I think you may also not fully grasp the original problem. You do not have different outcomes. There is only ONE outcome, WIN OR LOSE, the question is what the probability of winning is after taking the combined "signals" and their individual, uncorrelated, probability of winning into account.
sorry but you now disappoint me because what you said is bollocks, it has nothing whatsoever to do with the problem at hand. This is a mathematical problem, and while I do not claim my answer must be right I insist there is a mathematical exhaustive answer or else the clear answer that it cannot be calculated with the information at hand. But the answer has nothing to do with how wide stops are. Guaranteed!!!
so you are telling us the combine win prob is 36%? How can this be? I already disproved you. If strat1 wins 100/100 and strat2 60/100, you claim you have a win prob of 60%??? This is completely incorrect as it violates that strat1 wins always. Any comments?
Precisely. Each strat has a win rate of 60%, however both have to agree. When means that the strat is essentially one strat with two triggers. Again, there is a reason why one percent control 40 percent of the wealth
You have two coins that each come up heads 60% of the time. What are the chances of getting two heads when you flip them? Obviously it's 36%. Any other combination is a wash or a loser. If somebody subs a two-headed coin for one of the original coins, you still don't have a 100% win rate, it is only 60%.
There is no holy grail. Here is my superficial take: (1) Define "winning" as the sum of all profit and loss of the 2 methods being > 0; (2) Define "losing" as the sum being <= 0; (3) Methods A and B have winning probability of 0.6 each. (Known) (4) When a method wins, put a "+" sign in front of it, and "-" sign vice versa; (5) The winning rate of both methods is [probability of { (+A -B) > 0 }] or [probability of { (-A +B) > 0}] or [prob { +A } and prob {+B}] = 1 - (the reverse of the above) Unless there is definitive info on the relative size of the wins and losses of the two methods, the overall probability cannot be computed under the above definitions. But what is funny is [prob (+A-B)>0] and [prob (-A+B)>0] ; you want something and the reverse of something to have the same property (>0). Think about it. Putting more methods together doesn't seem to beat a coin toss. For technical analysis, here is something to consider: Evidence-Based Technical Analysis: Applying the Scientific Method and Statistical Inference to Trading Signals [Hardcover] David Aronson (Author) A quote form a review: "In this thought-provoking work, David Aronson tests more than 6,400 technical analysis rules and finds that none of them offer statistically significant returns when applied to trading the S&P 500. "
That`s true. There is no such thing as probability in trading. If you get the entry right you win - If you get the entry wrong you lose!
pbj, thanks for the correction. What a blunder of mine in the case of 3 systems. Of course the product space will be A x B x C