whoever is more risk averse and whoever has less to fall back on.

That is a satisfying answer to the wrong question. Real bargaining is a sequence: someone opens, someone counters, time passes, and every day of delay costs both sides money. The next lesson models the haggling itself, with impatient players making alternating offers, and finds that the procedure has a unique subgame perfect outcome which converges on the split derived here as the delay between offers shrinks. It also finds that people in laboratories do not play it.