a trading problem for mathematicians

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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This is not a probability problem.

Let's say that the probability that you'd walk on a straight line with your left eye covered is 90%, and the probability that you'd walk on a straight line with your right eye covered is 90% too. What do you think is the probability of you walking straight if you covered both your eyes?
 
OMG! Is this a joke thread that belongs in Chit Chat?

The question has been answered.

Mods, please close this thread.
 
Why don't we give some real world examples that might actually help traders.

1) Let's assume for sake of argument that a support or resistance working is only 50%. So using this alone does not give you a good enough edge to win at trading.

2) Let's assume for sake of argument that a break of a support or a resistance only gives you a 50% win rate, so that does not give you a good enough edge to trade.

3) However, let's assume you combine either one of these events with a non-lagging indicator, or you combine a support line for example with a fib line (Note: fib lines and indicators by themselves only have a 50% chance of being right).

4) You may also want to add in trend or RTM. For example you are going to either use the above with a trend, or a RTM strategy.

So we now have around 2 confirmations from non correlated signals which we use to either take or not take the trade, or at least give us an alert so we can watch candle patterns after the 2 confirmations for a 3rd confirmation.

What is the win% of the above system, its around 75% if you use good stops and targets based on either price action or support and resistance.

To work in real life, you also must control the urge to over trade / revenge trade which can hurt you even if you have an edge and are patient enough to wait for a trade.
 
Quote from kut2k2:

OMG! Is this a joke thread that belongs in Chit Chat?

The question has been answered.

Mods, please close this thread.
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These threads remind me why one percent control 42% of the wealth.:D
 
first of all you changed your original problem several times and this points to the fact you dont fully understand what you like to ask.

My firm reading of your problem is that you have two "signal engines". Each "signal engine" has to point to a trading signal in order to actually take a trade. From this it logically follows that you take 1/2 the number of trades that each signal engine would have independently generated. (1/2 for each engine alone -> 1/4 combined).

This is a completely independent problem and knowledge of this is IRRELEVANT to trying to determine the winning rate of the combined rate, GIVEN THE OUTCOMES ARE UNCORRELATED as you stated. Once both signal engine flag a trade we take a trade, and the real question now is what is the win rate of taking such trade, and it is clear it must be 60% because each individual engine generates trades at a 60% win rate and the engines are uncorrelated and thus generate i.i.d. outcomes.

Not sure what is so hard about this problem unless I am the one who does not understand your problem.

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...
 
Quote from asiaprop:

first of all you changed your original problem several times and this points to the fact you dont fully understand what you like to ask.

My firm reading of your problem is that you have two "signal engines". Each "signal engine" has to point to a trading signal in order to actually take a trade. From this it logically follows that you take 1/2 the number of trades that each signal engine would have independently generated. (1/2 for each engine alone -> 1/4 combined).

This is a completely independent problem and knowledge of this is IRRELEVANT to trying to determine the winning rate of the combined rate, GIVEN THE OUTCOMES ARE UNCORRELATED as you stated. Once both signal engine flag a trade we take a trade, and the real question now is what is the win rate of taking such trade, and it is clear it must be 60% because each individual engine generates trades at a 60% win rate and the engines are uncorrelated and thus generate i.i.d. outcomes.

Not sure what is so hard about this problem unless I am the one who does not understand your problem.
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Am I the only person here who can read this part of the OP? :

"the signal must be agreed by the two methods at the same time"

What, you people think that's irrelevant? BOTH methods have to agree.

If event A has a probability of p(A) and uncorrelated (aka independent) event B has a probability of p(B), then the joint probability A _AND_ B is

p(A and B) = p(A)*p(B)

This is Probability 101. Even the guy who claims he learned it in elementary school (yeah, right) and then proceeded to blow it with p(A or B or C) might have gotten it IF he could read the OP properly.
 
you are wrong bro, here is why:

(1) lets say strat1 has probability 1 of winning
(2) strat2 has probability 0.5 of winning
(3) only take a trade if both strats generate a signal
(4) => the win probability of the combined signal is 1!!!!!!



Quote from kut2k2:

Am I the only person here who can read this part of the OP? :

"the signal must be agreed by the two methods at the same time"

What, you people think that's irrelevant? BOTH methods have to agree.

If event A has a probability of p(A) and uncorrelated (aka independent) event B has a probability of p(B), then the joint probability A _AND_ B is

p(A and B) = p(A)*p(B)

This is Probability 101. Even the guy who claims he learned it in elementary school (yeah, right) and then proceeded to blow it with p(A or B or C) might have gotten it IF he could read the OP properly.
More...
 
why dont you model it in Excel? Generate 10000 random numbers between [1..10]. Simulate each a winning trade if 5<=x<=10 is generated, a losing trades else. It does not matter whether you calculate your combined probability over a space of 1 experiment or all 10000 you get to an avg. win probability of 60%.

I think the fallacy here is to think that if you have infinitely many of such signal generators you must almost surely converge to a combined win probability of 1 because you add up many systems with "edge", I beg to disagree.

Quote from jhiro:

There is no formula that will work it out, you'll have to independently test every time they correlate.. the fact they both give 60% in isolation means fuck all

If math could be applied perfectly to the markets everyone would be rich and we'd have a theory of everything explaining all 11 dimensions with a formula 3 inches long, gravity would be married to the electro-magnetic, weak and strong nuclear forces, the standard model would embrace relativity and Einstein and Heisenberg would kiss, fuck and make up
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Quote from kut2k2:


"the signal must be agreed by the two methods at the same time"

What, you people think that's irrelevant? BOTH methods have to agree.

If event A has a probability of p(A) and uncorrelated (aka independent) event B has a probability of p(B), then the joint probability A _AND_ B is

p(A and B) = p(A)*p(B)

This is Probability 101. Even the guy who claims he learned it in elementary school (yeah, right) and then proceeded to blow it with p(A or B or C) might have gotten it IF he could read the OP properly.
More...

This is flat out wrong, you are thinking something different, the AND condition - what is the probability of event A AND B both being true given each event has a 0.6 of being true individually. Then it's 0.6*0.6

What OP is asking is what is the win probability of the TRADE when two strategies with a 0.6 win probability BOTH give the trigger to place said trade. Think about it for a second, if each individual trigger has a 0.6 chance of win, how can waiting for both to trigger reduce your win to 0.36?

The question is valid and clear. It sounds simple until you try to quantify what "winning rate" is in OP's question.

For example using the loaded die template:

Strategy 1: win=0.6 lose=0.4
Strategy 2: win=0.6 lose=0.4

All possible outcome=4
win/win, win/lose, lose/win, lose/lose

All possible WIN outcome=3
win/win,win/lose,lose/win

Probability of each of these outcome occurring then sum it up.
P(WIN) = 0.6*0.6 + 0.6*0.4 + 0.4*0.6 = 0.84 -> 84%

But this is not correct because we are making an assumption it's an OR condition where win/lose and lose/win both equals to a win. Need a quant to fit this into a standard probability template then it will become easy.

Anyone else feeling stupid right now, raise your hand.
 
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