<p>Let
<span class="math-container">$$
w_s(\alpha) = \sum_{x \in X} x\,\,(\alpha f(x|s) +(1- \alpha) h(x|s)).
$$</span>
note that this function is continuous in <span class="math-container">$\alpha$</span>. Next, let
<span class="math-container">$$
w(\alpha) = \min_{s \in S} w_s(\alpha),
$$</span>
which is also continuous as it is a minimum of continuous functions.</p>
<p>Note that:
<span class="math-container">$$
w(1) = \min_s \sum_x x f(x|s), \text{ and } w(0) = \min_s \sum_x x h(x|s).
$$</span>
Also, by assumption <span class="math-container">$f \succeq_S g \succeq_S h$</span>, so<br />
<span class="math-container">$$
w(1) \ge \min_{s \in S} \sum_{x \in X} x g(x|s) \ge w(0).
$$</span>
By the intermediate value theorem, there is a <span class="math-container">$\gamma \in [0,1]$</span> such that:
<span class="math-container">$$
\sum_x x g(x|s) = w(\gamma) \equiv \min_{s \in S} \sum_{x \in X} x \,\,(\gamma f(x|s) + (1-\gamma) h(x|s)).
$$</span>
as such, <span class="math-container">$g \sim_S \gamma f + (1-\gamma) h$</span></p>

