a trading problem for mathematicians

that the two methods have no correlation means the two methods are independent. it does not mean one is long the other must be short.

Quote from the1:

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.
 
I stand corrected. Now I will have to think some more. Shit, I'd rather watch Nemo for the 84th time :D

Quote from zedDoubleNaught:

I'm sorry, but if the other system must lose, wouldn't that be a correlation of -1?
 
With 2 coins, there are 4 possible outcomes. Of these, 1 outcome has 2 heads.
Therefore the probability of this is 1/4 or 25% (rounded)

And in this case 60*1/4 or 15%

In other word BS strategy:D
 
doomed with the trading-signal-confirmation strategy.

Quote from SnakeEYE:

With 2 coins, there are 4 possible outcomes. Of these, 1 outcome has 2 heads.
Therefore the probability of this is 1/4 or 25% (rounded)

And in this case 60*1/4 or 15%

In other word BS strategy:D
 
Congratulations you found the holy grail. with 4 systems confirmed with each other, you have 100% losing strategy:).

Quote from SnakeEYE:

With 2 coins, there are 4 possible outcomes. Of these, 1 outcome has 2 heads.
Therefore the probability of this is 1/4 or 25% (rounded)

And in this case 60*1/4 or 15%

In other word BS strategy:D
 
Quote from the1:

I stand corrected. Now I will have to think some more. Shit, I'd rather watch Nemo for the 84th time :D

Sorry about that, after I posted I deleted my comment because I didn't want to be nitpicking over a detail, but looks like you've quoted it before that. Just wanted to note it because I forget it and get tripped up sometimes -- the range is from perfect negative correlation, to no correlation, to perfect positive correlation.

Anyways, for the thread, you might need to mention if the 2 strategies block each other or not -- is it that there can only be on one trade on at a time, or can both systems place a trade on different instruments at the same time?
 
Quote from trend2009:

Congratulations you found the holy grail. with 4 systems confirmed with each other, you have 100% losing strategy:).

With 4 it would be 3,75%:D
 
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.
Is it me or if I have 3 losing systems with 0.35, 0.40 and 0.45 it gives me a wining rate of 0.72!!! I dont think so.
IMO the winning rate is
(1/3 * 0.60 + 1/3 * 0.60 +1/3 * 0.60)/3 = 0.60
The sharpe would improve though. The more uncorrelated systems you have the less volatility there will be. The greatest gains being from 1 to 20-30. You can read Ray Dalio of Bridgewater for references. Or this is the reason why Buffet has about 30 stocks in his portfolio.
 
I guess you misunderstood the problem. The problem is that the trading signal from method A must be confirmed by method B so that the signal is valid.


Quote from total_keops:

Is it me or if I have 3 losing systems with 0.35, 0.40 and 0.45 it gives me a wining rate of 0.72!!! I dont think so.
IMO the winning rate is
(1/3 * 0.60 + 1/3 * 0.60 +1/3 * 0.60)/3 = 0.60
The sharpe would improve though. The more uncorrelated systems you have the less volatility there will be. The greatest gains being from 1 to 20-30. You can read Ray Dalio of Bridgewater for references. Or this is the reason why Buffet has about 30 stocks in his portfolio.
 
Quote from trend2009:

I guess you misunderstood the problem. The problem is that the trading signal from method A must be confirmed by method B so that the signal is valid.

If you have systems of prob p1 and p2, the combined is
2 * p1 * p2

If you have 3 systems, the combined is
4 * p1 * p2 * p3.

I know...you are going to ask why you can get > 1 prob.
Well, in that case, the systems are not independent, which is why the solution is nonsense.

And if they are all coin flips, then you have a 50% probability.
 
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