Random-settlement arbitration and the generalized Nash solution: one-shot and infinite-horizon cases

Abstract

We study bilateral bargaining á la Nash (Econometrica 21:128–140, 1953) but where players face two sources of uncertainty when demands are mutually incompatible. First, there is complete breakdown of negotiations with players receiving zero payoffs, unless with probability p, an arbiter is called upon to resolve the dispute. The arbiter uses the final-offer-arbitration mechanism whereby one of the two incompatible demands is implemented. Second, the arbiter may have a preference bias toward satisfying one of the players that is private information to the arbiter and players commonly believe that the favored party is player 1 with probability q. Following Nash’s idea of ‘smoothing,’ we assume that $1-p$ is larger for larger incompatibility of demands. We provide a set of conditions on p such that, as p becomes arbitrarily small, all equilibrium outcomes converge to the Nash solution outcome if $q=1/2$, that is when the uncertainty regarding the arbiter’s bias is maximum. Moreover, with $q\ne 1/2$, convergence is obtained on a special point in the bargaining set that, independent of the nature of the set, picks the generalized Nash solution with as-if bargaining weights q and $1-q$. We then extend these results to infinite-horizon where instead of complete breakdown, players are allowed to renegotiate.

Appendices

Appendix 1: Proofs

Proof of Theorem 1

Our proof consists of three steps.

Step 1 In this step, we will show that if $(\hat{x}_{1},\hat{x}_{2})$ is a Nash equilibrium of an NDG-RS that has settlement bias $q\in (0,1)$ (i.e., Player 1’s demand is chosen with probability q when the game moves to random settlement) and uses the breakdown probability $1-p$ where p satisfies Condition (ii) (i.e., $-\dfrac{q_{i}}{b_{i}^{S}}<\dfrac{\partial p(x_{1},x_{2})}{ \partial x_{i}}\le 0$ for any $(x_{1},x_{2})\in \partial ^{*}S$ with $x_{i}<b_{i}^{S}$, where $q_{i}=q$ for $i=1$ and $q_{i}=1-q$ for $i=2$), then $(\hat{x}_{1},\hat{x}_{2})$ must be a Phase 2 Nash equilibrium, that is $(\hat{x}_{1},\hat{x}_{2})\notin S$. Moreover, for any i, if $\hat{x}_{i}<b_{i}^{S}$, then $p(\hat{ x}_{1},\hat{x}_{2})q_{i}=-\dfrac{\partial p(\hat{x}_{1},\hat{x}_{2})}{ \partial x_{i}}(q_{i}\hat{x}_{i}+(1-q_{i})x_{i}^{*}(\hat{x}_{j}))$, and if $\hat{x}_{i}=b_{i}^{S}$, then $p(\hat{x}_{1},\hat{x}_{2})q_{i}\ge - \dfrac{\partial p(\hat{x}_{1},\hat{x}_{2})}{\partial x_{i}}(q_{i}\hat{x} _{i}+(1-q_{i})x_{i}^{*}(\hat{x}_{j}))$.

We now prove the above statements. There are two possible types of Nash equilibria: Phase 1 Nash equilibria and Phase 2 Nash equilibria. So, we have the following two cases.

• (i) Phase 1 Nash equilibria, i.e., the Nash equilibria where players’ demands are compatible with each other For this case, let $(\hat{x}_{1},\hat{x}_{2})\in S$ be a Phase 1 Nash equilibrium. Then, we must have $(\hat{x}_{1},\hat{x}_{2})\in \partial ^{*}S$ (if not, then $(\hat{x}_{1},\hat{x}_{2})$ cannot be a Nash equilibrium because one of the two players will have an incentive to deviate to a slightly higher demand, and the two players’ proposals are still compatible when the player makes the deviation). Suppose $\hat{x}_{1}<b_{1}^{S}$. Notice that Player 1’s payoff is $U_{1}(x_{1},x_{2})=p(x_{1},x_{2})(qx_{1}+(1-q)x_{1}^{*}(x_{2}))$ if the two players’ demands $(x_{1},x_{2})\in \partial S\cup S^{I}$. Taking the derivative of Player 1 ’s payoff with respect to $x_{1}$ at $(\hat{x}_{1},\hat{x}_{2})$ on the domain $\partial S\cup S^{I}$, we have¹⁶$\dfrac{\partial U_{1}(\hat{x}_{1},\hat{x}_{2})}{\partial x_{1}}=\dfrac{\partial p(\hat{x}_{1},\hat{ x}_{2})}{\partial x_{1}}(q\hat{x}_{1}+(1-q)x_{1}^{*}(\hat{x}_{2}))+p(\hat{x}_{1},\hat{x}_{2})q=\dfrac{\partial p(\hat{x}_{1},\hat{x} _{2})}{\partial x_{1}}\hat{x}_{1}+p(\hat{x}_{1},\hat{x}_{2})q= \dfrac{\partial p(\hat{x}_{1},\hat{x}_{2})}{\partial x_{1}}\hat{x}_{1}+ q\ge \dfrac{\partial p(\hat{x}_{1},\hat{x}_{2})}{\partial x_{1}} b_{1}^{S}+q>-q+q=0$, where the second equality follows from the fact that $x_{1}^{*}(\hat{x}_{2})=\hat{x}_{1}$ (which is due to the fact that $(\hat{x}_{1},\hat{x}_{2})\in \partial ^{*}S$), and the second inequality follows from Condition (ii). The above inequality then implies that Player 1 is strictly better off by making a slightly higher demand. If $\hat{x}_{1}=b_{1}^{S}$, then we must have $\hat{x }_{2}<b_{2}^{S}$ because $(\hat{x}_{1},\hat{x}_{2})\in \partial ^{*}(S)$ and $(b_{1}^{S},b_{2}^{S})\notin \partial ^{*}(S)$. In this case, it can be shown that Player 2 has an incentive to deviate to a slightly higher demand (the proof is similar to the case where $\hat{x}_{1}<b_{1}^{S}$). So, we have proved that $(\hat{x}_{1},\hat{x}_{2})$ cannot be a Nash equilibrium. Thus, there is no Phase 1 Nash equilibrium.

• (ii) Phase 2 Nash equilibria: i.e., the Nash equilibria where players’ demands are not compatible with each other: Let $(\hat{x}_{1},\hat{x}_{2})\notin S$ be a Phase 2 Nash equilibrium. Then, we must have $\hat{x}_{1}>x_{1}^{*}(\hat{x}_{2})$ and $\hat{x} _{2}>x_{2}^{*}(\hat{x}_{1})$. For Player i, we have either $\hat{x} _{i}<b_{i}^{S}$ or $\hat{x}_{i}=b_{i}^{S}$. We consider Player 1 first. We have the following two cases.

Case 1$\hat{x}_{1}<b_{1}^{S}$. Since $(\hat{x}_{1},\hat{x} _{2})$ is a Nash equilibrium, we must have:

$$\begin{aligned} \hat{x}_{1}= & {} \text{ argmax }_{x_{1}^{*}(\hat{x}_{2})<x_{1}< b_{1}^{S}} U_{1}(x_{1}, \hat{x}_{2})\nonumber \\= & {} \text{ argmax }_{x_{1}^{*}(\hat{x}_{2})<x_{1}< b_{1}^{S}} p(x_{1}, \hat{x}_{2})(qx_{1}+(1-q)x_{1}^{*}(\hat{x}_{2})) \end{aligned}$$

(4)

The first-order condition of the maximization problem in (4) is that $\dfrac{\partial p(\hat{x}_{1},\hat{x}_{2})}{\partial x_{1}}(q\hat{x}_{1}+(1-q)x_{1}^{*}(\hat{x}_{2}))+p(\hat{x}_{1},\hat{x} _{2})q=0$. That is, $p(\hat{x}_{1},\hat{x}_{2})q=-\dfrac{\partial p(\hat{x}_{1},\hat{x}_{2})}{\partial x_{1}}(q\hat{x}_{1}+(1-q)x_{1}^{*}(\hat{x} _{2}))$.

Case 2$\hat{x}_{1}=b_{1}^{S}$. In this case, it must be true that Player 1’s marginal utility at $(\hat{x}_{1},\hat{x}_{2})$ is nonnegative. So, we must have $p(\hat{x}_{1},\hat{x}_{2})q\ge -\dfrac{ \partial p(\hat{x}_{1},\hat{x}_{2})}{\partial x_{1}}(q\hat{x}_{1}+(1-q)x_{1}^{*}(\hat{x}_{2}))$.

