If T > c/2 and, in case (N, I), the pure strategy equilibrium is selected, learning
the own provision cost is strictly dominant, and thus the threshold T is (weakly)
smaller than c/2.15 However, if we focus on the mixed strategy equilibrium, T > c/2
for a small pH, and the value of information Vi1 can even be negative for all T ∈
(⅛/2,ch/2). Thus, the strategic value of remaining uninformed is not only present
in the case where an uninformed individual i has a dominant strategy not to concede
before T (as in Lemma 3a), but also when the individuals randomize their concession
time (as in Lemma 3b(ii)). The sufficiently high probability that the rival has a low
cost and volunteers immediately with positive probability makes it optimal for i to
disregard information that is available without cost. This strategic value disappears
only if the probability of having a high contribution cost, pH, is large, because, from
the point of view of the rival, an early concession of the individual who knows his
provision cost is less likely.
Example Consider the following example where ⅛ = 2, and ch = IO.16 As-
sumption 1 requires that 1 < T < 5.
(a) Suppose that pH = 0.75. If T → c/2 = 4 from below, the value of information
Vi1 is positive. Hence, the critical threshold T < c/2. Setting Vi1 (T) = 0 yields
T = 1.94. Thus, for all T < 1.94, only one individual learns his cost of provision,
and for all T > 1.94, both individuals learn their cost of provision.
(b) Now suppose that pH = 0.5. Vi1 is negative if T approaches c/2 = 3 (from below).
Hence, if in case (N, I) the pure strategy equilibrium is selected, T = c/2 = 3, and
if the mixed strategy equilibrium is selected, T > c/2. In the latter case, T = 3.56.
(c) If pH = 0.25, again Vi1 is negative if T approaches c/2 = 2, and T = c/2
if in case (N, I) the pure strategy equilibrium is selected. If the mixed strategy
equilibrium is selected, Vil is negative for all T satisfying Assumption 1, and thus
there is no equilibrium where both individuals find out about their cost of provision
with probability one.
15Concretely, if V°^3~1 is negative for all T < c/2, T? = c/2. Otherwise, T is defined as the
solution to Viσ^3~1 (t^ = 0.
16Details on this example are in Appendix B.
18