Trailing Stop Expectancy - A Mathematical Problem | Forex Factory

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Trailing Stop Expectancy - A Mathematical Problem

  • #1
  • First Post: Jul 24, 2010 10:14pm Jul 24, 2010 10:14pm
Dear All,

I have repeatedly encountered a problem when I backtesting an trade-entry method. In the method, I have a fixed stop loss of -1R. I see the maximum possible profit on the chart for each setup; sometimes, it is +1R, +3R, +20R and so on. The problem is that these positive value is the maximum possible values, what I really want is the expected value if I use trailling stopping.

I have attached a figure to illustrate the problem. In the figure, you can see once the buy (long) position is entered, the position went up with retracements, and achieve a maximum profit of +10R. As you can see, the best possible profit of this position is +10R, but this is not guaranteed. Normally, less than +10R is earned in this case. Should I say the expected profit is +5R in this case???

My backtesting is manual, and I don't want to go down to each case to check. Is there a mathematically approximated solution for this?

cheers,

kk
Attached Image
  • #2
  • Jul 25, 2010 3:47am Jul 25, 2010 3:47am
In a random market the expected reward would be 1R.

If you believe the market isn't random, and have experience on how the market behaves after you receive an entry signal, I'm thinking you could use statistics to estimate the expected return. This would require estimating a mean return and the standard deviation. The best you get with this method is an estimate, as it assumed a lognormal distribution, which is obviously not observed in the real market.
The nail that sticks out gets hammered back in
  • #3
  • Jul 25, 2010 5:39am Jul 25, 2010 5:39am
I don't see the point in back testing a strategy with only an entry. Why not work out exit conditions and apply them?

Alternatively, just take note of the maximum return for all trades, and then you can see where a fixed profit target (3R,5R etc) would of got you, and you can see which would of been most profitable.
  • #4
  • Jul 25, 2010 8:06am Jul 25, 2010 8:06am
Thanks for your replies.

I have a preliminary exiting method - a fixed take profit of +1R or +2R, but this cannot answer my question about how much I should expect. If I take profit at +1R and the position run to +10R, it seems a bad exiting method.

I don't want to use experimental method to do a lot of backtesting, and I don't even know whether I need thousands of setups to draw a meaningful estimation. If there exist a quick logical and mathematical rough answer to my question as stated, it will be ideal.
  • #5
  • Jul 25, 2010 8:34am Jul 25, 2010 8:34am
Trailing a stop in conjunction with a scaling-out exit strategy would be better.
i.e. enter with fixed SL. Exit 1/2 position at say 1R. Only start trailing when 2R is reached on the other half.

I've found that trailing does hurt the performance of various systems I've backtested with no scaling out (i.e. all in, all out). Fixed SL/TP seems to have better expectancy.

In manual trading it's tempting to tighten the SL as the trade progresses or move the SL to BE too soon. Unless you're moving the SL according to a tested strategy, you're just giving in to fear by moving the SL. Even though it seems like you're avoiding losing money by tightening, you're actually losing money long-term by cutting profits short prematurely.
You are in a maze of twisty little passages, all alike.
  • #6
  • Jul 25, 2010 9:36am Jul 25, 2010 9:36am
Quoting stevegee58
I've found that trailing does hurt the performance of various systems I've backtested with no scaling out (i.e. all in, all out). Fixed SL/TP seems to have better expectancy.
I agree. I have a similar experience too.
  • #7
  • Jul 25, 2010 9:46am Jul 25, 2010 9:46am
If it is assumed that stopping at 0R, 1R, 2R,... 10R has equal probability in the example, the expected profit will be 5R (i.e. taking the average). But is this a realistic assumption? If so, that means with trailling catching half of the maximum possible profit that I can see on the chart is the right expectation. Any idea?
  • #8
  • Jul 25, 2010 10:41am Jul 25, 2010 10:41am
Quoting kk007
If there exist a quick logical and mathematical rough answer to my question as stated, it will be ideal.

