Divide the dollar and conquer more: sequential bargaining and risk aversion

Abstract

We analyze the problem of dividing a fixed amount of a single commodity between two players on the basis of the Nash bargaining solution (NBS). For one-shot negotiations, a cornerstone result of Roth (Axiomatic models of bargaining. Springer, Berlin, 1979) establishes that the more risk averse player will obtain less than half the total amount. In the present paper, we assume that the bargaining procedure occurs over several rounds. In each round, an increasing share of the total amount is negotiated over in accordance with the NBS, the disagreement point being determined by the outcome of the previous round. In line with Roth’s result, the final amount received by the more risk averse player is still bounded by half the total amount. As a new feature, however, this player does not lose from bargaining for more rounds if his opponent exhibits non-increasing absolute risk aversion. What is more, both players’ risk profiles become essentially irrelevant if successive bargaining takes place over sufficiently small commodity increments. Each player then gets approximately half of the commodity.

Appendix

We first formally present the alternative approach to the sequential bargaining procedure introduced in Sect. 2. We then prove Theorems 1 and 2.

1.1 Alternative approach to the sequential bargaining procedure

To complement the point of view taken in the main body of our paper (Point of View 1: stock-based), we present here an equivalent alternative (Point of View 2: increment-based). To that end, with $y_0^{\mathcal {P}}=0,$ we define recursively for n,  with $1\le n\le N,$ the outcome

$$\begin{aligned} y^{{\mathcal {P}}}_n:=x^*(u_{A,n},u_{B,n};0,0;q_n-q_{n-1}), \end{aligned}$$

where $u_{A,n}, u_{B,n}$ are shifted utility functions defined by

$$\begin{aligned} u_{A,n}(y):= & {} u_A(y+\sigma _{n-1}) -u_A(\sigma _{n-1}),\end{aligned}$$

(7)

$$\begin{aligned} u_{B,n}(y):= & {} u_B(y + q_{n-1}-\sigma _{n-1})-u_B(q_{n-1}-\sigma _{n-1}), \end{aligned}$$

(8)

and $\sigma _n^{\mathcal {P}} := \sum _{i=1}^n y_n^{{\mathcal {P}}}$ sums the outcomes of all individual rounds up to n for player A.

Figure 3 illustrates the changing utility function of player A from Point of View 2 for the three-round bargaining protocol ${\mathcal {P}}^3$ used in Fig. 1.

Fig. 3

Left Utility functions of player A in each round of ${\mathcal {P}}^3.$ Player B’s utility function is simply linear. Right The bargaining sets and outcomes in each of the three rounds. As bargaining is now viewed as happening over increments, the nested structure of Fig. 1 is no longer present

From this second point of view, the two players bargain in the nth round over the amount $q_n-q_{n-1}.$ This may or may not be larger than the amount bargained over in any of the previous rounds. The focus is now on the incremental variable $y^{{\mathcal {P}}}_n,$ with the state variable $\sigma _{n}^{{\mathcal {P}}}$ playing merely an auxiliary role. If the two players do not come to an agreement, they obtain nothing in that round and the negotiation ends with what they obtained in all previous rounds, yielding the outcome vector $(\sigma ^{\mathcal {P}}_{n-1}, q_{n-1}-\sigma ^{\mathcal {P}}_{n-1}).$ This explains the shifts in Eqs. (7) and (8). As opposed to Point of View 1, the disagreement point is now the same in every round—namely zero—but the players’ attitude towards risk is not, as it depends on the outcome of the previous rounds. To be explicit about the evolution of the increments $y^{{\mathcal {P}}}_n,$ note that for all n,  with $1\le n \le N,$ Eq. (2) now takes the form

$$\begin{aligned}&u'_A\left( y_{n}^{{\mathcal {P}}}+\sigma _{n-1}^{{\mathcal {P}}}\right) \cdot \left[ u_B\left( q_{n}-y_{n}^{{\mathcal {P}}} -\sigma _{n-1}^{{\mathcal {P}}}\right) -u_B\left( q_{n-1} -\sigma _{n-1}^{{\mathcal {P}}}\right) \right] \nonumber \\&\quad =\left[ u_A\left( y_{n}^{{\mathcal {P}}}+ \sigma _{n-1}^{{\mathcal {P}}}\right) -u_A\left( \sigma _{n-1}^{{\mathcal {P}}}\right) \right] \cdot u'_B\left( q_{n}-y_{n}^{{\mathcal {P}}} -\sigma _{n-1}^{{\mathcal {P}}}\right) . \end{aligned}$$

(9)

Finally, we point out that from Eqs. (4) and (9), it follows that the two points of view are indeed equivalent, as for all n,  with $1\le n\le N,$ we have

$$\begin{aligned} x^{\mathcal {P}}_n=\sigma _n^{\mathcal {P}} \end{aligned}$$

and

$$\begin{aligned} y_n^{\mathcal {P}}=x_n^{\mathcal {P}}-x_{n-1}^{\mathcal {P}}. \end{aligned}$$

1.2 Proof of Theorem 1

Part (i) Using Point of View 1 (stock-based, cf. p. 7), it suffices to show that for all n,  with $1\le n \le N,$

$$\begin{aligned} x_n^{{\mathcal {P}}}\le \frac{q_n}{2}. \end{aligned}$$

(10)

To demonstrate (10) we use induction on n. The case $n=1$ follows immediately from Corollary 1. Now assume that $1<n\le N$ and $x_{n-1}^{{\mathcal {P}}}\le {q_{n-1}}/{2}.$ By Thomson’s theorem (cf. p. 6), we may assume without loss of generality that $x_{n-1}^{{\mathcal {P}}}= {q_{n-1}}/{2},$ so that the disagreement point in the $n\text {th}$ round is $({q_{n-1}}/{2}, {q_{n-1}}/{2}).$ An application of Corollary 1 then yields the claim.

