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I believe insights from repeated game theory have been used in designing the assignment scheme for security patrols in many occasions. In this matter, the main research projects with close connections to applications that I know of are (a) PROTECT for the US Coast Guard (b) TRUSTS for the Los Angeles Sheriff's Department (c) IRIS for the Federal Air ...

A couple hints. Regarding the lower bound on $\epsilon$: What happens if deviation occurs at stage $T$? In other words, there is no opportunity for your so-called "punishment stages". Regarding the upper bound on $\epsilon$: Suppose player 2 deviates at stage $T-1$ but player 1 does not. What must be true about $\epsilon$ in order for player 1 to ...

It is true because of the discounting. If the discount parameter were $\delta = 1$ then players 1 and 2 could alternate playing TL and MR in equilibrium and then reach the average payoff vector $1,1,5$. But if $\delta < 1$ someone will not play along. Suppose the players are supposed to start with TL. Then MR would follow in the next round then, then TL ...

There is also no NE which sustains coopration for more or less the same reason as in the SPNE case. Consider, a PD played twice. A strategy contains five actions, one for each decision node: one in the beginning (empty history) and one for each of the four period-2 histories (CC,CD,DC,DD). I claim that any strategy other than (D;D;D;D;D) is dominated. ...

Yes, a whole book has been written on Behavioral Game Theory. More specifically, standard solution concept such as Nash equilibrium requires that players best respond to a correct belief about other players' moves. The following are examples that relax one of these cognitive restrictions: Quantal response equilibrium allows for the possibility that ...

Why don't you attack the problem straighrforwardly and investigate the profitability of a deviation? Suppose there was a Nash equilibrium in which both players decide to stay forever. Both players obtain payoff $\frac{1}{1-d}$ on path. This really is a Nash equilibrium if anf only if there is no profitable deviation. Consider the deviation to quit in the ...

The payoff depends entirely on how you set the game up. Here's one example of the case where $2R \leq T + S$: Player B Cooperate | Defect Player Cooperate (1, 1) | (-2, 100) A Defect (100, -2) | (-1, -1) Here, $R = 1$, $T = 100$, and $S = -1$, and $2R < T ...

(i) In the 1 round case, tit-for-tat is not a NE. To see this notice that the tit-for-tat strategy, as you describe, dictates that the players play $(H,H)$ in the first (and only) round---as you point out, this is clearly not a NE, since either player can increase her payoff by changing her strategy from $H$ to $C$. Perhaps what you missed is that the ...

The problem you mentioned is why dominance is not really used frequently. It is a very weak concept in the sense that is usually has no "grip", i.e. not many strategies are eliminated. That is why we use Nash equilibria or even other concepts (e.g., the trembling hand perfect equilibrium which was mentioned by @desnesp, or, some form of subgame perfection). ...

A usual refinement concept used to deal with weakly dominated strategies is the trembling hand perfect equilibrium. (I do not know others but this one works quite well. The strategy in question is indeed weakly dominated by the following strategy In the first period, player plays $B$. In all other periods, player plays $A$.

Assume the players have to choose integers, otherwise a best response may not exist. Let the payoff of winning be $\alpha\cdot[\frac23\text{ of the average}]$, $\alpha>0$. Consider the two player (A and B) case, and let's verify whether choosing above $0$ is optimal. Suppose A chooses $x>0$. Then B can guarantee a win by choosing $x-1$, since $\...

I see two ways to produce histories with multi-state deviations (not including the degenerate case in which all states defect): Multiple states defect in different periods. This possibility is ruled out so long as Equation 3 (the necessary and sufficient condition for unilateral defection at time $t-1$ to be punished by all other states at time $t$) is ...

Finally, I believe it is easy to answer it. Well, in case where some player will not cooperate and the triger strategy is enabled, we have the following soloutions: Soppuse that p2 does not cooperate in the first round so, he will gain a payoof of 1, but in the second round p1 plays $a_2$ and he won't cooperate with p2 in perpetual. Thus, we have that p2 ...

In a finitely repeated game with a unique NE, the only SPNE is the repetition of the unique NE. The reason is that by backward induction the NE will be played in the last period and, hence, also in the penultimate period, and so on. Here, there are multiple NE so that the different NE can be used to reward and punish previous behavior. Your idea of how to ...

Though a bit unconventional to assume an infinitely repeated game without discounting, the short answer to your question is: yes. Any sequence of stage game Nash Equilibria is supportable as a SPNE. See, for example slide 11 of this deck, or these notes from Levin, starting on page 6. This is more directly addressed by the definition of SPNEs, found on page ...

What you are implying is thoroughly studied by what is called algorithmic game theory. There you can find the concept of a learning algorithm. In game theoretical scenarios where the mathematical equilibrium is difficult to evaluate, those algorithms are used to get an approximation. What they do (in a simplified way) 1) Player A starts with an array of ...

I do not want to do all the algebra, but I give you a hint. You are correct: You are asked about $\pi^C$ in your Condition (1) and you are also right that $\pi^D$ changes. In your parameter setting with $c=0$: $$\pi^C := \pi (q^C, q^C) = (1- 2 q^C) q^C,$$ where $q^C$ is the quantity they coordinate on. Then $\pi^D$ is the profit from best-responding to $q^...

@denesp nailed the answer on the head. This is not different from the original game in any meaningful way, therefore, the nash equilibrium remains the same. Size of payouts don't change the outcome of the game. This is because the strategy of choosing zero (assuming all parties are fully rational, a condition for Nash Equilibrium) dominates all other ...

I am going to construct pure strategies, taking average payoffs so far as a state variable, that achieve the payoffs $(4,4)$ in the infinitely repeated game. Call the row player's actions $T$, $M$, and $B$ for Top, Middle and Bottom respectively. Similarly, call the column player's actions $L$, $C$, and $R$. Define $$v_i^t = \frac{1 - \delta}{1 - \delta^t}...

Disclaimer: I only have a slight clue about repeated games and I have virtually no clue about coding (except the compulsory stuff I had to do in grad school). That being said, consider this stream of consciousness as how I would approach this problem (if I knew how to code). I am not sure this actually works, maybe someone could support this answer with a ...

Suppose all players play unconditioned Nash equilibria, except player $i$ who diverts and plays another strategy. The other strategy of player $i$ may depend on history. We have to show that player $i$'s payoff in the repeated game does not increase. For every time-period t, define the following random variables: $X^{t}$ - the utility of player $i$ in time ...

It seems to me that this will not make a difference if you only consider pure strategies. Consider that player $i$'s winnings is $x$ and from this he gets utility $U_i(x)$. As long as $U_i()$ is increasing in $x$ player $i$ will prefer outcomes with larger winnings to lower ones, which is exactly what he would do if he had constant marginal returns. Another ...