A.7 Proof of Corollary 1
Part (i) follows from Lemma 5(iii). For part (ii), the threshold T is larger than c/2
if and only if the value of information is negative as T approaches c/2. Moreover,
Vi1 may even be negative for all T. Note that c/2 — ClnpH > ol/2 — ol InpH, and
for small pH, we have ol/2 — ol lnpH > oh/2. With monotonicity of (16), it follows
that, for all T ∈ (ol/2,oh/2), V)j is smaller than
_ Pn 1_ cH 1 си
lim V/ (T ) = — cL — —(oh — cl) e2 2cr + рн ce 2 + 2c
T →ch/2 2
which is negative for small pH since the second and the third term approach zero if
pH → 0. For intermediate values of pH, we have T ∈ (c/2, oh/2).
A.8 Proof of Proposition 2
Suppose that i remains uninformed and j acquires information. If T < c/2, i’s
expected payoff is strictly higher than j,s expected payoff (compare the first row in
(6) to the first row in (7)).
Thus, if T < min {c/2,t},
in the equilibrium of the
game of sequential information acquisition, the first mover remains uninformed.
If T > c/2 and in case (N, I) the mixed strategy equilibrium is selected, by
continuity of the expected payoffs, E (π⅛) > E (πj) also holds for T = c/2 + δ, δ > 0
sufficiently small (compare the second rows in (6) and (7)). As T → c/2 — clnpH,
the payoff of the uninformed player i approaches υ — C, while the informed player j
gets strictly more than v — C.26 The difference in payoffs E (πj∙) — E (π⅛) is strictly
increasing in T, and there is a critical value Ts where E (πj∙) = E (πi). Depending
on the parameters of the model, Ts can be smaller or larger than the threshold T
for information acquisition. If Ts > T, in all equilibria where one player chooses
to remain uninformed, this will be the first mover. If Ts < T, there can also be
equilibria of the game with sequential information acquisition where the first mover
chooses to acquire information and the second mover remains uninformed.27
26By choosing qjL (0) = 1 and qjH (T) = 1, j can ensure a payoff of at least υ — Plc^ —
Ph (ch/2 + T), which is strictly larger than υ — c; thus, his equilibrium payoff cannot be smaller.
27This is the case, for instance, if ¾ = 2, ch = 10, and рн = 0.3. There, in case (N, I),
34
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