Complete Information Pivotal-Voter Model with Asymmetric Group Size

In this note, we characterize the equilibria of the standard pivotal-voter participation game between two groups of voters of asymmetric sizes, as originally proposed by Palfrey and Rosenthal [1983. A strategic calculus of voting. Public Choice. 41, 7-53].


Introduction
Pivotal-voter models were pioneered by the seminal contribution of Palfrey and Rosenthal (1983) (henceforth, PR). They analyze a complete information setting where two groups of individuals, each preferring one of two alternatives, simultaneously choose between abstaining or voting for their preferred alternative. Voting is costly and the winner is decided by majority rule. Despite the simplicity of this pivotal-voter game, when solving for its equilibria technical difficulties and multiplicity issues arise, and PR provide a partial characterization of these equilibria, some conjectures, and the analysis of special cases. 1 Nevertheless, though PR's work has been proven to be highly influential in recent decades, a complete characterization of the equilibria of their original game is still missing. The independent work of Nöldeke and Peña (2016) -henceforth, NP -characterizes the equilibria under symmetric group size and symmetric probability of voting across groups. In contrast, we allow the two groups of voters to have asymmetric size and the voters of different groups to vote with asymmetric probability, as initially proposed by PR.
An equilibrium is a probability of voting that each individual of a group follows with no profitable deviations. If the two groups are of the same size, as in NP, the equilibria are a mapping from the unique group size to a unique probability of voting that every individual follows, regardless of the group. If the groups are of different sizes, as in the present paper, the equilibrium is a mapping from the two group sizes to two probabilities of voting, one for the individuals of each group. NP confirm PR's conjectures on totally mixed equilibria (i.e., equilibria where all individuals vote with the same probability p ∈ (0, 1)) in the symmetric setting; namely, that there is a threshold in the cost of voting below which there is no such equilibrium, at which the unique such equilibrium is that everyone votes with a probability 0.5, and above which there exist exactly two such equilibria, one where everyone votes with a probability lower than 0.5 and one where everyone votes with probability greater than 0.5. The present note deals with the asymmetric setting, and we prove that 1) there are at most two totally mixed equilibria, 2 and 2) that for sufficiently low costs of voting there is also a "Partially-Mixed" equilibrium where the individuals of the minority (i.e., the smaller group) vote with probability 1 and the individuals of the majority (i.e., the bigger group) vote with a certain probability p ∈ (0, 1), which we show to be increasing in the size of the minority group and decreasing in the size of the majority group and in the cost of voting. This "Partially-Mixed" equilibrium resembles the equilibrium of the pivotal-voter model with private information on the cost of voting -see Taylor and Yildirim (2010) -where members of the minority group vote with a strictly greater probability than those in the majority do. This is often called the "underdog effect." 3 Thus, 1) and 2) show that for all costs of voting there are at most three equilibria, regardless of the number of individuals composing each group.
In Section 2 we describe the original PR's voter participation game. In Section 3 we analyze the equilibria, examining whether they entail pure strategies played by individuals of both groups (Subsection 3.1), one group only (Subsection 3.2), or no group (Subsection 3.3). Section 4 concludes.

Model
Consider a complete information setting with two groups of individuals of size m and n, with m, n ∈ N + . Throughout the paper we assume m > n > 1. The analysis of n = 1 is ruled out to avoid having to deal with trivial cases. The analysis of m = n produces different results -see NP. We use subindex i ∈ {m, n} to identify the group with an abuse of notation. The individuals are called to cast a vote between two alternatives, M and N . An individual of group m prefers alternative M , and an individual of group n prefers alternative N . That is, if M (N ) wins, the payoff of an individual m (n) equals 1, otherwise it equals 0. If an individual casts a vote, she faces a cost of voting, c > 0. Thus, the payoff for an individual when her preferred policy wins is 1−c if she voted, and 1 if she did not vote. Individuals simultaneously choose whether to vote for their preferred alternative or abstain, since voting for the non-preferred alternative is strictly dominated. The winning alternative is decided by majority rule, and ties are broken by a fair coin toss.
