A. Proof of Proposition 1

For any belief a _x , sequential rationality requires $$\beta _{{TW}} (x_{A} )=0$$ . The remaining assertions of the proposition are an immediate consequence of the following lemmata. The first three are preliminary results. The latter five each characterize the equilibria in one of the five regions of the parameter space.

(i) In any equilibrium of the game, we have

Consequently, the strategic Country A must accept any offer $$x_{A} > y_{A} $$ .

(ii) If $$\beta (s_{A} )=0$$ in equilibrium, then

(iii) If in equilibrium $$\alpha =0$$ , then $$\beta (r_{A} )=1$$ .

(iv) If in equilibrium $$\beta (y_{A} )> 0$$ , then $$\alpha _{x} (y_{A} )=0$$ .

Proof.

1. (i) At the node labeled $$A[b_{x} ]$$ , rejecting the offer $$(x_{A} )$$ gives the strategic Country A a payoff of y _A with probability b _x and 0 with probability $$1{ - }b_{x} $$ . On the other hand, accepting produces a payoff of x _A for sure. Therefore, it accepts for sure if $$b_{x} < x_{A} /y_{A} $$ , rejects for sure if $$b_{x} > x_{A} /y_{A} $$ , and is indifferent between accepting and rejecting if $$b_{x} =x_{A} /y_{A} $$ . Since it must always be that $$b_{x} \in [0,1]$$ , Country A will accept any offer $$x_{A} > y_{A} $$ .

2. (ii) Observe that if $$\beta (s_{A} )=0$$ then from Equation 4 in the main text,

   (A2) $$b_{x} (x_{A} )=\left\{ {\matrix{ 1 \hfill & {{\rm if}\,x_{A} \,\ne\,s_{A} {\rm .}} \hfill \cr 0 \hfill & {{\rm if}\,x_{A} =s_{A} } \hfill \cr } } \right.$$
   This, along with (A1), immediately gives (*).

3. (iii) If $$\alpha =0$$ , then $$a_{1} =0$$ . Therefore, Country B believes that Country A will reject any offer $$x_{A} < r_{A} $$ . So by making such an offer it expects to receive a payoff of y _B . On the other hand, the offer r _A would be accepted for sure and give Country B a payoff of $$1{ - }r_{A} > y_{B} $$ . Therefore, the strategic Country B will make offer the offer r _A with certainty.

4. (iv) By the result of (iii), if Country B makes the offer y _A with positive probability, then it must be that $$\alpha > 0$$ , consequently $$a_{1} > 0$$ . Suppose, for the sake of contradiction, that the strategic Country A rejects this offer with probability $$\delta > 0$$ . Then the expected payoff to Country B of making this offer is

But by making the offer $$y_{A} {\,+\,}{\varepsilon}$$ , where $$0< {\varepsilon}< \delta (1{ - }y_{A} { - }y_{B} )$$ , Country B would have an expected payoff of

since by Lemma 1(i) the offer $$y_{A} {\,+\,}{\varepsilon}$$ is accepted by the strategic Country A. One can then use the assumption that $${\varepsilon}< \delta (1{ - }y_{A} { - }y_{B} )$$ to verify that the payoff in Inequality A4 is greater than the payoff in Inequality A3.

Proof. Recall that Equation 19 in the main text states that $$1> r_{A} > y_{A} > s_{A} > 0$$ . So what we must show is that β does not put any probability mass on the intervals $$[0,s_{A} )$$ , $$(s_{A} ,y_{A} )$$ , $$(y_{A} ,r_{A} )$$ and $$(r_{A} ,1]$$ . Next, observe that if $$a_{1} =0$$ , then $$\alpha =0$$ and by Lemma 1(iii), only r _A is in support of β. Therefore, we can assume throughout the remainder of this proof that $$a_{1} > 0$$ .

Suppose β puts positive mass on $$(r_{A} ,1]$$ . By Lemma 1(i), any offer $$x_{A} \in (r_{A} ,1]$$ will be accepted by both the strategic type and the crazy type; so it will be accepted with certainty. But so will the offer r _A . Since $$1{ - }r_{A} > 1{ - }x_{A} $$ for all $$x_{A} \in (r_{A} ,1]$$ , Country B has a profitable deviation to the offer r _A . Therefore, β must put zero probability mass on $$(r_{A} ,1]$$ .

Suppose that β puts positive mass on $$(y_{A} ,r_{A} )$$ . Then, there exists $$x_{A} \in (y_{A} ,r_{A} )$$ in support of β. If such an offer x _A is made, then by Lemma 1(i) it is accepted with probability a ₁ and rejected with probability $$1{ - }a_{1} $$ . But again by Lemma 1(i), the offer $${{x_{A} {\,+\,}y_{A} } \over 2}$$ will also be accepted with probability a ₁ and rejected with probability $$1{ - }a_{1} $$ . Moreover, deviating to this offer is profitable for Country B. Therefore it cannot be that x _A is in support of β. Consequently, β cannot put positive probability on the interval $$(y_{A} ,r_{A} )$$ .