Similarly, for Player 2, if $\hat{x}_{2}<b_{2}^{S}$, then $p(\hat{x}_{1},\hat{x}_{2})(1-q)=-\dfrac{\partial p(\hat{x}_{1},\hat{x}_{2})}{ \partial x_{2}}((1-q)\hat{x}_{2}+qx_{2}^{*}(\hat{x}_{1}))$. If $\hat{x} _{2}=b_{2}^{S}$, then $p(\hat{x}_{1},\hat{x}_{2})(1-q)\ge -\dfrac{\partial p(\hat{x}_{1},\hat{x}_{2})}{\partial x_{2}}((1-q)\hat{x}_{2}+qx_{2}^{*}(\hat{x} _{1}))$.

Finally, if in addition we have that p it is a twice differentiable function and is (weakly) decreasing as well as strictly concave in $x_{i}$ (for $i=1,2$), then the necessary conditions above are also sufficient. Therefore a Nash equilibrium must exist.

Step 2 In this step, we show that if $\{(\hat{x}_{1}^{n},\hat{x}_{2}^{n})\}_{n=1}^{\infty }$ is a sequence of Nash equilibria where $(\hat{x}_{1}^{n},\hat{x}_{2}^{n})$ is a Nash equilibrium of the NDG-RS that has settlement bias q and uses the breakdown probability $1-p_{n}$ where $\{p_{n}\}_{n=1}^{\infty }$ is a regular sequence of probabilities, then the two players’ equilibrium proposals, $(\hat{x}_{1}^{n},x_{2}^{*}(\hat{x}_{1}^{n}))$ and $(x_{1}^{*}(\hat{x}_{2}^{n}),\hat{x}_{2}^{n})$, will converge to each other as n goes to infinity. In particular, we will show that $\lim _{n\rightarrow \infty }(\hat{x}_{1}^{n}-x_{1}^{*}(\hat{x}_{2}^{n}))=0$ and $\lim _{n\rightarrow \infty }(\hat{x}_{2}^{n}-x_{2}^{*}(\hat{x}_{1}^{n}))=0$.

Suppose that $\lim _{n\rightarrow \infty }(\hat{x} _{1}^{n}-x_{1}^{*}(\hat{x}_{2}^{n}))=0$ does not hold. Then, using the fact that $\hat{x}_{1}^{n}-x_{1}^{*}(\hat{x}_{1}^{n})>0$ for all n (noticing that for any n, $(\hat{x}_{1}^{n},\hat{x}_{2}^{n})$ must be a Phase 2 Nash equilibrium according to Step 1), there must exist a subsequence of $\{(\hat{x}_{1}^{n},\hat{x} _{2}^{n})\}_{n=1}^{\infty }$, say $\{(\hat{x}_{1}^{n_{k}},\hat{x} _{2}^{n_{k}})\}_{k=1}^{\infty }$, such that $\hat{x}_{1}^{n_{k}}-x_{1}^{*}(\hat{x}_{2}^{n_{k}})>\epsilon$ for all k and for some $\epsilon >0$. Notice that the sequence $\{\hat{x}_{1}^{n_{k}}-x_{1}^{*}(\hat{x} _{2}^{n_{k}})\}_{k=1}^{\infty }$ lies in a compact set (because $|\hat{x} _{1}^{n_{k}}-x_{1}^{*}(\hat{x}_{2}^{n_{k}})|\le b_{1}^{S}$ for any k), so it must have a convergent subsequence, say $\{\hat{x} _{1}^{n_{k_{l}}}-x_{1}^{*}(\hat{x}_{2}^{n_{k_{l}}})\}_{l=1}^{\infty }$. In addition, we must have $\lim _{l\rightarrow \infty }(\hat{x} _{1}^{n_{k_{l}}}-x_{1}^{*}(\hat{x}_{2}^{n_{k_{l}}}))=\epsilon ^{\prime }$ for some $\epsilon ^{\prime }\ge \epsilon$.

Since $(\hat{x}_{1}^{n_{k_{l}}},\hat{x}_{2}^{n_{k_{l}}})$ must be a Phase 2 Nash equilibrium for any l, it must be true that in the NDG-RS that uses the breakdown probability $p_{n_{k_{l}}}$ and has settlement bias q, making the demand $\hat{x}_{1}^{n_{k_{l}}}$ is not worse than making the demand $x_{1}^{*}(\hat{x}_{2}^{n_{k_{l}}})$ for Player 1 (the latter demand will make the two players’ demands just compatible). That is, $(q\hat{x}_{1}^{n_{k_{l}}}+(1-q)x_{1}^{*}(\hat{x}_{2}^{n_{k_{l}}}))p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x}_{2}^{n_{k_{l}}})\ge x_{1}^{*}(\hat{x}_{2}^{n_{k_{l}}})$. The fact that $\lim _{l\rightarrow \infty }(\hat{x}_{1}^{n_{k_{l}}}-x_{1}^{*}(\hat{x}_{2}^{n_{k_{l}}}))= \epsilon ^{\prime }>0$ and the fact that $p_{n}$ converges to 0 uniformly on any domain where an arbitrarily small neighborhood of $\partial S$ is excluded (i.e., Condition (iii)) imply that $p_{n_{k_{l}}}(\hat{x} _{1}^{n_{k_{l}}},\hat{x}_{2}^{n_{k_{l}}})$ converges to zero as $l\rightarrow \infty$. This implies that $x_{1}^{*}(\hat{x} _{2}^{n_{k_{l}}})\rightarrow 0$ as $l\rightarrow \infty$. So, $\hat{x} _{2}^{n_{k_{l}}}\rightarrow b_{2}^{S}$ as $l\rightarrow \infty$. Similarly, we have $x_{2}^{*}(\hat{x}_{1}^{n_{k_{l}}})\rightarrow 0$ as $l\rightarrow \infty$, and thus $\hat{x}_{1}^{n_{k_{l}}}\rightarrow b_{1}^{S}$ as $l\rightarrow \infty$.

Suppose that Condition (iv) holds for $i=1$. Let $\hat{ \epsilon }=\min \{\epsilon _{1},\epsilon _{2}\}$. Since $\hat{x} _{1}^{n_{k_{l}}}\rightarrow b_{1}^{S}$ and $\hat{x}_{2}^{n_{k_{l}}} \rightarrow b_{2}^{S}$, we have $|\hat{x}_{1}^{n_{k_{l}}}-b_{1}^{S}|<\hat{ \epsilon }$ and $|\hat{x}_{2}^{n_{k_{l}}}-b_{2}^{S}|<\hat{\epsilon }$ for all $l>L$ for some $L>N_{1}$. This implies that $(\hat{x}_{1}^{n_{k_{l}}},\hat{x} _{2}^{n_{k_{l}}})\in N_{\hat{\epsilon }}(b_{1}^{S},b_{2}^{S})\subseteq N_{\epsilon _{1}}(b_{1}^{S},b_{2}^{S})$ for all $l>L$. It also implies that $q\hat{x}_{1}^{n_{k_{l}}}+(1-q)x_{1}^{*}(\hat{x}_{2}^{n_{k_{l}}})\ge q\hat{x} _{1}^{n_{k_{l}}}\ge q(b_{1}^{S}-\hat{\epsilon })$ for all $l>L$. So, for any $l>L$, we have $-\dfrac{\partial p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x} _{2}^{n_{k_{l}}})}{\partial x_{1}}(q\hat{x}_{1}^{n_{k_{l}}}+(1-q)x_{1}^{*}(\hat{x}_{2}^{n_{k_{l}}}))\ge -\dfrac{\partial p_{n_{k_{l}}}(\hat{x} _{1}^{n_{k_{l}}},\hat{x}_{2}^{n_{k_{l}}})}{\partial x_{1}}q(b_{1}^{S}-\hat{ \epsilon })\ge -\dfrac{\partial p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x} _{2}^{n_{k_{l}}})}{\partial x_{1}}q(b_{1}^{S}-\epsilon _{2})>\dfrac{1}{ b_{1}^{S}-\epsilon _{2}}p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x} _{2}^{n_{k_{l}}})q(b_{1}^{S}-\epsilon _{2})=p_{n_{k_{l}}}(\hat{x} _{1}^{n_{k_{l}}},\hat{x}_{2}^{n_{k_{l}}})q$, where the first inequality follows from the fact that $q\hat{x}_{1}^{n_{k_{l}}}+(1-q)x_{1}^{*}(\hat{x} _{2}^{n_{k_{l}}})\ge q(b_{1}^{S}-\hat{\epsilon })$ and $-\dfrac{\partial p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x}_{2}^{n_{k_{l}}})}{\partial x_{1}}\ge 0$ (according to Condition (v)), the second inequality follows from the fact that $-\dfrac{\partial p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}}, \hat{x}_{2}^{n_{k_{l}}})}{\partial x_{1}}\ge 0$ and $b_{1}^{S}-\hat{\epsilon }\ge b_{1}^{S}-\epsilon _{2}$, and the last inequality follows from the fact that $-\dfrac{\partial p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x} _{2}^{n_{k_{l}}})}{\partial x_{1}}>\dfrac{1}{b_{1}^{S}-\epsilon _{2}} p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x}_{2}^{n_{k_{l}}})$ for any $(\hat{x}_{1}^{n_{k_{l}}},\hat{x}_{2}^{n_{k_{l}}})\in N_{\epsilon _{1}}(b_{1}^{S},b_{2}^{S})$ (according to Condition (iv)).

