3 Solution concepts
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% Basic notation - general % Basic notation - general % Special symbols % Basic functions %image %cardinality % Operators - Analysis % Linear algebra % Linear algebra - Matrices % Statistics %at time % Useful commands % Game Theory %T-stage game %strategy in a repeated game %successor function %player function%action function
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3.1 Domination of strategies
Let \(N = \left\{1, \dots, n\right\}\) be a finite set and for each \(i \in N\) let \(X_i\) be a set. Moreover, let \(X := \prod_{i \in N} X_i = \left\{(x_1, \dots, x_n) | x_j \in X_j, \; j \in N\right\}\). We shall now introduce the following notation:
- for \(i \in N\) we define \(X_{-i}:= \prod_{j \neq i} X_j\), i.e., \[ X_{-i}= \left\{(x_1, \dots, x_{i-1}, x_{i+1}, \dots, x_n) | x_j \in X_j, \forall j \neq i\right\}; \]
- an element of \(X_{-i}\) will be denoted by \[ \boldsymbol{x}_{-i}= (x_1, \dots, x_{i-1}, x_{i+1}, \dots, x_n); \] as a slight abuse of notation we write \((x_i, \boldsymbol{x}_{-i})\) to denote \((x_1, \dots, x_i, \dots, x_n) \in X\).
Definition 3.1 (Strictly dominated strategies) Let \(s_i, s_i' \in S_i\) be strategies of player \(i\) as per Definition 2.2. Then \(s_i'\) is strictly dominated by \(s_i\) (write \(s_i \succ s_i'\)) if for any possible combination of the other players’ strategies, \(\boldsymbol{s}_{-i}\in S_{-i}\), we have \[ u_i(s_i, \boldsymbol{s}_{-i}) > u_i(s'_i, \boldsymbol{s}_{-i}) \] for all \(\boldsymbol{s}_{-i}\in S_{-i}\).
Conjecture 3.1 An intelligent and rational player will never play a strictly dominated strategy.
Proof. Clearly, intelligence implies that the player should recognize dominated strategies, and rationality implies that the player will avoid playing them.
Definition 3.2 (Strictly dominant strategy) We say \(s_i \in S_i\) is strictly dominant if every other pure strategy of player \(i\) is strictly dominated by \(s_i\). Similarly, we say \(s_i \in S_i\) is strictly dominated in game \(G\) if there exists a pure strategy \(s_i' \in S_i\) such that \(s_i' \succ s_i\).
Observe that every player has at most one strictly dominant strategy and that strictly dominant strategies do not have to exist. Furthermore, we can make a similar claim to Conjecture 3.1:
Conjecture 3.2 Any rational player will play the strictly dominant strategy if it exists.
Definition 3.3 (Strictly dominant strategy equilibrium) A strategy profile \(\boldsymbol{s} \in S\) is a strictly dominant strategy equilibrium if \(s_i\) is strictly dominant for all \(i \in N\).
Corollary 3.1 If the strictly dominant strategy equilibrium exists, it is unique and rational players will play it.
3.1.1 Examples
Example 3.1 (Prisoner’s dilemma) In the Prisoner’s dilemma, \((C, C)\) is the strictly dominant strategy equilibrium.
| \(C\) | \(S\) | |
|---|---|---|
| \(C\) | \((-5,-5)\) | \((0,-20)\) |
| \(S\) | \((-20,0)\) | \((-1,-1)\) |
Example 3.2 (Battle of Sexes) In the Battle of Sexes, no strictly dominant strategies exist.
| \(O\) | \(F\) | |
|---|---|---|
| \(O\) | \((2,1)\) | \((0,0)\) |
| \(F\) | \((0,0)\) | \((1,2)\) |
Indiana Jones, his father, and the Nazis have all converged at the site of the Holy Grail. The two Joneses refuse to help the Nazis reach the last step. So the Nazis shoot Indiana’s dad. Only the healing power of the Holy Grail can save the senior Dr. Jones from his mortal wound. Suitably motivated, Indiana leads the way to the Holy Grail. But there is one final challenge. He must choose between literally scores of chalices, only one of which is the cup of Christ. While the right cup brings eternal life, the wrong choice is fatal. The Nazi leader impatiently chooses a beautiful gold chalice, drinks the holy water, and dies from the sudden death that follows from the wrong choice. Indiana picks a wooden chalice, the cup of a carpenter. Exclaiming “There’s only one way to find out” he dips the chalice into the font and drinks what he hopes is the cup of life. Upon discovering that he has chosen wisely, Indiana brings the cup to his father and the water heals the mortal wound.
