Two roads to the same formula

Two people divide a surplus of total size 1. If they fail to agree, they get $d_1$ and $d_2$ respectively ($d_1 + d_2 < 1$). How will they divide it? Economics has two very different approaches to this question.

Nash’s axiomatic approach does not describe the bargaining process at all. It asks instead what properties a “reasonable” division should have. In 1950 Nash proposed four axioms (invariance to linear transformations of utility, Pareto efficiency, symmetry and independence of irrelevant alternatives) and proved that the only solution satisfying them is the division that maximises the Nash product $(x_1 – d_1)(x_2 – d_2)$ (Nash 1950). Here $(d_1, d_2)$ is called the threat point or disagreement point. Relaxing the symmetry axiom gives the asymmetric Nash solution with weight $\alpha$: maximise $(x_1 – d_1)^{\alpha}(x_2 – d_2)^{1-\alpha}$.

Rubinstein’s strategic approach writes out the bargaining process explicitly. Player 1 makes an offer; player 2 accepts or rejects; after a rejection a period passes and player 2 makes an offer, and so on, alternately, with no deadline. Each period of delay multiplies player $i$’s payoff by a discount factor $\delta_i < 1$, representing the cost of waiting. In 1982 Rubinstein proved that this infinite-horizon game has a unique subgame-perfect equilibrium, in which agreement is reached immediately (Rubinstein 1982).

# (a) Asymmetric Nash: maximise (x - d1)^a (1 - x - d2)^(1-a) => x* = d1 + a(1-d1-d2).
# (b) Rubinstein, player 1 proposes first: x1* = (1 - delta2) / (1 - delta1 delta2); check
#     against 400 rounds of backward induction. (c) delta_i = exp(-r_i dt), dt -> 0.
import math

def nash(d1, d2, a, steps=100000):
    xs = [d1 + (1 - d1 - d2) * k / steps for k in range(1, steps)]
    return max(xs, key=lambda x: a * math.log(x - d1) + (1 - a) * math.log(1 - x - d2))

for d1, d2, a in ((0, 0, .5), (.3, .1, .5), (.3, .1, .7)):
    print(f"Nash d=({d1}, {d2}) a={a}: x1 = {nash(d1, d2, a):.4f}; formula "
          f"{d1 + a * (1 - d1 - d2):.4f}")

def backward(dl1, dl2, rounds=400):    # the last proposer (player 2) takes everything
    share = 1.0
    for k in range(rounds - 1):        # proposer offers the responder her continuation
        share = 1 - (dl2 if k % 2 == 0 else dl1) * share
    return share

for dl1, dl2 in ((.9, .9), (.95, .8), (.8, .95)):
    print(f"Rubinstein delta=({dl1}, {dl2}): x1 = {(1 - dl2) / (1 - dl1 * dl2):.4f};"
          f" backward induction {backward(dl1, dl2):.4f}")
r1, r2 = 0.05, 0.15
print(f"r1 = {r1}, r2 = {r2} per day; Nash weight r2/(r1+r2) = {r2 / (r1 + r2):.4f}")
for dt in (7, 1, 0.1, 0.01):
    a, b = math.exp(-r1 * dt), math.exp(-r2 * dt)
    print(f"  an offer every {dt:>4} days: player 1 gets {(1 - b) / (1 - a * b):.4f}")
# Nash d=(0, 0) a=0.5: x1 = 0.5000; formula 0.5000
# Nash d=(0.3, 0.1) a=0.5: x1 = 0.6000; formula 0.6000
# Nash d=(0.3, 0.1) a=0.7: x1 = 0.7200; formula 0.7200
# Rubinstein delta=(0.9, 0.9): x1 = 0.5263; backward induction 0.5263
# Rubinstein delta=(0.95, 0.8): x1 = 0.8333; backward induction 0.8333
# Rubinstein delta=(0.8, 0.95): x1 = 0.2083; backward induction 0.2083
# r1 = 0.05, r2 = 0.15 per day; Nash weight r2/(r1+r2) = 0.7500
#   an offer every    7 days: player 1 gets 0.8628
#   an offer every    1 days: player 1 gets 0.7684
#   an offer every  0.1 days: player 1 gets 0.7519
#   an offer every 0.01 days: player 1 gets 0.7502

Maximising the Nash product numerically agrees with the formula: with threat point (0.3, 0.1) and equal weights, player 1 gets 0.6, her threat point plus half of the remaining 0.6. Rubinstein’s formula and 400 rounds of backward induction also agree exactly. With both discount factors at 0.9, the first proposer gets 0.5263, a small advantage from moving first; when player 1 is more patient (0.95 against 0.8) she gets 0.8333, and in the reverse case only 0.2083. Patience is bargaining power.

