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Background: A game in strategic form is a triple G = ⟨N, (Si)i∈N , (ui)i∈N ⟩, where N is the set of players, Si is the set of pure strategies of player i and ui : ×i∈N Si → R is the utility function of player i. A game G is finite if the cardinality of N and ×i∈N Si are finite. The mixed extension of a finite game G is a triple Γ = ⟨N, (Σi)i∈N , (Ui)i∈N ⟩, where Σi := ∆(Si) is the set of mixed strategies of player i and Ui : ×i∈N Σi → R is the multilinear extension of ui. Unless stated otherwise, in the exercises it is assumed that players are rational, have common knowledge of rationality and of the game, have complete information, and can only randomize independently.
The typical pure strategy of player i is denoted by si and the typical mixed strategy by σi. The set of all pure strategy profiles is S := ×i∈N Si and has typical element s. The set of pure strategy profiles not including i’s strategy is S−i := ×j∈N {i}Sj and has typical element denoted by s−i. Similarly, Σ := ×i∈N ∆(Si) has typical element σ and Σ−i := ×j∈N {i}∆(Sj ) has typical element σ−i.
To ease notation, I will denote the expected utility of agent i from mixed profile σ by ui(σ) rather than Ui(σ). Also, I define ui(σi, s−i) :=
P
si∈Si σi(si)ui(si, s−i), and similarly for^ ui(si, σ−i).
Exercise 1. Show that in any normal form game, a player can have at most one strictly dominant strategy.
Background: Let G = ⟨N, (Si)i∈N , (ui)i∈N ⟩ be a game in strategic form and denote by Γ its mixed extension. A mixed strategy σi ∈ Σi of player i in Γ strictly dominates σ′ i ∈ Σi if
ui(σi, s−i) > ui(σ′ i, s−i) for all s−i ∈ S−i.
A strategy σi ∈ Σi is strictly dominant if it strictly dominates all strategies σ′ i ∈ Σi \ {σi}.
Answer: Let Γ = ⟨N, (Σi)i∈N , (ui)i∈N ⟩ be a game in strategic form. Suppose σi, σ′ i ∈ Σi, with σi ̸= σ i′ are strictly dominant strategies for player i. Then, by definition, σi strictly dominates σ i′, and σ i′ strictly dominates σi; that is,
ui(σi, s−i) > ui(σ′ i, s−i) for all s−i ∈ S−i
and ui(σ i′, s−i) > ui(σi, s−i) for all s−i ∈ S−i.
A contradiction.
1
Exercise 2. Find the set of Nash equilibria of the following game b 1 b 2 b 3 a 1 5 , 6 7 , 2 0 , 8 a 2 3 , 4 5 , 6 1 , 2 a 3 2 , 7 0 , 2 0 , 0
Background: For finite two-person strategic games, the set of strategies surviving iterated elimination of strictly dominated strategies coincides with the set of rationalizable strategies.
Given a finite two-person strategic game Γ = ⟨{ 1 , 2 }, (Σ 1 , Σ 2 ), (u 1 , u 2 )⟩, let Zi ⊆ Si be the set of rationalizable pure strategies of player i. Then, for all Nash equilibria (σ∗ 1 , σ∗ 2 ) ∈ Σ 1 × Σ 2 it holds supp(σ∗ 1 ) × supp(σ 2 ∗ ) ⊆ Z 1 × Z 2.
A strategy profile (σ 1 , σ 2 ) is a Nash equilibrium if and only if σi ∈ BRi(σ−i) for all i ∈ { 1 , 2 }, where BRi(σ−i) = argmaxσi∈Σi ui(σi, σ−i) is agent i’s best reply correspondence. This implies ui(si, σ−i) is constant over si ∈ supp(σi) for all i ∈ { 1 , 2 }. The latter is therefore a necessary condition for (σ 1 , σ 2 ) to be a Nash equilibrium. It is not sufficient as there might exist some i ∈ { 1 , 2 }, si ∈/ supp(σi) for which ui(si, σ−i) > ui(σi, σ−i). Nonetheless, it is useful for finding mixed Nash equilibria analytically.
