8
votes
Is a mixed strategy ever the best response to a pure strategy?
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 ...
4
votes
Incentive compatibility: Weakly dominant strategy versus Nash equilibrium?
A Nash equilibrium that consists of weakly dominant strategies is a stronger solution concept than a NE itself.
Consider the following simple matrix game where best replies have been marked with *
\...
- 2,373
4
votes
Accepted
Rationalizable action profiles in nice symmetric games
Isn't the mixed extension of matching pennies a good example of this? The strategies $p,q$ are both elements of $[0,1]$, and
$$
\begin{align*}
U_1(p,q) & = pq + (1-p)(1-q) \\
\\
U_2(p,q) & = p(...
- 29.7k
4
votes
Accepted
Game with an equilibrium in pure, but none in mixed strategies?
As I said in the comment, Nash's theorem shows the existence of a Nash equilibrium (possibly but not necessarily in mixed strategies). If you are interested in Nash equilibria is proper mixed ...
- 5,301
3
votes
Different payouts of pure strategies in mixed strategies
For the reason you mentioned, if a player randomizes in equilibrium over some pure strategies, then these strategies must give the same payoff against the strategy choices of the other players. The ...
- 13.7k
3
votes
Why is Dominant-Strategy Incentive Compatibility treated as a virtue, and are there mechanism design models which treat it as undesirable?
The games played in mechanisms are not "games in the colloquial sense". Whatever players really value is represented by their payoffs, so if they value strategic richness of a game, this is ...
- 7,450
3
votes
Differences between best response, dominant strategy and Nash equilibrium
Consider the following game between P1 (row player) and P2 (column player):
\begin{array}{|c|c|c|}\hline
& L & R \\\hline
T& 1,1 & 2,0 \\\hline
B& 0,0 & 1,1 \\\hline
\end{array}...
- 16.1k
3
votes
Fill out Normal-Form Game to obtain exactly one mixed equilibrium
To ensure that the game has a single, completely-mixed, Nash equilibrium there needs to be "cyclic unique best responses", i.e., either
BR(L) = O, BR(O) = R, BR(R) = U, BR(U) = L,
or
BR(L) = U, BR(...
- 286
2
votes
Accepted
Derivation of Equilibrium Strategy in 1st-price Auction?
I remember slaving over the notation in this book when I was a bad undergraduate. It brings up some interesting memories, some which may help you.
$F(x)$ is the cumulative distribution of a single ...
- 6,678
2
votes
Derivation of Equilibrium Strategy in 1st-price Auction?
I think you have a typo -- the equilibrium bidding strategy in the first-price auction you specify should be $$ \beta(x) = \frac{N-1}{N}x $$
Here's a hint that might help. The CDF of the order ...
- 2,156
2
votes
Nash Equilibrium and Dominant Strategy
Do you have a formal notion of what 'stable' means? Nash equilibria are often thought of informally as the strategies that support stable outcomes. If that's all that the term 'stable' means, then of ...
- 2,156
2
votes
Accepted
Nash Equilibrium and Dominant Strategy
You're on the right track here. You need to check every outcome for its potential to be a NE. You're correct in stating that outcomes (5,10) and (10,5) are NEs however you didn't identify that (5,5) ...
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2
votes
Is it possible to find a nash equilibrium that is not an equilibrium in weakly dominant strategy?
\begin{array}{|c|c|c|}
\hline
&L&R\\\hline
T&1,1&0,0\\\hline
B&0,0&0,0\\\hline
\end{array}
In the game above, there are two pure strategy Nash equilibria:
$(T,L)$ is an ...
- 16.1k
2
votes
Is a mixed strategy ever the best response to a pure strategy?
Repeated games and nonlinear utility
Let's assume a trivial two-player game where each player has two options A and B; and the payout is +1/-1 if players pick the same and -1/+1 if players pick ...
- 488
2
votes
Accepted
Term for the phenomenon when an undesirable but competitive strategy dominates
A game outcome that is Pareto optimal or Pareto efficient is one where no one player can be made better off without making at least one player worse off. So a Nash equilibrium can easily be Pareto sub-...