Now suppose that β puts positive mass on $$[0,s_{A} ){\cup}(s_{A} ,y_{A} )$$ . Then if Country B makes an offer $$x_{A} \in [0,s_{A} ){\cup}(s_{A} ,y_{A} )$$ , we have $$b_{x} (x_{A} )=1$$ by Equation 4 in the main text. Therefore, Country B’s expected payoff of making the offer x _A is equal to y _B . But if Country B deviates to the offer $$y_{A} {\,+\,}{\varepsilon}$$ , where $$0< {\varepsilon}< 1{ - }y_{A} { - }y_{B} $$ , the strategic Country A will accept, giving Country B an expected payoff of

where the inequality holds by Assumption 1c and $${\varepsilon}< 1{ - }y_{A} { - }y_{B} $$ . In other words, Country B has a profitable deviation to the offer $$y_{A} {\,+\,}{\varepsilon}$$ . Consequently, β cannot put positive probability mass on $$[0,s_{A} ){\cup}(s_{A} ,y_{A} )$$ .

1. (i) There is no equilibrium with $$\beta (r_{A} )=0$$ .

2. (ii) There is no equilibrium with $$\beta (y_{A} )> 0$$ and $$\beta (s_{A} )=0$$ .

3. (iii) If $$a_{0} < \underline{a} $$ , then in equilibrium we must have $$\beta (r_{A} )=1$$ . If $$a_{0} > \underline{a} $$ and $$b_{0} > \underline{b} $$ , then in equilibrium we must have $$\beta (r_{A} )< 1$$ .

4. (iv) If $$b_{0} < \underline{b} $$ , then in equilibrium we must have $$\alpha =0$$ . If $$b_{0} > \underline{b} $$ , then in equilibrium we must have $$\alpha > 0$$ .

5. (v) If $$b_{0} > \overline{b} $$ and $$a_{0} > \underline{a} $$ , then in equilibrium we must have $$\alpha _{x} (s_{A} )> 0$$ .

Proof.

1. (i) Suppose there is an equilibrium with $$\beta (r_{A} )=0$$ . Then by Lemma 2, β puts positive probability only on a subset of $$\{ s_{A} ,y_{A} \} $$ . But because $$0< s_{A} < y_{A} < z_{A} $$ by Assumptions 1a and 2b, Country A’s expected payoff from attacking must be less than z _A , its payoff to peace. Thus $$\alpha =0$$ , which by Lemma 1(iii) implies $$\beta (r_{A} )=1$$ . Contradiction.

2. (ii) Suppose there is an equilibrium with $$\beta (y_{A} )> 0$$ and $$\beta (s_{A} )=0$$ . By Lemma 1(iv), we must have $$\alpha _{x} (y_{A} )=0$$ . Since Lemma 3(i) implies $$\beta (r_{A} )> 0$$ , Country B’s expected payoff from the offer y _A must equal its expected payoff from the offer r _A :

which reduces to $$a_{1} =\bar{a}$$ . But note that by (*) we must have $$\alpha _{x} (s_{A} )=0$$ . Consequently, by deviating to the offer s _A , Country B can receive the expected payoff

where the inequality follows from substituting the expression for $$\bar{a}$$ , simplifying and using Assumptions 2a and 2b. Thus the deviation is profitable to Country B. Contradiction.

(iii) Suppose $$a_{0} < \underline{a} $$ . Country B’s maximum expected payoff from making the offer y _A or s _A is $$a_{1} (1{ - }s_{A} ){\,+\,}(1{ - }a_{1} )y_{B} $$ . It is easily verified that this expected payoff is strictly less than $$1{ - }r_{A} $$ when $$a_{1} < \underline{a} $$ . Combining this with Lemma 2 and the fact that $$a_{1} \,\equiv\,{{\alpha a_{0} } \over {1{ - }a_{0} {\,+\,}\alpha a_{0} }}\leq a_{0} $$ for all $$\alpha \in [0,1]$$ yields $$\beta (r_{A} )=1$$ .

On the other hand if $$a_{0} > \underline{a} $$ and $$\beta (r_{A} )=1$$ , then $$\alpha _{x} (s_{A} )=0$$ by (*). Consequently, by making the offer s _A , Country B has an expected payoff of $$a_{1} (1{ - }s_{A} ){\,+\,}(1{ - }a_{1} )y_{B} $$ , which is strictly less than $$1{ - }r_{A} $$ whenever $$a_{1} < \underline{a} $$ . The payoff to Country A of attacking is therefore $$b_{0} r_{A} {\,+\,}(1{ - }b_{0} )s_{A} > z_{A} $$ , since $$b_{0} > \underline{b} $$ . Consequently, $$\alpha =1$$ and $$a_{1} =a_{0} < \bar{a}$$ , establishing a contradiction.

(iv) If Country A chooses to attack, then its maximum expected payoff is $$(1{ - }b_{0} )s_{A} {\,+\,}b_{0} r_{A} $$ . One can easily verify that its expected payoff from peace, $$z_{A} $$ , is strictly greater than this payoff whenever $$b< \underline{b} $$ . Therefore, $$\alpha =0$$ .

On the other hand if $$b_{0} > \underline{b} $$ and $$\alpha =0$$ , then by Lemma 1(iii) we need $$\beta (r_{A} )=1$$ . Therefore, by attacking, Country A can get an expected payoff of $$b_{0} r_{A} {\,+\,}(1{ - }b_{0} )s_{A} $$ . Since $$b_{0} > \underline{b} $$ , this expected payoff is greater than z _A , establishing a contradiction.

