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Nash bargaining for log-convex problems

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Abstract

We introduce log-convexity for bargaining problems. With the requirement of some basic regularity conditions, log-convexity is shown to be necessary and sufficient for Nash’s axioms to determine a unique single-valued bargaining solution up to choices of bargaining powers. Specifically, we show that the single-valued (asymmetric) Nash solution is the unique solution under Nash’s axioms without that of symmetry on the class of regular and log-convex bargaining problems, but this is not true on any larger class. We apply our results to bargaining problems arising from duopoly and the theory of the firm. These problems turn out to be log-convex but not convex under familiar conditions. We compare the Nash solution for log-convex bargaining problems with some of its extensions in the literature.

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Notes

  1. Thomson (2009) provides an excellent survey of the literature on the axiomatic approach to bargaining as pioneered by Nash. An alternative approach is to model bargaining explicitly as a non-cooperative game. Nash (1953, p. 129) argued that the two approaches “are complementary; each helps to justify and clarify the other.” See Binmore et al. (1986) for a formal establishment of the relationship between the Nash axiomatic and the sequential non-cooperative approaches to bargaining.

  2. This interpretation can be traced to Shubik (1959, p. 50).

  3. By the Lemma in Roth (1977, p. 65), a bargaining solution is Pareto optimal whenever the solution satisfies IIA, INV, and strict individual rationality (SIR). Thus, IIA, INV, and PO can be replaced by IIA, INV, and SIR. Since bargaining problems in our class satisfy all the assumptions in Roth (1977) except for the convexity, we will consider IIA, INV, and SIR as the Nash’s axioms for the asymmetric Nash bargaining solution.

  4. Bargaining problems arising from the principal–agent problems need not be convex unless randomized contracts are allowed (see Ross 1973; Hougaard and Tvede 2003).

  5. Similarly, Hougaard and Tvede (2003, p. 82) argue that it is not reasonable to allow for randomized contracts in the context of a principal–agent problem.

  6. Zhou’s characterization of single-valued solutions does not specify how to select among multiple Nash product maximizers. In a recent paper, Qin et al. (2014) provide a complete characterization of exact single-valued solutions satisfying Nash axioms under more general conditions on bargaining problems.

  7. They replaced PO with weak Pareto optimality (WPO). See Peters and Vermeulen (2012, pp. 26–27) for discussions about their refinements of multi-valued extensions in Kaneko (1980), Mariotti (1998a) and Xu and Yoshihara (2006). For further references and discussions on non-convex bargaining problems, see the recent survey by Thomson (2009).

  8. See Mariotti (1998b) for an alternative characterization of Conley and Wilkie’s (1996) extension.

  9. We have recently become aware of the Boche and Schubert (2011) paper. Our analysis is carried out independently of their work.

  10. Unless noticed otherwise, bargaining solutions are single-valued throughout the rest of the paper.

  11. See Boche and Schubert (2011) for an application of log-convexity to extend the symmetric Nash solution with zero threat point.

  12. Indeed, it is straightforward to verify that \((S\cap S^{\prime },d)\) is regular and \(V(S\cap S^{\prime },d)\subseteq V(S,d)\cap V(S^{\prime },d)\). On the other hand, take any \(v\in V(S,d)\cap V(S^{\prime },d),\,u\in S\), and \(u^{\prime }\in S^{\prime }\) such that \(u_{i}>d_{i},\,u_{i}^{\prime }>d_{i},\,v_{i}\le \ln (u_{i}-d_{i})\), and \(v_{i}\le \ln (u_{i}^{\prime }-d_{i})\) for \(i=1,2\). Set \(\underline{u}_{i}=\min \{u_{i},u_{i}^{\prime }\}\). Then, \(v_{i}\le \ln (\underline{u}_{i}-d_{i})\) and \(d<<\underline{u}\le u,u^{\prime }\). By the regularity, \(\underline{u}\in S\cap S^{\prime }\). This shows \(v\in V(S\cap S^{\prime },d)\). Hence, \(V(S\cap S^{\prime },d)\supseteq V(S,d)\cap V(S^{\prime },d)\).

