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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.

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

1 Answer 1

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

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