# About Theorem 1.1 in Game Theory: Analysis of Conflict by Roger Myerson

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 $$$$\max_{x \in X} u(x, t) = 1\space\space \text{and}\space\space \min_{x \in X} u(x, t) = 0,\space\space \forall t \in \Omega; \tag1$$$$ $$$$p(R|T) = p(R|S)p(S|T),\space\space \forall R,\space\space \forall S,\space\space \text{and}\space\space \forall T\space\space \text{such that} \tag2$$$$ $$$$R \subseteq S \subseteq T \subseteq \Omega\space\space \text{and}\space\space S \neq \emptyset;$$$$ $$$$f \succsim_S g\space\space \text{if and only if}\space\space E_p(u(f)|S) \geq E_p(u(g)|S),\tag3$$$$ $$$$\forall f, g \in L, \forall S \in \Xi.$$$$ 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 $$$$\left(\frac{1}{|\Omega|}\right)f + \left(1 - \frac{1}{|\Omega|}\right)a_0 = \left(\frac{1}{|\Omega|}\right)\sum_{t \in \Omega}\sum_{x \in X}f(x|t)c_{x, t}.$$$$

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, $$$$L = \{f : \Omega \to \Delta(X)\}.$$$$

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, $$$$f(\cdot|t) = (f(x|t))_{x \in X} \in \Delta(X).$$$$

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 $$$$\Xi = \{S\hspace{0.2cm} |\hspace{0.2cm} S \subseteq \Omega\hspace{0.2cm} \text{and}\hspace{0.2cm} S \neq \emptyset\}.$$$$

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 $$$$c_{x, t}(\cdot|r) = [x](\cdot|r)\space\space \text{if}\space\space r = t,\\ c_{x, t}(\cdot|r) = a_0(\cdot|r)\space\space \text{if}\space\space r \neq t.$$$$ That is, $$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!

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.

• I really appreciate it! Thank you so much! Sep 6 at 15:08