  13. Since \(V(S, d)\) is comprehensive, it suffices to consider boundary points that are Pareto optimal.

  14. Since \(x_1 < y_1\), the condition is equivalent to \(y\) lying below the tangent line.

  15. Consider affine transformation \(\tau :\mathfrak {R}^2\longrightarrow \mathfrak {R}^2\):

    $$\begin{aligned} \tau (u) = \left( {\alpha \over z_1}u_1, {1-\alpha \over z_2}u_2\right) ,\ u\in \mathfrak {R}^2. \end{aligned}$$

    Then, by (10), \(\tau (S) = S^\circ \). Since \(f(S^\circ , d^\circ ) = (\alpha , 1-\alpha ),\,f(\tau (S), \tau (d^\circ )) = (\alpha , 1-\alpha )\). By INV, \(\tau (f(S, d^\circ ))=f(\tau (S), \tau (d^\circ ))\). Hence, \(f(S, d^\circ ) = \tau ^{-1}(\alpha , 1-\alpha ) = z\).

  16. Bilateral stability of the \(n\)-person symmetric Nash solution was introduced in Harsanyi (1959). Bilateral stability also holds for the asymmetric Nash solution. See Lensberg (1988) for general results on multilateral as well as bilateral stabilities of the symmetric Nash solution.

  17. Binmore (2007, pp. 480–481) illustrated the use of Nash bargaining solution as a tool for studying collusive behavior in a Cournot duopoly with asymmetric constant marginal costs. However, the profit choice set without side payments is drawn as if it is convex (Figure 16.13(a)), which is not with respect to the assumed demand and cost functions.

  18. Explicit transfers or briberies between firms may be too risky due to antitrust scrutiny.

  19. The non-convexity of \(\Pi \) was shown in Tirole (1988, p. 242, 271) under one extra assumption that the monopoly profit function for firm 2 is concave. We will show the non-convexity without such an assumption.

  20. If \(\psi \) is concave, then \(S\) is convex in which case both extended Nash solutions reduce to the standard Nash solution.

  21. The reason is as follows. Since \(\bar{v}\) is in \(\bar{V}\), there exists a sequence \(\{\bar{v}(n)\}_{n}\) in the convex hull of \(V\) such that \(\bar{v} (n)\rightarrow \bar{v}\). By Carathéodory Theorem, for each \(n\), there exist \(v^{1}(n),v^{2}(n),v^{3}(n)\) in \(V\) and \(\lambda (n)\in \mathfrak {R}_{+}^{3}\) satisfying (26) such that \(\bar{v}(n)=\lambda _{1}(n)v^{1}(n)+\lambda _{2}(n)v^{2}(n)+\lambda _{3}(n)v^{3}(n)\). Finally, (25) and (27) are established by setting \(\epsilon (n)=\bar{v}-\bar{v}(n)\) for all \(n\).

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Correspondence to Cheng-Zhong Qin.

Additional information

Professor Shuzhong Shi passed away before the completion of this paper. He was a pleasure and an inspiration to work with and will be deeply missed.

We gratefully acknowledge helpful comments from Rabah Amir, Kim Border, Juan Carrillo, Bo Chen, Matthew Jackson, Ehud Kalai, Joel Sobel, Walter Trockel, Simon Wilkie, Adam Wong, Lin Zhou, and seminar and conference participants at Beijing University, Ohio State University, Southwest Economic Theory Conference, the Third Congress of the Game Theory Society, Shanghai University of Finance and Economics, University of Arizona, University of Bielefeld, University of California at Irvine, Riverside, and San Diego, University of Southern California, and Zhejiang University. We also thank an anonymous referee for comments that helped to improve the paper.

Appendix: Proofs

Appendix: Proofs

Before proceeding with the proof of Theorem 1, we make the following observations. First, the logarithmic transformation \((v_{1},v_{2})=(\ln (u_{1}-d_{1}),\ln (u_{2}-d_{2}))\) is a homeomorphism from \(S\cap \{(u_{1},u_{2})\,|\,u_{1}>d_{1},u_{2}>d_{2}\}\) onto \(V(S,d)\). It follows that the boundary \(\partial V(S,d)\) of \(V(S,d)\) must be the homeomorphic image of that of \(S\cap \{(u_{1},u_{2})\,|\,u_{1}>d_{1},u_{2}>d_{2}\}\); that is,

$$\begin{aligned} \partial V(S,d)=\left\{ \big (\ln (u_{1}-d_{1}),\ln (u_{2}-d_{2})\big )\bigg | u\in \partial S: \begin{array}{cll} &{} u_{1}>d_{1},&{} \\ &{} u_{2}>d_{2}.&{} \end{array} \right\} . \end{aligned}$$

