Your example has 3 states: vanilla, chocolate, and neither.

But you only explicitly assigned utilities to 2 of them, although you implicitly assigned the state of 'neither' a utility of 0 initially. Then when you applied the transformation to vanilla and chocolate you didn't apply it to the 'neither' state, which altered preferences for gambles over both transformed and untransformed states.

E.g. if we initially assigned u(neither)=0 then after the transformation we have u(neither)=4, u(vanilla)=7, u(chocolate)=12. Then an action with a 50% chance of neither and 50% chance of chocolate has expected utility 8, while the 100% chance of vanilla has expected utility 7.

Quick puzzle about utility functions under affine transformations

by Liron 1 min read16th Jul 20165 comments

6


Here's a puzzle based on something I used to be confused about:

It is known that utility functions are equivalent (i.e. produce the same preferences over actions) up to a positive affine transformation: u'(x) = au(x) + b where a is positive.

Suppose I have u(vanilla) = 3, u(chocolate) = 8. I prefer an action that yields a 50% chance of chocolate over an action that yields a 100% chance of vanilla, because 0.5(8) > 1.0(3).

Under the positive affine transformation a = 1, b = 4; we get that u'(vanilla) = 7 and u'(chocolate) = 12. Therefore I now prefer the action that yields a 100% chance of vanilla, because 1.0(7) > 0.5(12).

How to resolve the contradiction?