If strict preferences over three outcomes x, y and z, do not satisfy transitivity, so that $x \succ y \succ z \succ x$, then these preferences cannot be represented by a utility function.
Is this statement true or false? Can anyone explain it to me?
$\begingroup$
$\endgroup$
$\begingroup$
$\endgroup$
In fact, we don not even call them preferences. In order for a binary relationship to be a preference relationship, it must be both complete and transitive. The main goal of modelling preferences is to understand choice. And we cannot understand choice unless preferences are a quasi-order, i.e., one can rank alternatives (with ties allowed). In the case you mention, $x \succ y \succ z \succ x$, we would not know how to rank them. Further, it is clear that no function $u(ยท)$ can reproduce this ranking, as transitivity of real numbers makes it impossible to have $u(x)>u(y)>u(z)>u(x)$.
Note that this is different from $x \sim y \sim z.$ In this case you could say "no ranking exists" (although, perhaps it would be more adequate to say that "any ranking is consistent with the preferences"). In these circumstances, the utility function does exist, as it only must happen that $u(x)=u(y)=u(z)$.
-
$\begingroup$ Thanks a lot! It really helps me. I also have another query which is referred to convex preference. A consumer with convex preferences weakly preference over \{R}^2_+ who is indifferent between the bundles (5, 1) and (11, 3), will like the bundle (8, 2) at least as well as either of the first two bundles. Is this statement reasonable? $\endgroup$ – huaxin Chen Mar 7 '19 at 0:01
-
$\begingroup$ @huaxinChen. If you liked the answer you should accept it. Also you should post the new question as a new question $\endgroup$ – Patricio Mar 7 '19 at 13:54