CC BY-SA 4.0
2023-10-14T07:17:23.627Z
2023-10-14T07:17:23.627Z
tdm
<p>I am self-studying game theory using Game Theory: Analysis of Conflict by Roger Myerson. Here is an exercise from the textbook. I tried it myself, but I am not sure if it is correct. I would really appreciate it if someone could help me check!</p>
<h2><strong>Question</strong></h2>
<blockquote>
<p>Suppose that the set of prizes <span class="math-container">$X$</span> is a finite subset of <span class="math-container">$\mathbb{R}$</span>, the set of real numbers, and a prize <span class="math-container">$x$</span> denotes an award of <span class="math-container">$x$</span> dollars. A decision-maker says that, if he knew that the true state of the world was in some set <span class="math-container">$T$</span>, then he would weakly prefer a lottery <span class="math-container">$f$</span> over another lottery <span class="math-container">$g$</span> (that is, <span class="math-container">$f \succsim_T g$</span>) if and only if
<span class="math-container">\begin{align*}
\min_{s \in T} \sum_{x \in X} xf(x|s) \geq \min_{s \in T} \sum_{x \in X} xg(x|s).
\end{align*}</span>
(That is, he prefers the lottery that gives the higher expected payoff in the worst possible state.) Does this preference relation satisfies the continuity axiom?</p>
</blockquote>
<h2><strong>Axiom</strong></h2>
<blockquote>
<p><strong>Continuity Axiom</strong><span class="math-container">$\quad$</span> <em>If <span class="math-container">$f \succsim_S g$</span> and <span class="math-container">$g \succsim_S h$</span>, then there exists some number <span class="math-container">$\gamma$</span> such that <span class="math-container">$0 \leq \gamma \leq 1$</span> and <span class="math-container">$g \sim_S \gamma f + (1 - \gamma)h$</span>.</em></p>
</blockquote>
<h2><strong>My Attempt</strong></h2>
<p>I am not completely sure whether this axiom is satisfied by the preference relation or not. I tried a couple of examples, but failed to show any violation. So I tried to prove that this preference relation does satisfy the continuity axiom, and here is my attempt:</p>
<blockquote>
<p><strong>Proof</strong><span class="math-container">$\quad$</span> Suppose <span class="math-container">$f \succsim_T g$</span> and <span class="math-container">$g \succsim_T h$</span>. Then,
<span class="math-container">\begin{align*}
\min_{s \in T} \sum_{x \in X} xf(x|s) \geq \min_{s \in T} \sum_{x \in X} xg(x|s) \geq \min_{s \in T} \sum_{x \in X} xh(x|s).\tag1
\end{align*}</span>
Let <span class="math-container">$s_1$</span> be any state of the world such that
<span class="math-container">\begin{align*}
\min_{s \in T} \sum_{x \in X} xg(x|s) \geq \sum_{x \in X} xh(x|s_1).\tag2
\end{align*}</span>
Let <span class="math-container">$\gamma \in [0,1]$</span>. By definition, we have
<span class="math-container">\begin{align*}
\sum_{x \in X}x(\gamma f + (1 - \gamma)h)(x|s) = \gamma \sum_{x \in X} xf(x|s) + (1 - \gamma) \sum_{x \in X} xh(x|s). 
\end{align*}</span>
We defined a procedure of finding a <span class="math-container">$\gamma \in [0,1]$</span> such that <span class="math-container">$g \sim_T \gamma f + (1 - \gamma)h$</span>, that is
<span class="math-container">\begin{align*}
\min_{s \in T} \sum_{x\ in X} x g(x|s) = \min_{s \in T} \left\{\gamma\sum_{x \in X}xf(x|s) + (1 - \gamma)\sum_{x \in X}xh(x|s)\right\}.
\end{align*}</span>
Let <span class="math-container">$s_1$</span> be the real state of the world. By (1) and (2), we have
<span class="math-container">\begin{align*}
\sum_{x \in X} xf(x|s_1) \geq \min_{s \in T} \sum_{x \in X} xg(x|s) \geq \sum_{x \in X} xh(x|s_1).
\end{align*}</span>
Thus, there exists a <span class="math-container">$\gamma_1 \in [0,1]$</span> such that
<span class="math-container">\begin{align*}
\min_{s \in T} \sum_{x \in X} xg(x|s) = \gamma_1 \sum_{x \in X} xf(x|s_1) + (1 - \gamma_1)\sum_{x \in X}xh(x|s_1).\tag3
\end{align*}</span>
If
<span class="math-container">\begin{align*}
\gamma_1 \sum_{x \in X} xf(x|s_1) + (1 - \gamma_1) \sum_{x \in X} xh(x|s_1) = \min_{s \in T} \sum_{x \in X} x(\gamma_1 f + (1 - \gamma_1)h)(x|s),
\end{align*}</span>
then we are done. If not, let <span class="math-container">$s_2$</span> denote the state of the world such that
<span class="math-container">\begin{align*}
\gamma_1 \sum_{x \in X} xf(x|s_2) + (1 - \gamma_1) \sum_{x \in X} xh(x|s_2) = \min_{s \in T} \sum_{x \in X} x(\gamma f + (1 - \gamma)h)(x|s).
\end{align*}</span>
Since, by (3),
<span class="math-container">\begin{align*}
\gamma_1 \sum_{x \in X} xf(x|s_2) + (1 - \gamma_1) \sum_{x \in X} xh(x|s_2) &lt; \min_{s \in T} \sum_{x \in X} xg(x|s),\tag4
\end{align*}</span>
and since, by (1),
<span class="math-container">\begin{align*}
\sum_{x \in X} xf(x|s_2) \geq \min_{s \in T} \sum_{x \in X} xg(x|s),
\end{align*}</span>
we must have
<span class="math-container">\begin{align*}
\sum_{x \in X} xh(x|s_2) &lt; \min_{s \in T} \sum_{x \in X} xg(x|s).
\end{align*}</span>
Then, there must exists a <span class="math-container">$\gamma_2 \in [0,1]$</span> such that
<span class="math-container">\begin{align*}
\gamma_2 \sum_{x \in X} xf(x|s_2) + (1 - \gamma_2) \sum_{x \in X} xh(x|s_2) = \min_{s \in T} \sum_{x \in X} xg(x|s).\tag5
\end{align*}</span>
Next, I prove that
<span class="math-container">\begin{align*}
\gamma_2 \sum_{x \in X} xf(x|s_1) + (1 - \gamma_2) \sum_{x \in X} xh(x|s_1) \geq \min_{s \in T} \sum_{x \in X} xg(x|s).\tag6
\end{align*}</span>
Assume to the contrary that
<span class="math-container">\begin{align*}
\gamma_2 \sum_{x \in X} xf(x|s_1) + (1 - \gamma_2) \sum_{x \in X} xh(x|s_1) &lt; \min_{s \in T} \sum_{x \in X} xg(x|s).
\end{align*}</span>
Then
<span class="math-container">\begin{align*}
\gamma_2 &lt; \frac{\min_{s \in T} \sum_{x \in X} xg(x|s) - \sum_{x \in X}xh(x|s_1)}{\sum_{x \in X} xf(x|s_1) - \sum_{x \in X}xh(x|s_1)} = \gamma_1,
\end{align*}</span>
where the equality comes from equation (3). But, by (4), we have
<span class="math-container">\begin{align*}
\gamma_1 &lt; \frac{\min_{s \in T} \sum_{x \in X} xg(x|s) - \sum_{x \in X}xh(x|s_2)}{\sum_{x \in X}xf(x|s_2) - \sum_{x \in X}xh(x|s_2)} = \gamma_2,
\end{align*}</span>
where the equality comes from equation (5). Therefore, we get a contradiction. So, the inequality (6) holds. Moreover, (6) implies <span class="math-container">$\gamma_2 \geq \gamma_1$</span>. Again, if
<span class="math-container">\begin{align*}
\gamma_2 \sum_{x \in X} xf(x|s) + (1 - \gamma_2) \sum_{x \in X} xh(x|s) = \min_{s \in T} \sum_{x \in X} x(\gamma_2 f + (1 - \gamma_2)h)(x|s),
\end{align*}</span>
then we are done. If not, we repeat the above process. By this procedure, we are always able to find a <span class="math-container">$\gamma \in [0,1]$</span> such that <span class="math-container">$g \sim_T \gamma f + (1 - \gamma)h$</span>. Therefore, this preference relation satisfies the continuity axiom.</p>
</blockquote>
<h2><strong>My Question</strong></h2>
<p>Could someone please help me check if my answer is correct? I really appreciate it!</p>
<p>If the continuity axiom is, in fact, violated, please share a counterexample. If the continuity axiom is indeed satisfied, but my proof has some flaws or there is a better way to prove it, please also consider sharing it as an answer. Thank you very much in advance!</p>
<h2><strong>Background Information</strong></h2>
<p>For more background information about notations, axioms, and so forth, please refer to <a href="https://economics.stackexchange.com/questions/56281/about-theorem-1-1-in-game-theory-analysis-of-conflict-by-roger-myerson">this post</a>.</p>