The last rows connect the two approaches. Suppose player 2’s daily “impatience rate” is three times player 1’s (0.15 against 0.05). With an offer every 7 days, player 1 gets 0.8628, which still includes the first-mover advantage; as the interval falls to 0.01 days she gets 0.7502, approaching $r_2/(r_1 + r_2) = 0.75$.

A BATNA: threat point or outside option?

In their 1981 best-seller Getting to Yes, Fisher and Ury coined the term BATNA, the “best alternative to a negotiated agreement”: the best thing you can do if no agreement is reached (Fisher and Ury 1981). Their advice was to know and improve your BATNA, because it sets the bottom line you can accept and also determines your negotiating power.

In Nash’s axiomatic model this holds exactly: treat the BATNA as the threat point, and each point added to $d_i$ adds $1 – \alpha$ points to your share. In alternating-offers models, however, a fallback works differently.

# Surplus 1, common discount factor 0.9, player 1 proposes first. Player 2's share under:
# (1) split the difference: outside option w2 as a Nash threat point, equal weights;
# (2) outside option principle (Sutton 1986; Binmore, Shaked, Sutton 1989): max(base, w2)
#     with base = delta / (1 + delta); (3) an INSIDE option: a flow paid to player 2
#     in each round WHILE talks go on, worth b2 if it lasted for ever (backward
#     induction).
D = 0.9
split = lambda w2: w2 + (1 - w2) / 2
principle = lambda w2: max(D / (1 + D), w2)

def inside(b2, rounds=600):
    v2 = 1.0                                   # last round: player 2 proposes, takes all
    for k in range(rounds - 1):
        if k % 2 == 0: v2 = (1 - D) * b2 + D * v2      # player 1 offers 2 his value
        else:          v2 = 1 - D * (1 - v2)           # player 2 offers 1 her value
    return v2                                  # round 1, player 1 proposing

print(f"no option: player 2 gets {D / (1 + D):.4f}")
print("  w2  | split the difference | outside option principle")
for w2 in (0.0, 0.2, 0.4, 0.47, 0.5, 0.6, 0.8):
    print(f" {w2:4.2f} |        {split(w2):.4f}        |          {principle(w2):.4f}")
print("inside flow b2 -> player 2's share: " +
      "  ".join(f"{b2:.1f}: {inside(b2):.4f}" for b2 in (0, .2, .4, .6)))
# no option: player 2 gets 0.4737
#   w2  | split the difference | outside option principle
#  0.00 |        0.5000        |          0.4737
#  0.20 |        0.6000        |          0.4737
#  0.40 |        0.7000        |          0.4737
#  0.47 |        0.7350        |          0.4737
#  0.50 |        0.7500        |          0.5000
#  0.60 |        0.8000        |          0.6000
#  0.80 |        0.9000        |          0.8000
# inside flow b2 -> player 2's share: 0.0: 0.4737  0.2: 0.5789  0.4: 0.6842  0.6: 0.7895

With no option at all, player 2 (the second mover) gets $\delta/(1+\delta) = 0.4737$. As the outside option rises from 0 to 0.47, the split-the-difference prediction raises his share steadily from 0.50 to 0.735; under the outside option principle his share stays at 0.4737 without moving. Only when the outside option exceeds 0.4737 (for example 0.5, 0.6 or 0.8) is it “activated”, and then he gets exactly the outside option, with no additional share of the surplus. The inside option is entirely different: every 0.2 added to it raises player 2’s share by about 0.105, just as a Nash threat point would.

Credible threats and where patience comes from

Two further points. First, an outside option must be credible. An “alternative” that you would not actually take when you reject an offer is, in the strategic model, as good as non-existent; conversely, an option the other side believes you will use affects the outcome even if you yourself do not like it. Second, “patience” is a discount rate in the model, but in reality it is often set by cash flow: a party with only three months of cash loses far more from every week of delay than a well-funded one. Book 19 of this series discussed cash buffers and the Kelly criterion; here it is enough to note that a buffer not only lowers the risk of ruin but also directly increases one’s patience in bargaining.

Chapter summary

Nash’s axiomatic bargaining solution maximises $(x_1 – d_1)^{\alpha}(x_2 – d_2)^{1-\alpha}$; Rubinstein’s alternating-offers model gives $x_1^* = (1 – \delta_2)/(1 – \delta_1\delta_2)$, so patience is bargaining power. As the interval between offers goes to zero the two coincide, with the Nash weight set by relative patience. A BATNA sets your bottom line, but whether it raises your share depends on the structure of the negotiation: as a Nash threat point it matters continuously; as an outside option in alternating offers it matters only when it exceeds the share you would get anyway; payoffs earned during the negotiation are what move shares continuously, like a threat point. The next chapter turns to another kind of information problem: one side knows the quality of the goods and the other does not.