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Trailing Stop Expectancy - A Mathematical Problem

  • #21
  • Jul 26, 2010 9:56am Jul 26, 2010 9:56am
Quoting kk007
...with hundreds of setups with real data are not feasible...
kk007,
got that. I'm suggesting that you identify some trades from historical data, and list these in a spreadsheet or similar. Then you can automate just the trailing stop strategy to find a range of parameters that work OK.

You will probably already have identified these anyway.

Statistically, you need 30 trades or more to have a valid test (though I don't know exactly why this is the magic number). You don't need hundreds, but even if you are doing it manually, you will still need at least 30.

Some entries are impossible to code, but trailing stops are childs play.

I am willing to help if you want to go this way, as I have done something similar with TradeStation.

Good luck either way.
  • #22
  • Jul 26, 2010 11:00am Jul 26, 2010 11:00am
Quoting hagenson
kk007,
got that. I'm suggesting that you identify some trades from historical data, and list these in a spreadsheet or similar. Then you can automate just the trailing stop strategy to find a range of parameters that work OK.

You will probably already have identified these anyway.

Statistically, you need 30 trades or more to have a valid test (though I don't know exactly why this is the magic number). You don't need hundreds, but even if you are doing it manually, you will still need at least 30.

Some entries are impossible to code, but trailing...
Thank hagenson,

Sample size is not a simple issue. 30 may be enough for estimate a one-dimensional problem. (i.e. only one variable change other variables keep constant). In real data, a few variables in active, (in this case drift rate, variable rate, max profit, etc) 30 data may not enough.

Nevertheless, many thanks for your opinions and willingness of help.

cheers,

k.k.
  • #23
  • Jul 26, 2010 11:22am Jul 26, 2010 11:22am
Quoting kk007
Thank hagenson,

Sample size is not a simple issue. 30 may be enough for estimate a one-dimensional problem. (i.e. only one variable change other variables keep constant). In real data, a few variables in active, (in this case drift rate, variable rate, max profit, etc) 30 data may not enough.

Nevertheless, many thanks for your opinions and willingness of help.

cheers,

k.k.
Why dont you work out the average move higher before it returns back to your entry? Then take the mean move higher. Then if you're really clever take into account volatility and the Average daily trading range over the past 2 weeks.

If the price has already moved 70% of the daily range in one direction from the previous days close. The trade only has 30% more potential.

I wouldn't make it too complicated though.
  • #24
  • Jul 26, 2010 11:32am Jul 26, 2010 11:32am
Quoting Brettus
Why dont you work out the average move higher before it returns back to your entry? Then take the mean move higher. Then if you're really clever take into account volatility and the Average daily trading range over the past 2 weeks.

If the price has already moved 70% of the daily range in one direction from the previous days close. The trade only has 30% more potential.

I wouldn't make it too complicated though.
Are you suggesting that the mean of maximum profit among the winning trades is the expected profit? But, this is obviously wrong.

Or, are you suggesting that the area of the "hill" in the figure divided by the width of the "hill"? If so, what is the logic?

I want simple answer too, but I need a right answer.
  • #25
  • Jul 26, 2010 11:36am Jul 26, 2010 11:36am
No. Not at all. For obvious reasons. I'm suggesting that any trades that reach the mean are profitable. Any that dont, are losing trades or breakeven.

If mean is 5R. 10 trades hit 5R, 5 trades 0 (breakeven) and 5 trades -2R. then expected return is 2R.
  • #26
  • Jul 26, 2010 11:39am Jul 26, 2010 11:39am
Quoting Brettus
Why dont you work out the average move higher before it returns back to your entry? Then take the mean move higher. Then if you're really clever take into account volatility and the Average daily trading range over the past 2 weeks.

If the price has already moved 70% of the daily range in one direction from the previous days close. The trade only has 30% more potential.

I wouldn't make it too complicated though.
What does "the average move higher" mean in your post???
  • #27
  • Jul 26, 2010 11:42am Jul 26, 2010 11:42am
Quoting Brettus
No. Not at all. For obvious reasons. I'm suggesting that any trades that reach the mean are profitable. Any that dont, are losing trades or breakeven.

