Optimized Risk vs Reward Equation

My question here is that if there would be "optimal" (I would be assume that this would mean high enought deviation from random) value for the variables defined here for random entry, wouldn't it prove that the market moves in certain kind of way all the time?

I could understand that there could be variables that would work for certain kind of entries (for example strong breakouts) but for random entries I really can't see how this could work.

Optimization Problems with Functions of Two Variables
http://www.analyzemath.com/calculus/...imization.html

The third example looks similar to our problem, I'll have to solve it a few times to better understand it.


Several optimization problems are solved and detailed solutions are presented. These problems involve optimizing functions in two variables using first and second order partial derivatives.

Problem 1: You decide to build a box that has the shape of a rectangular prism with a volume of 1000 cubic centimeters. Find the dimensions x, y and z of the box so that the total surface area of all 6 faces of the box is minimum.


Solution to Problem 1:

The total area A of all six faces of the prism is given by.

A = 2xy + 2yz + 2zx

http://www.analyzemath.com/calculus/..._problem_1.gif


The volume of the box is given; hence

xyz = 1000

Solve the above for z.

z = 1000 / (xy)

Substitute z in the expression of the area A to obtain.

A(x,y) = 2xy + 2y * 1000 / (xy) + 2x * 1000 / (xy) = 2xy + 2000 / x + 2000 / y

We now need to find x and y that minimize the area A. We first need to find the critical points and then test the second partial derivatives. The first order partial derivatives of A are given by

Ax(x,y) = 2y - 2000 / (x2)

Ay(x,y) = 2x - 2000 / (y2)

The critical points are found by setting Ax(x,y) = 0 and Ay(x,y) = 0 and solving the system obtained. Which gives

2y - 2000 / (x2) = 0 which gives 2yx2 = 2000

2x - 2000 / (y2) = 0 which gives 2xy2 = 2000

Solve the above to obtain

x = 10 and y = 10

We now need to find the second order partial derivatives.

Axx(x,y) = 4000/(x3)

Ayy(x,y) = 4000/(y3)

Axy(x,y) = 2

We now need to test the values of Axx, Ayy and Axy at the point (10,10) in order to use the theorem on minima and maxima of functions with 2 variables.

D = Axx(10,10) Ayy(10,10) - Axy2(10,10) = 4 * 4 - 4 = 12


D is positive and Axx(10,10) = 4 is positive and therefore the area A is minimum for

x = 10 cm

y = 10 cm

z = 1000/(xy) = 10 cm.
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Problem 2: Find the dimensions of a six-faced box that has the shape of a rectangular prism with the largest possible volume that you can make with 12 squared meters of cardboard.

Solution to Problem 2:

Using all available cardboard to make the box, the total area A of all six faces of the prism is given by.

A = 2xy + 2yz + 2zx = 12

The volume V of the box is given by

V = xyz

Solve the equation 2xy + 2yz + 2zx = 12 for z

z = (6 - xy) / (x + y)

Substitute z in the expression of the volume V to obtain.

V(x,y) = xy (6 - xy) / (x + y)

Let us find the critical points by first finding the first order partial derivatives

Vx(x,y) = -y2( x2 + 2xy - 6 ) / (x + y) 2

Vy(x,y) = -x2( y2 + 2xy - 6 ) / (x + y) 2

We now solve the system of equations given by Vx = 0 and Vy = 0. One obvious solution is given by the point (0,0) but is not physically possible. Other solutions are found by setting

x2 + 2xy - 6 = 0

and

y2 + 2xy - 6 = 0

Subtracting the equations term by term we obtain

x2 - y2 = 0

Solve to obtain

x = y and x = - y

The solution x = -y is not valid for this problem since both x and y are dimensions and cannot be negative. Use x = y in the equation x2 + 2xy - 6 = 0, we obtain

x2 + 2x2 - 6 = 0

Solve for x

x = sqrt(2)

Find y

y = x = sqrt(2)

Let us now find the second order partial derivatives

Vxx(x,y) = -2y2( y2 + 6 ) / (x + y) 3

Vyy(x,y) = -2x2( x2 + 6 ) / (x + y) 3

Vxy(x,y) = -2xy(x2 + 3xy + y2 - 6 ) / (x + y) 3

We now need the values of Axx, Ayy and Axy to find the value of D = Vxx(sqrt(2),sqrt(2)) Vyy(sqrt(2),sqrt(2)) - Vxy2(sqrt(2),sqrt(2)) in order to use the theorem on minima and maxima of functions with 2 variables.

