# Relative responsiveness of bargaining solutions to changes in status-quo payoffs.

Introduction

It is a well-known fact that several prominent bargaining solutions--such as the Nash, Kalai/Smorodinsky, equal gains, and equal sacrifice solutions, which will be defined rigorously in the next section--are responsive to changes in status-quo (i.e., disagreement or fallback) payoffs in that when an agent's status-quo payoff increases, his solution payoff either stays the same or increases) That is, when an Agent i's status-quo payoff increases while that of the other agent stays the same, Agent i is not worse off as a result; in many cases he is better off and consequently the other agent is worse off. The next natural question is "are some of the abovementioned bargaining solutions more responsive than others with respect to changes in status-quo payoffs."

In the next section, we will provide examples which will imply that a general result about these solutions' relative status-quo point responsiveness is impossible to establish. We will then show that, using a fairly general class of bargaining problems--namely, the 'Constant Elasticity of Substitution' class of bargaining problems--and regardless of the concavity of the Pareto frontier and the increase in the status-quo point, the equal gains solution is most responsive with respect to changes in status-quo payoffs, followed by the Nash solution. The equal sacrifice solution turns out to be the least responsive, followed by the Kalai/Smorodinsky solution.

The Model and Results

A bargaining problem is a pair (S,d) where the feasible set, S, is a subset of [R.sup.2.sub.+] and d[member of]S is the status-quo point such that there is some x[member of] S with x>d. (2) The agents receive d unless they unanimously agree on a compromise x in S. Let [[SIGMA].sup.2.sub.d] be the class of pairs (S,d), where S [subset or euqal to] [R.sup.2.sub.+] is convex and compact. Given [[SIGMA].sup.2.sub.d], a solution is a function F associating with every (S,d) in [[SIGMA].sup.2.sub.d] a point F(S,d) [member of] S. Let [partial derivative] S denote the Pareto frontier (or boundary) of S; i.e., [partial derivative] S [equivalent]- {x [member of] S|x' > x' [not member of] S}. Note that [partial derivative]S includes both strongly and weakly Pareto optimal points. That is, there can exist x[member of][partial derivative]S and x'[member of] S such that x'[greater than or equal to]x, but if so, then x' must also be in OS. E.g., consider S= ch {(0,0), (1,0), (1,1), (0,2)}, where "ch" stands for "convex hull of." Let x=(1,1) and x'=(1,0). Then any point in the segment (x, x'] is weakly dominated by x=(1,1). That is, any point in the segment (x, x'] would be weakly Pareto optimal while the segment [x", x] would be strongly Pareto optimal, where x"=(0,2). The Nash solution, N, is such that given any (S, d) [member of] [[SIGMA].sup.2.sub.d], its outcome N(S,d) maximizes ([N.sub.1](S,d)-[d.sub.1])([N.sub.2](S,d)-[d.sub.2]) (Nash 1950, 1953). The equal gains solution, E, is such that given any (S, d) [member of] [[SIGMA].sup.2.sub.d], its outcome E(S,d) is the maximal point in S for which [E.sub.1] (S, d) - [d.sub.1] = [E.sub.2](S, d) - [d.sub.2] holds (Kalai 1977; Roth 1979b).

Let [b.sub.1] = max{xl(x,y) [member of] (S,d)} and [b.sub.2] = max{y|(x,y) [member of] (S,d)}. That is, [b.sub.1] and [b.sub.2] are the highest utilities for Agents 1 and 2 in the quadrant of S that Pareto dominates d. The Kalai/Smorodinsky solution, KS, is such that given any (S,d) [member of] [[SIGMA].sup.2.sub.d], its outcome KS(S,d) is the maximal point in S for which ([KS.sub.2] (S, d) - [d.sub.2])/([KS.sub.1] (S, d) - [d.sub.1]) = ([b.sub.2] - [d.sub.2]) / ([b.sub.1] - [d.sub.1]) holds (Kalai and Smorodinsky 1975; Anbarci 1997).