If mean is 5R. 10 trades hit 5R, 5 trades 0 (breakeven) and 5 trades -2R. then expected return is 2R.
Are you talking about trailing stop at all? Does what you have said has anything to do with trailling?
  • #28
  • Edited 9:58pm Jul 26, 2010 9:17pm | Edited 9:58pm
Hi Noloqy,

This post is a follow up of your post and is not about trailing.
I have made a simple calculation for a no-edge/random price movement analysis with an example.

In a no-edge system, with a fixed SL of -1R, for each 10 losing trades, the corresponding max possible is roughly something like this:

One +10R trade
One + 5R trade
One + 3R trade
Two + 2R trade
Five + 1R trade

Here, no matter a fixed TP is set to +1R, +2R, +3R, +5R, or +10R, the overall profit is zero (after adding the losses).

If I add up the above sequence, I will get a max possible profit +27R (1x10+1x5+1x3+2x2+5x1). However, what can be expected to get out of this +27R max is only +10R to compensate for the lossing trades.

kk

P.S. This +27R may have fooled a lot of people of mistaking a no-edge trading method as a highly profitable trading method, and blaming their own poor skill in exiting.


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...
  • #29
  • Jul 26, 2010 9:41pm Jul 26, 2010 9:41pm
OK. I come to a conclusion for the problem. The expected profit of using trailing stop depend not only on the max possible profit.

The max possible profit only tell half of the story, the other half rests on the volitality of the price movement. If the volitatility tend to zero the profit expectation in the example is +9R. If the volitatility is extremely high compared to the trailing stop allowance, the profit expection is +0R.

In other words, the expected profit is a function of the volatility. As far as the volatility cannot be estimated, the expected profit cannot be estimated.

Thanks for all previous response, although there is no direct answer, but it did stimulate my mind.
  • #30
  • Jul 26, 2010 10:03pm Jul 26, 2010 10:03pm
To be honest, I can't see a valid way of determining an expected value unless you use empirical data. It would be nice to know what your entries are based on? Are they random? Expected values depend on the market conditions within a period of time and because pairs can move 'anywhere', we cannot predict accurate future expected values unless there is positive correlation from historical back-testing.

I have not dealt with forex in relation to the Wiener Process, and don't know how you would go about implementing this into a program that can accurately portray potential price movements, but I love the idea!

Something to ponder, somewhat unrelated, but the idea behind it is fundamentally similar to the process you are trying to implement in regards to expectance and market conditions...

Take a pair (CHF/JPY for example).
Add RSI (14).
Consider price sell/buy entries when overbought/oversold.
You'll notice that you could make +300 pips if you grab the max profit on one trade, but -50 if you grab the max on some other, but you cannot say that they each have equal probabilities of reaching an expected value (in it's simplest form, the mean pip yield of exit points) because you don't know the "forces behind the trend" and cannot predict such enough to expect a certain value to appear. But what can you predict without adding more variables? Minimums and Maximums over historical data. Gather the range between entry and exit points, which could be based on time (seems to be your focus with the Wiener Process) or like in my example, the time it takes from and overbought to an oversold position. If indeed your entries are random, and assuming that countries dollar values are random, the law of large numbers states that you should expect as time->infinity and # of trades->infinity every pair to be at parity. However, we know that this scenario doesn't exist. There are factors that move currencies. Know these factors (or apply the appropriate isomorphism), use them to predict ranges (max/min), and you'll have an advantage when a certain condition is met.

Since I don't know exactly what you are trying to achieve (I have an idea), I cannot give an further insight, but since qualitative methods are difficult to produce, start with one variable, and move from there.
  • #31
  • Jul 27, 2010 9:13pm Jul 27, 2010 9:13pm
Thanks. I am not looking for empirical answers, because of the infeasibility in collect the probable huge amount of setups as explained. I am looking for theorical answers based on simple assumptions for reference.

Quoting sqrt(-1)
To be honest, I can't see a valid way of determining an expected value unless you use empirical data. It would be nice to know what your entries are based on? Are they random? Expected values depend on the market conditions within a period of time and because pairs can move 'anywhere', we cannot predict accurate future expected values unless there is positive correlation from historical back-testing.