- 16.4k
2
votes
Rationalizable strategies and Weak Dominance
In 2-player games, the strategies that survive iterated elimination of strictly dominated strategies are called rationalizable. Note that even if no strategy is strictly dominant, there can be ...
- 5,301
2
votes
Accepted
Understanding the notations in Bayesian game definition
Set of players, I guess this is quite clear. As an example, take player $A$ and $B$
Actions. This just tells you what the two players can do. For example $A$ can choose $U$(p) or $D$(own) and $B$ can ...
- 12.8k
2
votes
How can we prove that an equilibrium in dominant strategies is a Nash equilibrium?
Your intuition about the proof is correct indeed. A more formal proof would involve examining the definitions of dominant strategy and Nash equilibrium:
A strategy $s_i^d$ is a dominant strategy for ...
- 16.1k
2
votes
Nash equilibrium in strictly mixed strategies
Nash's theorem says that every finite game has a NE in mixed strategies, but here mixed, coming without the qualifier strictly, implies the weak version that includes pure strategies. So your ...
- 7,450
2
votes
Accepted
Does Dominant Strategy Incentive Compatibility apply to this modification of Rock Paper Scissors?
DSIC is a possible property of a direct mechanism, not of an arbitrary game. A direct mechanism starts by players reporting their private information (i.e., types), so it is a very special kind of ...
- 7,450
2
votes
Accepted
Why is Dominant-Strategy Incentive Compatibility treated as a virtue, and are there mechanism design models which treat it as undesirable?
The answer by VARulle is perfectly fine. I just want to add something and may repeat his point for completeness.
A mechanism designer often has a policy goal such as efficiently allocating a good to ...
- 5,301
2
votes
Accepted
Can a mixed strategy that is strictly dominant exist if there is no strictly dominant pure strategy?
It is not possible for a mixed (non-pure) strategy to be strictly dominant. A mixed strategy can be weakly dominant, but only if all pure strategies in its support are weakly dominant.
This ...
- 136
1
vote
Accepted
Understanding mixed strategies
you would get a higher payoff by assigning a probability of 0 to the pure strategy that is dominated
This is not relevant, as no one is claiming that $\sigma_1$ is optimal in any way.
I can't find ...
- 29.7k
1
vote
Can game theoretic concepts be applied to any groups of strategies collectively partitioning the strategy space?
I see two ways to interpret the question.
Combinatorial game theory often studies (perfect information sequential) games with very large strategy spaces (like chess) often hoping to "solve" ...
- 215
1
vote
Strategic game with complete informaation
The given solution is as follows:
Suppose $a_1 < a_3$. Straightforward argument shows that the
set of actions that constitute pure best response for player 2 is $\{1, . . . , a_1\}$.
When $a_1 > ...
- 1,048
1
vote
Set of rationalizable strategies for this 4 x 4 matrix
The set of rationalizable strategies is the set of strategies that survive the iterated elimination of strictly dominated strategies, i.e., strategies that are never a best response. It is a weaker ...
- 5,301
1
vote
Accepted
Find value of $\beta$ for which there is a strictly dominant strategy
Hint
Let $h_i^*(h_{-i}$) be firm $i$'s best response to the other $N-1$ firms' strategy profile $h_{-i}$. If $h_i^*(\cdot)$ is a dominant strategy, then it must be independent of the other firms' ...
- 16.1k
1
vote
Is this market monopoly or duopoly?
The definition of a monopoly is one company having the exclusive control of a good or service. Since there are 2+ companies it does not fit the definition of a monopoly.
A duopoly is a situation ...
- 954
1
vote
Illustrating difficulty to find dominant strategies: series of simple one-shot two-players games
Start with the following matrix.
$$ \begin{matrix}1 && 0 && 0\\
1 && 1 && 0\\
1 && 1 && 1 \end{matrix}$$
You will choose a $1$ in the matrix and ...
- 356
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