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Representation in Multi-Issue Delegated Bargaining

  • Shourjo Chakravorty EMAIL logo
Veröffentlicht/Copyright: 18. April 2018

Abstract

Delegated bargaining over multiple issues of varying sizes is a frequent occurrence and happens during such contexts as real estate sales and political representation. Past work has shown that while simultaneously bargaining over two issues, it is in the best interest of an individual to choose a representative with identical preferences as herself. By allowing the sizes of the two issues to vary, this paper uses a two-issue, two-player Rubinstein bargaining game to show that an individual is best served by selecting a representative whose preferences closest resemble her bargaining rival’s preferences. It is also found that in some instances depending on the size of the issues the individual will be indifferent when choosing between a representative with identical preferences and another with different preferences to bargain on her behalf.

Acknowledgements

I am grateful to Steven Slutsky, David Sappington, Caprice Knapp and Sanford Berg for the guidance and support they provided while writing this paper. I also thank Dave Brown, Michelle Phillips and seminar participants at the Department of Economics at Istanbul Technical University for helpful comments.

Appendix

A Proof of Lemma 1

There are four candidate solutions for Equilibrium Condition eq. (4):

  1. xa<X, ya=0 and xb<X, yb=0,

  2. xa<X, ya=0 and xb=X,

  3. xa=X and xb<X, yb=0,

  4. xa=X and xb=X.

To prove Lemma 1, I examine which of the four candidate solutions (i), (ii), (iii) and (iv) solve Equilibrium Condition eq. (4).

Candidate solution (i): xa<X, ya=0andxb<X, yb=0

Candidate solution (i) implies that the Equilibrium Condition eq. (4) becomes:

(15)(Xxa)+Y=δ(Xxb+Y),
(16)xb=δxa.

Solving eqs (15) and (16) for xa and xb:

(17)xa=X+Y1+δ,
(18)xb=δ(X+Y)1+δ.

For Solutions eqs (17) and (18) to be a valid, it must be that 0xa<X and 0xb<X. Because δ(0,1) and X and Y are nonnegative numbers, it must be that xa0 and xb0. For xa<X, it must be that Y<δX, and for xb<X, it must be that Y<Xδ. Because δX<Xδ, Candidate solution (i) is a valid solution in the range Y<δX.

Candidate solution (ii): xa<X, ya=0 and xb=X

Candidate solution (ii) implies that the Equilibrium Condition eq. (4) becomes:

(19)(Xxa)+Y=δ(Yyb),
(20)X+αSyb=δxa.

Solving eqs (19) and (20) for xa and yb:

(21)xa=(δαS)XαS(1δ)Yδ2αS,
(22)yb=(1δ)Xδ(1δ)Yδ2αS.

For Solutions eqs (21) and (22) to be valid, it must be that 0xa<X and 0ybY. To check this validity, we examine the two possible subcases: (a) δ2αS0 and (b) δ2αS<0.

Subcase (a) δ2αS0. For xa<X, it must be Y>δXαS and for yb0, it must be YXδ. Therefore, for Candidate solution (ii) to be a valid solution in Subcase (a), it must be δXαS<Xδ. But that would imply δ2αS<0. But in Subcase (a), δ2αS0. Consequently, Candidate solution (ii) cannot exist in Subcase (a).

Subcase (b) δ2αS<0. For xa<X, it must be Y<δXαS and for yb0, it must be YXδ. Therefore, for Candidate solution (ii) to be a valid solution in Subcase (b), it must be δXαS>Xδ. But that would imply δ2αS>0. But in Subcase (b), δ2αS<0. Consequently, Candidate solution (ii) cannot exist in Subcase (b).

This means Candidate solution (ii) cannot be a valid solution for Equilibrium Condition eq. (4).

Candidate solution (iii): xa=X and xb<X, yb=0

Candidate solution (iii) implies that the Equilibrium Condition eq. (4) becomes:

(23)Yya=δ(Xxb+Y),
(24)xb=δ(X+αSya).

Solving eqs (23) and (24) for ya and xb:

(25)ya=(1δ)(YδX)1δ2αS,
(26)xb=δ[(1δαS)X+αS(1δ)Y]1δ2αS.

For Solutions eqs (25) and (26) to be a valid, it must be that 0yaY and 0xb<X. δ(0,1) and 0αS<1 imply that the denominator of Solutions eqs (25) and (26), 1δ2αS>0. By inspection, we also see that the numerator of Solution eq. (26) is nonnegative. This implies xb0. Looking at the numerator of Solution eq. (25), for ya0 it must be that YδX. For yaY, it must be that (1δ)(YδX)1δ2αSY. This inequality simplifies to (1δαS)Y+(1δ)X0. By inspection, we can see that this latter inequality always holds. Therefore, yaY. For xb<X, it must be that δ[(1δαS)X+αS(1δ)Y]1δ2αS<X. This inequality simplifies to Y<XδαS. Hence, Candidate solution (iii) is a valid solution in the range δXY<XδαS.

Candidate solution (iv): xa=X and xb=X

Candidate solution (iv) implies that the Equilibrium Condition eq. (4) becomes:

(27)Yya=δ(Yyb),
(28)X+αSyb=δ(X+αSya).

Solving eqs (27) and (28) for ya and yb:

(29)ya=αSYδXαS(1+δ),
(30)yb=δαSYXαS(1+δ).

