Solutions to Lecture 9 problems
Infinitely repeated Battle of Sexes
Part (1). The infinite outcome is (OO,SO,OO,OO, OO, ...). (We use a convention that Her action is denoted first and His action is second).
The payoff of She is
5 + 0δ + 5δ^{2} + 5δ^{3} + ...
=
5 + 5δ^{2}(1 + δ + δ^{2} + ...) = 5 + 5δ^{2}(1)/(1 − δ).
The payoff of He is
3 + 0δ + 3δ^{2} + 3δ^{3} + ...
=
3 + 3δ^{2}(1 + δ + δ^{2} + ...) = 3 + 3δ^{2}(1)/(1 − δ).
Part (2i). Her current payoff is equal to
5 + δ3 + δ^{2}5 + δ^{3}3 + ... = (5 + δ3)(1 + δ^{2} + δ^{4} + ...) = (5 + δ3)/(1 − δ^{2}).
If she deviates in any period, she receives 0 in that period (insted of 3 or 5) and at most the the same payoff in any other subsequent period as using the current strategy. Because 0 is smaller than 3 or 5, her current strategy is a best response.
His current payoff is
3 + δ5 + δ^{2}3 + δ^{3}5 + ... = (3 + δ5)(1 + δ^{2} + δ^{4} + ...) = (3 + δ5)/(1 − δ^{2}).
If he deviates in any period, he receives 0 in that period and at most 3 in any subsequent period (notice that following his deviation, she plays Opera forever). Thus, the deviation leads to weakly smaller payoff in each period than the current strategy. It follows that his behavior is a Nash equilibrium.
Part (2ii). Consider a history in the beginning of period 2 h = (OS). In particular, in period 1 she played Opera and he played Stadium. From now on, His strategy tells Him to alternate between Stadium (periods 2, 4, 6,...) and Opera (periods 3, 5, 7, ..). Her strategy tells her to play Opera forever. Her strategy is not abest response as she would like to met Him in every period.
Part (2iii). We can modify His strategy so that He always plays Opera after every history in which He and She did not met in the past.
Battle of Sexes with incomplete information
Part (a) Let
σ_{He}(dh) = S = σ_{He}(hM), σ_{He}(hA) = O,
σ_{She}(m) = S, σ_{She}(a) = O.
We will show that this strategy profile is a Bayesian Nash equilibrium for
p ≤ (5)/(8).
In the same way as in the class, we can check that his strategy is a best response to her strategy (remember that type “didn’t hear” consider both of the types of She equally likely).
We check whether Her strategy is a best response. If She has type “meet”, she knows for sure that He will choose S. Because She “meet” type wants to meet with Him, she will choose S as a best response.
On the other hand, She “avoid” type expects that He chooses O with probability p (i.e., when He is hA type) and He chooses S with probabiltiy 1 − p (i.e., when He is the “meet” type. If p ≤ (5)/(8), choosing O is a best response for Her.
Part (2) When p > (2)/(3), then the strategy profile described above is not an equilibrium because Her action if she is “avoid” type is not a best response. Intuitively, when She is pretty sure that he heard her conversation, She expect him to follow Her to the Opera.
We will find an equilibrium, in which Her “avoid” type randomizes. Let
σ_{He}(dh) = S = σ_{He}(hM), σ_{He}(OhA) = (5)/(8p),
σ_{She}(m) = S, σ_{She}(Oa) = (5)/(8).
We will show that this strategy profile is a Bayesian Nash equilibrium for
p > (5)/(8).
Notice that the incentives of His type “heard M” and Her type “meet” are the same as previously. His type “didn’t hear” expects to find her at the Stadium with probability more than 1 ⁄ 2, which only strenghtens his resolve to choose S.
We will check that the types “heard A of Him and “avoid” of Her are indifferent between playing O and S. Notice that She “avoid” expects him to choose Opera with the probability exactly (1 − p)0 + p(5)/(8p) = (5)/(8). This makes her indifferent between choosing O and S. Moreover, He “heard A” expects Her to be in the Opera with probability (2)/(3). Thus, He is indifferent as well.
Solutions to Lecture 10 problems
Cournot model with asymmetric beliefs about costs
Update (02.12.2014): The text of this solution was corrected to reflect the discussion that we had in class. The previous solution did not take into account the fact that the best response and the equilibrium actions have to be nonnegative.
Part (1). Define
E_{i}c = π_{i}c_{H} + (1 − π_{i})c_{L}. for each player i Then, using the same arguments as in the class, we can check that each player
i type
c best response quantity is equal to
where
E_{i}q_{ − i} = π_{ − i}q^{H}_{ − i} + (1 − π_{ − i})q^{L}_{ − i}.
Notice that the best response of the low cost type is always higher than the best response of the high cost type,
q^{BR}_{i}(q_{ − i}, c^{L}) ≥ q_{i}(q_{ − i}, c^{L})
Suppose that
(q_{i}(c))^{c = L, H}_{i = 1, 2} is a profile of equilibrium quantities. Then, for each
i and each cost
c, it must be that
q_{i}(c) = q^{BR}_{i}(q_{ − i}, c).
Given the best response formula (
2↑), for each player
i, there are three possible cases:

