<p>Let's first have a look at the left hand side of the equation. Take an outcome <span class="math-container">$y$</span> and a state <span class="math-container">$r$</span>. There are two cases:</p>
<ol>
<li><p>If <span class="math-container">$y$</span> is not the worst outcome in state <span class="math-container">$r$</span> then the lottery gives:
<span class="math-container">$$
\left(\frac{1}{|\Omega|}\right) f(y|r) + \left(1-\frac{1}{|\Omega|}\right) \underbrace{a_0(y|r)}_{=0} = \left(\frac{1}{|\Omega|}\right) f(y|r)
$$</span>
Notice that <span class="math-container">$a_0(y|r)$</span> is zero as the probability of getting <span class="math-container">$y$</span> in state <span class="math-container">$r$</span>  is zero (as <span class="math-container">$y$</span> is not the worst outcome).</p>
</li>
<li><p>If <span class="math-container">$y$</span> is the worst outcome in state <span class="math-container">$r$</span> then <span class="math-container">$a_0(y|r) = 1$</span> and we get instead
<span class="math-container">$$
\left(\frac{1}{|\Omega|}\right) f(y|r) + \left(1-\frac{1}{|\Omega|}\right) \underbrace{a_0(y|r)}_{=1}= \left(\frac{1}{|\Omega|}\right)f(y|r) + \left(1 - \frac{1}{|\Omega|}\right).
$$</span></p>
</li>
</ol>
<p>Now, let us focus on the right hand side and show that it is the same in both cases. First let us split up the sum into the different cases, case I where (<span class="math-container">$x \ne y$</span> and <span class="math-container">$t \ne r$</span>), case II where (<span class="math-container">$x \ne y$</span> and <span class="math-container">$t = r$</span>), case III where (<span class="math-container">$x = y$</span> and <span class="math-container">$t \ne r$</span>) and finally, case IV where (<span class="math-container">$t = r$</span> and <span class="math-container">$x = y$</span>).</p>
<p><span class="math-container">$$
\begin{align*}
\frac{1}{|\Omega|} \sum_{x \in X} \sum_{t \in \Omega} f(x|t)\,\,c_{x,t}(y|r)
=&amp; \frac{1}{|\Omega|} \sum_{x \ne y} \sum_{t \ne r} f(x|t) c_{x,t}(y|r), &amp;&amp; \text{case I}\\
&amp;+ \frac{1}{|\Omega|} \sum_{x \ne y} f(x|r) c_{x,r}(y|r), &amp;&amp; \text{case II}\\
&amp;+ \frac{1}{|\Omega|} \sum_{t \ne r} f(y|t) c_{y,t}(y|r), &amp;&amp; \text{case III}\\
&amp;+ \frac{1}{|\Omega|} f(y|r) c_{y,r}(y|r). &amp;&amp; \text{case IV}
\end{align*}
$$</span>
Substituting out <span class="math-container">$c_{x,t}(y|r)$</span> with <span class="math-container">$[x](y|r)$</span> if <span class="math-container">$t = r$</span> and <span class="math-container">$a_0(y|r)$</span> when <span class="math-container">$t \ne r$</span> gives
<span class="math-container">$$
\begin{align*}
\frac{1}{|\Omega|} \sum_{x \in X} \sum_{t \in \Omega} f(x|t)\,\,c_{x,t}(y|r)
=&amp; \frac{1}{|\Omega|} \sum_{x \ne y} \sum_{t \ne r} f(x|t) a_0(y|r),\\
&amp;+ \frac{1}{|\Omega|} \sum_{x \ne y} f(x|r) \underbrace{[x](y|r)}_{=0},\\
&amp;+ \frac{1}{|\Omega|} \sum_{t \ne r} f(y|t) a_0(y|r),\\
&amp;+ \frac{1}{|\Omega|} f(y|r) \underbrace{[y](y|r)}_{=1}.
\end{align*}
$$</span>
Notice that <span class="math-container">$[x](y|r)$</span> is zero if <span class="math-container">$x \ne y$</span> and equal to <span class="math-container">$1$</span> if <span class="math-container">$x = y$</span>. This gives:
<span class="math-container">$$
\begin{align*}
\frac{1}{|\Omega|} \sum_{x \in X} \sum_{t \in \Omega} f(x|t)\,\,c_{x,t}(y|r)
=&amp; \frac{1}{|\Omega|} \sum_{x \ne y} \sum_{t \ne r} f(x|t) a_0(y|r),\\
&amp;+ \frac{1}{|\Omega|} \sum_{t \ne r} f(y|t) a_0(y|r),\\
&amp;+ \frac{1}{|\Omega|} f(y|r).
\end{align*}
$$</span></p>
<p>To determine <span class="math-container">$a_0(y|r)$</span> we need again distinguish between the two cases.</p>
<ol>
<li><p>if <span class="math-container">$y$</span> is not the worst outcome in state <span class="math-container">$r$</span> then <span class="math-container">$a_0(y|r) = 0$</span>, so we get the final outcome
<span class="math-container">$$
\begin{align*}
\frac{1}{|\Omega|} \sum_{x \in X} \sum_{t \in \Omega} f(x|t)\,\,c_{x,t}(y|r)
=&amp; \frac{1}{|\Omega|} \sum_{x \ne y} \sum_{t \ne r} f(x|t) \underbrace{a_0(y|r)}_{=0},\\
&amp;+ \frac{1}{|\Omega|} \sum_{t \ne r} f(y|t) \underbrace{a_0(y|r)}_{=0},\\
&amp;+ \frac{1}{|\Omega|} f(y|r),\\
=&amp; \frac{1}{|\Omega|} f(y|r).
\end{align*}
$$</span>
This is the same as in point 1 above.</p>
</li>
<li><p>If <span class="math-container">$y$</span> is the worst outcome in state <span class="math-container">$r$</span> then <span class="math-container">$a_0(y|r) = 1$</span> and we obtain:
<span class="math-container">$$
\begin{align*}
\frac{1}{|\Omega|} \sum_{x \in X} \sum_{t \in \Omega} f(x|t)\,\,c_{x,t}(y|r)
=&amp; \frac{1}{|\Omega|} \sum_{x \ne y} \sum_{t \ne r} f(x|t) \underbrace{a_0(y|r)}_{=1},\\
&amp;+ \frac{1}{|\Omega|} \sum_{t \ne r} f(y|t) \underbrace{a_0(y|r)}_{=1},\\
&amp;+ \frac{1}{|\Omega|} f(y|r),\\
=&amp; \frac{1}{|\Omega|} \sum_{x \ne y} \sum_{t \ne r} f(x|t)
+ \frac{1}{|\Omega|} \sum_{t \ne r} f(y|t),\\
&amp;+ \frac{1}{|\Omega|} f(y|r),\\
=&amp; \frac{1}{|\Omega|} \sum_{t \ne r} \underbrace{\sum_{x \in X} f(x|t)}_{=1}
+ \frac{1}{|\Omega|} f(y|r),\\
=&amp; \frac{1}{|\Omega|} \underbrace{\sum_{t \ne r} 1}_{=|\Omega|-1} + \frac{1}{|\Omega|} f(y|r),\\
=&amp; \frac{|\Omega|-1}{|\Omega|}+ \frac{1}{|\Omega|} f(y|r),\\
=&amp; \left(1 - \frac{1}{|\Omega|}\right) + \frac{1}{|\Omega|} f(y|r).
\end{align*}
$$</span>
which coincides with point 2 above.</p>
</li>
</ol>

