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# 数学代写|离散数学代写Discrete Mathematics代考|Recalling the Strategy Definition

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## 数学代写|离散数学代写Discrete Mathematics代考|Recalling the Strategy Definition

Before beginning to analyze sequential-move games, it is critically important to make sure that you completely understand the definition of “strategy.” Otherwise, you’re in for some major discomfort as you try to learn the material in this and the next parts of this book.

Consider the ultimatum-offer bargaining game just described. In this game, player 1’s strategy is simply a number $p$, which we can assume is between 0 and 100. Thus, the strategy space for player 1 is $S_1=[0,100]$. Player 2’s strategy is from a more complicated space. Note that player 2 has an infinite number of information sets, one for each of the feasible offers of player 1. For instance, one information set corresponds to player 1 having just made the offer $p=28$; another information set follows the offer $p=30.75$; another follows the offer $p=62$; and so on. Because there is an infinite number of points in the interval $[0,100]$, player 2 has an infinite number of information sets.

Remember that a strategy for a player is a complete contingent plan. Thus, player 2’s strategy must specify player 2’s choice between Yes and No at every one of player 2’s information sets. In other words, player 2’s strategy describes whether she will accept an offer of $p=28$, whether she will accept an offer of $p=30.75$, whether she will accept an offer of $p=62$, and so on. Formally, player 2’s strategy in this game can be expressed as a function that maps player 1 ‘s price offer $p$ to the set {Yes, No}. That is, considering $p \in[0,100]$, we can write player 2’s strategy as some function $s_2:[0,100] \rightarrow{$ Yes, No $}$. Then, for whatever offer $p$ that player 1 makes, player 2’s response is $s_2(p)$.

Here are some examples of strategies for player 2 in the ultimatum-offer bargaining game. A really simple strategy is a constant function. One such strategy specifies $s_2(p)=$ Yes for all $p$; this strategy accepts whatever player 1 offers. Another type of strategy for player 2 is a “cutoff rule,” which would accept any price at or below some cutoff value $\underline{p}$ and otherwise would reject. For a given number $\underline{p}$, this strategy is defined by
$$s_2(p)= \begin{cases}\text { Yes } & \text { if } p \leq \underline{p} \ \text { No } & \text { if } p>\underline{p}\end{cases}$$

## 数学代写|离散数学代写Discrete Mathematics代考|Incredible Threats in the Stackelberg Duopoly Game

Here is another example. Consider the Stackelberg duopoly game described in exercise 6 of Chapter $14^2$ In this game, firm 1 selects a quantity $q_1 \in[0,12]$, which is observed by firm 2 , and then firm 2 selects its quantity $q_2 \in[0,12]$. Firm 1’s payoff is $\left(12-q_1-q_2\right) q_1$, and firm 2’s payoff is $\left(12-q_1-q_2\right) q_2$. Remember that firm 2’s strategy can be expressed as a function $s_2:[0,12] \rightarrow[0,12]$ that maps firm 1’s quantity into firm 2’s quantity in response. This is because each value of $q_1$ yields a distinct information set for firm 2. Part (c) of the exercise asked you to confirm that for any $x \in[0,12]$, there is a Nash equilibrium of the game in which $q_1=x$ and $s_2(x)=(12-x) / 2$.
Let us check this assertion for $x=0$, the case in which firm 1 is supposed to produce $q_1=0$ and firm 2 is supposed to follow with the quantity $q_2=6$. In words, by producing nothing, firm 1 leaves the entire market to firm 2 . First verify that $q_2=6$ is the payoff-maximizing quantity for firm 2 when firm 1 produces 0 ; clearly $q_2=6$ maximizes $\left(12-q_2\right) q_2$. Next, note that this calculation is not enough to verify Nash equilibrium because we have not yet specified the strategy for firm 2. We have so far only specified that $s_2(0)=6$; we have not yet defined $s_2\left(q_1\right)$ for $q_1 \neq 0$. Furthermore, we need to check whether firm 1 would have the incentive to deviate from $q_1=0$.

Consider the following strategy for firm $2: s_2(0)=6$ and $s_2\left(q_1\right)=12-q_1$ for every $q_1 \neq 0$. Note that if firm 1 produces a positive amount, then firm 2 will produce exactly the amount that pushes the price (and therefore firm 1’s payoff) down to 0 . Clearly, against strategy $s_2$, firm 1 cannot gain by deviating from $q_1=0$. Furthermore, against $q_1=0$, firm 2 has no incentive to deviate from $s_2$. To see this, observe that by changing the specification $s_2(0)=6$, firm 2’s payoff would decrease. Moreover, changing the specification of $s_2(x)$ for any $x \neq 0$ would have no effect on firm 2’s payoff, as player 1’s strategy is $q_1=0$. Thus, $\left(0, s_2\right)$ is a Nash equilibrium.

## 数学代写|离散数学代写Discrete Mathematics代考|Recalling the Strategy Definition

$$s_2(p)= \begin{cases}\text { Yes } & \text { if } p \leq \underline{p} \ \text { No } & \text { if } p>\underline{p}\end{cases}$$

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