Thus, \(V(S,d)\) is closed. Second, because \(S\) is bounded above, \( V(S,d) \) is bounded above and \(\partial V(S,d)\not =\emptyset \). Third, for any \( (u_{1},u_{2})\in S\cap \{(u_{1},u_{2})\,|\,u_{1}>d_{1},u_{2}>d_{2}\}\), the intervals \(((d_{1},u_{2}),(u_{1},u_{2})]\) and \(((u_{1},d_{2}),(u_{1},u_{2})]\) are transformed into \(((-\infty ,\ln (u_{2}-d_{2})),(\ln (u_{1}-d_{1}),\ln (u_{2}-d_{2}))]\) and \(((\ln (u_{1}-d_{1}),-\infty ),(\ln (u_{1}-d_{1}),\ln (u_{2}-d_{2}))]\) in \(V(S,d)\), respectively. As a result, \(V(S,d)=V(S,d)-\mathfrak {R}_{+}^{2}\). Thus, by moving the origin properly, we can assume, without loss of generality, that for some constant \(h>0\),

$$\begin{aligned} V(S,d)\subset -(h,h)-\mathfrak {R}_{+}^{2}, \end{aligned}$$
(23)

so that the closed cone generated by \(V(S,d)\) and any line passing through the origin with a normal vector in \(\mathfrak {R}_{+}^{2}\) has no point in common other than the origin.

Lemma 3

Let \(U\subseteq \mathfrak {R}^{2}\) be closed, convex, and comprehensive (i.e., \( U=U-\mathfrak {R}_{+}^{2}\)). If the boundary \(\partial U\) of \(U\) is nonempty, then \( U=\partial U-\mathfrak {R}_{+}^{2}\).

Proof

Notice \(\partial U-\mathfrak {R}_{+}^{2}\subseteq U-\mathfrak {R}_{+}^{2}=U\) is automatic. Conversely, let \(u\) be any interior point of \(U\). Then, there exists a number \(\epsilon >0\) such that \((u_{i}+\delta ,u_{j})\in U\) for \(i\not = j\) and for all \(\delta \in [0,\epsilon )\). Set \(\delta _{i}=\sup \{\delta |(u_{i}+\delta ,u_{j})\in U\}\). Since \(U\) is convex and comprehensive, \(\delta _{i}<\infty \) for at least one \(i\), for otherwise \( U=\mathfrak {R}^{2}\) which contradicts \(\partial U\not =\emptyset \). Assume without loss of generality \(\delta _{1}<\infty \). Then, \(u^{1}=(u_{1}+\delta _{1},u_{2})\in \partial U\) because \(U\) is closed. It follows that \( u=u^{1}-(\delta _{1},0)\in \partial U-\mathfrak {R}_{+}^{2}\). This shows \(U\subseteq \partial U-\mathfrak {R}_{+}^{2}\).

Proof of Theorem 1

As observed earlier, log-convexity is equivalent to conditions that \(V(S, d)\) is convex and it does not contain segments with normal vectors \((\alpha , 1-\alpha )\) for all \(\alpha \in (0, 1)\). To prove the necessity, we only need to show that \(V = V(S,d)\) is convex because the necessity of the rest is obvious.

Let \(\bar{V}\) denote the closure of the convex hull of \(V\). Notice \( \bar{V}\) also satisfies \(\bar{V}=\bar{V}-\mathfrak {R}_{+}^{2}\); hence, from Lemma 3, \( \bar{V}=\partial {\bar{V}}-\mathfrak {R}_{+}^{2}\). For any boundary point \(\bar{v}\in \partial {\bar{V}}\), it follows from the separation theorem and the comprehensiveness of \(\bar{V}\) that there exists \(a\in \mathfrak {R}^{2}_+\) such that

$$\begin{aligned} a\cdot \bar{v}=\max _{v\in \bar{V}}a\cdot v = \max _{v\in V}a\cdot v. \end{aligned}$$
(24)

Without loss of generality, we may take \(a=(\theta ,1-\theta )\) for some \( \theta \in [0,1]\).

