Game Theory with Applications to Finance and
Marketing
Lecture 1: Games with Complete Information, Part I Professor Chyi-Mei Chen, R1102
(TEL) x2964
(email) cchen@ccms.ntu.edu.tw
1. Definition 1: A game can be described by (i) the set of players, (ii) the strategies of players, and (iii) the payoffs of players as functions of their strategies. Players are assumed to maximize their payoffs and payoffs are generally interpreted as von Neumann-Morgenstern utilities. Players are rational in the sense that they will choose strategies to maximize their (expected) payoffs. A game described in this way is called a game in normal form, or a strategic game. (A game can also be described in extensive form; see below. These definitions are given by von Neumann and Morgenstern (1944).) An event is the players’ mutual knowledge if all players know the event, and an event is called the players’ common knowledge if all players know it, all players know that they all know it, all players know that they all know that they know it, and so on and so forth. If the norm form of a game (and the rationality of all players) is the players’ common knowledge, then the game is one with complete information.
Example 1: Consider the Cournot game in normal form. The set of players: two firms, i = 1, 2. The strategies: the two firms’ output quantities: qi ∈ Si ≡ [0, ∞), i = 1, 2. The payoff of each player i:
πi(qi, qj) = qi[P (qi+ qj) − c] − F,
where P (·) is the inverse demand function, c and F are respectively the variable and fixed costs.
Example 2: A two-player game with a finite number of actions (strate-gies) is usually represented by a bimatrix:
player 1/player 2 L R
U 0,1 -1,2
where there are two players in the game, who simultaneously choose actions, and action profiles (U,L), (U,R), (D,L) and (D,R) result in respectively payoff profiles (0, 1), (−1, 2), (2, −1) and (−2, −2) for the two players (where by convention the first coordinate in a payoff profile stands for player 1’s payoff).
2. Our purpose of learning the non-cooperatitve game theory is essen-tially practical. In application, we shall first describe an economical or business problem as a game in norm form (or more often in extensive form; see below), and then proceed to solve the game so as to generate useful predictions about what the major players involved in the original economical or business problem may do. For this purpose, we need to adopt certain equilibrium concepts or solution concepts. In the remain-der of this note we shall review the following solution concepts (and illustrate how those solutions may be obtained by considering a series of examples):
• Rational players will never adopt strictly dominated strategies; • Common knowledge about each player’s rationality implies that
rational players will never adopt iterated strictly dominated strate-gies;
• Rational players will never adopt weakly dominated strategies; • Common knowledge about each player’s rationality implies that
rational players will never adopt iterated weakly dominated strate-gies;
• Rational players will never adopt strategies that are never best responses, or equivalently, rational players will adopt only ratio-nalizable strategies;
• Rational players will adopt Nash equilibrium strategies;
• Rational players will adopt trembling-hand perfect equilibrium strate-gies;
• Rational players will adopt subgame-perfect Nash equilibrium strate-gies;
• Rational players will adopt proper equilibrium strategies; • Rational players will adopt strong equilibrium strategies;
• Rational players will adopt coalition-proof strategies.
In practice, the solution concept of Nash equilibrium is most widely accepted. In rare cases only will we turn to other equilibrium concepts. 3. Definition 2: A pure strategy is one like U, R, D, or L in example 2. It has to be a complete description of a player’s actions taken throughout the game. A mixed strategy is a prob. distribution over the set of pure strategies. Immediately, a pure strategy is a mixed strategy.
4. Definition 3: Consider a game in normal form,
G= (I ⊂ <; {Si; i ∈ I}; {ui : Πi∈ISi → <; i ∈ I}),
where I is the set of players (we are allowing an uncountably infinite number of players here), Si is the set of pure strategies feasible to
player i (also known as the pure strategy space of player i), and ui(·)
is player i’s payoff as a function of the strategy profile. If I and Si are
finite for all i ∈ I, then we call G a finite game. We shall represent Πi∈ISi by S (the set of all possible pure strategy profiles). If Si has
cardinality m, then the set of feasible mixed strategies for player i, denoted Σi, is a simplex of dimension m − 1. Each element of Σi is a
probability distribution over the set Si. We denote a generic element
of Si, Σi, S, and Σ ≡ Πi∈IΣi by respectively si, σi, s, and σ. Since
σi is a probability distribution over Si, we let σi(si) be the probability
assigned by σi to the pure strategy si. Note that being consistent with
the notion of non-cooperative games, the mixed strategies of players are un-correlated. More precisely, given a mixed strategy profile σ, player i’s payoff is ui(σ) ≡ X s∈S [ΠI j=1σj(sj)]ui(s),
where note that we have abused the notation a little to let ui(σ) denote
the expected value of ui(s) under the joint probability distribution σ.
5. Definition 4: Let σ−i be some element of
Σ−i ≡ Σ1× Σ2× · · · Σi−1 × Σi+1× · · · ΣI,
where I is the cardinality of I. Let ri(σ−i) ∈ Σi be the set of best
ui(σi, σ−i) for all σi ∈ Σi). A (mixed) strategy σi is weakly dominated
by σ0
i for player i if for some σi0 ∈ Σi,
ui(σi, σ−i) ≤ ui(σ0i, σ−i), ∀σ−i ∈ Σ−i,
and σi is strictly dominated if the above inequalities are all strict. Our
first equilibrium concept is that rational players will not use strictly dominated strategies. Consider the following normal form game, known as the prisoner’s dilemma:
player 1/player 2 Don’t Confess Confess
Don’t Confess 0,0 -3,1
Confess 1,-3 -2,-2
In this game, “Don’t Confess” is strictly dominated by “Confess,” and hence the unique undominated outcome is the one where both players confess the crime. Note that unlike Walrasian equilibrium, this equi-librium is Pareto inefficient. That equilibria in a game are generally inefficient is the first lesson to be learned here (and all kinds of economic theory starts from here).
6. Based on the implicit assumption that the whole normal form game is the players’ common knowledge, the above dominance argument can be extended further so that we shall be looking at outcomes that survive from the procedure of iterative deletion of strictly dominated strategies. The following is an example.
player 1/player 2 L M R
U 0,-1 0,0 1,1
M 2,3 3,1 32,-1
D 4,2 1,1 2, 3
2
Note that M is not strictly dominated by L from player 2’s perspective, but since U is strictly dominated by M from player 1’s perspective, and in the absence of U, M is strictly dominated by L from player 2’s perspective, we should not expect player 2 to use M. It follows that player 2 will use L and hence player 1 will use D.
Observe that we have repeatedly used the assumption that rationality of all players is the players’ common knowledge. For example, if player 2 is not sure about player 1’s being rational, then player 2 may not want to ignore strategy M.
7. Mixed strategies that assign strictly positive probabilities to strictly dominated pure strategies are themselves strictly dominated. A mixed strategy that assigns strictly positive probabilities only to pure strate-gies which are not even weakly dominated may still be strictly domi-nated. Consider also the following example:
player 1/player 2 L R
U 1,3 -2,0
M -2,0 1,3
D 0,1 0,1
The mixed strategy (0.5, 0.5, 0) for player 1 is strictly dominated by D. 8. It can be shown that iterated deletion of strictly dominated strategies will lead to the set of surviving outcomes which is independent of the order of deletion. The same is not true for iterated deletion of weakly dominated strategies. Consider the following normal form game: Example 3:
player 1/player 2 L R
U 1,1 0,0
M 1,1 2,1
D 0,0 2,1
If we delete U and then L, then we conclude that the payoff profile would be (2, 1). If we delete D and then R, then we conclude that the payoff profile would be (1, 1).
9. Definition 5: A normal form game is dominance solvable if all players are indifferent between all outcomes that survive the iterative procedure where all the weakly dominated actions of each player are eliminated
simultaneously at each stage. Is the game in example 3 dominance solvable?1
10. Consider the following two-player game. They each announce a natural number ai not exceeding 100 at the same time. If a1+ a2 ≤ 100, then
player i gets ai; if a1+ a2 >100 with ai < aj, then players i and j get
respectively ai and 100 − ai; and if a1 + a2 > 100 with ai = aj, then
each player gets 50. Determine if this game is dominance solvable. 11. Consider example 1 with the further specifications that c = F = 0,
P(q1+ q2) = 1 − q1− q2. Like the game of prisoner’s dilemma, there is
a unique outcome surviving from the iterated deletion of strictly dom-inated strategies. We say that this game has a dominance equilibrium. To see this, note that the best response for firm i as a function of qj
(called firm i’s reaction function) is
qi = ri(qj) =
1 − qj
2 .
