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Optimized Risk vs Reward Equation

  • #81
  • Nov 11, 2009 3:49am Nov 11, 2009 3:49am
No, i'm not familiar with the 4-minute mile and its history. I just did a quick search on it and it's being able to run a mile in under 4 minutes (as the name implies). It was first done by Roger Bannister in 1954. It has been done many times since, but it's like a standard benchmark these days.

I'm not sure how that answers my question. Maybe I missed some important part of the history of the 4-minute mile.
  • #82
  • Nov 11, 2009 5:19am Nov 11, 2009 5:19am
It took him 2 years to figure out the equation. I hope I will be able to replicate the result in less amount of time. As more people try to figure it out, the less time it will take them to achieve that.

How hard can it be? We know a lot of properties a function we are looking for must possess. This will narrow the search significantly.

Summa summarum, he won't tell you, at least not directly, but he might give you another clue.
The world is not based on reality, but on the perception of reality.
  • #83
  • Nov 11, 2009 6:08am Nov 11, 2009 6:08am
Quoting twoblink
mathematician.. are you familiar with the 4 minute mile and its history? if so, this will have answered your question..
The four-minute mile was considered unbreakable for a very long time. After Bannister broke it, a handful of men also broke it within months. The implication being that once others saw something was possible, they were able to do it as well. It speaks to the self-limitations we impose on ourselves.
  • #84
  • Edited 12:38pm Nov 11, 2009 7:12am | Edited 12:38pm
Quoting bigvlada
Summa summarum, he won't tell you, at least not directly, but he might give you another clue.
And IMHO he shouldn't [tell directly]. It's always better to tell people to think and give some clues than spoonfeed the answers. It also restricts the reward for those people who are really willing to work for it.

Myself... I wait until someone else has replicated this. I still don't think this is possible in a wider context - I assume the working result will be an abstraction of some special cases of market movements (which is still a great result if it stands the test of time).

(In other words: This is why in science there is peer review and I'm too lazy to do it myself )
  • #85
  • Nov 11, 2009 11:29am Nov 11, 2009 11:29am
I have to disagree here, there is a difference between dropping hints and posting a whole lot of obfuscated garbage. I consider myself fairly mathematically literate and after trying to read this thread a couple of times I still have no idea what is going on. The OP seems to be asserting that there is some magic distance at which he can place a second trade and maximize profit, how he can do this without referring to the characteristics of the underlying time series is beyond me, it makes no intuitive sense that such a relationship should exist. But then I can't rule out that all this second year university math and hand waving is simply beyond me.
The breaking of a wave cannot explain the whole sea.
  • #86
  • Nov 11, 2009 11:34am Nov 11, 2009 11:34am
Quoting Craig
The OP seems to be asserting that there is some magic distance at which he can place a second trade and maximize profit, how he can do this without referring to the characteristics of the underlying time series is beyond me, it makes no intuitive sense that such a relationship should exist.
Neither to me, that's why I'm waiting someone else to replicate this to at least some extent so I can dissect what's really going on.

I do assume that the result goes back to decision trees and some abstracted cases of market movements. That would make sense.
  • #87
  • Nov 11, 2009 11:41am Nov 11, 2009 11:41am
Hey mikkom, if you're interested their is quite an interesting thread on ET at the moment, search for 'can linear regression analysis...' the posts of 'MAESTRO' are particularly interesting, that is what I call hint dropping.
The breaking of a wave cannot explain the whole sea.
  • #88
  • Nov 11, 2009 11:46am Nov 11, 2009 11:46am
Quoting Craig
Hey mikkom, if you're interested their is quite an interesting thread on ET at the moment, search for 'can linear regression analysis...' the posts of 'MAESTRO' are particularly interesting, that is what I call hint dropping.
I did actually read a big part of that thread - have to read rest of it too.

ET is a strange place. There are some gems here and there and then there is so much annoying garbage that it's almost impossible to read anything there.
  • #89
  • Nov 11, 2009 12:02pm Nov 11, 2009 12:02pm
Quoting mikkom
I did actually read a big part of that thread - have to read rest of it too.

ET is a strange place. There are some gems here and there and then there is so much annoying garbage that it's almost impossible to read anything there.
Yep, judicious use of the ignore list is a must, there are some gems however.
The breaking of a wave cannot explain the whole sea.
  • #90
  • Nov 12, 2009 1:05am Nov 12, 2009 1:05am
I'm gonna give this a try, not sure if it's correct or not...

If the problem is described by http://www.forexfactory.com/showpost...1&postcount=18, the distances between A, B and C are determined by phi. So I will use phi to determine the position size as well. Is this optimal? Not sure...

