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Since the choice of $q_i$ can be conditioned on $(s_i,s_j)$, strategies in this game are of the form $(\hat s_i, \hat q_i(s_i,s_j))$. For the given values $\alpha=3$ and $k=1$, the SPNE can be calculated as the profile where $s^*_i=1/3$ and $q^*_i\equiv 1$. Indeed, production levels $q_i^*=1$ are the unique NE in all subgames, independently of the chosen $... 1 I think this is a typo in the paper. As far as I see, the houses$\{h_5, h_6, h_7\}$are not occupied, so it should be that$H_V = \{h_5, h_6, h_7\}$. 0 I'll assume that$S$is a constant. By the way, the solution for the SPNE is$q_i = \frac{k}{\alpha - 2}$which is derived from$q_i = (q_j + k)/(\alpha - 1) = ((q_i + k)/(\alpha - 1) + k)/(\alpha - 1)$. To answer your question, one should know what SPNE does. It is actually a refinement of NE by eliminating non-credible threats. Knowing this, the idea ... 1 This is a static Bayesian game. The two players share a common prior over the type space which is characterized by the uniform distribution. For Bayesian games, you have to specify the belief of each player given his/her type, and the belief of each player determines their action. Basically, the strategy is a mapping from one's type to an action through ... 4 Claim: If choice sets$T, M,$and$A$are finite, then an assessment$\{\beta^*_{r}, \beta^*_{s}, \mu^*\}$is a WPBE (weak perfect Bayesian equilibrium) of the two-stage signalling game between receiver$r$and sender$s$if and only if it is a SE (sequential equilibrium). Proof: SE$\implies$WPBE is trivial since SEs are PBEs by construction, and thus are ... 0 Let's first consider the case with the discount factor$\rho = 1$. In any finitely repeated case, the NE is to play (cheat, cheat) simply because of backward induction. A sophisticated rational player would think that, in the last round, the other player has no chance to punish him by playing cheat in the next round because there's no next round, therefore,... 0 Edit: This is an incorrect answer. Please disregard and see comments below. I don't think tit-for-tat is NE for infinitely repeated games. Tit-for-tat simply is, on average, the best strategy against many other strategies. Tit-for-tat will lose out, for example, if the only other strategy is to deviate all the time. Read Dawkin's The Selfish Gene for more ... 3 (i) In the 1 round case, tit-for-tat is not a NE. To see this notice that the tit-for-tat strategy, as you describe, dictates that the players play$(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$. Perhaps what you missed is that the ... 0 Here is a proof that the response given by Schelling in Eric’s answer is (as good as) the best A can do. A can only do so much in retaliation for B and C teaming up against him. He can threaten to shoot B with certainty, threaten to shoot C with certainty, or some mix of the two. Let’s call y the probability he will shoot C if they team up against him, ... -1 If only one player is allow to make a statement, we have the following Theorems: Theorem 1: If$N\gt 3$, then survival probability for that contender must be strictly less than$\frac{N-1}N$. Proof: WLOG, suppose only player$1$can make a statement and players$2$~$N$must keep silent. Notice that by killing player$1$in their turns, players$2$~$N$... 0 Schelling himself gave an answer for Q1 in which A can achieve a surviving probability arbitrarily close to 5/6. I quote the book below. The wording is painfully tortuous at times. But the answer seems right. I don't know if it's THE BEST result A can achieve. Professor Schelling offered a solution that guarantees Anderson near certainty of survival, by ... 1 A caveat, this is not really an answer, more of an extended comment--that's why I've put it as community wiki. This is a cool game, one of commitment. I'll try to formalize the problem. This is a first pass, so people should edit as they see fit. It's probably easiest to solve the case where,$T$, the number of rounds before a forced execution is$1$. Let ... 2 As a first point, in general, the dotted line indicates that the nodes are in the same information set. So in your example, there is a single information set for the Red player and as such she has 2 actions---since she cannot tell which of the 3 nodes she is in, she cannot condition her action on this information, and therefore must take the same action (... 2 The text says domination is "in the game remaining after one stage of elimination". In stage one, column L can be eliminated. As this is exercise b., I am guessing that was exercise a. 1 Let me answer here putting together all the hints in the comments. You've figured out that the equilibrium bid for a type$v$bidder is$b(v) = \frac{2}{3}v$. The bids are strictly increasing in$v$, so bidder$i$wins whenever$b(v_i) \geq b(v_j)$or$v_i \geq v_j$. The expected payoff for a type$v$bidder is thus $$(v -b(v))P(b_1 \geq b_2)$$ or $$(... 1 That is a fine proof. Though you probably want to show your last statement "The other inequalities don't require \epsilon_n to be restricted". It is fairly obvious, but won't hurt to be super clear. Perhaps a simpler proof would be that any convex combination of the two strategies already found will also work. That is \lambda(0,1/2,1/2)+(1-\lambda)(0,1/3,... 1 Pick an arbitrary cdf F that is supported on [-a,a]. The median, m, must satisfy$$\int_{-a}^{m}dF \geq \frac{1}{2}, \quad \text{and} \quad \int_{m}^{a}dF \geq \frac{1}{2}$$Note that$m$is not generally$0$. The phrase you mention, "the unique NE is where both candidates chooses the policy associated with the median voter," simply means that the ... 2 You are along the right track but are making certain errors in your proof, Your notation is confusing. You haven't defined what$S^{'D}$is. I would suggest sticking with the defined strategy space and not creating unnecessary notation in the proof. When you write$v(s'_{i},s'_{-i}) \geq v(s_{i},s_{-i})$, you are not only changing player i's strategy but ... 3 Let$\pi_L(p,q) = 3pq + p(1-q) + 2(1-p)(1-q)$denote the expected payoff of the lecturer. The best response is defined as$BR_L(q) = \arg\max_{p \in [0,1]}\pi_L(p,q)\$. Now, note that \begin{align} \frac{\partial \pi_L(p,q)}{\partial p} \begin{cases} > 0 \quad& \text{for } q > 1/4,\\ = 0 & \text{for } q = 1/4,\\ < 0 & \text{for } q < ...