Answer:
There are no pure strategy Nash equilibria: we must look for for mixed strategy Nash equilibria.
Notice that a 2 strictly dominates a 3. Therefore, we can restrict the support of σ 1 to {a 1 , a 2 } and the support of σ 2 to {b 1 , b 2 , b 3 }.
Player 1 cannot make player 2 indifferent over all her pure strategies (requires some algebra). It follows that in a Nash equilibrium supp(σ 2 ) ⊂ {b 1 , b 2 , b 3 }.
- Candidate equilibrium strategies σ 2 = (q, 1 − q, 0) and σ 1 = (p, 1 − p, 0) with p, q ∈ [0, 1].
Notice that BR 1 (σ 2 ) = {(1, 0 , 0)} for all q ∈ [0, 1]. But σ 1 = (1, 0 , 0) does not make player 2 indifferent over {b 1 , b 2 }. Then, there is no mixed Nash equilibrium where supp(σ 2 ) = {b 1 , b 2 }.
- Candidate equilibrium strategies σ 2 = (0, q, 1 − q) and σ 1 = (p, 1 − p, 0) with p, q ∈ [0, 1].
Player 1 is indifferent over {a 1 , a 2 } just in case q = 1/3. Similarly, player 2 is indifferent over {b 2 , b 3 } just in case p = 2/5. The strategy profile (σ∗ 1 , σ∗ 2 ) = ((2/ 5 , 3 / 5 , 0), (0, 1 / 3 , 2 /3)) is a candidate Nash equilibrium. However, u 2 (σ∗ 1 , b 1 ) > u 2 (σ∗ 1 , σ∗ 2 ). It follows that there is no mixed Nash equilibrium where supp(σ 2 ) = {b 2 , b 3 }.
- Candidate equilibrium strategies σ 2 = (q, 0 , 1 − q) and σ 1 = (p, 1 − p, 0) with p, q ∈ [0, 1].
Player 1 is indifferent over {a 1 , a 2 } just in case q = 1/3. Similarly, player 2 is indifferent over {b 1 , b 3 } just in case p = 1/2. The strategy profile (σ∗ 1 , σ∗ 2 ) = ((1/ 2 , 1 / 2 , 0), (1/ 3 , 0 , 2 /3)) is a candidate Nash equilibrium. Since u 2 (σ∗ 1 , b 2 ) < u 2 (σ 1 ∗ , σ∗ 2 ), the profile (σ 1 ∗ , σ 2 ∗ ) is a Nash equilibrium.
The set of Nash equilibria is {((1/ 2 , 1 / 2 , 0), (1/ 3 , 0 , 2 /3))}.
Exercise 4. There are two identical firms producing a homogeneous good whose demand curve is q = 100 − p. Firms simultaneously choose prices. Neither firm can produce more than K < 100 units of output. If both firms choose the same price p, then they share the market equally subject to the capacity constraint of K. If the prices are unequal, say pi < pj , then the low-price firm i sells qi(pi, pj ) = min(100 − pi, K). If qi(pi, pj ) < K, then firm j sells nothing. If qi(pi, pj ) = K, then firm j sells min(100 − qi(pi, pj ) − pj , K). The cost of production for each firm is 10 per unit.
Background: Let G = ⟨N, (Si)i∈N , (ui)i∈N ⟩ be a game in strategic form. Suppose there exists a strategy for player i that guarantees her a payoff of g; that is, ∃si ∈ Si such that ui(si, s−i) ≥ g for all s−i ∈ S−i. Then, if s∗^ is an equilibrium, ui(s∗) ≥ g. Proof: Suppose s∗^ is an equilibrium with ui(s∗) < g and that ˆsi guarantees player i a payoff of g. Then ui(ˆsi, s∗−i) = g > ui(s∗) implying s∗^ is not an equilibrium.
Suppose s∗^ ∈ S is a Nash equilibrium. Then maxsi∈Si ui(si, s∗−i) exists for all i ∈ N. Proof: Suppose not, then s∗ i does not exist either.
Question (i): Derive the payoff function of either firm.