Note that q1 > 1−02 = r1(0) is strictly dominated for firm 1.2 Since
rationality of firm 1 is the two firms’ common knowledge, firm 2 will re-alize that any q2 <
1−1 2
2 = r2(r1(0)) is strictly dominated by r2(r1(0)).3
1Definition 5 can be found in Osborne and Rubinstein’s Game Theory. A related
definition is the following: A game is solvable by iterated strict dominance, if the iterated deletion of strictly dominated strategies leads eventually to a unique undominated strategy profile; see Fudenberg and Tirole’s Game Theory, Definition 2.2.
2
Observe that (i) given any qj ∈ <+, Πi(·, qj) is a strictly decreasing function on
[ri(qj), +∞); and (ii) ri(·) is a strictly decreasing function on <+. Thus if qi> 1−02 = ri(0),
then for all qj ∈ <+, we have
qi> ri(0) ≥ ri(qj) ⇒ Πi(ri(0), qj) > Πi(qi, qj),
proving that any such qi is strictly dominated by ri(0), regardless of firm j’s choice qj. 3Note that Π
2(q1,·) is a strictly increasing function on [0, r2(q1)]. Since by common
knowledge about firm 1’s rationality, firm 2 believes that any q1 chosen by firm 1 must
satisfy q1≤ r1(0), we must have
q2< r2(r1(0)) ≤ r2(q1),
and hence
But then, common knowledge about rationality again implies that any q1 >
1−1 4
2 is strictly dominated for firm 1, which in turn implies that
q2 < 1−3
8
2 is strictly dominated for firm 2. Repeating the above
argu-ment, we conclude that firm 1 will find any q1 > 1−5
16
2 strictly
domi-nated. Thus we obtain a sequence qn 1 = 4n −(1+4+42 +···+4n −1) 2×4n of
undom-inated output levels at the n-th stage of deleting strictly domundom-inated strategies. Since the sequence converges uniquely to 1
3, we conclude
that the only output level surviving from the iterated dominance pro-cess is 13 for both firms. Thus, (q∗
1, q2∗) = (13, 1
3) is the (unique)
domi-nance equilibrium.
12. (Market Share Competition)
player 1/player 2 Don’t Promote Promote
Don’t Promote 1,1 0,2-c
Promote 2-c,0 1-c,1-c
In the above, two firms can spend c > 0 on promotion. One firm would capture the entire market by spending c if the other firm does not do so. Show that when c is sufficiently small, this game has a dominance equilibrium.
13. Consider also the following moral-hazard-in-team problem: Two work-ers can either work (s=1) or shirk (s=0). They share the output 4(s1+ s2) equally. Working however incurs a private cost of 3. Show
that this game has a dominance equilibrium.
14. Definition 6: A pure strategy Nash equilibrium (NE) for a game in normal form is a set of pure strategies (called a pure strategy profile from now on), one for each player, such that if other players play their specified strategies, a player also finds it optimal to play his specified strategy. In this case, these strategies are equilibrium strategies. For instance, (U,R) and (D,L) are two pure strategy NEs for the game
proving that any such q2 is strictly dominated by r2(r1(0)) from firm 2’s perspective, no
in example 2. A mixed strategy NE is defined similarly, where the equilibrium strategies are mixed strategies.4
More formally, a profile σ ∈ Σ is a (mixed strategy) NE if and only if ui(σ) ≥ ui(si, σ−i), ∀i ∈ I, ∀si ∈ Si.
Observe that this definition asks us to check unilateral deviations in pure strategy only! Show that, however, it implies that
ui(σ) ≥ ui(σ0i, σ−i), ∀i ∈ I, ∀σi0 ∈ Σi.
Show that the following is an equivalent definition for (mixed strategy) NE: a profile σ ∈ Σ is an NE if and only if for all i ∈ I, for all si, s0i ∈ Si,
ui(si, σ−i) < ui(s0i, σ−i) ⇒ σi(si) = 0.
This new equivalence condition says that if a pure strategy siis assigned
by σi with a strictly positive probability, then si has to be a best
response against σ−i.
15. A single auction consists of a seller facing more than one buyer (or a buyer facing more than one seller) with some object(s) to sell (respec-tively, to buy), where the multiple buyers (respec(respec-tively, the multiple sellers) submit bids to compete for the object(s). The object is of private value if learning other bidders’ valuations for the object does not change one’s own valuation, and it is of common value if bidders have an (unknown) identical valuation. The four popular auction rules are English auction, Dutch auction, the first-price sealed-bid auction, and the second-price sealed-bid auction. With an English auction, the seller-auctioneer will start with a minimum acceptable price and raise the price gradually to attain a situation where there is exactly one bidder remains interested in the object, and in the latter situation the object is rendered to that bidder in exchange for the price at which the
4The following interpretation may be helpful. In example 2, if player 2 expects player
1 to play U, then his best response is to play R, and if player 1 correctly expects that the above is player 2’s belief, then he cannot gain by picking an action that proves that player 2’s belief is incorrect. An NE is by definition a situation where, if the two players expect their strategy profile, they cannot gain by making unilateral deviations.
winning bidder is selected. With a Dutch auction, the seller-auctioneer will start with a very high price and then reduce the price gradually to attain a situation where at least one bidder shows interest in the object at that price, and in the latter situation the object is sold to the first bidder who reveals his interest. In the two sealed-bid auctions mentioned above, all bidders must submit secretly their bids at the same time to an auctioneer, and the object is delivered to the bidder submitting the highest bid. The difference between the two sealed-bid auctions is that in the first-price auction, the winnder bidder must pay his own bid, but in the second-price auction, the winning bidder will pay the highest losing bid. See Milgrom and Weber (1982).
Now suppose that an indivisible object will be sold to one of the N bidders whose private valuations for the object are respectively v1, v2,· · · , vN. Show that a second-price sealed-bid auction with
pri-vate values for an indivisible object has an NE where all bidders bid their valuations for the object. (Hint: Consider bidder 1’s problem. Let the random variable ˜B denote the maximum of the bids submitted by bidders 2, 3, · · · , N . Compare any bid b submitted by bidder 1 with the bid v1 we suggest. Note that with the bidding strategy b, bidder
1’s expected payoff is
E[(v1− ˜B)1[b> ˜B]],
where (i) we have ignored the probability of a tie, and (ii) 1A is an
indicator function for event A, so that it equals one if A happens and zero if otherwise. Show that b and v1 make a difference only when
v1 > ˜B > b or v1 < ˜B < b (again ignoring the probability of a tie).
Deduce that b is weakly dominated by v1.)
16. Consider N bidders competing for one indivisible object, for which they attach values v1 > v2 > · · · > vN respectively. Show that if the seller
adopts a first-price sealed-bid auction, then in all NE’s for this game, player 1 gets the object.
17. A game is called a zero-sum game ifP
i∈Iui(σ) is independent of σ ∈ Σ.5
5Thus we might better call it a constant-sum game. However, recall that a von
Neumann-Mogenstern utility function is determined only up to a positive affine trans-form (see my note in Investments, Lecture 2, Part I), and hence deducting the payoffs by the same number really changes nothing relevant.
A zero-sum game may have multiple NE’s, but players obtain the same payoff profile in all the NE’s. To see this, suppose that there are two players, and that σ and σ0 are both NE’s for the zero-sum game. It
must be that u1(σ1, σ2) ≥ u1(σ10, σ2) and u2(σ10, σ 0 2) ≥ u2(σ10, σ2) so that u1(σ1, σ2) ≥ u1(σ10, σ02).
The same argument shows that
u1(σ1, σ2) ≤ u1(σ10, σ20)
and hence we conclude that
u1(σ1, σ2) = u1(σ10, σ20)
implying that
u2(σ1, σ2) = u2(σ10, σ20)
as well.
18. Unlike dominance equilibrium, which may not exist for a game, a Nash equilibrium always exists for a finite normal form game. This is stated in theorems 1 and 2 below.
Theorem 1(Wilson, 1971; Harsanyi, 1973): Almost every finite strate-gic game has an odd number of NE’s in mixed strategies; more precisely, with the set of players and their strategy spaces fixed, the set of pay-off functions that result in a strategic game having an even number of NE’s has zero Lebesgue measure.
