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shokwave comments on Bayes Slays Goodman's Grue - Less Wrong Discussion

0 Post author: potato 17 November 2011 10:45AM

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Comment author: shokwave 19 November 2011 06:38:44PM *  2 points [-]

Let's say that the green premises brings the probability of "A new unobserved emerald is green." to 99%. In the skeptic's hypothesis, by symmetry it should also bring the probability of "A new unobserved emerald is grue." to 99%. But of course after time T, this would mean that the probability of observing a green emerald is 99%, and the probability of not observing a green emerald is at least 99%, since these sentences have no intersection, i.e., they cannot happen together, to find the probability of their disjunction we just add their individual probabilities. This must give us a number at least as big as 198%...

Let's do actual Bayesian math on this problem. Let Green_n be "the green premises 1 through n", and so on.

Pr( An emerald is grue | Emerald is green, it is before time T ) = ~1.

Pr( An emerald is grue | Emerald is green, it is after time T ) = ~0.

Pr( An emerald is grue | Emerald is blue, it is before time T ) = ~0.

Pr( An emerald is grue | Emerald is blue, it is after time T ) = ~1.

These are our grue axioms - the probabilistic representation of "grue iff green before time T or blue after time T".

Pr( New Emerald is green | Green_n ) = 0.99 (this is our first sentence axiom)

Pr( New Emerald is grue | Emerald is green ) = undefined. We need to know if we are pre-T or post-T. Without the prior probability for being pre-T (from which we can derive its complement, post-T, or vice versa).

But that is wussing out; Bayesian agents should always be able to assign some level of credence. Assume maximum ignorance about T: it is equally likely to be pre-T or post-T.

We can find Pr( New Emerald is grue | Emerald is green ) by finding Pr( New Emerald is grue | Emerald is green, pre-T ) and adding it to Pr( New Emerald is grue | Emerald is green, post-T ) then normalising.

The pre-T case: Pr( New Emerald is grue | Green_n, pre-T ) = Pr( emerald is grue | emerald is green, pre-T ) * Pr( emerald is green). We know that Pr( emerald is green ) is 0.99. We have Pr( emerald is grue | emerald is green, pre-T ) as an axiom of ~1 above. So 0.99 * ~1 = ~0.99.

The post-T case: Pr( New Emerald is grue | Green_n, post-T ) = Pr( emerald is grue | emerald is green, post-T ) * Pr( emerald is green). We know that Pr( emerald is green ) is 0.99. We have Pr( emerald is grue | emerald is green, post-T ) as an axiom of ~0 above. So 0.99 * ~0 = ~0.

Normalising gives us: Pr( New Emerald is grue | Emerald is green ) = ~.495.

This is for the case where we don't know if pre-T or post-T. When you say

But of course after time T

we can ask a better question than "Pr( New Emerald is grue | Emerald is green ) ?". We can ask Pr( An emerald is grue | Emerald is green, it is after time T ). This is an axiom from before! We now know it's ~0, which resolves the problem: the probabilities sum to 0.99 + epsilon, which is below 1, which conserves the Kolmogorov axioms.

Whence the problem? The error creeps in when you use the pre-T case to get one .99, and then you use the complement of the post-T case to get another .99, and then add them together. If you specify pre-T or post-T, but then swap T in calculating some of the posterior probabilities, of course you can violate probability theory! You're already violating it by varying the state of T within the calculation!

If I am not mistaken, this is my independent formulation of the formal Bayesian resolution of the Goodman Grue paradox.