Answer: The payoff function of firm i is πi(pi, pj ) = (pi−10)qi(pi, pj ) where qi(pi, pj ) = max{q˜i(pi, pj ), 0 } and
q˜i(pi, pj ) =
min{ 100 − pi, K} if pi < pj min{(100 − pi)/ 2 , K} if pi = pj min{ 100 − K − pi, K} if pi > pj
Question (ii): Suppose K = 40. Show that there is no pure strategy Nash equilibrium in this game.
Answer: Firm i could guarantee itself a non-negative profit by setting pi ≥ 10 and always incurs negative profits by setting pi < 10. It follows that in equilibrium pi ≥ 10 for all i ∈ { 1 , 2 }.
Consider a candidate equilibrium (p 1 , p 2 ) with 10 < p 1 < p 2. Within those range of prices, the firms’ profit functions are restricted to π 1 (p 1 , p 2 ) = (p 1 − 10) min{ 100 − p 1 , 40 } and π 2 (p 1 , p 2 ) = (p 2 −
- max{min{ 60 −p 2 , 40 }, 0 }. Note that for any p 2 ≥ 100 −K = 60 firm 2 obtains zero profit and would rather set a price 10 < p 2 < 60. However, maxp 1 ∈[10,p 2 ) π 1 (p 1 , p 2 ) exists only if p 2 ≥ 100 − K = 60 — see figures. That is, in the range of prices where firm 1’s profit function has a maximum, firm 2 would rather deviate. Because the existence of a maximum at π 1 (·, p 2 ) is a necessary condition for p 2 to be a Nash equilibrium strategy, there exists no Nash equilibrium (p 1 , p 2 ) with 10 < p 1 < p 2 ≤ 100 (and by symmetry with 10 < p 2 < p 1 ≤ 100). If a pure equilibrium exists, it must satisfy p 1 = p 2.
10 p 2 60 100
0
p 1 < p 2
π^1
(p
, p 1
10 60 p 2 100
0
p 1 < p 2
π^1
(p
, p 1
Consider candidate equilibria (p 1 , p 2 ) = (p, p) with 10 ≤ p < 100. I will split the problem into four cases (with K = 40)
(a) 10 ≤ p < 100 − 2 K → (100 − p)/ 2 > K at capacity, ↑ (b) p = 100 − 2 K → (100 − p)/2 = K exactly at capacity, ↑ (c) 100 − 2 K < p ≤ 100 − K → 100 − K − p < K below capacity, ↓ at capacity (d) 100 − K < p ≤ 100 → 100 − p < K below capacity, ↓ below capacity
Case (a): consider a candidate equilibrium (p 1 , p 2 ) = (p, p) with 10 ≤ p < 100 − 2 K = 20. Firm i’s profits are πi(p, p) = (p − 10)K = (p − 10)40 < 400. If firm i sets an higher price than firm j ̸= i by deviating to p′^ = 100 − 2 K = 20, it obtains profit πi(p′, p) = (p′^ − 10) min{ 100 − K − p′, 40 } = 400, which is profitable.
Case (b): consider a candidate equilibrium (p 1 , p 2 ) = (p, p) with p = 100 − 2 K = 20. Firm i’s profits are πi(p, p) = (p − 10)40 = 400. If firm i sets an higher price p′^ = 20 + ϵ with ϵ > 0 it obtains profits π(p + ϵ, p) = (p + ϵ − 10) min{ 100 − K − p − ϵ, K} = (20 + ϵ − 10) min{ 60 − 20 − ϵ, 40 } = (10 + ϵ)(40 − ϵ), which is higher than 400 for some ϵ > 0.
Case (c): consider a candidate equilibrium (p 1 , p 2 ) = (p, p) with 20 = 100 − 2 K < p ≤ 100 − K = 60. Firm i’s profits are πi(p, p) = (p − 10)(100 − p)/2. If firm i undercuts firm j ̸= i by deviating to p − ϵ with ϵ > 0, it obtains profit πi(p − ϵ, p) = (p − ϵ − 10) min{ 100 − p + ϵ, 40 } = (p − ϵ − 10)40. The deviation is profitable πi(p − ϵ, p) > πi(p, p) if (p − ϵ − 10)40 > (p − 10)(100 − p)/2. That is if
(p − ϵ − 10)80 > (p − 10)(100 − p) → ϵ 80 < p^2 − 30 p + 200 → ϵ < (p^ −^ 20)( 80 p −^ 10),
where the righthand side is positive since we are restricted to 20 < p ≤ 60.