D = Vxx(sqrt(2),sqrt(2)) Vyy(sqrt(2),sqrt(2)) - Vxy2(sqrt(2),sqrt(2)) = 5/2


D is positive and Vxx(sqrt(2),sqrt(2)) = -sqrt(2) is negative and therefore the volume V is maximum for

x = sqrt(2) meters

y = sqrt(2) meters

z = (6- xy) / (x + y) = sqrt(2) meters.

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Problem 3: Find the distance from the point (1,2,-1) to the plane given by the equation x - y + z = 3.

Solution to Problem 3:

One way to find the distance from a point to a plane is to take a point (x,y,z) on the plane; find the distance between this point and the given point and minimize it. Because the distance involves the square root, it would be better to minimize the square of the distance. Let the square of the distance between the given point and the point (x,y,z) on the plane be f.

f(x,y,z) = (x - 1)2 + (y - 2)2 + (z + 1)2

We now solve the equation of the plane for z to obtain

z = 3 - x + y

Substitute z in f by 3 - x + y in f.

F(x,y) = (x - 1)2 + (y - 2)2 + (- x + y + 4)2

We now find the first order partial derivatives

Fx(x,y) = 2(x - 1) + 2(-1)(-x + y + 4)

Fy(x,y) = 2(y - 2) + 2(-x + y + 4)

We now need to find the critical points by setting the first partial derivatives equal to zero.

2(x - 1) + 2(-1)(-x + y + 4) = 0 and 2(y - 2) + 2(-x + y + 4) = 0

We now solve the system of equations to obtain

(8/3 , 1/3 , 2/3)

We now calculate the second order derivatives

Fxx(x,y) = 4

Fyy(x,y) = 4

Fxy(x,y) = -2

We now need to find the sign of D = Fxx(8/3,1/3) Fyy(8/3,1/3) - Fxy2(8/3,1/3) in order to use the theorem on minima and maxima of functions with 2 variables

D = 4 * 4 - (-2)2 = 12

Since D is positive and Fxx is positive, F has a minimum at the point (8/3,1/3) which correspond to a point on the plane given by

(8/3,-1/3,2/3)

The distance d between the given point and the plane is given by

d = sqrt [ (1 - 8/3)2 + (2 - 1/3)2 + (-1 - 2/3)2 ]

= 5 / sqrt(3)

I believe that this, third example, is the correct way to implement the equation, once we figure out how it does look like.

TwoBlink isn't talking about optimizing based on market conditions at all as far as I know. He's talking about a purely mathematically based optimization point. I'm with you though... everything I've tried leads to the conclusion that no such point exists. The results I get always point to the "optimization strategy" of buying as much as possible in the first entry, never adding on, and letting the profit run.

I do understand that he's not talking abut optimizing based on market conditions but if you find 3 points that are optimal for the variables you mentioned, you can easily derive the way market moves or derive a way market does not move from those 3 points.

I just see no way to optimize entry points without fitting them in some way to market movements - there simply can not be optimal way otherwise because every entry/exit pair for complete trade set removes spread + commission from the final result so each new entry must be much better than 50/50 to counter that. The percent effect of spread/commission also grows based on how many entries/exits you have and how small final profit you aim to.

This Mathematica program could be useful as a starting point in our function finding problem.

Many famous sequences occur as the coefficients of a series for an exponential function, which is called the exponential generating function of the sequence.

Here's the exponential function that generates Fibonacci numbers:

http://img707.imageshack.us/img707/6...cifunction.jpg


I had to zip it, the site won't accept nbp files.

Even though all of the optimization schemes I've tried yield divergent results (i.e. there is no optimum add-on point, so just go for broke with a single entry), there is a compelling non-formal argument FOR the existence of such a point.

First of all, adding on (or pyramiding) has the advantage of using accumulated profits to add to a position, increasing future profit potential while all the while locking in more profit as each new add-on point is reached. But is there an optimum point at which to add?

Well, would we add on to our trade if it was only 1 pip in profit? Probably not, because we wouldn't have accumulated enough profit yet to justify the additional risk. So the optimum point is NOT at the 1-pip profit point.

Would we then wait for our trade to go into profit by 1000 pips before adding on? Again, probably not because we would be giving up all the additional profit we could have accumulated by adding at points prior to the 1000 pip profit point. So the optimum add-on point is not there either.

So if the optimum point is not at these two extremes, must there not be SOME point between these extremes that DOES balance out our conflicting goals (don't add too soon but don't wait too long)?

The trick is to quantify the effects of the add-on variables on these conflicting goals.

