# Quick puzzle about utility functions under affine transformations

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?

• You as­sume util­ity of get­ting nei­ther is 0 both be­fore and af­ter the trans­for­ma­tion. You need to trans­form that util­ity too, eg from 0 to 4.

• 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.

• What they said about the U(-)=0 prob­lem. But the way I think about it re­solves more con­tra­dic­tions, more eas­ily, IMO.

• Utility is no more than a math­e­mat­i­cal ar­ti­fact, do not phrase ques­tions in terms of utility

Utility func­tions are equiv­a­lent un­der pos­i­tive af­fine trans­forms. This is a huge clue that think­ing about util­ity will lead to ma­jor in­tu­ition prob­lems. In­stead, use quan­tities that are not am­bigu­ous. You’re gonna have to get rid of the a and b in au(x)+b, so you’re go­ing to need three states of the world, always, be­fore you’re al­lowed to use in­tu­ition. You can com­bine them in differ­ent ways, but I like

r = [U(x)-U(z)] /​ [U(y)-U(z)]

Mere differ­ences in util­ity are not pinned down, be­cause of scale. Ra­tios of differ­ences in util­ity are great, though. It’s 2.67x as good to go from noth­ing to choco­late as to go from noth­ing to vanilla. 0 and 3 and 8, or 4 and 7 and 12, those are just there for com­pu­ta­tional con­ve­nience in some cir­cum­stances and can be ig­nored.

• You for­got to spec­ify the util­ity of the other out­come when your 50% chance of choco­late fails. As­sum­ing the out­come (“failed at­tempt at choco­late”) has a u util­ity of 0, it will then have a u’ util­ity that is nonzero, which you failed to in­clude in your sec­ond calcu­la­tion.

Nice try, buster ;)

BTW: Up voted, I would strongly like to see more of this type of con­tent.

• In the origi­nal VNM the­o­rem, lot­ter­ies are taken over global states of the world, mean­ing that prefer­ences are ex­pressed over mu­tu­ally ex­clu­sive states of the world.
As­sum­ing there’s no other states of the world be­sides vanilla (V) and choco­late (C), your origi­nal lot­ter­ies are:

0.5*u(V) + 0.5u(c) = 1.5 + 4 = 5.4

against

1u(v) + 0u(c) = 3

so your prefer­ence goes to the first lot­tery. In the sec­ond set of lot­ter­ies you have

0.5*u(V) + 0.5u(c) = 3.5 + 6 = 9.4

against

1u(v) + 0u(c) = 7

You con­tinue to pre­fer the first lot­tery.