In this paper, it said that given $\omega$ is the current state of the world and $P_1(\omega)$ is the unique element from the information partition $\textit{P_1}$ of actor 1 in which the player 1 is informed that $P_1(\omega)$ contains $\omega$. According to the aforementioned paper (p.1237), it says that "To say that 1 knows that 2 knows (Event) E means that E includes all $P_2$ in the information partition $\textit{P_2}$ that intersect $P_1(\omega)$." I wonder why is the aforementioned intersection in this passage so important, since intersection here only mean that two information partition elements share some states.

I think there are two thing I need to understand: The interpretation of the shared states of two information partition elements from different people; the interpretation of the case when two informational partition elements intersect.

Also why is it seem to be inapplicable for the case when the partition elements mentioned are partially in $E$


2 Answers 2


I'm not an expert in epistemic game theory but let me see if I can help with an example.

Suppose the state space is $\Omega = \{1,2,3,4,5\}$, and there are two players with information partitions $$ \mathcal{P}_1 = \left\{\{1,2,3\},\{4,5\} \right\} \\ \mathcal{P}_2 = \left\{\{1,2\},\{3,4\},\{5\}\right\} $$ Suppose the true state is $\omega = 5$. Let us consider the event $E = \{4,5\}$.

Player 1 "knows" $E$, since $P_1 = \{4,5\} \subset E$.

Player 2 "knows" $E$, since $P_2 = \{5\} \subset E$.

However, does Player 1 know that Player 2 knows $E?$ No.

What are the elements of $\mathcal{P}_2$ that intersect $P_1$? $\{3,4\}$ and $\{5\}$.

However, $\{3,4\} \not \subset E$, so Player 1 does not know that Player 2 knows $E$.

On the other hand, does Player 2 know that Player 1 knows $E$? Yes.

What are the elements of $\mathcal{P}_1$ that intersect $P_2$? $\{4,5\}$, and $\{4,5\} \subset E$, so Player 2 knows that Player 1 knows $E$.

  • $\begingroup$ I understand the concept algebraically. However, I still do not understand intuitively why (in your example) {3,4} also has to belong to E for Player 1 to know that Player 2 knows E. Also what about the case when a partition element is partially in $E$? $\endgroup$
    – Aqqqq
    Commented Nov 17, 2019 at 14:50

The intersection itself is not terribly interesting, what matters here is only that it is not empty- that it contains some state.

To know event $E$ means that one does not consider it possible that the event $E$ does not obtain and that $E$ does indeed obtain. Since the true state is always considered possible, that $i$ knows $E$ at $\omega$ means that $\mathcal{P}_i(\omega)\subseteq E$- all states considered possible by $i$ at $\omega$ are in $E$. Now, that states at which $i$ knows $E$ are an event itself. Write $K_i(E)$ for this event. Then $$K_i(E)=\{\omega\in\Omega\mid \mathcal{P}_i(\omega)\subseteq E\}$$ is the set of states at which $i$ knows $E$. We are now looking at the event $$K_1\big(K_2(E)\big),$$ the event that $1$ knows that $2$ knows that $E$ holds. Now, $$K_1\big(K_2(E)\big)=\{\omega\in\Omega\mid \mathcal{P}_1(\omega)\subseteq K_2(E)\}$$ $$=\{\omega\in\Omega\mid \mathcal{P}_1(\omega)\subseteq K_2(E)\}=\big\{\omega\in\Omega\mid \mathcal{P}_1(\omega)\subseteq \{\omega'\in\Omega\mid \mathcal{P}_2(\omega')\subseteq E\}\big\}$$ $$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}_1(\omega)\text{ then } \mathcal{P}_2(\omega')\subseteq E\big\}$$ $$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}_1(\omega), P\in\mathcal{P_2}, \omega'\in P\text{ then } P\subseteq E\big\}$$ $$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}_1(\omega)\cap P, P\in\mathcal{P_2}\text{ then } P\subseteq E\big\}$$ $$=\big\{\omega\in\Omega\mid \text{ if }\mathcal{P}_1(\omega)\cap P\neq\emptyset, P\in\mathcal{P_2}\text{ then } P\subseteq E\big\}.$$


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