CC BY-SA 4.0
2023-09-06T15:08:47.903Z
I really appreciate it! Thank you so much!
Beerus
CC BY-SA 4.0
2023-09-06T06:11:21.853Z
2023-09-06T06:11:21.853Z
tdm
<p>I am self-studying game theory using Myerson's <em>Game Theory: Analysis of Conflict</em>. I got some trouble understanding his proof of Theorem 1.1, the Expected-Utility Maximization Theorem. The Theorem goes like this:</p>
<blockquote>
<p><strong>Theorem 1.1</strong><span class="math-container">$\space\space$</span> Axioms 1.1AB, 1.2, 1.3, 1.4, 1.5AB, 1.6AB, and 1,7 are jointly satisfied if and only if there exists a utility function <span class="math-container">$u : X \times \Omega \to \mathbb{R}$</span> and a conditional probability function <span class="math-container">$p : \Xi \to \Delta(\Omega)$</span> such that
<span class="math-container">\begin{equation}
\max_{x \in X} u(x, t) = 1\space\space \text{and}\space\space \min_{x \in X} u(x, t) = 0,\space\space \forall t \in \Omega; \tag1
\end{equation}</span>
<span class="math-container">\begin{equation}
p(R|T) = p(R|S)p(S|T),\space\space \forall R,\space\space \forall S,\space\space \text{and}\space\space \forall T\space\space \text{such that} \tag2
\end{equation}</span>
<span class="math-container">\begin{equation}
R \subseteq S \subseteq T \subseteq \Omega\space\space \text{and}\space\space S \neq \emptyset;
\end{equation}</span>
<span class="math-container">\begin{equation}
f \succsim_S g\space\space \text{if and only if}\space\space E_p(u(f)|S) \geq E_p(u(g)|S),\tag3
\end{equation}</span>
<span class="math-container">\begin{equation}
\forall f, g \in L, \forall S \in \Xi.
\end{equation}</span>
Furthermore, given these Axioms 1.1AB - 1.7, Axiom 1.8 is also satisfied if and only if conditions (1) - (3) here can be satisfied with a state-omdependent utility.</p>
</blockquote>
<p>In his proof, there is one step, which is a one-line equation that I cannot understand:</p>
<blockquote>
<p>... (Here, <span class="math-container">$|\Omega|$</span> denotes the number of states in the set <span class="math-container">$\Omega$</span>.) Notice that
<span class="math-container">\begin{equation}
\left(\frac{1}{|\Omega|}\right)f + \left(1 - \frac{1}{|\Omega|}\right)a_0 = \left(\frac{1}{|\Omega|}\right)\sum_{t \in \Omega}\sum_{x \in X}f(x|t)c_{x, t}.
\end{equation}</span></p>
</blockquote>
<p>Could someone please help me explain this equation? <strong>Some background information and notation</strong>:</p>
<blockquote>
<p><strong>Definition</strong><span class="math-container">$\space\space$</span> Let <span class="math-container">$X$</span> denote the set of possible <em>prizes</em> that the decision-maker could ultimately get. Let <span class="math-container">$\Omega$</span> denote the set of possible <em>states</em>, one of which will be the <em>true state of the world</em>. Suppose <span class="math-container">$X$</span> and <span class="math-container">$\Omega$</span> are finite.</p>
<p><strong>Definition</strong><span class="math-container">$\space\space$</span> A lottery is any function <span class="math-container">$f$</span> that specifies a nonnegative real number <span class="math-container">$f(x|t)$</span>, for every prize <span class="math-container">$x$</span> in <span class="math-container">$X$</span> and every state <span class="math-container">$t$</span> in <span class="math-container">$\Omega$</span>, such that <span class="math-container">$\sum_{x \in X} f(x|t) = 1$</span> for every <span class="math-container">$t$</span> in <span class="math-container">$\Omega$</span>. Let <span class="math-container">$L$</span> denote the set of all such lotteries; that is,
<span class="math-container">\begin{equation}
    L = \{f : \Omega \to \Delta(X)\}.
\end{equation}</span></p>
<p><strong>Definition</strong><span class="math-container">$\space\space$</span> For any state <span class="math-container">$t$</span> in <span class="math-container">$\Omega$</span> and any lottery <span class="math-container">$f$</span> in <span class="math-container">$L$</span>, <span class="math-container">$f(\cdot|t)$</span> denotes the probability distribution over <span class="math-container">$X$</span> designated by <span class="math-container">$f$</span> in state <span class="math-container">$t$</span>; that is,