There exist sequences \(\{v^{k}(n)\}_{n}\) in \(V\) for \(k=1,2,3,\,\{\lambda (n)\}_{n}\) in \(\mathfrak {R}_{+}^{3}\), and \(\{\epsilon (n)\}_{n}\) in \(\mathfrak {R}\) such thatFootnote 21

$$\begin{aligned}&\bar{v}=\lambda _{1}(n)v^{1}(n)+\lambda _{2}(n)v^{2}(n)+\lambda _{3}(n)v^{3}(n)+\varepsilon (n),\end{aligned}$$
(25)
$$\begin{aligned}&\lambda _{1}(n)+\lambda _{2}(n)+\lambda _{3}(n)=1,\end{aligned}$$
(26)
$$\begin{aligned}&\varepsilon (n)\rightarrow 0. \end{aligned}$$
(27)

By the nonnegativity of \(\lambda (n)\) and (26), we may assume

$$\begin{aligned} \lambda (n)\rightarrow \bar{\lambda }\in \mathfrak {R}_{+}^{3}\ {\text{ with }}\ \bar{ \lambda }_{1}+\bar{\lambda }_{2}+\bar{\lambda }_{3}=1. \end{aligned}$$
(28)

By (23), the sequences \(\{\lambda _{k}(n)v^{k}(n)\}_{n},\,k=1,2,3\), are bounded above and by (25), (27), and (28), they are also bounded below. Hence, we may assume

$$\begin{aligned} \lambda _{k}(n)v^{k}(n)\rightarrow \bar{w}^{k}\in \mathfrak {R}^{2},\quad k=1,2,3\Rightarrow \bar{v}=\bar{w}^{1}+\bar{w}^{2}+\bar{w}^{3}. \end{aligned}$$
(29)

Suppose first \(\bar{\lambda }_{k}>0\) for \(k =1, 2, 3\). Then, since \(V\) is closed, (25) and (27)–(29) imply

$$\begin{aligned} v^{k}(n)\rightarrow \bar{v}^{k}\in V,\quad \ k=1,2,3,\ \quad \bar{v}=\bar{\lambda } _{1}\bar{v}^{1}+\bar{\lambda }_{2}\bar{v}^{2}+\bar{\lambda }_{3}\bar{v}^{3}. \end{aligned}$$
(30)

Suppose now \(\bar{\lambda }_{k}=0\) for some \(k\) but \(\bar{\lambda }_{k^{\prime }}>0\) for \(k^{\prime }\not =k\). Without loss of generality, assume \(\bar{ \lambda }_{1}=0,\,\lambda _2 > 0\), and \(\lambda _3 > 0\). In this case,

$$\begin{aligned} \bar{v}=\bar{w}^{1}+\bar{\lambda }_{2}\bar{v}^{2}+\bar{\lambda }_{3}\bar{v} ^{3},\quad \bar{\lambda }_{2},\bar{\lambda }_{3}>0,\quad \bar{\lambda }_{2}+\bar{\lambda }_{3}=1. \end{aligned}$$
(31)

By (24) and (29),

$$\begin{aligned} a\cdot \bar{w}^{1}=\lim _{n\rightarrow \infty }a\cdot \lambda _{1}(n)v^{1}(n)=\lim _{n\rightarrow \infty }\lambda _{1}(n)a\cdot v^{1}(n)\le \lim _{n\rightarrow \infty }\bar{\lambda }_{1}a\cdot \bar{v}=0. \end{aligned}$$

On the other hand, by (24) and (31),

$$\begin{aligned} a\cdot \bar{w}^{1}=a\cdot \bar{v}-\left[ \bar{\lambda }^{2}(a\cdot \bar{v}^{2})+ \bar{\lambda }^{3}(a\cdot \bar{v}^{3})\right] \ge 0. \end{aligned}$$

It follows that \(a\cdot \bar{w}^{1}=0\). This shows that \(\bar{w}^{1}\) lies on the line having normal vector \(a = (\theta ,1-\theta )\in \mathfrak {R}_{+}^{2}\) and passing though the origin. Furthermore, as the limit of \( \{\lambda _{1}(n)v^{1}(n)\},\,\bar{w}^{1}\) is in the closed cone generated by \(V\). Thus, by (23), we must have \(\bar{w}^{1}=0\). Suppose finally \(\bar{ \lambda }_{k}=1\) for some \(k\). Assume without loss of generality \(\bar{\lambda }_{3}\not =0\). In this case, a similar proof as before shows \(\bar{w}^{1}= \bar{w}^{2}=0\).

