16
Two areas that have been profoundly affected by game theoretic research stemming from Nash's contribution are
Oligopoly theory
There are actually a few examples of what would come to be known as Nash equilibrium in the industrial organization literature that predate Nash's work (for example, Cournot's 1838 analysis of oligopoly competition). However, until ...
11
You're not alone in your skepticism of the relevance of game theory.
Some of the greats, including Gary Becker, were at times dismissive of the practical/empirical importance of game theory (see the introduction/preface of his Economic Theory book). No doubt it is in a way foundational to the economic sciences (see Myerson's great essay on Nash's ...
10
This is only half a joke : Nash-equilibrium gives a very good prediction on the relative size of groups of foraging ducks on a pond when two food sources are established at opposite sides of the pond.
A very good explanation can be found at https://headbiotech.wordpress.com/nash-equilibrium-example-on-ducks/, among other places (https://headbiotech....
8
Fixing the strategy of the opponent, a mixed strategy never yields a strictly higher utility if you are expected utility-maximizing.
The reason is that the expected utility from a mixed strategy is at most as high as the highest utility from the pure strategies which this mixed strategy plays with positive probability.
That is not to say that a mixed ...
7
Yes, there is always a pure Nash equilibrium. See:
I Milchtaich (1996). Congestion games with player-specific payoff functions.
Games and economic behavior 13 (1), 111-124.
You are interested in the special case of singleton congestion games with player-specific payoff functions.
And yes, they can be computed in polynomial time. See Corollary 7 in:
...
7
Let's review the definitions of the two concepts. Let $\sigma$ be a profile of strategies and $\mu$ a system of beliefs.
A pair $(\sigma,\mu)$ is a weak perfect Bayesian equilibrium (WPBE)
if
$\sigma$ is sequentially rational given $\mu$, and
$\mu$ is derived from $\sigma$ using Bayes rule whenever applicable.
and
A pair $(\sigma,\mu)...
6
Yes there absolutely is. It is generally agreed upon that utility is (at least) ordinal. That means I can compare utility levels for a single person (not necessarily across people) and the numbers have meaning in this sense. So for a single person:
$U(down, right) = 4 > 1 = U(Up, Left)$.
This means (Down,Right) is strictly better for each agent than (...
6
The following paper compares the efficiency of the Bertrand and Cournot game in the case of product differentiation. However, their utility function is more general than the Dixit/Stiglitz case. You also do not have an infinite number of firms competing in each differentiated product but just one firm per differentiated product (the number of differentiated ...
6
A leader-leader Stackelberg is a situation in the Stackelberg model where both firms believe they are leaders. This leads to global production being much higher than expected by both firms, as they anticipate small production by the other firm in response to their high production.
6
I have used gambit (python) in the past and could recommend it. It also includes a small GUI which makes things very intuitive.
R also has GameTheory if you prefer this platform instead
6
It is actually assumed that $b_i(v_i)$ is of the form $\alpha_i+\beta_i \cdot v_i$. So it is an affine function. Linearity only works if the bottom of the uniform distribution is 0.
A somewhat intuitive reasoning is that if the valuations are uniformly distributed over $[c,d]$ and $b_i(v_i)$ is a symmetric equilibrium, then $b_i(v_i)+k$ should define a ...
6
There are three players, 1,2,3, and the action spaces of players 1 and 2 are both $[0,1]\times\mathbb{N}$, the action space of player 3 is $[0,1]$.
The generic action of player 1 is written as $(x,m)$, the generic action of player 2 is written as $(y,n)$, and the generic action of player 3 is written as $(x')$.
The payoff of player 3 is simply $1$ if $x'=x$...
6
One reason why there exists no ultimate overview is that these issues are still under debate. A great entry point would be the survey "Foundations of Strategic Equilibrium" by John Hillas and Elon Kohlberg in Volume 3 of the Handbook of Game Theory with Economic Applications. A preprint version of the survey can be found here.
6
There is no guarantee that symmetric games have symmetric equilibria.
See this paper for concrete examples.
There is also no guarantee that symmetric games have pure-strategy equilibria. For example, the following game is symmetric and has no pure-strategy equilibria
$$
\begin{array}{cccc}
& \mathrm{a} & \mathrm{b} & \mathrm{c}\\
\mathrm{a} &...
6
There is a series of papers that address precisely this question. The most famous ones are probably Walker and Wooders (2001) and Chiappori, Levitt, and Groseclose (2002) that deal with penalty kicks and tennis serves. Both papers conclude that the behavior of professional athletes is consistent with them playing g a mixed strategy equilibrium. A more recent ...
5
The conventional definition of Bayesian Nash equilibrium (BNE) is as follows:
A pure strategy BNE is a profile of
type-contingent strategies $$(s_i(\theta_i),s_{-i}(\theta_{-i}))=(s_1(\theta_1),\dots,s_{i-1}(\theta_{i-1}),s_i(\theta_i),s_{i+1}(\theta_{i+1}),\dots,s_m(\theta_m))$$
such that for each $i\in P$, $$ s_i(\theta_i)\in\arg\max_{s_i'\in
S_i} ...
