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Chapter 15
Exercise 15.10.1
Bob’s strategies are functions from the card he observes into actions he takes, so
that he has four possible strategies: CC,CF, FC, FF (C stands for “Call” and F for
“Fold”, so that for example the strategy CF means call if he observes H and to fold
if he observes L, and the strategy FF instructs him to fold in either case). Alice’s
strategies are functions from observed Bob’s actions into her actions, so she has four
strategies cc, cf, fc, ff (e.g., strategy fc means to fold if call and call if fold).
To illustrate how to compute the payoffs in the strategic form, take for instance
the profile (FC, cf). With prob = 12
Bob will observe either of H and L. In case he
observes H he will fold, and Alice will then fold as well, resulting in the payoff of
0 to each. If he observes L he will call and Alice will call, so that Alice gets 1 and
Bob gets −1. So the payoffs for (FC, cf) are 12(0, 0) + 1
2(−1, 1) = (−1
2, 1
2). The other
payoffs are obtained similarly. The resulting strategic form of the game is
cc cf fc ff
CC0 0 0 0
0 0 0 0
CF−1
2−1
20 0
12
12
0 0
FC12
12
0 0
−12
−12
0 0
FF0 0 0 0
0 0 0 0
Nash equilibria are constituted by any strategy profile where Bob plays any of
CC,CF, FF , and Alice plays either fc or ff . Strategy CF weakly dominates all
the other strategies for Bob.
1
Chapter 15
If Bob observes the card only with probability p the situation doesn’t change
much. Then Bob has 8 strategies (i.e., strategy tells him what he does if he observes
H, observes L, or doesn’t observe anything), and as before we would compute the
strategic form. But when Bob calls if he doesn’t observe the card, his expected payoff
is 0 regardless of what Alice does, so that this case will never contribute anything
to Bob and Alice’s payoffs. Similarly, when Bob folds if he doesn’t observe, their
expected payoffs are 0. Therefore, the strategic form would now be just two payoff
matrices from the previous case, both multiplied by p, and stacked on top of each
other and the same conclusions hold as before.
2
Chapter 15
Exercise 15.10.2
As in exercise 15.10.1 we can compute the strategic form which is now
cc cf fc ff
CC0 0 −1 −1
0 0 1 1
CF−1
2−1
20 0
12
12
0 0
FC32
32
0 0
−32
−32
0 0
FF1 1 1 1
−1 −1 −1 −1
Nash equilibrium is given by Bob’s mixed strategy (13, 2
3, 0, 0), and Alice’s (β, 2
3−
β, α, 13− α), where α ∈ [0, 1
3], β ∈ [0, 2
3]. Clearly, Alice will sometimes bet when Bob
is sure to win.
3
Chapter 15
Exercise 15.10.3
Bob’s analysis is incorrect because he counts the money in the pot as being his
whereas it actually belongs to whomever will win the pot. That is, his contribution to
the pot is sunk. Bob incurs an opportunity cost because he foregoes the opportunity
of being a winner.
4
Chapter 15
Exercise 15.10.4
The strategic form is (B stands for Bet, and F stands for Fold, so that e.g., BF
stands for Bet if L and Fold if H),
bb bf fb ff
BB0 −3a− b b− a −a
0 3a+ b a− b a
BF3a+ b 0 a+ b −2a
−3a− b 0 −a− b 2a
FBa− b −a− b 0 −2a
b− a a+ b 0 2a
FFa 2a 2a 0
−a −2a −2a 0
After successive deletion of strongly dominated strategies we are left with BB,FB
and bb, fb, and the claim follows.
5
Chapter 15
Exercise 15.10.5
In the casting stage the Nature chooses which of the roles a player will get to play,
in the Quiche, there are 2 possible roles (Tough or Wimpy) for Bob and a single
role (Nasty) for Alice. Once the casting stage is over, we then fill in the appropriate
payoffs into the outcome nodes of the two possible games that the players play,
depending on what has happened in the casting stage. These two games are depicted
in figure 15.9. Together (adding the casting move by Nature, and appropriately
drawing Alice’s information sets) these two games constitute the game of imperfect
information that Alice and Bob end up playing.
6
Chapter 15
Exercise 15.10.6
The strategic form is (e.g., dh stands for dove if N and hawk if Y )
dd dh hd hh
DD5 13
292
6
5 52
52
0
DH52
174
114
92
132
174
174
2
HD52
174
114
92
92
114
114
1
HH0 2 1 3
6 92
92
3
Now hh strongly dominates hd and dd, and after eliminating these for both players,
hh strongly dominates dh. Hence the only rationalizable strategy profile is (HH, hh),
contrary to Rousseau.