For Solutions eqs (29) and (30) to be a valid, it must be that 0yaY and 0ybY. yaY and ybY imply δ(αSY+X)0 and αSY+X0 respectively. Because δ(0,1) and 0αS<0, we know that these latter two inequalities always hold by inspection. Therefore, yaY and ybY always hold. For ya0 and yb0, it must be YδXαS and YXδαS respectively. Because XδαSδXαS, Candidate solution (iv) is a valid solution in the range YXδαS.

This completes the proof of Lemma 1.

B Proof of Lemma 2

The proof of Lemma 2 proceeds just like the proof of Lemma 1. There are four candidate solutions for Equilibrium Condition eq. (4):

  1. xa=0, ya<Y and xb=0, yb<Y,

  2. xa=0, ya<Y and yb=Y,

  3. ya=Y and xb=0, yb<Y,

  4. ya=Y and yb=Y.

To prove Lemma 2, I examine which of the four candidate solutions (i), (ii), (iii) and (iv) solve Equilibrium Condition eq. (4).

Candidate solution (i): xa=0, ya<Y and xb=0, yb<Y

Candidate solution (i) implies that the Equilibrium Condition eq. (4) becomes:

(31)X+(Yya)=δ(X+Yyb),
(32)αLyb=δαLya.

Solving eqs (31) and (32) for ya and yb:

(33)ya=X+Y1+δ,
(34)yb=δ(X+Y)1+δ.

For Solutions eqs (33) and (34) to be a valid, it must be that 0ya<Y and 0yb<Y. Because δ(0,1) and X and Y are nonnegative numbers, it must be that ya0 and yb0. For ya<Y, it must be that Y>Xδ, and for yb<Y, it must be that Y>δX. Because δXXδ, Candidate solution (i) is a valid solution in the range Y>Xδ.

Candidate solution (ii): xa=0, ya<Y and yb=Y

Candidate solution (ii) implies that the Equilibrium Condition eq. (4) becomes:

(35)X+(Yya)=δ(Xxb),
(36)xb+αLY=δαLya.

Solving eqs (35) and (36) for ya and xb:

(37)ya=(1δ)X+(1δαL)Y1δ2αL,
(38)xb=αL(1δ)(δXY)1δ2αL.

For Solutions eqs (37) and (38) to be valid, it must be that 0ya<Y and 0xbX. To check this validity, we examine the two possible subcases: (a) 1δ2αL0 and (b) 1δ2αL<0.

Subcase (a) 1δ2αL0. For ya<Y, it must be Y>XδαL and for xb0, it must be YδX. Therefore, for Candidate solution (ii) to be a valid solution in Subcase (a), it must be δX>XδαL. But that would imply 1δ2αL<0. But in Subcase (a), 1δ2αL0. Consequently, Candidate solution (ii) cannot exist in Subcase (a).

Subcase (b) 1δ2αL<0. For ya<Y, it must be Y<XδαL and for xb0, it must be YδX. Therefore, for Candidate solution (ii) to be a valid solution in Subcase (b), it must be δX<XδαL. But that would imply 1δ2αL>0. But in Subcase (b), 1δ2αL<0. Consequently, Candidate solution (ii) cannot exist in Subcase (b).

This means Candidate solution (ii) cannot be a valid solution for Equilibrium Condition eq. (4).

Candidate solution (iii): ya=Y and xb=0, yb<Y

Candidate solution (iii) implies that the Equilibrium Condition eq. (4) becomes:

(39)Xxa=δ(X+Yyb),
(40)αLyb=δ(xa+αLY).

Solving eqs (39) and (40) for xa and yb:

(41)xa=αL(1δ)(XδY)αLδ2,
(42)yb=δ[(1δ)X+(αLδ)Y]αLδ2.

For Solutions eqs (41) and (42) to be a valid, it must be that 0xaX and 0yb<Y. δ(0,1) and αL>1 imply that the denominator of Solutions eqs (41) and (42), αLδ2>0. By inspection, we also see that the numerator of Solution eq. (42) is nonnegative. This implies yb0. Looking at the numerator of Solution eq. (41), for xa0 it must be that YXδ. xaX implies αL(1δ)(XδY)αLδ2X. This inequality simplifies to (αLδ)XαL(1δ)Y which always holds. Therefore, xaX. yb<Y implies δ[(1δ)X+(αLδ)Y]αLδ2<Y. This latter inequality simplifies to Y>δXαL. Hence, Candidate solution (iii) is a valid solution in the range δXαL<YXδ.

Candidate solution (iv): ya=Y and yb=Y

Candidate solution (iv) implies that the Equilibrium Condition eq. (4) becomes:

(43)Xxa=δ(Xxb),
(44)xb+αLY=δ(xa+αLY).

Solving eqs (43) and (44) for xa and xb:

(45)xa=XδαLY1+δ,
(46)xb=δXαLY1+δ.

For Solutions eqs (45) and (46) to be a valid, it must be that 0xaX and 0xbX. Both xaX and xbX imply αLY+X0. Because αL>1 and X and Y are nonnegative numbers, we know that this latter inequality always holds by inspection. Therefore, xaX and xbX always hold. For xa0 and xb0, it must be YXδαL and YδXαL respectively. Because δXαLXδαL, Candidate solution (iv) is a valid solution in the range YδXαL.

This completes the proof of Lemma 2.

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Published Online: 2018-04-18

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