Since $(\hat{x}_{1}^{n_{k_{l}}},\hat{x}_{2}^{n_{k_{l}}})$ is a Phase 2 Nash equilibrium, according to Step 1, we must have $p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x}_{2}^{n_{k_{l}}})q\ge - \dfrac{\partial p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x} _{2}^{n_{k_{l}}})}{\partial x_{1}}(q\hat{x}_{1}^{n_{k_{l}}}+(1-q)x_{1}^{*}(\hat{x}_{2}^{n_{k_{l}}}))$ (and $p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x} _{2}^{n_{k_{l}}})(1-q)\ge -\dfrac{\partial p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x} _{2}^{n_{k_{l}}})}{\partial x_{2}}((1-q)\hat{x}_{2}^{n_{k_{l}}}+qx_{2}^{*}(\hat{x}_{1}^{n_{k_{l}}}))$) for any l. However, we have shown that $- \dfrac{\partial p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x} _{2}^{n_{k_{l}}})}{\partial x_{1}}(q\hat{x}_{1}^{n_{k_{l}}}+(1-q)x_{1}^{*}( \hat{x}_{2}^{n_{k_{l}}}))>p_{n_{k_{l}}}(\hat{x}_{1}^{n_{k_{l}}},\hat{x} _{2}^{n_{k_{l}}})q$ when l is sufficiently large, a contradiction.

So, $\lim _{n\rightarrow \infty }(\hat{x}_{1}^{n}-x_{1}^{*}(\hat{x}_{2}^{n}))=0$ must hold. Similarly, we can show that $\lim _{n\rightarrow \infty }(\hat{x}_{2}^{n}-x_{2}^{*}(\hat{x} _{1}^{n}))=0$ must hold. Hence, we have proved that the two players’ equilibrium ‘proposals’ must converge to each other.

Step 3 In this step, we show that if $\{( \hat{x}_{1}^{n},\hat{x}_{2}^{n})\}_{n=1}^{\infty }$ is a sequence of Nash equilibria where $(\hat{x}_{1}^{n},\hat{x}_{2}^{n})$ is a Nash equilibrium of the NDG-RS that has settlement bias q and uses the breakdown probability $1-p_{n}$ where $\{p_{n}\}_{n=1}^{\infty }$ is a regular sequence of probabilities, then the sequence $\{(\hat{x}_{1}^{n}, \hat{x}_{2}^{n})\}_{n=1}^{\infty }$ must converge to the generalized Nash solution outcome with bargaining weight q as $n\rightarrow \infty$, i.e., $\lim _{n\rightarrow \infty }\hat{x} _{1}^{n}=x_{1}^{N^{q}}$ and $\lim _{n\rightarrow \infty }\hat{x} _{2}^{n}=x_{2}^{N^{q}}$ where $(x_{1}^{N^{q}},x_{2}^{N^{q}})$ is the generalized Nash solution outcome with bargaining weight q.

Suppose that $\lim _{n\rightarrow \infty }\hat{x} _{1}^{n}=x_{1}^{N^{q}}$ does not hold. Then there exists a subsequence of $\{( \hat{x}_{1}^{n},\hat{x}_{2}^{n})\}_{n=1}^{\infty }$, say $\{(\hat{x} _{1}^{n_{i}},\hat{x}_{2}^{n_{i}})\}_{i=1}^{\infty }$, such that $|\hat{x} _{1}^{n_{i}}-x_{1}^{N^{q}}|>\epsilon$ for all i and for some $\epsilon >0$. Notice that the sequence $\{\hat{x}_{1}^{n_{i}}\}_{i=1}^{\infty }$ lies in a compact set, so it must have a convergent subsequence, say $\{\hat{x} _{1}^{n_{i_{j}}}\}_{j=1}^{\infty }$. Let $x_{1}^{\prime }$ be the limit of this subsequence, then we must have $|x_{1}^{\prime }-x_{1}^{N^{q}}|\ge \epsilon$. Since $\lim _{n\rightarrow \infty }(\hat{x}_{2}^{n}-x_{2}^{*}( \hat{x}_{1}^{n}))=0$ and $x_{2}^{*}(x_{1})$ is a continuous function, we have $\lim _{j\rightarrow \infty }\hat{x}_{2}^{n_{i_{j}}}=\lim _{j\rightarrow \infty }x_{2}^{*}(\hat{x}_{1}^{n_{i_{j}}})=x_{2}^{*}(x_{1}^{\prime })$. Observing that $x_{1}^{\prime }=b_{1}^{S}$ and $x_{2}^{*}(x_{1}^{\prime })=b_{2}^{S}$ cannot hold simultaneously because $(x_{1}^{\prime },x_{2}^{*}(x_{1}^{\prime }))\in \partial ^{*}S$ and $(b_{1}^{S},b_{2}^{S})\notin \partial ^{*}S$, we thus have the following three cases.

Case 1$x_{1}^{\prime }=b_{1}^{S}$ and $x_{2}^{*}(x_{1}^{\prime })<b_{2}^{S}$. Since $\lim _{j\rightarrow \infty }\hat{x} _{2}^{n_{i_{j}}}=x_{2}^{*}(x_{1}^{\prime })<b_{2}^{S}$, we have $\hat{x} _{2}^{n_{i_{j}}}<b_{2}^{S}$ when j is sufficiently large. So, we have $p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})(1-q)=-\dfrac{ \partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{ \partial x_{2}}((1-q)\hat{x}_{2}^{n_{i_{j}}}+qx_{2}^{*}(\hat{x} _{1}^{n_{i_{j}}}))$ when j is sufficiently large. On the other hand, we have $p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})q\ge - \dfrac{\partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x} _{2}^{n_{i_{j}}})}{\partial x_{1}}(q\hat{x}_{1}^{n_{i_{j}}}+(1-q)x_{1}^{*}( \hat{x}_{2}^{n_{i_{j}}}))$ (notice that this is true regardless of whether $\hat{x}_{1}^{n_{i_{j}}}=b_{1}^{S}$ or $\hat{x}_{1}^{n_{i_{j}}}<b_{1}^{S}$). So, $\dfrac{\partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x} _{2}^{n_{i_{j}}})}{\partial x_{1}}(q\hat{x}_{1}^{n_{i_{j}}}+(1-q)x_{1}^{*}( \hat{x}_{2}^{n_{i_{j}}}))\ge \dfrac{\partial p_{n_{i_{j}}}(\hat{x} _{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{\partial x_{2}}((1-q)\hat{x} _{2}^{n_{i_{j}}}+qx_{2}^{*}(\hat{x}_{1}^{n_{i_{j}}}))\dfrac{q}{1-q}$ when j is sufficiently large. So, $\dfrac{\dfrac{\partial p_{n_{i_{j}}}(\hat{x} _{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{\partial x_{1}}}{\dfrac{\partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{\partial x_{2}}}\le \dfrac{(1-q)\hat{x}_{2}^{n_{i_{j}}}+qx_{2}^{*}(\hat{x} _{1}^{n_{i_{j}}})}{q\hat{x}_{1}^{n_{i_{j}}}+(1-q)x_{1}^{*}(\hat{x} _{2}^{n_{i_{j}}})}\dfrac{q}{1-q}$ when j is sufficiently large (observing that $\dfrac{ \partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{ \partial x_{2}}<0$ when j is sufficiently large because $(\hat{x} _{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})\in S^{I}\cap N_{\epsilon _{3}}(\partial S)$ when j is sufficiently large, according to Condition (v)). We now consider the following two subcases.