The equal sacrifice solution, ES, is such that given any (S, d) [member of] [[SIGMA].sup.2.sub.d], its outcome ES(S,d) is the maximal point in S such that [b.sub.2] - [ES.sub.2](S, d) = [b.sub.1] - [ES.sub.1] (S, d) holds (O'Neill 1982; Aumann and Maschler 1985).

Although all bargaining solution concepts above are responsive to changes in status-quo points, (3) it is not straightforward to see whether some of these bargaining solutions are more responsive to those changes than others. To that end, we will introduce relative status-quo point responsiveness (RSPR): Consider all (S,d), (S,d') and solution concepts F and G such that [F.sub.i] (S,d) = [G.sub.i] (S,d), i= 1,2. For all [d.sub.i] > [d.sub.i]/and [d'.sub.j] = [d.sub.j], ij= 1,2, i[not equal to]j, we have [F.sub.i] (S,d')[greater than or equal to][G.sub.i]{S,d'). Then F is relatively more (status-quo point) responsive than G.

First, we will provide examples to indicate that one cannot possibly establish any general ranking of these solution concepts in terms of their relative status-quo point responsiveness. In all examples in this paragraph, consider d=(0,0) and d'=(0,[d.sub.2]) where [d.sub.2]>0. Consider S=ch { {(0,0), (0,k), (k,0)}. Observe that F(S,d)=(k/2, k2) for any of the solutions considered here. Then note that F(S,d') will still coincide for all of those solution concepts. Consider S=ch {(0,0), (0,k), (k,0), (k,k)}. Observe that F(S,d)=(k,k) for any of the solutions considered here. Note that F(S,d')=F(S,d)=(k,k) will still coincide for all of those solution concepts, except the equal gains solution; for the latter, [E.sub.1](S,d')<k, and [E.sub.2](S,d')=k will hold. Finally consider S=ch {(0,0), (0, ak), (ak, O), (k,k)}, where, say, a[member of](1,3/2]. Then observe that F(S,d)=(k,k) for any of the solutions considered here. Note that [F.sub.1](S,d')<[F.sub.1](S,d) will hold for all of those solution concepts, except the Nash solution; for the latter, N(S,d')=(k,k) will still hold as long as [d.sub.2] < (2 - a)k. (4) These examples indicate that there cannot be any general ranking of these solution concepts in terms of their relative status-quo point responsiveness. Thus, in order to establish such a ranking, one needs to consider a particular class of bargaining problems instead.

Consider the class of constant-elasticity of substitution (CES) bargaining problems with Pareto frontier y = [(1 - [x.sup.[alpha]).sup.1/[alpha]] (where x is Agent l's payoff and y is Agent 2's payoff) such that 1 <[alpha]<[infinity]. Since each such feasible set F is symmetric and all solution concepts here satisfy the Symmetry and Pareto Optimality axioms, they will have symmetric solution outcomes, F(S,d) when [d.sub.1] = [d.sub.2]; in that case, [b.sub.1] = 1 and [b.sub.2] = 1. See Fig. 1 which depicts a few CES Pareto frontiers; it also depicts the two extreme cases, [alpha] = l and [alpha] = [infinity]. Both of those cases were considered in the examples in the previous paragraph; as those examples indicate our results do not hold for those cases, and thus they are not admitted in our model. Observe that, using the Pareto frontier y = (1 - [x.sup.[alpha]).sup.1/[alpha]], we have [b.sub.2] = l and [b.sub.1] = (1 - [d.sup.[alpha]).sup.1/[alpha]].

The interpretation of [alpha] could follow from the risk aversion levels of the parties. If, ceteris paribus, both parties are risk-neutral, then [alpha] will be one. As both parties get more risk-averse at the same rate, then [alpha] will increase. The interpretation of could also follow from the level of progressivity of a country's tax system. Given risk-neutrality, if, ceteris paribus, there is no progressivity in the tax system, then a will be one. As the progressivity of the tax system increases, then a too will increase. Note that, in either interpretation of a, as a increases, gains from cooperation will increase as well. When [alpha] = 1, it is a constant-sum situation; i.e., S is the unit simplex. When [alpha][right arrow][infinity], S tends to a rectangle; i.e., to a conflict-free problem. In particular, when [alpha]=2, S is the North-east quadrant of a circle.

