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经济代写|博弈论代考Game theory代写|ECON7062 Knowledge

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经济代写|博弈论代考Game theory代写|Knowledge

The hats riddle is a motivating example for studying knowledge. Consider $n$ players, each of which is wearing a hat that is either red (r) or blue (b). The players each observe the others’ hats, but do not observe their own.

An outside observer announces in the presence of all the players that “At least one of you has a red hat.” They now play the following (non-strategic) game: a clock is set to ring every minute. At each ring, anyone who knows the color of their hat announces it, in which case the game ends. Otherwise the game continues.
Exercise 4.1. What happens?
We will formally analyze what transpires, after introducing some concepts and notation.
4.1.2 Knowledge spaces
Consider a situation in which there is uncertainty regarding one or more variables of interest. The set of all possible combinations of values of these variables is called the set of states of the world, so that knowing the state of the world is equivalent to knowing all there is to know and thus having no uncertainty.

We refer to a player’s type as the information (or the type of information) this player has regarding the state of the world. Thus, for example, one player may know one variable of interest, another player may know another variable, and a third could know the sum of the first two variables. A fourth could just know whether the first variable is non-zero.
Formally, a knowledge space ${ }^5$ is a tuple $\left(N, \Omega,\left{T_i\right}_{i \in N},\left{t_i\right}_{i \in N}\right)$ where

• $N$ is a set of players.
• $\Omega$ is the space of the states of the world. We will assume here that $\Omega$ is finite. ${ }^6$
• $T_i$ the space of possible types of player $i$.
• $t_i: \Omega \rightarrow T_i$ is player $i$ ‘s private signal or type.
We define the information partition function $P_i: \Omega \rightarrow 2^{\Omega}$ by
$$P_i(\omega)=\left{\omega^{\prime}: t_i\left(\omega^{\prime}\right)=t_i(\omega)\right} .$$

经济代写|博弈论代考Game theory代写|Knowledge

Let $\left(N, \Omega,\left{T_i\right}_{i \in N},\left{t_i\right}_{i \in N}\right)$ be a knowledge space. Let $A \in 2^{\Omega}$ be an event. We say that player $i$ knows $A$ at $\omega$ if $P_i(\omega) \subseteq A$, or, equivalently, if $t_i\left(\omega^{\prime}\right)=t_i(\omega)$ implies that $\omega^{\prime} \in A$. Intuitively, when $\omega$ occurs, $i$ does not know that – she only knows that $P_i(\omega)$ occurred. She thus knows that $A$ occurred only if $P_i(\omega)$ is contained in $A$.
Given an event $A \in 2^{\Omega}$, let $K_i A$ be the set of states of the world in which $i$ knows $A$ :
$$K_i A=\left{\omega: P_i(\omega) \subseteq A\right} .$$
Equivalently,
$$K_i A=\bigcup\left{P_i(\omega): P_i(\omega) \subseteq A\right} .$$
Theorem 4.2 (Kripke’s S5 system). 1. $K_i \Omega=\Omega$. A player knows that some state of the world has occurred.

$K_i A \cap K_i B=K_i(A \cap B)$. A player knows $A$ and a player knows $B$ if and only if she knows $A$ and $B$.

Axiom of knowledge. $K_i A \subseteq A$. If a player knows $A$ then $A$ has indeed occurred.

Axiom of positive introspection. $K_i K_i A=K_i A$. If a player knows $A$ then she knows that she knows A.

Axiom of negative introspection. $\left(K_i A\right)^c=K_i\left(\left(K_i A\right)^c\right)$. If a player does not know A then she knows that she does not know $A$.

Proof. 1. This follows immediately from the definition.
2.
\begin{aligned} K_i A \cap K_i B & =\left{\omega: P_i(\omega) \subseteq A\right} \cap\left{\omega: P_i(\omega) \subseteq B\right} \ & =\left{\omega: P_i(\omega) \subseteq A, P_i(\omega) \subseteq B\right} \ & =\left{\omega: P_i(\omega) \subseteq A \cap B\right} \ & =K_i(A \cap B) . \end{aligned}

By definition, if $\omega \in K_i A$ then $P_i(\omega) \subseteq A$. Since $\omega \in P_i(\omega)$, it follows that $\omega \in A$. Hence $K_i A \subseteq A$

By the previous part we have that $K_i K_i A \subseteq K_i A$.
To see the other direction, let $\omega \in K_i A$, so that $P_i(\omega) \subseteq A$. Choose any $\omega^{\prime} \in P_i(\omega)$. Hence $P_i\left(\omega^{\prime}\right)=P_i(\omega)$, and it follows that $\omega^{\prime} \in K_i A$. Since $\omega^{\prime}$ was an arbitrary element of $P_i(\omega)$, we have shown that $P_i(\omega) \subseteq K_i A$. Hence, by definition, $\omega \in K_i K_i A$.

The proof is similar to that of the previous part.
Interestingly, if a map $L: 2^{\Omega} \rightarrow 2^{\Omega}$ satisfies the Kripke $S 5$ system axioms, then it is the knowledge operator for some type: there exists a type space $T$ and a function $t: \Omega \rightarrow T$ such that $L$ is equal to the associated knowledge operator given by
$$K A={\omega: P(\omega) \subseteq A},$$
where $P(\omega)=t^{-1}(t(\omega))$.

博弈论代写

经济代写|博弈论代考Game theory代写|Knowledge

4.1. 2 知识空间 的状态就等同于知道所有可以知道的，因此没有不确定性。 一个玩家可能知道另一个变量，第三个玩家可能知萛前两个变量的总和。第四个可能只知萺第一个变量是否非零。

$N$ 是一组玩家。

$\Omega$ 是世界状态的空间。我们将在这里假设 $\Omega$ 是有限的。 ${ }^6$

$T_i$ 可能类型玩家的空间 $i$.

$t_i: \Omega \rightarrow T_i$ 是玩家 $i$ 的私人信号或类型。

\left 缺少或无法识别的分隔符

经济代写|博弈论代考Game theory代写|Knowledge

\left 缺少或无法识别的分隔符

〈eft 缺少或无法识别的分隔符

$K_i A \cap K_i B=K_i(A \cap B)$.一个玩家知道 $A$ 一个玩家知道 $B$ 当且仅当她知道 $A$ 和 $B$.

2.
〈left 缺少或无法识别的分隔符

$$K A=\omega: P(\omega) \subseteq A,$$

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