5
No. In static games, common knowledge of rationality is equivalent to rationalizability. Bernheim, in "Rationalizable Strategic Behavior" (Econometrica, July 1984) gives an example on page 1012 in Figure 1 of a normalform game in which there is a unique Nash equilibrium, yet multiple strategies are rationalizable.
5
Suppose $(\sigma_1^*, \sigma_2^*)$ is the Nash Equilibrium of the game $G$ consisting of players $\{1,2\}$ having strategy sets $A_1$ and $A_2$ and the payoff functions $u_1:A_1\times A_2 \rightarrow \mathbb{R}$ and $u_2:A_1\times A_2 \rightarrow \mathbb{R}$. Consider the game $G'$ in which everything else is same as $G$ except that the payoff functions are $...
5
Player 2
+---+-------+-------+-------+
| | A | B | C |
+---+-------+-------+-------+
Player 1 | A | (1,1) | (0,0) | (0,0) |
| B | (0,0) | (2,1) | (1,2) |
| C | (0,0) | (1,2) | (2,1) |
+---+-------+-------+-------+
If we allow $3\times 3$ games, then above is an example of the ...
5
Game theory explorer (GTE) offers a web-based solver that searches for Nash equilibria of the inputted games. Its documentation highlights the main differences between GTE and Gambit:
Gambit has been developed over the course of nearly 25 years and
presents a library of solution algorithms, formats for storing games,
ways to program the creation of ...
5
(i) $A$ & $B$
If player 1 play $A$ with probability $p$ and $B$ with probability $(1-p)$, where $0<p<1$, then player 2's expected payoff from playing
$D$ is $4p+4(1-p) = 4$
$E$ is $6p + 2(1-p) = 4p + 2$
$F$ is $6p + 4(1-p) = 2p + 4$
Since payoff from playing $F$ is more than the payoff from playing any other strategy for player 2, he will ...
5
Bayesian Nash equilibrium is a set of strategies $\{\sigma_i\}$ one for each player and some beliefs $\{\mu_i\}$ also one for each player such that $\sigma_i$ is a best response for player $i$ given his belief, $\mu_i$, and the beliefs are Bayesian for all players, given their information.
Each strategy $\sigma_i$ is a function from the set of types to a (...
5
Is there any more information about why von Neumann had this attitude?
There must be, but I have never seen it.
Or can we infer a reasonable answer?
Here is one possibility based on hearsay among the Game Theory community.
Von Neumann thought that there was a sharp distinction between zero-sum games and other games.
In zero-sum games, there was no ...
5
Level-k reasoning in the stag hunt game is analyzed in
Gracia-Lázaro, Carlos, Luis Mario FlorÃa, and Yamir Moreno. "Cognitive hierarchy theory and two-person games." Games 8.1 (2017): 1.
The idea that playing $s$ guarantees its payoff is discussed in
Aumann, Robert "Nash equilibria are not self-enforcing, in ‘‘Economic Decision-Making: Games, ...
5
The following two claims hold in the general $n$-shop case.
Claim 1. In equilibrium a shop closest to an edge (0 or 1) cannot be alone.
Proof. Such a shop could gain customers by moving slightly inward.
Claim 2. In equilibrium at most two shops can be in any location.
Proof. Assume there is an equilibrium where there are three or more shops in a location. ...
4
Well the intuition is quite straightforward: as you have mentioned yourself, player 2 is indifferent between playing any action in pure strategies, or randomizing in any possible way. So, for a mixed strategy equilibrium to exist, player 2 needs to play L w/ probability 4/7. If this is the case, it is irrelevant what Player 1 player---player 2's outcome is ...
4
So you want to find a maximum of $\sigma(s,z)$. If $\sigma$ is diagonally strictly concave you can do so by starting at any point and just following the gradient $g(s,z)$ until you find the maximum and no matter where you start, you will always end up at the same point (Start at the lower black points and follow the direction of the gradient (the direction ...
4
It will be useful to spell out the relevant definitions.
Let $A_i$ be the set of possible actions for player $i$, $A_{-i}$ be the set of possible joint actions of all players except for player $i$, $A$ be the set of possible joint actions of all the players. For $a=(a_i,a_{-i})\in A$, let $u_i(a)$ denote the payoff to player $i$ from action $a_i$, given ...
4
You are right. The problem here is a corner solution.
Let us define the optimal combination $X^* = \beta_1e_1^* + \beta_2e_2^*$.
The first derivative gives you
$$X^* = \dfrac{l_1}{2\beta_1} $$
whereas the second gives you
$$X^* = \dfrac{l_2}{2\beta_2}$$
They can only be true "by chance". This is, the existence of an interior solution cannot be ...
4
You could add up the equations
\begin{align}
\delta(1-\underline{S})S_j=e_i(S_i+S_j)^2 \\
\delta(1-\underline{S})S_i=e_j(S_i+S_j)^2
\end{align}
to get
\begin{align}
\delta(1-\underline{S})(S_i+S_j)=(e_i+e_j)(S_i+S_j)^2
\end{align}
which simplifies to
\begin{align}
\delta(1-\underline{S})=(e_i+e_j)(S_i+S_j).
\end{align}
This gives you $(S_i+S_j)$. From the ...
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