The first subcase is that $x_{2}^{*}(x_{1}^{ \prime })=x_{2}^{*}(b_{1}^{S})=0$. In this case, we have $\lim _{j\rightarrow \infty } \dfrac{(1-q)\hat{x}_{2}^{n_{i_{j}}}+qx_{2}^{*}(\hat{x}_{1}^{n_{i_{j}}})}{ q\hat{x}_{1}^{n_{i_{j}}}+(1-q)x_{1}^{*}(\hat{x}_{2}^{n_{i_{j}}})}\dfrac{q}{1-q}=\dfrac{(1-q)\times 0+q\times 0}{ qb_{1}^{S}+(1-q)b_{1}^{S}}\dfrac{q}{1-q}=0$. On the other hand, we have

$$\begin{aligned} \lim _{j\rightarrow \infty } \dfrac{\dfrac{\partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x} _{2}^{n_{i_{j}}})}{\partial x_{1}}}{\dfrac{\partial p_{n_{i_{j}}}(\hat{x} _{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{\partial x_{2}}}= & {} \lim _{j\rightarrow \infty } \left( -\dfrac{\partial x_{2}^{*}\left( \hat{x} _{1}^{n_{i_{j}}},p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x} _{2}^{n_{i_{j}}})|p_{n_{i_{j}}}\right) }{\partial x_{1}}\right) \\= & {} -\dfrac{\partial x_{2}^{*}(b_{1}^{S}-) }{\partial x_{1}}>0. \end{aligned}$$

The first equality in the above equation follows from the fact that by differentiating $p_{n}(x_{1},x_{2}^{*}(x_{1},c|p_{n}))=c$ (for $c<1$) with respect to $x_{1}$ we have $\dfrac{\dfrac{\partial p_{n}(x_{1},x_{2}^{*}(x_{1},c|p_{n}))}{ \partial x_{1}}}{\dfrac{\partial p_{n}(x_{1},x_{2}^{*}(x_{1},c|p_{n}))}{ \partial x_{2}}}=-\dfrac{\partial x_{2}^{*}(x_{1},c|p_{n})}{\partial x_{1}}$, which implies that $\dfrac{\dfrac{\partial p_{n}(x_{1},x_{2})}{\partial x_{1}}}{\dfrac{\partial p_{n}(x_{1},x_{2})}{\partial x_{2}}}=-\dfrac{ \partial x_{2}^{*}(x_{1},p_{n}(x_{1},x_{2})|p_{n})}{\partial x_{1}}$ by setting $c=p_{n}(x_{1},x_{2})$. The second equality in the above equation follows from the facts that $\lim _{j\rightarrow \infty } \hat{x} _{1}^{n_{i_{j}}}=x_{1}^{\prime }=b_{1}^{S}$ and $\lim _{j\rightarrow \infty } \hat{x}_{2}^{n_{i_{j}}}=\lim _{j\rightarrow \infty } x_{2}^{*}(\hat{x} _{1}^{n_{i_{j}}})=x_{2}^{*}(x_{1}^{\prime })=x_{2}^{*}(b_{1}^{S})$ and the facts that $\{\dfrac{\partial x_{2}^{*}(x_{1},p_{n}(x_{1},x_{2})|p_{n}) }{ \partial x_{1}}\}_{n=1}^{\infty }$ is uniformly continuous at any point on $\partial ^{*}S$ and $\dfrac{\partial x_{2}^{*}(x_{1},p_{n}(x_{1},x_{2})|p_{n}) }{\partial x_{1}}=\dfrac{\partial x_{2}^{*}(x_{1}) }{\partial x_{1}}$ for any $(x_{1},x_{2})\in \partial ^{*}S$ and any $p_{n}$. So, we have shown that $\lim _{j\rightarrow \infty }\dfrac{\dfrac{\partial p_{n_{i_{j}}}(\hat{x} _{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{\partial x_{1}}}{\dfrac{\partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{\partial x_{2}}}> \lim _{j\rightarrow \infty }\dfrac{(1-q)\hat{x}_{2}^{n_{i_{j}}}+qx_{2}^{*}( \hat{x}_{1}^{n_{i_{j}}})}{q\hat{x}_{1}^{n_{i_{j}}}+(1-q)x_{1}^{*}(\hat{x} _{2}^{n_{i_{j}}})}\dfrac{q}{1-q}$. This is a contradiction with the fact that $\dfrac{ \dfrac{\partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x} _{2}^{n_{i_{j}}})}{\partial x_{1}}}{\dfrac{\partial p_{n_{i_{j}}}(\hat{x} _{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{\partial x_{2}}}\le \dfrac{(1-q)\hat{x }_{2}^{n_{i_{j}}}+qx_{2}^{*}(\hat{x}_{1}^{n_{i_{j}}})}{q\hat{x} _{1}^{n_{i_{j}}}+(1-q)x_{1}^{*}(\hat{x}_{2}^{n_{i_{j}}})}\dfrac{q}{1-q}$ for sufficiently large j.

The second subcase is that $x_{2}^{*}(x_{1}^{\prime })=x_{2}^{*}(b_{1}^{S})>0$. In this case, there must exist a vertical segment from $(b_{1}^{S},0)$ to $(b_{1}^{S},x_{2}^{*}(b_{1}^{S}))$ on the Pareto frontier. In addition, we have $\lim _{j\rightarrow \infty }\dfrac{ (1-q)\hat{x}_{2}^{n_{i_{j}}}+qx_{2}^{*}(\hat{x}_{1}^{n_{i_{j}}})}{q\hat{x} _{1}^{n_{i_{j}}}+(1-q)x_{1}^{*}(\hat{x}_{2}^{n_{i_{j}}})}\dfrac{q}{1-q}=\dfrac{(1-q)x_{2}^{*}(b_{1}^{S})+qx_{2}^{*}(b_{1}^{S})}{qb_{1}^{S}+(1-q)b_{1}^{S}}\dfrac{q}{1-q}=\dfrac{ x_{2}^{*}(b_{1}^{S})}{b_{1}^{S}}\dfrac{q}{1-q}$. On the other hand, we have $\lim _{j\rightarrow \infty }\dfrac{\dfrac{\partial p_{n_{i_{j}}}(\hat{x} _{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{\partial x_{1}}}{\dfrac{\partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{\partial x_{2}}}=\lim _{j\rightarrow \infty }(-\dfrac{\partial x_{2}^{*}(\hat{x} _{1}^{n_{i_{j}}},p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x} _{2}^{n_{i_{j}}})|p_{n_{i_{j}}})}{\partial x_{1}})=-\dfrac{\partial x_{2}^{*}(b_{1}^{S}-)}{\partial x_{1}}$. So, we have $-\dfrac{\partial x_{2}^{*}(b_{1}^{S}-)}{\partial x_{1}}\le \dfrac{x_{2}^{*}(b_{1}^{S})}{b_{1}^{S}}\dfrac{q}{1-q}$. This implies that $(b_{1}^{S},x_{2}^{*}(b_{1}^{S}))$ must be the generalized Nash solution outcome with bargaining weight q. This is a contradiction with the fact that $x_{1}^{\prime }\ne x_{1}^{N^{q}}$.

Case 2$x_{1}^{\prime }<b_{1}^{S}$ and $x_{2}^{*}(x_{1}^{ \prime })=b_{2}^{S}$. This case cannot hold. The analysis is similar to Case 1 and is omitted.