Part (ii) To establish the desired inequality, it suffices to restrict our attention to a comparison of the outcomes of two bargaining protocols, where one is obtained from the other by splitting one round into two rounds, and then to apply a recursive argument to establish the desired relation between $x^{\mathcal {P}}_N$ and $x^{{\mathcal {P}}'}_{N'}.$ Without loss of generality, we may assume that ${\mathcal {P}}=(q_n)_{0\le n\le N}=: {\mathcal {P}}^N$ consists of $N+1$ elements and that ${{\mathcal {P}}'}=(q'_n)_{0\le n\le N+1}=:{\mathcal {P}}^{N+1}$ consists of $N+2$ elements, with ${\mathcal {P}}^N={\mathcal {P}}^{N+1}\setminus \{q\}$ (as sets) for some $q\in {\mathcal {P}}^{N+1}.$ Our proof is based on an induction on N.

$\underline{\textit{Base case}~(N=1){:}}$

Let ${\mathcal {P}}^1=(0,Q)$ and ${\mathcal {P}}^{2}=(0,q,Q),$ with $0<q<Q.$ We need to show that

$$\begin{aligned} x_1^{{\mathcal {P}}^1} =y_1^{{\mathcal {P}}^1} \le y_1^{{\mathcal {P}}^2}+y_2^{{\mathcal {P}}^2}=x_2^{{\mathcal {P}}^2}. \end{aligned}$$

(11)

Using Point of View 2, we start by writing down the governing equations for ${\mathcal {P}}^1$ and ${\mathcal {P}}^2.$ On the one hand, regarding ${\mathcal {P}}^1,$ the bargaining happens in one single round and thus Eq. (9) simplifies to

$$\begin{aligned} u'_A\left( y_{1}^{{\mathcal {P}}^1}\right) \cdot u_B\left( Q-y_{1}^{{\mathcal {P}}^1}\right) = u_A\left( y_{1}^{{\mathcal {P}}^1}\right) \cdot u'_B\left( Q-y_{1}^{{\mathcal {P}}^1}\right) . \end{aligned}$$

(12)

On the other hand, as for ${\mathcal {P}}^2,$ where the bargaining takes place in two consecutive rounds, Eq. (9) translates into the following pair of coupled equations for $y_1^{{\mathcal {P}}^2}$ and $y_2^{{\mathcal {P}}^2}$ (recall that we have assumed $u_A(0)=u_B(0)=0$):

$$\begin{aligned} u'_A\left( y_{1}^{{\mathcal {P}}^2}\right) \cdot u_B\left( q-y_{1}^{{\mathcal {P}}^2}\right) = u_A\left( y_{1}^{{\mathcal {P}}^2}\right) \cdot u'_B\left( q-y_{1}^{{\mathcal {P}}^2}\right) \end{aligned}$$

(13)

and

$$\begin{aligned}&u'_A\left( y_{2}^{{\mathcal {P}}^2}+y_{1}^{{\mathcal {P}}^2}\right) \cdot \left[ u_B\left( Q-y_{2}^{{\mathcal {P}}^2} -y_{1}^{{\mathcal {P}}^2}\right) -u_B\left( q -y_{1}^{{\mathcal {P}}^2}\right) \right] \nonumber \\&\quad =\left[ u_A\left( y_{2}^{{\mathcal {P}}^2}+ y_{1}^{{\mathcal {P}}^2}\right) -u_A\left( y_{1}^{{\mathcal {P}}^2}\right) \right] \cdot u'_B\left( Q-y_{2}^{{\mathcal {P}}^2} -y_{1}^{{\mathcal {P}}^2}\right) . \end{aligned}$$

(14)

To manipulate these equations efficiently, it will come in handy to introduce the inverse boldness function

$$\begin{aligned} g_i : = \frac{u_i}{u'_i} =: \frac{1}{b_i },\quad i= A,B. \end{aligned}$$

(15)

Note that $g_i$ is well-defined, positive, and continuous on (0, Q] because $u_i$ is a strictly increasing element of $C^2((0,Q])\cap C([0,Q])$ ($i=A,B$). Equation (12) can then be rewritten as

$$\begin{aligned} g_A\left( y_{1}^{{\mathcal {P}}^1}\right) =g_B\left( Q-y_{1}^{{\mathcal {P}}^1}\right) . \end{aligned}$$

(16)

Similarly, Eqs. (13) and (14) become

$$\begin{aligned} g_A\left( y_{1}^{{\mathcal {P}}^2}\right) =g_B\left( q-y_{1}^{{\mathcal {P}}^2}\right) \end{aligned}$$

