The name is absent

3 Expenditure, allocative efficiency and the
cost of delay

The tug-of-war with m> 2 resolves the allocation problem along a sequence
of states, where a violent battle may, but need not take place at each state.
Only in the tipping states is positive effort expended with positive probability.
Once the process leaves the tipping states, the war moves to a terminal state,
without further effort being expended. A tug-of-war that starts in a tipping
state will therefore be called "violent". A tug-of-war that starts outside a
tipping state will be called "peaceful".

Compared to the standard all-pay auction, the tug-of-war could be inter-
preted as an institution that saves cost of effort in the problem of allocating
a prize between rivals who are prepared to expend resources in fighting for
the prize. Suppose for instance that Z_a and Z_b are independent random
draws from a continuous distribution with support [0,κ], where к < K. Sup-
pose that these values are known to the contestants but are not observable
by the designer of the institution at the time that it must be implemented.
Consider the following tug-of-war as an anonymous mechanism in the case in
which m is an even number, so that m is integer valued. Start the tug-of-war
in the symmetric state m and assume, as we have throughout, that player
A attempts to move the state to j = 0 and player B attempts to move the
state to j = m. Then the following result derives the probability of peaceful
resolution:

Proposition 5 Let Γ(g) be the continuous cumulative distribution function
of g ≡ Z_a∕Z_b , with support [0, ∞]. The allocation is peaceful in the Markov
perfect equilibrium with a probability Γ(δ²) + (1 — Γ(^2)), and lims→₁(Γ(δ²) +
(1 — Γ( ⅛ ))) = 1.

Proof. For a proof we show that Г(5²) + (1 —Γ( j₂)) is the probability that
the symmetric state m/2 is not a tipping state. Suppose that g > ɪ. Then
Z_a > Z_b and δ~⁺¹Z_a > δ^m '~⁺¹⁾Z_b. Hence, j₀ — 1 > m. By Propositions
1-3 this implies that the tug-of-war that starts in m/2 consecutively moves
to j = 0 with no effort being expended. Let g < δ². Then Z_a < Z_b.
Applying the results in Propositions 1-3 with A and B and j = 0 and j = m
switching roles shows that the tug-of-war that starts in m/2 moves to j = m
with no effort being expended. Suppose now that g ∈ (1, j2 ). In this case

20

<<   18   19   20   21   22   23   24   >>

More intriguing information