CC BY-SA 4.0
2023-10-13T10:03:47.377Z
There might be a much easier proof using the intermediate value theorem.
tdm
CC BY-SA 4.0
2023-10-13T15:10:09.260Z
@tdm Thank you so much for your comment! I didn't quite get it how the intermediate value theorem would help prove this, because it assumes the function to be continuous. Could you please share your method as an answer? I really appreciate it!
Beerus
<microeconomics><mathematical-economics><game-theory><preferences><decision-theory>
Determine Whether A Preference Relation Satisfies The Continuity Axiom - from Exercise 1.1 in Game Theory: Analysis of Conflict by Roger Myerson
Beerus
# Determine Whether A Preference Relation Satisfies The Continuity Axiom - from Exercise 1.1 in Game Theory: Analysis of Conflict by Roger Myerson

(Asked by: username_0 on 2023-10-11)

I am self-studying game theory using Game Theory: Analysis of Conflict by Roger Myerson. Here is an exercise from the textbook. I tried it myself, but I am not sure if it is correct. I would really appreciate it if someone could help me check!

## **Question**

> Suppose that the set of prizes $X$ is a finite subset of $\mathbb{R}$, the set of real numbers, and a prize $x$ denotes an award of $x$ dollars. A decision-maker says that, if he knew that the true state of the world was in some set $T$, then he would weakly prefer a lottery $f$ over another lottery $g$ (that is, $f \succsim\_T g$) if and only if
> \begin{align\*}
> \min\_{s \in T} \sum\_{x \in X} xf(x|s) \geq \min\_{s \in T} \sum\_{x \in X} xg(x|s).
> \end{align\*}
> (That is, he prefers the lottery that gives the higher expected payoff in the worst possible state.) Does this preference relation satisfies the continuity axiom?