(17)

and

$$\begin{aligned}&u'_A\left( y_{2}^{{\mathcal {P}}^2}+y_{1}^{{\mathcal {P}}^2}\right) \cdot \left[ u'_B\left( Q-y_{2}^{{\mathcal {P}}^2} -y_{1}^{{\mathcal {P}}^2}\right) \right. \nonumber \\&\qquad \left. \cdot \, g_B\left( Q-y_{2}^{{\mathcal {P}}^2} -y_{1}^{{\mathcal {P}}^2}\right) -u'_B\left( q -y_{1}^{{\mathcal {P}}^2}\right) \cdot g_B \left( q -y_{1}^{{\mathcal {P}}^2}\right) \right] \nonumber \\&\quad =\left[ u'_A\left( y_{2}^{{\mathcal {P}}^2}+y_{1}^{{\mathcal {P}}^2}\right) \cdot g_A\left( y_{2}^{{\mathcal {P}}^2}+y_{1}^{{\mathcal {P}}^2}\right) \right. \nonumber \\&\qquad \left. -\,u'_A\left( y_{1}^{{\mathcal {P}}^2}\right) \cdot g_A\left( y_{1}^{{\mathcal {P}}^2}\right) \right] \cdot u'_B\left( Q-y_{2}^{{\mathcal {P}}^2} -y_{1}^{{\mathcal {P}}^2}\right) , \end{aligned}$$

(18)

respectively. Since $u_A', u_B'>0,$ Eq. (18), can be further rewritten with the help of Eq. (17) as

$$\begin{aligned}&g_B\left( Q-\left( y_{1}^{{\mathcal {P}}^2}+y_{2}^{{\mathcal {P}}^2} \right) \right) -g_A\left( y_{1}^{{\mathcal {P}}^2}+y_{2}^{{\mathcal {P}}^2}\right) \\&\qquad -\, g_A\left( y_{1}^{{\mathcal {P}}^2}\right) \cdot \left[ \frac{u'_B\left( q-y_{1}^{{\mathcal {P}}^2}\right) }{u'_B\left( Q- \left( y_{1}^{{\mathcal {P}}^2}+y_{2}^{{\mathcal {P}}^2}\right) \right) }-\frac{u'_A\left( y_{1}^{{\mathcal {P}}^2}\right) }{u'_A \left( y_{1}^{{\mathcal {P}}^2}+y_{2}^{{\mathcal {P}}^2}\right) }\right] =0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} h \left( y_{1}^{{\mathcal {P}}^2}+y_{2}^{{\mathcal {P}}^2} \right) =0, \end{aligned}$$

where $h\in C((0,Q])$ is defined by

$$\begin{aligned} h(w):= g_B(Q-w)-g_A(w)-g_A\left( y_{1}^{{\mathcal {P}}^2}\right) \cdot \left[ \frac{u'_B\left( q-y_{1}^{{\mathcal {P}}^2}\right) }{u'_B(Q-w)} -\frac{u'_A\left( y_{1}^{{\mathcal {P}}^2}\right) }{u'_A(w)}\right] .\nonumber \\ \end{aligned}$$

(19)

By uniqueness of the NBS, $h(w)=0$ admits only one solution in (0, Q],  namely $w=y_{1}^{{\mathcal {P}}^2}+y_{2}^{{\mathcal {P}}^2}.$

To complete the proof of the base case ($N=1$), it remains to show the following claim.

Claim Assuming $y_1^{{\mathcal {P}}^1}\ne y_1^{{\mathcal {P}}^2}+y_2^{{\mathcal {P}}^2},$ the following inequalities hold:

$$\begin{aligned}&h\left( y_{1}^{{\mathcal {P}}^1}\right) >0,\end{aligned}$$

(20)

$$\begin{aligned}&h\left( y_1^{{\mathcal {P}}^2}+Q-q\right) <0, \end{aligned}$$

(21)

$$\begin{aligned}&y_1^{{\mathcal {P}}^1}<y_1^{{\mathcal {P}}^2}+Q-q. \end{aligned}$$

(22)

Indeed, if $y_1^{{\mathcal {P}}^1}= y_1^{{\mathcal {P}}^2}+y_2^{{\mathcal {P}}^2},$ (11) trivially holds. Otherwise, (20)–(22) allow the application of the Intermediate Value Theorem. The zero of h,  namely $w=y_{1}^{{\mathcal {P}}^2}+y_{2}^{{\mathcal {P}}^2}$, then has to lie in the interval $(y_1^{{\mathcal {P}}^1},y_1^{{\mathcal {P}}^2}+Q-q) \subset (y_1^{{\mathcal {P}}^1},Q],$ and thus the base case ($N=1$) follows.

We now turn to the proof of the three inequalities of the claim.

Proof of (20)

From Eqs. (16) and (19), we obtain

$$\begin{aligned} h(y_{1}^{{\mathcal {P}}^1})= & {} g_B\left( Q-y_{1}^{{\mathcal {P}}^1}\right) -g_A\left( y_{1}^{{\mathcal {P}}^1}\right) \nonumber \\&-\,g_A\left( y_{1}^{{\mathcal {P}}^2}\right) \cdot \left[ \frac{u'_B\left( q-y_{1}^{{\mathcal {P}}^2}\right) }{u'_B\left( Q-y_{1}^{\mathcal { P}^1}\right) }-\frac{u'_A\left( y_{1}^{{\mathcal {P}}^2}\right) }{u'_A\left( y_{1}^{{\mathcal {P}}^1}\right) }\right] \\= & {} -\,g_A\left( y_{1}^{{\mathcal {P}}^2}\right) \cdot \left[ \frac{u'_B\left( q-y_{1}^{{\mathcal {P}}^2}\right) }{u'_B \left( Q-y_{1}^{{\mathcal {P}}^1}\right) }-\frac{u'_A \left( y_{1}^{{\mathcal {P}}^2}\right) }{u'_A \left( y_{1}^{{\mathcal {P}}^1}\right) }\right] . \end{aligned}$$

