You are too kind! It is my pleasure to be here and share my fascination with the subject! Cheers, MAESTRO
My answer is that you are applying Bayes without any regard to assumptions made regarding the random variables involved: http://arxiv.org/abs/1103.6136 As far as the Monty Hall problem, its application and even relevance to stock trading is not even clear. Bayes' formula deals with independent events. In stock trading events may not be independent. Certain events cause other events to take place. For example, a certain stock dropping more than a certain amount causes money to flow in other stocks. This may not be reflected in your statistics. Trading is not as easy as contidtional probabilities.
unless you bunch things together ,apples,oranges,bananas= fruit....trend of djt,spx,nq,vs movement in currencies, bonds =market...don't know shite about formulas but from an untrained eye these( >90%) correlate together as one
I have a question on how these numbers got inserted: If I write out Bayes formula in words, I get (~ means not, obs = observation, hypo = hypothesis): 1) P(hypo | obs) = P(obs | hypo)*P(hypo) / [P(obs | hypo)*P(hypo) + P(obs | ~hypo)*P(~hypo)] 2) P(will hit high | at 90%) = P(at 90% | will hit high)*P(will hit high) / [P(at 90% | will hit high)*P(will hit high) + P(at 90% | will not hit high)*P(will not hit high)] Is my phrasing correct? if so, the for the numerator, is it: P(at 90% | will hit high) = .90 P(will hit high) = .80 or is it the reverse? I was confused on the example as initially worded, I don't see where "P(will hit high)" is stated. thanks
I don't think the answer given is correct, because when finding the numerator you need to know the probability of hitting the high: p(high |.80 range)*p(.8 range)+p(high|not .80 rane)*p(not .80 range). Since we don't know the probability of .80 range, we can't solve it without added info (the distribution type if any) Maestro do you not see this? What are your assumptions of the distribution?
MAESTRO, In "The Illusion of Certainty: Risk, Probability, and Chance" video Josh Tenenbaum talks about 00101 sequence. Out of the 16 combinations presented, two of them are particularly interesting where 001 ends with 00 and the second one is 00101 and has the highest count. Is it related at all to the rigid structures you were talking about trying to explain the phase shift in chaos, and to my question to you about "third time is a charm" thing? It all seem to be related to "the patterns in random walk". Thank you.
Yes, it assumes the normal distribution. Please do not forget, this example was given just for the illustration purpose only. It, of course, has huge assumptions built in. However, if one takes this general approach a very useful method could be constructed. Please do not forget that I have to watch very carefully what I write so that I do not by mistake disclose any sensitive info here. I will always keep the conversations here at ET at a very abstract levels. The purpose of my posts is education only. It is inadvisable to treat the examples that I give literally. However, you can use them to ignite your interest in the subject discussed here. P.S. The answer that was given earlier (0.53) is very close to correct value.
If you are to be very careful about what you write it can mean two things: 1. You have an edge already, so I see no point you looking for one. You should be making enough money to live a lavish lifestyle away from these websites. 2. You do not have an edge and as such you have no sensitive information. You use this common trick as a way of trying to get sensitive information from others. At any rate, I don't like it when people I talk with admit they are hidding information from me and I won't participate in this non-exchange of sensitive information, don't worry about a thing.
MAESTRO and others might find this video interesting. Autonomous Flying Robots Flock Like Birds http://www.wired.com/wiredscience/2011/09/flocking-robots/