Case (d): consider a candidate equilibrium (p 1 , p 2 ) = (p, p) with 60 = 100 − K < p < 100. Firm i’s profits are πi(p, p) = (p − 10)(100 − p)/2. If firm i undercuts firm j ̸= i by deviating to p − ϵ with ϵ ∈ (0, 60 − p), it obtains profit πi(p − ϵ, p) = (p − ϵ − 10) min{ 100 − p + ϵ, 40 } = (p − ϵ − 10)(100 − p + ϵ). Note that there is a profitable deviation since πi(p − ϵ, p) > πi(p, p) for some ϵ ∈ (0, 60 − p).
Therefore, there is no pure strategy Nash equilibrium.
Question (ii): Suppose K = 30. Show that there is a pure strategy Nash equilibrium.
By the same argument used to answer (i), any pure strategy Nash equilibrium must have p 1 = p 2.
Case (a) consider a candidate equilibrium (p 1 , p 2 ) = (p, p) with 10 ≤ p < 40. Firm i’s profits are πi(p, p) = (p − 10)30 < 900. If firm i sets an higher price thatn firm j ̸= i by deviating to p′^ = 40, it obtains profit πi(40, p) = (40 − 10) min{ 100 − 30 − 40 , 30 } = 900, which is profitable.
Case (b): consider a candidate equilibrium (p 1 , p 2 ) = (p, p) with p = 40. Firm i’s profits are πi(p, p) = (p − 10)30 = 900. If firm i sets an higher price p′^ = 40 + ϵ with ϵ > 0 it obtains profits
π(p + ϵ, p) = (40 + ϵ − 10) min{ 70 − 40 − ϵ, 30 } = (30 + ϵ)(30 − ϵ),
which is smaller than 900 for all ϵ > 0. Deviating to a lower price cannot be profitable for {p ≤ 40 } because firms are selling at capacity, and it would only reduce their margin. This implies (p 1 , p 2 ) = (40, 40) is a Nash equilibrium.
Exercise 6. Consider an electoral contest between two political parties which have to announce (si- multaneously) party positions in the policy space given by the interval [0, 1]. Voters’ ideal points are uniformly distributed over the interval [0, 1]. Each voter votes for the party whose announced position is closest to his ideal point. In case the two parties’ announced point is equidistant from a voter’s ideal point, then the voter votes for each party with probability 1/2. Each party wants to maximize its vote share.
Question (i): Show there is a unique pure strategy Nash equilibrium in which each party locates at 1/2.
Answer: Let i denote a player in { 1 , 2 } and denote its opponent by j = 2 − i. Then, let si ∈ [0, 1] be a pure strategy of player i and sj be a pure strategy of player i’s opponent.
Consider a pair of strategies (si, sj ) with si = sj < 1 /2. Note that ui(si, sj ) = 1/2. Consider a devia- tion of player i to sj + 2ε, where ε > 0. The deviation is profitable if ui(sj + 2ε, sj ) = 1 − sj − ε > 1 /2; that is, for any ε ∈ (0, 1 / 2 − sj ). Note that (0, 1 / 2 − sj ) is not empty, implying there is no equilibrium with si = sj < 1 /2. The case where si = sj > 1 /2 is symmetric (check deviation of player i to sj − 2 ε).
Consider a pair of strategies (si, sj ) with si < sj. This cannot be an equilibrium because i could deviate to (si + sj )/2 obtaining utility ui((si + sj )/ 2 , sj ) = (3sj + si)/ 4 > (sj + si)/2 = ui(si, sj ).
So, neither si < sj , nor si = sj ̸= 1/2 can be supported in equilibrium. Thus remains si = sj = 1/2, in which case any deviation of player i (sj ± ε) is not profitable.
Question (ii): Suppose there are three parties. Show that there is no pure strategy Nash equilibrium.