In this scene, Indy behaved “suboptimally (irrationally)”, because he overlooked his strictly dominant strategy, which would be to give the water to his father without tasting. Here are the possible outcomes:
- If Indiana has chosen the right cup, his father is still saved.
- If Indiana has chosen the wrong cup, then his father dies but Indiana is spared.
Testing the cup before giving it to his father doesn’t help, since if Indiana has made the wrong choice, there is no second chance – Indiana dies from the water and his father dies from the wound.
3.2 Iterated Strict Dominance in Pure Strategies
We know that, by Conjecture 3.1, no rational player ever plays strictly dominated strategies. As each player knows that each player is rational, each player knows that his opponents will not play strictly dominated strategies and thus all opponents know that effectively they are facing a “smaller” game. As rationality is common knowledge, everyone knows that everyone knows that the game is effectively smaller. Thus everyone knows, that nobody will play strictly dominated strategies in the smaller game (and such strategies may indeed exist). Because it is common knowledge that all players will perform this kind of reasoning again, the process can continue until no more strictly dominated strategies can be eliminated.
This principle (or reasoning) yields the Iterated Elimination of Strictly Dominated Strategies.
Definition 3.4 (Iterated Elimination of Strictly Dominated Strategies) Define a sequence \(D_i^0, D_i^1, D_i^2, \dots\) of strategy sets of player \(i\). Also denote by \(G_{DS}^{k}\) the game obtained from \(G\) by restricting to \(D_i^k\), \(i \in N\). We shall call the following algorithm “Iterated Elimination of Strictly Dominated Strategies”:
- Initialize \(k = 0\) and \(D_i^0 = S_i\) for each \(i \in N\).
- For all players \(i \in N\): Let \(D^{k+1}_i\) be the set of all pure strategies of \(D_i^k\) that are not strictly dominated in \(G_{DS}^{k}\).
- If \(D^{k+1}_i = D^{k}_i\) for all players \(i \in N\), then stop. Otherwise, let \(k := k+1\) and go to 2.
We say that \(s_i \in S_i\) survives IESDS if \(s_i \in D_i^k\) for all \(k = 0, 1, 2,\dots\) (or until stop).
I modified the algorithms (or definitions) 3.4 and 3.9 to stop when nothing changes across two iterations. In the slides, this condition is not present.
This change was motivated to make each of these processes stop at some point when no further iterations are necessary. But just to be sure, maybe use/remember the original/official definitions.
Definition 3.5 A strategy profile \(\boldsymbol{s} = (s_1, \dots, s_n) \in S\) is an IESDS equilibrium if each \(s_i\) survives IESDS. A game is IESDS solvable if it has a unique IESDS equilibrium.
Remark. If all \(S_i\) are finite, then in 2. we may remove only some of the strictly dominated strategies (not necessarily all). The result is not affected by the order of elimination since strictly dominated strategies remain strictly dominated even after removing some other strictly dominated strategies.
3.2.1 Examples
Example 3.3 In the Prisoner’s dilemma, the strategy profile \((C, C)\) is the only one surviving the first round of IESDS.