In summary, by letting \(\bar{v}^{k}\) be arbitrary element in \(V\) when \(\bar{ \lambda }_{k}=0\), the preceding analysis establishes

$$\begin{aligned} \bar{v}=\bar{\lambda }_{1}\bar{v}^{1}+\bar{\lambda }_{2}\bar{v}^{2}+\bar{ \lambda }_{3}\bar{v}^{3}. \end{aligned}$$
(32)

When \(\bar{\lambda }_{k}>0\), (24) and (32) together imply

$$\begin{aligned} a\cdot \bar{v}^{k}=\max _{v\in V}a\cdot v, \end{aligned}$$

which implies \(\bar{v}^{k}\in \partial V\). If \(a\in \mathfrak {R}_{++}^{2}\), then the above equality and the assumption that (4) has a unique solution imply \(\bar{ v}^{k}=\bar{v}\) whenever \(\bar{\lambda }_{k}>0\). Hence, \(\bar{v}\in \partial V\subset V\). If \(a=(1,0)\), then

$$\begin{aligned} \bar{v}_{1} = a\cdot \bar{v} = a\cdot \bar{v}^{k} = \bar{v}_{1}^{k}. \end{aligned}$$

Assume without loss of generality \(\bar{\lambda }_{1}>0\) and \(\bar{v} _{2}^{1}=\max \{\bar{v}_{2}^{k}\,|\,\bar{\lambda }_{k}>0\}\). Then, by (32),

$$\begin{aligned} \bar{v}_{2}=\bar{\lambda }_{1}\bar{v}_{2}^{1}+\bar{\lambda }_{2}\bar{v} _{2}^{2}+\bar{\lambda }_{3}\bar{v}_{2}^{3}\le \bar{v}_{2}^{1}. \end{aligned}$$

Thus, \(\bar{v}\le \bar{v}^{1}\). This shows \(\bar{v}\in V\). If \( a=(0,1) \), then a similar proof establishes \(\bar{v}\in V\).

We have shown that \(\bar{v}\in V\) for any \(\bar{v}\in \partial \bar{V}\). Thus, by Lemma 3,

$$\begin{aligned} V\subseteq \bar{V}=\partial \bar{V}-\mathfrak {R}_{+}^{2}\subseteq V-\mathfrak {R}_{+}^{2}=V. \end{aligned}$$

This concludes \(V=\bar{V}\) which implies that \(V\) is convex. \(\square \)

Proof of Proposition 2

To show the non-convexity of \(\Pi \), notice that the PF of \(\Pi \) is given by the function

$$\begin{aligned} \pi _{1}=\psi (\pi _{2})=h(p(\pi _{2}),\pi _{2})),\quad \pi _{2}\in [0,\bar{\pi }_{2}]. \end{aligned}$$

By (13),

$$\begin{aligned} h_{2}(p,\pi _{2})=-{\frac{p-c_{1}}{p-c_{2}},}\ h_{22}(p,\pi _{2})=0,\quad h_{12}(p,\pi _{2})={\frac{c_{2}-c_{1}}{(p-c_{2})^{2}}}, \end{aligned}$$
(33)

and

$$\begin{aligned} h_{11}(p,\pi _{2})=\bar{\pi }_{1}^{\prime \prime }(p)-{\frac{ 2(c_{2}-c_{1})\pi _{2}}{(p-c_{2})^{3}}},\ p^{\prime }(\pi _{2})=-{\frac{ h_{12}(p(\pi _{2}),\pi _{2})}{h_{11}\left( p(\pi _{2}),\pi _{2}\right) }}. \end{aligned}$$
(34)

It follows from the envelope theorem, (33), and (34) that

$$\begin{aligned} \psi ^{\prime }(\pi _{2})=h_{2}(p(\pi _{2}),\pi _{2}), \end{aligned}$$
(35)

and

$$\begin{aligned} \psi ^{\prime \prime }(\pi _{2})=h_{12}(p(\pi _{2}),\pi _{2})p^{\prime }(\pi _{2})=-{\frac{[h_{12}(p(\pi _{2}),\pi _{2})]^{2}}{h_{11}(p(\pi _{2}),\pi _{2})}>0,} \end{aligned}$$
(36)

where the strict inequality holds since \(h_{11}(p,\pi _{2})<0\) due to the strict concavity of \(\bar{\pi }_{1}(p)\).