Each individual i chooses her probability of voting, denoted by p i , that maximizes her expected payoff, given the choices of all other individuals. We consider Quasi-Symmetric Nash Equilibria (QSNE), that is, individuals of group i follow the same equilibrium strategy p * i . Besides being used in PR, 4 the QSNE has been used in private-information pivotal-voter models to obtain that individuals adopt cut-off strategies in the cost of voting (e.g., Börgers, 2004;Taylor and Yildirim, 2010).
A pair (p * i , p * j ) is a QSNE if an individual of group i does not want to deviate from p * i if she expects every other individual of group i to also play p * i and all individuals of group j to play p * j . A QSNE can be of one of the following three types: Define A i as the probability that the vote of an individual of group i is pivotal: (1) for i, j ∈ {m, n} and i = j. 5 We now explain how the expression (1) is constructed. A single individual m, who computes her probability of being pivotal, takes as given the probabilities of voting (p m , p n ) of all other individuals. The individual m is pivotal when her vote either breaks a tie or when it creates one. In (1) the first summation is her probability of breaking a tie, and the second of creating a tie. She can break a tie with her vote, if the number of individuals that vote for m equals the number of individuals that vote for n. Let us call this number s. Out of m − 1 other m-individuals, exactly s vote with probability m−1 On the other hand, out of n n-individuals, exactly s vote with probability n s p s n (1 − p n ) n−s . The second summation of (1) is similarly constructed: individual m can create a tie with her vote, if the number of individuals that vote for m (which is again called s) is one less than the number of individuals that vote for n.
An individual i casts a vote if her expected utility from casting the pivotal vote is greater than her cost voting. Since ties are broken by a fair coin toss, if the vote of a pivotal individual creates (breaks) a tie, her expected utility 4 PR's definition of QSNE allows for asymmetries across group members if they play pure strategy. In other words, only players playing a totally mixed strategy (i.e., p i ∈ (0, 1)) are supposed to follow the same equilibrium strategy p * i , and some players within the same group who play pure strategy play p * i = 0 and others play p * i = 1. This gives rise to what PR call "k-equilibria." 5 Notice that, in order to lighten the notation, we use A i rather than A i (m, n, pm, pn).

3
Electronic copy available at: https://ssrn.com/abstract=2874414 increases from 0 to 1/2 (from 1/2 to 1). In both cases the increase in utility is 1/2. Thus, the condition for individual i to vote reads, 3 Computing the equilibria 3.1 "Pure" equilibria , then individual i's strictly dominant strategy is respectively to abstain or to vote with certainty (i.e., pure strategy). If A i = 2c, the individuals of group i are indifferent between voting or not (i.e., mixed strategy). Therefore, 2c can be interpreted as the minimum probability of being pivotal such that an individual of group i will vote. For this reason, if c > 1/2 no individual votes in equilibrium. 6 This turns out to be true even if c = 1/2. We formalize these results in the following lemma.
Proof. Fix i ∈ {m, n} and let p * i > 0. First, if c > 1/2, by (2) we have A i > 1, which is a contradiction, since A i is a probability. Second, if c = 1/2, by (2) A i = 1. Then p * i ∈ (0, 1) because otherwise no individual in i would be pivotal with certainty, so A i < 1, which leads to a contradiction. Therefore, we only need to rule out c = 1/2 and p * i = 1, which we do in the remainder of the proof. For this we need to distinguish the following cases: Case 1. If p * m = 1 then alternative M wins regardless of p * n because m > n. Then, no n individual would want to face the cost of voting, so p * n = 0. Thus, for a single m-individual a deviation to p m = 0 would be profitable, leading to a contradiction. Case 2. If p * n = 1 then it is necessary that A n ≥ 2c = 1, and thus A n = 1. This is the case if the n individuals are certain that either n or n − 1 of the m individuals vote. However, this cannot happen either if p * m ∈ (0, 1) (because this would imply a stochastic number of votes cast by group m) or if p * m = {0, 1} (because this would imply m or 0 votes cast by group m). Thus, we reach a contradiction.
The previous lemma shows that if the cost of voting is high enough, the only equilibrium that exists is the "Pure" one in which nobody votes. In the remainder of the paper we analyze the more interesting case of a lower cost of voting (c < 1/2), so that individuals may vote with positive probability. This causes strategic interactions that will generate multiple equilibria.