This last expression is positive if the expression in square brackets is negative. Using Eqs. (12) and (13), the inequality in (20) is thus equivalent to

$$\begin{aligned} \frac{u_B\left( q-y_{1}^{{\mathcal {P}}^2}\right) }{u_A\left( y_{1}^{{\mathcal {P}}^2}\right) }< \frac{u_B\left( Q-y_{1}^{{\mathcal {P}}^1}\right) }{u_A\left( y_{1}^{{\mathcal {P}}^1}\right) }. \end{aligned}$$

(23)

Consider now the function k defined on the interval [q, Q] by

$$\begin{aligned} k(r):=\frac{u_B\left( r-y_{1}(r)\right) }{u_A\left( y_{1}(r)\right) }. \end{aligned}$$

(24)

Here, $y_{1}(r)$ denotes the outcome for player A when bargaining over an amount r in a single round takes place in accordance with the NBS with disagreement point (0, 0)—as captured by the bargaining problem $(u_A,u_B;0,0;r).$ In particular, we have $y_1(q)=y_1^{{\mathcal {P}}^2}$ and $y_1(Q)=y_1^{{\mathcal {P}}^1},$ connecting Eq. (24) with Eq. (23). The defining equation for $y_1(r)$ is derived from Eq. (2) using Definition (15):

$$\begin{aligned} b_B(r-y_1(r))=\frac{u_B'(r-y_1(r))}{u_B(r-y_1(r))}= \frac{u'_A(y_1(r))}{u_A(y_1(r))}=b_A(y_1(r)). \end{aligned}$$

(25)

Using Definitions (3) and (15), and because $b_i(x)>0$ for $x\in [0,Q),$ a direct computation shows that

$$\begin{aligned} b'_i(x)=-a_i (x)b_i(x) -b_i^2(x)<0,\quad \text {for } x\in (0,Q] \text { and } i=A,B. \end{aligned}$$

(26)

Hence the Implicit Function Theorem can be applied to Eq. (25), yielding that $y_1$ and therefore also k are elements of $C^2([q,Q]).$

To show (23), it is sufficient to demonstrate that the derivative of k is non-negative and that k is not a constant function. If it were a constant on [q, Q],  we would have an equality in (23) instead of a strict inequality, meaning that $h(y_1^{{\mathcal {P}}^1})=0.$ In this case, $y_1^{{\mathcal {P}}^1}$ would be a zero of h,  thereby implying that $y_1^{{\mathcal {P}}^1}=y_1^{{\mathcal {P}}^2}+y_2^{{\mathcal {P}}^2},$ which would contradict the assumption that $y_1^{{\mathcal {P}}^1}\ne y_1^{{\mathcal {P}}^2}+y_2^{{\mathcal {P}}^2}.$ To compute the derivative of k,  we use the chain rule as well as Eq. (25) and arrive at

$$\begin{aligned} k'(r)= \frac{u'_B(r-y_1(r)) }{u_A\left( y_{1} (r)\right) } \left( 1- 2 y_1'(r)\right) ,\quad \hbox {for }r\in [q,Q]. \end{aligned}$$

To demonstrate (23), it suffices to show that

$$\begin{aligned} y_1'(r)\le \frac{1}{2},\quad \hbox {for }r\in [q,Q]. \end{aligned}$$

(27)

For all $r\in [q,Q],$ we thus compute

$$\begin{aligned} y_1'(r)= & {} \frac{b'_B(r-y_1(r))}{b'_B(r-y_1(r))+b'_A(y_1(r)) } \\= & {} \frac{-a_B(r-y_1(r))b_B(r-y_1(r)) -b^2_B(r-y_1(r))}{-a_B(r-y_1(r))b_B(r-y_1(r))-b^2_B(r-y_1(r))-a_A(y_1(r)) b_A(y_1(r))-b^2_A(y_1(r))} \\= & {} \frac{a_B(r-y_1(r))b_A(y_1(r)) +b^2_A(y_1(r))}{ (a_B(r-y_1(r))+a_A(y_1(r))) b_A(y_1(r))+2 b^2_A(y_1(r))} \\= & {} \frac{a_B(r-y_1(r)) +b_A(y_1(r))}{ a_B(r-y_1(r))+a_A(y_1(r))+2 b_A(y_1(r))}, \end{aligned}$$

where the first and third equality hold by Eq. (25) and the Implicit Function Theorem, the second by Eq. (26), and the last because $b_A(y_1(r))>0$ for all $r\in [q,Q].$ Hence, (27) is equivalent to

$$\begin{aligned} a_A \left( y_1(r)\right) \ge a_B \left( r-y_1(r)\right) ,\quad \hbox {for }r\in [q,Q]. \end{aligned}$$

But the latter condition must hold because for all $r\in [q,Q],$

$$\begin{aligned} a_A \left( y_1(r)\right) \ge a_B \left( y_1(r)\right) \ge a_B \left( r-y_1(r)\right) . \end{aligned}$$

The first inequality is due to Condition (RA). The second inequality holds because player B satisfies (N-IARA) and because Corollary 1 guarantees that $y_1(r)\le r-y_1(r)$ for all $r\in [q,Q].$

Proof of (21)