| \(C\) | \(S\) | |
|---|---|---|
| \(C\) | \((-5,-5)\) | \((0,-20)\) |
| \(S\) | \((-20,0)\) | \((-1,-1)\) |
Example 3.4 In the Battle of Sexes, all strategies survive all rounds (i.e. IESDS says: “anything may happen, sorry”)
| \(O\) | \(F\) | |
|---|---|---|
| \(O\) | \((2,1)\) | \((0,0)\) |
| \(F\) | \((0,0)\) | \((1,2)\) |
Exercise 3.1 Presume the game is given by the Table 3.1:
| \(L\) | \(C\) | \(R\) | |
|---|---|---|---|
| \(L\) | \((4,3)\) | \((5,1)\) | \((6,2)\) |
| \(C\) | \((2,1)\) | \((8,4)\) | \((3,6)\) |
| \(R\) | \((3,0)\) | \((9,6)\) | \((2,8)\) |
Solution. Surely, from the Definition 3.4 follows \[ D_1^0 = S_1 = \left\{L, C, R\right\}, \qquad D_2^0 = S_2 = \left\{L, C, R\right\}. \]
Notice that for player 2, playing \(C\) is strictly dominated by \(R\). On the other hand, in this iteration, there are no strictly dominated strategies for the first player. Therefore next iteration reads \[ D_1^1 = D_1^0 = \left\{L, C, R\right\}, \qquad D_2^1 = \left\{L, R\right\} \] and the table transforms to \[ \begin{bmatrix} (4,3) & (6,2) \\ (2,1) & (3,6) \\ (3,0) & (2,8) \end{bmatrix}. \]
Now the strategy \(L\) dominates both \(C, R\) for player 1. For player 2, no strategy is strictly dominant. Now the second iteration has the form \[ D_1^2 = \left\{L\right\}, \qquad D_2^2 = \left\{L, R\right\} \] with the table \[ \begin{bmatrix} (4,3) & (6,2) \end{bmatrix}. \]
In this case, the \(R\) strategy is strictly dominated by \(L\) for the second player and we get \[ D_1^3 = \left\{L\right\}, \qquad D_2^3 = \left\{L\right\} \]
Here, the iterations stop and the strategies \(L\) are the only one that survives IESDS (for both players). Hence the \((L, L)\) is the IESDS equilibrium of this game.
3.2.2 Political Science Example: Median Voter Theorem
In this example of a two-player game by Hotelling (1929) and Downs (1957), we have \[ N = \left\{1,2\right\}, \qquad S_i = \left\{1, 2, \dots, 10\right\}, \] where each player (a candidate) plays his position on a political and ideological spectrum, see Figure 3.1 – 10 voters belong to each position (or in real life, 10% of voters belong to each position). Voters then vote for their closest candidate and if there is a tie, then half the votes go to each candidate. The payoff in this game is the number of voters for the candidate, and each candidate (selfishly) strives to maximize this number.
Here in \(G = G_{DS}^{0}\) surely, 1 and 10 are the (only) strictly dominated strategies \(\implies D_1^1 = D_2^1 = \left\{2, \dots, 9\right\}\). Then again, in \(G_{DS}^{1}\) the 2 and 9 are the (only dominated strategies) \(\implies D_1^2 = D_2^2 = \left\{3,\dots, 8\right\}\). If we repeat this, only strategies 5 and 6 emerge not dominated and survive IESDS.
3.3 Belief & Best Response
IESDS eliminated apparently unreasonable behavior (leaving “reasonable” behavior implicitly untouched). What if we rather want to actively preserve reasonable behavior? But a question arises, what is reasonable? The simple answer is what we believe is reasonable :). To build an intuition, consider the following situation
Imagine that your colleague did something stupid
What would you ask him? Usually something like “What were you thinking?”
The colleague may respond with a reasonable description of his belief in which his action was (one of) the best he could do
You may, of course, question the reasonableness of the belief
To formalize this kind of reasoning, we start with the next few definitions.
Definition 3.6 (Belief) A belief of player \(i\) is a pure strategy profile \(\boldsymbol{s}_{-i}\in S_{-i}\) of his opponents.
Definition 3.7 (Best Response) A strategy \(s_i \in S_i\) of player \(i\) is a best response to a belief \(\boldsymbol{s}_{-i}\in S_{-i}\) if \[ u_i(s_i, \boldsymbol{s}_{-i}) \geq u_i(s'_i, \boldsymbol{s}_{-i}) \] for all \(s'_i \in S_i\).
Conjecture 3.3 A rational player who believes that his opponents will play \(\boldsymbol{s}_{-i}\in S_{-i}\) always chooses a best response to \(\boldsymbol{s}_{-i}\in S_{-i}\).