Next, we show that \((\Pi ,d)\in {\mathcal {L}}\) for all \(d\ge 0\) that are strictly Pareto dominated in \(\Pi \). By the discussions following Proposition 1, it suffices to show that \(\psi ^{\prime }(\pi _{2})<0\) and \(e(\pi _{2})\equiv \frac{(\pi _2-d_2)\psi ^{\prime }(\pi _{2})}{\psi (\pi _{2})-d_1}\) is strictly decreasing in \(\pi _2\) over \( (d_2, \bar{\pi }_2)\).

By (33) and (35), \(\psi ^{\prime }(\pi _{2})<0\) if \(p(\pi _{2})>c_{2}\). Suppose \(p(\pi _{2})<\bar{p}_{1}\). The strict concavity of \(\bar{\pi }_{1}(p)\) implies that \(\bar{\pi }_{1}^{\prime }(p)>0\) for \(p\in [p(\pi _{2}), \bar{p}_{1}]\), which in turn implies \(h_{1}(p,\pi _{2})>0 \) for \(p\in [p(\pi _{2}),\bar{p}_{1}]\). Consequently, \(p(\pi _{2})\) cannot be Pareto optimal. Thus, \(p(\pi _{2})\ge \bar{p}_{1}>c_{2}\). By the strict concavity of \(\bar{\pi }_{1}(p)\) and (34),

$$\begin{aligned} -h_{11}(p,\pi _{2})>{\frac{2(c_{2}-c_{1})\pi _{2}}{(p-c_{2})^{3}}>0} \end{aligned}$$
(37)

for any \(\pi _{2} \in (0,\bar{\pi }_{2})\). It then follows from (33)-(37) that

$$\begin{aligned} \psi ^{\prime }(\pi _{2})+\pi _{2}\psi ^{\prime \prime }(\pi _{2})&= h_{2}(p(\pi _{2}),\pi _{2})+\pi _{2}{\frac{[h_{12}(p(\pi _{2}),\pi _{2})]^{2}}{-h_{11}(p(\pi _{2}),\pi _{2})}} \\&< -{\frac{p(\pi _{2})-c_{1}}{p(\pi _{2})-c_{2}}+\frac{c_{2}-c_{1}}{2[p(\pi _{2})-c_{2}]}} \\&= -\frac{2p(\pi _{2})-c_{1}-c_{2}}{2[p(\pi _{2})-c_{2})]} \\&< 0 \end{aligned}$$

for any \(\pi _{2}\in (0,\bar{\pi }_{2})\), where the last inequality holds since \(p(\pi _{2})\ge \bar{p}_{1}>c_{2}>c_{1}\). Since the derivative of \(e(\pi _2)\) has the same sign as

$$\begin{aligned} \left[ \psi (\pi _2)-d_1\right] \left[ \psi ^{\prime }(\pi _{2})+\pi _{2}\psi ^{\prime \prime }(\pi _{2}) - d_2\psi ^{\prime \prime }(\pi _{2})\right] - (\pi _2-d_2)[\psi ^{\prime }(\pi _2)]^2 <0, \end{aligned}$$

where the inequality follows from \(\pi _2 \ge d_2\ge 0,\,\psi (\pi _2)\ge d_1,\,\psi ^{\prime \prime }(\pi _2)>0\), and \(\psi ^{\prime }(\pi _{2})+\pi _{2}\psi ^{\prime \prime }(\pi _{2}) <0\) for any \(\pi _2\in (d_2,\bar{\pi }_2)\). \(\square \)

Proof of Proposition 3

The first-order condition for maximization problem (22) is

$$\begin{aligned} (\rho - g)x'\phi '+ x = 0 \Leftrightarrow \mu _x (y+\phi (g)) - (\rho - g)\phi '(g) = 0. \end{aligned}$$
(38)

