IE54500 – Problem Set 4
Dr. David Johnson
Fall 2020
1. Pure and Mixed Nash Equilibria
Find all of the Nash equilibria in the following 2-player game. In the payoff matrix, (𝑥, 𝑦) indicates a payoff of 𝑥
for Player A and a payoff of 𝑦 for Player B.
Player B
Left
Middle
Right
Top
(3,3)
(4,8)
(5,4)
Player A Middle
(4,7)
(2,6)
(2,8)
Bottom (6,4)
(1,2)
(0,1)
In this game, the Nash equilibria will be at points where the first value is the maximum among those in the same
column (i.e., given Player B’s strategy, it is the maximum payoff among possible actions by Player A), and where
the second value is the maximum among those in the same row. For convenience, I will denote the strategies by
their payoffs because all of the payoff combinations are unique.
The potential pure equilibria in each row satisfying that criterion are therefore (4,8), (2,8), and (6,4). The
potential pure equilibria in each column are (6,4), (4,8), and (5,4). Two of these correspond to the same
strategies, so the Nash equilibria are (6,4) and (4,8). It is easy to verify that for each of these pairs of strategies,
neither player can do better predicated on the other player playing the same action.
Checking for mixed strategy equilibria, examine the expected payoff function for Player B, where we denote 𝑝1 as
the probability Player A plays Top, 𝑝2 as the probability Player A plays Middle, 𝑞1 as the probability Player B plays
Left, and 𝑞2 the probability Player B plays Middle:
𝑝1 (3𝑞1 + 4𝑞2 + 5(1 − 𝑞1 − 𝑞2 )) + 𝑝2 (4𝑞1 + 2𝑞2 + 2(1 − 𝑞1 − 𝑞2 )) + (1 − 𝑝1 − 𝑝2 )(6𝑞1 + 𝑞2 )
This has first-order conditions for 𝑝1 and 𝑝2 , respectively:
3𝑞1 + 4𝑞2 + 5 − 5𝑞1 − 5𝑞2 − 6𝑞1 − 𝑞2 = 5 − 8𝑞1 − 2𝑞2 = 0 ⇒ 8𝑞1 + 2𝑞2 = 5
4𝑞1 + 2𝑞2 + 2 − 2𝑞1 − 2𝑞2 − 6𝑞1 − 𝑞2 = 2 − 4𝑞1 − 𝑞2 = 0 ⇒ 4𝑞1 + 𝑞2 = 2
These two equations represent distinct parallel lines, so they do not admit a simultaneous solution for 𝑞1 and 𝑞2 .
This indicates that there is no mixed strategy over all three actions, but we should look for possibilities that mix
two of the actions. For example, consider the case where 𝑝2 = 0 for Player A. (For the intuition behind exploring
this option, observe that Middle is never the best response for Player A to any of Player B’s actions.) In this case,
the probability of playing Bottom is 1 − 𝑝1 . The expected payoff function for Player A becomes
𝑝1 (3𝑞1 + 4𝑞2 + 5(1 − 𝑞1 − 𝑞2 )) + (1 − 𝑝1 )(6𝑞1 + 𝑞2 )
which has FOC
3𝑞1 + 4𝑞2 + 5 − 5𝑞1 − 5𝑞2 − 6𝑞1 − 𝑞2 = 5 − 8𝑞1 − 2𝑞2 ⇒ 4𝑞1 + 𝑞2 = 5/2
Now, we notice that if Player A is playing a strategy where 𝑝2 = 0, then Middle dominates Right for Player B.
This tells us that Player B would never play Right as part of their best response function; in other words, we know
that, given this strategy from Player A, 1 − 𝑞1 − 𝑞2 = 0. Solving this and the equation above simultaneously, we
find that 𝑞1 = 𝑞2 = 1/2.
The payoff function for Player B, on the other hand, becomes
𝑞1 (3𝑝1 + 7𝑝2 + 4 − 4𝑝1 − 4𝑝2 ) + (1 − 𝑞1 )(8𝑝1 + 2 − 2𝑝1 )
which has FOC
3𝑝1 + 4 − 4𝑝1 − 8𝑝1 − 2 + 2𝑝1 = 2 − 7𝑝1 = 0 ⇒ 𝑝1 = 2/7
Therefore, Player A would have a probability of 5/7 of playing Bottom. A mixed-strategy Nash equilibrium is thus
(2/7, 0, 5/7) for Player A and (1/2, 1/2, 0) for Player B. Checking other possibilities shows that this is the only
mixed-strategy Nash equilibrium of the game.
