Decision theory is not one of my strengths, and I have a question about it.
Is there a consensus view on how to deal with the problem of "rival formalizations"? Peterson (2009) illustrates the problem like this:
Imagine that you are a paparazzi photographer and that rumour has it that actress Julia Roberts will show up in either New York (NY), Los Angeles (LA) or Paris (P). Nothing is known about the probability of these states of the world. You have to decide if you should stay in America or catch a plane to Paris. If you stay and [she] shows up in Paris you get $0; otherwise you get your photos, which you will be able to sell for $10,000. If you catch a plane to Paris and Julia Roberts shows up in Paris your net gain after having paid for the ticket is $5,000, and if she shows up in America you for some reason, never mind why, get $6,000. Your initial representation of the decision problem is visualized in Table 2.13.
Table 2.13
P | LA | NY | |
Stay | $0 | $10k | $10k |
Go to Paris | $5k | $6k | $6k |
Since nothing is known about the probabilities of the states in Table 2.13, you decide it makes sense to regard them as equally probable [see Table 2.14].
Table 2.14
P (1/3) | LA (1/3) | NY (1/3) | |
Stay | $0 | $10k | $10k |
Go to Paris | $5k | $6k | $6k |
The rightmost columns are exactly parallel. Therefore, they can be merged into a single (disjuntive) column, by adding the probabilities of the two rightmost columns together (Table 2.15).
Table 2.15
P (1/3) | LA or NY (2/3) | |
Stay | $0 | $10k |
Go to Paris | $5k | $6k |
However, now suppose that you instead start with Table 2.13 and first merge the two repetitious states into a single state. You would then obtain the decision matrix in Table 2.16.
Table 2.16
P | LA or NY | |
Stay | $0 | $10k |
Go to Paris | $5k | $6k |
Now, since you know nothing about the probabilities of the two states, you decide to regard them as equally probable... This yields the formal representation in Table 2.17, which is clearly different from the one suggested above in Table 2.15.
Table 2.17
P (1/2) | LA or NY (1/2) | |
Stay | $0 | $10k |
Go to Paris | $5k | $6k |
Which formalisation is best, 2.15 or 2.17? It seems question begging to claim that one of them must be better than the other — so perhaps they are equally reasonable? If they are, we have an example of rival formalisations.
Note that the principle of maximising expected value recommends different acts in the two matrices. According to Table 2.15 you should stay, but 2.17 suggests you should go to Paris.
Does anyone know how to solve this problem? If one is not convinced by the illustration above, Peterson (2009) offers a proof that rival representations are possible on pages 33–35.
The question being asked here is really "what priors should I assign?". I don't think I have the answer, but allow me to restate the problem:
"The subject is known to be headed for one of three places, A, B and C. What is the probability they turn up in each place?" To which the answer is 1/3 each by indifference.
Now why did it look ok to us to merge B and C into one option, (B or C)? Because (in the original problem) B and C were cities located in the same country, the U.S., and that prior geographical information had been incorporated into the problem. When we condition on the knowledge that A is in one country (France) and B and C are in another (the U.S.) the problem is, well, no longer symmetrical. And I confess I'm now actually unsure how or if indifference or maximum entropy can be applied to this now asymmetric problem.
You have to use the information about the asymmetry, which in this case involves an actress and geopolitical boundaries. This isn't a case where there's an elegant ignorance prior, you just have to actually use your knowledge.