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Another interesting situation is one which only player 1 is able to learn. Player 2 has no access to the exogenous learning process. Then, at any time t, player 1 has three possible types: a sure type who is sure that v = vL, a learning type who is still unsure and a

commitment type. I use γ1t to denote the posterior belief that player 1 is a learning type,

β1t to denote the posterior belief that player 1 is a rational type, pt to denote player 1’s

posterior belief that v = vL given he is a sure type at time t and finally γ2t to denote the

belief that player 2 is a normal type.

The next result shows that conditional on at least one of the two players being normal, the one-sided learning model has the same longest delay as the model without learning.

Theorem 3.5. In the one-sided learning model, conditional on at least one of the two players being normal, the longest delay is always T =−(vH−p0vL) log(1−γ0)

c .

5_{The change in posterior beliefs also plays an important role here. If the players update beliefs naively,} then the lower bound on longest delay with learning is

˜

T =−(vH−vL) log(1−x)

c −

(vH−γ_γ0₀p0₋−_xxvL) log(1−γ0+x)

c ,

Proof. To prove the theorem, I need to consider two separate cases λγ0p0 ≤ _v c

H−p0vL and

λγ0p0 > _v c

H−p0vL. The unique sequential equilibrium when λγ0p0 ≤ c

vH−p0vL is characterized

by the following proposition.

Proposition 3.2. If λγ0p0 ≤ c

vH−p0vL, there exists a unique sequential equilibrium in the one-sided learning model satisfying:

(1) for player 1, the learning type concedes with probability zero at time 0 and at a positive rate between time 0 and T; the sure type player concedes with probability one upon receiving the first Poisson signal;

(2) the normal type player 2 concedes with strictly positive probability at time 0 and at a positive rate between time 0 and T;

(3) T =−(vH−p0vL) log(1−γ0)

c and after time T, only the commitment type player stays; (4) at time t ∈ (0, T] , player 1’s expected concession rate is _v c

H−p0vL and player 2’s expected concession rate is

c(1−p0+p0e−λt) vH(1−p0+p0e−λt)−p0e−λtvL

.

Proof. The proof of the equilibrium properties is very similar to the proof in the two-sided learning model and is omitted. The assumption that λγ0p0 ≤ _v c

H−p0vL guarantees that the

learning type player 1 must randomize and the sure type player 1 must concede immediately in equilibrium. Therefore, denote x1t (x2t) to be the equilibrium concession rate of the

learning type player 1 (normal type player 2). The indifference conditions imply:

(γ2tx2t+λpt)ptvL=−c+γ2tx2tvH +λptvL−λpt(1−pt)vL.

and

(γ1tx1t+γ1tλpt)p0vL=−c+γ1tx1tvH +γ1tλptvH.

˙ γ1t=−(λpt+x1t)γ1t(1−γ1t) and γ˙2t=−x2tγ2t(1−γ2t). Therefore, we have: ˙ γ1t=− c vH −p0vL (1−γ1t) and γ˙2t=− c vH −ptvL (1−γ2t). Since pt = p0e −λt

p0e−λt+1−p0 < p0, the expected concession rate of player 2 is smaller than the expected concession rate of player 1. Also the learning type of player 1 and the normal type of player 2 have to stop conceding at the same timeT. As a result, the normal type of player 2 has to concede with a strictly positive probability at time 0.

T is determined by the shortest time of concessionT =−(vH−p0vL) log(1−γ0)

c and γ2t satis- fies: 1−γ₀0 1−γ2t =e−vHct vH −p0vL (vH −vL)p0e−λt+ (1−p0)vH −_λvH(cvL_vH−_vL) .

At time 0, the probability of concession by the normal type of player 2 is chosen such that:

γ₀0 = 1−e−vHcT vH −p0vL (vH −vL)p0e−λT + (1−p0)vH −_λvH(cvL_vH−_vL) .

If λγ0p0 ≤ _vH−cp0vL, then initially the sure type of player 1 will randomize such that the

expected concession rate is always c

vH−p0vL. Then at time 0, it is impossible for player 1 to

concede with a positive probability. Next, I will show that there cannot exist two disjoint time intervals (t0, t1) and (t2, t3) such that the sure type is indifferent on both intervals and strictly prefers conceding for t ∈ (t1, t2). If there exists such an equilibrium, at t2, it must be the case that: λγt2pt2 ≥

c

vH−p0vL. Since γt1pt1 > γt2pt2, λγt1pt1 > c

vH−p0vL. Therefore, if

a sure type player concedes with probability one at t1, his normal opponent must stay for sure, which leads to a contradiction. Therefore, on the equilibrium path, the sure type will first randomize until T1 and the learning type randomizes afterwards.

At t < T1, denote γt to be the belief that player 1 is a learning type at time t and βt

to be the belief that a player is a sure type at time t. Suppose the existing sure type has a concession rate of xt and a new sure type will concede with probability yt. The indifference

of player 2 means that:

(βtxt+γtλptyt)p0vL =−c+ (βtxt+γtλptyt)vH.

The laws of motion for βt and γt are such that:

˙

βt=−xtβt(1−βt)−γtλptyt(1−βt) +γtλpt

and

˙

γt=−γtλpt+γt(βtxt+γtλptyt).

The above equations imply that: ˙ βt+ ˙γt=− c vH −p0vL (1−βt−γt) and hence βt+γt= 1−(1−γ0)e c vH−p0vL_.

Notice fort ≥T1,βt= 0 and the expected concession rate for the learning type of player

is also _v c

H−p0vL. Therefore, beginning from β0+γ0 =γ0,βt+γt satisfies:

βt+γt = 1−(1−γ0)e

c vH−p_0vL

for all t ≥0. There is also no discontinuity in βt+γt for any t >0. As a result, it must be

the case thatT =−(vH−p0vL) log(1−γ0)

c .

The above result implies that allowing only one player to learn is better than allowing both players to learn in terms of delay regardless of what the initial parameters are. Compared to a model without learning, one-sided learning does not increase the longest time of waiting if either player 1 or player 2 is normal. In particular, the expected equilibrium concession rate of player 1 is exactly the same as the case without learning.