a trading problem for mathematicians

T

trend2009

Suppose I have two trading methods with no correlation to each other. each of the two methods, if applied alone to the market, has a winning rate of 60%. the question is: if the two methods applied at the same time to the market (the signal must be agreed by the two methods at the same time), what is the winning rate?

Any mathematician here gives a try? and show how you get the result?
 
Quote from trend2009:

Suppose I have two trading methods with no correlation to each other. each of the two methods, if applied alone to the market, has a winning rate of 60%. the question is: if the two methods applied at the same time to the market (the signal must be agreed by the two methods at the same time), what is the winning rate?

Any mathematician here gives a try? and show how you get the result?
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I am not a mathematician but I know from elementary school that the particular winning rate will be the union of the probability spaces:

w = .6 +.6 - .6 x.6 = 0.84 -> 84%

Edit: For the proof you will have to offer something in exchange but it is trivial right from the axioms of probability.
 
I doubt it is the answer. if we have 3 systems that have 60% each, what would be the winning rate then? would it be over 100%?

Quote from intradaybill:

I am not a mathematician but I know from elementary school that the particular winning rate will be the intersection of the probability spaces:

w = .6 +.6 - .6 x.6 = 0.84 -> 84%

Edit: For the proof you will have to offer something in exchange but it is trivial right from the axioms of probability.
More...
 
Quote from trend2009:

I doubt it is the answer. if we have 3 systems that have 60% each, what would be the winning rate then? would it be over 100%?
More...

DO you want to learn? I don't like people who have doubts.

P(A+B) = P(A)+P(B) - P(A)P(B)

P(A+B+C) = P(A)+P(B) +P(C) - P(A)P(B) - P(A)P(C) - P(B)P(C)

= 0.72

I guarantee to you that if the events are independent the union will have P <1.
 
Quote from intradaybill:

DO you want to learn? I don't like people who have doubts.

P(A+B) = P(A)+P(B) - P(A)P(B)

P(A+B+C) = P(A)+P(B) +P(C) - P(A)P(B) - P(A)P(C) - P(B)P(C)

= 0.72

I guarantee to you that if the events are independent the union will have P <1.
More...



For 2 independent systems at 60% win rate each, the probability of both being wrong is (0.4)*(0.4) = 0.16, so the win rate when both agree is 1 - 0.16 = 84%.

For 3 independent systems at 60% win rate each, the probability of all three being wrong is (0.4)*(0.4)*0.4 = 0.064, so the win rate when all three agree is 1 - 0.064 = 93.6%.
 
If the two methods have no correlation to one another then if one system wins, by definition, the other system MUST lose and if that is the case then when one system goes long the other system must be going short. If this weren't true then at some point in the game both systems would either be long or short at the same time, giving them a correlation factor above zero. The gain in one system will be erased by the loss in the other system, thereby generating a win rate of 0%, assuming simultaneous exit.

This problem isn't as easy as 0.6 x 0.6 = 36% win, or 0.4 x 0.4 = 16% lose because 36 + 16 is not equal to 100.

Quote from trend2009:

Suppose I have two trading methods with no correlation to each other. each of the two methods, if applied alone to the market, has a winning rate of 60%. the question is: if the two methods applied at the same time to the market (the signal must be agreed by the two methods at the same time), what is the winning rate?

Any mathematician here gives a try? and show how you get the result?
More...
 
when you have two methods, the result is 0.86; with three, the result is 0.72?

do not you think it is incorrect?

Quote from intradaybill:

DO you want to learn? I don't like people who have doubts.

P(A+B) = P(A)+P(B) - P(A)P(B)

P(A+B+C) = P(A)+P(B) +P(C) - P(A)P(B) - P(A)P(C) - P(B)P(C)

= 0.72

I guarantee to you that if the events are independent the union will have P <1.
More...
 
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