Using Eq. (17) and the fact that $u_A$ is strictly increasing, we obtain, after some straightforward algebra, that

$$\begin{aligned} h\left( y_1^{{\mathcal {P}}^2}+Q-q\right)= & {} g_B\left( q-y_1^{{\mathcal {P}}^2} \right) -g_A\left( y_1^{{\mathcal {P}}^2}+Q-q\right) \\&-\,g_A\left( y_1^{{\mathcal {P}}^2}\right) \cdot \left[ 1-\frac{u'_A\left( y_{1}^{{\mathcal {P}}^2}\right) }{u'_A\left( y_1^{{\mathcal {P}}^2}+Q-q)\right) }\right] \\= & {} \frac{u_A\left( y_1^{{\mathcal {P}}^2}\right) -u_A \left( y_1^{{\mathcal {P}}^2}+Q-q\right) }{u_A'\left( y_1^{{\mathcal {P}}^2}+Q-q\right) } < 0. \end{aligned}$$

Proof of (22)

Consider the function $v\in C^2([q,Q]),$ where

$$\begin{aligned} v(r):=y_1(r)+Q-r. \end{aligned}$$

It follows immediately from (27) that

$$\begin{aligned} v'(r)< 0,\quad \hbox {for all }r\in [q,Q]. \end{aligned}$$

Hence, because $q<Q,$

$$\begin{aligned} y_1^{{\mathcal {P}}^1}= & {} y_1(Q)=y_1(Q)+Q-Q = v(Q) < v(q)= y_1(q)+Q-q\\= & {} y_1^{{\mathcal {P}}^2}+Q-q, \end{aligned}$$

as desired.

$\underline{{ Induction~step}~(N \rightarrow N+1){:}}$

We assume now that for any pair of utility functions $u_A$ and $u_B$ satisfying the assumptions of Part (ii) of Theorem 1, we have

$$\begin{aligned} x_{N-1}^{\mathcal {\tilde{P}}} =\sum _{n=1}^{N-1} y_{n}^{\mathcal {\tilde{P}}}\le \sum _{n=1}^{N} y_{n}^{\mathcal {\tilde{P'}}} =x_{N}^{\mathcal {\tilde{P'}}}, \end{aligned}$$

where $\mathcal {\tilde{P}}=(\tilde{q}_n)_{0\le n \le N-1}$ and $\mathcal {\tilde{P}'}=(\tilde{q}'_n)_{0\le n \le N}$ are arbitrary protocols over an arbitrary amount $\tilde{Q}>0,$ with the sole restriction that $\tilde{{\mathcal {P}}}'$ is finer than $\tilde{{\mathcal {P}}}.$ Using the definitions introduced in the beginning of the proof of Part (ii) of Theorem 1, we distinguish two cases.

Case 1: $q_1\ne q'_1.$

In this case, the first round of protocol ${\mathcal {P}}={\mathcal {P}}^N$ corresponds to the first two rounds of protocol ${{\mathcal {P}}'}={\mathcal {P}}^{N+1}.$ Figure 4 illustrates this.

Fig. 4

Illustration of case 1

For both bargaining protocols ${\mathcal {P}}$ and ${{\mathcal {P}}'},$ the negotiation for the last $N-1$ rounds may be described by a single protocol $\tilde{{\mathcal {P}}}:=(\tilde{q}_n)_{0\le n \le N-1}$ with the total amount $\tilde{Q}:=Q-q_1=Q-q'_2>0,$ where $\tilde{q}_n:=q_{n+1}=q'_{n+2}$ for $n>1.$¹⁴ However, the initial disagreement point $x_0^{\tilde{{\mathcal {P}}}}$ differs: it is $x_0^{\tilde{{\mathcal {P}}}}=x_1^{\mathcal {P}}$ in the case of ${\mathcal {P}}$ and $x_0^{\tilde{{\mathcal {P}}}}=x_2^{{\mathcal {P}}'}$ in the case of ${{\mathcal {P}}'}.$ Since we know from the base case ($N=1$) that $x_1^{\mathcal {P}}\le x_2^{{\mathcal {P}}'},$ an iterated application of Thomson’s theorem (cf. p. 6) yields $x_N^{\mathcal {P}} \le x_{N+1}^{{\mathcal {P}}'}.$

Case 2: $q_1= q'_1.$

In this case, the first round for both protocols ${\mathcal {P}}={\mathcal {P}}^N$ and ${{\mathcal {P}}'}={\mathcal {P}}^{N+1}$ is the same, i.e., $q_1= q_1'.$ This implies that

$$\begin{aligned} x_1^{\mathcal {P}}=y_1^{\mathcal {P}}= y_1^{{\mathcal {P}}'}=x_1^{{\mathcal {P}}'}. \end{aligned}$$

It is useful to think of the divisions of the remaining amount $\tilde{Q}:=Q-q_1=Q-q_1'$ as protocols in their own right, so that ${\mathcal {P}}$ is replaced by $\mathcal {\tilde{P}}=(\tilde{q}_n)_{0\le n \le N-1}$, where $\tilde{q}_0=0$ and $\tilde{q}_n:=q_{n+1}-q_1$ for $1\le n\le N-1.$ Similarly, ${{\mathcal {P}}'}$ is replaced by $\mathcal {\tilde{P}'}=(\tilde{q}'_n)_{0\le n \le N},$ where $\tilde{q}'_0=0$ and $\tilde{q}'_n:=q'_{n+1}-q'_1$ for $1\le n\le N.$ By construction, both $\mathcal {\tilde{P}}$ and $\mathcal {\tilde{P}'}$ are protocols over $\tilde{Q}$ and, moreover, $\mathcal {\tilde{P}'}$ is finer than $\mathcal {\tilde{P}}$ by exactly one subdivision. Figure 5 illustrates this.