Let X0 be the first positionsize @ A. The first add on @ B will be called Y1, and Y1 = X0/phi. Ephemeral lot size @ C would be: X0 + Y1 = X0 + X0/phi = X0*phi, and is called Z1.
Now treat C as A, so we have positionsize X0*phi @ A. The second add on position, Y2, @ B (the new B that is) is (X0*phi)/phi = X0. Ephemeral lotsize @ C (new C) is X0*phi + X0 = X0*phi^2, called Z2.

This will give these equations for the positionsizes:
X0 @ A = is given by the first entry which is based on your own trendfollowing system
Yn @ Bn = X0*phi^(n-2), given for n = 1, 2, 3, ..., where Bn is all add on positions
Zn @ Cn = X0*phi^n, given for n = 1, 2, 3 ..., where Cn is all ephemeral positions
n = how many add on positions we have

The profit is always:
Profit = Zn*Bn_SL/phi, where n equals how many add on positions we have
Bn_SL is the size of the SL given to the latest add on position
The profit will always be known due to this condition. This also means that the bigger SL I have the bigger the profit for those add on positions. But the lesser my SL is the more positions I can add on, which also will increase the profits. There's a tradeoff here, and I'm not sure yet how to calculate it.


Given A and lotsize X0, you have to know either:

 

  1. distance to the next add on position B, or
  2. SL for next add on position B

Say I have found an add on entry at 40 pips away from current ephemeral position (or A), then I know that my SL size needs to be @ 40/(phi^2) or about 15 pips.
Or I have found an add on entry with SL size of 15 pips. Then this add on needs to be 15*phi^2 = 40 pips away from current ephemeral postion.

I find this add on technique that is based on math only to be not to my liking. I would need the add on entry and the SL size to be based on my entry rules. So the question is: what happens if the SL size of the entry position is different from twoblink's graph? Or is the whole math that I've presented here all wrong? Easy to say that this is confusing...

But in this post http://www.forexfactory.com/showpost...6&postcount=25 twoblink says this among other things:
There are 4 variables and I can determine the other 3 anytime one is given to me. I can even calculate the phantom 5th variable.

Var 1) Lot size X
Var 2) A distance to B
Var 3) Lot size Y
Var 4) B distance to SL
Phantom Var 5) SL point of A

If I treat A as a B, since I know the SL for B, and because I'm basically taking an integral of B to get A, and if I treat A as B, I will be able to get SL for A by going backwards, I actually get 5 variables total from 1 equation; not only that, fully optimized in risk reward.

So my equations thus far can only give me the positionsize based on the add on position's distance to ephemeral position or the add on position's stoploss size. And this means that there is an equation for determining the lotsizes based on the entries stoploss size and distance to ephemeral position, which is what I'm looking for.

Didn't make the mile in 4 minutes the first try...

  • #91
  • Nov 12, 2009 10:40pm Nov 12, 2009 10:40pm
With the reference to the 4-minute mile, Dopey is correct. Once you know someone can do it, the "magic" is gone, and everybody starts doing it.

Craig.. I post my research in hopes that it will greatly cut down the time for other people who are pursuing the same thing. I have theories, be they true or not, then I test them. if they work, great, if not, oh well.. I hope others can learn from my mistakes. I also posted my theory in a diagram, and said that it didn't work, the equations turned out to be divergent and never converged on an optimized point. So i've had tests and plenty of failures. oh well.. that's life.

As for how I can optimize without referencing to the underlining timeframe; I can give you two answers:

1) i am optimizing against pips, and so it really doesn't matter if there was a 100 pip movement in the eurusd, and it took 1 hour or 1 day. Since my optimizations are distance references only, the time is not referenced and need not be referenced.

If I said, when you are +80 pips, add 1 more lot.. does what timeframe I use really matter?? so perhaps i am confused about your question of why timeframe is needed for profit and risk optimizations, both of which have not T component.

2) Rubik's cube. In 2008, Tomas Rokicki said he can solve any rubik's cube (well, his program) in 22 moves. How is it possible that he can make such claims without even looking at how messed up the cube itself is? Because it's a mathematical problem, and how messed up the cube is in its initial state is not really part of the equation, and need not be.

If my posts help, then i'm happy that i can help. If they hinder, then you should stop reading them... and spend more time in thinking.

Quote
Reading, after a certain age, diverts the mind too much from its creative pursuits. Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
(Albert Einstein)
google:
  • #92
  • Edited 11:57pm Nov 12, 2009 11:45pm | Edited 11:57pm
Quoting twoblink
I also posted my theory in a diagram, and said that it didn't work, the equations turned out to be divergent and never converged on an optimized point. So i've had tests and plenty of failures. oh well.. that's life.
So the diagram that you posted isn't the one used for your equation... That would mean that everything discussed until now is wrong, and C ins't based on phi?

