Game Theory Basics - Department of Mathematics

132 CHAPTER 5. BARGAININGuv = [(1 − ε)U + εu][(1 − ε)V + εv]= [U + ε(u −U)][V + ε(v −V )]= UV + ε[u + v − (U +V ) + ε(u −U)(v −V )].Now, because u + v > 2, but U +V = 2, the term u + v − (U +V ) is positive, and ε(u −U)(v −V ) can be made smaller than that term in absolute value, for suitably small ε > 0,even if (u−U)(v−V ) is any negative number. Consequently, uv > 1 =UV for sufficientlysmall positive ε, which shows that UV was not the maximum Nash product in S, whichcontradicts the construction of (U,V ). This proves that, indeed, S ⊆ T , and hence N(S) =N(T ) as claimed.In order to show that the point (U,V ) = (1,1) is unique, we can again look at thehyperbola, and argue similarly as before. Namely, suppose that (u,v) is a point in Sdifferent from (1,1) so that u · v = 1. Then this point and (1,1) are two points on thehyperbola, and points on the line segment joining these two points belong to S because Sis convex. However, they have a larger Nash product. For example, the mid-point 1/2 ·(1,1)+1/2·(u,v) has the Nash product (1/2+u/2)(1/2+v/2) = 1/4+u/4+v/4+u·v.Because u · v = 1, this is equal to 1/2 + (u + v)/4, which is larger than 1 if and only if(u+v)/4 > 1/2 or u+v > 2. The latter condition is equivalent to u+1/u > 2 or, becauseu > 0, to u 2 +1 > 2u or u 2 −2u+1 > 0, that is, (u−1) 2 > 0, which is true because u ≠ 1.So, again because S is convex, the Nash product has a unique maximum in S.The assumption that S has a point (u,v) with u > u 0 and v > v 0 has the followingpurpose: Otherwise, for (u,v) ∈ S, all utilities u of player I or v of player II (or both) fulfilu = u 0 or v = v 0 , respectively. Then the Nash product (U − u 0 )(V − v 0 ) for (U,V ) ∈ Swould always be zero, so it has no unique maximum. In this rather trivial case, thebargaining set S is either just the singleton {(u 0 ,v 0 )} or a line segment {u 0 } × [v 0 ,V ] or[u 0 ,U] × {v 0 }. In both cases, the Pareto-frontier consists of a single point (U,V ), whichdefines the unique Nash solution N(S).⇒ If the bargaining set is a line segment as discussed in the preceding paragraph, whyis its Pareto-frontier a single point? Is this always so if the line segment is notparallel to one of the coordinate axes?It can be shown relatively easily that the maximisation of the Nash product as describedin theorem 5.3 gives a solution (U,V ) that fulfils the axioms in definitions 5.1and 5.2.⇒ In exercise 5.2, you are asked to verify this for axiom (f).We describe how to find the bargaining solution (U,V ) for the bargaining set S infigure 5.1. The threat point (u 0 ,v 0 ) is (1,1). The Pareto-frontier consists of two linesegments, a first line segment joining (1,3.5) to (2,3), and a second line segment joining(2,3) to (4,1). The Nash product (u − u 0 )(v − v 0 ) for a point (u,v) on the Pareto-frontieris the size of the area of the rectangle between the threat point (u 0 ,v 0 ) and the point (u,v).In this case, the diagram shows that the second line segment creates the larger rectanglein that way. A point on the second line segment is of the form (1 − p)(2,3) + p(4,1) for0 ≤ p ≤ 1, that is, (2 + 2p,3 − 2p). Because (u 0 ,v 0 ) = (1,1), the resulting Nash product