# Determine Whether A Preference Relation Satisfies The Continuity Axiom - from Exercise 1.1 in Game Theory: Analysis of Conflict by Roger Myerson

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

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.

• There might be a much easier proof using the intermediate value theorem.
– tdm
Oct 13 at 10:03
• @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! Oct 13 at 15:10

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, $$g \sim_S \gamma f + (1-\gamma) h$$