What is the cumulative probability? - page 7
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We have two oracles! The first one says: Price will cross or touch 1.5000 with probability 0.6 within the current day.
The second oracle disagrees and says: Price will cross or touch 1.5000 with probability 0.2 within the current day.
What is the final probability that the price will cross or touch 1.5000 within the current day ???????????.
Note that if the first oracle's prediction were the same as the second: p1=p2=0.2, the final probability would be 0.2. How simple it is.
But if the first oracle still gives p1=0.6 ? Then how to calculate the final probability ???????
If predictions weights are the same for all aracles and in case p1=p2=0.2 then probably we should add all predictions of all aracles and divide by their number according to the principle of average. Ie if one gives 0.2 and second 0.6 prediction then (0.2+0.6)/2=0.4, i.e. probability increases. If a third arakul is added, his opinion will also be correctly taken into account. But it's only in case of equal strength of their forecasts. IMHO of course, but I think so.
I have already suggested that the problem should not be regarded as statistical, but as a problem of expert judgement.
And coaster does not rank the experts (oracles), i.e. does not say anything about the reliability of their predictions, considering it the same apparently.
I know they use Kemeny's median and Kemeny's average.
The median is the estimate, the sum of distances from which to the estimates of all experts is minimal.
The Kemeney mean is the same, only for squares of distances. In this case, min((P-0.2)^2 + (P-0.6)^2)) is just in the middle. P=0.4
But this is not probability. It is the confidence of the committee's assessment (with the 1st saying "yes" with 6 points confidence, the 2nd being 2 points confident in its assessment of "yes").
(In the simplest case, the experts only vote "yes" or "no", the decision is made by a simple majority).
If predictions weights are the same for all aracles and in case p1=p2=0.2 then probably we should add all predictions of all aracles and divide by their number according to the principle of average. Ie if one gives 0.2 and second 0.6 prediction then (0.2+0.6)/2=0.4, i.e. probability increases. If a third arakul is added, his opinion will also be correctly taken into account. But it's only in case of equal strength of their forecasts. Of course, it's just the way I see it.
That's what I thought at the beginning. But when I think about weights of forecasts I understand that they increase as far as forecast values go from 0.5. So the closer your forecast value is to 100% or 0%, the more weight it has. The thing is that those 100% numbers are not coming from the ground but from statistics, and 50/50, for example, means that the Forecaster can not make even a weak prediction, so it naturally has more weight.
I have already suggested that the problem should not be regarded as statistical, but as a problem of expert judgement.
And coaster does not rank the experts (oracles), i.e. does not say anything about the reliability of their predictions, considering it the same apparently.
I know they use Kemeny's median and Kemeny's average.
The median is the estimate, the sum of distances from which to the estimates of all experts is minimal.
The Kemeney mean is the same, only for squares of distances. In this case, min((P-0.2)^2 + (P-0.6)^2)) is just in the middle. P=0.4
But this is not probability. It is the confidence of the committee's assessment (with the 1st saying "yes" with 6 points confidence, the 2nd being 2 points confident in its assessment of "yes").
(In the simplest case, the experts only vote "yes" or "no", the decision is made by a simple majority).
All have a 100 per cent rating. They are all brothers. And they all have the same mother: statistics.