Definition 3.8 A strategy \(s_i \in S_i\) is never best response if it is not a best response to any belief \(\boldsymbol{s}_{-i}\in S_{-i}\).
Clearly, a rational player never plays any strategy that is never best response.
Proposition 3.1 If \(s_i\) is strictly dominated for the player \(i\), then it is never best response.
The opposite, though, does not have to be true in pure strategies. Consider the game given by Table 3.2.
| \(X\) | \(Y\) | |
|---|---|---|
| \(A\) | \((1,1)\) | \((1,1)\) |
| \(B\) | \((2,1)\) | \((0,1)\) |
| \(C\) | \((0,1)\) | \((2,1)\) |
Here \(A\) is never best response but it is not strictly dominated either by \(B\), or by \(C\).
3.3.1 Elimination of Stupid Strategies
Using similar iterated reasoning as for IESDS, see Definition 3.4, strategies that are never best response can be iteratively eliminated.
Definition 3.9 (Rationalizable) Define a sequence \(R_i^0, R_i^1, \dots\) of strategy sets of player \(i\). Also, denote by \(G_{Rat}^{k}\) the game obtained from \(G\) by restricting to \(R_i^k\) for \(i \in N\). Consider the following algorithm
- Initialize \(k = 0\) and \(R_i^0 = S_i\) for each \(i \in N\).
- For all players \(i \in N\): Let \(R_i^{k+1}\) be the set of all strategies of \(R_i^k\) that are best responses to some beliefs in \(G_{Rat}^{k}\).
- If \(R^{k+1}_i = R^{k}_i\) for all players \(i \in N\), then stop. Otherwise, let \(k := k+1\) and go to 2.
We say that \(s_i \in S_i\) is rationalizable if \(s_i \in R_i^k\) for all \(k = 0, 1, 2, \dots\) (or until stop).
Definition 3.10 A strategy profile \(\boldsymbol{s} = (s_1, \dots, s_n) \in S\) is a rationalizable equilibrium if each \(s_i\) rationalizable.
We say that a game is solvable by rationalizability if it has a unique rationalizable equilibrium.
For some reason, rationalizable strategies are almost always defined using mixed strategies!
3.3.2 Examples
Example 3.5 In the Prisoners’ dilemma, the strategy profile \((C, C)\) is the only rationalizable equilibrium.
| \(C\) | \(S\) | |
|---|---|---|
| \(C\) | \((-5,-5)\) | \((0,-20)\) |
| \(S\) | \((-20,0)\) | \((-1,-1)\) |
Example 3.6 In the Battle of Sexes, all strategies are rationalizable.
| \(O\) | \(F\) | |
|---|---|---|
| \(O\) | \((2,1)\) | \((0,0)\) |
| \(F\) | \((0,0)\) | \((1,2)\) |
3.3.3 Cournot Doupoly
Consider a game \(G = (N, (S_i)_{i \in N}, (u_i)_{i \in N})\) with \(N = \left\{1,2\right\}\), \(S_i = [0, \infty)\) and payoffs of form \[ \begin{aligned} u_1(q_1, q_2) &= q_1(\kappa - q_1 - q_2) - q_1 c_1 = (\kappa - c_1)q_1 - q_1^2 - q_1 q_2 \\ u_2(q_1, q_2) &= q_2(\kappa - q_1 - q_2) - q_2 c_2 = (\kappa - c_2)q_2 - q_2^2 - q_1 q_2 \\ \end{aligned} \]
For simplicity, we shall assume \(c_1 = c_2 = c\) and denote \(\theta= \kappa - c\).
Cournot competition is an economic model used to describe an industry structure in which companies compete on the amount of output they will produce, which they decide on independently of each other and at the same time. It is named after Antoine Augustin Cournot (1801–1877) who was inspired by observing competition in a spring water duopoly. It has the following features:
- There is more than one firm and all firms produce a homogeneous product, i.e., there is no product differentiation;
- Firms do not cooperate, i.e., there is no collusion;
- Firms have market power, i.e., each firm’s output decision affects the good’s price;
- The number of firms is fixed;
- Firms compete in quantities rather than prices; and
- The firms are economically rational and act strategically, usually seeking to maximize profit given their competitors’ decisions.