## **Axiom**

> **Continuity Axiom**$\quad$ *If $f \succsim\_S g$ and $g \succsim\_S h$, then there exists some number $\gamma$ such that $0 \leq \gamma \leq 1$ and $g \sim\_S \gamma f + (1 - \gamma)h$.*

## **My Attempt**

I am not completely sure whether this axiom is satisfied by the preference relation or not. I tried a couple of examples, but failed to show any violation. So I tried to prove that this preference relation does satisfy the continuity axiom, and here is my attempt:

> **Proof**$\quad$ Suppose $f \succsim\_T g$ and $g \succsim\_T h$. Then,
> \begin{align\*}
> \min\_{s \in T} \sum\_{x \in X} xf(x|s) \geq \min\_{s \in T} \sum\_{x \in X} xg(x|s) \geq \min\_{s \in T} \sum\_{x \in X} xh(x|s).\tag1
> \end{align\*}
> Let $s\_1$ be any state of the world such that
> \begin{align\*}
> \min\_{s \in T} \sum\_{x \in X} xg(x|s) \geq \sum\_{x \in X} xh(x|s\_1).\tag2
> \end{align\*}
> Let $\gamma \in [0,1]$. By definition, we have
> \begin{align\*}
> \sum\_{x \in X}x(\gamma f + (1 - \gamma)h)(x|s) = \gamma \sum\_{x \in X} xf(x|s) + (1 - \gamma) \sum\_{x \in X} xh(x|s).
> \end{align\*}
> We defined a procedure of finding a $\gamma \in [0,1]$ such that $g \sim\_T \gamma f + (1 - \gamma)h$, that is
> \begin{align\*}
> \min\_{s \in T} \sum\_{x\ in X} x g(x|s) = \min\_{s \in T} \left\{\gamma\sum\_{x \in X}xf(x|s) + (1 - \gamma)\sum\_{x \in X}xh(x|s)\right\}.
> \end{align\*}
> Let $s\_1$ be the real state of the world. By (1) and (2), we have
> \begin{align\*}
> \sum\_{x \in X} xf(x|s\_1) \geq \min\_{s \in T} \sum\_{x \in X} xg(x|s) \geq \sum\_{x \in X} xh(x|s\_1).
> \end{align\*}
> Thus, there exists a $\gamma\_1 \in [0,1]$ such that
> \begin{align\*}
> \min\_{s \in T} \sum\_{x \in X} xg(x|s) = \gamma\_1 \sum\_{x \in X} xf(x|s\_1) + (1 - \gamma\_1)\sum\_{x \in X}xh(x|s\_1).\tag3
> \end{align\*}
> If
> \begin{align\*}
> \gamma\_1 \sum\_{x \in X} xf(x|s\_1) + (1 - \gamma\_1) \sum\_{x \in X} xh(x|s\_1) = \min\_{s \in T} \sum\_{x \in X} x(\gamma\_1 f + (1 - \gamma\_1)h)(x|s),
> \end{align\*}
> then we are done. If not, let $s\_2$ denote the state of the world such that
> \begin{align\*}
> \gamma\_1 \sum\_{x \in X} xf(x|s\_2) + (1 - \gamma\_1) \sum\_{x \in X} xh(x|s\_2) = \min\_{s \in T} \sum\_{x \in X} x(\gamma f + (1 - \gamma)h)(x|s).
> \end{align\*}
> Since, by (3),
> \begin{align\*}
> \gamma\_1 \sum\_{x \in X} xf(x|s\_2) + (1 - \gamma\_1) \sum\_{x \in X} xh(x|s\_2) < \min\_{s \in T} \sum\_{x \in X} xg(x|s),\tag4
> \end{align\*}
> and since, by (1),
> \begin{align\*}
> \sum\_{x \in X} xf(x|s\_2) \geq \min\_{s \in T} \sum\_{x \in X} xg(x|s),
> \end{align\*}
> we must have
> \begin{align\*}
> \sum\_{x \in X} xh(x|s\_2) < \min\_{s \in T} \sum\_{x \in X} xg(x|s).
> \end{align\*}
> Then, there must exists a $\gamma\_2 \in [0,1]$ such that
> \begin{align\*}
> \gamma\_2 \sum\_{x \in X} xf(x|s\_2) + (1 - \gamma\_2) \sum\_{x \in X} xh(x|s\_2) = \min\_{s \in T} \sum\_{x \in X} xg(x|s).\tag5
> \end{align\*}
> Next, I prove that
> \begin{align\*}
> \gamma\_2 \sum\_{x \in X} xf(x|s\_1) + (1 - \gamma\_2) \sum\_{x \in X} xh(x|s\_1) \geq \min\_{s \in T} \sum\_{x \in X} xg(x|s).\tag6
> \end{align\*}
> Assume to the contrary that
> \begin{align\*}
> \gamma\_2 \sum\_{x \in X} xf(x|s\_1) + (1 - \gamma\_2) \sum\_{x \in X} xh(x|s\_1) < \min\_{s \in T} \sum\_{x \in X} xg(x|s).
> \end{align\*}
> Then
> \begin{align\*}
> \gamma\_2 < \frac{\min\_{s \in T} \sum\_{x \in X} xg(x|s) - \sum\_{x \in X}xh(x|s\_1)}{\sum\_{x \in X} xf(x|s\_1) - \sum\_{x \in X}xh(x|s\_1)} = \gamma\_1,
> \end{align\*}
> where the equality comes from equation (3). But, by (4), we have
> \begin{align\*}
> \gamma\_1 < \frac{\min\_{s \in T} \sum\_{x \in X} xg(x|s) - \sum\_{x \in X}xh(x|s\_2)}{\sum\_{x \in X}xf(x|s\_2) - \sum\_{x \in X}xh(x|s\_2)} = \gamma\_2,
> \end{align\*}
> where the equality comes from equation (5). Therefore, we get a contradiction. So, the inequality (6) holds. Moreover, (6) implies $\gamma\_2 \geq \gamma\_1$. Again, if
> \begin{align\*}
> \gamma\_2 \sum\_{x \in X} xf(x|s) + (1 - \gamma\_2) \sum\_{x \in X} xh(x|s) = \min\_{s \in T} \sum\_{x \in X} x(\gamma\_2 f + (1 - \gamma\_2)h)(x|s),
> \end{align\*}
> then we are done. If not, we repeat the above process. By this procedure, we are always able to find a $\gamma \in [0,1]$ such that $g \sim\_T \gamma f + (1 - \gamma)h$. Therefore, this preference relation satisfies the continuity axiom.

