It took me a while, but here’s what I gather about this (I’m pretty sure it’s correct, but I’m not an expert).
A calibrated forecast means that if you say there is a 20% chance of rain, then it actually rains 20% of the time. It’s a desired feature, but not the only one (e.g. you could be calibrated by stating: Chick-fil-A is open every day except Monday, but your forecast will always be wrong on Sunday and Monday).
So if
1. you are Bayesian (you state your beliefs)
2. and coherent (the laws of probability apply, so e.g. if P(A) = 0.4, then P(not A) cannot be anything other than 0.6):
and you are predicting something (e.g rain tomorrow), then if you believe it will rain with probability 0.7 but you are 80% sure of your belief, you won’t say 0.7; you will say something else: 0.8 × 0.7 + 0.2 × something_you_believe = 0.58. Coherence forces you to collapse your uncertainty into your probability at each forecast.
This theorem shows that, over many forecasts, in your belief system you are certain to be producing a calibrated forecast: your current beliefs assign probability 1 to the proposition that your future forecasts will be calibrated.
But that can’t be, which is the paradox. So Bayesianism is too strong compared to how scientists reason, because scientists always think their model can have an error.
I didn't see a real Bayesian point of view in that article.
A Bayesian does not give you a probability estimate they give you a probability distribution for the probability!
Like in Star Trek Spock is always saying something like "Captain, we have a 15.31% chance of surviving this mission" which is a ridiculous example of precision without accuracy. [1]
If you observe a coin flipped 100 times and it came up heads 65 times it is not a crazy point estimate to say it has a 65% chance of coming up heads but this is just one sample and if you did it another time maybe it comes up 61 or 68 times. You are better saying that the probability distribution of the probability is β(65,35) or maybe β(65.5,35.5) or β(66,36) since that has the "error bars" built in, can be updated if you get more samples, etc.
[1] ... and you know he underestimates survival probabilities the same way Scotty overestimates how long it will take to fix the engines
The \pi_i in the paper is not the estimate of a latent parameter. It is the predictive probability of the event, which is a single number by necessity in a binary challenge. It's the integration of a distribution function which can contains very complex distributions: in my example something_you_believe can be a probability distribution.
So everything in the paper is distribution and when you forecast for a binary event, you give a number which is the expectation of that distribution. This is a probabilistic forecast.
If you were to give a probabilistic forecast for a continuous quantity, then yes you would give in a distribution, as in section 4.2
Emphasizing this response. Bayesian models can always produce simple probabilities if you ask them to. E.g., given this data, what is the probability that the next flip is heads?
The fact that the model is represented as a distribution over Bernoulli parameter p doesn't contradict this: you just integrate over the posterior.
We've got a coin here! Let's ask a frequentist and a bayesian what they think about the probability that flipping it would make it land heads up?
Frequentist: How would I know? I haven't seen it flipped once, nor do I know how you've selected it from all the other coins that exist.
Bayesian: It's 50%!
Then we flip the coin 10,000 times and observes that it landed up heads exactly 5,000 times.
Frequentist: Huh, that's weird. You see, if the probability was 1/2, the expected deviation from the mean in this case would've been 50, so I'd expected to see either about 4'950, or 5'050 heads... still, MLE provides the answer of 1/2 bu-u-ut...
Bayesian: It's 50%!
This two-strawmen thought experiment clearly demonstrates the superiority of the Bayesian approach in learning useful information from the real-world observations.
It's really kinda shame that both personal certainties and physical probabilities follow the same algebraical rules while having entirely different nature; most of the time, you are not very interested in how much is someone is certain of some outcome, you're much more interested in the actual outcome or at least the actual probability of that outcome. Granted, most of the time you can only readily access someone's certainty of an outcome, but this is just a proxy for the quantity you're actually interested in knowing.
> You are better saying that the probability distribution of the probability is β(65,35) or maybe β(65.5,35.5) or β(66,36)
s/probability/your personal certainty/g. The probability of the coin landing heads up is what it is, and it usually doesn't depend on any of your knowledge.
>and you know he underestimates survival probabilities the same way Scotty overestimates how long it will take to fix the engines
NOT SPOCK!!!
Was Spock poorly calibrated?
If he was well calibrated there is no way they would have made it through 79 episodes!
Spock had not realized that James Kirk emitted a psionic reality distortion field through higher dimensional "luck", if he had he would have been a bit more relaxed.
> if you believe it will rain with probability 0.7 but you are 80% sure of your belief
What does this even mean...?
If I believe it will rain with probability 0.7, that 0.7 figure should already be taking into account the sum total of all of my uncertainty over all of my beliefs: my trust in the weather forecast, my past experience with the local area in this season, my certainty that the earth will continue to exist tomorrow.
Bayesians of course accept that their models can have errors, and if they're doing a good job they'll factor all of the most influential ones into the probability calculation itself.
> What does this even mean...?
You are making the same point that I did, if you read the rest. 0.7 at 80% is an intermediate step in formulating a forecast, and the point is to show that you cannot stop there and have to include everything in your number, like you have said. In your words, the intermediate steps are what you list (trust in weather forecasts, past experience,…).
No, what I'm saying is that 0.7 is the ending point of the calculation, and there is no 80% associated with it. The 0.7 number already factors in my uncertainty in all of my knowledge about whether it will rain tomorrow. Saying that you're 80% sure of your 0.7 probability forecast throws a TypeErrorException.
yes and in my example the ending point of the calculation is 0.58.
I make an intermediate step explicit in this derivation because it is important for the understanding of the problem.
This kind of intermediate step happens a lot in forecasting competitions, where participants are asked for their forecasts and their confidence in their forecasts. I want to show here that you need to include that in your forecast (all you belief) and not keep it separated.
But again this seems like a TypeError.
If you're asking me for my forecast and my confidence in my forecast, I would say this:
Forecast: It will rain tomorrow. Confidence: 0.7
If you say "No no, I mean what is your confidence in the 0.7 number" then I have no idea what you're talking about. 0.7 is my confidence. It's not valid to attach a confidence to that number. When the next day comes, we simply find out whether it rains or does not rain, and whoever put the highest probability on the correct outcome wins.
If you're doing this with intermediate steps then I have to ask: where did the 0.7 probability come from as an intermediate step, and what is the epistemic meaning of the 0.8 attached to it? If you only felt 80% sure that you were 70% sure that it would rain tomorrow, then what is the proposition that you are 70% confident about?