a trading problem for mathematicians

Discussion in 'Trading' started by trend2009, Jun 15, 2011.

  1. 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.

     
    #11     Jun 15, 2011
  2. the1

    the1

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

     
    #12     Jun 15, 2011
  3. 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
     
    #13     Jun 15, 2011
  4. doomed with the trading-signal-confirmation strategy.

     
    #14     Jun 15, 2011
  5. Congratulations you found the holy grail. with 4 systems confirmed with each other, you have 100% losing strategy:).

     
    #15     Jun 15, 2011
  6. 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?
     
    #16     Jun 15, 2011
  7. With 4 it would be 3,75%:D
     
    #17     Jun 15, 2011
  8. 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.
     
    #18     Jun 15, 2011
  9. 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.


     
    #19     Jun 15, 2011
  10. noddyboy

    noddyboy

    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.
     
    #20     Jun 15, 2011