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There is no point in talking about the data if it doesn't tell you useful things about the population or type of thing you are studying, if it's not for buildin

There is no point in talking about the data if it doesn't tell you useful things about the population or type of thing you are studying, if it's not for building a predictive model of the subjects. This is the entire reason for research. Also P(A|B) = P(B|A) if the prior probabilities are equal, often a reasonable prior.

nonbel 9y ago
>"There is no point in talking about the data if it doesn't tell you useful things about the population or type of thing you are studying" Ok, but I fail to see the relevance of this to anything. >"Also P(A|B) = P(B|A) if the prior probabilities are equal, often a reasonable prior." Here, A = Hypothesis and B = data (or vice versa). You are claiming that the "probability of seeing your data whether or not the hypothesis is correct", is often near the "probability the hypothesis is correct independent of your data"? Based on what? What principle leads you to think these two probabilities should be near each other? BTW, if you actually have one this is like the holy grail of statistics.
- Bromskloss 9y ago
  > You are claiming that the "probability of seeing your data whether or not the hypothesis is correct", is often near the "probability the hypothesis is correct independent of your data"? Shouldn't that be "the probability of seeing the data, _given_ that the hypothesis is correct" and "the probability of the hypothesis being correct, _given_ that the data has been observed"?
  - nonbel 9y ago
    No, the other poster is referring to Bayes' rule P(A|B) = P(A)*P(B|A)/P(B). The claim is that it is reasonable to assume P(A) ~ P(B), so that P(A|B) ~ P(B|A). I am referring to P(A) and P(B) in the equation.
    - Bromskloss 9y ago
      Ah, OK, got it.
      - nonbel 9y ago
        Its hard to tell over the internet, but if you are being sarcastic please let me know. I have no problem with clarifying further.
        Bromskloss 9y ago
        No sarcasm! I believe I understand what you were saying now.
  - deleted 9y ago
    [deleted]
- BenoitEssiambre 9y ago
  vaguely P(observations|pure noise generator) ~ P(pure noise generator | observations) I agree there are debates surrounding priors, especially when you get into higher order, meta or purely hypothetical territory (there is no actual noise generator here) like this, but I think these are unavoidable.
  - nonbel 9y ago
    >"vaguely P(observations|pure noise generator) ~ P(pure noise generator | observations)" For this to be true, P(observations) ~ P(pure noise generator) as you said earlier. Anyway, you can look at it from either perspective, doesn't really matter. On what basis do you claim this is reasonable to expect? I suspect you have none whatsoever... in fact simple thought experiments will lead us to the opposite conclusion, and there is plenty of data telling us the same.

nonbel 9y ago

>"There is no point in talking about the data if it doesn't tell you useful things about the population or type of thing you are studying" Ok, but I fail to see the relevance of this to anything. >"Also P(A|B) = P(B|A) if the prior probabilities are equal, often a reasonable prior." Here, A = Hypothesis and B = data (or vice versa). You are claiming that the "probability of seeing your data whether or not the hypothesis is correct", is often near the "probability the hypothesis is correct independent of your data"? Based on what? What principle leads you to think these two probabilities should be near each other? BTW, if you actually have one this is like the holy grail of statistics.

Bromskloss 9y ago
> You are claiming that the "probability of seeing your data whether or not the hypothesis is correct", is often near the "probability the hypothesis is correct independent of your data"? Shouldn't that be "the probability of seeing the data, _given_ that the hypothesis is correct" and "the probability of the hypothesis being correct, _given_ that the data has been observed"?
- nonbel 9y ago
  No, the other poster is referring to Bayes' rule P(A|B) = P(A)*P(B|A)/P(B). The claim is that it is reasonable to assume P(A) ~ P(B), so that P(A|B) ~ P(B|A). I am referring to P(A) and P(B) in the equation.
  - Bromskloss 9y ago
    Ah, OK, got it.
    - nonbel 9y ago
      Its hard to tell over the internet, but if you are being sarcastic please let me know. I have no problem with clarifying further.
      - Bromskloss 9y ago
        No sarcasm! I believe I understand what you were saying now.
- deleted 9y ago
  [deleted]
BenoitEssiambre 9y ago
vaguely P(observations|pure noise generator) ~ P(pure noise generator | observations) I agree there are debates surrounding priors, especially when you get into higher order, meta or purely hypothetical territory (there is no actual noise generator here) like this, but I think these are unavoidable.
- nonbel 9y ago
  >"vaguely P(observations|pure noise generator) ~ P(pure noise generator | observations)" For this to be true, P(observations) ~ P(pure noise generator) as you said earlier. Anyway, you can look at it from either perspective, doesn't really matter. On what basis do you claim this is reasonable to expect? I suspect you have none whatsoever... in fact simple thought experiments will lead us to the opposite conclusion, and there is plenty of data telling us the same.