I am actually thinking about a dual ema bot, but at this point I am just collecting intraday ES data and analyzing it at the end of day. Instead of using crossovers, it would consider the relationship, slope, of the two emas to make decisions. If the fast ema is under the slow ema and both slopes are down, sell. It might use a sentiment guage to avoid sell or buy entries at certain times.
Frosty's trading bot goes live part 2
- Thread starter frostengine
- Start date
Do you really think a system is effective taking $1000 in heat to make $300. If it cut losses quickly it might find another entry to get in and make even more.
I respectifully disagree that taking such a bad risk/reward ratio will lead to positive results over time. I think it is one reason why the bots is down so much, without knowing more about the system that is actually being used.
I didn't realize this is one just one contract. Sorry about that.
My post which you quoted was in response to this statement from walterjennings (which I quoted):
That is a false statement and I provided a simple hypothetical example of why it is a false statement. Nothing more.
I'm not sure why you are taking my post and example out of context to discuss a different subject altogether (how a stop affects the overall profitability of the system, not a single trade)
After our last exchange on this very subject Frosty replied that he backtested with different stops and found them to reduce the overall system profitability. If you don't understand how that can be or if it is true then please take it up with him. I've already done enough of my own ATS system development/backtesting to understand why that is not at all unusual.
I was nto trying to take something out of context so apologies if that is how it seemed. I still cannot imagine that in general allowing a trade to take so much heat for the chance of a small profit will lead to positive results over time and Frosty's REAL results so far make me still believe that to be true. Backtests may show one thing but real money is the ultimate proof.
Frosty, just take some time to look at each trade and determine if there are patterns of loss or gain you can refine. Sometimes all a system needs is a minor tweak.
I just ran a backtest. The data set was 5-minute ES bars from 2002 to 2006. I varied the length of both short term and long term EMAs from 5 to 200 bars, with an increment of 5. That gave me 1600 strategies to test. The slippage was assumed to be 0.125 ES points and the commissions were assumed to be $2.40 per side. One contract was traded through the test period.
Results: All 1600 tested strategies ended up losing money, with the best performing strategy losing $43,000.
Here are a few typical days (green circles indicate long entries, red circles indicate short entries):
umm. maybe you didn't understand me. if you read my post again you'll see i asked if frost meant he found stops created bigger daily drawdowns or bigger specific trade drawdowns, nothing else. theoretically neither is the case, practically you can see something different.
as for my statement, it is absolutely correct and i believe you are the one who is mistaken.
proof1:
take specific trade T. the prices after entering T (ignoring why we entered) have a range from p1 <= price <= p2. so giving any system S, running on trade T, it can exit at any price between p1 and p2, let minprofit = min(S(T, p1),S(T,p2)), and maxprofit = max(S(T,p1),S(T,p2)), where any other S(T, price) gives a profit between minprofit and maxprofit, since simple long or short positions give monotonic P/L. so minprofit <= S(T, exitprice) <= maxprofit
now lets add a stop at stopprice, we have two cases,
case 1: stopprice < p1 OR stopprice > p2 (stopprice outside of range), the stop does not get hit => stop does not effect the maximum draw down.
case 2: p1 <= stopprice <= p2 (stopprice inside of range), so let stopprofit = S(T,stopprice), given p1 <= stopprice <= p2, we know minprofit <= stopprofit <= maxprofit (remember S(T,price) is monotonic). we also know that max drawdown of T is defined as the minimum possible profit of a given system running on trade T, so max drawdown of T = minprofit. since minprofit <= stopprofit.
=> stop does not decrease (make worse) the maximum drawdown.
/\ adding a stop will not decrease (make worse) the maximum drawdown on any trade T, given system S, with prices ranging between p1 and p2.
so thats for a specific trade. we can expand that proof to apply to daily drawdown. ill write a corollary to proof1 to prove that adding a stop wont increase the max daily drawdown if you dont believe me.
(hint: take a set of trades, T1, T2 ... Tn in a day, each has a max drawdownN=minprofitN, apply above proof to each to show stops dont effect the minprofits or sum of minprofit1+minprofit2+.. etc).
unless my logic is flawed. there you go
Not to discourage you but we tried similar systems a while back and while they were profitable for a little bit, they almost all required too much curve fitting and started losing money a few months after implementation. We switched to a different system that doesn't use emas and have had fantastic results. Be careful about over fitting with ema's - there are often too many numbers to manipulate.
I didn't have much time to go through your "proof" to find the flaw, but I believe GTS already refuted your theorem before you even came up with it. Here it is again:
Price action: 10, 11, 9, 10, 12 (close of the day)
Strategy 1 (without stop): enters long at 10, exits at 12. Result: gain of 2 points.
Strategy 1 (with a 1 point stop): enters long at 10, exits at 9. Result: loss of 1 point.
i fail to see how that has anything to do with max possible drawdown of a trade in that price action. if P/L gets to a level to be stopped out, it has the 'possibility' of not getting any better, causing the system to have the same or worse possible drawdown without the stop. ie. worse possible drawdown than the stop. si?
i never said in 'every case' a stop will result in more profit, i doubt anyone here is stupid enough to say something like that... so giving a counter example like that is meaningless since it does'nt apply to the max possible drawdown, just the drawdown in one case (otherwise known as trade P/L). im talking about the general idea of a 'max drawdown' and the effects of stops on it. maybe try understanding what someone is saying before you try to 'refute' it. thanks
Theorem still stands.
Jennings' ATS Stop Theorem: Given any system, adding a stoploss to it will not increase the maximum possible drawdown.