Following theorem 1, we guess that there is at least one mixed strategy NE for the game in example 2. Let us find them. The key observation here is that, if a mixed strategy assigns strictly positive probabilities to more than one pure strategy, then the player must feel indifferent
about these pure strategies. Let x be the prob. that player 1 uses U and y the prob. that player 2 uses L. We must have
1 · x + (−1) · (1 − x) = 2 · x + (−2) · (1 − x) ⇒ x = 1 2;
0 · y + (−1) · (1 − y) = 2 · y + (−2) · (1 − y) ⇒ y = 1 3.
These mixed strategies are said to be totally mixed, in the sense that they assign strictly positive prob.’s to each and every pure strategy. Here we have only one mixed strategy NE. If we have two totally mixed NE’s, then we naturally have a continuum of mixed strategy NE’s (why?). From now on, we denote the set of totally mixed strate-gies of player i by Σ0
i. Observe that Σ0i is simply the interior of Σi,
when Σi is endowed with the usual Euclidean topology.
19. (Matching Pennies) Note that some games do not have pure strategy Nash equilibrium:
player 1/player 2 H T
H 1,-1 -1,1
T -1,1 1,-1
20. Theorem 2 (Nash, 1950): Every finite game in normal form has a mixed strategy equilibrium.
Theorem 2 is actually a special version of the following more general theorem (see Fudenberg and Tirole’s Game Theory, Theorem 1.2). Theorem 20: (Debreu-Glicksberg-Fan) Consider a strategic game
G= {I, S = (S1, S2,· · · , SI), (ui : S → <; i ∈ I)}
with I being a finite set (the game has a finite number of players). If for all i ∈ I, Si is a nonempty compact convex subset of some Euclidean
space, and if for all i ∈ I, ui is continuous in s and quasi-concave in si
when given s−i, then G has an NE in pure strategy.
21. Suppose that r : Σ → Σ0 is a correspondence (a multi-valued function),
we say that r(·) is non-empty. If r(σ) is a convex subset of <n for all
σ ∈ Σ, then we say that r(·) is convex. If for all σ ∈ Σ, for all sequences {σn; n ∈ Z+} in Σ converging to σ, each sequence {σn0 ∈ r(σn); n ∈ Z+}
has a convergent subsequence converging to some element in r(σ), then we say that r(·) is upper hemi-continuous (u.h.c.). If for all σ ∈ Σ and for all σ0 ∈ r(σ), whenever {σ
n; n ∈ Z+} is a sequence in Σ converging
to σ there must be a sequence {σ0
n; n ∈ Z+} in Σ0 converging to σ0 such
that σ0
n ∈ r(σn), then we say that r(·) is lower hemi-continuous (l.h.c.).
The correspondence r(·) is said to be continuous if it is both upper and lower hemi-continuous.
22. Here we give some examples of upper hemi-continuous correspondences. First consider the following strategic game
player 1/player 2 L R
U 1,1 0,0
D 0,0 z,2
Let us consider all mixed strategy NE’s of this game. Any mixed strat-egy NE can be represented by (p, q), where p is the probability that player 1 adopts U and q the probability that player 2 adopts L. Simple calculations give the following results. For z > 0, the game has three NE’s, (p, q) = (1, 1), (p, q) = (0, 0), and (p, q) = (23,1+zz ); for z < 0, iterated deletion of strictly dominated strategies implies that the game has a unique NE, which is (p, q) = (1, 1); and for z = 0, the game has an (uncountably) infinite number of NE’s: besides (1, 1), (0, 0), and (2
3,0), which are obtained from the above three NE’s by letting z ↓ 0,
any (p, 0) with 0 ≤ p ≤ 23 are also NE’s. The idea is that as long as player 2 is willing to play R with probability one, given z = 0, player 1 is genuinely indifferent about U and D, and hence can randomize with any probability p, and for p ≤ 23, player 1 really prefers R to L.
Apparently, player 1’s equilibrium mixed strategy p depends on z (which is the sole parameter here to distinguish one strategic game from an-other). This dependence defines a correspondence p(z), called the Nash equilibrium correspondence. This correspondence is easily seen to be upper hemi-continuous (and if you draw the graph of p(·), then you
will realize why we also refer to upper hemi-continuity by the closed graphproperty), but it fails to be lower hemi-continuous. Observe that any p ∈ (0,23) is contained in p(z), but it is not a limit of a sequence of NE’s corresponding to some zn → 0. Note that except for the case
where z = 0 (which is an event of zero Lebesgue measure on <), this game has an odd number of NE’s, verifying Wilson’s theorem.
Next, consider the following strategic game
player 1/player 2 L R
U 1,1 0,0
D 0,0 z,z
Again, any mixed strategy NE can be represented by (p, q), where p is the probability that player 1 adopts U and q the probability that player 2 adopts L. Simple calculations give the following results. For z > 0, the game has three NE’s, (p, q) = (1, 1), (p, q) = (0, 0), and (p, q) = ( z
1+z, z
1+z); for z < 0, iterated deletion of strictly dominated strategies
implies that the game has a unique NE, which is (p, q) = (1, 1); and for z = 0, the game has two NE’s: (1, 1) and (0, 0). Again, consider the correspondence p(z). It is still upper hemi-continuous, although the number of NE’s drops in the limit as z ↓ 0. Note also that this game has an odd number of NE’s except when z = 0.
23. Theorem 3 (Kakutani, 1941): Let r : Σ → Σ be a nonempty, convex, upper hemi-continuous correspondence where Σ ⊂ <n is nonempty,
convex and compact. Then, there exists σ∗ ∈ Σ such that r(σ∗) = σ∗.
24. We now sketch the proof of theorem 2 using theorem 3. Let Si and
Σi be respectively player i’s pure strategy space and mixed strategy
space. Assume there are n players. Then, Σi is a simplex which is
compact. Denote Σ = Πn
i=1Σi. Let σi be a typical element of Σi,
and let σ = (σ1, σ2,· · · , σn) and σ−i = (σ1, σ2,· · · , σi−1, σi+1,· · · , σn).
Let ri(σ−i) be player i’s best response against other players’ mixed
strategies σ−i, and define r(σ) as the Cartesian product of ri(σ−i).
Now r : Σ → Σ is a correspondence defined on Σ! If we can show the existence of a fixed point of r, then we are done.
Since players’ payoff functions are linear (hence continuous) in mixed strategies, and since the simplex Σi is compact, there exist solutions
to players’ best response problems (Weierstrass theorem). That is, r is nonempty. Moreover, if σi1, σi2 ∈ ri(σ−i), then clearly for all λ ∈ [0, 1],
λσi1+ (1 − λ)σi2 ∈ ri(σ−i),
proving that riis convex, which in turn implies that r is convex. Finally,
we claim that r is u.h.c. To see this, suppose instead that r fails to be upper hemi-continuous at some profile σ ∈ Σ, then there must exist some player i, a sequence {σ−i,n; n ∈ Z+} in Σ−i, and a sequence
{σ0
i,n; n ∈ Z+} in Σi converging to some σ0 ∈ Σi with σi,n0 ∈ ri(σ−i,n) for
all n ∈ Z+ such that σ0i ∈ ri(σi)c. This would mean that there exists
some σ00
i ∈ Σi such that ui(σi00) > ui(σi0). Since ui(·) is continuous, and
since the sequence {σ0
i,n; n ∈ Z+} converges to σ0, this would mean that
ui(σi00) > ui(σi,n0 ) for n sufficiently large, a contradiction to the fact that
σ0
i,n∈ ri(σ−i,n) for all n ∈ Z+.
Now since r(·) is empty, convex, and u.h.c., and since Σ is non-empty, convex, and compact in <n, by theorem 3 there must exists
some σ∗ ∈ Σ such that r(σ∗) = σ∗.
25. From the proof of theorem 2, it is understandable why discontinuous payoff functions may result in non-existence of NE’s. When payoff functions fail to be continuous, it may happen that the best response correspondence is not non-empty, or fails to be upper hemi-continuous. For example, consider the single-player game where his payoff function f : < → < is defined as f (x) = −|x| · 1[x6=0]− 1[x=0], then the player
has no best response. Even if all best responses are well defined, the best response correspondence my fail to be upper hemi-continuous, as the following example shows. Suppose we have a two-player strategic game where S1 = S2 = [0, 1], u1(s) = −(s1− s2)2, and
u2(s) = −(s1− s2− 1 3) 2· 1 [s1≥13]− (s1− s2+ 1 3) 2· 1 [s1<13].