(v) Suppose for the sake of contradiction that $$\alpha _{x} (s_{A} )=0$$ . By Lemma 1(i), this implies $$b_{x} (s_{A} )\leq s_{A} /y_{A} $$ , or equivalently

which follows from noting that $$b_{x} (s_{A} )$$ is given by Equation 6 in the main text. Since $$b_{0} > \overline{b} > \underline{b} $$ , Lemma 3(iv) implies $$\alpha > 0$$ ; consequently $$a_{1} > 0$$ . So, if the strategic Country B makes the offer s _A , it gets $$a_{1} (1{ - }s_{A} ){\,+\,}(1{ - }a_{1} )y_{B} > a_{1} (1{ - }y_{A} ){\,+\,}(1{ - }a_{1} )y_{B} $$ by Assumption 2b. Therefore, $$\beta (y_{A} )=0$$ . By Lemma 3(i) and (iii), we know that $$\beta (r_{A} ),\beta (s_{A} )> 0$$ . This implies the indifference condition

or, equivalently $$a_{1} =\underline{a} $$ . Substituting $$a_{1} $$ from Equation 2 in the main text, this reduces to $$\alpha ={{1{ - }a_{0} } \over {a_{0} }}{{\underline{a} } \over {1{ - }\underline{a} }}\in (0,1)$$ , where the strict inclusion holds because $$a_{0} > \underline{a} $$ by assumption. But $$\alpha \in (0,1)$$ implies the indifference condition

Combining $$\beta (s_{A} )$$ in Inequality A9 with Inequality A7 implies $$b_{0} \leq \overline{b} $$ . This contradicts our assumption that $$b_{0} > \overline{b} $$ .

Proof. Lemma 3(iv) implies $$\alpha =0$$ . Then Lemma 1(iii) implies $$\beta (r_{A} )=1$$ . Then Lemma 1(ii) implies that $$\alpha _{x} $$ is given by (*). Moreover, it is easy to verify that the given specifications for α, β and $$\alpha _{x} $$ are all sequentially rational given the starting beliefs, $$a_{0} $$ and $$b_{0} $$ , and the updated beliefs, $$a_{1} $$ and $$b_{x} $$ , that they imply.

Proof. Lemma 3(iii) implies that $$\beta (r_{A} )=1$$ . Then by Lemma 1(ii), $$\alpha _{x} $$ is given by (*). Finally, it is easy to verify that Country A’s expected payoff from attack is $$b_{0} r_{A} {\,+\,}(1{ - }b_{0} )s_{A} $$ , which is strictly greater than its payoff to peace, z _A , whenever $$b_{0} > \underline{b} $$ . Thus $$\alpha =1$$ . Moreover, it is easy to verify that the given specifications for α, β and $$\alpha _{x} $$ are all sequentially rational given the starting beliefs, $$a_{0} $$ and $$b_{0} $$ , and the updated beliefs, $$a_{1} $$ and $$b_{x} $$ , that they imply.

$$\alpha ={{1{ - }a_{0} } \over {a_{0} }} \cdot {{1{ - }r_{A} { - }y_{B} } \over {r_{A} { - }s_{A} }}$$ , $$\alpha _{x} $$ is given by (*) above, $$\beta (r_{A} )={{z_{A} { - }s_{A} } \over {b_{0} (r_{A} { - }s_{A} )}}$$ , $$\beta (s_{A} )=1{ - }{{z_{A} { - }s_{A} } \over {b_{0} (r_{A} { - }s_{A} )}}$$ , and $$\beta (y_{A} )=0$$ .

Proof.

Step (1): First we show that $$\beta (y_{A} )=0$$ . Suppose, for the sake of contradiction, that $$\beta (y_{A} )> 0$$ . Lemma 1(iv) implies that $$\alpha _{x} (y_{A} )=0$$ . Lemma 3(ii) implies $$\beta (s_{A} )> 0$$ as well. Therefore, we need the indifference condition

Since Lemma 3 $$(iv)$$ implies $$\alpha > 0$$ , which in turn implies $$a_{1} > 0$$ , we solve Inequality A10 for

where the strict inclusion follows from Assumptions 2a and 2b. This then implies $$b_{x} (s_{A} )=s_{A} /y_{A} $$ by Lemma 1(i). Using Equation 4 in the main text, we solve for

Since $$\alpha > 0$$ and Inequality A11 implies that Country A must have the same expected payoff from accepting and rejecting the offer s _A , we need

in which we can substitute (A12) and $$\beta (r_{A} )=1{ - }\beta (y_{A} ){ - }\beta (s_{A} )$$ , and solve to get

This expression on the right hand side of Inequality A14 is non-negative if and only if $$b_{0} \geq \overline{b} $$ . But this contradicts the premise of the Lemma.