<span class="math-container">\begin{equation}
    f(\cdot|t) = (f(x|t))_{x \in X} \in \Delta(X).
\end{equation}</span></p>
<p><strong>Definition</strong><span class="math-container">$\space\space$</span> The information that the decision-maker might have about the true state of the world can be described by an <em>event</em>, which is a nonempty subset of <span class="math-container">$\Omega$</span>. Let <span class="math-container">$\Xi$</span> denote the set of all such events, so that
<span class="math-container">\begin{equation}
    \Xi = \{S\hspace{0.2cm} |\hspace{0.2cm} S \subseteq \Omega\hspace{0.2cm} \text{and}\hspace{0.2cm} S \neq \emptyset\}.
\end{equation}</span></p>
<p><strong>Definition</strong><span class="math-container">$\space\space$</span> For any two lotteries <span class="math-container">$f$</span> and <span class="math-container">$g$</span> in <span class="math-container">$L$</span>, and for any event <span class="math-container">$S$</span> in <span class="math-container">$\Xi$</span>, we write <span class="math-container">$f \succsim_S g$</span> if and only if the lottery <span class="math-container">$f$</span> would be at least as desirable as <span class="math-container">$g$</span>, in the opinion of the decision-maker, if he learned that the true state of the world was in the set <span class="math-container">$S$</span>. That is, <span class="math-container">$f \succsim_S g$</span> if and only if the decision-maker would be willing to choose the lottery <span class="math-container">$f$</span> when he has to choose between <span class="math-container">$f$</span> and <span class="math-container">$g$</span> and he knows only that the event <span class="math-container">$S$</span> has occurred.</p>
<p><strong>Definition</strong><span class="math-container">$\space\space$</span> For any number <span class="math-container">$\alpha$</span> such that <span class="math-container">$0 \leq \alpha \leq 1$</span>, and for any two lotteries <span class="math-container">$f$</span> and <span class="math-container">$g$</span> in <span class="math-container">$L$</span>, <span class="math-container">$\alpha f + (1 - \alpha)g$</span> denotes the lottery in <span class="math-container">$L$</span> such that
<span class="math-container">\begin{align*}
(\alpha f + (1 - \alpha)g)(x|t) = \alpha f(x|t) + (1 - \alpha)g(x|t),\quad \forall x \in X,\quad \forall t \in \Omega.
\end{align*}</span></p>
<p><strong>Axiom 1.1A (Completeness)</strong><span class="math-container">$\space\space$</span> <span class="math-container">$f \succsim_S g$</span> or <span class="math-container">$g \succsim_S f$</span></p>
<p><strong>Axiom 1.1B (Transitivity)</strong><span class="math-container">$\space\space$</span> If <span class="math-container">$f \succsim_S g$</span> and <span class="math-container">$g \succsim_S h$</span>, then <span class="math-container">$f \succsim_S h$</span></p>
<p><strong>Axiom 1.2 (Relevance)</strong><span class="math-container">$\space\space$</span> If <span class="math-container">$f(\cdot|t) = g(\cdot|t)$</span> <span class="math-container">$\forall t\in S$</span>, then <span class="math-container">$f \sim_S g$</span></p>
<p><strong>Axiom 1.3 (Monotonicity)</strong><span class="math-container">$\space\space$</span> If <span class="math-container">$f \succ_S h$</span> and <span class="math-container">$0 \leq \beta &lt; \alpha \leq 1$</span>, then <span class="math-container">$\alpha f + (1 - \alpha)h \succ_S \beta f + (1 - \beta)h$</span></p>
<p><strong>Axiom 1.4 (Continuity)</strong><span class="math-container">$\space\space$</span> 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></p>