We conclude this section proving that when c < 1/2 no "Pure" equilibria exist.
Proof. Fix i ∈ {m, n} and j = i. Assume p * i = 0. Then p * j = 0 cannot be a QSNE because A j = 1 so a deviation to voting for a j individual would be profitable. Also p j = 1 cannot be a QSNE because a deviation fo abstantion of a j-individual would not affect the outcome of the election and save her cost of voting. Finally, p * m = p * n = 1 cannot be a QSNE because the n individuals lose for sure and they would thus be better-off not voting.

"Partially-Mixed" equilibria
The next lemma establishes the existence of a unique "Partially-Mixed" equilibrium where the members of the minority group n vote with certainty exist. Note that PR do not analyze this case. 7 Lemma 3. There exists aĉ < 1/2 such that: if c >ĉ, there exists no "Partially-Mixed" QSNE, and if c ≤ĉ, there exists a unique "Partially-Mixed" QSNE. This is given by p * n = 1 and a certain p * m ∈ (0, 1) which is increasing in n and decreasing in m and c.
Proof. Fix i ∈ {m, n} and j = i.
Step 1. We show that there is no "Partially-Mixed" QSNE which involves members of one group abstaining, p * i = 0, and of the other group playing a mixed strategy, p j ∈ (0, 1). If individuals i abstain, an individual j is pivotal only if none of her groupmates happen to vote. Thus, we can write the probabilities of being pivotal as In order to sustain any such "Partially-Mixed" QSNE, it has to be that A i ≤ 2c and A j = 2c, and thus A i ≤ A j , or equivalently Since p j ∈ (0, 1), we can divide the above condition by (1−p j ) j−1 and obtain j ≤ 1, which is a contradiction.
Step 2. We show that there is no "Partially-Mixed" QSNE which involves members of the majority voting with certainty: p * m = 1 and p * n ∈ (0, 1). Since the m > n, the m group wins with certainty, so the n individuals are better off abstaining (p * n = 0) so as to save the cost c without affecting their probability of victory.
Step 3. The last case to analyze is that of a "Partially-Mixed" QSNE with p * n = 1 and p * m ∈ (0, 1). Then we have In order to sustain any such "Partially-Mixed" QSNE, it has to be that A m = 2c and A n ≥ 2c. Thus, A m ≤ A n .
We divide the remainder of the proof of this lemma into two steps. We will use Figure 1 to provide the intuition behind the two steps.  Figure 1, which helps follow the proof. Thus, in the interval where the "Partially-Mixed" QSNE has to lie, namely, p m ∈ (0, p * * m ] (see Step 3.1), A m increases in p m . Therefore, settinĝ c = 1 2 A m | pm=p * * m provides the unique cut-off value for c below (above) which a "Partially-Mixed" equilibrium does (not) exist, and thus the result follows. Step Divide by (m−1)! and multiply by n!(m−n+1)! both sides of the inequality and obtain which can be rewritten as: In order to find the roots of the quadratic equation (4) we use a non-standard method which significantly simplifies the computation. In particular, the solution of a general equation Ax 2 + Bx + C = 0 can be written as: 8 Formula (5) is equivalent to the standard formula for the roots of a quadratic equation, but it is more convenient in our case because: i) it gives one valid root when A = 0, which could happen in our case, and ii) the root is easier to compare withp m which we will discuss in Step 3.2, since they are both roots of quadratic equations with identical coefficients B and C, but with different A's.
By (5), the roots of (4) are Next we simplify the discriminant: and thus The root with "-" in (6) belongs to the interval [0, 1]. If m − 2n + 1 > 0, the root with "-" is negative and the function defined by (4) is convex, thus (4) is satisfied at all p m 's greater than its root with "+". If m − 2n + 1 < 0, the root with "-" is greater than 1 and the function defined by (4) is concave, thus, once again, (4) is satisfied at all p m 's greater than its root with "+". If m = 2n − 1, trivially p * * m = 1 2 and the function defined by (4) is increasing. Thus p m ≤ p * * m , with Step 3.1 of the proof.