Case 3$x_{1}^{\prime }<b_{1}^{S}$ and $x_{2}^{*}(x_{1}^{ \prime })<b_{2}^{S}$. The facts that $\lim _{j\rightarrow \infty }\hat{x} _{1}^{n_{i_{j}}}=x_{1}^{\prime }<b_{1}^{S}$ and that $\lim _{j\rightarrow \infty }\hat{x}_{2}^{n_{i_{j}}}=x_{2}^{*}(x_{1}^{\prime })<b_{2}^{S}$ imply that $\hat{x}_{1}^{n_{i_{j}}}<b_{1}^{S}$ and $\hat{x} _{2}^{n_{i_{j}}}<b_{2}^{S}$ when j is sufficiently large. So, we have $p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})= \dfrac{ \partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{ \partial x_{1}}(q\hat{x}_{1}^{n_{i_{j}}}+(1-q)x_{1}^{*}(\hat{x}_{2}^{n_{i_{j}}}))\dfrac{1}{q} = \dfrac{\partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x} _{2}^{n_{i_{j}}})}{\partial x_{2}}((1-q)\hat{x}_{2}^{n_{i_{j}}}+qx_{2}^{*}(\hat{x} _{1}^{n_{i_{j}}}))\dfrac{1}{1-q}$ when j is sufficiently large. So, $\dfrac{\dfrac{ \partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{ \partial x_{1}}}{\dfrac{\partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x }_{2}^{n_{i_{j}}})}{\partial x_{2}}}= \dfrac{(1-q)\hat{x} _{2}^{n_{i_{j}}}+qx_{2}^{*}(\hat{x}_{1}^{n_{i_{j}}})}{q\hat{x} _{1}^{n_{i_{j}}}+(1-q)x_{1}^{*}(\hat{x}_{2}^{n_{i_{j}}})}\dfrac{q}{1-q}$ when j is sufficiently large. However, noticing that $\lim _{j\rightarrow \infty } \dfrac{\dfrac{\partial p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x} _{2}^{n_{i_{j}}})}{\partial x_{1}}}{\dfrac{\partial p_{n_{i_{j}}}(\hat{x} _{1}^{n_{i_{j}}},\hat{x}_{2}^{n_{i_{j}}})}{\partial x_{2}}} = \lim _{j\rightarrow \infty } (-\dfrac{\partial x_{2}^{*}(\hat{x} _{1}^{n_{i_{j}}},p_{n_{i_{j}}}(\hat{x}_{1}^{n_{i_{j}}},\hat{x} _{2}^{n_{i_{j}}})|p_{n_{i_{j}}}) }{\partial x_{1}})=-\dfrac{\partial x_{2}^{*}(x_{1}^{\prime }) }{\partial x_{1}}$ and $\lim _{j\rightarrow \infty } \dfrac{(1-q)\hat{x}_{2}^{n_{i_{j}}}+qx_{2}^{*}(\hat{x}_{1}^{n_{i_{j}}})}{q\hat{x} _{1}^{n_{i_{j}}}+(1-q)x_{1}^{*}(\hat{x}_{2}^{n_{i_{j}}})}=\lim _{j\rightarrow \infty }\dfrac{(1-q)x_{2}^{*}(x_{1}^{\prime })+qx_{2}^{*}(x_{1}^{\prime })}{ qx_{1}^{\prime }+(1-q)x_{1}^{\prime }}=\dfrac{x_{2}^{*}(x_{1}^{\prime })}{ x_{1}^{\prime }}\dfrac{q}{1-q}$, we must have $-\dfrac{\partial x_{2}^{*}(x_{1}^{\prime }) }{ \partial x_{1}}=\dfrac{x_{2}^{*}(x_{1}^{\prime })}{x_{1}^{\prime }}\dfrac{q}{1-q}$. This implies that $x_{1}^{\prime }=x_{1}^{N^{q}}$, which is a contradiction with the assumption that $x_{1}^{\prime }\ne x_{1}^{N^{q}}$.

So, we have shown that $\lim _{n\rightarrow \infty }\hat{x} _{1}^{n}=x_{1}^{N^{q}}$ must hold. This implies that $\{(\hat{x}_{1}^{n},\hat{x} _{2}^{n})\}_{n=1}^{\infty }$ must converge to the generalized Nash solution outcome with bargaining weight q. $\square$

Proof of Proposition 1

For the first part of Proposition 1, it is sufficient to show that the sequence of the nth MAP-based probabilities is regular.

Fig. 6

Impact of an increase in Player 1’s demand

Let $\hat{p}(x)=1-\alpha (A_{m(x)})$. So, the nth MAP-based initiation probability $p_{n}=\hat{p} ^{n}$. Notice that $\dfrac{ \partial \hat{p}^{n}(x_{1},x_{2})}{\partial x_{i}}=n\hat{p} ^{n-1}(x_{1},x_{2})\dfrac{\partial \hat{p}(x_{1},x_{2})}{ \partial x_{i}}=0$ for any $(x_{1},x_{2})\in \partial ^{*}S$ with $x_{i}<b_{i}^{S}$ because $\dfrac{\partial \hat{p}(x_{1},x_{2})}{\partial x_{i}}=0$ for any $(x_{1},x_{2})\in \partial ^{*}S$ with $x_{i}<b_{i}^{S}$.¹⁷ So, $p_{n}$ satisfies Condition (ii) for any n. Condition (iii) is satisfied because as $n\rightarrow \infty$, $p_{n}$ converges to zero at any $(x_{1},x_{2})\in S^{I}\backslash N_{\epsilon }(\partial S)$ for any $\epsilon >0$ and $\{p_{n}\}_{n=1}^{\infty }$ is a monotonically decreasing sequence of functions. For Condition (iv), notice that when the initiation probability is $\hat{p}$, if the two players’ demands $(x_{1},x_{2})$ are incompatible and Player 1 increases his demand by $\Delta x_{1}$, then the initiation probability p will decrease by $\Delta p\approx \frac{1}{2}(\Delta x_{1})(|\dfrac{ \partial x_{2}^{*}(x_{1})}{\partial x_{1}}|\Delta x_{1})+(|\dfrac{ \partial x_{2}^{*}(x_{1})}{\partial x_{1}}|\Delta x_{1})(x_{1}-x_{1}^{*}(x_{2}))$ (see the right figure of Fig. 6). So, $\dfrac{\partial \hat{p}(x_{1},x_{2})}{ \partial x_{1}}= \dfrac{\partial x_{2}^{*}(x_{1})}{\partial x_{1}} (x_{1}-x_{1}^{*}(x_{2}))$. If $(x_{1},x_{2})$ is close to $(b_{1}^{S},b_{2}^{S})$, then $\dfrac{\partial \hat{p}(x_{1},x_{2})}{ \partial x_{1}}\approx \dfrac{\partial x_{2}^{*}(x_{1})}{\partial x_{1}} (b_{1}^{S}-x_{1}^{*}(b_{2}^{S}))$. Notice that $\dfrac{\partial p_{n}(x_{1},x_{2})}{\partial x_{1}}=n\hat{p} ^{n-1}\dfrac{\partial \hat{p} (x_{1},x_{2})}{\partial x_{1}}<-\dfrac{1}{ b_{1}-\epsilon _{2}}\hat{p}^{n}$ if and only if $n\dfrac{\partial \hat{p} (x_{1},x_{2})}{\partial x_{1}}<- \dfrac{1}{b_{1}-\epsilon _{2}}\hat{p}$. The latter inequality holds if $(x_{1},x_{2})$ is in a sufficiently small neighborhood of $(b_{1}^{S},b_{2}^{S})$ because $\dfrac{\partial \hat{p} (x_{1},x_{2})}{ \partial x_{1}}\approx \dfrac{\partial x_{2}^{*}(x_{1})}{ \partial x_{1}}(b_{1}^{S}-x_{1}^{*}(b_{2}^{S}))$ is bounded away from zero when $(x_{1},x_{2})$ is sufficiently close to $(b_{1}^{S},b_{2}^{S})$ (which implies that $n\dfrac{\partial \hat{p}(x_{1},x_{2})}{\partial x_{1}}$ goes to negative infinity as $n\rightarrow \infty$). So, Condition (iv) is satisfied. Condition (v) is satisfied obviously. For Condition (vi), note that

$$\begin{aligned} \dfrac{\partial x_{2}^{*}(x_{1},p_{n}(x_{1},x_{2})|p_{n})}{\partial x_{1} }= & {} -\dfrac{\dfrac{ \partial p_{n}(x_{1},x_{2})}{\partial x_{1}}}{\dfrac{ \partial p_{n}(x_{1},x_{2})}{\partial x_{2}}}\\= & {} -\dfrac{n\hat{p}^{n-1} \dfrac{\partial \hat{p}(x_{1},x_{2})}{\partial x_{1}} }{n\hat{p}^{n-1} \dfrac{\partial \hat{p}(x_{1},x_{2})}{\partial x_{2}}}=- \dfrac{\dfrac{ \partial \hat{p}(x_{1},x_{2})}{\partial x_{1}}}{\dfrac{ \partial \hat{p} (x_{1},x_{2})}{\partial x_{2}}}\\= & {} -\dfrac{\dfrac{\partial x_{2}^{*}(x_{1}) }{\partial x_{1}}(x_{1}-x_{1}^{*}(x_{2}))}{\dfrac{ \partial x_{1}^{*}(x_{2})}{\partial x_{2}}(x_{2}-x_{2}^{*}(x_{1}))}. \end{aligned}$$