In this context, \(q_i \in S_i\) is spent manufacturing time of \(i\)-th firm, \(c_i\) its fixed costs per one manufactured product and \(\kappa\) scales the variable costs based on the total manufactured quantity.
What is a best response of player 1 to a given \(q_2\)?
Solve \(\frac {\partial u_1} {\partial q_1} = \theta- 2q_1 - q_2 = 0\), which gives that \(q_1 = (\theta- q_2)/2\) is the only best response of player 1 to \(q_2\). Similarly, \(q_2 = (\theta- q_1)/2\) is the only best response of player 2 to \(q_1\). Since \(q_2 \geq 0\) (and therefore \((\theta- q_2)/2 \leq \theta/ 2\)), we obtain that \(q_1\) is never best response iff \(q_1 > \theta/2\) and just as well, \(q_2\) is never best response iff \(q_2 > \theta/2\). Thus \(R_1^1 = R_2^1 = [0, \theta/2]\). Now, in \(G_{Rat}^{1}\), we still have that \(q_1 = (\theta- q_2)/2\) is the best response to \(q_2\), and \(q_2 = (\theta- q_1)/2\) the best response to \(q_1\). Since \(q_2 \in R_2^1 = [0, \theta/2]\), we obtain that \(q_1\) is never best response iff \(q_1 \in [0, \theta/4)\). Similarly \(q_2\) is never best response iff \(q_2 \in [0, \theta/4)\). In general, after \(2k\) iterations we have \(R_1^{2k} = R_2^{2k} = [l_k, r_k]\), where
- \(r_k = (\theta- l_{k-1})/2\) for \(k \geq 1\)
- \(l_k = (\theta- r_{k})/2\) for \(k \geq 1\) and \(l_0 = 0\)
Solving the recurrence we obtain
- \(l_k = \theta/3 - \left( \frac 1 4 \right)^k \theta/3\)
- \(r_k = \theta/3 + \left( \frac 1 4 \right)^{k-1} \theta/6\)
Hence, \(\lim_{k \to \infty} l_k = \lim_{k \to \infty} r_k = \theta/3\) and thus \((\theta/3, \theta/3)\) is the only rationalizable equilibrium. But does this mean, that \(q_i = \theta/3\) provide the best outcomes possible? \[ u_1(\theta/3, \theta/3) = u_2(\theta/3, \theta/3) = \theta^2/9 \]
No, when we consider \(q_i = \theta/4\), we get \[ u_1(\theta/4, \theta/4) = u_2(\theta/4, \theta/4) = \theta^2/8, \] but this would require cooperation between players (firms).
3.4 IESDS vs Rationalizability in Pure Strategies
Theorem 3.1 Assume that \(S\) is finite. Then for all \(k\) we have that \(R_i^k \subseteq D_i^k\) (from Definition 3.4 and Definition 3.9). That is, in particular, all rationalizable strategies survive IESDS.
The opposite inclusion does not have to hold in pure strategies. As a counter-example consider a game given by a table
| \(X\) | \(Y\) | |
|---|---|---|
| \(A\) | \((1,1)\) | \((1,1)\) |
| \(B\) | \((2,1)\) | \((0,1)\) |
| \(C\) | \((0,1)\) | \((2,1)\) |
Recall, from Table 3.2, that \(A\) is never best response but it is not strictly dominated by either \(B\), or \(C\). That is, \(A\) survives IESDS but is not rationalizable.
Lemma 3.1 If \(s_i\) is a best response to \(\boldsymbol{s}_{-i}\) in \(G_{Rat}^{k}\), then \(s_i\) is a best response to \(\boldsymbol{s}_{-i}\) in \(G\).
Proof. Let us prove Lemma 3.1 by induction on \(k\). For \(k = 0\) we have \(G_{Rat}^{k} = G_{Rat}^{0} = G\) and the claim holds trivially.