7
Chapter 15
Exercise 15.10.7
We refer to figure 15.6(a) for the picture. When Player II gets to move, he compares
his expected payoff from firing with what he gets if he chickens out, which is b.
Thus, he will fire if and only if b ≤ 12. Given this strategy of II, Player I compares his
payoff of chickening out with his expected payoff if he fires. Denote by Pr(b ≤ 12|a)
the conditional probability that Player I assigns to the event that II will fire, given
that his own type is a. Now, the expected payoff of I if his type is a and he fires
is 23
(Pr(b ≤ 1
2|a)(1
21 + 1
2a) + Pr(b > 1
2|a)1
)+ 1
30. We have to consider 2 different
cases. First, if a ≤ 12, then Pr(b ≤ 12|a) = 1
212
+ 121 = 3
4(because half of the time b
is equal to a, and since a is assumed to be less than or equal 12, b will be as well, and
the other half of the time, b is independent, so the probability of it being less than
12
is then equal to 12). Thus, the expected payoff of firing for I is then 1
12(5 + 3a),
and he will fire if this is more than a, which is always the case for a ≤ 12. Similarly,
when a ≥ 12, I’s expected payoff from firing is 1
12(7 + a), so that he then fires if and
only if a ≤ 711
.
8
Chapter 15
Exercise 15.10.8
As in section 10.2.2, Alice’s and Bob’s profits are given by πA(a, b) = (K−A−a−b)a
and πB(a, b) = (K−B−a−b)b, except that A andB are now random variables. So the
expected profit of Alice if she is of type A is EπA(a, b) = E [(K − A− a− b)a] = (K−
A−a− b)a, because πA(a, b) is a linear function of b so that when we take expectation
over b, we can distribute the expectation over the product and the sum (in fancy
language, expectation is a linear operator). Here b is Bob’s average output, which is
at this point still unknown. Similarly, we have for Bob EπB(a, b) = (K−B− a− b)b.
Now we can compute Alice’s best-reply function by maximizing her expected profit
with respect to her output, while treating her type A as a constant (Alice knows
her type). This gives Alice’s output as a function of her type A and Bob’s average
output b, a = K−A−b2
. Now, b is a constant, so we can take the expectation over a to
obtain a = K−c1−b2
. Similarly, we get for Bob b = K−B−a2
and b = K−c2−a2
. Together,
the equations for a and b give a = K−2c1+c23
and b = K−2c2+c13
.
9
Chapter 15
Exercise 15.10.9
Notice that Pr(B = L|A = H) = 1 so that Alice knows for sure that Bob is of type
L. Bob, however, only thinks that Alice is of type H with Pr(A = H|B = L) = 19, so
that Alice knows Bob is badly mistaken. Alice can exploit this fact by producing more
than she would be able to produce otherwise. The reason is that Bob is convinced
she is of type L which produces more than type H, so Bob will produce less than he
would if he knew Alice was H. Alice, knowing this, will be able to cash in. Formal
argument needs a bit of accounting: different types have different beliefs about the
other player. Let aH denote Alice’s output when she is of type H, aL is her output
when she is type L, and a = 19aH + 8
9aL. Similarly for Bob. Then, from maximization
of Alice’s and Bob’s profit, we get the following equations for Alice’s output (and
symmetrically for Bob): aH = 12(K−H−bL), aL = 1
2(K−L− b), and a = 1
9aH + 8
9aL.
From these three equations and from the three symmetric equations for Bob, we get
aH = bH = 151
(17K + 9L− 26H), aL = bL = 151
(H + 17K − 18L). It is easy to check
that when A = H, Alice is playing a best reply to Bob’s production level, as opposed
to Bob, who is under-producing given Alice’s production.
10
Chapter 15
Exercise 15.10.10
We use the expressions for α(A) and β(B) derived in section 15.5.1. We need to
inductively compute ¯A, ¯B,¯A, and so on. Since A = cB, we have ¯A = cB = cdA.
Similarly, we get ¯B = cdB. Now we can compute¯A = cA = c2dB, whence we can
already observe the pattern. Next we plug this into the formulae for α(A) and β(B).