Fig. 5

Illustration of case 2

To stay within the framework in which we can apply our induction hypothesis, we need to shift the corresponding utility functions appropriately. Namely, we use $u_{i,2}(y)$ instead of $u_i=u_{i,1}(y),$ $i=A,B$ (cf. Point of View 2 (increment-based)). This means that we can apply both $\mathcal {\tilde{P}'}$ and $\mathcal {\tilde{P}}$ to the bargaining problem defined by $(u_{A,2},u_{B,2};0,0;\tilde{Q}).$ The following relations thus hold by construction:

$$\begin{aligned} x_n^{{\mathcal {P}}}= & {} y_1^{{\mathcal {P}}}+x_{n-1}^{\tilde{{\mathcal {P}}}} \quad \text {for } 2\le n \le N,\end{aligned}$$

(28)

$$\begin{aligned} x_n^{{{\mathcal {P}}'}}= & {} y_1^{{\mathcal {P}}}+x_{n-1}^{\tilde{{\mathcal {P}}}'} \quad \text {for } 2\le n \le N+1. \end{aligned}$$

(29)

Obviously, ${u}_{B,2}$ satisfies Condition (N-IARA). As for Condition (RA), note that

$$\begin{aligned} -\frac{u_{A,2}''(y)}{u_{A,2}'(y)}= & {} -\frac{u_A''(y+y^{\mathcal {P}}_1)}{u_A'(y+y^{\mathcal {P}}_1)} =a_A(y+y^{\mathcal {P}}_1) \ge a_B(y+y^{\mathcal {P}}_1) \ge a_B(y+q_1-y^{\mathcal {P}}_1) \\= & {} -\frac{u_B''(y+q_1-y^{\mathcal {P}}_1)}{u_B'(y+q_1-y^{\mathcal {P}}_1)} =-\frac{u_{B,2}''(y)}{u_{B,2}'(y)}. \end{aligned}$$

Here, we have used Condition (RA) for $u_A$ and $u_B$ in the first inequality and Condition (N-IARA) in the second inequality, together with the fact that $y_1^{\mathcal {P}}\le q_1-y_1^{\mathcal {P}},$ which holds by Corollary 1. As a consequence, the induction hypothesis ensures that

$$\begin{aligned} x_{N-1}^{\tilde{{\mathcal {P}}}} \le x_{N}^{\tilde{{\mathcal {P}}}'}. \end{aligned}$$

Together with Eqs. (28) and (29), this inequality, in turn, implies that

$$\begin{aligned} x_N^{{\mathcal {P}}} \le x_{N+1}^{{{\mathcal {P}}'}}, \end{aligned}$$

thereby completing the proof of the theorem. $\square$

1.3 Proof of Theorem 2

Consider the protocol ${\mathcal {P}}^{\varepsilon }=(q^\varepsilon _n)_{0\le n\le N}$ over the amount $Q>0,$ where $q^\varepsilon _0=0,$ $q^\varepsilon _1=\varepsilon /2,$ $q^\varepsilon _N=Q$ and $q^\varepsilon _2, q^\varepsilon _3, \dots ,q^\varepsilon _{N-1}$ will be specified later. The proof is based on Taylor expansions around $\sigma _{n-1}^{{\mathcal {P}}^{\varepsilon }}$ and $q^\varepsilon _{n-1}-\sigma _{n-1}^{{\mathcal {P}}^{\varepsilon }},$ respectively, of the following four terms in Eq. (9) (to simplify notation, dependencies on $\varepsilon$ and ${\mathcal {P}}^{{\varepsilon }}$ are henceforth suppressed):

$$\begin{aligned}&\overbrace{u'_A\left( y_{n}+\sigma _{n-1}\right) }^{(\text {I})} \cdot \left[ \overbrace{u_B\left( q_{n}-y_{n} -\sigma _{n-1}\right) -u_B\left( q_{n-1} -\sigma _{n-1}\right) }^{(\text {II})}\right] \nonumber \\&\quad =\left[ \overbrace{u_A\left( y_{n}+\sigma _{n-1}\right) -u_A\left( \sigma _{n-1}\right) }^{(\text {III})}\right] \cdot \overbrace{u'_B\left( q_{n}-y_{n} -\sigma _{n-1} \right) }^{(\text {IV})}. \end{aligned}$$

(30)

Using that $u_A$ and $u_B$ have two continuous derivatives on (0, Q],  we obtain the following Taylor expansions for any n,  with $2\le n \le N$:

$$\begin{aligned} (\text {I})= & {} u'_A\left( \sigma _{n-1}\right) + \xi _\mathrm{{I}} \left( y_{n}\right) ,\end{aligned}$$

(31)

$$\begin{aligned} (\text {II})= & {} u'_B\left( q_{n-1}-\sigma _{n-1}\right) \cdot \left( q_n-q_{n-1}-y_{n}\right) + \xi _\text {II}\left( q_n-q_{n-1}-y_{n}\right) ,\end{aligned}$$

