Wrong again.
The theory as I described it works and the above formula is -functionally- correct.
To better the results of the standard theories the problem seems to be the choice of the form of the q' function.
* I 'm sorry for writing some random things in my last post, but it's because I was...
No, the difficulty here is to approximate q, for the realistic situations.
q' = 0.5 + L . (q - 0.5)
is wrong - good only for two predictions.
Here is a paper that says something:
http://www.math.canterbury.ac.nz/research/ucdms2003n6.pdf
The solution of the problem appears to be as I described it for the 2x2 case.
One of course has to take measurements to see what departure there is from the ideal case when the two predictions are not 100% independent.
But what if we have more than 2 predictions and more than 2 predictors ?
If...
We established the probability formula for the case of two independent probability forecasts:
f(p,q) = p*q / (p*q + (1-p)*(1-q))
In practice the two forecasts are never independent.
If p >= q > 0.5 then the "true" q is somewhere in between.
That is we have to replace q in the formula...
The argument used by mathpages.com to derive the formula is hellishly tortuous.
Let me try to do this differently:
Each of the two forecasters, weatherman and Indian, follows his own set of methods.
We can't tell to what extent those methods are the same.
The ultimate case of independence...
I have some clues and some lines from you and I 'm trying to put it together.
There are some extremes in the problem:
The bad extremes:
- First predictor makes reasonably good predictions, the other one just flips a coin. Then p2 = 0.5 and the formula from mathpages (the nature of which I...
Let's call the statement "rain", statement A and the statement "no rain", statement B.
Prior to the event there are four possible predictive statements A-A, A-B, B-A, and B-B, each with its respective probabilities (p1, 1-p1), (p2,1-p2) etc of positive and negative outcome.
So if we measure...
My answers are not coherent maybe because I have reached my present limits of knowledge about the problem, which after all I sort of made up in the first place without knowing the answer. I 'll think more about it in the light of the day.
I did have some examples with figures showing a certain...
It adds something, like the 75% that became 90% a few posts above.
But some other times the weatherman is unfairly penalized.
The way I see it the Indian is useless if he says random things.
Above this extreme it should be true that one opinion adds to the other.
If you have M guessers and N states of matter they are guessing, it becomes difficult.
The system (sun-rain, weatherman-indian) is only 2x2 so it looks easy.
Among the N states of matter, or types of weather if you like, some are likely to be pretty infrequent, so how long do we say to our...
I see.
You break the problem into components and say "correct weather station + correct indian = x%" and so on. And if Indian is infact worthless then it should turn out x = p.
Perhaps I was thinking along the lines of probability systems with lots of such components + observers, where the...
The scientific method is the first formula ?
I don't believe so.
It is scientific but under the conditions it specifies.
Here is an obvious case of failure:
Predictor A is a learned journalist, predictor B is Paul the Octopus (& don't spoil by saying you believed the octopus !).
In my...
Does it make any sense to do that:
f(p,q) = p^a*q^b/( p^a*q^b + (1-p)^a*(1-q) ^b)
where a, b are experimental constants ?
The sum of f(p,q) over the various outcomes is not 1 but I could normalize f(p,q) to make it equal to 1. It's kind of heuristic.
If a=1, b=0 then f(p,q) = p and the...
It does n't look outrageous to me if f(p,q) = 0.9, if the two pieces of evidence are uncorrelated.
Suppose police receive a UFO call. They say "it's another loonie this time of the night'. But when they receive a second call, they jump to attention. Don't they ?
That's because they believe the...