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- pursuit replied Oct 7, 2011
Really? You missed the not so subtle sarcasm?
- pursuit replied Oct 5, 2011
Impossible! XYZ here traded it on demo for a day and had 7 wins and 1 loss!
- pursuit replied Oct 4, 2011
How long did you backtest for?
- pursuit replied Oct 4, 2011
Will you demo it for 3 years or more? If not, that's worthless.
- pursuit replied Oct 4, 2011
No, because I don't have data. I don't understand how you guys could be trading it, even on a demo without backtesting properly. This is not serious. This is gambling.
- pursuit replied Sep 26, 2011
so did anyone backtest this on at least a few years of data?
- pursuit replied Sep 20, 2011
I think this can be summed up as: discretionary systems that lack mechanical rules for critical parts of the system (entry, exit) are difficult (impossible?) to teach. If entry rule is: when A > B and A +C < D and... etc. then Enter Long. There's no ...
- pursuit replied Sep 18, 2011
This corresponds with my observations. Can we use a stoploss to kill the black swan and still retain enough edge? That's the question.
- pursuit replied Sep 18, 2011
Why not? I believe ONLY in backtesting.
- pursuit replied Sep 17, 2011
I think this system is not feasible without a stoploss. Can we test different stop sizes in for EA?
- pursuit replied Sep 17, 2011
I see. I have to be honest with myself and say: I'm afraid this method is not for me. Too much discretion.
- pursuit replied Sep 17, 2011
Cool method. What I'm having trouble with is figuring out exact rules for when to enter. I realized that this is discretionary method but I'm personally uncomfortable with uncertainty and discretion. I guess it's just my personality. Has anyone ...
- pursuit replied Sep 16, 2011
So there's no stop loss but we close losing positions if a set up the opposite way occurs?
- pursuit replied Sep 4, 2011
I saw another possibility here. Mark the rectangle zone across the chart. Then as the price approaches the lower border from below (rising price) - go long and exit at the top of the rectangle. Stop = 2X of the rectangle size. So basically this ...
- Posts by Member Search: 'pursuit'