I'm new to this post and all the arguments involved. I'm not a mathematician...yet I try to look at the market from mathematical
view point. My premise if from a cartisian coordinate...or rectangular
coordinate. After all...the chart is exactly that... so why fight it. You
mathematicans on this post are of immense interest to me...even though
I have to struggle in following you. Since I've only happen to trip on
this thread...my response will lack certain points to yours. But I hope to
try catching up to you fast sprinters as yours truly is an octogenarian
who made it in commodities years ago. First of all: Price is a function
of time....P=f(t). Somewhere on this post there was a mention of F(X)1
to F(X)2....which means a price to the next price. O.K....fine, time only
goes in one direction. We look back to see the foot prints left in the
snow. We can look back as far as we want...in other words the 'grave
yard'. Even look at the inscription on the tombstone...year of the death
etc. This...to determine the probability of 'tomorrow'. One of you started
out this thread with just that premise....envoking the mention of the
Pascal Triangle. Ah yes...Blaise Pascal...who was asked by an enthus-
iastic gambler...Chevalier De Mere to solve a problem for him. Without
getting too far in doing a Tarzan cry and then swinging thru the jungle
via vines...Pascal communicated his solution to Pierre Fermat. All kinds
of mathematical geniuses with names ending like Andre Nikolaevich Kolmogorov...if only the Pascal Triangle could help your tongue with
that one...came after. But to the point...regardless all the math. The
'tick' (pip) is the only truth. It appears in all the 'time frames'. If
anything is going to happen,...it's going to begin with the 'tick'.
But the market action appears only during the 'Daily'....yet the 'tick'
shows up on all the time frame going back as far as the 'yearly'...and
beyond if you want to. A 'tick' in the yearly?....it's like a ping-ball in
the Pacific Ocean. There are all kinds of math parodies in the market
ready to give you a run for your money. And the odds are that you're
going to lose. Yes the market follows a trend...and that trend is a
tangent angle....the eponential and logarithmic functions along with
the trigonometric functions as well. The resistance in the market is
logarithmic while the support is eponential...but in the lower time
frames you have more of them than in the higher time frames. Regard-
less...in that lower time frame with more curves each being its own
expo and log....and in between your linear, the inflection. Wonderful...
isn't it? But you still have to know how to 'read the ticks'...know what
the 'sitting of the ducks' in the ticks...then up to the minute to see
the 'range'. Do the 'ducks' start to fly? Which way? and then look at
the 'petit soldats' in the minutes to see which way are they are march-
ing? If they're going to 'trend up'...they have keep breaking its own
range. But they don't have to do it all in one scoop. They can retreat
in the lower time frames...and then fool around a bit, and then make
another attach. This is where most of the traders start to get all
fouled up...especially the newbies. Anyway...I hope you get the idea.
I look forward to studying you guys....
I mentioned that I'll try to catch up to you sprinters....that is if I don't
kick the bucket before then. The undertaker keeps mumbling something
about,"You may linger a little longer...but I'll get you in the end". Quite
funny....isn't he? He probably has his own show on Broadway..."The
Dead Souls"....starring Humprey Pushcart and Kirk Dogface.
ciao
Tonyj

Twoblink, let me see If I got this right:

I'm using price as my z-axis, price as a function of stop-loss and lot size.

so the coordinates in 3D are (x,y, f(x,y)) and we are looking for f(x,y), the function that gives us optimal price for our stop loss and lot size numbers.

Attached are some examples of two variable function graphs.

I have another question, are the variables mutually exponentially proportional ?

Meaning: A variable y is exponentially proportional to a variable x, if y is directly proportional to the exponential function of x, that is if there exists a non-zero constant k that satisfies the equation http://upload.wikimedia.org/math/1/8...a576220fbf.png

Here's the ultra detailed explanation of how to draw a graph of two variable function.

Calculus: How to sketch functions of two variables. Chris Tisdell UNSW Sydney

So, if I'm right, the price is the distance from domain of the function to the surface (in our case, the distance to the surface that has saddle point).

http://img260.imageshack.us/img260/2383/3dfunction.jpg


http://upload.wikimedia.org/wikipedi...ddle_point.png

I'm in Tennessee right now.. visiting a friend, but heading home in 2 hours.. so let me clarify something that I thought was blatantly obvious, but I guess I was wrong.

Give A lots @ X point, there exists an infinite amount of B lots @ Y points I can add to my position. When I add to my position however, I can now calculate every single price point possibility, and thus measure my risk/reward.