Step 3.2. The fact that A m = 0 if p m = 0 is trivial, since p * n = 1. Consider A m = 2c. Callp m the maximum of the function A m in p m ∈ [0, 1], which we will prove to be unique. In order to computep m , take the derivative of A m with respect to p m , which equals 0 if and only if Again, apply the quadratic formula (5), and obtain If m = 2n, triviallyp m = 1 2 (see Figure 1) which is a case broadly analyzed in PR. The discriminant can be simplified to: where we discard the root with "+" in front of the square root in (7) because it is not in [0, 1]. 9 Notice that m = n + 1 =⇒p m = 1, as can be seen in Figure  1.
A straightforward comparison ofp m and p * * m yields thatp m > p * * m , and the reasoning before Step 3.1 concludes the proof of the first part of the Lemma.
We are left to show the comparative statics of p * m with respect to c, m, n. We already proved that A m increases in p m and thus the equilibrium p * m , which solves A m = 2c decreases in c. By the same argument, it is sufficient to conclude the comparative statics exercise to show that A m increases if one substitutes m + 1 instead of m, and that it decreases if one substitutes n + 1 instead of n. The former condition coincides with inequality (3), which has to hold in a "Partially-Mixed" equilibrium, whereas the latter condition is equivalent to: where the last inequality is similar to (4). In fact, the only difference is that the coefficient of the term p 2 m is now smaller, and thus (4) implies (8), which concludes the proof.

"Totally-Mixed" equilibria
Lemmas 1, 2, and 3 completely characterized the "Pure" and "Partially-Mixed" equilibria. In this Subsection we analyze the "Totally-Mixed" equilibria, which may exist only when c < 1/2. In any "Totally-Mixed" equilibrium the voting conditions (2) for the two groups hold with equality. That is, the mixing condition for the m individuals is A m = 2c (9) and that of the n individuals is In a QSNE all individuals within a group employ the same strategy. Thus, in order to analyze the "Totally-Mixed" equilibria we look for pairs (p m , p n ) ∈ (0, 1) 2 which satisfy both (9) and (10). In particular, we focus on the p m solving (9) for every p n ∈ (0, 1), and on the p n solving (10) for every p m ∈ (0, 1). The intersections between these two sets in the space (p m , p n ) ∈ (0, 1) 2 are the "Totally-Mixed" equilibrium pairs (p * m , p * n ), which are what we are after in this Subsection.
The "Totally-Mixed" case -in contrast to the "Pure" and "Partially-Mixed" cases -entails solving a system of two polynomial equations of a potentially large degree -expressions (9) and (10) -, and thus it is presumably impossible to find an algebraic solution for equilibrium strategies. For this reason, we use a number of results about the space (p m , p n ) ∈ (0, 1) 2 where these equilibria lie on.
Conditions (9) and (10) require A m = A n . The set of pairs (p m , p n ) ∈ (0, 1) 2 where A m = A n is composed of a decreasing line, namely p m + p n = 1, and an increasing line. Throughout the rest of the paper, we will refer to these two lines as "decreasing line" and "increasing line", for the sake of brevity. We depict this set in Figure 2 for the special case of m = 3 and n = 2. All "Totally-Mixed" equilibria must fall into the set defined by those two lines. In the Appendix we characterize the set A m = A n , through a series of lemmas. First, we find the four points where these two lines reach the edges of the (p m , p n )-space (Lemma 4). Second (Lemma 5), we prove that for every p m (p n ) there are at most two p n (p m ) that solve A m = A n . Third, we characterize the decreasing line, namely p m + p n = 1 (Lemma 6). Finally, we characterize the increasing line (Corollary 2). 10 We summarize this series of lemmas in Proposition 1.
Proposition 1. The only points (p m , p n ) ∈ (0, 1) 2 satisfying condition are the points along the line p m + p n = 1 and along a strictly increasing continuous line that goes from (0, 0) to (p * * m , 1) where p * * m = n(n−1) .
Proof. See Appendix.
A consequence of Proposition 1 is the following corollary. 11 Corollary 1. If c < 1/2, there exist at most two "Totally-Mixed" equilibria.

