The name is absent

these points leads to different continuation games and allocates a considerable
rent between A and B. This makes competition particularly strong at these
states. We call these states "tipping states" because success of an advantaged
player at each of these two states "tips" the game so that victory is obtained
without further effort. A loss by the advantaged player throws the system
back into a competitive state where the player becomes disadvantaged.

Proposition 1 also shows that the allocation of a prize in a tug-of-war
leads to a seemingly peaceful outcome whenever the conflict starts in a state
other than a tipping state. This will be important for drawing conclusions
in section 3 about the efficiency properties of a tug-of-war as an allocation
mechanism.

Proposition 1 does not consider all possible parameter cases. Before turn-
ing to the remaining cases, note that the case j₀ = 1 cannot emerge, as this
requires SZ_a < S^m^¹Z_s, and this contradicts Z_a ≥ Z_b for m > 2. However,
player A’s dominance could be sufficiently large that no interior j₀ exists that
has the properties defined in Proposition 1. This leads to

Proposition 2 Suppose that S^{mm 1} Z_a > SZ_b. Then a unique Markov perfect
equilibrium exists with v_b(j) = 0 and v_A(j) = S^jZ_a for all j ∈ {1,...,m — 2},
and v_A(m — 1) = S^tu-¹Z_a — SZ_b and v_b(m — 1) = 0 at j = m — 1.

Proof. We show that the following effort choices constitute an equilib-
rium and yield the payoffs described in the proposition. Uniqueness follows
the argument in the Appendix.

Effort is a(j) = b(j) = 0 for all j ∈ M^mt∖{m — 1} and for j = m — 1 efforts
are chosen according to the following cumulative distribution functions:

Fm→(α) = ½    ^sf ^a ∈ fff^A           (20)

[1 for a > SZ_b

G ι(b) = ½ ^{(1 — sm}^^zA ) ^{for b} ∈ ^[0’^SZb]            (21)

^b ɪ 1 for   b>SZs.            ^{( )}

Note that this behavior yields the payoffs that are characterized in Proposi-
tion 2. For states j = 1, 2,..., (m —2), A wins after j further battles, and none
of the players expends effort. This confirms v_A(j) = S^jZ_a and v_b(j) = 0 for
all j = 1, ...m — 2. For j = m, the payoffs are v_A(m) = 0 and v_s(m) = Z_b.
Finally, for j = m — 1, given the mixed strategies described by (20) and (21),
the payoffs are v_A(m — 1) = S^tu-¹Z_a > SZ_b and v_b(m — 1) = 0.

16

<<   14   15   16   17   18   19   20   >>

More intriguing information