Recall that \(D(y) = x(y+\phi (g(y)))/(\rho - g(y))\). Thus, simple calculation shows that (38) implies

$$\begin{aligned} D'(y) = {x'(y+\phi (g(y)))\over \rho - g(y)}\ \mathrm{and}\ {D(y)\over D'(y)} = {x(y+\phi (g(y)))\over x''(y+\phi (g(y)))}. \end{aligned}$$

Consequently, assumptions (i) and (iii) in Lemma 2 are automatically satisfied due to the assumptions in Proposition 3. Next, since \(g(y) < \rho \) and \(\mu _x (y + \phi (g(y))) = (\rho - g(y))\phi '(g(y))\), there exists \(\bar{y}\) such that \(\mu _x(\bar{y} + \phi (g(\bar{y}))) - \bar{y} < 0\). This together with \(\mu (y) = \mu _x(y + \phi (g(y)))\) as shown above implies that assumption (ii) in Lemma 2 is also automatically satisfied.

Let \(\epsilon (y) = -yD'(y)/D(y)\). Then, from the derivation of \(D(y)\) it follows that

$$\begin{aligned} \epsilon (y) = -{yx'(y)\over x(y)} = {y\over \mu _x(y+\phi (g(y)))}. \end{aligned}$$

Hence,

$$\begin{aligned} \epsilon '(y) > 0 \Leftrightarrow \mu _x > y{d\mu _x\over dy}. \end{aligned}$$
(39)

By (38),

$$\begin{aligned} {d\mu _x\over dy} = \left[ (\rho -g)\phi '' - \phi '\right] {g'} \end{aligned}$$

and

$$\begin{aligned} {g'} = {\mu '_x\over (\rho - g)\phi '' - \phi ' - \mu '_x\phi '}. \end{aligned}$$

Notice that the dominator on the right-hand side of the last equation is positive under the assumptions. Combining these last two equations with (39) would then imply

$$\begin{aligned} \epsilon '(y) > 0 \Leftrightarrow (\mu _x - y\mu '_x)[(\rho - g)\phi '' - \phi '] - \mu _x\mu '_x\phi ' > 0 \end{aligned}$$

which holds under the assumed conditions. This completes the proof that all the conditions in Lemma 2 are satisfied. \(\square \)

Proof of Proposition 4

Notice that \((S, d^\circ )\in {\mathcal {R}}\) and by Proposition 1, \((S, d)\in {\mathcal {L}}\). Thus, by Theorem 2, there is a unique Nash solution given by (2), where the first-order condition is given by

$$\begin{aligned} -\frac{u_{1}\psi ^{\prime }(u_{1})}{\psi (u_{1})}=\frac{\alpha }{1-\alpha }. \end{aligned}$$
(40)

The Nash solution \((N^\alpha _1(S,d^{\circ } ),N^\alpha _2(S,d^{\circ }))\) is determined by (40) and \(u_{2}=\psi (u_{1})\). It follows that the ratio of the two payoff gains is

$$\begin{aligned} R^{N}(\alpha )\equiv \frac{N^\alpha _2(S,d^{\circ })}{ N^\alpha _1(S,d^{\circ })}=-\frac{1-\alpha }{\alpha }\psi ^{\prime }(N^\alpha _1(S,d^{\circ })). \end{aligned}$$

Moreover, the left-hand side of (40) is strictly increasing in \(u_{1}\) by \( (C)\) and the right-hand side is strictly increasing in \(\alpha \). It follows that \(N^\alpha _1(S,d^{\circ })\) is strictly increasing in \(\alpha \) for any interior solution.

The frontier of the convex hull of \(S\) is a straight line given by

$$\begin{aligned} u_{2}=\frac{\bar{u}_{2}}{\bar{u}_{1}}(\bar{u}_{1}-u_{1}). \end{aligned}$$

Applying Nash solution on the convex hull yields a solution \(u=(\alpha \bar{u}_{1},(1-\alpha )\bar{u}_{2})\). This solution and the Conley–Wilkie’s extended Nash solution have the same ratio of the payoff gains for the two players, which is given by

$$\begin{aligned} R^{CW}(\alpha )\equiv \frac{CW^\alpha (S,d^{\circ })}{ CW^\alpha _1(S,d^{\circ })}=\frac{(1-\alpha )\bar{u}_2 }{\alpha \bar{u}_1 }. \end{aligned}$$