Discussion
In the present note we have characterized the possibly asymmetric equilibria of the possibly asymmetric version of the pivotal-voter model proposed in PR's seminal contribution. Compared to the symmetric setting, which has already been analyzed in NP, our asymmetric model gives rise to an extra equilibrium where the minority votes with certainty and the majority plays a mixed strategy in participation. Besides providing such characterization, we show that the existence of at most two totally mixed equilibria carries over to the asymmetric version of the model. While NP provide their result through an elegant application of mathematical results on polynomials in Bernstein form to the conditional pivot probability, we prove our result in the asymmetric setting through an application of Descartes' rule of signs to the unconditional pivotal probability. 12 As one might expect, these two mathematical tools are not orthogonal, and in fact results on the number of roots of polynomials in Bernstein form partially rely on Descartes' rule of signs. Thus, despite a vast range of results being at hand when a polynomial can be written in Bernstein form (see Farouki, 2012), Descartes' rule of signs has a broad applicability -being more primitive -, and has in fact long been used in a variety of fields of economics; among others, by Hirshleifer (1958) . 10 In the case of m = n analyzed in NP the increasing line corresponds to the 45 degree line, pm = pn. 11 Trivially, for c ≥ 1/2 there is no "Totally-Mixed" equilibrium since the mixing conditions (9) and (10) are violated. 12 A version of Descartes' rule of signs states that if the terms of a polynomial in x are ordered by descending exponent of x, then the number of positive roots in x of the polynomial is at most equal to the number of sign differences between consecutive nonzero coefficients.

A Appendix
A.1 Proof of Proposition 1 Proposition 1 characterizes the set of pairs (p m , p n ) which solves A m = A n , see Figure 2. Its proof is divided into four lemmas. Lemma 4 characterizes the four pairs (p m , p n ) at the edges of the unit box (p m , p n ) which solve A m = A n . Lemma 5 shows that for each p m (p n ) there exist at most two p n (p m ) that satisfy A m = A n . Lemma 6 shows that p m + p n = 1 always solves A m = A n ; this line connects two of the four points of Lemma 4 in the (p m , p n )-space. Corollary 2 shows that there is a unique, strictly increasing and continuous line in the (p m , p n )-space which satisfies A m = A n and connects the remaining two points of Lemma 4. This concludes the proof of Proposition 1. .
Proof. By continuity of A m and A n in p m and p n , in order to analyze the pairs (p m , p n ) solving A m = A n we analyze its limit for A m (12) It is easy to see that (12) is the equation corresponding to (3) in Lemma 3, and thus its unique solution is p m = p * * m . Lemma 5. Fix m > n > 1, c < 1/2 and i ∈ {m, n} with j = i. For a given p i ∈ (0, 1), there exist at most two values of p j ∈ (0, 1) which satisfy A m = A n .
Proof. Take p n as given and consider A m = A n as a polynomial in p m . 13 In order to prove that there are at most two roots of A m = A n in p m we will show that the coefficients of (p 0 m , p 1 m , p 2 m , ...) change sign at most twice, and thus the result follows by Descartes' rule of signs. 14 Descartes' rule provides an upper bound in the number of positive roots in p m of a polynomial in p m . Since our focus is on the roots where p m ∈ (0, 1), we use Jacobi's substitution 15 and we rewrite the polynomial in terms of y = pm 1−pm , so that p m ∈ (0, 1) implies y ∈ (0, +∞). Since y is a one-to-one mapping (0, 1) → (0, +∞), the upper bound in the number of roots of A m = A n in y ∈ (0, +∞) corresponds to the upper bound in the number of roots of A m = A n in p m ∈ (0, 1).
Simplify ( In (13) three summations stop at s = n − 1. Add to those summations the n th element, which equals 0. This procedure will allow us to merge all summations. Next, in the two summations on the left-hand side, separate the elements multiplying y s and y s+1 , and obtain Next, we merge all summations with y s and all summations with y s+1 , and add a 0 element to both summations (s = n + 1 in the first, and s = −1 in the second). 13 A similar proof can be carried out taking pm as given making Am = An a polynomial in pn.