2. Asymmetric Marginal Costs
Consider a Cournot duopoly model where inverse demand is 𝑃(𝑄) = 𝑎 − 𝑄 but firms have asymmetric marginal
costs. Firm 1 has marginal costs 𝑐1 and Firm 2 has marginal costs 𝑐2 .
a) What is the Nash equilibrium if 0 < 𝑐𝑖 < 𝑎/2 for each firm?
b) What is the Nash equilibrium if 𝑐1 < 𝑐2 < 𝑎 but 2𝑐2 > 𝑎 + 𝑐1 ?
Denote the profit for Firm 1 as 𝜋1 and for Firm 2 as 𝜋2 . Then for 𝑖 ≠ 𝑗,
𝜋𝑖 = 𝑞𝑖 (𝑎 − 𝑞𝑖 − 𝑞𝑗 ) − 𝑐𝑖 𝑞𝑖
which has FOC
𝑎 − 𝑐𝑖 − 𝑞𝑗
𝜕𝜋𝑖
= 𝑎 − 𝑐𝑖 − 𝑞𝑗 − 2𝑞𝑖 = 0 ⇒ 𝑞𝑖∗ =
𝜕𝑞𝑖
2
Subbing in 𝑞𝑗∗ (and switching to subscripts of 1 and 2 for clarity going forward), we can solve
𝑞1∗ =
𝑎 − 𝑐1 1 𝑎 − 𝑐2 − 𝑞1
𝑎 − 2𝑐1 + 𝑐2 + 𝑞1∗
𝑎 − 2𝑐1 + 𝑐2
−
⇒ 2𝑞1∗ =
⇒ 3𝑞1∗ = 𝑎 − 2𝑐1 + 𝑐2 ⇒ 𝑞1∗ =
2
2
2
2
3
We also know by symmetry that 𝑞2∗ =
𝑎−2𝑐2 +𝑐1
.
3
However, all of this does not take the physical system described in the problem into account, which carries
implicit constraints that 𝑞1 ≥ 0, 𝑞2 ≥ 0, and 𝑎 − 𝑞1 − 𝑞2 ≥ 0.
When 0 < 𝑐𝑖 < 𝑎/2, this implies that 𝑎 − 𝑐2 − 2𝑐1 > 0, so 𝑞1∗ > 0. Similarly, 𝑞2∗ > 0.
Further,
𝜋1 = 𝑞1 (𝑎 − 𝑞1 − 𝑞2 − 𝑐1 ) = 𝑞1 (𝑎 −
2𝑎 − 𝑐1 − 𝑐2
𝑎 + 𝑐2 − 2𝑐1
− 𝑐1 ) = 𝑞1 (
)>0
3
3
Thus, if 0 < 𝑐𝑖 < 𝑎/2, all of the implicit constraints clear “as usual” and we end up with both firms producing
and receiving positive profits for doing so.
In the case where 𝑐1 < 𝑐2 < 𝑎 and 2𝑐2 > 𝑎 + 𝑐1 , then we find that 𝑎 + 𝑐1 − 2𝑐2 < 0. However, we found that
𝑞2∗ =
𝑎 − 2𝑐2 + 𝑎1
3
so this implies that 𝑞2∗ < 0. Effectively, Firm 2’s higher costs push them out of the market because Firm 1 can use
their lower costs to undercut them. Therefore, the optimal strategy for Firm 2 is not to produce by setting 𝑞2∗ = 0,
and this gives Firm 1 a monopoly.
At that point, Firm 1 sees 𝑃 = 𝑎 − 𝑞1 , and 𝜋1 = (𝑎 − 𝑞1 )𝑞1 − 𝑐1 𝑞1 . Taking the first-order condition here, we get
the usual monopoly result that 𝑞1∗ = (𝑎 − 𝑐1 )/2, and therefore 𝑃 = (𝑎 + 𝑐1 )/2. But we are told by assumption
that 𝑐2 > (𝑎 + 𝑐1 )/2, so Firm 2’s costs would still be higher than 𝑃 under the monopoly. Therefore, even at the
monopoly pricing, Firm 2 has no incentive to reenter the market and 𝑞1∗ = (𝑎 − 𝑐1 )/2, 𝑞2∗ = 0.