A few questions if you care to answer:

My original theory was, that there exists a point in space, in which a risk graph with respect to position, and a trans-axel'ed profit graph with respect to position, would meet at a point on the X/Y, Y/Z axis, thus creating a "mini-max" AKA saddlepoint in Calculus terms.
- What is meant by the word position here? And isn't the risk and profit equations 1-dimensional, I mean what parametres would give an n-dimensional risk and profit equation, which will give the saddlepoints? Could you perhaps give some more hints on these equations.

Given original entry @ A with lotsize X, you want to find an optimized add on at B with lotsize Y, which can give you the ephemeral position C given by C = (AX + BY)/(X + Y). Thus the two unknown parametres are B and Y and these are determined based on pure math, or can you have an add on position at a certain distance with a certain SL and then calculate the optimal lotsize for that add on? Basically is the add on pure mathematical or is it situation-based?

As above, if add on position has the stoploss @ B_SL then the profit would be given by: Profit = (position @ C) * (distance between C and B_SL). Is it true that the profit is always given by this equation, and is it this equation (or some other profit equation) that should be optimized?


I'm really not looking for a free lunch, ignore my questions if you think otherwise. I would just like to understand more...
  • #93
  • Nov 13, 2009 9:49am Nov 13, 2009 9:49am
If i buy 2.4 lots @ 91.58 in jpy, then I have a "long position" in jpy. so that's what i mean by position.

Now, as an example, I can buy 1.5 more lots at 92.14 or 1.8 lots at 92.45 (just making up numbers) what should my decision be??? well, since lot size alters my risk, and price point in which I add to my position alters my risk, then i have 2 factors, lot size, and distance from X to Y.. yes? So given the lot size to be B and the distance to be (Y-X) then mathematically, I need to find an optimized B/Y pairing. Yes?

We have 2 variables with rates of change. Classic calculus problem, take a first derivative and optimize.

Quote
or can you have an add on position at a certain distance with a certain SL and then calculate the optimal lotsize for that add on? Basically is the add on pure mathematical or is it situation-based?

The equation tells you how much you should add when you add given an initial size A@X, and yes, since you know what the optimum situation should look like, you can calculate the optimal lotsize for that add-on. And of course, the magic happens that once you do add on another position, you've created an ephemeral position to lather, rinse, and repeat.
google:
  • #94
  • Nov 15, 2009 2:27pm Nov 15, 2009 2:27pm
Quoting Craig
Hey mikkom, if you're interested their is quite an interesting thread on ET at the moment, search for 'can linear regression analysis...' the posts of 'MAESTRO' are particularly interesting, that is what I call hint dropping.
That's a cool thread. I love these types of threads - not because I necessarily agree with the main posters on everything - but because it provokes my own understanding of how the markets (and nature in general) works.

Keep'em coming.
  • #95
  • Edited 6:39pm Nov 15, 2009 5:35pm | Edited 6:39pm
Guess I must be missing something.

Given the diagram in twoblink's post #18:

Profit if stop is hit = (A+B)X - BY

By the golden ratio:
B/A = A/(A+B)
B(A+B) = A^2
BA + B^2 = A^2
set B=1
A+1=A^2
A^2-A-1=0
using the quadratic equation then,
A= 1.618B (standard derivation of phi)
B=0.618A

Substituting into the profit equation:

P = (A+0.618A)X - 0.618AY = 1.618AX - 0.618AY

When a new lot is purchased, the average price of the entire position moves from the first entry toward the second in proportion to the ratio of the 2nd entry size to the sum of the entry sizes (easily provable and should be familiar to most traders)...so:

Distance from A to ephC = (A+B)Y/(X+Y)
But according to the diagram, this distance also equals B, so:

(A+B)Y/(X+Y) = B

Substituting again:

1.618AY/(X+Y) = 0.618A

1.618AY = 0.618AX + 0.618AY

AY = 0.618AX

Y = 0.618X

Profit equation now becomes:

1.618AX - 0.618A(0.618X) = 1.618AX - 0.382AX = 1.236AX

But mazimizing this simply means maximizing either A, X or both.

Which was obvious right at the start without all the mathematical calisthenics. And maximizing profit has nought to do with where the second entry is or how big it is. Maximizing profit depends on price moving in your favor by the largest possible amount after entering with the maximum possible amount.

Glad we got that straightened out.


EDIT: ahhhh, and now as I catch up on posts I see that twoblink reached the same conclusion; that diagram doesn't work and the solution is divergent. OK.
  • #96
  • Nov 15, 2009 5:53pm Nov 15, 2009 5:53pm
Quoting twoblink
If i buy 2.4 lots @ 91.58 in jpy, then I have a "long position" in jpy. so that's what i mean by position.