[FIGURE 1 OMITTED]

The following lemma holds for any bargaining problem:

Lemma 1 For any (S, d) [member of] [[SIGMA].sup.2.sub.d], [E.sub.i] > [KS.sub.i] if and only if [E.sub.j]<[KS.sub.j]<[ES.sub.j].

Proof of Lemma 1 The Kalai/Smorodinsky solution is defined as ([KS.sub.2] - [d.sub.2])/ ([b.sub.1] - [d.sub.1])([KS.sub.1] - [d.sub.1])= ([b.sub.2] - [d.sub.2])/( [b.sub.1] - [d.sub.1]). We will keep manipulating the RHS (i.e., ([b.sub.2] - [d.sub.2])/([b.sub.1] - [d.sub.1])) to obtain the desired result. Let KS abbreviate the LHS; i.e., KS = ([KS.sub.2] - [d.sub.2])/([KS.sub.1] - [d.sub.1]). The RHS can be rewritten as [([b.sub.2] - [ES.sub.2])+ ([ES.sub.2] - [d.sub.2])]/[([b.sub.1] - [ES.sub.1]) + ([ES.sub.1] - [d.sub.1])]. Meanwhile, since ([b.sub.2] - [ES.sub.2]) = ([b.sub.1] - [ES.sub.1]) holds for the equal sacrifice solution, we have [([b.sub.2] - [ES.sub.2]) + ([ES.sub.2] - [d.sub.2])]/ [([b.sub.2] - [ES.sub.2]) + ([ES.sub.1] - [d.sub.1])]. This can be rewritten as [1 + ([ES.sub.2] - [d.sub.2])/( [b.sub.2] [ES.sub.2])]/[1 + ([ES.sub.1] - [d.sub.1])/([b.sub.1], - [ES.sub.1])]. Multiply the numerator and the denominator of the latter expression by ([b.sub.2] - [ES.sub.1])/([ES.sub.1] - [d.sub.1]) and use simple algebra to obtain [([b.sub.1] - [ES.sub.1])/([ES.sub.1] - [d.sub.1]) + ([ES.sub.2] - [d.sub.2])/([ES.sub.1] - [d.sub.1])/([b.sub.1] - [d.sub.1])/([ES.sub.1] - [d.sub.1])]. Multiply the numerator and the denominator of the latter expression by ([ES.sub.1] - [d.sub.1])/([b.sub.1] - [d.sub.1]) to obtain [([b.sub.1] - [ES.sub.1])/([b.sub.1] - [d.sub.1]) + ([ES.sub.1] - [d.sub.1])/([b.sub.1] - [d.sub.1])][([ES.sub.2] - [d.sub.2])/([ES.sub.1] - [d.sub.1])]. Since ([E.sub.2] - [d.sub.2])/([E.sub.1] - [d.sub.1]), plug it in the latter expression to get [([b.sub.1] - [ES.sub.1])/ ([b.sub.1] - [d.sub.1])][([E.sub.2] - [d.sub.2])/([E.sub.1] - [d.sub.1])] + [([ES.sub.1] - [d.sub.1])/([b.sub.1] - [d.sub.1])][([ES.sub.2] - [d.sub.2])/([ES.sub.1] - [d.sub.1])].

Let ES = ([b.sub.2] - [ES.sub.2])/([b.sub.1] - [ES.sub.1]) and E = ([E.sub.2] - [d.sub.2])([E.sub.1] - [d.sub.1]). Then the last expression of the RHS becomes [([b.sub.1] - [ES.sub.1])/([b.sub.1] - [d.sub.1])]E+ [([ES.sub.1] - [d.sub.1])/ ([b.sub.1]- [d.sub.1])]ES which is equal to our initial LHS, KS. Thus, KS is the weighted average of E and ES where E's weight is ([b.sub.1] - [ES.sub.1])/([b.sub.1] - [d.sub.1]) and ES's weight is ([ES.sub.1] - [d.sub.1])/([b.sub.1] - [d.sub.1]). This implies that given some d, [E.sub.i]>[KS.sub.i]>[ES.sub.i] if and only if [E.sub.j]<[KS.sub.j]<[ES.sub.j] ~.