As $(x_{1},x_{2})$ converges to some point on the strong Pareto frontier, the last term in the above equation will converge to the slope of the Pareto frontier at that point (because $\dfrac{x_{1}-x_{1}^{*}(x_{2})}{ x_{2}-x_{2}^{*}(x_{1})}$ will converge to $\dfrac{\partial x_{1}^{*}(x_{2})}{\partial x_{2}}$). That is, $\dfrac{\partial x_{2}^{*}(x_{1},p_{n}(x_{1},x_{2})|p_{n})}{\partial x_{1}}$ is continuous at any point on the strong Pareto frontier. Since $\dfrac{ \partial x_{2}^{*}(x_{1},p_{n}(x_{1},x_{2})|p_{n})}{\partial x_{1}}$ is the same for all n, this implies that $\dfrac{\partial x_{2}^{*}(x_{1},p_{n}(x_{1},x_{2})|p_{n})}{\partial x_{1}}$ is uniformly continuous at any point on the strong Pareto frontier. $\square$

We now prove the second part of Proposition 1. The Nash solution outcome $(x_{1}^{N},x_{2}^{N})$ must be such that $(x_{1}^{N},x_{2}^{N})\in \partial ^{*}S$ and $\dfrac{x_{2}^{N}}{ x_{1}^{N}}= -\dfrac{\partial x_{2}^{*}(x_{1}^{N})}{\partial x_{1}}$. This implies that if $(x_{1}^{\prime },x_{2}^{\prime })\in \partial S$ is such that $\dfrac{ x_{2}^{\prime }}{x_{1}^{\prime }}>-\dfrac{\partial x_{2}^{*}(x_{1}^{\prime })}{\partial x_{1}}$, then $(x_{1}^{\prime },x_{2}^{\prime })$ must lie on the upper left of the Nash solution outcome on the Pareto frontier or coincide with the Nash solution outcome. If $(x_{1}^{\prime },x_{2}^{\prime })\in \partial S$ is such that $\dfrac{ x_{2}^{\prime }}{ x_{1}^{\prime }}<-\dfrac{\partial x_{2}^{*}(x_{1}^{\prime })}{\partial x_{1}}$, then $(x_{1}^{\prime },x_{2}^{\prime })$ must lie on the lower right of the Nash solution outcome on the Pareto frontier or coincide with the Nash solution outcome.

Let $(\hat{x}_{1}^{n},\hat{x }_{2}^{n})$ be a Phase 2 Nash equilibrium of the NDG-RS that uses the nth MAP-based breakdown probability $1-p_{n}$, where $p_{n}=\hat{p} ^{n}$ with $\hat{p}(x)=1-\alpha (A_{m(x)})$. If $\hat{x} _{1}^{n}=b_{1}^{S}$, then it is obvious that $(\hat{x } _{1}^{n},x_{2}^{ *}(\hat{x}_{1}^{n}))$ must lie on the lower right of the Nash solution outcome on the Pareto frontier or coincide with the Nash solution outcome. If $\hat{x}_{1}^{n}<b_{1}^{S}$, then according to Step 1 of the proof of Theorem 1, we have $\hat{p}^{n}(\hat{x}_{1}^{n},\hat{x} _{2}^{n})=-\dfrac{\partial \hat{p}^{n}(\hat{x}_{1}^{n},\hat{x}_{2}^{n}) }{ \partial x_{1}}(\hat{x} _{1}^{n}+x_{1}^{*}(\hat{x}_{2}^{n}))=-n \hat{p} ^{n-1}\dfrac{\partial \hat{p}(\hat{x}_{1}^{n},\hat{x}_{2}^{n})}{ \partial x_{1}}(\hat{x} _{1}^{n}+x_{1}^{*}(\hat{x}_{2}^{n}))=-n\hat{p} ^{n-1}\dfrac{ \partial x_{2}^{*}(\hat{x}_{1}^{n})}{\partial x_{1}}( \hat{x} _{1}^{n}-x_{1}^{ *}(\hat{x}_{2}^{n}))(\hat{x} _{1}^{n}+x_{1}^{*}(\hat{x }_{2}^{n}))$ (where the last equality is because $\dfrac{\partial \hat{p} (x_{1},x_{2}) }{\partial x_{1}}=\dfrac{ \partial x_{2}^{*}(x_{1})}{ \partial x_{1}} (x_{1}-x_{1}^{*}(x_{2}))$ if $(x_{1},x_{2})$ are incompatible; see also the proof for the first part of Proposition 1). For Player 2, we have $\hat{p} ^{n}(\hat{x}_{1}^{n},\hat{x}_{2}^{n})\ge - \dfrac{\partial \hat{p} ^{n}( \hat{x}_{1}^{n},\hat{x}_{2}^{n})}{\partial x_{2}} (\hat{x} _{2}^{n}+x_{2}^{*}(\hat{x}_{1}^{n}))=-n\hat{p}^{n-1}\dfrac{ \partial \hat{p}(\hat{x}_{1}^{n},\hat{x}_{2}^{n})}{\partial x_{2}}(\hat{x} _{2}^{n}+x_{2}^{*}(\hat{x}_{1}^{n}))=-n\hat{p}^{n-1}\dfrac{ \partial x_{1}^{*}(\hat{x}_{2}^{n})}{\partial x_{2}}(\hat{x} _{2}^{n}-x_{2}^{*}(\hat{x}_{1}^{n}))(\hat{x}_{2}^{n}+x_{2}^{*}( \hat{ x}_{1}^{n}))$, regardless of whether $\hat{x}_{2}<b_{2}^{S}$ or $\hat{x} _{2}=b_{2}^{S}$. So, we have $-\dfrac{\partial x_{2}^{*}(\hat{ x} _{1}^{n})}{\partial x_{1}} (\hat{x}_{1}^{n}-x_{1}^{*}(\hat{x} _{2}^{n}))(\hat{x}_{1}^{n}+x_{1}^{*}(\hat{x}_{2}^{n}))\ge -\dfrac{ \partial x_{1}^{*}(\hat{x}_{2}^{n})}{ \partial x_{2}}(\hat{x} _{2}^{n}-x_{2}^{*}(\hat{x}_{1}^{n}))(\hat{x} _{2}^{n}+x_{2}^{*}( \hat{x}_{1}^{n}))$. That is, $\dfrac{\hat{x} _{2}^{n}+x_{2}^{*}( \hat{x} _{1}^{n})}{\hat{x}_{1}^{n}+x_{1}^{*}(\hat{x }_{2}^{n})}\le \dfrac{ \dfrac{\partial x_{2}^{*}(\hat{x}_{1}^{n})}{ \partial x_{1}}(\dfrac{ \partial x_{1}^{*}(\hat{x}_{2}^{n})}{\partial x_{2}})^{-1}}{\dfrac{\hat{ x}_{2}^{n}-x_{2}^{*}(\hat{x}_{1}^{n})}{\hat{x} _{1}^{n}-x_{1}^{*}( \hat{x}_{2}^{n})}}$. Noting that $\dfrac{\hat{x} _{2}^{n}-x_{2}^{*}( \hat{x}_{1}^{n})}{\hat{x}_{1}^{n}-x_{1}^{*}( \hat{x }_{2}^{n})}\ge -( \dfrac{\partial x_{1}^{*}(\hat{x}_{2}^{n}) }{\partial x_{2}})^{-1}$ (which is due to the fact that the two players’ demands $\hat{x }_{1}^{n}$ and $\hat{x}_{2}^{n}$ are incompatible and the fact that the Pareto frontier is concave because S is convex) and $\dfrac{\hat{x} _{2}^{n}+x_{2}^{*}(\hat{x}_{1}^{n})}{\hat{x} _{1}^{n}+x_{1}^{*}( \hat{x }_{2}^{n})}>\dfrac{x_{2}^{*}(\hat{x} _{1}^{n})}{\hat{x}_{1}^{n}}$, we must have $\dfrac{x_{2}^{*}(\hat{x} _{1}^{n})}{\hat{x}_{1}^{n}}<- \dfrac{ \partial x_{2}^{*}(\hat{x} _{1}^{n})}{\partial x_{1}}$. So, $( \hat{x} _{1}^{n},x_{2}^{*}(\hat{x }_{1}^{n}))$ must be on the lower right of the Nash solution outcome on the Pareto frontier or coincide with the Nash solution outcome.