14 Page 28, Corollary 1, Prasolov (2001). The version of Descartes' rule of signs we apply states that the number of positive roots of the polynomial is at most equal to the number of sign differences between consecutive nonzero coefficients. 15 Page 28, Prasolov (2001).   We are ready to apply Descartes's rule of signs to the polynomial (14) in y s+1 . The term p s n (1−pn) −s−2 (m−1)!(n−1)! (s+1)!(s+2)!(m−s−1)!(n−s)! is always positive. 16 The coefficients of y n+1 (s = n) and y 0 (s = −1) are both positive, and the coefficient of y n (s = n − 1) has the sign of n 2 − m. In the remainder of the proof we focus on the number of sign changes of the term in the square brackets as s goes from −1 to n − 1, which affects the number of sign changes of the polynomial (14). The term in the square brackets is a polynomial of degree three in s, thus it has potentially three real roots in s. However, in what follows we prove that it has at most two roots in s ∈ (−1, n − 1).
Then plug these factors in each addend of the term in the square brackets of (14).
And then collect the constant term.
nmp n (1 − p n )(1 + n) + n 2 p 2 n (n + 1)m − p n mn(n + 1) = nmp n (n + 1)(1 − p n + np n − 1) = nm(n + 1)p 2 n (n − 1) = nm(n 2 − 1)p 2 n The terms that are multiplied by x are not needed for our argument, which uses only the signs of the coefficient of x 3 , x 2 , and the one of the intercept.
The intercept is positive. 17 The coefficient of x 3 takes the sign of n 2 − m. Therefore, we have three cases to check.
Case I: n 2 > m. Then both the coefficients of x 3 and the coefficient of y n in (14) are positive. 18 Therefore, when x → −∞, the polynomial in x tends to −∞. Since the intercept is positive, at least one root is negative, and thus the polynomial (of degree 3) in x crosses the positive axis at most twice. Therefore, it has at most two positive roots in x (and hence in s), which proves that the coefficients of y (and hence of p m ) change sign at most twice.
Case II: n 2 < m. Then both the coefficients of x 3 and the coefficient of y n in (14) are negative. Therefore, one change of sign occurs when moving from n to n + 1. This means that for s ∈ (−1, n − 1) we need to have at most one sign change for having in total at most two roots. Let us focus on the polynomial of degree three in x. We know that when x → −∞(+∞) the polynomial in x tends to +∞(−∞), and that its intercept is positive. Thus, in order to show that it crosses the positive part of the horizontal axis at most once, we study its derivative. In fact, it is sufficient to show that the derivative is a parabola with negative horizontal coordinate of the vertex. Since at x = 0 the polynomial is positive, this implies that for x ≥ 0, the slope of the original polynomial changes sign at most once. The horizontal coordinate of the vertex of a quadratic of the form Ax 2 + Bx + C is −B/(2A). We know that A is negative when n 2 < m . Thus, it suffices to show that B is also negative. The sign of B coincides with the sign of the coefficient of x 2 in the original polynomial of degree 3. That is: where the right-hand side is increasing in p, and thus proving that it holds for p = 0 is sufficient. This is equivalent to proving that m 2n + n 2 + 3 > n 2 + m which is trivially true. Case III: n 2 = m. In this case the coefficients of x 3 and y n in (14) are zero. The latter implies that no change of sign occurs when moving from n to n + 1. 19 The former implies that the original polynomial of degree three in x is of degree two, and thus there are at most two positive roots in x (and s), which again proves that the coefficients of y (and p m ) change sign at most twice.
where in the second-to-last step we used the symmetry rule for binomial coefficients, and in the last step we used Vandermonde's identity. 20 Lemma 6 characterizes the decreasing line in the (p m , p n )-space which solves A m = A n and connects (p m , p n ) = (0, 1) and (p m , p n ) = (1, 0). The remaining pairs (p m , p n ) which solve A m = A n are characterized by the following corollary, and this concludes the proof of Proposition 1.

Corollary 2.
There exists a unique, strictly increasing and continuous line in the (p m , p n )-space which satisfies A m = A n and connects (0, 0) and (p * * m , 1).