I have not dealt with forex in relation to the Wiener Process, and don't know how you would go about implementing this into a program that can accurately...
  • #32
  • Jul 28, 2010 3:16am Jul 28, 2010 3:16am
Quoting kk007
In other words, the expected profit is a function of the volatility. As far as the volatility cannot be estimated, the expected profit cannot be estimated
The volatility can and is estimated every day by option traders.

For example like this: http://papers.ssrn.com/sol3/papers.c...ract_id=306002
  • #33
  • Jul 28, 2010 7:13am Jul 28, 2010 7:13am
Quoting Adal
The volatility can and is estimated every day by option traders.

For example like this: http://papers.ssrn.com/sol3/papers.c...ract_id=306002
What can be estimated is the historical volatility before entering, but what affects trailling is the volatility after entering.
  • #34
  • Jul 28, 2010 11:04am Jul 28, 2010 11:04am
Quoting kk007
What can be estimated is the historical volatility before entering, but what affects trailling is the volatility after entering.
I have a feeling that if you can work this out, you are effectively proving chaos theory.
  • #35
  • Jul 28, 2010 9:04pm Jul 28, 2010 9:04pm
Quoting Brettus
I have a feeling that if you can work this out, you are effectively proving chaos theory.
Joking. At the time of entering, the historical volatility is easy to calculate (This is only simple maths.), but you cannot know what the volatility will change to after the entry.
  • #36
  • Aug 13, 2010 9:06pm Aug 13, 2010 9:06pm
Inspired by ForexQuant in another thread, I have something to share:

In a random walk scenario, ignoring spread cost and TP above +10R, setting stop loss to -1R, the maximum no. of R that can be got with hindsight on average is about +2R. That is the double the expected value without hindsight, +1R. See attached.

With further calculation, if only TP above +100R is ignored, the maximum no of R that can be got with hindsight on average is about +4R. And if TP above +5R is ignored, the maximum no of R that can be got with hindsight on average reduces to +1.45R.
Attached Image
  • #37
  • Aug 13, 2010 11:03pm Aug 13, 2010 11:03pm
Seems something wrong in the logic of the calculation. Let me rethink.

Quoting kk007
Inspired by ForexQuant in another thread, I have something to share:

In a random walk scenario, ignoring spread cost and TP above +10R, setting stop loss to -1R, the maximum no. of R that can be got with hindsight on average is about +2R. That is the double the expected value without hindsight, +1R. See attached.

With further calculation, if only TP above +100R is ignored, the maximum no of R that can be got with hindsight on average is about +4R. And if TP above +5R is ignored, the maximum no of R that can be got with hindsight on average reduces...
  • #38
  • Aug 13, 2010 11:14pm Aug 13, 2010 11:14pm
Anyone has a better idea about how to compare the max possible in hindsight and the expected?

Quoting kk007
Seems something wrong in the logic of the calculation. Let me rethink.
  • #39
  • Aug 14, 2010 12:47am Aug 14, 2010 12:47am
Just did a simulation the Max Vs. Expected ratio is 4 times. (rather than 2 times as said above) for up to +10 R.

Here a deep illustration of hindsight chart reading!
Attached File(s)
File Type: xls MaxVsExp2.xls   64 KB | 393 downloads
  • #40
  • Aug 14, 2010 3:08am Aug 14, 2010 3:08am
Oh! I know what! The expected value should be +0.5R, and so +2R is four times the expected value, and thus close to the simulation result!

Quoting kk007
Inspired by ForexQuant in another thread, I have something to share:

In a random walk scenario, ignoring spread cost and TP above +10R, setting stop loss to -1R, the maximum no. of R that can be got with hindsight on average is about +2R. That is the double the expected value without hindsight, +1R. See attached.

With further calculation, if only TP above +100R is ignored, the maximum no of R that can be got with hindsight on average is about +4R. And if TP above +5R is ignored, the maximum no of R that can be got with hindsight on average reduces...
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