# 经济代写|博弈论代写Game Theory代考|The Strategic-Form Representation of Extensive-Form Games

#### Doug I. Jones

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## 经济代写|博弈论代写Game Theory代考|The Strategic-Form Representation of Extensive-Form Games

Our next step is to relate extensive-form games and equilibria to the strategic-form model. To define a strategic form from an extensive form, we simply let the pure strategies $s \in S$ and the payoffs $u_i(s)$ be exactly those we defined in the extensive form. A different way of saying this is that the same pure strategies can be interpreted as either extensive-form or strategic-form objects. With the extensive-form interpretation, player $i$ “waits” until $h_i$ is reached before deciding how to play there; with the strategic-form interpretation, he makes a complete contingent plan in advance.

Figure 3.8 illustrates this passage from the extensive form to the strategic form in a simple example. We order player 2’s information sets from left to right, so that, for example, the strategy $s_2=(\mathrm{L}, \mathrm{R})$ means that he plays $L$ after L $_{\text {ind }} \mathrm{R}$ after D.

As another example, consider the Stackelberg game illustrated in figure 3.3. We will again order player 2’s information sets from left to right, so that player 2’s strategy $\hat{s}_2=(4,4,3)$ means that he plays 4 in response to $q_1=3$, plays 4 in response to 4 , and plays 3 in response to 6. (This strategy happens to be player 2 ‘s Cournot reaction function.) Since player 2 has three information sets and three possible actions at each of these sets, he has 27 pure strategies. We trust that the reader will forgive our not displaying the strategic form in a matrix diagram!

There can be several extensive forms with the same strategic form, as the example of simultaneous moves shows: Figures $3.4 \mathrm{a}$ and $3.4 \mathrm{~b}$ both correspond to the same strategic form for the Cournot game.

At this point we should note that the strategy space as we have defined it may be unnecessarily large, as it may contain pairs of strategies that are “equivalent” in the sense of having the same consequences regardless of how the opponents play.

## 经济代写|博弈论代写Game Theory代考|The Equivalence between Mixed and Behavior Strategies in Games of Perfect Recall

The equivalence between mixed and behavior strategies under perfect recall is worth explaining in some detail, as it also helps to clarify the workings of the extensive-form model. Any mixed strategy $\sigma_i$ of the strategic form (not of the reduced strategic form) generates a unique behavior strategy $h_i$ as follows: Let $R_i\left(h_i\right)$ be the set of player $i$ ‘s pure strategies that do not preclude $h_i$, so that for all $s_i \in R_i\left(h_i\right)$ there is a profile $s_{-i}$ for player $i$ ‘s opponents that reaches $h_i$. If $\sigma_i$ assigns positive probability to some $s_i$ in $R_i\left(h_i\right)$, define the probability that $b_i$ assigns to $a_i \in A\left(h_i\right)$ as
$$b_i\left(a_i \mid h_i\right)=\sum_{i s_i \in R_i\left(h_i\right) \text { and } s_i\left(h_i\right)=a_i ;} \sigma_i\left(s_i\right) / \sum_{\left{s_i \in R_i\left(h_i\right)i\right.} \sigma_i\left(s_i\right) \text {. }$$ If $\sigma_i$ assigns probability 0 to all $s_i \in R_i\left(h_i\right)$, then set $$b_i\left(a_i \mid h_i\right)=\sum{\left{s_i\left(h_i\right)=a_i i\right.} \sigma_i\left(s_i\right)^?$$
In either case, the $b_i(\cdot \mid \cdot)$ are nonnegative, and
$$\sum_{a_i \in A\left(h_i\right)} b_i\left(a_i \mid h_i\right)=1,$$
because each $s_i$ specifies an action for player $i$ at $h_i$.
Note that in the notation $b_i\left(a_i \mid h_i\right)$, the variable $h_i$ is redundant, as $a_i \in A\left(h_i\right)$, but the conditioning helps emphasize that $a_i$ is an action that is feasible at information set $h_i$.

# 博弈论代考

## 经济代写|博弈论代写Game Theory代考|The Equivalence between Mixed and Behavior Strategies in Games of Perfect Recall

$$b_i\left(a_i \mid h_i\right)=\sum_{i s_i \in R_i\left(h_i\right) \text { and } s_i\left(h_i\right)=a_i ;} \sigma_i\left(s_i\right) / \sum_{\left{s_i \in R_i\left(h_i\right)i\right.} \sigma_i\left(s_i\right) \text {. }$$如果$\sigma_i$为所有$s_i \in R_i\left(h_i\right)$分配概率0，则设置$$b_i\left(a_i \mid h_i\right)=\sum{\left{s_i\left(h_i\right)=a_i i\right.} \sigma_i\left(s_i\right)^?$$

$$\sum_{a_i \in A\left(h_i\right)} b_i\left(a_i \mid h_i\right)=1,$$

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