Observe that in general lims1↓13 u2(s) 6= lims1↑13 u2(s), proving that u2
is not continuous in s. However, r2(·) is a well-defined correspondence;
(single-valued) function. Still, r2(·) is not upper hemi-continuous (when
the correspondence is single-valued, upper hemi-continuity becomes the continuity of the single-valued function). You can check that r2(s1) and
r1(s2) do not intersect, and hence this two-player game has no NE at
all.
26. Example 1 (see the section on dominance equilibrium) can be used to understand the above theorems 1 and 2. Define hi(·) ≡ ri(rj(·)).
Apparently, this function can be restricted to the domain of defintion [0,12], which is a non-empty, convex, compact subset of <, and more-over, the functional value of hiis also contained in [0,12]. If this function
intersects with the 45-degree line, then the intersection defines a pure strategy NE. (This game has no mixed strategy NE because a firm’s profit is a strictly concave function of his own output level given any output choice of its rival.) Now it follows from Brouwer’s fixed point theorem (a special version of theorem 3) that if hi(·) is continuous
then there exists an NE for the game, and moreover, as one can verify, generically a continuous hi(·) will intersect the 45-degree line in an odd
number of times.6
27. Theorem 4: Consider a two-player game where each player i simul-taneously chooses a number si in the unit interval, and where player
i’s payoff function ui(s1, s2) is continuous in (s1, s2). This game has a
mixed strategy NE.
Proof Call this game Γ. Consider a sequence of modified games {Γn}
in normal form where the set of players and the payoff functions are the same as in Γ, but the players’ common pure strategy space is
Sn = {0, 1 n, 2 n,· · · , n− 1 n ,1}.
Theorem 2 implies that for all n ∈ Z+, Γn has an NE (denoted σn)
in mixed strategy. Since the set of probability measures on [0, 1] is weakly compact (see my note on “distribution functions” in the course of stochastic processes), and the products of compact spaces are com-pact (see my note on “basic ideas in topology” in the course of stochas-tic processes), the sequence {σn} has a convergent subsequence in weak
6
Brouwer’s fixed point theorem says that if f : A → A is continuous, where A ⊂ <n is
topology, for which let σ be the limiting probability measure. Let the convergent subsequence be {σnk; k ∈ {Z}
+}. We claim that σ is a
mixed strategy NE for Γ. The proof proceeds in two steps. First, every prob. measure on [0, 1] is the limit (in weak topology) of a sequence of prob. measures of which each term is a prob. measure on Snk (recall
Helly’s convergence theorem). Second, let σ0
i be any prob. measure on
[0, 1], then by step 1, there exists a sequence {[σnk
i ]0} that converges
weakly to it. Fix i ∈ {1, 2}. For each k ∈ Z+, we have by definition,
ui(σn k i , σ nk j ) − ui([σn k i ] 0 , σnk j ) ≥ 0,
which, by the continuity of ui in (s1, s2) and the definition of weak
convergence of prob. measures on [0, 1], implies that ui(σi, σj) − ui(σi0, σj) ≥ 0,
proving that σ is an NE in mixed strategy for Γ. 28. Theorem 4 is a special case of the following theorem:
Theorem 40: (Glicksberg, 1952) Consider a strategic game G= {I, S = (S1, S2,· · · , SI), (ui : S → <; i ∈ I)}
with I being a finite set (the game has a finite number of players). If for all i ∈ I, Si is a nonempty compact subset of some (common)
metric space M , and if for all i ∈ I, ui is continuous, then G has an
NE in mixed strategy.
29. A strategic game is symmetric if ui = uj and Si = Sj for all i, j ∈ I.
Theorem 5: A finite symmetric game has a symmetric Nash equilib-rium in mixed strategy.
Proof Suppose that in an I-person finite game, Si = S1 and ui(·) =
u1(·) for all i = 1, 2, · · · , I. This implies that Σi = Σ1 for all i. We
show this game has a symmetric NE. For any profile σ, we write ui(σ) = ui(σi, σ−i).
For any σ1 ∈ Σ1, define
Ri(σ1) = arg max σi∈Σ1
subject to
σ−i = σ1I−1.
It is clear that Ri(·) is independent of i. Since Σ1 and Ri : Σ1 →
Σ1 satisfy all the requirements for Kakutani’s theorem to apply, we
conclude that for some σ1 ∈ Σ1, σ1 ∈ Ri(σ1), where the profile σ1I is
obviously an NE.
30. Definition 7: A game can be described in extensive form, which needs to specify the timing of players’ moves and what they know when they move, in addition to the things specified in normal form. A game in extensive form is also known as an extensive game. Usually, a game in extensive form is depicted as a game tree:
(I)— U p— (II)— Right— (0, 1) Lef t— (−1, 2) ... Down— (II)— Right— (2, −1) Lef t— (−2, −2)
Note that player II’s information set is denoted by the two nodes con-nected by a dotted line, which says that player II, when determining her own moves, does not know whether player I has moved up or down. Formally, an information set is a set of decision nodes for a player, who cannot tell which node in the information set he is currently on. A game where at least one player has incomplete knowledge about the history of the game up to the point when he or she is ready to move is called a game with imperfect imformation.
31. Because the extensive form specifies two more things than the normal form, there does not exist a one-to-one correspondence between the two. In fact, a normal form can correspond to more than one extensive form. Consider the following extensive game. Player 1 first chooses among pure strategies A, B, and C. If A is chosen, then player 1 gets 2 and player 2 gets 1. Otherwise, upon seeing A’s choice of action, B can react by choosing either L or R. The payoffs resulting from these
strategies are summarized in the following bimatrix.
player 1/player 2 L R
B 4,2 1,0
C 0,1 32,0
Draw the extensive form for this game.
32. So, how do we construct the strategic form from its extensive coun-terpart? The complication is that we need to define pure and mixed strategies for the normal form game, while all we know is what each player can do at each of his information sets in the extensive game. First, we define a pure strategy for player i as a complete description of which action player i will take (with probability one) at each of his information sets. Second, if two pure strategies of player i always (“al-ways” means when put together with any profile σ−i ∈ Σ−i) generate
the same payoff for player i, then they are said to be equivalent, and a reduced normal formgame is obtained from the original extensive game if for each player, equivalent pure strategies are identified (remove all of them but one).7 A mixed strategy for player i is then defined as a
probability distribution over all feasible pure strategies in the reduced normal form game. To fix terminology, mixed strategies in the original extensive game are referred to as behavior strategies, and each behavior strategy of player i specifies a set of probability distributions for player i, where each probability distribution corresponds to one information set of player i, indicating how player i may randomly choose among the actions feasible at that particular information set.
33. What are the relations between mixed strategies in the strategic form and behavior strategies in the extensive form? The answer is that, each mixed strategy generates a unique behavior strategy, but a be-havior strategy can be generated by more than one mixed strategy. Kuhn’s theorem tells us that in any game of perfect recall, every mixed
7For example, consider an extensive game where player 1 first chooses between A and B,
and if A is chosen then the game ends; or else, following B, players 1 and 2 simultaneously choose between a and b. Player 1 has four pure strategies, and two of them, (A,a) and (A,b), are equivalent strategies; for another example, see the game Battle of Sex below.
strategy is equivalent to the behavior strategy that it generates, and every behavior strategy is equivalent to all the mixed strategies that generate it; see Fudenberg and Tirole’s Game Theory for details.8
34. Consider the Cournot game in example 1. Let
P(Q) = 1 − Q, Q = q1+ q2, c= F = 0.
This is a game with simultaneous moves, and hence a game with imper-fect information. Let us solve for the pure strategy NE.9 By definition,
it is a pair (q∗
1, q2∗), such that given q2∗, q1∗ is profit maximizing for firm
1, and given q∗
1, q∗2 is profit maximizing for firm 2. The procedure is
first to find the reaction function for firm i given any qj firm j might
choose:
max
qi f(q1) = qi(1 − qi− qj),
where f (·) is concave and hence the first-order condition is necessary and sufficient for optimal solution. This gives
q∗i =
1 − qj
2 .
An NE, by definition, is the intersection of the two firms’ reaction functions. Solving simultaneously
q∗1 = 1 − q ∗ 2 2 and q∗2 = 1 − q ∗ 1 2 we have q∗ 1 = q2∗ = P∗ = 13.
