(b) For рн = 0.5, cl∕2 — cl Inрн = 1 — 2 In 0.5 < 3 = c∕2. With (15),
Iim
T →c∣2
V-1
Z
—2 — (0.5)2 (3+10
2 + 21n0.5
)+α5(2+3)
—0.153.
Hence, if in case (N, I) the pure strategy equilibrium is selected, T = c∕2 = 3. For
the mixed strategy equilibrium, with (17), the value of information for T > c∕2 is
(0.5)2 (τ + 120 — 2 + 21n0.5^
+ 16e- 1+ï
2
which is equal to zero for rT = 3.56.
(c) For рн = 0.25, cl∕2 — cl 1nрн = 1 — 21n0.25 > 2 = c∕2. At T = c∕2,
0.25 , x 1 2 /4 ∖
_ (10 — 2) e2- 2 +0.25 Ç2 + 2J
1.606.
Thus VZ1 is negative for all T < c∕2. This implies that T = c∕2 = 2 for the pure
strategy equilibrium in case (N, I). Now consider the mixed strategy equilibrium in
case (N, I) and suppose that T approaches ch∕2 = 5. Since 1 — 21n0.25 < 5, Vzi is
given by (17) and approaches
(0.25)2 (δ+y
1n 0.25^
1 . 1.5 _____
+ -4e-2+ 4 ≈ —0.2097.
By monotonicity, Vzi is negative for all T ∈ (1, 5).
References
[1] Amann, E., Leininger, W., 1996. Asymmetric all-pay auctions with incomplete
information: the two-player case. Games and Economic Behavior 14, 1-18.
[2] Bilodeau, M., Slivinski, A., 1996. Toilet cleaning and department chairing: vol-
unteering a public service. Journal of Public Economics 59, 299-308.
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