Probability theory question... - page 3
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Suppose we have 5 indicators, each with probability Dn at each moment of time shows the correct entry direction. Then let calculate the probability that most of them (3 out of 5) show the right direction. That is reached by combinations [1,2,3], [1,2,4],...[3,4,5]. You accumulate the probabilities corresponding to these combinations and get the required probability. That is D1*D2*D3 + ... D3*D4*D5.
Everything is brilliantly simple...
Пусть у нас есть 5 индикаторов, каждый с вероятностью Dn в каждый момент времени показывает правильное направление входа. Тогда посчитаем вероятность того, что большинстов (3 из 5) показывает правильное нарпавление. Это дотсигается комбинациями [1,2,3], [1,2,4],...[3,4,5]. Собираете соответствующие этим собятиям вероятности и получаете искомую вероятность. То есть, D1*D2*D3 + ... D3*D4*D5.
vizit писал (а) >>Independent.
How about this?
D1*D2*D3*(1-D4)*(1-D5)+...
Although I have MORE doubts about independence....
Suppose we have 5 indicators, each with probability Dn at each moment of time shows the correct entry direction. Then let calculate the probability that most of them (3 out of 5) show the right direction. That is reached by combinations [1,2,3], [1,2,4],...[3,4,5]. You accumulate the probabilities corresponding to these combinations and get the required probability. That is D1*D2*D3 + ... D3*D4*D5.
Got it!!!
Thank you very much!!!
Independent.
But if they are independent and the probability of error is the same for all, then it is not difficult to count.
And LeoV's version deserves attention.
Got it!!!
Thank you so much!!!
But it remains an open question to determine the probability of each indicator...
How about this?
D1*D2*D3*(1-D4)*(1-D5)+...
Yes, you're right, the formula is not quite right. You need to add guessed fours and one fives (and probabilities)
2 Mischek about vizit - 'forewarned is forearmed'. :)
(And about twins and 'price derivatives')
vizit, don't bother, it's useless. It's much easier to run the whole system in a tester. The vast majority of indicators are highly correlated - and one should not be fooled by the fact that the signals of each are confirmed by the others. If the whole crowd is running in one direction from an enraged bull, you should not look for confirmation from your neighbours running next to you. The behaviour of an individual in a crowd is strongly correlated with the behaviour of others in the same crowd, because people in such a situation usually behave very similarly, and the main acting factor is the same for all (the bull is the price). But this does not mean that they all act correctly.
We can speak about confirmation only in case signals of individual indicators are uncorrelated (or better yet, independent). By the way, even if the correlation coefficient between each two indicators is about 0.6-0.7, the estimate of the degree of reliability of predictions is much worse than it seems. Even if you put a thousand of these indicators, there will be a non-zero limit of error probability.
Again: The task is not about indicators. This is for the sake of example.
vizit - you want an answer not about indicators, ask not about indicators :)