<p><strong>Axiom 1.5A (Objective Substitution)</strong><span class="math-container">$\space\space$</span> If <span class="math-container">$e \succsim_S f$</span> and <span class="math-container">$g \succsim_S h$</span> and <span class="math-container">$0 \leq \alpha \leq 1$</span>, then <span class="math-container">$\alpha e + (1 - \alpha)g \succsim_S \alpha f + (1 - \alpha)h$</span></p>
<p><strong>Axiom 1.5B (Strict Objective Substitution)</strong><span class="math-container">$\space\space$</span> If <span class="math-container">$e \succ_S f$</span> and <span class="math-container">$g \succ_S h$</span> and <span class="math-container">$0 &lt; \alpha \leq 1$</span>, then <span class="math-container">$\alpha e + (1 - \alpha)g \succ_S \alpha f + (1 - \alpha)h$</span></p>
<p><strong>Axiom 1.6A (Subjective Substitution)</strong><span class="math-container">$\space\space$</span> If <span class="math-container">$f \succsim_S g$</span> and <span class="math-container">$f \succsim_T g$</span> and <span class="math-container">$S \cap T = \emptyset$</span>, then <span class="math-container">$f \succsim_{S \cup T} g$</span></p>
<p><strong>Axiom 1.6B (Strict Subjective Substitution)</strong><span class="math-container">$\space\space$</span> If <span class="math-container">$f \succ_S g$</span> and <span class="math-container">$f \succ_T g$</span> and <span class="math-container">$S \cap T = \emptyset$</span>, then <span class="math-container">$f \succ_{S \cup T} g$</span></p>
<p><strong>Axiom 1.7 (Interest)</strong><span class="math-container">$\space\space$</span> For every state <span class="math-container">$t$</span> in <span class="math-container">$\Omega$</span>, there exist prizes <span class="math-container">$y$</span> and <span class="math-container">$z$</span> in <span class="math-container">$X$</span> such that <span class="math-container">$[y] \succ_{\{t\}} [z]$</span></p>
<p><strong>Axiom 1.8 (State Neutrality)</strong><span class="math-container">$\space\space$</span> For any two states <span class="math-container">$r$</span> and <span class="math-container">$t$</span> in <span class="math-container">$\Omega$</span>, if <span class="math-container">$f(\cdot|r) = f(\cdot|t)$</span> and <span class="math-container">$g(\cdot|r) = g(\cdot|t)$</span> and <span class="math-container">$f \succsim_{\{r\}} g$</span>, then <span class="math-container">$f \succsim_{\{t\}} g$</span></p>
<p><strong>Definition</strong><span class="math-container">$\space\space$</span> Let <span class="math-container">$a_1$</span> be a lottery that gives the decision-maker one of the best prizes in every state; and let <span class="math-container">$a_0$</span> be a lottery that gives him one of the worst prizes in every state. That is, for every state <span class="math-container">$t$</span>, <span class="math-container">$a_1(y|t) = 1 = a_0(z|t)$</span> for some prizes <span class="math-container">$y$</span> and <span class="math-container">$z$</span> such that, for every <span class="math-container">$x$</span> in <span class="math-container">$X$</span>, <span class="math-container">$y \succsim_{\{t\}} x \succsim_{\{t\}} z$</span>. Such best and worst prizes can be found in every state because the preference relation <span class="math-container">$\succsim_{\{t\}}$</span> forms a transitive ordering over the finite set <span class="math-container">$X$</span>.</p>
<p><strong>Definition</strong><span class="math-container">$\space\space$</span> For any prize <span class="math-container">$x$</span> and any state <span class="math-container">$t$</span>, let <span class="math-container">$c_{x, t}$</span> be the lottery such that
<span class="math-container">\begin{equation}
    c_{x, t}(\cdot|r) = [x](\cdot|r)\space\space \text{if}\space\space r = t,\\
    c_{x, t}(\cdot|r) = a_0(\cdot|r)\space\space \text{if}\space\space r \neq t.
\end{equation}</span>
That is, <span class="math-container">$c_{x, t}$</span> is the lottery that always gives the worst prize, except in state <span class="math-container">$t$</span>, when it gives prize <span class="math-container">$x$</span>.</p>
</blockquote>
<p>I really appreciate any help!</p>

