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FrameBenignly comments on Open thread, Dec. 21 - Dec. 27, 2015 - Less Wrong Discussion
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If I have 3 options, A, B, and C, and I'm 40% certain the best option is A, 30% certain the best option is B, and 30% certain the best option is C, would it be correct to say that I've confirmed option A instead of say my best evidence suggests A? This can sort of be corrected for with the standard Bayesian confirmation model, but the problem becomes larger as the number of possibilities increases to the point where you can't get a good read on your own certainty, or to the point where the number of possibilities is unknown.
I don't understand your question. Is this about maintaining beliefs over hypotheses or decision-making?
I'm arguing that Bayesian confirmation theory as a philosophy was originally conceived as a model using only two possibilities (A and ~A), and then this model was extrapolated into problems with more than two possibilities. If it had been originally conceived using more than two possibilities, it wouldn't have made any sense to use the word confirmation. So explanations of Bayesian confirmation theory will often entail considering theories or decisions in isolation rather than as part of a group of decisions or theories.
So if there are 20 possible explanations for a problem, and there is no strong evidence suggesting any one explanation, then I will have 5% certainty of the average explanation. Unless I am extremely good at calibration, then I can't confirm any of them, and if I consider each explanation in isolation from the other explanations, then all of them are wrong.
It doesn't matter whether we're talking about hypotheses or decision-making.
I'm not sure whether this is true, but it's irrelevant. Bayesian confirmation theory works just fine with any number of hypotheses.
If by "confirm" you mean "assign high probability to, without further evidence", yes. That seems to me to be exactly what you'd want. What is the problem you see here?
You sound confused. The "confirmation" stems from
(source)
So what if p(H) = 1, p(H|A) = .4, p(H|B) = .3, and p(H|C) = .3? The evidence would suggest all are wrong. But I have also determined that A, B, and C are the only possible explanations for H. Clearly there is something wrong with my measurement, but I have no method of correcting for this problem.
H is Hypothesis. You have three: HA, HB, and HC. Let's say your prior is that they are equally probable, so the unconditional P(HA) = P(HB) = P(HC) = 0.33
Let's also say you saw some evidence E and your posteriors are P(HA|E) = 0.4, P(HB|E) = 0.3, P(HC|E) = 0.3. This means that evidence E confirms HA because P(HA|E) > P(HA). This does not mean that you are required to believe that HA is true or bet your life's savings on it.
That's a really good explanation of part of the problem I was getting at. But that requires considering the three hypotheses as a group rather than in isolation from all other hypotheses to calculate 0.33.
No, it does not.
Let's say you have a hypothesis HZ. You have a prior for it, say P(HZ) = 0.2 which means that you think that there is a 20% probability that HZ is true and 80% probability that something else is true. Then you see evidence E and it so happens that the posterior for HZ becomes 0.25, so P(HZ|E) = 0.25. This means that evidence E confirmed hypothesis HZ and that statement requires nothing from whatever other hypotheses HA,B,C,D,E,etc. might there be.
How would you calculate that prior of 0.2? In my original example, my prior was 1, and then you transformed it into 0.33 by dividing by the number of possible hypotheses. You wouldn't be able to do that without taking the other two possibilities into account. As I said, the issue can be corrected for if the number of hypotheses is known, but not if the number of possibilities is unknown. However, frequently philosophical theories of bayesian confirmation theory don't consider this problem. From this paper by Morey, Romeijn, and Rouder:
You need to read up on basic Bayesianism.
Priors are always for a specific hypothesis. If your prior is 1, this means you believe this hypothesis unconditionally and no evidence can make you stop believing it.
You are talking about the requirement that all mutually exclusive probabilities must sum to 1. That's just a property of probabilities and has nothing to do with Bayes.
Yes, it can. To your "known" hypotheses you just add one more which is "something else".
Really, just go read. You are confused because you misunderstand the basics. Stop with the philosophy and just figure out how the math works.
You don't need to know the number, you need to know the model (which could have infinite hypotheses in it).
Your model (hypothesis set) could be specified by an infinite number of parameters, say "all possible means and variances of a Gaussian." You can have a prior on this space, which is a density. You update the density with evidence to get a new density. This is Bayesian stats 101. Why not just go read about it? Bishop's machine learning book is good.
Not really. A hypothesis's prior probability comes from the total of all of your knowledge; in order to determine that P(HA)=0.33 Lumifer needed the additional facts that there were three possibilities that were all equally likely.
It works just as well if I say that my prior is P(HA)=0.5, without any exhaustive enumeration of the other possibilities. Then evidence E confirms HA if P(HA|E)>P(HA).
(One should be suspicious that my prior probability assessment is a good one if I haven't accounted for all the probability mass, but the mechanisms still work.)
Which is one of the other problems I was getting at
If you start with inconsistent assumptions, you get inconsistent conclusions. If you believe P(H)=1, P(A&B&C)=1, and P(H|A) etc. are all <1, then you have already made a mistake. Why are you blaming this on Bayesian confirmation theory?
Wait, how would you get P(H) = 1?
Fine. p(H) = 0.5, p(H|A) = 0.2, p(H|B) = 0.15, p(H|C) = 0.15 It's not really relevant to the problem.
The relevance is that it's a really weird way to set up a problem. If P(H)=1 and P(H|A)=0.4 then it is necessarily the case that P(A)=0. If that's not immediately obvious to you, you may want to come back to this topic after sleeping on it.
Fair enough.
\sum_i p(H|i) need not add up to p(H) (or indeed to 1).
No, it doesn't.
Edit - I'm agreeing with you. Sorry if that wasn't clear.