Step (2): We now show that $$\alpha _{x} (s_{A} )=0$$ . Suppose $$\alpha _{x} (s_{A} )> 0$$ . Then we need $$b_{x} (s_{A} )\geq s_{A} /y_{A} $$ by Lemma 1(i). If $$\alpha _{x} (s_{A} )=1$$ then the expected payoff to Country B from making the offer s _A would be y _B . But by Lemma 1(i), Country B’s expected payoff of offering r _A is $$1{ - }r_{A} > y_{B} $$ by Assumption 2a. Therefore, we need $$\beta (r_{A} )=1$$ , which contradicts $$b_{x} (s_{A} )\geq s_{A} /y_{A} $$ .

Now, suppose that $$\alpha _{x} (s_{A} )\in (0,1)$$ . This implies $$b_{x} (s_{A} )=s_{A} /y_{A} $$ , so that $$\beta (s_{A} )$$ is given by Inequality A12. Since we showed in Step (1) that $$\beta (y_{A} )=0$$ , and we know from Lemma 3(iv) that $$\alpha > 0$$ , we need

But this inequality reduces to $$b_{0} \geq \overline{b} $$ , which is again a contradiction.

Step (3): We now calculate the equilibrium values of α, β and $$\alpha _{x} $$ . We have shown that $$\alpha _{x} (s_{A} )=0$$ , and $$\beta (y_{A} )=0$$ , which implies that $$\alpha _{x} $$ is given by (*) above. Since $$a_{0} > \underline{a} $$ , Lemma 3(iii) implies $$\beta (r_{A} )< 1$$ . Therefore, by Lemmas 2 and 3(i), we need $$\beta (r_{A} ),\beta (s_{A} )> 0$$ . These results imply the indifference condition Inequality A8, which in turn implies $$a_{1} =\underline{a} $$ . This implies that $$\alpha ={{1{ - }a_{0} } \over {a_{0} }}{{\underline{a} } \over {1{ - }\underline{a} }}\in (0,1)$$ , where the strict inclusion follows from the fact that $$a_{0} > \underline{a} $$ by assumption. Next, $$\alpha \in (0,1)$$ requires the indifference condition in Inequality A9 to be satisfied. The expressions for $$\beta (s_{A} )$$ and $$\beta (r_{A} )$$ follow. Moreover, it is easy to verify that any behavioral strategy profile satisfying the specifications in the statement of Lemma 6 constitutes an equilibrium, given the assumptions on $$a_{0} $$ and $$b_{0} $$ , and the updated beliefs $$a_{1} $$ and $$b_{x} $$ implied by the behavioral strategy profile.

$$\alpha =1$$ , $$\beta (r_{A} )=1{ - }{{1{ - }b_{0} } \over {b_{0} }}{{s_{A} } \over {y_{A} { - }s_{A} }}$$ , $$\beta (s_{A} )={{1{ - }b_{0} } \over {b_{0} }}{{s_{A} } \over {y_{A} { - }s_{A} }}$$ , $$\beta (y_{A} )=0$$

Proof.

Step (1): We first show that $$\beta (s_{A} )={{1{ - }b_{0} } \over {b_{0} }}{{s_{A} } \over {y_{A} { - }s_{A} }}$$ . By Lemma 3(v), we need $$\alpha _{x} (s_{A} )> 0$$ . Observe that $$\alpha _{x} (s_{A} )=1$$ implies $$\beta (s_{A} )=0$$ . (The argument is the same as in Step (2) of Lemma 6). Thus, according to Lemma 3(ii), $$\beta (y_{A} )=0$$ , and this contradicts Lemma 3(iii). Therefore, we conclude that $$\alpha _{x} (s_{A} )\in (0,1)$$ . By Lemma 1(i), this implies that $$b_{x} (s_{A} )=s_{A} /y_{A} $$ , from which $$\beta (s_{A} )={{1{ - }b_{0} } \over {b_{0} }}{{s_{A} } \over {y_{A} { - }s_{A} }}$$ follows.

Step (2): We now show that $$\beta (y_{A} )=0$$ in equilibrium. Suppose that $$\beta (y_{A} )> 0$$ . In Step (1) we showed that $$\beta (s_{A} )> 0$$ . Therefore, the indifference condition in Inequality A5 must be satisfied; thus $$a_{1} =\overline{a} $$ and $$\alpha ={{1{ - }a_{0} } \over {a_{0} }}{{\bar{a}} \over {1{ - }\bar{a}}}$$ . Since $$a_{0} < \bar{a}$$ , by assumption, we have $$\alpha > 1$$ , which is absurd.

Step (3): We now establish the values of the other choice variables in equilibrium. By Step (2), we have $$\beta (y_{A} )=0$$ . By Lemma 3(v) and the argument in Step (1), we have $$\alpha _{x} (s_{A} )\in (0,1)$$ and $$\beta (s_{A} )={{1{ - }b_{0} } \over {b_{0} }}{{s_{A} } \over {y_{A} { - }s_{A} }}$$ . Therefore, Country A is indifferent between accepting the offer s _A and rejecting it. These observations imply that the expected payoff to Country A from attacking is

which is greater than z _A whenever $$b_{0} > \overline{b} $$ . Therefore, $$\alpha =1$$ . Thus $$a_{1} =a_{0} $$ . Then Country B must be indifferent between the offers r _A and s _A ; that is

This implies that $$\alpha _{x} (s_{A} )$$ takes the value stated in the Lemma. Moreover, it is easy to verify that any behavioral strategy profile satisfying the specifications in the statement of Lemma 6 constitutes an equilibrium, given the assumptions on a ₀ and b ₀, and the updated beliefs a ₁ and b _x implied by the behavioral strategy profile.