<microeconomics><mathematical-economics><game-theory><decision-theory><expected-utility>
About Theorem 1.1 (the Expected-Utility Maximization Theorem) in Game Theory: Analysis of Conflict by Roger Myerson
Beerus
# About Theorem 1.1 (the Expected-Utility Maximization Theorem) in Game Theory: Analysis of Conflict by Roger Myerson

(Asked by: username_0 on 2023-09-05)

I am self-studying game theory using Myerson's *Game Theory: Analysis of Conflict*. I got some trouble understanding his proof of Theorem 1.1, the Expected-Utility Maximization Theorem. The Theorem goes like this:

> **Theorem 1.1**$\space\space$ Axioms 1.1AB, 1.2, 1.3, 1.4, 1.5AB, 1.6AB, and 1,7 are jointly satisfied if and only if there exists a utility function $u : X \times \Omega \to \mathbb{R}$ and a conditional probability function $p : \Xi \to \Delta(\Omega)$ such that
> \begin{equation}
> \max\_{x \in X} u(x, t) = 1\space\space \text{and}\space\space \min\_{x \in X} u(x, t) = 0,\space\space \forall t \in \Omega; \tag1
> \end{equation}
> \begin{equation}
> p(R|T) = p(R|S)p(S|T),\space\space \forall R,\space\space \forall S,\space\space \text{and}\space\space \forall T\space\space \text{such that} \tag2
> \end{equation}
> \begin{equation}
> R \subseteq S \subseteq T \subseteq \Omega\space\space \text{and}\space\space S \neq \emptyset;
> \end{equation}
> \begin{equation}
> f \succsim\_S g\space\space \text{if and only if}\space\space E\_p(u(f)|S) \geq E\_p(u(g)|S),\tag3
> \end{equation}
> \begin{equation}
> \forall f, g \in L, \forall S \in \Xi.
> \end{equation}
> Furthermore, given these Axioms 1.1AB - 1.7, Axiom 1.8 is also satisfied if and only if conditions (1) - (3) here can be satisfied with a state-omdependent utility.