Now, as an example, I can buy 1.5 more lots at 92.14 or 1.8 lots at 92.45 (just making up numbers) what should my decision be???

Choice A:

Stop would be 92.14 - 0.618(92.14-91.58) = 91.79
Combined position would be 3.9 lots at an average entry of:
91.58 + (1.5/3.9)(92.14-91.58) = 91.80
Profit if hit would be 1 pip x 3.9 lots = 3.9 (or $39 or whatever)

Choice B:

Stop would be 92.45 - 0.618(92.45 - 91.58) = 91.91
Combined position would be 4.2 lots at an average entry of:
91.58 + (1.8/4.2)(92.45 - 91.58) = 91.95
Profit if hit would be -4 pips x 4.2 lots = -16.8 (or -$168 etc.)

So choice A is superior.


As I understand it, you're saying there's some such choice that maximizes profits, but I don't see how. Suppose I choose to add 0.1 lots at 150.57? Or maybe 69,567,345.23? Where's the maximum? It's potentially infinite.
  • #97
  • Nov 16, 2009 1:07am Nov 16, 2009 1:07am
Yep, so my first try was divergent as well..

If you can calculate the total sum of a combo A+B position, for all known B's, and the potential profit against the potential SL, and you worked out an equation that was convergent.. then you will note that given a X@A, there exists one and only 1 distance for a combination Y@B with SL@Z.
google:
  • #98
  • Nov 16, 2009 7:53pm Nov 16, 2009 7:53pm
Quoting twoblink
Yep, so my first try was divergent as well..

If you can calculate the total sum of a combo A+B position, for all known B's, and the potential profit against the potential SL, and you worked out an equation that was convergent.. then you will note that given a X@A, there exists one and only 1 distance for a combination Y@B with SL@Z.
By "for all known B's" do you mean all possible values of B? Because that's a pretty big domain. Or does a "known B" mean some TA-derived point on a chart?

By "the potential profit AGAINST the potential SL" are you talking about the profit IF the SL is hit? Or are you talking about comparing the potential loss of an SL hit to some other potential profit number? If it's the latter, then that "potential profit" is what's unlimited and creates the divergence.
  • #99
  • Nov 17, 2009 4:31am Nov 17, 2009 4:31am
Quote
Because we can calculate every scenario and calculate the risk and reward of every position.. A divergent result will give you an infinity (aka non-answer) a convergent result will give you a maximized reward, or minimized risk. A combo of intersection will give you a saddlepoint and thus a mini-max point.

Quote
If i buy 2.4 lots @ 91.58 in jpy, then I have a "long position" in jpy. so that's what i mean by position.Now, as an example, I can buy 1.5 more lots at 92.14 or 1.8 lots at 92.45 (just making up numbers) what should my decision be??? well, since lot size alters my risk, and price point in which I add to my position alters my risk, then i have 2 factors, lot size, and distance from
X to Y.. yes? So given the lot size to be B and the distance to be (Y-X) then mathematically, I need to find an optimized B/Y pairing. Yes?

Quote
The equation tells you how much you should add when you add given an initial size A@X, and yes, since you know what the optimum situation should look like, you can calculate the optimal lotsize for that add-on. And of course, the magic happens that once you do add on another position, you've created an ephemeral position to lather, rinse, and repeat.

Quote
If you can calculate the total sum of a combo A+B position, for all known B's, and the potential profit against the potential SL, and you worked out an equation that was convergent.. then you will note that given a X@A, there exists one and only 1 distance for a combination Y@B with SL@Z.


lot size/length diagram

first series: lot size (how much you invest): x1, x2, x3....xn
second series: length (entry points): y1, y2, y3...yn

- the series is convergent if and only if it has a limit, if the series does not have a limit, it is divergent.
- if the series has a limit, it is unique.
- if the series is convergent, it is limited.

If I am correct, We always look only the fraction of the function, the length between first and second positions, or between combined first and second and third position, or between combined second and third and fourth position. The first limit for both series is zero, we cannot enter 0 lots and any currency pair cannot have the value of zero.

The process should go like this:

- enter a position at X1 (using any method you fancy) with A1 lots (according to your MM)
- use the equation to find the optimal distance from first position (saddle point) to enter a second position Y1
- use the equation to find the optimal lot size for second position B1
- put a stoploss at inflection point
- find a limit of a series between X1 and Y1
- find a limit of a series between A1 and B1
- those limits are values for X2 and A2 (third position)
- punch X2 and A2 into equation to get Y2 and B2.

Is this making any sense?
The world is not based on reality, but on the perception of reality.
  • #100
  • Nov 17, 2009 4:47am Nov 17, 2009 4:47am
No.
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