The intuition of the above result is very straightforward: E(S,d) is the intersection of the Pareto frontier and the line that goes through d with a slope of one. ES(S,d) is the intersection of the Pareto frontier and the line that goes through b with a slope of one. As a result, KS(S,d), which is the intersection of the Pareto frontier and the line that goes through d and b, lies between E(S,d) and ES(S,d) as a weighted average of them.

Our main result is as follows:

Theorem 1 For any (S,d) [member of] [[SIGMA].sup.2.sub.d] with [partial derivative]S such that y = [(1 - [x.sup.[alpha]).sup.1/[alpha]], d=(0,0) and [d.sub.2]>0,

(i) When [alpha] = 1, [N.sub.1] = [ES.sub.1] = [KS.sub.1] = [E.sub.1].

(ii) When [alpha] > 1, [ES.sub.1]>[KS.sub.1]>[N.sub.1]>[E.sub.1].

Proof of Theorem 1 The outcome of the equal sacrifice solution solves [b.sub.2]-y=[b.sub.1]-X. Since [b.sub.2]= 1 and [b.sub.1] = [(1 - [d.sup.[alpha].sub.2]).sup.1/[alpha]], it is 1 - y = [(1 - [d.sup.[alpha].sub.2]).sup.1/[alpha]]-x; i.e.:

[(1 -[x.sup.[alpha]]).sup.1/[alpha]] - x - 1 + [(1 - [d.sup.[alpha].sub.2]).sup.1/[alpha]] = O. (1)

Let LHS(ES) denote the LHS of Eq. 1; i.e., LHS(ES) = [(1 - [x.sup.[alpha]]).sup.1/[alpha]] - x - 1+ [(1 - [d.sup.[alpha].sub.2]).sup.1/[alpha]].

The outcome of the Kalai/Smorodinsky solution solves (y - [d.sub.2])/x = (1 - [d.sub.2])/ [(1 - [d.sup.[alpha].sub.2]).sup.1/[alpha]]. That is, it solves [((1 - [x.sup.[alpha]).sup.1/[alpha]] - [d.sub.2])(1 - [d.sup.[alpha].sub.2]).sup.1/[alpha]]-x(1 - [d.sub.2]) = 0. That can be rewritten as:

[(1 - [x.sup.[alpha]]).sup.1/[alpha]] - x(1 - [d.sub.2])/[(1 - [d.sup.[alpha].sub.2]).sup.1/[alpha]] - [d.sub.2] = 0. (2)

Let LHS(KS) denote the LHS of Eq. 2; i.e., LHS(KS) = [(1 - [x.sup.[alpha]]).sup.l/[alpha]] - x(1 - [d.sub.2])/ [(1 - [d.sup.[alpha].sub.2]).sup.1/[alpha]] - [d.sub.2].

The outcome of the equal gains solution is y - [d.sub.2] =x; i.e.:

[(1 - [x.sup.[alpha]).sup.1/[alpha]] - x - [d.sub.2] = 0. (3)

Let LHS(E) denote the LHS of Eq. 3; i.e., LHS(E) = [(1 - [x.sup.[alpha]]).sup.1/[alpha]] -x - [d.sub.2].

The outcome of the Nash solution maximizes (y-[d.sub.2]) x; i.e., N maximizes [[(1 - [x.sup.[alpha]]).sup.1/[alpha]] - [d.sub.2]]x. First-order conditions yield [(1 - [x.sup.[alpha]]).sup.1/[alpha]](1 - [2x.sup.[alpha]])/(1 - [x.sup.[alpha]]) - [d.sub.2] = 0. Thus, the Nash solution solves:

[(1 - [x.sup.[alpha]]).sup.1/[alpha]](1 - [2x.sup.[alpha]])/(1 - [x.sup.[alpha]]) - [d.sub.2] = 0 (4)