(32)

$$\begin{aligned} (\text {III})= & {} u'_A\left( \sigma _{n-1}\right) \cdot y_{n}+ \xi _\mathrm{{III}} \left( y_{n}\right) ,\end{aligned}$$

(33)

$$\begin{aligned} (\text {IV})= & {} u'_B\left( q_{n-1}-\sigma _{n-1}\right) + \xi _\mathrm{{IV}}\left( q_n-q_{n-1}-y_{n}\right) , \end{aligned}$$

(34)

where $\xi _\mathrm{{I}},$ $\xi _\mathrm{{I}},$ $\xi _\mathrm{{III}}$ and $\xi _\mathrm{{IV}}$ are the error terms. Since $\sigma _{n-1} \ge \sigma _1>0$ and $q_{n-1}-\sigma _{n-1} \ge q_1-\sigma _{1}>0,$ both the leading order and the error terms in Eqs. (31)–(34) can be estimated by using bounds on the first and second derivatives of the utility functions $u_A$ and $u_B,$ which are bounded on any compact subinterval of (0, Q] by the Extreme Value Theorem. More specifically, we define

$$\begin{aligned} M^{\#}_A:= & {} \sup _{x\in [\sigma _1,Q]}|u_A^\#(x)|<\infty ,\quad (\#=','') \\ \nonumber \end{aligned}$$

(35)

$$\begin{aligned} M^{\#}_B:= & {} \sup _{x \in [q_1-\sigma _1,Q]}|u_B^{\#}(x)|<\infty . \quad (\#=','') \end{aligned}$$

(36)

Note that since $q_1=\frac{\varepsilon }{2},$ the outcome of the first bargaining round in accordance with protocol ${\mathcal {P}}^{\varepsilon ,N}$ is independent of N,  and hence so are the constants $M_i^{\#}$ ($i=A,B$).¹⁵ Using Taylor’s theorem, we can estimate the error terms as follows:

$$\begin{aligned}&|\xi _\mathrm{{I}}(y_n)|\le M^{\prime \prime }_A|y_n|, \\&|\xi _\mathrm{{II}}(q_n-q_{n-1}-y_{n})|\le \frac{1}{2} M^{\prime \prime }_B |q_n-q_{n-1}-y_{n}|^2, \\&|\xi _\mathrm{{III}}(y_n)|\le \frac{1}{2} M^{\prime \prime }_A |y_n|^2, \\&|\xi _\mathrm{{IV}}(q_n-q_{n-1}-y_{n})|\le M^{\prime \prime }_B |q_n-q_{n-1}-y_{n}|. \end{aligned}$$

Using the fact that for all $1\le n \le N$ we have $y_n \le q_n-q_{n-1}$ as well as $q_n-q_{n-1}-y_n \le q_n-q_{n-1},$ Eq. (30) implies that for any n,  with $2\le n \le N,$

$$\begin{aligned}&\left| u'_A\left( \sigma _{n-1}\right) \cdot u'_B\left( q_{n-1}-\sigma _{n-1}\right) \cdot \left( q_n-q_{n-1}-2y_n \right) \right| \nonumber \\&\quad \le \frac{3}{2} \left( M_A^{\prime \prime }M_B^{\prime } +M_A^{\prime }M_B^{\prime \prime }\right) \left( q_n-q_{n-1}\right) ^2 + M_A^{\prime \prime }M_B^{\prime \prime } \left( q_n-q_{n-1}\right) ^3. \end{aligned}$$

(37)

Moreover, since $u_i''\le 0$ ($i=A,B$),

$$\begin{aligned} |u'_A\left( \sigma _{n-1}\right) \cdot u'_B\left( q_{n-1}-\sigma _{n-1}\right) |\ge m_A' m_B', \end{aligned}$$

where

$$\begin{aligned} m_A':= & {} u_A'(Q-(q_1-y_1))>0, \\ m_B':= & {} u_B'(Q-y_1)>0. \end{aligned}$$

To see that $m_A'$ and $m_B'$ are indeed strictly positive, note that the derivatives of $u_i$ are non-negative and non-increasing, since $u_i$ itself is strictly increasing and concave ($i=A,B$). Thus, $u'_i(x)=0$ can only occur at the boundary value $x=Q,$ if at all. The positivity of $m_A'$ and $m_B'$ then follows from the fact that $0<y_1<q_1\le Q.$¹⁶ In conclusion, we obtain that for any n with $2\le n \le N,$

$$\begin{aligned}&\left| y_n-\frac{q_n-q_{n-1}}{2}\right| \nonumber \\&\quad \le \frac{3\left( M_A^{\prime \prime }M_B^{\prime } +M_A^{\prime }M_B^{\prime \prime }\right) }{4m_A'm_B'} \left( q_n-q_{n-1}\right) ^2+ \frac{M_A^{\prime \prime }M_B^{\prime \prime }}{2m_A'm_B'} \left( q_n-q_{n-1}\right) ^3.\nonumber \\&\quad =: C_1 (q_n-q_{n-1})^2+C_2(q_n-q_{n-1})^3 \nonumber \\&\quad \le \left( C_1 { \cdot } \mathop {\hbox {max}}\limits _{2\le { l}\le N}|q_{{ l}}-q_{{ l}-1}| + C_2 { \cdot } \mathop {\hbox {max}}\limits _{2\le { l} \le N}|q_{{l}}-q_{{ l}-1}|^2 \right) (q_n-q_{n-1}), \end{aligned}$$

(38)

where $C_1,$ $C_2$ are independent of N by definition.

