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

<p>Suppose the state space is <span class="math-container">$\Omega = \{1,2,3,4,5\}$</span>, and there are two players with information partitions
<span class="math-container">$$
\mathcal{P}_1 = \left\{\{1,2,3\},\{4,5\} \right\} \\
\mathcal{P}_2 = \left\{\{1,2\},\{3,4\},\{5\}\right\}
$$</span>
Suppose the true state is <span class="math-container">$\omega = 5$</span>.
Let us consider the event <span class="math-container">$E = \{4,5\}$</span>.</p>

<p>Player 1 "knows" <span class="math-container">$E$</span>, since <span class="math-container">$P_1 = \{4,5\} \subset E$</span>.</p>

<p>Player 2 "knows" <span class="math-container">$E$</span>, since <span class="math-container">$P_2 = \{5\} \subset E$</span>.</p>

<p>However, does Player 1 know that Player 2 knows <span class="math-container">$E?$</span> <strong>No</strong>.</p>

<p>What are the elements of <span class="math-container">$\mathcal{P}_2$</span> that intersect <span class="math-container">$P_1$</span>? <span class="math-container">$\{3,4\}$</span> and <span class="math-container">$\{5\}$</span>.</p>

<p>However, <span class="math-container">$\{3,4\} \not \subset E$</span>, so Player 1 does not know that Player 2 knows <span class="math-container">$E$</span>.</p>

<p>On the other hand, does Player 2 know that Player 1 knows <span class="math-container">$E$</span>? <strong>Yes</strong>.</p>

<p>What are the elements of <span class="math-container">$\mathcal{P}_1$</span> that intersect <span class="math-container">$P_2$</span>? <span class="math-container">$\{4,5\}$</span>, and <span class="math-container">$\{4,5\} \subset E$</span>, so Player 2 knows that Player 1 knows <span class="math-container">$E$</span>.</p>

CC BY-SA 4.0
2019-11-17T14:50:26.030Z
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 $E$?
Aqqqq
CC BY-SA 4.0
2019-11-16T05:50:13.107Z
2019-11-16T05:50:13.107Z
Walrasian Auctioneer
<p>The intersection itself is not terribly interesting, what matters here is only that it is not empty- that it contains some state.</p>
<p>To know event <span class="math-container">$E$</span> means that one does not consider it possible that the event <span class="math-container">$E$</span> does not obtain and that <span class="math-container">$E$</span> does indeed obtain. Since the true state is always considered possible, that <span class="math-container">$i$</span> knows <span class="math-container">$E$</span> at <span class="math-container">$\omega$</span> means that <span class="math-container">$\mathcal{P}_i(\omega)\subseteq E$</span>- all states considered possible by <span class="math-container">$i$</span> at <span class="math-container">$\omega$</span> are in <span class="math-container">$E$</span>. Now, that states at which <span class="math-container">$i$</span> knows <span class="math-container">$E$</span> are an event itself. Write <span class="math-container">$K_i(E)$</span> for this event. Then <span class="math-container">$$K_i(E)=\{\omega\in\Omega\mid \mathcal{P}_i(\omega)\subseteq E\}$$</span> is the set of states at which <span class="math-container">$i$</span> knows <span class="math-container">$E$</span>. We are now looking at the event <span class="math-container">$$K_1\big(K_2(E)\big),$$</span> the event that <span class="math-container">$1$</span> knows that <span class="math-container">$2$</span> knows that <span class="math-container">$E$</span> holds. Now,
<span class="math-container">$$K_1\big(K_2(E)\big)=\{\omega\in\Omega\mid \mathcal{P}_1(\omega)\subseteq K_2(E)\}$$</span>
<span class="math-container">$$=\{\omega\in\Omega\mid \mathcal{P}_1(\omega)\subseteq K_2(E)\}=\big\{\omega\in\Omega\mid \mathcal{P}_1(\omega)\subseteq \{\omega'\in\Omega\mid \mathcal{P}_2(\omega')\subseteq E\}\big\}$$</span>
<span class="math-container">$$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}_1(\omega)\text{ then } \mathcal{P}_2(\omega')\subseteq E\big\}$$</span>
<span class="math-container">$$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}_1(\omega), P\in\mathcal{P_2}, \omega'\in P\text{ then } P\subseteq E\big\}$$</span>
<span class="math-container">$$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}_1(\omega)\cap P, P\in\mathcal{P_2}\text{ then } P\subseteq E\big\}$$</span>
<span class="math-container">$$=\big\{\omega\in\Omega\mid \text{ if }\mathcal{P}_1(\omega)\cap P\neq\emptyset, P\in\mathcal{P_2}\text{ then } P\subseteq E\big\}.$$</span></p>

CC BY-SA 4.0
2021-01-07T12:38:57.527Z
2021-01-07T12:38:57.527Z
Michael Greinecker
<p>In this <a href="https://projecteuclid.org/download/pdf_1/euclid.aos/1176343654" rel="nofollow noreferrer">paper</a>, it said that given <span class="math-container">$\omega$</span> is the current state of the world and <span class="math-container">$P_1(\omega)$</span> is the unique element from the information partition <span class="math-container">$\textit{P_1}$</span> of actor 1 in which the player 1 is informed that <span class="math-container">$P_1(\omega)$</span> contains <span class="math-container">$\omega$</span>. 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 <span class="math-container">$P_2$</span> in the information partition <span class="math-container">$\textit{P_2}$</span> that intersect <span class="math-container">$P_1(\omega)$</span>."
 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. </p>

<p>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.</p>

<p>Also why is it seem to be inapplicable for the case when the partition elements mentioned are partially in <span class="math-container">$E$</span></p>

<game-theory><information>
How to interpret the intersected information partition element of different actors?
Aqqqq
# How to interpret the intersected information partition element of different actors?

(Asked by: username_0 on 2019-11-15)

In this [paper](https://projecteuclid.org/download/pdf_1/euclid.aos/1176343654), 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 $E$

**Tags:** game-theory, information


---

## 2 answers

### Answer by username_1 on 2019-11-16. Score: 1

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 $E$.

> **Comments:**
> **username_0**: 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 $E$?  

---

### Answer by username_2 on 2021-01-07. Score: 1

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,
$$K\_1\big(K\_2(E)\big)=\{\omega\in\Omega\mid \mathcal{P}\_1(\omega)\subseteq K\_2(E)\}$$
$$=\{\omega\in\Omega\mid \mathcal{P}\_1(\omega)\subseteq K\_2(E)\}=\big\{\omega\in\Omega\mid \mathcal{P}\_1(\omega)\subseteq \{\omega'\in\Omega\mid \mathcal{P}\_2(\omega')\subseteq E\}\big\}$$
$$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}\_1(\omega)\text{ then } \mathcal{P}\_2(\omega')\subseteq E\big\}$$
$$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}\_1(\omega), P\in\mathcal{P\_2}, \omega'\in P\text{ then } P\subseteq E\big\}$$
$$=\big\{\omega\in\Omega\mid \text{ if }\omega'\in\mathcal{P}\_1(\omega)\cap P, P\in\mathcal{P\_2}\text{ then } P\subseteq E\big\}$$
$$=\big\{\omega\in\Omega\mid \text{ if }\mathcal{P}\_1(\omega)\cap P\neq\emptyset, P\in\mathcal{P\_2}\text{ then } P\subseteq E\big\}.$$


---