The LHS(N) denote LHS of Eq. 4; i.e., LHS(N) = [(1 - [x.sup.a]).sup.1/a](1 - [2x.sup.a])/(1 - [x.sup.a]) - [d.sub.2]

If some x solved, say, both Eqs. 1 and 2, then LHS(ES)=LHS(KS) would hold. If, however, LHS(ES) is greater LHS(KS), then a higher x is needed to solve Eq. 1 than to solve Eq. 2. Similarly, if LHS(KS) is greater LHS(ES), then a higher x is needed to solve Eq. 2 than to solve Eq. 1. This is because all LHS(ES), LHS(KS), LHS(E), and LHS(N) are decreasing in x.

(i) Note that when [alpha] = 1, then Eqs. 1, 2, 3, and 4 all reduce to (1 - x) - x -[d.sub.2] = 0. Thus, all [N.sub.1] = [ES.sub.1] = [KS.sub.1] = [E.sub.1] = (1 - [d.sub.2])/2.

(ii) [ES.sub.1] > [KS.sub.1] > [E.sub.1]: We only need to show that LHS(ES)>LHS(E). Then Lemma 1 will imply that [ES.sub.1] > [KS.sub.1] > [E.sub.1] LHS(ES) = [(1 - [x.sup.a]).sup.1/a] - x - 1 + (1 - [d.sup.a.sub.2]).sup.1/a], and LHS(E) = [(1 - [x.sup.a]).sup.1/a] - x - [d.sub.2]. LHS(ES) > LHS(E) reduces to [(1 - [d.sup.a.sub.2]).sup.1/a] > (1 - [d.sub.2]), which can rewritten as (1 - [d.sup.a.sub.2]) > [(1 - [d.sub.2]).sup.1/a]. This holds since [d.sub.2] [member of] (0, 1) and [alpha] > 1.

[N.sub.1] > [E.sub.1]: LHS(N) = [(1 - [x.sup.a]).sup.1/a](1 - [2x.sup.a])/(1 - [x.sup.a]) - [d.sub.2], and LHS(E) = [(1 - [x.sup.a]).sup.1/a] x - [d.sub.2]. LHS(N) > LHS(E) reduces to [(1 - [x.sup.a]).sup.1/a]/(1 - [x.sup.a]) > [[(l - [x.sup.a]).sup.1/a] - x]/ (1 - [x.sup.a] - [x.sup.a]), which holds since x > [x.sup.a] (i.e., the numerator of LHS is x greater than that of the RHS and the denominator of LHS is [x.sub.a] greater than that of the RHS).

To show [KS.sub.1] > [N.sub.1] we need the following lemma:

Lemma 2 x < [(1 - [d.sup.a.sub.2]).sup.1/a] for N and KS. (5)

Proof of Lemma 2 First observe that x < (1 - [d.sup.a.sub.2]).sup.1/a] is equivalent to [d.sub.2] < (1 - [x.sup.a]).sup.1/a]. We will prove it by showing that when we subtract (1 - [x.sup.a]).sup.1/a] - [d.sub.2] from LHS(KS) and LHS(N), each will be negative.

Subtract [(1 - [x.sup.a]).sup.1/a] - [d.sub.2] from LHS(KS) to observe that LHS(KS)= - x(1 - [d.sub.2])/[(1 - [d.sup.a.sub.2]).sup.1/a] < 0. Subtract [(1 - [x.sup.a]).sup.1/a] - [d.sub.2]] from LHS(N) to observe that - [(1 - [x.sup.a]).sup.1/a] [x.sup.a]/(1 - [x.sup.a]) < 0. That proves Lemma 2.

