New answers tagged game-theory
0
There is a complete characterization of equilibria in Malueg (2010) [1].
The structure of these equilibria is in a sense a generalization of the property mentioned in the question - that demands should sum up to 100. Instead for every demand $x_s$ that Sarah makes with positive probability, there is some probability that Ruth would demand exactly the ...
1
This is probably an old question. However there is a book on PDE's and game theory.
Game Theory and Partial Differential Equations, Pablo Blanc and Julio Daniel Rossi.
If you found other books on this matter, I would like to know as well.
0
From the buyer's point of view, the "best" supplier is the one who offers them the highest utility. Using the utility function for scoring suppliers seems quite natural then. (In this paper the utility function is assumed to be quasilinear in money, which is just for simplicity.)
3
$BA'$ doesn't work because $A'$ is not a best response to $X$, which is a best response to $B$.
3
There are three classes of equilibria of this game.
The first class is sequential:
\begin{equation}
(s_1,s_2)=(y,r)
\end{equation}
and the beliefs are
\begin{equation}
\mu_1(a)=\mu_1(b)=\mu_2(a\mid y)=\mu_2(b\mid y)=\frac12.
\end{equation}
The second class is not sequential, but weak perfect Bayesian:
\begin{equation}
(s_1,s_2)=(x,l)
\end{equation}
and the ...
0
In the standard theory of games, simultaneous and sequential games are distinguished by the means of something called "Information Sets". Alluding to the origins of Game Theory (von Neumann and Morgenstern 1944), simultaneous move games can be thought of as special cases of sequential move games. In their landmark work, vN-M give an explicit method ...
1
Given that the system does converge, i.e. $a_i^t=a_i^{t+1}=a_i^*$ for all $i$ after some $T<\infty$, and that $a_i^{t+1}\in BR_i(a_{-i}^{t})$, it follows that $a_i^*\in BR_i(a_{-i}^*)$ for all $i$. Hence $a^*$ is a Nash equilibrium of the underlying game.
2
A payoff pair $(x,y)$ is Pareto efficient if it is not Pareto dominated. This means that there does not exist another payoff pair $(x', y')$ such that $x < x'$ and $y < y'$. (Depending on the exact definition, sometimes of of the two inequalities can be weak).
All payoffs in the payoff matrix except for $(C,\beta), (B, \beta)$ and $(C, \delta)$ are ...
7
No, there is not. Consider a game with two players, Ann and Bob. Both choose such vectors with entries $0$ or $1$ of the form $(a_1,a_2,\ldots,a_J)$ or $(b_1,b_2,\ldots,b_J)$, respectively. If $\sum_{i=1}^J a_i+b_i$ is odd, Ann wins and Bob loses. If the number is even, Ann loses and Bob wins. Clearly, one of them loses in every profile of pure strategies, ...
8
Not really. There are many compact metrizable topologies you can put on this space, but none that relate meaningfully to the structure of the problem.
Let's look first at the case $A_1=[0,1]$ and $A_2=\{0,1\}$. Consider the elevation function $e:A_1\times\Sigma\to\Delta(A_2)=[0,1]$ given by $e(a,\eta)=\eta(a)$. If you want the ultimate action choice of ...
3
In an experimental setting, how could you prevent the players from adopting a mixed strategy?
I don't think you can. Restricting access to mixed strategies is essentially
banning the use of any private randomization devices. But since there are various ways to perform mental coin-flips, not all of which are readily observable, it would be prohibitively ...
3
Consider the following game between P1 (row player) and P2 (column player):
\begin{array}{|c|c|c|}\hline
& L & R \\\hline
T& 1,1 & 2,0 \\\hline
B& 0,0 & 1,1 \\\hline
\end{array}
$T$ is P1's dominant strategy
$T$ is P1's best response to both of P2's strategies $L$ and $R$
$L$ is P2's best response to P1's strategy $T$
$R$ is P2's ...
3
Looking for the subgame perfect equilibrium here is hopeless; there really is more than one (in spite of what the assignment says).
When you use backward-induction, you have to use both. There is a subgame perfect equilibrium in which player $2$ chooses $d$ at the correspond decision node and a subgame perfect equilibrium in which player $2$ chooses $e$ at ...
4
Having spent time with a few eager students of game theory myself, I sense that the "real" question you might have wanted to ask was:
What is the best way to motivate people to get inoculated, given that they have varying attitudes towards the vaccine or vaccination in general.*
To answer that question from a normative perspective, i.e. assuming ...
4
You seem to be mistaken in what payoff goes to which player.
By convention, the first value goes to the Row player (here P1), and the second value goes to the Column player (here P2).
6
The Nash equilibrium is indeed (down, right).
Note that your chart has helpfully underlined the max value among all possible strategies each player can play, conditional on what the other player plays. The steps on the graph are useful as well: If no numbers are underlined in a cell, that strategy set is strictly dominated by something else. If one number ...
1
You have the right idea, make a 4 by 3 payoff matrix with fruit at the top and at the side weather/water yes/no, giving 12 different cells, then put the respective dollar payoffs.
You could also make another table and find the expected payoff by multiplying each payoff by its probability, although since all the probabilities are the same, the table wouldn’t ...
4
Decision under uncertainty is sometimes called a "game against chance", and can thus be modeled as a two-player normal form game: the decision-maker vs Nature/Chance. The possible states would form a set of pure strategies for Nature, and Nature commits to a publicly known mixed strategy that randomizes over those pure strategies (assume Nature is ...
5
The diagonally strict concavity property is better known as the strict monotonicity property of the pseudo-gradient.
An operator $\Psi:\mathbb{R}^n \to \mathbb{R}^n$ is strictly monontone if the following holds true:
$$\forall x_0, x_1 \in \mathbb{R}^n: \left< x_0 - x_1, \Psi(x_0) - \Psi(x_1) \right> >0. $$
In your case $\Psi$ would be $g$.
The ...
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