42
  • #9
  • Jul 25, 2010 10:55am Jul 25, 2010 10:55am
Quoting Dutch_trader
42

Always love this answer.
  • #10
  • Jul 25, 2010 8:57pm Jul 25, 2010 8:57pm
Quoting flyer415
Always love this answer.
Any better answers?
  • #11
  • Jul 25, 2010 10:16pm Jul 25, 2010 10:16pm
Quoting kk007
Any better answers?
Assuming price movement does follow Brownian Motion then I strongly believe that this problem can be answered by pascal triangle because you can easily find out how many and possible steps needed to reach your retracement point. However price movement does not follow Brownian Motion but Fractional Brownian Motion therefore the solution is complicated, IMHO.
  • #12
  • Jul 25, 2010 11:58pm Jul 25, 2010 11:58pm
Quoting stevegee58
I've found that trailing does hurt the performance of various systems I've backtested with no scaling out (i.e. all in, all out). Fixed SL/TP seems to have better expectancy.
Quoting ForexQuant
I agree. I have a similar experience too.
I think it is a misunderstanding about trailing stop. If you see the maximum possible take profit is +10R, like that in the example, you cannot expect to capture all +10R. The expected profit is much lower than this maximum, no matter which stopping method you use, and so trailing stop is not the one to be blamed, and there is no obvious reason to say that fixed SL/TP can capture more.
  • #13
  • Jul 26, 2010 12:31am Jul 26, 2010 12:31am
Quoting kk007
I think it is a misunderstanding about trailing stop. If you see the maximum possible take profit is +10R, like that in the example, you cannot expect to capture all +10R. The expected profit is much lower than this maximum, no matter which stopping method you use, and so trailing stop is not the one to be blamed, and there is no obvious reason to say that fixed SL/TP can capture more.
So is that means you have an obvious reason to say that fixed SL/TP capture less or trailing SL/TP capture more?
  • #14
  • Jul 26, 2010 12:36am Jul 26, 2010 12:36am
Quoting ForexQuant
So is that means you have an obvious reason to say that fixed SL/TP capture less or trailing SL/TP capture more?
No. I didn't say that.
  • #15
  • Jul 26, 2010 3:51am Jul 26, 2010 3:51am
Quoting kk007
If it is assumed that stopping at 0R, 1R, 2R,... 10R has equal probability in the example, the expected profit will be 5R (i.e. taking the average). But is this a realistic assumption? If so, that means with trailling catching half of the maximum possible profit that I can see on the chart is the right expectation. Any idea?
You should give up on this if you aren't willing to crunch the numbers, in order to quantify your edge.

The assumption you make is not correct, and the expected return remains 1R no matter how high the ultimate TP is (10, 20 or any other number).
The nail that sticks out gets hammered back in
  • #16
  • Jul 26, 2010 5:50am Jul 26, 2010 5:50am
Quoting Noloqy
You should give up on this if you aren't willing to crunch the numbers, in order to quantify your edge.
Let me wait a bit longer. Perhaps, some wise person has a good suggestion.

Quoting Noloqy
The assumption you make is not correct, and the expected return remains 1R no matter how high the ultimate TP is (10, 20 or any other number).
Yes the assumption is most likely wrong. Just wanted to see if someone will give better assumption.

You are so sure it is +1R; do you have any supporting reason? I thought it is what is to be expected if a fixed +1R TP is used.
  • #17
  • Jul 26, 2010 6:38am Jul 26, 2010 6:38am
Ok I'll give you a numerical example, and then a theoretical explanation.
Lets assume you set your ultimate TP at 4R, and trail the stop once you reach a multiple of R. (Note that this assumption is made for the sake of simplicity, as the same theory applies for a continuous trailing stop)
The payoff scheme is as follows
___0
-1___1
___0___2
______1___3
_________2___4
You see there is a chance to move to the next integer, or get stopped out at the previous integer, where you set your stop. At 4R you take the full profit.

Now we introduce the random market probabilities.
___1
0.5___0.5
___0.25___0.25
______0.125___0.125
_________0.0625___0.0625

Now we multiply the probabilities with the returns, to get the expected payoff.
___1
-0.5___0.5
___0_____0.5
______0.125___0.375
_________0.125___0.25

If you add up all the bold numbers in the expected payoff tree, you will always get 0, no matter where you set your ultimate TP. If you set the probability of the initial loss equal to the value of 1R, the sum of all positive returns also equals 1R.

