0
$\begingroup$

If X is finite, define this function $u : X \rightarrow \mathbb{R}$ by $u(x) = |\{z\in X:z \prec x \}|$. Prove that $u$ is a utility function for $\succsim$.


Is it sufficient to prove that the relation is transitive and complete?
By Lemma: If $\succsim$ has a utility function, then it is transitive and complete.

$\endgroup$
  • 1
    $\begingroup$ You're given a very specific utility function and are asked to prove that this particular function represents the preference over a finite choice set. In other words, you're asked to prove $u(x)\ge u(y)\;\Leftrightarrow\; x\succsim y$ for all $x,y\in X$, where $u$ is as defined in the question. $\endgroup$ – Herr K. Oct 8 '18 at 17:51
  • $\begingroup$ ok thank you. I still get confused on how to prove things like this. $\endgroup$ – Zhang_anlan Oct 8 '18 at 21:01
1
$\begingroup$

You're asked to prove that $u(x)\ge u(y)\;\Leftrightarrow\;x\succsim y$ for any $x,y\in X$, where $u(x)=|\{z\in X:z\prec x\}|$, i.e. the utility of $x$ is measured by the number of other alternatives that rank strictly below it. Since $X$ is finite, let's suppose without loss of generality that $X=\{1,2,\dots,N\}$ where $N$ is some finite number.

I'll prove the case in which there is no indifference among the alternatives, say, $1\succ2\succ\cdots\succ N$. I'll let you finish the proof by establishing the case where there are indifferences among subsets of alternatives.


Step 1. Establishing $u(x)>u(y)\;\Rightarrow\;x\succ y$.

Suppose $u(x)>u(y)$. By the definition of $u$, the number of alternatives strictly worse than $x$ is larger than the number of alternatives strictly worse than $y$. If $y\succsim x$, this would simply contradict the previous statement. Hence, we must have $x\succ y$.

Step 2. Establishing $x\succ y\;\Rightarrow\;u(x)>u(y)$.

Suppose $x\succ y$. Since we assume no indifference among alternatives, the set of strictly worse alternatives to $x$, $\{z\in X:x\succ z\}$, must contain more elements than the set of strictly worse alternatives to $y$, $\{z\in X:y\succ z\}$. In other words, $|\{z\in X:x\succ z\}|>|\{z\in X:y\succ z\}|$. Therefore, we obtain $u(x)>u(y)$ as a result.

Taken together, steps 1 and 2 demonstrate that $x\succ y\;\Leftrightarrow\;u(x)>u(y)$ for any arbitrary $x,y\in X$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.