<h3>Showing the left-hand inequality</h3>

<p>Let's rearrange the equation,
<span class="math-container">$$
\begin{align*}
x &amp;= r_i \frac{\mu\left( B^p_i\left( E \right) \right)}{\mu\left( E \right)} - \frac{\mu\left( X \cap \left[ B^p_i\left( E \right) \setminus E \right] \right)}{\mu\left( E \right)} \\
&amp; \iff \underbrace{x \mu\left( E \right) + \mu\left( X \cap \left[ B^p_i\left( E \right) \setminus E \right] \right)}_{\text{LHS}} = r_i \mu\left( B^p_i\left( E \right) \right).
\end{align*}
$$</span>
Since all terms in <span class="math-container">$\text{LHS}$</span> are weakly positive, we can omit <span class="math-container">$\mu( X \cap [ B^p_i( E ) \setminus E ] )$</span> from <span class="math-container">$\text{LHS}$</span>, giving
<span class="math-container">$$
x \mu\left( E \right) \leq r_i \mu\left( B^p_i\left( E \right) \right).
$$</span>
Since <span class="math-container">$\mu( E ) \geq p \mu( B^p_i( E ) )$</span>, this directly implies <span class="math-container">$x p \leq r_i$</span>.</p>

<h3>Showing the right-hand inequality</h3>

<p>Since <span class="math-container">$\mu( E ) \leq \mu( B^p_i( E ) )$</span> and <span class="math-container">$x \geq 0$</span>, we have
<span class="math-container">$$
\begin{align*}
&amp;\frac{1}{\mu\left( E \right)} \left[ r_i \mu\left( B^p_i\left( E \right) \right) - \mu\left( X \cap \left[ B^p_i\left( E \right) \setminus E \right] \right) \right] \\
&amp;\geq \frac{1}{\mu\left( B^p_i\left( E \right) \right)}\left[ r_i \mu\left( B^p_i\left( E \right) \right) - \mu\left( X \cap \left[ B^p_i\left( E \right) \setminus E \right] \right) \right] \\
&amp;= \underbrace{r_i - \frac{\mu\left( X \cap \left[ B^p_i\left( E \right) \setminus E \right] \right)}{\mu\left( B^p_i\left( E \right) \right)}}_{\text{RHS}}.
\end{align*}
$$</span>
It must be that <span class="math-container">$X \cap [ B^p_i( E ) \setminus E ] \subseteq B^p_i( E ) \setminus E$</span>, so
<span class="math-container">$$
\text{RHS} \geq r_i - \frac{\mu\left( B^p_i\left( E \right) \setminus E \right)}{\mu\left( B^p_i\left( E \right) \right)} = r_i - \frac{\mu\left( B^p_i\left( E \right) \right) - \mu\left( E \right)}{\mu\left( B^p_i\left( E \right) \right)} = r_i - \left( 1 - \mu\left( \left. E \right| B^p_i\left( E \right) \right) \right).
$$</span>
By definition (and this is inequality (9) in the text) <span class="math-container">$\mu( E | B^p_i( E ) ) \geq p$</span>, thus
<span class="math-container">$$
\text{RHS} \geq r_i - \left( 1 - p \right).
$$</span>
Noting that the original equality was <span class="math-container">$x = \text{RHS}$</span>, we have
<span class="math-container">$$
x \geq r_i - \left( 1 - p \right) \;\; \iff \;\; r_i \leq x + 1 - p.
$$</span></p>

CC BY-SA 4.0
2018-11-09T15:36:35.807Z
2018-11-09T15:36:35.807Z
kyle
<p>I'm trying to understand Monderer and Samet's 1989 paper on approximate common knowledge.</p>

<p>I'm stuck in the last part of the proof of the "agreeing to disagree" Theorem A, where the upper bound of the posterior is established. The lower bound seems clear but I just cannot understand how one of the terms is written as (1-p). Am pasting a picture below of the part where I'm lost:</p>

<p><a href="https://i.sstatic.net/910xH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/910xH.png" alt="enter image description here"></a></p>

<p>The last line which establishes the upper bound of r_i is unclear to me. Any help would be greatly appreciated. The link of the paper is : </p>

<p><a href="https://ie.technion.ac.il/~dov/cpb_monderer_samet.pdf" rel="nofollow noreferrer">https://ie.technion.ac.il/~dov/cpb_monderer_samet.pdf</a></p>

<p>(Theorem A is in page 180-181)</p>

<game-theory>
Approximating common knowledge with common beliefs (Monderer and Samet, 1989)
Swaminathan B
# Approximating common knowledge with common beliefs (Monderer and Samet, 1989)

(Asked by: username_0 on 2018-11-08)

I'm trying to understand Monderer and Samet's 1989 paper on approximate common knowledge.

