In spirit I agree with "the real rules have no exceptions". I believe this applies to physics just as well as it applies to decision-making.
But, while the foundational rules of physics are simple and legible, the physics of many particles -- which are needed for managing real-world situations -- includes emergent behaviours like fluid drag and turbulence. The notoriously complex behaviour of fluids can be usefully compressed into rules that are simple enough to remember and apply, such as inviscid or incompressible flow approximations, or tables of drag coefficients. But these simple rules are built on top of massively complex ones like the Navier-Stokes equation (which is itself still a simplifying assumption... (read more)
You're welcome! And I'm sorry if I went a little overboard. I didn't mean it to sound confrontational.
X and ~X will always receive the same score by both the logarithmic and least-squares scoring rules that I described in my post, although I certainly agree that the logarithm is a better measure. If you dispute that point, please provide a numerical example.
Because of the 1/N factor outside the sum, doubling predictions does not affect your calibration score (as it shouldn't!). This factor is necessary or your score would only ever get successively worse the more predictions you make, regardless of how good they are. Thus, including X and ~X in the enumeration neither hurts nor helps your calibration score (regardless of whether using the log or the least-squares rule).
I agree... (read more)
The calibration you get, by the way, will be better represented by the fact that if you assigned 50% to the candidate that lost, then you'll necessarily have assigned a very low probability to the candidate that won, and that will be the penalty that will tell you your calibration is wrong.
The problem is the definition of more specific. How do you define specific? The only consistent definition I can think of is that a proposition A is more specific than B if the prior probability of A is smaller than that of B. Do you have a way to consistently tell whether one phrasing of a proposition is more or less... (read 475 more words →)
Yes, but only because I don't agree that there was any useful information that could have been obtained in the first place.
I don't understand why there is so much resistance to the idea that stating "X with probability P(X)" also implies "~X with probability 1-P(X)". The point of assigning probabilities to a prediction is that it represents your state of belief. Both statements uniquely specify the same state of belief, so to treat them differently based on which one you wrote down is irrational. Once you accept that these are the same statement, the conclusion in my post is inevitable, the mirror symmetry of the calibration curve becomes obvious, and given that symmetry, all lines must pass through the point (0.5,0.5).
Imagine the following conversation:
A: "I predict with 50% certainty that Trump will not... (read more)
"Candidate X will win the election with 50% probability" also implies the proposition "Candidate X will not win the election with 50% probability". If you propose one, you are automatically proposing both, and one will inevitably turn out true and the other false.
If you want to represent your full probability distribution over 10 candidates, you can still represent it as binary predictions. It will look something like this:
Candidate 1 will win the election: 50% probability
Candidate 2 will win the election: 10% probability
Candidate 3 will win the election: 10% probability
Candidate 4 will win the election: 10% probability
Candidate 5 will win the election: 10% probability
Candidate 6 will win the election: 2% probability
...
Candidate 1 will not win the election: 50% probability
Candidate 2 will not win the election: 90% probability
...
The method described in my post handles this situation perfectly well. All of your 50% predictions will (necessarily) come true 50% of the time, but you rack up a good calibration score if you do well on the rest of the predictions.
Agreed.
I'm really not so sure what a frequentist would think. How would they express "Jeb Bush will not be the top-polling Republican candidate" in the form of a repeated random experiment?
It seems to me more likely that a frequentist would object to applying probabilities to such a statement.
Thanks!
I think my point is different, although I have to admit I don't entirely grasp your objection to Nostalgebraist's objection. I think Nostalgebraist's point about rules being gameable does overlap with my example of multi-agent systems, because clear-but-only-approximately-correct rules are exploitable. But I don't think my argument is about it being hard to identify legitimate exceptions. In fact, astrophysicists would have no difficulty identifying when it's the right time to stop using Newtonian gravity.
But my point with the physics analogy is that sometimes, even if you actually know the correct rule, and even if that rule is simple (Navier-Stokes is still just one equation), you still might accomplish a lot more by... (read more)