Proof.

Step (1): We begin by showing that $$\beta (y_{A} )> 0$$ . Suppose for the sake of contradiction that $$\beta (y_{A} )=0$$ . Lemma 3(iii) implies $$\beta (s_{A} ),\beta (r_{A} )> 0$$ . Then, the exact argument as in Step (1) of Lemma 7 establishes that $$\beta (s_{A} )={{1{ - }b_{0} } \over {b_{0} }}{{s_{A} } \over {y_{A} { - }s_{A} }}$$ . The exact argument as in Step (3) of Lemma 7 establishes that $$\alpha =1$$ , hence $$a_{1} =a_{0} $$ . Now, the assumption that $$a_{0} > \bar{a}$$ can be re-written as $$a_{0} (1{ - }y_{A} { - }y_{B} )> 1{ - }r_{A} { - }y_{B} $$ . Observe from this that we can find $${\varepsilon}> 0$$ small enough so that

Since we stated above that $$a_{1} =a_{0} $$ , and we know that Lemma 1(i) states that the strategic Country A must accept any offer greater than y _A , the term on the left-hand side of Inequality A17 is Country B’s expected payoff from making the offer $$y_{A} {\,+\,}{\varepsilon}$$ , while the term on the right-hand side is its expected payoff from the offer r _A . Since we need $$\beta (r_{A} )> 0$$ by Lemma 3(i), we have a contradiction. Therefore $$\beta (y_{A} )> 0$$ .

Step (2): We now establish the value of the choice variables in equilibrium. Step (1) shows that $$\beta (y_{A} )> 0$$ , and by Lemma 3(i) and (ii), we need $$\beta (r_{A} ),\beta (s_{A} )> 0$$ as well. These imply a number of indifference conditions as follows. With the help of Lemma 1(iv) and Lemma 3(i), we need the indifference condition Inequality A5 to be met. This implies $$a_{1} =\bar{a}$$ , thus $$\alpha ={{1{ - }a_{0} } \over {a_{0} }}{{1{ - }r_{A} { - }y_{B} } \over {r_{A} { - }y_{A} }}\in (0,1)$$ , where the strict inclusion follows from $$a_{0} > \bar{a}> \underline{a} $$ . We also need the indifference condition Inequality A10, which implies $$\alpha _{x} (s_{A} )={{y_{A} { - }s_{A} } \over {1{ - }s_{A} { - }y_{B} }}\in (0,1)$$ as in Inequality A11. Obviously, the stated expression for $$\alpha _{x} (x_{A} )$$ when $$x_{A} \,\ne\,s_{A} $$ follows from Lemma 1(i) and (iv). This in turn implies $$\beta (s_{A} )={{1{ - }b_{0} } \over {b_{0} }}{{s_{A} } \over {y_{A} { - }s_{A} }}$$ as in Inequality A12. Finally, because we showed that $$\alpha \in (0,1)$$ and Country A is indifferent between accepting and rejecting the offer s _A (recall that $$\alpha _{x} (s_{A} )\in (0,1)$$ ), we also need the indifference condition

which we can solve using $$\beta (s_{A} )={{1{ - }b_{0} } \over {b_{0} }}{{s_{A} } \over {y_{A} { - }s_{A} }}$$ and $$\beta (r_{A} ){\,+\,}\beta (y_{A} ){\,+\,}\beta (s_{A} )=1$$ to get

Furthermore, it is easy to verify that because $$b_{0} > \overline{b} $$ , the expressions for $$\beta (y_{A} )$$ and $$\beta (r_{A} )$$ given by Inequality A19 are strictly positive. Moreover, it is easy to verify that these choice variables are sequentially rational given the assumptions on $$a_{0} $$ , $$b_{0} $$ , and the implied updated beliefs $$a_{1} $$ and $$b_{x} $$ .

B. Proof of Proposition 3

Let $$h=(z_{A} ,z_{B} ,s_{A} ,r_{A} ,y_{A} ,y_{B} )$$ , $$h^{{\ast}} =(z_{A} ,z_{B} ,s_{A} ,r'_{A} ,y_{A} ,y_{B} )$$ and $$h^{\dag } =(z_{A} ,z_{B} ,s'_{A} ,r_{A} ,y_{A} ,y_{B} )$$ be three parameter profiles, each satisfying Assumptions 1 and 2, and suppose that $$r'_{A} > r_{A} $$ and $$s'_{A} > s_{A} $$ . Notice that a change from h to $$h^{{\ast}} $$ or from h to $$h^{\dag } $$ will affect both the equilibrium behavior within the five regions described in Proposition 1 as well as the boundaries of these regions. Let $$\underline{a} ,\bar{a},\underline{b} $$ and $$\overline{b} $$ denote the boundaries of the regions under parameter profile h; $$\underline{a} ^{{\ast}} ,\bar{a}^{{\ast}} ,\underline{b} ^{{\ast}} $$ and $$\overline{b} ^{{\ast}} $$ the boundaries under parameter profile $$h^{{\ast}} $$ ; and $$\underline{a} ^{\dag } ,\overline{a} ^{\dag } ,\underline{b} ^{\dag } $$ and $$\overline{b} ^{\dag } $$ the boundaries under parameter profile $$h^{\dag } $$ .