Comparing the two solutions yields

$$\begin{aligned} \frac{R^{N}(\alpha )}{R^{CW}(\alpha )}=\frac{-\psi ^{\prime }\left( N^\alpha _1(S,d^{\circ })\right) }{\bar{u}_{2}/\bar{u}_{1}}. \end{aligned}$$

The strict convexity of \(\psi \) and the monotonicity of \(N^\alpha _1(S,d^{ \circ })\) at the interior solution imply that \(-\psi ^{\prime }(N^\alpha _1(S,d^{\circ }))\) is strictly decreasing in \(\alpha \). Thus, there exists \(\hat{\alpha }\in (0,1)\) such that (a) \(R^{N}(\hat{\alpha } )=R^{CW}(\hat{\alpha })\), (b) \(R^{N}(\alpha )>R^{CW}(\alpha )\) if \(\alpha < \hat{\alpha }\), and (c) \(R^{N}(\alpha )<R^{CW}(\alpha )\) if \(\alpha >\hat{ \alpha }\). The claims then follow since both solutions are on the frontier. \(\square \)

Proof of Proposition 5

Notice first that \((S, d^\circ )\in {\mathcal {R}}\cap \mathcal L\). Set \(z = N^\alpha (S, d^\circ )\). Then, by Theorem 1, \(z\) is the only tangent point between the indifference curve of Nash product \(u_1^\alpha u_2^{1-\alpha }\) and the Pareto frontier of \(S\). The corresponding tangent line is represented by

$$\begin{aligned} u_{2}=\psi (z_{1})+\psi ^{\prime }(z_{1})(u_{1}-z_{1}). \end{aligned}$$

It intersects the horizontal axis at \(I_{1}=(z_{1}+\psi (z_{1})/\psi ^{\prime }(z_{1}),0)\) and it intersects the vertical axis at \(I_{2}=(0,\psi (z_{1})-z_{1}\psi ^{\prime }(z_{1}))\). The distances \(\Vert z - I_1\Vert \) from \(z\) to \(I_{1}\) and \(\Vert z - I_2\Vert \) from \(z\) and \(I_{2}\) are given by

$$\begin{aligned} \left\| z-I_{1}\right\| ^{2}&= (\psi (z_{1})/\psi ^{\prime }(z_{1}))^{2}+(\psi (z_{1}))^{2}=\frac{(\psi (z_{1}))^{2}[1+(\psi ^{\prime }(z_{1}))^{2}]}{(\psi ^{\prime }(z_{1}))^{2}} \end{aligned}$$

and

$$\begin{aligned} \left\| z-I_{2}\right\| ^{2}&= (z_{1})^{2}+(z_{1}\psi ^{\prime }(z_{1}))^{2}=(z_{1})^{2}\left[ 1+(\psi ^{\prime }(z_{1}))^{2}\right] . \end{aligned}$$

It follows that

$$\begin{aligned} \frac{\left\| z-I_{2}\right\| ^{2}}{\left\| z-I_{1}\right\| ^{2}} =\left( \frac{z_{1}\psi ^{\prime }(z_{1})}{\psi (z_{1})}\right) ^{2}. \end{aligned}$$
(41)

Recall that the proportional distance property states that

$$\begin{aligned} \frac{\left\| z-I_{2}\right\| }{\left\| z-I_{1}\right\| }=\frac{ 1-\alpha }{\alpha }. \end{aligned}$$

By (41), the preceding equality is equivalent to

$$\begin{aligned} \left( \frac{1-\alpha }{\alpha }-\frac{z_{1}\psi ^{\prime }(z_{1})}{\psi (z_{1})}\right) \left( \frac{1-\alpha }{\alpha }+\frac{z_{1}\psi ^{\prime }(z_{1})}{\psi (z_{1})}\right) =0. \end{aligned}$$

Since \(\psi ^{\prime }(z_{1})<0\), the above equality in turn is equivalent to (40), which is automatic because it is the first-order condition for Nash bargaining solution \(z = N^\alpha (S, d^\circ )\). \(\square \)

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Qin, CZ., Shi, S. & Tan, G. Nash bargaining for log-convex problems. Econ Theory 58, 413–440 (2015). https://doi.org/10.1007/s00199-015-0865-z

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