Oscar_Cunningham comments on Utility is unintuitive

Oscar_Cunningham 10 Dec 2010 10:02 UTC
0 points

there is no reason as far as I can tell why u([E1 and E2]) has to have anything to do with u(E1)+u(E2), for instance. Among other things, u is only defined up to an additive constant and so not only is there no reason to be true, it will be completely false for almost all possible utility functions

If I have a bet that loses me 1 util on heads, and 2 utils on tails, I wont take it; it’s a sure loss. Add a million to those values and we get a bet that I would take. So clearly u changes on addition of constants.

Multiplicative constants won’t have an effect, but they don’t change additivity, either.
- JGWeissman 10 Dec 2010 19:59 UTC
  2 points
  Parent
  If I have a bet that loses me 1 util on heads, and 2 utils on tails, I wont take it; it’s a sure loss. Add a million to those values and we get a bet that I would take. So clearly u changes on addition of constants.
  
  If you were really adding a constant to the utility function, you would also add a million to the utility of not taking the bet.
  
  Let U be your original utility function and V be U + 1,000,000, differring by an additive constant.
  
  Then
```
EU(refuse bet) = 0
EU(take bet) = .5 * (-1) + .5 * (-2) = -1.5
```
  So, using U, you would refuse the bet.
  
  And
```
EV(refuse bet) = 1,000,000
EV(take bet) = .5 * (999,999) + .5*(999.998) = 999,998.5
```
  So, using V, you would also refuse to take the bet.
  
  As expected, adding a constant to the utility function did not change the decision.
- nshepperd 10 Dec 2010 10:25 UTC
  0 points
  Parent
  A sure loss compared to what? Not taking the bet has a utility too.
  
  A bet that loses 1 util on heads and 2 on tails compared to the status quo would have utilities like:
  - u(heads) = −1
  - u(tails) = −2
  - u(status quo) = 0
  Add a constant to all these and the status quo will still be higher.
  - Oscar_Cunningham 10 Dec 2010 10:27 UTC
    2 points
    Parent
    Correct, my mistake.