8A game of perfect recall is an extensive game with special restrictions on its information
sets: players never forget what they knew in the past. In application, almost all games we shall encounter are of perfect recall.
9For any mixed strategy σ of firm j, firm i’s payoff function π
i(qi; σ) = qi(1−qi−Eσ[qj])
is strictly concave. This implies that the set of best responses must be a singleton, and hence this game has no mixed strategy equilibria.
35. Let us reconsider example 1 by assuming that firm 1 moves first by setting q1, and firm 2 observes q1 before determining its q2. Now this
is a game with sequential moves, and hence a game with perfect infor-mation. (This game has a name called Stackelberg game.) The key difference here is that a pure strategy of firm 2 is not a quantity q2;
rather, it is a function q2(q1) which means that for different firm 1’s q1
observed, q2 may vary. Before we solve the pure strategy NE for this
game, we need a definition.
36. Definition 8: A subgame is the remainder of a game tree starting from some singleton information set (an information set composed of only one node).
37. Definition 9: (Selten, 1965) A subgame perfect Nash equilibrium (SPNE) is an NE for a game in extensive form, which specifies NE strategies in each and every subgame. (This definition is further gen-eralized into sequential rationality by Kreps and Wilson (1982).)
38. A game where all information sets are singletons is one of perfect in-formation. Players must move sequentially in a game of perfect infor-mation. The way to solve the SPNE’s of a finite game with perfect information is referred to as backward induction: we first consider the (last) subgames of which the immediate successors are the penultimate nodes, and since each such subgame has a single player facing a finite number of actions, there will be an optimal action for him in this last subgame; then, we can move backward on the game tree to the im-mediate predecessors of these last subgames, and in solving the NE’s for the subgames starting from these predecessors, we should assume that the players know which optimal actions will be taken if one of those last subgames is reached; and then we move backward on the game tree again, and so on and so forth. Because we are given a fi-nite extensive game with perfect information, the above procedure will eventually reach the beginning of the game tree, thereby determining an equilibrium path, which by definition is a pure strategy SPNE. This procedure is usually referred to as Kuhn-Zermelo algorithm. Observe
that if no two terminal nodes give any player the same payoff, the ob-tained SPNE is unique.
Example 4: Every NE in the above game tree is also an SPNE. Example 5: Chen is the owner-manager of a firm. The debt due one year from now has a face value equal to $10. The total assets in place worth only $8. But, just now, Chen found an investment opportunity with NPV=x > 1, which does not need any extra investment other than the current assets in the firm. Chen comes to his creditor and asks the latter to reduce the face value of debt by $1. He claims (he is really bad) that he will not take the investment project unless the creditor is willing to reduce the face value as he wants.
(i) Suppose x > 2. Show that there is an NE in which the creditor agrees to reduce the face value of debt and Chen makes the invest-ment.
(ii) Show that the NE in (i) is not an SPNE, because it involves incred-ible threat from Chen.
(iii) How may your conclusion about (ii) change if x ∈ (1, 2]?
(iv) Define bankruptcy as a state where the firm’s net worth drops to zero. In case of (iii), conclude that Chen’s company has not gone bankrupt.
39. Example 5 shows that in a dynamic game, only SPNEs are reasonable outcomes among NEs. Now we apply the procedure of backward induc-tion to solve the SPNE for the Stackelberg game. First, consider the subgame starting with firm 2’s decision. A subgame is distinguished by firm 1’ choice of q1. Given q1, firm 2 has infinite possible
strate-gies q2(·), but which one is the best? Of course, the profit maximizing
strategy is the best, and hence
q2 =
1 − q1
2 .
Now go back to firm 1’s problem. Firm 1 will max
q1 q1(1 − q1− q2(q1)).
Check the concavity first and solve for the optimal q1. We have q∗1 = 12,
q∗2 = 1 4, P
∗ = 1
be the leader (the one who moves first) is beneficial in quantity-setting games; (ii) In Stackelberg game price is socially more efficient.
Note that with sequential moves firm 1 (the leader) can affect firm 2’s choice of q2 by committing to any q1 it wants, and since r2(q1) is
decreasing in q1 (when firm 1 expands output firm 2 will respond by
cutting back its own output level in order to prevent the price from dropping too much; in this sense the firms’ output choices are strategic substitutes), firm 1 has more incentives to expand output than in the simultaneous (Cournot) game.10 This explains why the leader enjoys
a higher equilibrium supply quantity, and why the equilibrium price is lower than that in the (simultaneous) Cournot game. More precisely, observe that in the Cournot game firm 1 believes that the product price it is faced with is 1−q1−q2, meaning that one unit of output expansion
will result in a dollar price reduction, while in the Stackelberg game, firm 1 believes that the price it is faced with is 1 − q1− r2(q1), meaning
that the price will drop less when it increases its output by one unit. 40. Now consider the famous Bertrand game. Two firms produce a
homo-geneous good and sell it to homohomo-geneous consumers. Producing one unit of the good costs c, where 0 < c < 1. Each consumer is willing to pay as much as 1 dollar to buy 1 unit of the good. The total population of consumers is (normalized to) 1. The firms are competing in price. Consumers maximize consumer surplus and firms maximize profits. A pure strategy Nash equilibrium is (p1, p2), where pi is the price chosen
by firm i, i = 1, 2. Assume that firms get the same market share if they choose the same price. Show that there is a unique Nash equilibrium (called the Bertrand outcome) where p1 = p2 = c.
Proof Let F (p) and G(p) be the distribution functions for p1 and p2
chosen by firm 1 and firm 2 respectively. Thus F and G define com-pletely a mixed strategy profile for this game. We shall assume that (F, G) is an NE, and from here we shall derive a series of necessary conditions on F and G (to be summarized in the following steps). At first, let SF and SG be respectively the supports of F and G.
Step 1: SF ⊂ [c, 1], SG ⊂ [c, 1], inf SF = inf SG = p, and sup SF =
10
Strategic substitutes and complements were first defined by Bulow, Geanakoplos, and Klemperer (1985).
sup SG = p.
Since pricing below c results in sure losses and pricing above 1 will sure push consumers away, we have SF ⊂ [c, 1], SG ⊂ [c, 1]. We claim
that the two sets SF and SG have the same infimum (greatest lower
bound) and supremum (least upper bound). To see this, note that if at P F(P ) = 1, then any price q > P is weakly dominated for G, because facing q, regardless of p1, consumers would rather purchase from firm
1. Similarly, if at p F (p) = 0 then there is no reason that G will ran-domize over q < p: q is dominated by q+p2 , because both q and q+p2 will attract all consumers for sure but apparently the price q+p2 generates a higher profit. Thus for some p and p, we have
inf SF = inf SG= p, sup SF = sup SG = p.
In particular, we have SF ⊂ [p, p] ⊂ [c, 1] and SG ⊂ [p, p] ⊂ [c, 1].
To ease notation, let p = p and P = p.
Step 2: Fix any x ∈ (c, 1). Then neither F nor G can have a jump at x.
To see this, suppose instead that x ∈ (c, 1) is a point of jump of F . Thus firm 1 may randomize on the price x with a strictly positive probability. In this case, there exists > 0 small enough such that all prices contained in the interval [x, x + ) are dominated for firm 2 by some price q < x with q sufficiently close to x. Since (F, G) is by assumption an NE, it must be that G assigns zero probability to the interval [x, x + ). Again, since F is the best response against G, this must imply that x is not a best response for firm 1 (the price x +
2, for
example, is strictly better than x), and firm 1 should not have assigned a strictly positive probability to x, a contradiction. We conclude that F and G are continuous except possibly at p and P (when p = c or P = 1).
Step 3: If P > p then every point in (p, P ) is contained in both SF
and SG.
Suppose instead that, say, F is flat on an interval [a, b] ∈ (p, P ). Since p is the infimum of SF, F (b) > 0. Moreover, since b < P , F (b) < 1.