3. Subgame Perfect Nash Equilibria
In the following dynamic game of complete and perfect information, find all of the Nash equilibria. Which of
them are subgame perfect? (Player 1 has action space {𝑇𝑜𝑝, 𝐵𝑜𝑡𝑡𝑜𝑚}; Player 2 has action space {𝐿𝑒𝑓𝑡, 𝑅𝑖𝑔ℎ𝑡}.)
Using backwards induction, we can find a subgame perfect Nash equilibrium as 1: {𝐵}, 2: {𝑇 → ℛ; 𝐵 → ℒ},
where 𝑇 → ℛ means “If player 1 plays 𝑇, play ℛ.” We can look for other equilibria that are not subgame perfect
by examining other strategies for each player. One is 1: {𝐵}, 2: {𝑇 → ℒ; 𝐵 → ℒ}, and another is
1: {𝑇}, 2: {𝑇 → ℛ; 𝐵 → ℛ}. These are not subgame perfect, but it is clear to see that neither player has incentive
to deviate.
4. Repeated Bertrand Duopoly
Recall the static Bertrand duopoly model: two firms produce homogeneous products and choose their prices
simultaneously. Demand for firm 𝑖’s product is 𝑎 − 𝑝𝑖 if 𝑝𝑖 < 𝑝𝑗 , is 0 if 𝑝𝑖 > 𝑝𝑗 , and is (𝑎 − 𝑝𝑖 )/2 if 𝑝𝑖 = 𝑝𝑗 .
Marginal costs for both firms are 𝑐 < 𝑎. Now consider an infinitely repeated game based on this model. Show
that the firms can use trigger strategies (i.e., a cooperative strategy that switches forever to the single-stage
game Nash equilibrium after any deviation) to sustain the monopoly price level in a subgame-perfect Nash
equilibrium if and only if each firm has a discount rate 𝛿 ≤ 1. (Assume that the discount rate is defined such that
the present value at time period 0 of profit 𝜋𝑡 in time period 𝑡 is 𝜋𝑡 /(1 + 𝛿)𝑡 .)
When cooperating, each firm charges the monopoly price but splits the demand in half. Each firm can initiate a
trigger strategy by undercutting the monopoly price 𝑝𝑀 by any amount; thus, in the limit they can charge 𝑝𝑀 − 𝜖
for 𝜖 > 0 arbitrarily small. They would then claim the full demand in that cycle but would be retaliated against
by the other firm setting a price of 𝑐 to ensure they cannot be undercut further.
In this model, the payoff in each period is the profit 𝜋𝑖 = 𝑄𝑖 (𝑝𝑖 − 𝑐) =
𝜋1𝑇 =
𝑎−𝑝𝑀
2
∙ (𝑝𝑀 − 𝑐), so across all periods,
(𝑎 − 𝑝𝑀 )
(𝑎 − 𝑝𝑀 )
(𝑎 − 𝑝𝑀 )
1 (𝑎 − 𝑝𝑀 )
(𝑝𝑀 − 𝑐) + 𝑑1 ∙
(𝑝𝑀 − 𝑐) + 𝑑12 ∙
(𝑝𝑀 − 𝑐) + ⋯ =
(𝑝𝑀 − 𝑐)
2
2
2
1 − 𝑑1
2
If the trigger is utilized, then the payoff becomes
𝜋1𝐷 = (𝑎 − 𝑝𝑀 ) ∙ (𝑝𝑀 − 𝑐) + 𝑑1 ∙
𝑎−𝑐
(𝑐 − 𝑐) = (𝑎 − 𝑝𝑀 )(𝑝𝑀 − 𝑐)
2
where we denote 𝑑1 = 1/(1 + 𝛿1 ).
The strategy of cooperation is sustainable if 𝜋1𝑇 > 𝜋1𝐷 , so
1 𝑎 − 𝑝𝑀
1 1
1
(𝑝𝑀 − 𝑐) ≥ (𝑎 − 𝑝𝑀 )(𝑝𝑀 − 𝑐) ⇒
≥ 1 ⇒ 𝑑1 ≥ ⇒ 𝛿1 ≤ 1
1 − 𝑑1 2
1 − 𝑑1 2
2
By symmetry, 𝛿2 ≤ 1 as well. By generality, this same argument holds no matter what time period the trigger is
activated (because up to that point, the profits are identical to cooperation).
