<p>If you take the standard Aumann model, but allow for states occuring with probability zero, you can taje some nonempty event $N$ such that for the agents prior $p$, one has $p(N)=0$. Let $K\cap N=\emptyset$ and $p(K)&gt;0$. If the agent knows $K\cup N$ but not $K$, then she assigns probability $1$ to $K$, even though she does not know $K$.</p>

CC BY-SA 3.0
2015-05-11T00:19:57.907Z
2015-05-11T00:19:57.907Z
Michael Greinecker
<p>I have some difficulties parsing your question (especially the "this" in "it could be the case, an event is a proper subset of the event that player i knows this event obtains with probability one, but never the case for knowledge operator with absolute certainty") so maybe what I am proposing here is "trivial" (and therefore not what you are looking for). But I don't have enough reputation to comment so this ends up an "answer" anyway. See it as a question asking for clarification.</p>

<p>Let $X\sim Unif([0,5])$ be a random variable.</p>

<p>Let $E_1=\{X\in [0,5]\}$. The agent knows with certainty $E_1$, since the agent knows with certainty that $X\sim Unif([0,5])$.</p>

<p>Let $E_2 = E_1\setminus \{3.14\}$. That is, $E_2$ is the event that $X$ is not $3.14$. Now, $P(E_2)=1$ so the agent knows with probability $1$ that $E_2$. However, the agent does not know $E_2$ with absolute certainty since it is possible (but only with probability $0$) that $X$ is in fact $3.14$.</p>

<p>Is this what you have in mind?</p>

CC BY-SA 3.0
2015-02-11T08:11:28.683Z
Thank you. What you are proposing is correct. But I found them quite straightforward in light of definitions. I want to know why in literature, these two operators are distinguished, i.e. why interchange  one operator with the other could generate some interesting result.
Metta World Peace
CC BY-SA 3.0
2015-02-12T09:03:20.043Z
@MettaWorldPeace I would convert this to a comment but most of it would get cut off. I'm not sure what to do. I could either leave it here, because it's useful clarification. Or you could incorporate it into the original question as clarification.
jmbejara
CC BY-SA 3.0
2015-02-12T12:44:44.040Z
@jmbejara Thanks for attention. In that case, I prefer to leave it here, because it's Monkeynomics' contribution.
Metta World Peace
CC BY-SA 3.0
2015-02-10T21:22:15.177Z
2015-02-10T21:22:15.177Z
Monkeynomics
<p>An event with a probability $1$ happens <strong>almost surely</strong>. There is the same difference between that and <strong>absolute certainty</strong> that in English. One cannot definitively say that these outcomes will never occur, but can for most purposes assume this to be true. The distinction is important because if $A_n, n \in \mathbb{N} $, with for each $n$ $P(A_n)=1$, then you cannot assume $P(\cap_{n \in \mathbb{N}} A_n)=1 $, which you could if the events were absolutely certain, which is why one need to pay attention to the distinction, even though it has no impact on the immediate consequences of the statement.</p>

CC BY-SA 3.0
2015-05-11T12:58:29.730Z
2015-05-11T12:58:29.730Z
VicAche
<p>In the literature of interactive epistemology, for a player, knowing an event obtains with probability one and knowing an event obtains with absolute certainty are different. Is there a nontrivial (say, it could be the case, an event is a proper subset of the event that player $i$ knows this event obtains with probability one, but never the case for knowledge operator with absolute certainty) proposition that could serve as an example to show they're not interchangeable in the language which consists of two corresponding knowledge operators, a set of states of world, set relations, set operations, and all logical connectives.</p>

CC BY-SA 3.0
2015-02-06T13:47:08.300Z
Interesting question, could you please provide an example paper or book from this literature to orient others?
BKay
CC BY-SA 3.0
2015-02-06T14:09:47.643Z
@BKay  With pleasure. http://healy.econ.ohio-state.edu/aumannbrandenburger.pdf
Metta World Peace
<mathematical-economics><game-theory><academic-graduate>
Interchangeability between knowing an event obtains with probability 1 and knowing an event obtains with absolute certainty?
Metta World Peace
# Interchangeability between knowing an event obtains with probability 1 and knowing an event obtains with absolute certainty?

(Asked by: username_0 on 2015-02-06)

In the literature of interactive epistemology, for a player, knowing an event obtains with probability one and knowing an event obtains with absolute certainty are different. Is there a nontrivial (say, it could be the case, an event is a proper subset of the event that player $i$ knows this event obtains with probability one, but never the case for knowledge operator with absolute certainty) proposition that could serve as an example to show they're not interchangeable in the language which consists of two corresponding knowledge operators, a set of states of world, set relations, set operations, and all logical connectives.

**Tags:** mathematical-economics, game-theory, academic-graduate
> **Comments:**
> **username_1**: Interesting question, could you please provide an example paper or book from this literature to orient others?  
> **username_0**: @BKay  With pleasure. http://healy.econ.ohio-state.edu/aumannbrandenburger.pdf  

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## 3 answers

### Answer by username_2 on 2015-05-11. Score: 4

If you take the standard Aumann model, but allow for states occuring with probability zero, you can taje some nonempty event $N$ such that for the agents prior $p$, one has $p(N)=0$. Let $K\cap N=\emptyset$ and $p(K)>0$. If the agent knows $K\cup N$ but not $K$, then she assigns probability $1$ to $K$, even though she does not know $K$.


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### Answer by username_3 on 2015-02-10. Score: 2

I have some difficulties parsing your question (especially the "this" in "it could be the case, an event is a proper subset of the event that player i knows this event obtains with probability one, but never the case for knowledge operator with absolute certainty") so maybe what I am proposing here is "trivial" (and therefore not what you are looking for). But I don't have enough reputation to comment so this ends up an "answer" anyway. See it as a question asking for clarification.

Let $X\sim Unif([0,5])$ be a random variable.

Let $E\_1=\{X\in [0,5]\}$. The agent knows with certainty $E\_1$, since the agent knows with certainty that $X\sim Unif([0,5])$.

Let $E\_2 = E\_1\setminus \{3.14\}$. That is, $E\_2$ is the event that $X$ is not $3.14$. Now, $P(E\_2)=1$ so the agent knows with probability $1$ that $E\_2$. However, the agent does not know $E\_2$ with absolute certainty since it is possible (but only with probability $0$) that $X$ is in fact $3.14$.

Is this what you have in mind?

> **Comments:**
> **username_0**: Thank you. What you are proposing is correct. But I found them quite straightforward in light of definitions. I want to know why in literature, these two operators are distinguished, i.e. why interchange  one operator with the other could generate some interesting result.  
> **username_4**: @MettaWorldPeace I would convert this to a comment but most of it would get cut off. I'm not sure what to do. I could either leave it here, because it's useful clarification. Or you could incorporate it into the original question as clarification.  
> **username_0**: @jmbejara Thanks for attention. In that case, I prefer to leave it here, because it's Monkeynomics' contribution.  

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### Answer by username_5 on 2015-05-11. Score: 1

An event with a probability $1$ happens **almost surely**. There is the same difference between that and **absolute certainty** that in English. One cannot definitively say that these outcomes will never occur, but can for most purposes assume this to be true. The distinction is important because if $A\_n, n \in \mathbb{N} $, with for each $n$ $P(A\_n)=1$, then you cannot assume $P(\cap\_{n \in \mathbb{N}} A\_n)=1 $, which you could if the events were absolutely certain, which is why one need to pay attention to the distinction, even though it has no impact on the immediate consequences of the statement.


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