Similarly, we can show that Player 2’s equilibrium proposal $(x_{1}^{*}(\hat{x}_{2}^{n}),\hat{x} _{2}^{n})$ must lie on the upper left of the Nash solution outcome on the Pareto frontier or coincide with the Nash solution outcome.

Proof of Proposition 2

(i) Phase 1 Nash equilibria: let $(\hat{x}_{1},\hat{x}_{2})$ be a Phase 1 Nash equilibrium. Then, we must have $(\hat{x}_{1},\hat{x} _{2})\in S$. In addition, we must have $(\hat{x}_{1},\hat{x}_{2})\in \partial ^{*}S$ (if not, then $(\hat{ x}_{1},\hat{x}_{2})$ cannot be a Nash equilibrium because one of the two players will have an incentive to deviate to a higher demand, and the two players’ proposals are still compatible when the player makes the deviation). We will next show that any $(x_{1},x_{2})\in \partial ^{*}S$ is a Nash equilibrium. Let $(\hat{x} _{1},\hat{x}_{2})$ be a given point on $\partial ^{*}S$. Let Player 2 ’s demand $\hat{x}_{2}$ be given. If Player 1 makes the demand $\hat{x} _{1}$, then the two players’ demands are just compatible and Player 1’s payoff is thus $\hat{x }_{1}$. Suppose Player 1 makes the demand $x_{1}^{\prime }$ with $x_{1}^{\prime }>\hat{x}_{1}$; Player 1 is never better off by decreasing his demand, so the only possible deviation for Player 1 is to increase his demand. Then Player 1’s payoff is $\dfrac{ \hat{x}_{1}+x_{1}^{\prime }}{2}p$, where p is the initiation probability when Player 1’s demand is $x_{1}^{\prime }$ and Player 2’s demand is $\hat{x}_{2}$. Notice that $p=\alpha (B_{m(x_{1}^{\prime },\hat{x}_{2})})=\alpha (B(x_{1}^{*}(\hat{x }_{2}),x_{2}^{*}(x_{1}^{\prime })))=\alpha (B( \hat{x}_{1},x_{2}^{*}(x_{1}^{\prime })))=\hat{x}_{1}x_{2}^{*}(x_{1}^{\prime })$, where the third equality follows from the fact that $( \hat{x}_{1},\hat{x}_{2})\in \partial ^{*}S$. So, $\dfrac{\hat{x} _{1}+x_{1}^{\prime }}{2}p=\dfrac{ \hat{x}_{1}+x_{1}^{\prime }}{2}\hat{x} _{1}x_{2}^{*}(x_{1}^{\prime })= \hat{x}_{1}\dfrac{\hat{x}_{1}x_{2}^{*}(x_{1}^{\prime })+x_{1}^{\prime }x_{2}^{*}(x_{1}^{\prime })}{2}\le \hat{x}_{1}\dfrac{2\alpha (S)}{2}= \hat{x}_{1}$ (the inequality is strict if $\hat{x}_{1}>0$). So, Player 1 is better off by making the demand $\hat{x} _{1}$ (which is just compatible with Player 2’s demand) than deviating to any higher demand. Similarly, Player 2 is better off by making the demand $\hat{x}_{2}$ (which is just compatible with Player 1’s demand) than deviating to any higher demand. So, we have proved that any $(x_{1},x_{2})\in \partial ^{*}S$ must be a Nash equilibrium.

(ii) Phase 2 Nash equilibria: let $(\hat{x}_{1}, \hat{ x}_{2})$ be a Phase 2 Nash equilibrium. Then we must have $(\hat{x} _{1}, \hat{x}_{2})\notin S$. This implies that $\hat{x}_{1}>x_{1}^{*}( \hat{x} _{2})$ and $\hat{x}_{2}>x_{2}^{*}(\hat{x}_{1})$.

If Player 2 ’s demand $\hat{x}_{2}$ is such that $x_{1}^{*}(\hat{x}_{2})>0$, then according to the proof of (i), Player 1 is strictly better off by making the demand $x_{1}^{*}(\hat{x}_{2})$, rather than $\hat{x}_{1}$. So, $(\hat{x}_{1},\hat{x}_{2})$ cannot be a Nash equilibrium.

If Player 2’s demand $\hat{x}_{2}$ is such that $x_{1}^{*}(\hat{x}_{2})=0$ (which must imply that $\hat{x } _{2}=b_{2}^{S}$), then for Player 1, any $\hat{x}_{1}\in [0,b_{1}^{S}]$ is a best response (because the initiation probability p will be zero regardless of Player 1 ’s demand). However, only if $\hat{x }_{1}$ is such that $x_{2}^{*}(\hat{x }_{1})=0$ (which must imply that $\hat{x} _{1}=b_{1}^{S}$), will $\hat{x} _{2}=b_{2}^{S}$ be a best responses of Player 2. So, $(b_{1}^{S},b_{2}^{S})$ is the only possible Phase 2 Nash equilibrium, and it is a Phase 2 Nash equilibrium if and only if $x_{1}^{*}(b_{2}^{S})=0$ and $x_{2}^{*}(b_{1}^{S})=0$ (note that, if $x_{1}^{*}(b_{2}^{S})>0$—which occurs when there is a horizontal segment in the Pareto frontier—or $x_{2}^{*}(b_{1}^{S})>0$—which occurs when there is a vertical segment in the Pareto frontier—then there is no Phase 2 Nash equilibrium). In addition, there is no other Phase 2 Nash equilibria. $\square$

Appendix 2

Proof of Remark 1

Since $\{p_{n}\}_{n=1}^{\infty }$ is regular, $(\hat{x}_{1}^{n},\hat{x} _{2}^{n})$ can only be a Phase 2 Nash equilibrium for any n (according to Step 1 in the proof of Theorem 1). This implies that $(q\hat{x}_{1}^{n}+(1-q)x_{1}^{*}(\hat{ x}_{2}^{n})) p_{n}(\hat{x}_{1}^{n},\hat{x}_{2}^{n})\ge x_{1}^{*}(\hat{ x}_{2}^{n})$ and $((1-q)\hat{x}_{2}^{n}+qx_{2}^{*}(\hat{x}_{1}^{n}))p_{n}(\hat{x}_{1}^{n},\hat{x}_{2}^{n})\ge x_{2}^{*}(\hat{x}_{1}^{n})$ for all n (because for Player i, making the demand $\hat{x}_{i}$ cannot be strictly worse off than making the demand that is just compatible with the opponent’s demand). Let $(x_{1}^{\prime },x_{2}^{\prime })\in \partial ^{*}S$ be the limit point of $\{(\hat{x}_{1}^{n},\hat{x} _{2}^{n})\}_{n=1}^{\infty }$, i.e., $\lim _{n\rightarrow \infty }\hat{x} _{1}^{n}=x_{1}^{\prime }$ and $\lim _{n\rightarrow \infty }\hat{x} _{2}^{n}=x_{2}^{\prime }$. Since $(x_{1}^{\prime },x_{2}^{\prime })\in \partial ^{*}S$, we must have either $x_{1}^{\prime }>0$ or $x_{2}^{\prime }>0$. We thus have the following two cases.