Proof. Fix i ∈ {m, n} and j = i. Lemma 5 proves that for each p i , there are at most two p j that solve A i = A j and Lemma 6 proves that for every p i , one of the solutions p j must lie on p i + p j = 1. Therefore, for each p i , there is another solution p j not belonging to p i + p j = 1. Since A m = A n is a polynomial, the extra set of solutions lies on a continuous line, and since there is only one p j per p i , this line is unique. To see why it is also strictly increasing, suppose it is not. Then for some p n there exist at least four p m that solve A m = A n : one belonging to p m + p n = 1 and at least three belonging to the other continuous line, which is a contradiction to Lemma 5. The last part of the Corollary follows from Lemma 4.
Therefore, for a given p m ∈ (0, p * * m ), there exist exactly two values of p n ∈ (0, 1) which satisfy A m = A n . For a given p n ∈ (0, 1), there exist exactly two values of p m ∈ (0, 1) which satisfy A m = A n .

A.2 Proof of Corollary 1
Once A m = A n is fully characterized (Proposition 1), it is easy to bound the number of equilibria. We now provide a loose description of how the two equilibria behave as c changes, and in the five steps below we provide a proof of what we describe. Consider Figure 3, where the crossing between the increasing and decreasing lines is called X, we change c and plot the two mixing conditions (red and blue lines). When c → 0, there are two equilibria arbitrarily close to points (0, 1) and (1, 0). As c increases, the two equilibria move towards point X along the decreasing line, they reach X, and then they move along the increasing line (one upward and one downward) towards (0, 0) and (p * * m , 1). They reach these two points for sufficiently high c.  Step 1. The only pairs (p m , p n ) which solve A m = A n for c → 0 are (0, 1) and (1, 0).
When c → 0, in a "Totally-Mixed" equilibria A m = A n → 0. For any pair (p m , p n ) ∈ (0, 1) 2 , at least one between p m and p n has to tend to 0 or 1 for A m = A n → 0 to hold. We have already characterized the limits of A m and A n as p m → {0, 1} and p n → {0, 1} at the beginning of the proof of Lemma 4, where it is easy to see that if we fix i ∈ {m, n} with j = i the unique solution in p j of lim pi→0 A m = lim pi→0 A n = 0 is p j = 1, and symmetrically, the unique solution in p j of lim pi→1 A m = lim pi→1 A n = 0 is p j = 0. Thus, when c → 0, the only existing "Totally-Mixed" equilibria (p m , p n ) converge to (0, 1) and (1, 0) and lie along the decreasing line.
Step 2. ∃! c ∈ (0, 1/2) such that for c > c (respectively, c = c, c < c) which there is no (respectively, one, two) equilibrium on the decreasing line. It is easy to see that (15) is a strictly quasi-concanve function of p m , that is, A m is increasing if and only if p ≤ p m . Define c ≡ A m ( p m ). Then c is the minimum cost for which the equilibria on the decreasing line are defined. If c > c, there is no equilibrium on the decreasing line. If c = c, there is exactly one equilibrium on the decreasing line. If c < c, there are exactly two equilibria on the decreasing line, one with p m > p m and one with p m < p m .
Step 3. ∃!c ∈ (0, 1/2) such that the equilibrium is arbitrarily close to the point (0, 0), or to the point (p * * m , 1). This is needed to rule out that, as c changes, the equilibria "return" to one of these two extreme points, because if so they might cease to exist.
From the limits of A m and A n at the beginning of the proof of Lemma 4 it is easy to see that lim A n ∈ (0, 1).
Step 4. c is the unique value of c for which the equilibria on the increasing line are at X. Also, there is no other value of c for which either equilibrium reaches X.
Consider the two equilibria on the extreme of the increasing line (0, 0), and (p * * m , 1). As c changes, they move continuously (in (p m , p n )) along the increasing line, and when they reach X they need to satisfy the condition p n + p m = 1, and thus the reasoning of Step 2 holds, pinning down the unique value c.
Step 5. Along the increasing line the two mixing conditions considered as functions of p m have slope different than the slope of the increasing lines, otherwise new equilibria might show up other than the two which originated at (0, 0), and (p * * m , 1). 22 Figure 4 shows the the number of equilibria might vary