Optimized Risk vs Reward Equation

"be larger when right, smaller when wrong". That's true, for the most part. (I haven't read the entire thread yet)

This talk of a saddle point...I use something similar to my own risk management, if it in fact refers to what I think it does. First of all, the existence of a point where risk may be more concentrated necessitates the existence of a preset performance benchmark target, equidistant in nature to a preset performance drawdown point. For example, 10% gain or 10% loss. Now assuming multiple trades must be taken within this framework, if the question is about some R/R multiple, that depends on the MAX desired (predetermined) percent risk allowed for any one trade.

So back to the example. Let's say I want to gain either gain 10% or lose 10% over a series of trades (Why I want to do this in the first place is because if I happen to gain rather than lose, I could then double my leverage, for instance. That would be the only point). And let's then also say I am only willing to risk UP TO 5% on any one trade. How then best to incorporate that risk into the target percent/equidistant drawdown percent framework? The answer lies in the creation of a "midpoint" between the target percent point and the drawdown percent point. At the very middle is where it is best to risk the most (up to, but not more than my initial allowed max risk per trade of 5%). As I get closer to EITHER side of the extremes, that risk percent has to come down, independently of absolute gain or loss. The closer I get to either my 10% target or my 10% drawdown point, that 5% per trade may then become 3% per trade, or even 1% per trade if I'm really close to EITHER "target". The optimal R/R multiple is based on that initial predetermined risk percent (in our example it was 5%), ASSUMING that from the first trade the best possible outcome would be achievement of the target benchmark in one single trade. Thus, in this example it would be 2:1 reward/risk. For instance, if my maximum desired risk on any given trade were 3% instead, the optimum R/R would then be 3.3/1. So basically as (or if) I get closer to the "middle point" my risk increases, and as i get closer to the extreme my risk decreases. Why would I want to decrease my risk percent as I get closer to my target benchmark? Well because if I know my max desired risk percent, and then my extension the optimum R/R corresponding to it, as I get closer to my target percent, I don't need to take AS big a risk to get there - rather just enough to get there on one trade. Perhaps that's what a saddle point may refer to?

Almost. What you're talking about is a minimum OR maximum point on a curve in a 2D plane. A saddle point refers to a minimum AND a maximum point on a surface in 3D space. It's a point at which the height of the surface is at a maximum on one axis (north/south let's say), and thus can only fall in both directions. The same point is at a minimum on the other axis (east/west) so it's value can only rise from that point in those two directions.

I don't see that a saddle point has any bearing here however, which is why I basically skip over those posts. We're not looking for a minimum AND maximum for ONE value. We're looking for a MINIMUM of RISK and a MAXIMUM of POTENTIAL PROFIT. So we're looking to minimize one value while maximizing another. We're also not doing it in 3D space, but in 4D space, because we have 3 independent variables.

So to sum up, our 3 independent variables are:

X: The amount to add to our position (measured in units, lots, whatever) as a factor of our original position size. An X value of 0.45 would mean to add 45% of the original position.

Y: The distance from the original entry point to the new entry point as a factor of ?????. This is measured in pips. What it's a factor of though is a mystery. It can't be a factor of the original distance from entry to stop, because that distance will move from a negative number, to zero, to (hopefully) a large positive number. But anyway, this variable is a distance on a chart.

Z: The distance from the new average entry point after adding (TwoBlink calls this the ephemeral point) to the new stop loss point. This is measured in pips and is a factor of ?????. Again, that part is unclear.

So we have 3 independent variables, X, Y, and Z. And from those we can calculate DEPENDENT values for RISK and POTENTIAL REWARD. So there are five variables total; 3 independent and 2 dependent. We're trying to maximize one dependent variable while minimizing the other at the same point.

By "the same point" I mean the same 3 values for X, Y, and Z. So if we find out that risk is minimized AND reward is maximized at X=0.76, Y=2.4, and Z=1.5 for example, then we've found the "magic numbers" that TwoBlink says he has found.

Maybe we should play a game of 20 questions with TwoBlink to at least find his values for these numbers... "Is X greater than 0.5? Is it less than 0.75? Is Y greater than 2?"

Actually, minimizing risk and maximizing potential gain is the same thing as maximizing the ratio PG/R. So now we're down to 3 independent variables and only 1 dependent variable.

Two of the independent variables are ill defined however. Also, the means of calculating potential gain and risk for ALL possible points sounds like a task for an NSA supercomputer. Failing that, the method must involve some kind of "area under the curve" calculation. If that's the case, I'm probably going to leave it there as I'm a big fan of discrete mathematics... Infinitesimal analysis? Not so much.

capitalist, I think that we will be even more confused with the numbers, there's always a way to get specific set of numbers. The problem is, you won't know if it is applicable to any set of numbers. I've said in the beginning It is very difficult to replicate someone's thought process.

Here's the article about phi, with part about phi as a continued fraction.
http://www.maths.surrey.ac.uk/hosted...nacci/phi.html

I'm reading the article about continued fractions in general, interesting stuff.
http://www.maths.surrey.ac.uk/hosted...i/cfINTRO.html

this could prove useful, continued fraction calculator
http://www.maths.surrey.ac.uk/hosted...ci/cfCALC.html

this section explains how to calculate convergent continued fractions

http://img710.imageshack.us/img710/4...tinuedfrac.jpg

here's the one that has both phi and e

http://img187.imageshack.us/img187/3...inuedfract.jpg

Question.. vs what? A physical exercise??

