I really enjoy this sequence but there’s a sticking point that’s making me unable to continue until I figure it out. It seems to me rather obvious that… utility functions are not shift-invariant! If I denominate option A at 1 utilon and option B at 2 utilons, that means I am indifferent between a certain outcome of A and a 50% probability of B—and this is no longer the case if I shift my utility function even slightly. Ratios of utilities mean something concrete and are destroyed by translation. Since your entire argument seems to rest on that inexplicably not being the case, I can’t see how any of this is useful.
If you have U0(A)=1 and U0(B)=2, then if we shift it by some constant c to get a new utility function: Uc(A)=1+c and Uc(B)=2+c then we can still get the same result.
If we look at the expected utility, then we get: Certainty of A:
1.0∗U0(A)=U0(A)=1
1.0∗Uc(A)=Uc(A)=1+c
50% chance of B, 50% chance of nothing:
0.5∗U0(B)=0.5∗2=1 (so you are indifferent between certainty of A and a 50% chance of B by U0)
0.5∗Uc(B)=0.5∗(2+c)=1+0.5∗c
I think this might be where you got confused? Now the expected values are different for any nonzero c!
The issue is that it is ignoring the implicit zero. The real second equation is:
0.5∗U0(B)+0.5∗U0(∅)=0.5∗U0(B) + 0 = 1$
0.5∗Uc(B)+0.5∗Uc(∅)=0.5∗(2+c)+0.5∗c=1+0.5c+0.5c=1+c
Which results in the same preference ordering.
I really enjoy this sequence but there’s a sticking point that’s making me unable to continue until I figure it out. It seems to me rather obvious that… utility functions are not shift-invariant! If I denominate option A at 1 utilon and option B at 2 utilons, that means I am indifferent between a certain outcome of A and a 50% probability of B—and this is no longer the case if I shift my utility function even slightly. Ratios of utilities mean something concrete and are destroyed by translation. Since your entire argument seems to rest on that inexplicably not being the case, I can’t see how any of this is useful.
Utility functions are shift/scale invariant.
If you have U0(A)=1 and U0(B)=2, then if we shift it by some constant c to get a new utility function: Uc(A)=1+c and Uc(B)=2+c then we can still get the same result.
If we look at the expected utility, then we get:
Certainty of A:
1.0∗U0(A)=U0(A)=1
1.0∗Uc(A)=Uc(A)=1+c
50% chance of B, 50% chance of nothing:
0.5∗U0(B)=0.5∗2=1 (so you are indifferent between certainty of A and a 50% chance of B by U0)
0.5∗Uc(B)=0.5∗(2+c)=1+0.5∗c
I think this might be where you got confused? Now the expected values are different for any nonzero c!
The issue is that it is ignoring the implicit zero. The real second equation is:
0.5∗U0(B)+0.5∗U0(∅)=0.5∗U0(B) + 0 = 1$
0.5∗Uc(B)+0.5∗Uc(∅)=0.5∗(2+c)+0.5∗c=1+0.5c+0.5c=1+c
Which results in the same preference ordering.
Holy heck I have been enlightened. And by contemplating nothingness too! Thanks for the clarification, it all makes sense now.