First consider Statement (1). Notice that $$\underline{a} ,\overline{a} ,\underline{b} $$ and $$\overline{b} $$ are all decreasing in $$r_{A} $$ , holding other parameters constant. From the characterization of Proposition 1, it follows immediately that the result holds as long as the change $$r_{A} $$ does not result in a change in the region in which $$a_{0} $$ and $$b_{0} $$ fall; and, in particular, the relationship is strict within regions (iii) and (v). Now, consider the case in which a change from h to $$h^{{\ast}} $$ leads to a change in the region in which $$a_{0} $$ and $$b_{0} $$ fall. Several cases are possible. If $$b_{0} \in [\underline{b} ^{{\ast}} ,\underline{b} ]$$ , then $$\beta (r'_{A} )\leq 1$$ , while $$\beta (r_{A} )=1$$ . The same is true if $$a_{0} \in [\underline{a} ^{{\ast}} ,\underline{a} ]$$ and $$b_{0} > \underline{b} ^{{\ast}} $$ hold simultaneously. Now consider the case in which $$a_{0} \in [\underline{a} ,\overline{a} ^{{\ast}} ]$$ and $$b_{0} \in [\overline{b} ^{{\ast}} ,\overline{b} ]$$ . This corresponds to the case in which $$\beta (r_{A} )={{z_{A} { - }s_{A} } \over {b_{0} (r_{A} { - }s_{A} )}}$$ and $$\beta (r_{A} )=1{ - }{{1{ - }b_{0} } \over {b_{0} }}{{s_{A} } \over {y_{A} { - }s_{A} }}$$ ; indeed, the characterization of equilibrium behavior is given by region (iii) under h and by region (iv) under $$h^{{\ast}} $$ . In this case, $$\beta (r_{A} )\geq \beta (r'_{A} )$$ follows immediately from noticing that the highest value of $$b_{0} $$ is given by $$\overline{b} $$ and from rearranging terms. If, instead, $$a_{0} \geq \bar{a}^{{\ast}} $$ and $$b_{0} \in [\overline{b} ^{{\ast}} ,\overline{b} ]$$ , then $$\beta (r_{A} )={{z_{A} { - }s_{A} } \over {b_{0} (r_{A} { - }s_{A} )}}$$ and $$\beta (r'_{A} )={{z_{A} { - }b_{0} y_{A} } \over {b_{0} (r_{A} { - }y_{A} )}}$$ . From the fact that $$b_{0} \geq \overline{b} ^{{\ast}} $$ , we can conclude that $$\beta (r_{A} )\geq \beta (r'_{A} )$$ if and only if $${{r'_{A} { - }s_{A} } \over {r_{A} { - }s_{A} }}\geq {{r'_{A} { - }y_{A} } \over {r'_{A} { - }y_{A} }}$$ , which holds since $$r'_{A} > r_{A} $$ . Finally, suppose that $$a_{0} \in [\bar{a}^{{\ast}} ,\bar{a}]$$ and $$b_{0} \geq \overline{b} $$ . In this case, $$\beta (r_{A} )={{z_{A} { - }s_{A} } \over {b_{0} (r_{A} { - }s_{A} )}}$$ and $$\beta (r'_{A} )={{z_{A} { - }b_{0} y_{A} } \over {b_{0} (r_{A} { - }y_{A} )}}$$ . The result holds, as the lowest possible value that $$b_{0} $$ can take is $$\overline{b} $$ .

Now consider Statement (2). Note that $$\underline{a} $$ and $$\overline{b} $$ are increasing in $$s_{A} $$ , while $$\underline{b} $$ and $$\bar{a}$$ are respectively decreasing and constant in $$s_{A} $$ . From Proposition 1, if an increase in $$s_{A} $$ (that is, a decrease in $$1{ - }s'_{A} $$ ) does not change the relevant region, $$\alpha _{x} (s_{A} )$$ will weakly increase and will strictly increase in regions (iv) and (v). If the relevant region changes, the equilibrium value of $$\alpha _{x} (s_{A} )$$ will either stay constant at 0 or increase from 0 to some positive value (such an increase will happen when either $$a_{0} \geq \underline{a} ^{\dag } $$ and $$b_{0} \in [\overline{b} ,\overline{b} ^{\dag } ]$$ , or when $$b_{0} \geq \overline{b} ^{\dag } $$ and $$a_{0} \in [\underline{a} ,\underline{a} ^{\dag } ]$$ ). This concludes the proof.

C. Proof of Proposition 4

Note that $$y_{B} =\bar{y}{ - }y_{A} $$ , so changes in $$y_{A} $$ will affect $$y_{B} $$ when keeping $$\bar{y}$$ fixed. So, in what follows, whenever we refer to an “increase in y _A ” (for instance), we mean an “increase in y _A holding $$\bar{y}$$ fixed,” which actually results in an equal decrease in y _B .