We show how to compute the expression for α(A). The expression is
α(A) =1
2(K − A)− 1
4(K − dA) +
1
8(K − cdA)− 1
16(K − cd2A) + ...
We rearrange this into
α(A) = K(1
2− 1
4+
1
8− ...)− A(
1
2− 1
4d+
1
8cd− 1
16cd2 + ...).
We already know from section 15.5.1 that the coefficient in front of K equals 13. To
compute the coefficient multiplying A denote it by x and observe that
x =1
2− 1
4d+
1
4cdx.
From here we can immediately compute x = 2−d4−cd
, so that α(A) = 13K − 2−d
4−cdA. The
expression for β(B) is computed in the same manner.
11
Chapter 15
Exercise 15.10.11
If their beliefs are consistent then what Bob believes about Alice must be the same
as what Bob believes about what Alice believes that Bob believes about Alice. That
is, it must be that¯A = A, which gives cd = 1.
12
Chapter 15
Exercise 15.10.12
We have that b = K−B−a2
, implying that b = 13(K − 2c2 + A). Since a = K−A−b
2, so
that a = K−A−b2
. Now we can insert the expression for b, to obtain a = 16(2K+ 2c2−
3A− A).
In case Alice is type L < c2 but A > c2 Bob will overproduce, and knowing that,
Alice will best-reply to his production level and produce less.
13
Chapter 15
Exercise 15.10.13
Let [L,H] be the interval of possible types for Alice, and let l be her type such that
she reveals her cost for all types A ≤ l, but doesn’t reveal her type for A > l. The
production level for type A < l is then a(A) = 13(K−2A+c2), and Bob’s production
is b(A) = 13(K−2c2+A). On the other hand, if Alice is of type A′ > l, her production
is given as in exercise 15.10.12, except that we need to take conditional expectation
over her types, A(l) = E[A | A > l]. Thus, α(A) = 16(2K + 2c2 − 3A − A(l)) and
b = 13(K − 2c2 + A(l)). Inserting this into the expression for Alice’s profit, it follows
that for A > l, close to l, Alice would be better off revealing her cost.
14
Chapter 15
Exercise 15.10.14
We have b = K−B−a2
, so that b = K−B−¯a2
. Next, a = K−A−b2
, so that a = K−A(B)−b2
,
and taking the expectation we obtain ¯a = K− ¯A−b2
. Using these to solve for α(A) and
β(B), we obtain the desired equations.
Another type B of Bob might hold a belief that A(B) is high so that ¯A would
then be high and Alice would produce less.
15
Chapter 15
Exercise 15.10.15
Pr(I = A | II = D) = 0 whereas Pr(I = A) = 0.01, so that choices aren’t indepen-
dent. Pr(I = B | II = C) = .9. A and D will know for sure.
16
Chapter 15
Exercise 15.10.16
(a) A,C: (d,R); A,D: (d, L); B,C: (u, L); B,D : (u,R).
(b) After successive deletion of dominated strategies, the solution is A plays u,
B plays u, C plays L, D plays R (note that everyone except actor A has a
strongly dominant strategy, given the probabilities in figure 15.11(b)).
(c) Probabilities, payoffs to actor C and nothing else, since all actors except actor
A are playing a strongly dominated strategy.
(d) See answer to part (c).
(e) The difference reflects the fact that the actors have to optimize against the
average opponent rather than against a specific opponent as in part (a). Given
that C plays the strategy which is best on average, A can get 8 instead of −4.
17
Chapter 15
Exercise 15.10.17
(a) From left to right: (8,−8), (−8, 8), (0, 0), (−4, 4); (−4, 4), (8,−8), (0, 0),
(12,−12); (−8, 8), (12,−12), (−12, 12), (16,−16); (4,−4), (−4, 4), (0, 0), (−8, 8).
(b) Von Neumann’s pure strategies are s1s1, s1s2, s2s1, s2s2, similarly for Morgen-
stern.
(c) For instance the payoff vector for the profile (s1s1, t1t2) is (8α−8β−4γ,−8α+
8β + 4γ), where α = 0.01, β = 0.09, γ = 0.9. Similarly, we compute the rest of
the strategic form. The resulting game is clearly zero sum.
(d) The saddle point is (s1s1, t1t2).
(e) t1t2 strongly dominates all the other strategies for II, after deleting these, s1s1
is optimal for I.
18
Chapter 15
Exercise 15.10.18
An actor always knows his identity but also knows that the other actors only know
his “expected identity”, meaning that as perceived by the actor (through his belief
and the belief about others’ beliefs) the game is not zero sum.