Now assume that the claim is true for some \(k\) and that \(s_i\) is a best response to \(\boldsymbol{s}_{-i}\) in \(G_{Rat}^{k+1}\). Let \(s'_i\) be a best response to \(\boldsymbol{s}_{-i}\) in \(G_{Rat}^{k}\). Then \(s'_i \in G_{Rat}^{k+1}\), since \(s'_i\) is not eliminated from \(G_{Rat}^{k}\). However, since \(s_i\) is a best response to \(\boldsymbol{s}_{-i}\) in \(G_{Rat}^{k+1}\), we get \(u_i(s_i, \boldsymbol{s}_{-i}) \geq u_i(s'_i, \boldsymbol{s}_{-i})\). Thus, by Definition 3.7, \(s_i\) is a best response to \(\boldsymbol{s}_{-i}\) in \(G_{Rat}^{k}\).
By the induction hypothesis, \(s_i\) is a best response to \(\boldsymbol{s}_{-i}\) in \(G\) and the lemma has been proven.
The proof of Theorem 3.1 is not necessary for the final exam!
Proof. Let us now focus on Theorem 3.1 and prove \(R_i^k \subseteq D_i^k\) for all players \(i\) by induction on \(k\).
For \(k = 0\) we have \(R_i^0 = S_i = D_i^0\) by definition (see Definition 3.4 and Definition 3.9).
Assume that \(R_i^k \subseteq D_i^k\) for some \(k \geq 0\) and prove that \(R_i^{k+1} \subseteq D_i^{k+1}\). Let \(s_i \in R_i^{k+1}\), then there must be \(\boldsymbol{s}_{-i}\in R^k_{-i}\) such that \(s_i\) is a best response to \(\boldsymbol{s}_{-i}\) in \(G_{Rat}^{k}\), as \(s_i\) must have not been eliminated in \(G_{Rat}^{k}\). By the Lemma 3.1, \(s_i\) is a best response to \(\boldsymbol{s}_{-i}\) in \(G\) as well. Also, by the induction hypothesis, \(s_i \in R^{k+1}_i \subseteq R^k_i \subseteq D^k_i\) and \(\boldsymbol{s}_{-i}\in R^k_{-i}\subseteq D^k_{-i}\).
However, then \(s_i\) is a best response to \(\boldsymbol{s}_{-i}\) in \(G_{DS}^{k}\), as this follows from the fact, that the “best response” relationship of \(s_i\) and \(\boldsymbol{s}_{-i}\) is preserved when removing arbitrarily many other strategies. Thus \(s_i\) is not strictly dominated in \(G_{DS}^{k}\) and \(s_i \in D^{k+1}_i\).
This concludes the proof.
3.5 Nash equilibria
We may raise certain criticisms of previous approaches:
- Strictly dominant strategy equilibria often do not exist;
- IESDS and rationalizability may not remove any strategies.
A typical example is the Battle of Sexes:
| \(O\) | \(F\) | |
|---|---|---|
| \(O\) | \((2,1)\) | \((0,0)\) |
| \(F\) | \((0,0)\) | \((1,2)\) |
Here all strategies are equally reasonable according to the above concepts. But are all strategy profiles really equally reasonable? Assume that each player has a belief about the strategies of other players. By Conjecture 3.3, each player plays a best response to his beliefs, so is \((O, F)\) as reasonable as \((O, O)\) in this respect? Note that if player 1 believes that player 2 plays \(O\), then playing \(O\) is reasonable, and if player 2 believes that player 1 plays \(F\), then playing \(F\) is reasonable. But such beliefs cannot be correct together! Hence \((O, O)\) can be obtained as a profile where each player plays the best response to his belief and the beliefs are correct.
Nash equilibrium can be defined as a set of beliefs (one for each player) and a strategy profile in which every player plays a best response to his belief and each strategy of each player is consistent with the beliefs of his opponents.
Definition 3.11 (Nash Equilibrium) A pure-strategy profile \(\boldsymbol{s}^* = (s^*_1, \dots, s^*_n) \in S\) is a (pure) Nash equilibrium if \(s^*_i\) is a best response to \(\boldsymbol{s}^*_{-i}\) for each \(i \in N\), that is \[ u_i(s^*_i, \boldsymbol{s}^*_{-i}) \geq u_i(s_i, \boldsymbol{s}^*_{-i}) \] for all \(s_i \in S_i\) and all \(i \in N\).