q^{L}_{i} > q^{H}_{i} > 0. In this case, it must be that

q^{L}_{i} > q^{H}_{i} = 0. In this case, it must be that

q^{L}_{i} = q^{H}_{i} = 0. In this case, it must be that
We need to consider all possible combination of cases. For instance, consider cases
1^{(1)}1^{(2)} (here, the superscript denotes the player). Then all quatities solve equations (
3↑) . We can solve the equations in similar way to the equations that we solved in class. The solutions are equal to
q_{i}(c) = (α − c − E_{ − i}q_{ − i})/(2) = (α − c − (α − 2E_{ − i}c + E_{i}c)/(3))/(2).
The above profile is an equilibrium of the game with incomplete information if and only if all the above quantities are positive. The last requirement delivers conditions on parameters under which the above solution is an equilibrium.
For another example, consider the case
3^{(1)}1^{(2)}. Here, we have
q^{H}_{2}
=
(1)/(2)(α − c^{H}), q^{L}_{2} = (1)/(2)(α − c^{L}),
q^{H}_{1}
=
q^{L}_{1} = 0.
The above profile is a Bayesian Nash equilibrium if and only if inequalities (
3↑) for player 2 and (
5↑) for player 1 are satisfied, i.e.,
α − c^{L}
<
π_{2}⎛⎝(1)/(2)(α − c^{H})⎞⎠ + (1 − π_{2})⎛⎝(1)/(2)(α − c^{L})⎞⎠,
α − c^{L}
>
α − c^{L} > 0.
All the other cases are dealt accordingly.
Part (2). The expected equilibrium payoff of type
c is
q_{i}(c)(α − c − q_{i}(c) − E_{ − i}q_{ − i}) =
⎛⎜⎝(α − c − (α − 2E_{ − i}c + E_{i}c)/(3))/(2)⎞⎟⎠⎛⎜⎝α − c − (α − c − (α − 2E_{ − i}c + E_{i}c)/(3))/(2) − (α − 2E_{ − i}c + E_{i}c)/(3)⎞⎟⎠
=
⎛⎜⎝(α − c − (α − 2E_{ − i}c + E_{i}c)/(3))/(2)⎞⎟⎠⎛⎝(1)/(2)(α − c) + (1)/(2)(α − 2E_{ − i}c + E_{i}c)/(3) − (α − 2E_{ − i}c + E_{i}c)/(3)⎞⎠
=
(1)/(4)⎛⎝α − c − (α − 2E_{ − i}c + E_{i}c)/(3)⎞⎠^{2}.
The higher the beliefs
π_{i}, the higher the expected cost of player
i, E_{i}c, and the lower the profits of player
i’s types. This makes sense  if player
i is likely to have high cost, player
− i expects that player
i chooses small quantity. This makes it best response for player
− i to choose large quantity (remember that in Cournot duopoly, player
− ibest response decreases with player
i’s quantity.) But large quantity of player
− i reduces the profits of player
i's types.
Thus, player i would like player − i to belief that i has low costs, or that π_{i} is very low.
Attack on the city
Part (1). There are two players. Each player has two types, w, s. Type shas probability π_{i} and type w has probability 1 − π_{i}.
Part (2). The payoff of type
w of player 1 is equal to
(1 − π_{ − i})⋅1 + π_{ − i}⋅( − 1) = 1 − 2π_{ − i}
if she attackes, and 0 if she does not attack.
The payoff of type s of player 1 is equal to − 1if she attacks and 0 if she does not attack.
Part (3). Such an equilibrium always exists.
Part (4). If 2π_{i} ≤ 1 for each player i, then there exists an equilibrium, in which the generals attack if and only if the see weak fortifications. The probabiloity that the attack is successful is equal to the probability that both sides of the fortifications are weak, which is equal to (1 − π_{1})(1 − π_{2}).
Cleaning room game I
There is a unique BNE in which both high value types always clean and the low value types never clean. (To see why, notice that clkeaning is a best response for the high type given that the low type of the other guy does not clean and no matter what the high type of the other guy is doing.)
Romeo and Juliet
Part 1. No. (If p > (1)/(4), then Romeo’s best response is Casino. In such a case, Lucky’s JUliet action is not optimal. If p < (1)/(4), Rome’s best response is Boxing, in which case Unlucky Juliet’s action is not optimal.
Part 2. The payoff from Casino is equal to
p(α_{L}(1 + 2) + (1 − α_{L})1) + (1 − p)1.
The payoff from Boxing Match is equal to
p(α_{L}0 + (1 − α_{L})2) + (1 − p)2.
This implies
α_{L} = (1)/(4p)
Part 3. Let
α_{L} be the probability that l ucky Juliet goes to the Casino and let
β be the probability that Romeo goes to the Casino. Because lucky Juliet must be indifferent in equilibrium, we have
− β + 3(1 − β) = β − (1 − β),
which implies
β = (2)/(3). Unlucky Juliet must prefer Boxing, so that
3β − (1 − β) > − β + 1 − β,
which holds if
β = (2)/(3). Because Romeo must also be indifferent, we have
p(α_{L}(1 + 2) + (1 − α_{L})1) + (1 − p)1 = p(α_{L}0 + (1 − α_{L})2) + (1 − p)2.
This implies
α_{L} = (1)/(4p).
In particular,
α_{L} is a welldefined probability if
p ≥ (1)/(4).
Part 4. We consider two separate games with perfect information. The two games are solved separately. With Lucky Juliet, there is a unique equilibrium. In the equilibrium, Juliet goes to Casino with prob. (1)/(4), and Romeo goes to Casin with prob. (2)/(3). When Juliet is Unlucky, it is, respectively, (1)/(4) and (1)/(3).