New answers tagged nash-equilibrium
2
There is no apriori reason for the Best Response map to be a contraction in general. Here's a simple example (since Battle of Sexes has been my go-to for the past few days):
$$
\begin{array}{c|lcr}
\text{Player1/Player 2$\rightarrow$} & \text{F} & \text{T} \\
\hline
\text{F} & 3,1 & 0,0 \\
\text{T} & 0,0 & 1,3 \\
\end{array}
$$
Denote ...
1
To prove that that some given combination of strategies is a Nash equilibrium you don't need to use a fixed-point theorem (such as Brouwer's or the fixed-point theorem for contractions on Banach spaces).
What you do have to do, is check that they are best responses to one another.
This true for mixed and pure strategies.
You also seem to be asking how you ...
2
The best way to visualise what's going on would be to use Harsanyi's transformation. I'm not drawing the game tree here (but I think Tirole has it in his example).
Let's set up notations first. We will denote player 1's strategy by $x=Pr(T)$. We will call the decision of player 2 following the realisation of game $i$ by $y_i=Pr(T)$ - i.e. player 2, following ...
2
Matrix looks correct. To final all pure strategy BNEs, you'll have to discuss cases based on the value of $p$.
For example, if $p\in(0,1)$, then $FT$ is player 2's unique best response to $F$. Thus, to have a BNE, you'd want $F$ to be player 1's best response to $FT$ as well, meaning that you'd require $3p>1-p$, or $p>\frac14$. Hence, $(F,FT)$ is a BNE ...
0
Notice that if a player is indifferent between two strategies, she gets the same payoff from either strategy. This means that anything is optimal (a best response): Playing either pure strategy or any mixed strategy. This includes the mixed strategy which makes her opponent indifferent.
Of course, if the player is not indifferent, she will choose either one ...
0
The question being investigated by the video is the existence of Nash equilibria, not the optimal choices by the players.
There are two obvious pure Nash equilibrium joint strategies, namely both play B or both play F, since in either case a deviation from the strategy by one of the players brings a negative expected effect for that play is the other goes ...
3
Suppose player $i$ plays the mixed strategy $\mathbb{P}_i(B)= p_i$, and assume for now that the support of $\mathbb{P}_i$ is $\{B,F\}$ (i.e. player 1 plays a fully mixed strategy). For both $B$ and $F$ to be in 1's support, he must obtain the same expected payoff from either strategy (otherwise, he would put all the weight on the strategy with the higher ...
1
why would a player want to balance out the payoffs of another player
I don't think anyone is saying that a player wants to do this. But in mixed equilibrium their strategy is such that this property holds. Without this property, any mixed strategy of the other player would be suboptimal.
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. ...
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