We can now estimate the difference between the outcome $x_N=\sum _{n=1}^N y_n$ for player A in accordance with protocol ${\mathcal {P}}^{\varepsilon ,N}$ and half the total amount Q,  namely,

$$\begin{aligned}&\left| x_N-\frac{Q}{2} \right| = \left| \sum _{n=1}^N y_n - \frac{Q}{2} \right| \nonumber \\&\quad \le y_1 +\left| \sum _{n=2}^N \left( y_n-\frac{q_n-q_{n-1}}{2}\right) \right| + \left| \sum _{n=2}^N \frac{q_n-q_{n-1}}{2}-\frac{Q}{2}\right| \end{aligned}$$

(39)

$$\begin{aligned}&\quad \le y_1+ \left( C_1 \cdot \mathop {\hbox {max}}\limits _{2\le l \le N}|q_l-q_{l-1}| + C_2 \cdot \mathop {\hbox {max}}\limits _{2\le l \le N}|q_{l}-q_{l-1}|^2 \right) \cdot \,\left| \sum _{n=2}^N (q_n-q_{n-1}) \right| + \frac{q_1}{2}\nonumber \\&\quad \le \frac{\varepsilon }{2} + \left( C_1 \cdot \mathop {\hbox {max}}\limits _{2\le l \le N}|q_{l}-q_{ l-1}| + C_2 \cdot \mathop {\hbox {max}}\limits _{2\le {l} \le N}|q_{l}-q_{ l-1}|^2 \right) \cdot \,(Q -q_1)+\frac{\varepsilon }{4}, \end{aligned}$$

(40)

where we have used (38) and the fact that $y_1=\varepsilon /2.$ Choosing N and $q_2, \dots , q_N,$ with $q_2\le \dots \le q_N,$ such that

$$\begin{aligned} \left( C_1 \cdot \mathop {\hbox {max}}\limits _{2\le {l} \le N}|q_{l}-q_{l-1}| + C_2 \cdot \mathop {\hbox {max}}\limits _{2\le {l} \le N}|q_{l}-q_{l-1}|^2 \right) \cdot Q< \frac{\varepsilon }{4} \end{aligned}$$

concludes the proof of the theorem. $\square$

Remark 1

In the proof of Theorem 2, the fact that the derivatives of $u_i$ ($i=A,B$) at zero might be singular required us to treat the first bargaining round separately, by imposing $q_1=\varepsilon /2,$ which is independent of N. If, however, we assume that $u_i\in C^2([0,Q])$ and $u'_i(Q)>0$ ($i=A,B$), this is no longer necessary. Indeed, in this case, the suprema in Eqs. (35) and (36) remain finite when evaluated over the entire interval [0, Q],  thereby enabling us to replace (38) by

$$\begin{aligned}&\left| y_n-\frac{q_n-q_{n-1}}{2}\right| \\&\quad \le \left( C_1 \cdot \mathop {\hbox {max}}\limits _{1\le l \le N}|q_{ l}-q_{l-1}| + C_2 \cdot \mathop {\hbox {max}}\limits _{1\le l \le N}|q_{l}-q_{l-1}|^2 \right) \cdot \,(q_n-q_{n-1}). \end{aligned}$$

We stress that this inequality is now valid for all $1\le n \le N,$ with constants $C_1$ and $C_2$ independent of $\varepsilon .$ This leads to the following estimate of the difference between the outcome $x_N$ and half the total amount Q:

$$\begin{aligned} \left| x_N-\frac{Q}{2} \right| \le \left( C_1 \cdot \mathop {\hbox {max}}\limits _{1\le l \le N}|q_{l}-q_{l-1}| + C_2 \cdot \mathop {\hbox {max}}\limits _{1\le l \le N}|q_{l}-q_{l-1}|^2 \right) \cdot Q. \end{aligned}$$

(41)

As a consequence, the assertion of Corollary 2 follows for $u_i'(Q)>0$ and extends to all $u_i \in C^2([0,Q])$ by a simple approximation argument at Q ($i=A,B$).¹⁷

Remark 2

Assume again that $u_i'(Q)>0$ ($i=A,B$). Then (41) can be used to determine the required granularity of a particular protocol more explicitly than (40), given any error margin $\varepsilon >0.$ This is possible because $C_1$ and $C_2$ in (41) are now independent of $\varepsilon .$ As an example, consider the case of equidistant $q_n,$ i.e., $q_n:= n\cdot Q/N$ for $1\le n\le N.$ Then, using (41), we obtain that $| x_N-\frac{Q}{2} |<\varepsilon$ if

$$\begin{aligned} \frac{C_1Q^2}{N} + \frac{C_2Q^3}{N^2} <\varepsilon . \end{aligned}$$

It then follows that the number of required rounds $N^\varepsilon$ grows at most inversely proportional to the error margin $\varepsilon >0.$ Namely,

$$\begin{aligned} N^\varepsilon := {\left\{ \begin{array}{ll} \lceil C_1 Q^2 / \varepsilon \rceil &{}\text {if } C_2=0, \\ \left\lceil \frac{1}{\sqrt{\frac{\varepsilon }{C_2Q^3}+\frac{C_1^2}{(4C_2^2Q^2)}}-\frac{C_1}{2C_2Q}} \right\rceil &{}\text {if } C_2>0. \end{array}\right. } \end{aligned}$$

We note that $C_2=0$ if one player has linear utility.