In his proof, there is one step, which is a one-line equation that I cannot understand:

> ... (Here, $|\Omega|$ denotes the number of states in the set $\Omega$.) Notice that
> \begin{equation}
> \left(\frac{1}{|\Omega|}\right)f + \left(1 - \frac{1}{|\Omega|}\right)a\_0 = \left(\frac{1}{|\Omega|}\right)\sum\_{t \in \Omega}\sum\_{x \in X}f(x|t)c\_{x, t}.
> \end{equation}

Could someone please help me explain this equation? **Some background information and notation**:

> **Definition**$\space\space$ Let $X$ denote the set of possible *prizes* that the decision-maker could ultimately get. Let $\Omega$ denote the set of possible *states*, one of which will be the *true state of the world*. Suppose $X$ and $\Omega$ are finite.
>
> **Definition**$\space\space$ A lottery is any function $f$ that specifies a nonnegative real number $f(x|t)$, for every prize $x$ in $X$ and every state $t$ in $\Omega$, such that $\sum\_{x \in X} f(x|t) = 1$ for every $t$ in $\Omega$. Let $L$ denote the set of all such lotteries; that is,
> \begin{equation}
> L = \{f : \Omega \to \Delta(X)\}.
> \end{equation}
>
> **Definition**$\space\space$ For any state $t$ in $\Omega$ and any lottery $f$ in $L$, $f(\cdot|t)$ denotes the probability distribution over $X$ designated by $f$ in state $t$; that is,
> \begin{equation}
> f(\cdot|t) = (f(x|t))\_{x \in X} \in \Delta(X).
> \end{equation}
>
> **Definition**$\space\space$ The information that the decision-maker might have about the true state of the world can be described by an *event*, which is a nonempty subset of $\Omega$. Let $\Xi$ denote the set of all such events, so that
> \begin{equation}
> \Xi = \{S\hspace{0.2cm} |\hspace{0.2cm} S \subseteq \Omega\hspace{0.2cm} \text{and}\hspace{0.2cm} S \neq \emptyset\}.
> \end{equation}
>
> **Definition**$\space\space$ For any two lotteries $f$ and $g$ in $L$, and for any event $S$ in $\Xi$, we write $f \succsim\_S g$ if and only if the lottery $f$ would be at least as desirable as $g$, in the opinion of the decision-maker, if he learned that the true state of the world was in the set $S$. That is, $f \succsim\_S g$ if and only if the decision-maker would be willing to choose the lottery $f$ when he has to choose between $f$ and $g$ and he knows only that the event $S$ has occurred.
>
> **Definition**$\space\space$ For any number $\alpha$ such that $0 \leq \alpha \leq 1$, and for any two lotteries $f$ and $g$ in $L$, $\alpha f + (1 - \alpha)g$ denotes the lottery in $L$ such that
> \begin{align\*}
> (\alpha f + (1 - \alpha)g)(x|t) = \alpha f(x|t) + (1 - \alpha)g(x|t),\quad \forall x \in X,\quad \forall t \in \Omega.
> \end{align\*}
>
> **Axiom 1.1A (Completeness)**$\space\space$ $f \succsim\_S g$ or $g \succsim\_S f$
>
> **Axiom 1.1B (Transitivity)**$\space\space$ If $f \succsim\_S g$ and $g \succsim\_S h$, then $f \succsim\_S h$
>
> **Axiom 1.2 (Relevance)**$\space\space$ If $f(\cdot|t) = g(\cdot|t)$ $\forall t\in S$, then $f \sim\_S g$
>
> **Axiom 1.3 (Monotonicity)**$\space\space$ If $f \succ\_S h$ and $0 \leq \beta < \alpha \leq 1$, then $\alpha f + (1 - \alpha)h \succ\_S \beta f + (1 - \beta)h$
>
> **Axiom 1.4 (Continuity)**$\space\space$ If $f \succsim\_S g$ and $g \succsim\_S h$, then there exists some number $\gamma$ such that $0 \leq \gamma \leq 1$ and $g \sim\_S \gamma f + (1 - \gamma)h$
>
> **Axiom 1.5A (Objective Substitution)**$\space\space$ If $e \succsim\_S f$ and $g \succsim\_S h$ and $0 \leq \alpha \leq 1$, then $\alpha e + (1 - \alpha)g \succsim\_S \alpha f + (1 - \alpha)h$
>
> **Axiom 1.5B (Strict Objective Substitution)**$\space\space$ If $e \succ\_S f$ and $g \succ\_S h$ and $0 < \alpha \leq 1$, then $\alpha e + (1 - \alpha)g \succ\_S \alpha f + (1 - \alpha)h$
>
> **Axiom 1.6A (Subjective Substitution)**$\space\space$ If $f \succsim\_S g$ and $f \succsim\_T g$ and $S \cap T = \emptyset$, then $f \succsim\_{S \cup T} g$
>
> **Axiom 1.6B (Strict Subjective Substitution)**$\space\space$ If $f \succ\_S g$ and $f \succ\_T g$ and $S \cap T = \emptyset$, then $f \succ\_{S \cup T} g$
>
> **Axiom 1.7 (Interest)**$\space\space$ For every state $t$ in $\Omega$, there exist prizes $y$ and $z$ in $X$ such that $[y] \succ\_{\{t\}} [z]$
>
> **Axiom 1.8 (State Neutrality)**$\space\space$ For any two states $r$ and $t$ in $\Omega$, if $f(\cdot|r) = f(\cdot|t)$ and $g(\cdot|r) = g(\cdot|t)$ and $f \succsim\_{\{r\}} g$, then $f \succsim\_{\{t\}} g$
>
> **Definition**$\space\space$ Let $a\_1$ be a lottery that gives the decision-maker one of the best prizes in every state; and let $a\_0$ be a lottery that gives him one of the worst prizes in every state. That is, for every state $t$, $a\_1(y|t) = 1 = a\_0(z|t)$ for some prizes $y$ and $z$ such that, for every $x$ in $X$, $y \succsim\_{\{t\}} x \succsim\_{\{t\}} z$. Such best and worst prizes can be found in every state because the preference relation $\succsim\_{\{t\}}$ forms a transitive ordering over the finite set $X$.
>
> **Definition**$\space\space$ For any prize $x$ and any state $t$, let $c\_{x, t}$ be the lottery such that
> \begin{equation}
> c\_{x, t}(\cdot|r) = [x](\cdot|r)\space\space \text{if}\space\space r = t,\\
> c\_{x, t}(\cdot|r) = a\_0(\cdot|r)\space\space \text{if}\space\space r \neq t.
> \end{equation}
> That is, $c\_{x, t}$ is the lottery that always gives the worst prize, except in state $t$, when it gives prize $x$.