[KS.sub.1] > [N.sub.1]: By Lemma A.1, x < [(1 - [d.sup.a.sub.2]).sup.1/a]. Next, note that LHS(KS) > LHS(N) reduces to [(1 - [d.sup.a.sub.2]).sup.1/a]/(1 - [d.sub.2]) > x/[x.sup.a] By using Lemma 2 (i.e., x < [(1 - [d.sup.a.sub.2]).sup.1/a]), we will obtain the desired condition [(1 - [d.sup.a.sub.2]).sup.1/a]/(1 - [d.sub.2]) > x/[x.sup.a]. Observe that, given x < [(1 - [d.sup.a.sub.2]).sup.1/[alpha]] as well as x [member of][0, 1), 0 - [d.sup.a.sub.2]) [member of] [0, 1) and [alpha] > 1, [x.sup.a] is bigger fraction of x than (1 - [d.sup.[alpha].sub.2]) is of [(1 - [d.sup.[alpha].sub.2]).sup.1/[alpha]]. Therefore, so far we have [(1 - [d.sup.[alpha].sub.2]).sup.1/[alpha]]/(1 - [d.sup.[alpha].sub.2]) > x/[x.sup.[alpha]]. But since (1 - [d.sub.2]) < (1 - [d.sup.[alpha].sub.2]), we have [(1 - [d.sup.[alpha].sub.2]).sup.1/[alpha]]/ (1 - [d.sub.2]) > x/[x.sup.2.sub.[alpha]].

Concluding Remarks

Given the class of CES bargaining problems our main result is very robust. It holds regardless of the values of d and [alpha]. In other words, whether improvement in one agent's fallback position is small or big does not matter. Similarly, whether the gains from cooperation are small or big does not matter.

It is easy to see that E relies mainly on d. E(S,d) goes through d with slope one. ES, on the other hand, goes through b with slope one. Since b relies on d indirectly, it is straightforward to see why E responds more to changes in d than ES does. Our Lemma 1 provides the basic intuition as to why KS should rely less on d than E and more on d than ES: KS must lie between E and ES.

The hardest part to see is why and how N responds more to changes in d than KS does. KS relies equally on d and b (where b in turn relies on d indirectly). N, on the other hand, relies on d and a curvature condition on the Pareto frontier; this curvature condition is that the Nash solution payoff ratio [x.sub.2]/[x.sub.1] is sandwiched between the absolute values of the right-hand and left-hand derivatives of the Pareto frontier at ([X.sub.1], [X.sub.2]). Hence N's greater response to changes in d implies that the curvature condition of N relies on d more significantly than b relies on d.

Acknowledgment I would like to thank an anonymous referee for many useful comments and suggestions.

Published online: 24 July 2008

References

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Anbarci, N., Skaperdas, S., & Syropoulos, C. (2002). Comparing bargaining solutions in the shadow of conflict: How norms against threats can have real effects. Journal of Economic Theory, 106, 1-16.

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N. Anbarci ([mail]) School of Accounting, Economics and Finance, Deakin University, 221 Burwood Highway, Burwood, Victoria 3125, Australia e-mail: nejat.anbarci@deakin,edu.au

(1) See, for instance, Anbarci et al. 2002, which considers a setup where the Pareto frontier depends on the changes in the status-quo payoffs.

(2) Vector inequalities: Given x, x' [member of] [R.sup.2.sub.+], x[??]x' means [x.sub.i] [greater than or equal to] [x'.sub.i] for all i= 1,2; x[greater than or equal to]x' means x[greater than or equal to]x' and x[not equal to]x'; x>x' means [x.sub.i] > [x'.sub.i] for all i= 1,2. See Roth 1979a, for an excellent introduction to bargaining theory.

(3) Given the definitions of N(S,d), E(S,d) and KS(S,d) it is straightforward to see that they are responsive to changes in status-quo payoffs. To see that ES is responsive to changes in status-quo payoffs, first observe that given any S an increase in [d.sub.i] will leave [b.sub.i] unchanged but will either decrease [b.sub.j] or leave it unchanged (an increase in [d.sub.i] will decrease [b.sub.j] if the slope of the Pareto frontier at (O,[b.sub.j]) is different than zero or infinity; it will leave [b.sub.j] unchanged otherwise). ES is responsive to b. Therefore an increase in [d.sub.i] will either increase [ES.sub.i] or leave [ES.sub.i] unchanged.

(4) I thank the referee for pointing this out.

(5) This result also holds for E and ES too but we do not need those results here.
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