If you assume a random market, then there will never be a way to get a positive expected payoff. This means that under these conditions, also the exit strategy is not capable of capturing an edge. A comforting thought is that you cannot do anything wrong either.

I believe in a great extent of randomness in the market, and am therefore quite critical to all the self-proclaimed market psychologists that claim that you will make money if you have enough discipline.
The nail that sticks out gets hammered back in
  • #18
  • Jul 26, 2010 6:54am Jul 26, 2010 6:54am
Thanks for your explanation. I think I get what you mean. Here you assume price is random, and no edge is possible. While your elaborations are right on its own, it is not an answer for the thread topic. The thread topic is that for a fixed (but unknown beforehand) maximum profit, how much should I expect to get.

Quoting Noloqy
Ok I'll give you a numerical example, and then a theoretical explanation.
Lets assume you set your ultimate TP at 4R, and trail the stop once you reach a multiple of R. (Note that this assumption is made for the sake of simplicity, as the same theory applies for a continuous trailing stop)
The payoff scheme is as follows
___0...
  • #19
  • Jul 26, 2010 7:28am Jul 26, 2010 7:28am
You don't state exactly what you mean by trailing stops. There are different strategies to trail stops. But lets assume you mean when price goes up x pips, trail to y pips behind.

Try this for a model:
Each time you move the stop, work out the probablity the price will move up another x pips before it retraces y pips.

By multiplying each of these probabilities, you will get an ever decreasing probability as the price appproaches the ultimate target.

This way the probability of hitting even R1 will be lower than just a single stop and target. Lets say you have a really good system, with an 80% chance the price will move up x before it retraces to y, and that this is constant. Your probability will decrease like this:
0.8, 0.64, 0.51, 0.41, 0.33, 0.26, 0.21, 0.17...

Then take the profit levels of each of the stops (stop price-entry price), and multiply by the probability of that level, add that all up and you've got the expected profit. There is a mathematical shortcut for this, but I can't remember it. You could do it in a spreadsheet by repeating formula cells anyway.

So far so good, now to fill in some of the hard bits.
Markets don't move in linear price cycles, so even if you can figure out the probability of moving up x before you move back y it will not be constant. You could use something like the Black-Scholes model used for pricing options to come up with some kind of estimate. I don't know enough about that to know if that would work. I understand it is based on historical volatility and use a normal distribution to work out the probability an instrument will reach a given price.

Volatility is function of price change over time, so you will then have to estimate the time it takes to reach each stop level, and feed that into your model. But wait, markets don't move in linear time cycles either, so you are probably going to have to reference the work of W.D. Gann of the correlation between markets and astrology to figure out how to model that.

You can go on and on, but in the end it will prove nothing.

Unless you define your trailing stop strategy and work through it trade to see how it performs, and then do that again on an out sample of untested data, you will have no idea how your system will work in the real world.

It would be quicker and more meaningful to automate your strategy and back test it that way. If you can't automate the entries for whatever reason, you could put the entry dates and times in a little database and just test the trailing stop strategy on pre-identified trades. This would make it much easier to test an out sample too.
  • #20
  • Jul 26, 2010 8:11am Jul 26, 2010 8:11am
Hi hagenson,

Thanks for your reply. However, as I said in my first post

Quoting kk007
My backtesting is manual, and I don't want to go down to each case to check. Is there a mathematically approximated solution for this?
My setup is rare, need to look at a lot of bars in order to identify one, and criteria is a bit qualitative, and thus difficult to quantitative for programming, and backtesting with hundreds of setups with real data are not feasible.

What I may do now could be doing simulation. I am thinking to model price movement using Generalized Wiener Process, with the imcomplete assumption build-in the Process. I will set the x and y. and simulate using different variance rate and drift rate of the process. and different max possible profit.

So there are 5 variables:

x, y, variable rate, drift rate, max profit

Perhaps, the result can tell something?



Quoting hagenson
You don't state exactly what you mean by trailing stops. There are different strategies to trail stops. But lets assume you mean when price goes up x pips, trail to y pips behind.

Try this for a model:
Each time you move the stop, work out the probablity the price will move up another x pips before it retraces y pips.

By multiplying each of these probabilities, you will get an ever decreasing probability as the price appproaches the ultimate target.

This way the probability of hitting even R1 will be lower than just a single stop and target....
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