(If instead F (b) = 1 then P 6= sup SF, a contradiction.) There exists
a smallest x ∈ [c, 1] such that F (x) = F (b) so that 0 < F (x) < 1: By the right continuity of F , we have x = inf{y ∈ (p, P ) : F (y) = F (b)}. We claim that x is a best response for firm 1. Either x = p so that
F(x) > 0 implies that it is a point of jump or x > p but for all > 0, F(x − ) < F (x) (by definition of x) so that there exists a best response yn in each term of the sequence of intervals {(x − 1n, x]; n ∈ Z+} and
limn→∞yn = x. In the latter case, note that each best response yn
gives rise to the same expected profit for firm 1 and by Step 2, firm 1’s expected profit is continuous in (p, P ). We conclude that x attains the equilibrium expected profit of firm 1, and hence a best response of firm 1. On the other hand, if this were an equilibrium, then G would never randomize over (x, b), but then x could not be a best response for firm 1: it is weakly dominated by the price x +b−x
2 . Hence we have
a contradiction. Together with Step 2, we conclude that F and G are strictly increasing and continuous on (p, P ). Moreover, each point in (p, P ) is a best response for both firms.
We just reached the conclusion that SF = SG = [p, P ]. We next show
that p = P . Step 4: p = P .
Suppose instead that P > p, and consider any x ∈ (p, P ) for firm 1. We claim that G cannot have a jump at P if P > p: If G does, then P is not a best response for firm 1, and firm 1 will price below P with probability one; but then pricing at P yields zero profits for G which is dominated by any price in (p, P ). Now, since all x ∈ (p, P ) are best responses for firm 1 and they generate the same expected profit for firm 1, the following is a “constant” function on (p, P ):
x[1 − G(x)].
It follows from the fact that G does not jump at P that lim
x↑1x[1 − G(x)] = 0.
But then, for all x ∈ (p, P ),
x[1 − G(x)] = 0, ∀x ∈ (p, P ) ⇒ G(x) = 1, ∀x ∈ (p, P ) ⇒ G(p) = G(p+) = inf
x∈(p,P )G(x) = 1 ⇒ p = P.
Thus this game only has symmetric pure strategy Nash equilibria. It is now easy to show that there exists a unique Nash equilibrium for this
game, where p = P = c so that F (x) = G(x) = 1[c,+∞)(x), ∀x ∈ <+.11
41. Continue with the Bertrand game discussed in the preceding section but with the following modifications. Suppose that the two firms can first announce a “best price in town” policy, which promises their cus-tomers that they will match the lowest price a consumer can find in this town. For example, if p1 = 13 > p2 = 14, then everyone having
bought the product from firm 1 will receive an amount 1 3−
1
4 from firm
1. Assume that both firms have announced the “best price in town” policy. Reconsider the price competition between the two firms. Show that in one NE the best-price-in-town policy gives rise to the worst price for the consumers: In equilibrium the consumers are left with no surplus.
Proof Let pm be the optimal price chosen by a monopolistic firm.
Then setting price at pm by both firms forms an NE in the presence of
the best-price-in-town policy: Apparently, no firm can gain from uni-laterally raising the price, but if one firm uniuni-laterally lowering the price from pm, it does not gain sales volume because consumers purchasing
from the other firm will automatically get the difference in prices, and they do not want to switch to the firm charging the lowest price. This proves that no firms have incentives to unilaterally deviate from the price pm, and hence pm defines a symmetric NE.
Note that in this NE, consumers have the lowest possible consumer surplus (as if they were faced with a monopolistic firm, or a perfectly colluding cartel). Note also that this game has other pure strategy NE’s as well. In fact, any price contained in the interval [c, pm] defines
a symmetric NE. The NE where both pick the price pm is the Pareto
undominated one.
42. Continue with the above extensive game. Assume now that the two firms can first simultaneously choose to or not to announce the best-price-in-town policy, and then upon observing the two firms’ announce-ments, the two firms engage in the price competition. Show that in the unique subgame perfect NE both firms will announce the
best-price-in-11What do you think may happen if the two firms must move sequentially? Show that
town policy in the first stage, and then both price at pm.12
43. Suppose there are two firms, 1 and 2, with symmetric demand func-tions:
q1 = 1 − p1+ 0.5p2, q2 = 1 − p2+ 0.5p1.
Assume no costs for either firm. Suppose firms compete in price. Find the NE for the cases of (i) simultaneous moves; and (ii) sequential moves. The firms’ prices are strategic complements, and hence with sequential moves the equilibrium output level will be less efficient; see the previous discussions about the relationship between the Cournot game and the Stackelberg game and argue analogously.
44. Two heterogeneous goods are produced by firm 1 and firm 2 respec-tively, and the demands for the two goods are
Di(pi, pj) = max(0, 1 − api+ bpj), a, b > 0, i 6= j, i, j ∈ {1, 2}.
Find the pure strategy NE for this game assuming that the firms are competing in price. What restrictions do you need to impose on the parameters a, b so that firms will not get unbounded profits? Now we solve the prices from the above system of equations to get the inverse demands for the two firms:
Pi(qi, qj) = max(0, α − βqi+ γqj), i 6= j, i, j ∈ {1, 2}.
Now suppose instead that the two firms are competing in quantity. In which case (competing in price or quantity) do the two firms obtain higher profits in an NE? Why?
Solution Suppose that the two firms are competing in quantity. In an NE of this quantity setting game, a firm considering raising its quantity always assumes that its rival will not react. Since the game is really one of price-setting, what the quantity-setting conjectural variation says is really that when a firm lowers its price (so that its quantity is expanded) it expects its rival to move the latter’s price in such a manner that with
12
Here we assume that in the price competition stage, the two firms will always coordi-nate on Pareto undomicoordi-nated equilibria. Note that committing to the best-price-in-town policy essentially allows a firm to convince its rival that it will react to the latter’s price cut in no time.
the new prices the rival’s quantity does not change. This implies that one’s lowering its price is expected to be reacted right away by its rival by also lowering the latter’s price. Thus the firms in a quantity-setting game (again, they are really playing the price-setting game) have less incentives of lowering prices and expanding outputs. As a consequence firms have higher profits in equilibrium. (To fully understand the above argument, you are invited to provide a similar argument for a price-setting game in the context where the two firms are really playing the quantity setting game given Pi(qi, qj).)
45. Having considered firms’ imperfect competition in quantity and price, let us consider location (or spatial) competition. (Think of product positioning.) Two firms must each choose (simultaneously) a location on the Hotelling main street (the [0, 1] interval). Thus S1 = S2 = [0, 1].
For any pure strategy profile (xl, xr), where 0 ≤ xl ≤ xr ≤ 1, the firm
choosing xl has payoff
xl+
xr− xl
2 ;
and the firm choosing xr has payoff
xr− xl
2 + (1 − xr).
Show that the unique pure strategy NE of this game is (12,12). Show that the same is true if the two firms move sequentially. Now consider the same game but assume that there are three firms. A profile (xl, xm, xr)
generates xl + xr−x2 l for the firm choosing xl, xm2−xl + xr−x2 m for the
firm choosing xm, and xr−x2 m + (1 − xr) for the firm choosing xr. Show
that this game has no pure strategy NE. What if the three firms move sequentially?
Let Li ∈ [0, 1] be the location of firm i. An innocuous assumption is
that L3 = 12(L1+ L2) whenever firm 3 decides to locate at somewhere
between L1 and L2. (This assumption does matter, because firms 2 and
3 when deciding their own locations must form expectations about how firm 3 will locate. It is innocuous because firm 3 should be indifferent about any locations in between L1 and L2 once staying in between L1
and L2 is its decision. This seems to be the natural choice of firm 3.)
First we consider firm 3’s reaction function L∗ 3(L1, L2). It can be easily verified that L∗3(L1, L2) = LR+ , if LR ≤ min(2+L3L,1 − LL) (3 − 1) LR−LL 2 , if LR≥ max(3LL, 2+LL 3 ) (3 − 2) LL− , if LR ∈ [1 − LL,3LL] (3 − 3).
Moving backwards, we now derive L∗
2(L1) using firm 2’s correct
expec-tations regarding L∗
3(L1, L2). Since possible L1 and L2 are ∈ [0, 1], we
must consider the following six regions: (a) L2 ≥ L1, L2 ≤ min(2+L3 1,1 − L1); (b) L2 ≥ L1, L2 ≥ max(2+L3 1,3L1); (c) L2 ≥ L1, L2 ∈ [1 − L1,3L1]; (d) L2 ≤ L1, L1 ≤ min(2+L3 2,1 − L2); (e) L2 ≤ L1, L1 ≥ max(2+L3 2,3L2); (f) L2 ≤ L1, L1 ∈ [1 − L2,3L2].