19
Chapter 15
Exercise 15.10.19
(a) It does not have to be modified.
(b) In the approach of exercise 15.10.16, each player would, after observing his type,
decide with what probability to play either of his two actions, and then throw
an appropriate die. In the approach of exercise 15.10.17, each player would
in advance decide the probability with which to play each of his 4 strategies,
then throw an appropriate die, and then observe his type and implement the
appropriate action.
20
Chapter 15
Exercise 15.10.20
(a) The belief of I is (12, 1
2) regardless of the actor, while the belief of II is either
(12, 1
2) or (0, 1). I’s beliefs imply that C and D would have to be equally likely
and Pr(I = A, II = B) > 0. However, belief of D implies that Pr(I = A, II =
B) = 0 so that these beliefs are inconsistent.
(b) The model is the same as figure 15.12, but the casting move has different
probabilities for each type.
(c) For example, the payoffs to the strategy profile (s1s1, t2t2) are (3.96, 3.4).
21
Chapter 15
Exercise 15.10.21
(a) Nash equilibria are (Pay, Free−ride), (Free−ride, Pay) and a mixed equilib-
rium given by (4−c24, 4−c1
4). If c1 = c2 = 1 then the public good will be provided
with probability 1516
; if c1 = c2 = 3 the probability is 716
.
(b) Compute the payoffs, Ui(Pay | ci) = p(4 − ci) + (1 − p)(4 − ci) = 4 − ci, and
Ui(Free− ride | ci) = 4(1− p). Given that p ∈ [14, 3
4] it is now immediate that
(Free− ride, Pay) is an equilibrium. The probability that the public good is
provided is 1− p2.
(c) In a symmetric equilibrium, one of the types, depending on p, now plays a
mixed strategy. If p ∈ [34, 1], denote α = Pr(Pay | High). Then, as before,
Ui(Pay | ci) = 4 − ci, and Ui(Free − ride | ci) = 4 (αp+ (1− p)). For High
(ci = 3) to be indifferent, 1 = 4 (αp+ (1− p)), so that α = 4p−34p
; note that
α ∈ [0, 14]. It is easy to verify that Low prefers Pay.
If p ∈ [0, 14] then denote β = Pr(Pay | Low), and as before, we obtain β =
34(1−p)
∈ [34, 1].
(d) Let α and β be as in (c). Again, Ui(Pay) = 4−ci, but Ui(Free−ride | High) =
4 (0.2α + 0.8β), and Ui(Free − ride | Low) = 4 (0.8α + 0.2β). If α, β ∈ (0, 1)
then both types must be indifferent, hence 14
= 0.2α+0.8β and 34
= 0.8α+0.2β.
Solving these we obtain α = 1112
and β = 112
, and the probability of providing
the good is 0.1× α2 + 2× 0.4× αβ + 0.1× β2 = 1172
.
22
Chapter 15
Exercise 15.10.22
Let KA and KE be the knowledge operators of Adam and Eve, respectively, and let
CK be the common knowledge operator. If n is even, then we have KA({n}) = {n−
1, n}, KE({n}) = {n, n + 1}, so that KA({n}) ∩KE({n}) = {n}, hence CK({n}) =
{n}. The event that there are finitely many periods is E = ∪∞n=1{n}, so that E =
CK(E), hence in every state of the world, E is common knowledge.
23
Chapter 15
Exercise 15.10.23
In order for grim-trigger to be an equilibrium it has to be the case that in each state
of the world, both players believe that it is twice as likely that there will be more
repetitions rather than less. In order to construct a common prior P which would
make these beliefs consistent, it would thus have to be the case that P ({n}) = q2n,
for some constant q > 0. Since P is a probability measure, we would need to pick q
such that∑∞
n=1 q2n = 1, but that is impossible since
∑∞n=1 2n =∞.
24
Chapter 15
Exercise 15.10.24
Consider an actor of type n. If both players follow the prescribed strategies, than
n’s payoff is 23(2(n− 2)− 1) + 1
3(2(n− 1) + 3) = 2(n− 1), if n deviates and instead
plays hawk in period n−2, then his payoff is 2(n−3)+3 = 2(n−1)−1. The actors’
beliefs now come from a common prior P (N) = 12N , i.e., actor n obtains a signal that
the state of the world is in {n, n+ 1}, so that after updating, he puts probability 23
on state n.
25
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