## **My Question**

Could someone please help me check if my answer is correct? I really appreciate it!

If the continuity axiom is, in fact, violated, please share a counterexample. If the continuity axiom is indeed satisfied, but my proof has some flaws or there is a better way to prove it, please also consider sharing it as an answer. Thank you very much in advance!

## **Background Information**

For more background information about notations, axioms, and so forth, please refer to [this post](https://economics.stackexchange.com/questions/56281/about-theorem-1-1-in-game-theory-analysis-of-conflict-by-roger-myerson).

**Tags:** microeconomics, mathematical-economics, game-theory, preferences, decision-theory

> **Comments:**
> **username_1**: There might be a much easier proof using the intermediate value theorem.  
> **username_0**: @username_1 Thank you so much for your comment! I didn't quite get it how the intermediate value theorem would help prove this, because it assumes the function to be continuous. Could you please share your method as an answer? I really appreciate it!  

---

## 1 answers

### Answer by username_1 on 2023-10-14. Score: 3 | Accepted answer

Let
$$
w\_s(\alpha) = \sum\_{x \in X} x\,\,(\alpha f(x|s) +(1- \alpha) h(x|s)).
$$
note that this function is continuous in $\alpha$. Next, let
$$
w(\alpha) = \min\_{s \in S} w\_s(\alpha),
$$
which is also continuous as it is a minimum of continuous functions.

Note that:
$$
w(1) = \min\_s \sum\_x x f(x|s), \text{ and } w(0) = \min\_s \sum\_x x h(x|s).
$$
Also, by assumption $f \succeq\_S g \succeq\_S h$, so  
$$
w(1) \ge \min\_{s \in S} \sum\_{x \in X} x g(x|s) \ge w(0).
$$
By the intermediate value theorem, there is a $\gamma \in [0,1]$ such that:
$$
\sum\_x x g(x|s) = w(\gamma) \equiv \min\_{s \in S} \sum\_{x \in X} x \,\,(\gamma f(x|s) + (1-\gamma) h(x|s)).
$$
as such, $g \sim\_S \gamma f + (1-\gamma) h$


---