Note that this definition is equivalent to the previous one in the sense that \(\boldsymbol{s}^*_{-i}\) may be considered as the (consistent) belief of player \(i\) to which he plays a best response \(s^*_i\).
Example 3.7 In the Prisoner’s dilemma, \((C, C)\) is the only Nash equilibrium.
| \(C\) | \(S\) | |
|---|---|---|
| \(C\) | \((-5,-5)\) | \((0,-20)\) |
| \(S\) | \((-20,0)\) | \((-1,-1)\) |
Example 3.8 In the Battle of Sexes, only \((O, O)\) and \((F, F)\) are Nash equilibria.
| \(O\) | \(F\) | |
|---|---|---|
| \(O\) | \((2,1)\) | \((0,0)\) |
| \(F\) | \((0,0)\) | \((1,2)\) |
In Cournot duopoly, see Section 3.3.3, \((\theta/3, \theta/3)\) is the only Nash equilibrium, as best response relations \(q_1 = (\theta- q_2)/2\) and \(q_2 = (\theta- q_1)/2\) are both satisfied only by \(q_1 = q_2 = \theta/3\).
3.5.1 Stag Hunt
Two (in some versions more than two) hunters, players 1 and 2, can each choose to hunt
- stag (\(S\)) = a large tasty meal,
- hare (\(H\)) = also tasty but small,
but hunting stag is much more demanding and the forces of both players need to be joined (hare can be hunted individually). In a strategy-form game model, we have \(N = \left\{1,2\right\}, S_1 = S_2 = \left\{S,H\right\}\) and payoff of form
| \(S\) | \(H\) | |
|---|---|---|
| \(S\) | \((5,5)\) | \((0,3)\) |
| \(H\) | \((3,0)\) | \((3,3)\) |
Clearly, there are two Nash equilibria – \((S,S)\) and \((H,H)\), where the former is strictly better for each player than the latter! Which one is more reasonable?
If each player believes that the other one will go for a hare, then \((H, H)\) is a reasonable outcome \(\implies\) a society of individualists who do not cooperate at all. On the other hand, if each player believes that the other will cooperate, then this anticipation is self-fulfilling and results in what can be called a cooperative society.
This is supposed to explain that in the real world, there are societies that have similar endowments, access to technology and physical environment but have very different achievements, all because of self-fulfilling beliefs (or norms of behavior).
Another point of view might be to notice that \((H,H)\) is less risky. The minimum secured by playing \(S\) is 0 as opposed to 3 by playing \(H\) (We will get to this minimax principle later). So it seems to be rational to expect \((H, H)\)?
Theorem 3.2
- If \(\boldsymbol{s}^*\) is a strictly dominant strategy equilibrium, then it is the unique Nash equilibrium.
- Each Nash equilibrium is rationalizable and survives IESDS.
- If \(S\) is finite, neither rationalizability nor IESDS creates new Nash equilibria.
Corollary 3.2 Assume that \(S\) is finite. If rationalizability or IESDS results in a unique strategy profile, then this profile is a Nash equilibrium.
3.5.2 Interpretations of Nash Equilibria
Although we have seen the rigorous definition of a Nash equilibrium, it can be more intuitively interpreted in the following two ways.
When the goal is to give advice to all of the players in a game (i.e., to advise each player on what strategy to choose), any advice that was not an equilibrium would have the unsettling property that there would always be some player for whom the advice was bad, in the sense that, if all other players followed the parts of the advice directed to them, it would be better for some player to do differently than he was advised. If the advice is an equilibrium, however, this will not be the case, because the advice to each player is the best response to the advice given to the other players.
On the other hand, when the goal is prediction rather than prescription, a Nash equilibrium can also be interpreted as a potential stable point of a dynamic adjustment process in which individuals adjust their behavior to that of the other players in the game, searching for strategy choices that will give them better results.