We shall first compute regionally optimal Li
2(L1) for each region i ∈
{ a,b,c,d,e,f}, and then obtain the globally optimal L∗
2(L1) from the
six regionally optimal Li
2(L1)’s. Take region (a) for example. Given
L1, if firm 2 confines its location choice to the set of L2’s such that
(a) is satisfied, what will be the optimal La
2(L1) in this region? Note
that for (L1, L2) in region (a), firm 3 will subsequently choose to locate
arbitrarily close to firm 2 from the right (simply because (3 − 1) is satisfied). In this case, firm 2’s payoff is L2−L1
2 , an increasing function
of L2. Thus, we have
La2(L1) = min(
2 + L1
3 ,1 − L1). Similarly, one can derive other regional optima:
Lb2(L1) = max( 2 + L1 3 ,3L1), Lc2(L1) = max(1 − L1, L1), Ld2(L1) = min(1 − L1, L1), Le2(L1) = min(3L1− 2, L1 3 ),
Lf2(L1) = max(1 − L1,
L1
3 ).
Next we compare firm 2’s payoffs at these six regional optima, and deduce that L∗2(L1) = 2+L1 3 , if L1 ≤ 1 4; (2 − 1) 1 − L1, if L1 ∈ [14,34]; (2 − 2) L1 3 , if L1 ≥ 3 4 (2 − 3).
Now, we consider firm 1’s optimal decision, L∗
1. For L1such that (2−1)
holds, from the above analyses,
L∗2(L1) = 2 + L1 3 , L ∗ 3(L1, L∗2(L1)) = 1 − L1 3 ,
which implies that firm 1’s payoff is increasing in L1, and hence the
best L1 in this region is 14. On the other hand, if firm 1 chooses any L1
such that (2 − 2) holds, then
L∗2(L1) = 1 − L1, L∗3(L1), L∗2(L1)) = L1− ,
which implies that firm 1’s payoff is decreasing in L1. Once again the
regionally optimal L1 is 14. What about those L1 that satisfy (2 − 3)?
In this case, (3 − 2) holds, and hence firm 1’s payoff is (1 − L1) + L1−L13
4 , decreasing in L1. Hence, the regionally optimal L1 = 3
4, which
is obviously a strategic equivalent of L1 = 14. Thus, this game has two
subgame perfect Nash equilibria (assuming that firm 3 always locates itself at the midpoint whenever applicable):
(L∗ 1, L∗2, L∗3) = ( 1 4, 3 4, 1 2), and (L∗ 1, L∗2, L∗3) = ( 3 4, 1 4, 1 2).
46. Find all mixed strategy NEs for the following strategic game:
Player 1/Player 2 L M R
U 4, 0 0, 1 −1, −100
SolutionIt is easy to show that (D,R) is a pure strategy NE. Consider mixed strategy NE’s. Let π be the equilibrium probability that player 1 chooses U, and p and q respectively the equilibrium probabilities that player 2 chooses L and M. It is easy to verify that there are two mixed strategy NE’s of this game, which are
(π, p, q) = (1 2, 1 3, 2 3), (π, p, q) = ( 1 101, 1 3,0).
Note that in the first mixed strategy NE, given player 1’s mixed strat-egy, R is not a best response for player 2, and that is why player 2 only randomizes over L and M. In the second mixed strategy NE, given player 1’s mixed strategy, M is not a best response for player 2, and that is why player 2 randomizes only over L and R.
47. Mr. A and B are asked to simultaneously name a number in {1, 2, · · · , 100}, and if the two numbers are identical they each get 1 dollar, or else they get zero. Find all the NE’s. (Hint: Each non-empty subset E ⊂ {1, 2, · · · , 100} defines a mixed strategy NE where A and B both assign probability #(E)1 to each and every element in E, where #(E) is the number of elements in E.)
48. Consider the following strategic game:
Player 1/Player 2 L M R
U 5, 5 −1, 6 −2, −2
M 6, −1 0, 0 −2, −2
D −2, −2 −2, −2 −6, −6
Now assume this game is played twice, and each player maximizes the sum of his payoffs in the two rounds of play. Each player observes ev-eryone’s first-period action before the second-period game starts. This is called a repeated game, where the above stage game is played twice. (a) What are the highest symmetric payoffs in any NE?
(b) What are the highest symmetric payoffs in any SPNE?
players playing (M,M) in both rounds.13 For part (a), note first that
each player has 3×39 pure strategies, and in each pure strategy a player
must specify which among (U,M,D) (or (L,M,R)) will be chosen at the first stage, and at the beginning of stage 2, in each of the 9 possible subgames which among (U,M,D) (or (L,M,R)) will be chosen. Now show that the following pure strategy profile constitutes an NE, which is not subgame-perfect: player 1 plays U at the first stage and will play M at the second stage if (U,L) is the outcome of the first stage and he will play D at the second stage if the first-stage outcome is not (U,L); player 2 plays L at the first stage and will play M at the second stage if (U,L) is the outcome of the first stage and he will play R at the second stage if the first-stage outcome is not (U,L).)
49. Consider two firms engaged in Cournot competition. The inverse de-mand is
p= 3 − q1 − q2.
Assume that both firms have marginal cost equal to 1, but only firm 1 has the chance to spend F and brings the marginal cost down to zero. Firm 1 moves first by deciding to invest F or not to, which is unobservable to firm 2 (imperfect information). The two firms then play the Cournot game by selecting their own outputs. Compute the set of pure strategy Nash equilibria for this imperfect information game for F ∈ R+. (Hint: for different ranges of F , the equilibria may differ.)
Solution Suppose that firm 2 believes (correctly) that in equilibrium firm 1 makes the investment with probability 1 − π, where 0 ≤ π ≤ 1. Let the equilibrium quantity of firm 1 be q∗
1(c) if firm 1’s marginal cost
is c ∈ {0, 1}. Let q∗
2 be firm 2’s equilibrium quantity. Then, as part of
the Nash equilibrium, we must have q1∗(1) = arg max q1 q1(3 − q1− q ∗ 2 − 1), q1∗(0) = arg max q1 q1(3 − q1− q ∗ 2 − 0),
13The reader may get the wrong impression that an SPNE of a finitely repeated game
is nothing but a repetition of the NE in the stage game. In fact, strategic games that have a unique NE are quite unusual. If the stage game has more than one NE, then a repeated game will have a lot of SPNE’s which differ from any series of NE’s obtained from the static game, as long as the number of repetitions is large enough. We shall have more to say on this in a subsequent Lecture.
q2∗ = arg max q2 q2(3 − E[q ∗ 1] − q2 − 1), where E[q∗1] = πq1∗(1) + (1 − π)q∗1(0).
The (necessary and sufficient) first-order conditions of these three max-imizations give three equations with three unknowns. Solving, we have
q∗1(1) = 5 − π 6 , q ∗ 1(0) = 8 − π 6 , q ∗ 2 = 1 + π 3 .
Let Yi(c; π) be the equilibrium profit of firm i before subtracting the
expenditure on F , given that firm 1’s marginal cost is c and firm 2’s beliefs are such that firm 1 spends F with probability 1 − π. Then, one can show that
Y1(1; π) = [ 5 − π 6 ] 2, Y 1(0; π) = [ 8 − π 6 ] 2.
Suppose that there exist pure strategy NE’s. In a pure strategy NE, firm 1 either spends F with prob. 1 or does not spend F with prob. 1. Suppose first that there is a pure strategy NE in which firm 1 invests. Then this is expected correctly by firm 2, and hence π = 0. From the preceding discussion, this requires, for firm 1 to follow its equilibrium strategy,
Y1(0; 0) − F ≥ Y1(1; 0),
or F ≤ 3936.
Next, suppose instead that there is a pure strategy NE in which firm 1 does not invest. The earlier discussion implies that π = 1 and in equilibrium it must be that
Y1(0; 1) − F ≤ Y1(1; 1),
or F ≥ 33 36.
Finally, consider mixed strategy NE’s. For firm 1 to randomize between to and not to invest with respectively prob. (1 − π) and π,14
Y1(0; π) − F = Y1(1; π),
14
Note again that neither firm 1 nor firm 2 can randomize in the quantity-setting stage (why?), and hence a mixed strategy NE can at most involve firm 1 randomly making investment decisions.
or F ∈ (33 36,
39
36). Note that for all F ∈ R+, this game has an odd
number of NE’s.