(i) $x_{1}^{\prime }>0$. In this case, $(q\hat{x}_{1}^{n}+(1-q)x_{1}^{*}(\hat{ x}_{2}^{n})) p_{n}(\hat{x}_{1}^{n},\hat{x}_{2}^{n})\ge x_{1}^{*}(\hat{x} _{2}^{n})$ implies that $p_{n}(\hat{x}_{1}^{n},\hat{x}_{2}^{n})\ge \dfrac{ x_{1}^{*}(\hat{x}_{2}^{n})}{q\hat{x}_{1}^{n}+(1-q)x_{1}^{*}(\hat{ x}_{2}^{n})}$ when n is sufficiently large (using the fact that $q\hat{x} _{1}^{n}+(1-q)x_{1}^{*}(\hat{x}_{2}^{n})>0$ when n is sufficiently large because $\lim _{n\rightarrow \infty }\hat{x}_{1}^{n}=x_{1}^{\prime }>0$). We have $\lim _{n\rightarrow \infty }\dfrac{x_{1}^{*}(\hat{x}_{2}^{n})}{q\hat{x}_{1}^{n}+(1-q)x_{1}^{*}(\hat{ x}_{2}^{n})}=1$ because $\lim _{n\rightarrow \infty } \hat{x}_{1}^{n}=x_{1}^{\prime }$ and $\lim _{n\rightarrow \infty } x_{1}^{*}( \hat{x}_{2}^{n})=x_{1}^{*}(x_{2}^{\prime })=x_{1}^{\prime }$ (where the last equality holds because $(x_{1}^{\prime },x_{2}^{\prime })\in \partial ^{*}S$). In addition, $p_{n}(\hat{x}_{1}^{n},\hat{x}_{2}^{n})\le 1$ for all n. So, we have $\lim _{n\rightarrow \infty }p_{n}(\hat{x}_{1}^{n},\hat{x}_{2}^{n})=1$.

(ii) $x_{2}^{\prime }>0$. The proof is similar to case (i) and is omitted.

We thus have proved that $\lim _{n\rightarrow \infty }p_{n}(\hat{x}_{1}^{n}, \hat{x}_{2}^{n})=1$ if $\{(\hat{x}_{1}^{n},\hat{x}_{2}^{n})\}_{n=1}^{\infty }$ converges to some point on $\partial ^{*}S$ where $(\hat{x}_{1}^{n}, \hat{x}_{2}^{n})$ is a Nash equilibrium in the NDG-RS that uses the breakdown probability $1-p_{n}$ where $\{p_{n}\}_{n=1}^{\infty }$ is regular and that the probability that Player 1’s demand is chosen is $q\in (0, 1)$ when the game moves to random settlement. $\square$

Euclidean divergence and non-convergence We next consider an intuitively appealing p function that uses the Euclidean distance between the two players’ demands to determine the probability of moving to settlement stage. Even though such a p is continuous at the boundary of S, the resulting NDG-RS fails to have the Nash solution outcome as the limit equilibrium outcome.

When players make their demands, one can convert these demands into proposals by using the Pareto frontier of S. Given two such resulting proposals, a breakdown probability is based on Euclidean divergence if the breakdown probability decreases as the Euclidean distance between the two proposals falls. Noting that the maximum possible distance between any two proposals in S is $\sqrt{ (b_{1}^{S})^{2}+(b_{2}^{S})^{2}}$, an Euclidean divergence breakdown probability $1-p_{d}$ is given by

$$\begin{aligned} 1-p_{d}(x_{1},x_{2})= \dfrac{\sqrt{ (x_{1}-x_{1}^{*}(x_{2}))^{2}+(x_{2}-x_{2}^{*}(x_{1}))^{2}}}{\sqrt{ (b_{1}^{S})^{2}+(b_{2}^{S})^{2}}} \end{aligned}$$

for any $(x_{1},x_{2})\in S^{I}$ (and $1-p_{d}(x_{1},x_{2})=0$ for any $(x_{1},x_{2})\in \partial S$). It turns out that this otherwise intuitively appealing breakdown probability does not lead to Nash solution convergence. To see this, notice that $(x_{1},x_{2})\in \partial ^{*}S$ is a Phase 1 Nash equilibrium if and only if no player has an incentive to deviate to some higher demand. This means that for Player 1, if he deviates to some higher demand, say $x_{1}^{\prime }$, then his payoff $p_{d}(x_{1}^{\prime },x_{2})\dfrac{ x_{1}^{\prime }+x_{1}}{2}$, must not be greater than $x_{1}$ (the payoff when Player 1 chooses not to deviate). That is,

$$\begin{aligned} \left( 1-\dfrac{ \sqrt{(x_{1}^{\prime }-x_{1}^{*}(x_{2}))^{2}+(x_{2}-x_{2}^{*}(x_{1}^{\prime }))^{2}}}{ \sqrt{ (b_{1}^{S})^{2}+(b_{2}^{S})^{2}}}\right) \dfrac{x_{1}^{\prime }+x_{1}}{2} \le x_{1}, \end{aligned}$$

i.e.,

$$\begin{aligned} \left( \dfrac{\sqrt{ (x_{1}^{\prime }-x_{1})^{2}+(x_{2}^{*}(x_{1})-x_{2}^{*}(x_{1}^{\prime }))^{2}}}{ \sqrt{ (b_{1}^{S})^{2}+(b_{2}^{S})^{2}}}\right) \dfrac{x_{1}^{\prime }+x_{1}}{2} \ge \dfrac{x_{1}^{\prime }-x_{1}}{2} \end{aligned}$$

since $(x_{1},x_{2})\in \partial ^{*}S$. That is, we have

$$\begin{aligned} x_{1}^{\prime }+x_{1}\ge \dfrac{\sqrt{(b_{1}^{S})^{2}+(b_{2}^{S})^{2}} }{ \sqrt{1+\left( \dfrac{x_{2}^{*}(x_{1}^{\prime })-x_{2}^{*}(x_{1})}{ x_{1}^{\prime }-x_{1}}\right) ^{2}}} \end{aligned}$$

for any $x_{1}^{\prime }>x_{1}$. Similarly, Player 2 has no incentive to deviate to some higher demand $x_{2}^{\prime }>x_{2}$ if and only if

$$\begin{aligned} x_{2}^{\prime }+x_{2}\ge \dfrac{ \sqrt{(b_{1}^{S})^{2}+(b_{2}^{S})^{2}}}{ \sqrt{1+\left( \dfrac{ x_{1}^{*}(x_{2}^{\prime })-x_{1}^{*}(x_{2})}{ x_{2}^{\prime }-x_{2}}\right) ^{2}}}. \end{aligned}$$

Suppose that $b_{1}^{S}=b_{2}^{S}=1$ and that $S=\{(x_{1},x_{2})|0\le x_{1}\le 1,0\le x_{2}\le 1 \text{ and } 0\le x_{1}+x_{2}\le k\}$ for some $k\in [1,\frac{3}{2})$. Then, it can be verified that any $(x_{1},x_{2})\in \partial ^{*}S$ with $\dfrac{ 1}{2} \le x_{1}\le 1$ and $\dfrac{1}{2}\le x_{2}\le 1$ is a Phase 1 Nash equilibrium because if $x_{1}\ge \dfrac{1}{2}$ and $x_{2}\ge \dfrac{1 }{2}$, then for any $x_{1}^{\prime }>x_{1}$, we have

$$\begin{aligned} x_{1}^{\prime }+x_{1}>\dfrac{\sqrt{2} }{\sqrt{1+\left( \dfrac{x_{2}^{*}(x_{1}^{\prime })-x_{2}^{*}(x_{1}) }{x_{1}^{\prime }-x_{1}}\right) ^{2}}} =1 \end{aligned}$$

and for any $x_{2}^{\prime }>x_{2}$, we have

$$\begin{aligned} x_{2}^{\prime }+x_{2}>\dfrac{\sqrt{2} }{\sqrt{1+\left( \dfrac{x_{1}^{*}(x_{2}^{\prime })-x_{1}^{*}(x_{2}) }{x_{2}^{\prime }-x_{2}}\right) ^{2}}} =1. \end{aligned}$$

Let $p_{n}=p_{d}^{n}$. Note that as n increases, $p_{n}$ decreases for any given incompatible demands (i.e., for any $(x_{1},x_{2})\in S^{I}$). This implies that all Phase 1 Nash equilibria when the initiation probability is $p_{d}$ will remain as Phase 1 Nash equilibria when the initiation probability is $p_{n}$ for any $n>1$, because as n increases, a player will have less incentive to deviate from a demand that is compatible with his opponent’s demand to some higher demand. So, if we use $\{p_{n}\}_{n=1}^{\infty }$ as the sequence of probabilities, then the Nash solution convergence result cannot hold.