I'm stuck in the last part of the proof of the "agreeing to disagree" Theorem A, where the upper bound of the posterior is established. The lower bound seems clear but I just cannot understand how one of the terms is written as (1-p). Am pasting a picture below of the part where I'm lost:

[![enter image description here](https://i.sstatic.net/910xH.png)](https://i.sstatic.net/910xH.png)

The last line which establishes the upper bound of r\_i is unclear to me. Any help would be greatly appreciated. The link of the paper is :

<https://ie.technion.ac.il/~dov/cpb_monderer_samet.pdf>

(Theorem A is in page 180-181)

**Tags:** game-theory

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## 1 answers

### Answer by username_1 on 2018-11-09. Score: 0 | Accepted answer

### Showing the left-hand inequality

Let's rearrange the equation,
$$
\begin{align\*}
x &= r\_i \frac{\mu\left( B^p\_i\left( E \right) \right)}{\mu\left( E \right)} - \frac{\mu\left( X \cap \left[ B^p\_i\left( E \right) \setminus E \right] \right)}{\mu\left( E \right)} \\
& \iff \underbrace{x \mu\left( E \right) + \mu\left( X \cap \left[ B^p\_i\left( E \right) \setminus E \right] \right)}\_{\text{LHS}} = r\_i \mu\left( B^p\_i\left( E \right) \right).
\end{align\*}
$$
Since all terms in $\text{LHS}$ are weakly positive, we can omit $\mu( X \cap [ B^p\_i( E ) \setminus E ] )$ from $\text{LHS}$, giving
$$
x \mu\left( E \right) \leq r\_i \mu\left( B^p\_i\left( E \right) \right).
$$
Since $\mu( E ) \geq p \mu( B^p\_i( E ) )$, this directly implies $x p \leq r\_i$.

### Showing the right-hand inequality

Since $\mu( E ) \leq \mu( B^p\_i( E ) )$ and $x \geq 0$, we have
$$
\begin{align\*}
&\frac{1}{\mu\left( E \right)} \left[ r\_i \mu\left( B^p\_i\left( E \right) \right) - \mu\left( X \cap \left[ B^p\_i\left( E \right) \setminus E \right] \right) \right] \\
&\geq \frac{1}{\mu\left( B^p\_i\left( E \right) \right)}\left[ r\_i \mu\left( B^p\_i\left( E \right) \right) - \mu\left( X \cap \left[ B^p\_i\left( E \right) \setminus E \right] \right) \right] \\
&= \underbrace{r\_i - \frac{\mu\left( X \cap \left[ B^p\_i\left( E \right) \setminus E \right] \right)}{\mu\left( B^p\_i\left( E \right) \right)}}\_{\text{RHS}}.
\end{align\*}
$$
It must be that $X \cap [ B^p\_i( E ) \setminus E ] \subseteq B^p\_i( E ) \setminus E$, so
$$
\text{RHS} \geq r\_i - \frac{\mu\left( B^p\_i\left( E \right) \setminus E \right)}{\mu\left( B^p\_i\left( E \right) \right)} = r\_i - \frac{\mu\left( B^p\_i\left( E \right) \right) - \mu\left( E \right)}{\mu\left( B^p\_i\left( E \right) \right)} = r\_i - \left( 1 - \mu\left( \left. E \right| B^p\_i\left( E \right) \right) \right).
$$
By definition (and this is inequality (9) in the text) $\mu( E | B^p\_i( E ) ) \geq p$, thus
$$
\text{RHS} \geq r\_i - \left( 1 - p \right).
$$
Noting that the original equality was $x = \text{RHS}$, we have
$$
x \geq r\_i - \left( 1 - p \right) \;\; \iff \;\; r\_i \leq x + 1 - p.
$$


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