I really appreciate any help!

**Tags:** microeconomics, mathematical-economics, game-theory, decision-theory, expected-utility


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## 1 answers

### Answer by username_1 on 2023-09-06. Score: 3 | Accepted answer

Let's first have a look at the left hand side of the equation. Take an outcome $y$ and a state $r$. There are two cases:

1. If $y$ is not the worst outcome in state $r$ then the lottery gives:
   $$
   \left(\frac{1}{|\Omega|}\right) f(y|r) + \left(1-\frac{1}{|\Omega|}\right) \underbrace{a\_0(y|r)}\_{=0} = \left(\frac{1}{|\Omega|}\right) f(y|r)
   $$
   Notice that $a\_0(y|r)$ is zero as the probability of getting $y$ in state $r$ is zero (as $y$ is not the worst outcome).
2. If $y$ is the worst outcome in state $r$ then $a\_0(y|r) = 1$ and we get instead
   $$
   \left(\frac{1}{|\Omega|}\right) f(y|r) + \left(1-\frac{1}{|\Omega|}\right) \underbrace{a\_0(y|r)}\_{=1}= \left(\frac{1}{|\Omega|}\right)f(y|r) + \left(1 - \frac{1}{|\Omega|}\right).
   $$

Now, let us focus on the right hand side and show that it is the same in both cases. First let us split up the sum into the different cases, case I where ($x \ne y$ and $t \ne r$), case II where ($x \ne y$ and $t = r$), case III where ($x = y$ and $t \ne r$) and finally, case IV where ($t = r$ and $x = y$).

