The well-calibrated Bayesian [pdf] (1982)
fitelson.org
fitelson.org
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.
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
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
The fact that the model is represented as a distribution over Bernoulli parameter p doesn't contradict this: you just integrate over the posterior.
One problem is aggregating information over multiple steps in the reasoning chain the other problem is the powerset problem. The point probability estimate from integrating over the posterior wasn't useful for the first, so I didn't want it. The second problem is impossible in theory but possible in practice, as the existence of intelligence proves.
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.
NOT SPOCK!!!
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.
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,…).
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.
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?
For example, if it rains tonight then I am 70% sure it will rain tomorrow too. I am also 80% sure it will rain tonight. And I am 100% sure that if it doesn't rain tonight, it also will not rain tomorrow. Then I can chain these together: The chance of rain tomorrow is P(Rain tomorrow) = P(Rain tomorrow | rain tonight) * P(Rain tonight) = 0.8*0.7 = 0.56.
You can assign conditional probabilites to the correctness of your model itself. Then the phrasing is more like "I am 70% confident that it will rain tomorrow, conditional on my understanding of meteorology being correct" and "I am 80% sure that I understand meteorology correctly". Then you also need to add in a term for P(Rain | not understanding correctly).
you may replace "my model is wrong" by e.g " my certainty that no one is using cloud-activation technology".
Now, I'm not native in English and I see how someone (especially with a math background) might find this sentence confusing:
> 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.
Turns out it's very hard about math non-formally for generalist audience and might not have done the best job about it. It should read
> you are predicting something (e.g rain tomorrow), then your model believe it will rain with probability 0.7 but you are 80% sure of your model, then you won’t report 0.7; you will report : 0.8 × 0.7 + 0.2 × something_you_believe = 0.58. Coherence forces you to collapse your uncertainty into your probability at each forecast.
They propose a bet. If they flip it 100 times and the proportion of heads is within [0.4, 0.6], you win $100. If it's not, you pay $100. Do you take that bet?
Explanation: absent the magic store scenario, a `rational' person would take the bet. Your prior belief is that most coins are roughly unbiased. Given that they walked out of a magic store, you now have additional information. Maybe the coin is a trick coin. In that case, your belief that the coin is unbiased should be weaker, even if you don't know which direction the coin is biased in.
This illustrates two things: one, additional information (magic store) can update your beliefs. Two, a strong prior and a weak prior, in this case about the coin's bias, can lead to materially different decisions.
So I would certainly consider it likely, that they are trying to trick me. But the probability I would assign to this, would still be rooted in some frequency, somewhere under the hood I would try to estimate the possible situations leading to such an offer and in which fraction of them I will be tricked.
If I am doing a good job with that, then repeatedly being in this situation should result in me getting tricked with the probability I cooked up. If I am bad at figuring out the possible states and their probabilities, then I they will not match.
Calculated frequency is a point-estimate of bias. A Bayesian estimate is a distribution of belief over possible values of the bias, integrating all available information.
Where does that come from? It is not some intrinsic property of the coin, it comes from varying initial conditions. If you had enough precision when controlling your hand movements, you could in principle force an outcome with high probability.
But assuming you can not or at least do not do that, there is a certain set of initial states, some will lead to heads, some to tails, and each toss will start from a randomly selected initial state. So given my ignorance of the exact initial state, the coin will land heads with a probability equal to the number of initial states leading to heads divided by the number of initial states compatible with my observations of the initial state. [1]
Repeatedly tossing a coin will sample the set of initial states and the result will match the proportion of the number of states. At least as long as I am not wrong about the set of initial states.
The same applies to something like an election. I have imperfect knowledge about the state of the world but there is a set of states compatible with my knowledge about the world and certain subsets of them will lead to certain candidates to win.
[1] Maybe adjusted by some probability distribution over the initial states if they are not equally likely to be picked.