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Vladimir_Nesov comments on Bayesian Utility: Representing Preference by Probability Measures - Less Wrong

10 Post author: Vladimir_Nesov 27 July 2009 02:28PM

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Comment author: Vladimir_Nesov 27 July 2009 11:12:00PM *  1 point [-]

Let A be the action (set of possibilities consistent with taking the action), and O set of possible outcomes (each one rated by the utility function, assuming for simplicity that every concrete outcome is considered, not events-outcomes). We can assume . Then:

Comment author: Peter_de_Blanc 28 July 2009 04:54:37PM 0 points [-]

How do you calculate P(A)?

Comment author: Vladimir_Nesov 28 July 2009 08:52:04PM 0 points [-]

Trick question? P(A) is just a probability of some event, so depending on the problem it could be calculated in any of the possible ways. "A" can for example correspond to a value of some random variable in a (dynamic) graphical model, taking observations into account, so that its probability value is obtained from belief propagation.

Comment author: JGWeissman 27 July 2009 11:41:56PM 0 points [-]

As I already explained, that only works for actions that exclude some outcomes and renormalize the probabilities of remaining outcomes, preserving the ratios of their probabilities.

Suppose O had 2 elements, x1 and x2, such that p(x1) = p(x2) = .5. If you take action A, then you have conditional probabilities p(x1|A) = .2 and p(x2|A) = .8. In this case, your transformation of P(x|A) = P(x, A)/P(A) does not work. Because A did not remove x1 as a possibility, it just made it less likely.

Comment author: Vladimir_Nesov 27 July 2009 11:58:10PM *  0 points [-]

P(x|A) = P(x,A)/P(A) is by definition of conditional probability. You are trying to interpret x1 and x2 as events, while in grandparent comment x are elements of the sample space. If you want to consider non-concrete outcomes, compose them from smaller elements. For example, you can have P(O1)=P(O2)=.5, P(O1|A)=.2, P(O2|A)=.8, if O1={x1,x2}, O2={x3,x4}, A={x1,x3}, and p(x1)=.1, p(x2)=.4, p(x3)=.4, p(x4)=.1.