To prove the proposition, we must take into account the fact that a change in y _A may simultaneously change the thresholds $$\underline{a} $$ , $$\bar{a}$$ , $$\overline{b} $$ and $$\underline{b} $$ and the equilibrium behavior characterized by α, α _x and β. Note that, conditional on remaining in the interior of each of the five regions defined in Proposition 1 (and Figure 3), the comparative statics of the equilibrium behavior with respect to marginal changes in y _A are as follows:

• (B1) In regions (i) and (ii), the equilibrium behavior is constant in y _A .

• (B2) In region (iii), α is increasing in y _A , while $$\beta (x_{A} )$$ and $$\alpha _{x} (x_{A} )$$ are constant with respect to y _A for all $$x_{A} \,\ne\,y_{A} $$ .

• (B3) In region (iv), α is constant in y _A , $$\beta (r_{A} )$$ is increasing in y _A , $$\beta (s_{A} )$$ is decreasing in y _A , $$\beta (x_{A} )$$ is constant in y _A for all $$x_{A} \,\ne\,r_{A} ,s_{A} $$ , $$\alpha _{x} (s_{A} )$$ is increasing in y _A and $$\alpha _{x} (x_{A} )$$ is constant in y _A for all $$x_{A} \,\ne\,y_{A} ,s_{A} $$ .

• (B4) In region (v), α and $$\alpha _{x} (s_{A} )$$ are increasing in $$y_{A} $$ , $$\alpha _{x} (x_{A} )$$ is constant in y _A for all $$x_{A} \,\ne\,s_{A} ,y_{A} $$ , $$\beta (s_{A} )$$ and $$\beta \left( {r_{A} } \right)$$ are decreasing in y _A , $$\beta (y_{A} )$$ is increasing in y _A and $$\beta (x_{A} )$$ is constant in y _A for all $$x_{A} \,\ne\,r_{A} ,y_{A} ,s_{A} $$ .

We prove each of the five results stated in the proposition separately. Some results are straightforward and follow immediately from the equilibrium characterization in Proposition 1 and the observations in (B1)–(B4). So, we focus on cases that are not trivially implied by these results.

(1) Given the change in the thresholds resulting from an increase in y _A , and since α is non-decreasing in y _A in each of the five regions, it is easy to verify that $$\alpha '\geq \alpha $$ with strict inequality when $$\alpha \,\ne\,0,1$$ .

(2) The result is straightforward in all cases except the case in which $$a_{0} > \underline{a} '$$ and $$b_{0} \in [\overline{b} ',\overline{b} ]$$ . In this case, we have $$\beta (s_{A} )=1{ - }{{z_{A} { - }s_{A} } \over {b_{0} (r_{A} { - }s_{A} )}}$$ , while $$\beta '(s_{A} )={{1{ - }b_{0} } \over {b_{0} }}{{s_{A} } \over {y'_{A} { - }s_{A} }}$$ . One can verify that in the case we are analyzing, $$\left( {\beta (s_{A} ){ - }\beta '(s_{A} )} \right)$$ is increasing in $$b_{0} $$ , and converges to 0 as $$b_{0} \to\underline{b} '$$ . Therefore, it must be that $$\beta (s_{A} )> \beta '(s_{A} )$$ .

(3) The result is straightforward in all cases except the case in which $$a_{0} \in (\underline{a} ',\overline{a} ')$$ and $$b_{0} \in (\bar{b}',\overline{b} )$$ , and the case in which $$a_{0} > \bar{a}'$$ and $$b_{0} \in (\overline{b} ',\overline{b} )$$ . In the first case, $$\beta (r_{A} )={{z_{A} { - }s_{A} } \over {b_{0} (r_{A} { - }s_{A} )}}$$ and $$\beta (r'_{A} )={{b_{0} y'_{A} { - }s_{A} } \over {b_{0} (y'_{A} { - }s_{A} )}}$$ . The result follows from noticing that the difference $$\left( {\beta '(r_{A} ){ - }\beta (r_{A} )} \right)$$ increases in $$b_{0} $$ and it is equal to 0 when $$b_{0} =\bar{b}'$$ . In the second case, $$\beta (r_{A} )={{z_{A} { - }s_{A} } \over {b_{0} (r_{A} { - }s_{A} )}}$$ and $$\beta '(r_{A} )={{z_{A} { - }b_{0} y'_{A} } \over {b_{0} (r_{A} { - }y'_{A} )}}$$ . One can verify that in this case, $$\left( {\beta (r_{A} ){ - }\beta '(r_{A} )} \right)$$ is increasing in b ₀, and converges to 0 as $$b_{0} \to\bar{b}'$$ . Therefore, it must be that $$\beta (r_{A} )> \beta '(r_{A} )$$ .

(4) Since $$\beta (y_{A} )=0$$ except when $$a_{0} > \bar{a}$$ and $$b_{0} > \overline{b} $$ , the result is straightforward in all cases.