Answer: Consider a strategy profile a = (x, x, x) where x ∈ [0, 1]. Note that ui(a) = 1/3 for all i ∈ { 1 , 2 , 3 }. If x ≤ 1 /2, player i can deviate to x + 2ε and obtain ui(x + 2ε, x, x) = 1 − x − ε which is greater than 1/3 for all ε ∈ (0, 2 / 3 − x). If x > 1 /2 player i can deviate to x − 2 ε and obtain ui(x + 2ε, x, x) = x + ε which is greater than 1/3 for all ε ∈ (0, x − 1 /3). This concludes there are no symmetric pure strategy Nash equilibria.
Consider a strategy profile a = (al, ac, ar) where al < ac ≤ ar and {l, c, r} = { 1 , 2 , 3 }. If player l were to deviate to (ac + al)/2 it would obtain (3ac + al)/ 4 > (ac + al)/2 = u 1 (a). Note that a symmetric argument holds for al ≤ ac < ar where player r could deviate to (ac + ar)/2. This, joint with the previous paragraph, implies that there are no pure strategy Nash equilibria.
Exercise 7. Each of n people chooses whether or not to become a political candidate, and if so which position to take in [0, 1]. There is a continuum of citizens, each of whom has a favourite position in [0, 1], the distribution of favourite positions is given by a density function. Each citizen votes for the candidate closest to her ideal point. If k candidates choose the same position, then each gets the fraction 1/k of the votes that the position attracts. The winner of the competition is the candidate who receives the most votes. Each candidate’s preference is the following:
- she prefers to be the unique winner than to tie for first place
- she prefers to tie for first place than to stay out of the competition
- she prefers to stay out of the competition than to enter and lose
Question (i): Formulate this as a normal form game.
Answer: Assuming players cannot use mixed strategies (we have ordinal preferences) the normal form representation of the game is ⟨N, (Si)i∈N , (ui)i∈N ⟩ where N = { 1 ,... , n}, Si = { 0 , 1 } × [0, 1] and ui : ×i∈N Si → { 0 , 1 , 2 , 3 } with the utility numbers appropriately assigned given the description of the players’ preferences. It does not help toward the solution to write the utility functions analytically.
Question (ii): Find the set of Nash equilibria when n = 2.
Answer: Note that there are no equilibria where a player loses the election as the losing candidate would rather stay out. Note that there are no equilibria where both players stay out, as one player could become a candidate and be the unique winner. There are also no equilibria where one player is a candidate and the other is not, as the non-candidate player could become a candidate with the same position as the candidate-player and win the election in a tie. This implies that all equilibria have both players as winning candidates. Let F be the cumulative density function of the citizens’ preferences and F −^1 (y) = {x ∈ [0, 1] : F (x) = y} be the preimage of y under F. Then F −^1 (1/2) is the set of median positions. By an argument similar to the previous exercise, the set of pure strategy Nash equilibria is {((1, x 1 ), (1, x 2 )) : x 1 , x 2 ∈ F −^1 (1/2)}.
Question (iii): Show that there is no Nash equilibrium when n = 3.
Answer: There are no equilibria just in case F −^1 (1/2) is a singleton, i.e., there is a unique median voter (f (x) > 0 , ∀x ∈ [0, 1] is a sufficient condition). In that case, there are no equilibria where a player loses the election as the losing candidate would rather stay out. There are no equilibria where all players stay out, as one player could become a candidate and be the unique winner. There are also no equilibria where some players are candidates and some are not, as the non-candidate player could become a candidate with the same position as the winning-candidate(s)-player and win the election in a tie. This implies that equilibria must have all players as winning candidates. By reasoning similar to the previous exercise, there is no such pure strategy Nash equilibrium.
If F −^1 (1/2) is a closed interval rather than a singleton, there exist equilibria where two players become candidates at positions min F −^1 (1/2) and max F −^1 (1/2) and the remaining candidate stays out. Note that the two candidates must take positions at the extrema of F −^1 (1/2) otherwise the non-candidate player could become a candidate (either slightly to the left of the leftmost candidate or the right of the rightmost candidate) and be the unique winner of the election. When the two candidates take positions at the extrema of F −^1 (1/2), the non-candidate player cannot become a candidate and win the election (not even with a tie).