50. Two firms are competing in a declining industry in continuous time. If at time t ∈ [0, +∞) both stay, firm i gets profit density πi
d(t); if only
firm i stays, it gets πi
m(t). (d and m stand for respectively ‘duopoly’
and ‘monopoly’.) It is given that
πd1 = 2 − 2t, πd2 = 1 − 2t, πm1 = 11 4 − t, πm2 = 7 4 − t.
Show that there is a unique SPNE for this game. (Hint: Let Ti be
the equilibrium point in time firm i chooses to exit. First observe that T1 ∈ [1,114] and T2 ∈ [12,74]. Now repeatedly apply common knowledge
about the two firms’ rationality to make the following inferences: if at any t ∈ [5
4, 11
4] firm 1 is still operating, then firm 1 will stay till 11
4, and
knowing this, if at any t ∈ [5 4,
11
4] firm 2 is still operating, then firm 2
should leave immediately at t. Firm 1, knowing that firm 2 is rational and is able to make the above inference, will not leave until 11
4 if it is
still operating at any t ∈ [1,5
4]. Being able to make this last inference,
firm 2 will leave immediately at t = 12. Conclude thereby that T1 = 114
and T2 = 12 constitute the unique SPNE.)15
51. Three cowboys A, B, and C will take turn to shoot at one of their opponents (everyone can take one shot; A first, and then B, and then C, and then A (if still alive), and so on). The shooter’s feasible strategies consist of deciding his targets. With probability 0.8, 0.2 and zero, A,
15This game does have other NE’s which are not subgame perfect. Show that T 1= 1
and T2 = 74 constitute one such NE. To see that this NE is not subgame perfect, note
that in the subgame at time t ≥ 5
4 where neither firm has exited, T1 = 11
4 is a dominant
strategy for firm 1, to which firm 2’s best response is not T2= 74. That is, the NE does not
specify NE strategies for firm 2 in each and every subgame! This example and Example 5 above are due to Chyi-Mei Chen. The following sequential shooting problem is due to Lien-Kuo Hu.
B, and C may miss his target. Assume that if the target is hit, he dies. The game ends when there is exactly one cowboy remaining alive. The last survivor gets payoff 1, and others get zero. Find an SPNE.
Solution We shall look for a special kind of SPNE’s, called a stationary equilibrium. In such an equilibrium, a player’s strategy in a subgame depends only on the state at the current stage, but not on the history that reaches this subgame. In this game, the state of the current stage concerns how many opponents of the shooter are still alive. In a Markov perfect equilibrium of this game, therefore, in all the subgames where it is A’s turn to shoot and both B and C are alive, A will adopt the same equilibrium strategy.
(a) In every subgame where there are two cowboys alive, the shooter’s strategy is, trivially, to aim at his last opponent.
(b) Consider the subgame where a shooter is facing two opponents. If the shooter is C, then after C shoots the preceding (a) will apply, and hence in C’s interest, C prefers taking out B to A.
(c) Consider the subgame where the shooter is B and B is facing two opponents. After B shoots, if the target is missed, then the above (b) will apply, and B will be dead regardless who B’s current target is. On the other hand, if the target is not missed, then B is better off facing A. The conclusion is therefore B should aim at C. (d) Consider the subgame where the shooter is A and A is facing two
opponents. Again, we do not have to discuss the case where A misses his target. In case the target is hit, then A would rather let the last surviving opponent be B (so that A has the chance to shoot again). The conclusion is again that A should shoot at C. Thus our conclusion is that both A and B should try to shoot at C, and C should try to shoot at B, until someone is hit and dead, and from then on the remaining two start shooting at each other.
52. Consider the following voting game. An incumbent manager I is cur-rently managing an all-equity firm. The firm’s earnings will be YIunder
manager I’s control, and manager I himself will receive a private ben-efit ZI if he has the control. A raider C appeared just now to compete
C will receive a private benefit ZC if he defeats I and becomes the new
manager. There are two classes of stocks issued by the firm. Class i stockholders are entitled to a fraction Si of the firm’s earnings, and are
given a fraction Vi of the votes in managerial election, where i = A, B.
Assume that investors do not hold both stocks A and B at the same time. Investors are rational and can predict the outcome of control contest correctly. There are many small shareholders for each class of stock, so that no shareholder considers himself privotal. Suppose that the firm’s charter states that the rival C gets control if he obtains a fraction α of the votes.
The game proceeds as follows. The rival C first makes an offer to stock-holders, and then the incumbent manager can make another offer given C’s offer. Given the two offers, the stockholders then make tendering decisions.
(i) Suppose that SA= SB = 50%, VA = 100%, VB = 0%. Suppose that
it is common knowledge that YI = 200, YC = 180, and ZI = 0 < ZC.
Suppose that offers must be in integers. Show that if ZC ≥ 11, C can
purchase all class A stock at the price of 101, and the market value of the firm becomes 191.
(ii) Suppose that SA= 75%, SB = 25%, VA= 100%, VB = 0%.
More-over, YI, YC, ZI, ZC are as assumed in part (i). Show that the market
value of the firm becomes 196 in equilibrium.
(iii) Suppose that SA= 100%, SB = 0%, VA= 100%, VB = 0%.
More-over, YI, YC, ZI, ZC are as assumed in part (i). Show that the market
value of the firm becomes 201 in equilibrium.
(iv) From now on, assume ZI, ZC > 0. Suppose that SA = 100%,
SB = 0%, VA = 100%, VB = 0%. Moreover, YI = 10, YC = 100,
ZI = 1 < ZC. Show that the market value of the firm becomes 100 in
equilibrium.
(v) Suppose that SB = 100%, SA = 0%, VA = 100%, VB = 0%.
More-over, YI = 10, YC = 100, ZI = 1 < ZC. Show that the market value of
the firm becomes 101 in equilibrium.
53. Players 1 and 2 are bargaining over 1 dollar. The game proceeds in 2N periods. In period i, where i is odd, player 1 can make an offer (xi,1 − xi) to player 2, where xi is player 1’s share, and player 2 can
is so divided; or else, the game moves on to the i+1st period. In period j, where j is even, player 2 can make an offer (xj,1 − xj) to player 1,
where xj is again player 1’s share, and player 1 can either accept or
reject that offer. The game ends here if acceptance is player 1’s decision (with the dollar divided as player 2 proposed); or else, the game moves on to the j + 1st period. The dollar will be gone if agreement is not reached by the end of period 2N . We assume that waiting is costly, because both players have a (common) discount factor δ ∈ (0, 1). Show that this game has a unique subgame perfect Nash equilibrium (SPNE) where player 1 proposes (1−δ1+δ2N,δ+δ1+δ2N) and player 2 accepts it in the first period.16
54. Now reconsider Example 5, and modify the game by assuming that Chen will first optimally choose y, where y is the amount that Chen will ask the creditor to reduce from the face value of debt, and then Chen and the creditor play the game stated in Example 5. What is the optimal y chosen by Chen in the SPNE?
55. Two firms are asked to deliver respectively A > 0 and B > 0 units of a homogeneous good at time 1. They do not have inventory at time 0. At each point in time, holding one unit of inventory incurs a cost α > 0 (respectively β > 0) for firm A (respectively, firm B). Purchasing one unit of the good at time t incurs a unit cost c(t), where
c(t) = 1 + a(t) + b(t),
16
Thus our theory predicts that if N = 1 then player 2, when he gets the chance to move, will offer player 1 with zero payoff. This prediction is at odds with existing experimental evidence. Usually, the player who makes the offer behaves a lot more generous in those experiments than our theory would describe. One possibility is that player 1 (in the experiment) is not an expected utility maximizer; rather, his behavior may be consistent with Kahneman and Tversky’s (1979) prospect theory (see my note in Investments, Lecture 2, Part II). Say player 1 thinks that it is simply fair that he receives at least 0.3 dollars (called a reference point), and he is ready to reject any offer that gives him a payoff less than 0.3 dollars. If player 2 recognizes this, then player 2’s optimal strategy is to offer player 1 0.3 dollars. This example does not necessarily imply any flaws of the game theory per se (I am not saying that the game theory is flawless though); rather, it points out the importance of specifying correctly the payoffs for the players. Mis-specified payoffs lead to incorrect predictions about the final outcome of the game. Since our purpose of learning the game theory is essentially to make correct predictions, the importance of specifying correctly the normal form for the game cannot be overstated.