So let's say I buy JPY 1.5 @ 88.54. say I randomly pick a 2nd point, say i buy again 1.3 lots @ 90.23. What kind of risk/reward am I looking at if the price went to 100.54? what kind of risk/reward am I looking at if the price dipped to 87.00? where should I put the stoploss so that I can optimize upside potential while curbing downside potential?

So let me spell it out for you;

Given A lots @ X, find B lots @ Y, with a Z SL such that in every single conceivable price movement, risk is minimized, while profit is maximized.

B lot size is a factor
Z location of the SL is a factor
Y point (actually, distance from X to Y to be exact) is a factor
Hint: B __CANNOT__ be bigger than A, otherwise you'd have a divergence.

The kicker is the stacking that occurs. After you put in B@Y, you've created ephemeral C, and since we know that C __HAS TO BE__ between A and B, We also know that the equation needs to be fractal such that Given we know C position after the fact, a D position is greater (because the lot size of C is greater than A and thus the distance between C to D has to be greater than A to B) and applicable to the same equation.

Sounds impossible, but it's not.. The equation is fractal (should be fractal) and so if it's divergent, it's wrong.

It is also timeframe agnostic; since we are dealing wholly with distance, and lot size. 100 pips is 100 pips, regardless if you are using the 5m charts or the daily.

Capitalist; I don't know if you are a gun owner; but I think of it like this... the longer the barrel, the more pressure build up behind a bullet (up to a certain point) and the more accurate the bullet will be due to flight speed and spin stability; but at some point, the metallurgy doesn't hold up, and the flex in the longer barrel causes more harm than good. so for every round of bullet, there is an optimum length in which most of the gas is burned inside the barrel, but the barrel is short enough that it remains harmonically stiff so there is no flex.

No different than FX. I buy too big of a lot from the get go, I have huge up front risk. I buy too little, and don't have enough upside potential. I buy too near my original position, I increase risk. I buy too far away, I decrease potential. I buy too much on the add-on, i increase my risk. I buy too little, I decrease potential. etc.. anytime there are infinite possibilities to finite possible outcomes, there is a convergent answer..

Oh.. I've been living in Memphis for the past 4+ years.



I am. I know that in cases where there are competing advantages and disadvantages to any given value of a parameter, there must be a maximumally effective value of that parameter at some point, as I mentioned in my previous post.

I'm glad to know that the 3 variables to be derived are distance to new entry, size of new entry, and stop distance (from ephemeral point?). This is the direction in which I've been looking.

u guys make alotta money from this stuff?

he wasnt finished yet

Absolutely correct, imo.

If I'm understanding your post correctly...the only optimal point would be by an incidental repeat of the same or qualitatively better "conditions" that caused the initial entry. That can occur at any time, or it may not occur at all. However, each point should be viewed individually. So if there is an optimal add-on point, that point itself has its own best exit, which is entirely separate and distinct from where the initial entry point's best exit would be.

Because of that I believe this line of thinking is merely an intellectual exercise. If one cannot be successful without adding, one cannot be successful. The only benefit to this might be that under certain risk parameters, it might lower downside volatility somewhat, on aggregate.

I'm inclined to agree for the most part. On the subject of market "conditions" though, I think TwoBlink dispensed with this issue early in the thread by making it clear that this whole optimization discussion is predicated on the assumption that the trader has managed (somehow) to choose the correct direction over the pyramiding period. The phrase he used was something like "be larger when right, smaller when wrong."

As far as it being an intellectual exercise, I again agree. That's why I'm not allowing TwoBlink's riddle to hold up my development of a trading method. I'm a fan of long-tail, low risk high potential reward trading which involves pyramiding, but my add-on method is of the simple "use half the accumulated locked in profit" variety. It may not be "optimized" in a narrow sense, but it's optimized in the sense that I can use this method NOW, and not in 17 years when I find TwoBlink's equation.

hmm, z is not the price but the stop loss.

Oh well, back to the drawing board.

But this is encouraging.


If I interpreted everything correctly, these are the rules:

- for every price of the first position (for example 105,28; 1,0716; 0.9245...) there exist
appropriate stop loss and lot size number that satisfies the saddle point requirement.

- there exists the equation that will give the first add-on point that also satisfies the saddle point requirement with two limitations - second lot size must not be larger than first and distance between first and second positions is always smaller than distance between first and second ephemeral points.

- these two limitations are applied to every subsequent add-on position with clarification that every next lot size must not be bigger than sum of all previous lot sizes.

- all subsequent positions must also satisfy the saddle point requirement

- the risk is never bigger than the initial risk, regardless the number of add-on positions