(5) The result is straightforward in all cases except the case in which $$a_{0} \in (\bar{a},\bar{a}'')$$ and $$b_{0} > \overline{b} $$ . In this case, $$\alpha _{x} (s_{A} )={{y_{A} { - }s_{A} } \over {1{ - }s_{A} { - }y_{B} }}$$ and $$\alpha '_{x} (s_{A} )=1{ - }{{1{ - }r_{A} { - }y'_{B} } \over {a_{0} (1{ - }s_{A} { - }y'_{B} )}}$$ . One can verify that the quantity $$\alpha _{x} (s_{A} ){ - }\alpha '(s_{A} )$$ is strictly decreasing in $$a_{0} $$ on the interval $$(\underline{a} ,1)$$ and is equal to 0 when $$a_{0} =\tilde{a}$$ . The result follows instantly.

D. Proof of Proposition 6

Recall from the proof of Proposition 3 that $$\underline{a} ,\bar{a},\underline{b} ,\bar{b}$$ are all decreasing in $$r_{A} $$ ; $$\underline{a} $$ and $$\overline{b} $$ are increasing in s _A ; and $$\underline{b} $$ and $$\overline{b} $$ are, respectively, decreasing and constant in s _A . Furthermore, our assumptions imply $$r_{A} > s_{A} $$ , so that that V _A is larger in region (ii) than in region (iv), and larger in region (iv) than in regions (i), (iii) and (v).

Consider first the payoffs of Country A. V _A is constant in both r _A and s _A in regions (i), (iii) and (v), and increasing in both of these parameters in regions (ii) and (iv). Therefore the statement of the proposition follows immediately from these results and the changes in the boundaries of the five regions reported above.

Now consider Country B and recall that $$V_{B} =z_{B} { - }(1{ - }a_{0} {\,+\,}a_{0} \alpha )[z_{B} { - }(1{ - }r_{A} )]$$ . Therefore, V _B depends on r _A both directly and indirectly through the effect that r _A may have on α. As a result, we can immediately claim that $$V_{B}^{{\ast}} < V_{B} $$ whenever $$\alpha ^{{\ast}} \geq \alpha $$ . This happens if either $$a_{0} \leq \underline{a} $$ , or $$b_{0} \leq \underline{b} ^{{\ast}} $$ , or $$a_{0} \in [\underline{a} ^{{\ast}} ,\bar{a}^{{\ast}} ]$$ and $$b_{0} \geq \overline{b} ^{{\ast}} $$ . Instead, if $$a_{0} \in [\underline{a} ^{{\ast}} ,\underline{a} ]$$ and $$b_{0} \in [\underline{b} ,\overline{b} ^{{\ast}} ]$$ , $$V_{B} =1{ - }r_{A} $$ and $$V_{B}^{{\ast}} =z_{B} { - }(1{ - }a_{0} )\left[ {1{\,+\,}{{1{ - }r'_{A} { - }y_{B} } \over {r'_{A} { - }s_{A} }}} \right][z_{B} { - }(1{ - }r_{A} )]$$ . Thus using the lower bound for $$a_{0} $$ , we can conclude that $$V_{B}^{{\ast}} > 1{ - }r_{A} =V_{B} $$ . A similar reasoning leads to the same conclusion if $$a_{0} \in [\bar{a}^{{\ast}} ,\bar{a}]$$ and $$b_{0} \geq \overline{b} $$ . If instead, $$a_{0} \geq \underline{a} $$ and $$b_{0} \in [\underline{b} ,\overline{b} ^{{\ast}} ]$$ , one can check that V _B is decreasing in r _A , as $$V_{B} =z_{B} { - }(1{ - }a_{0} )\left( {{{1{ - }s_{A} { - }y_{B} } \over {r_{A} { - }s_{A} }}} \right)[z_{B} { - }(1{ - }r_{A} )]$$ , which is decreasing in r _A by Assumption 2b. If $$a_{0} \geq \bar{a}$$ and $$b_{0} \geq \overline{b} $$ , $$V_{B} =z_{B} { - }(1{ - }a_{0} )\left( {{{1{ - }y_{A} { - }y_{B} } \over {r_{A} { - }y_{A} }}} \right)[z_{B} { - }(1{ - }r_{A} )]$$ , which is increasing or decreasing in r _A , depending on whether $$1{ - }z_{B} < y_{A} $$ or $$1{ - }z_{B} \geq y_{A} $$ , respectively. Finally, if $$a_{0} \geq \bar{a}^{{\ast}} $$ and $$b_{0} \in [\overline{b} ^{{\ast}} ,\overline{b} ]$$ , we have that $$V_{B} =z_{B} { - }(1{ - }a_{0} ){{1{ - }s_{A} { - }y_{B} } \over {r_{A} { - }s_{A} }}[z_{B} { - }(1{ - }r_{A} )]$$ and $$V_{B}^{{\ast}} =z_{B} { - }(1{ - }a_{0} ){{1{ - }y_{A} { - }y_{B} } \over {r'_{A} { - }y_{A} }}[z_{B} { - }(1{ - }r'_{A} )]$$ . Thus it is immediate to verify that $$V_{B}^{{\ast}} $$ will be greater or lower than V _B depending on whether z _B is greater or lower than

Finally, consider a change in the aggressiveness of Country B as measured by $$1{ - }s_{A} $$ . Notice that V _B will increase (respectively, decrease) as α decreases (respectively, increases) after we change from h to $$h^{\dag } $$ . The result follows immediately by characterizing the regions in which these changes take place.