$$
\begin{align\*}
\frac{1}{|\Omega|} \sum\_{x \in X} \sum\_{t \in \Omega} f(x|t)\,\,c\_{x,t}(y|r)
=& \frac{1}{|\Omega|} \sum\_{x \ne y} \sum\_{t \ne r} f(x|t) c\_{x,t}(y|r), && \text{case I}\\
&+ \frac{1}{|\Omega|} \sum\_{x \ne y} f(x|r) c\_{x,r}(y|r), && \text{case II}\\
&+ \frac{1}{|\Omega|} \sum\_{t \ne r} f(y|t) c\_{y,t}(y|r), && \text{case III}\\
&+ \frac{1}{|\Omega|} f(y|r) c\_{y,r}(y|r). && \text{case IV}
\end{align\*}
$$
Substituting out $c\_{x,t}(y|r)$ with $[x](y|r)$ if $t = r$ and $a\_0(y|r)$ when $t \ne r$ gives
$$
\begin{align\*}
\frac{1}{|\Omega|} \sum\_{x \in X} \sum\_{t \in \Omega} f(x|t)\,\,c\_{x,t}(y|r)
=& \frac{1}{|\Omega|} \sum\_{x \ne y} \sum\_{t \ne r} f(x|t) a\_0(y|r),\\
&+ \frac{1}{|\Omega|} \sum\_{x \ne y} f(x|r) \underbrace{[x](y|r)}\_{=0},\\
&+ \frac{1}{|\Omega|} \sum\_{t \ne r} f(y|t) a\_0(y|r),\\
&+ \frac{1}{|\Omega|} f(y|r) \underbrace{[y](y|r)}\_{=1}.
\end{align\*}
$$
Notice that $[x](y|r)$ is zero if $x \ne y$ and equal to $1$ if $x = y$. This gives:
$$
\begin{align\*}
\frac{1}{|\Omega|} \sum\_{x \in X} \sum\_{t \in \Omega} f(x|t)\,\,c\_{x,t}(y|r)
=& \frac{1}{|\Omega|} \sum\_{x \ne y} \sum\_{t \ne r} f(x|t) a\_0(y|r),\\
&+ \frac{1}{|\Omega|} \sum\_{t \ne r} f(y|t) a\_0(y|r),\\
&+ \frac{1}{|\Omega|} f(y|r).
\end{align\*}
$$

To determine $a\_0(y|r)$ we need again distinguish between the two cases.

1. if $y$ is not the worst outcome in state $r$ then $a\_0(y|r) = 0$, so we get the final outcome
   $$
   \begin{align\*}
   \frac{1}{|\Omega|} \sum\_{x \in X} \sum\_{t \in \Omega} f(x|t)\,\,c\_{x,t}(y|r)
   =& \frac{1}{|\Omega|} \sum\_{x \ne y} \sum\_{t \ne r} f(x|t) \underbrace{a\_0(y|r)}\_{=0},\\
   &+ \frac{1}{|\Omega|} \sum\_{t \ne r} f(y|t) \underbrace{a\_0(y|r)}\_{=0},\\
   &+ \frac{1}{|\Omega|} f(y|r),\\
   =& \frac{1}{|\Omega|} f(y|r).
   \end{align\*}
   $$
   This is the same as in point 1 above.
2. If $y$ is the worst outcome in state $r$ then $a\_0(y|r) = 1$ and we obtain:
   $$
   \begin{align\*}
   \frac{1}{|\Omega|} \sum\_{x \in X} \sum\_{t \in \Omega} f(x|t)\,\,c\_{x,t}(y|r)
   =& \frac{1}{|\Omega|} \sum\_{x \ne y} \sum\_{t \ne r} f(x|t) \underbrace{a\_0(y|r)}\_{=1},\\
   &+ \frac{1}{|\Omega|} \sum\_{t \ne r} f(y|t) \underbrace{a\_0(y|r)}\_{=1},\\
   &+ \frac{1}{|\Omega|} f(y|r),\\
   =& \frac{1}{|\Omega|} \sum\_{x \ne y} \sum\_{t \ne r} f(x|t)
   + \frac{1}{|\Omega|} \sum\_{t \ne r} f(y|t),\\
   &+ \frac{1}{|\Omega|} f(y|r),\\
   =& \frac{1}{|\Omega|} \sum\_{t \ne r} \underbrace{\sum\_{x \in X} f(x|t)}\_{=1}
   + \frac{1}{|\Omega|} f(y|r),\\
   =& \frac{1}{|\Omega|} \underbrace{\sum\_{t \ne r} 1}\_{=|\Omega|-1} + \frac{1}{|\Omega|} f(y|r),\\
   =& \frac{|\Omega|-1}{|\Omega|}+ \frac{1}{|\Omega|} f(y|r),\\
   =& \left(1 - \frac{1}{|\Omega|}\right) + \frac{1}{|\Omega|} f(y|r).
   \end{align\*}
   $$
   which coincides with point 2 above.

> **Comments:**
> **username_0**: I really appreciate it! Thank you so much!  

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