5. Static Bayesian Nash Equilibria
Find all of the pure-strategy Bayesian Nash equilibria for the following static Bayesian game:
i)
Nature determines whether the two players’ payoffs are given by Game 1 or Game 2, with each game
being equally likely.
ii) Player 1 learns whether nature has drawn Game 1 or Game 2, but Player 2 does not.
iii) Player 1 chooses either 𝑇𝑜𝑝 or 𝐵𝑜𝑡𝑡𝑜𝑚. Player 2 simultaneously chooses either 𝐿𝑒𝑓𝑡 or 𝑅𝑖𝑔ℎ𝑡.
iv) Payoffs are given by the game drawn by nature:
Player 2
Left Right
Player 1 Top
(1,1) (0,0)
Bottom (0,0) (0,0)
Game 1
Player 2
Left Right
Player 1 Top
(0,0) (0,0)
Bottom (0,0) (2,2)
Game 2
(Note that this is not the only approach for this problem; we could also put the game in its extensive form.)
Because we know that each game is equally likely according to Nature, we can write the game in normal
form with each cell containing the expected payoffs for each player as follows:
Player 2
L
T,T 1/2, 1/2
Player 1 T,B 1/2, 1/2
B,T 0, 0
B,B 0, 0
R
0, 0
1, 1
0, 0
1, 1
In this table, (T, B) indicates that Player 1’s strategy would be to play Top if Nature has given Game 1 and
Bottom if Game 2, etc. The expected payoffs are the average of each player’s payoff in each game, since
each game is known to be equally likely. Now, visual inspection can find the pure-strategy Bayesian Nash
equilibria, where each player has no incentive to deviate. These are at [(T,T); L], [(T,B); R] and [(B,B); R]. In all
of the equilibria, we must also specify the Players’ beliefs. Player 1 knows what Nature has selected, so they
believe with probability 1 that the game is whichever Nature has drawn. Because Player 2 acts
simultaneously and there are no further stages in the game, they have no additional information to revise
their beliefs; as such, they believe that it is Game 1 with p=0.5 and Game 2 with p=0.5.
6. Pooling Perfect Bayesian Nash Equilibrium
Consider the following dynamic signaling game that begins by nature drawing one of three types for Player 1
with equal probability. Player 1 knows their type, but Player 2 does not. Player 1’s action space is {𝐿𝑒𝑓𝑡, 𝑅𝑖𝑔ℎ𝑡},
and Player 2’s action space is {𝑈𝑝, 𝐷𝑜𝑤𝑛}. Specify the conditions under which a pooling perfect Bayesian
equilibrium exists where all three Player 1 types play ℒ.
(We are not asked this, but we can first observe that L dominates R for type 𝑡1 , so there can be no pooling
equilibrium where all Player 1 types play ℛ.)
If Player 1 plays (L, L, L) (in other words, L for each of their three possible types), then Player 2 has gained no new
information and therefore believes 𝑃(𝑡𝑖 |ℒ) = 1/3 for all types. If Player 2 plays Up regardless of observing ℒ or
ℛ, then Player 1 would have no incentive to deviate since ℒ dominates ℛ for all types if Player 2’s strategy is
always to play Up.
Player 2 has expected payoff from this strategy of
1
𝔼2 [𝑢|ℒ] = (1 + 1 + 1) = 1
3
Their expected payoff from a strategy of playing Down is 𝔼2 [𝑢|ℒ] = 0, so playing Up is optimal. (A pure Up
strategy is clearly also optimal compared to any mixed strategy, given Player 1’s pooling strategy on ℒ.)
However, we require that Player 2’s strategy of playing Up if faced with ℛ be consistent with his beliefs about
Player 1’s type if faced with ℛ, so we need for 𝔼2 [𝑢|ℛ] ≥ 𝔼2 [𝑑|ℛ].
∴ 1 ∙ [𝜇(𝑡1 |ℛ) + 𝜇(𝑡2 |ℛ)] ≥ 1 ∙ 𝜇(𝑡3 |ℛ)
and we know 1 − 𝜇(𝑡3 |ℛ) = 𝜇(𝑡1 |ℛ) + 𝜇(𝑡2 |ℛ), so 1 − 𝜇(𝑡3 |ℛ) ≥ 𝜇(𝑡3 |ℛ) ⇒ 𝜇(𝑡3 |ℛ) ≤ 1/2.
Therefore, (L,L,L) and (u,u) is an equilibrium with a pooling strategy on L provided that 𝜇(𝑡3 |ℛ) ≤ 1/2.