A profit graph can be created for all infinite points after the position.. (discounting the bid/ask). A risk profile can be created for all infinite points also... As profit approaches infinity, and risk approaches infinity, a subtraction of the two yields a finite area under the curve.. and not only that, the area under the curve has a point on it in which profit is highest.

http://www.seykota.com/tribe/risk/fi...4_image001.gif
This is an image of the risk graph against a Kelly MM. What you will find is that for the most part; MM consists of Win% and Payrate. One moves the graph up/down, the other moves the graph left/right. (I teach this in my FX class in the MM portion of the class)

This is the risk graph of a given position with a given win% and a given payrate. If a secondary position were to be added, you would have created an ephemeral profile for the A+B combination. Each can also be assigned a risk profit, not just a profit profile. When a profit graph drops below 0, then it's not worth continuing (as it's not profitable) and when a risk graph does not yield any risk (OPM territory) then it's pointless to continue the risk graph.

When two profiles are combined, the optimized profit graph might not be a point on the risk graph, and the optimized point on the risk graph might not be a point on the profit graph.

For those who think buying a whole bunch and never pyramiding is the way to go; that's true in a game like roulette.. Where each additional bet is gains a house-erosion of 5.26%. But in FX, if you are correct, (FX Newtonian 1st law) says that you should add to it, as an object (in this case, a price) in motion tends to stay in motion unless stopped by an outside force.

When you buy and hold, you have maximized profit, but you also have maximized risk. If you are ok with that, that is fine.

I talk not about how to pick the correct direction here; I am merely talking about a mathematical optimization.

I also don't know what the problem is.. I operated under the assumption that such a point exists... and such a equation exists.. For 2+ years! So you all get the benefit of knowing that someone has already done it. That is MUCH more than what I had going into this..

As far as made much money.. If you dig around for my journal, my "Fist" method was cranking some 9000+ pips a WEEK. That's more than any professional I know of.. That's more than any institution I know of..

BTW... I have this programmed into an excel spreadsheet, so I have it as a lookup chart.. Quite wonderful. This takes at least the guess work out of where to add the next position and how much.

What this DOESN'T answer though, is how much for the initial position... or the direction...

Vs. a practical exercise. Searching for 3 magic numbers (2 of which aren't really defined) and which may or may not exist is an interesting intellectual pursuit, but it isn't practical to spend too much time on it at the expense of actual real-world trading tasks.

In other words, I'm not going to all of a sudden cease all trading until I solve the optimization riddle. That's what makes solving the riddle a purely intellectual exercise instead of a practical one.

well, I'm using the daily and weekly, so I can invest 99% of my time into solving the equation. And it is not intellectual, it is very down to earth. I want to be a better trader and compounding is the only realistic way of earning substantial sums (I do not have the million dollars to start with). I do not have the problem with discipline, or with the mm, been using stop loss since my first transaction on the demo and I have never used market orders on live account. My success rate went up from 16% in the beginning to above 50% although I am a trend trader (because of compounding, you get 4,5 positive trades instead of one). Compounding is also my nemesis, giving me more than one scarf (interestingly, more than 85% of my losses are on the gbp/chf pair). It's like fire, If you know how to control it, it is the best servant.

I'm spending 12 hours a day since the start of our discussion trying to crack the damn thing, and I'll do it, although I barely touched the math in years.

Summa summarum, the success is 1% intellect, 99% hard work. (I do not know if I translated this correctly, but you get the idea).

When a graph of two variables is a straight line passing through the origin, the variables are directly proportional to each other. As one variable increases, the other variable increases/decreases at a constant rate. The constant rate is determined by the gradient of the straight line.

Divide the change in height by the change in horizontal distance

Gradient = Change in Y/Change in X


http://www.mathsisfun.com/images/gradient0.gif

http://www.mathsisfun.com/gradient.html

hmm, the two variable function that has saddle and inflection points represented as an convergent continued fraction looks suspicious.

In that case, the function's equation changes whenever we want to add new entry, and we are always looking for the saddle and inflection points of every new function.

http://img515.imageshack.us/img515/7795/43888144.jpg
http://img143.imageshack.us/img143/8205/65482271.jpg
http://img205.imageshack.us/img205/567/25436626.jpg

http://home.att.net/~numericana/answer/fractions.htm

I'm reading the Handbook of Continued Fractions for Special Functions to see if this is true.

There is a vital component, one that is so minute yet is the influencer here..

I will tell you something important about the risk.. Let's say there is a machine in which, every time you give it a nickel, it spits out a dime. Do it long enough, and you'll be a millionaire. Your risk however, is the coin getting jammed as you insert it. But barring jams or the machine running out of dimes.. The risk is almost 0.. Something to think about.

Is it frequency? The risk is that the market doesn't allow for more trades.

hmm, are we looking for something like this, only with two variables?

http://img9.imageshack.us/img9/6779/82354910.jpg

http://img33.imageshack.us/img33/3157/79394159.jpg

http://demonstrations.wolfram.com/AR...laOfRamanujan/


pdf shows the properties of this function and expansion at x=0. I wonder if we are looking for the series expansion at saddle point.

We talked about infinite series. They are convergent if they have a limit.
Here's how to calculate the sum of infinite series.

We want to calculate all possible distances between a and b and all possible lot sizes between x (starting lot) and 0 (zero, y lot size is somewhere between 0 and x).

hehe sorry My bad..

But I dont know what is your insentive to mention the time value money Formula and commanding an all to every body I think every body aware of this formulation if i am not wrong


it is soo easy you can do in just a second I just CASIO v200 calculator And that is make those CFA formula very easy ..


keep your good work

Even Tho Srinivasan Ramajun Didn't have Normal School study On his Child Hood .. But in India The Britisher Rule In india so they just Have found S.Ramajun have The Extra ordinary Power of Mathematics .

The area under a curve as a sum

Here's the curve of some function called f(X) between X=0 and X=10. I've divided the area under the curve into columns of width dX.

http://www.ucl.ac.uk/Mathematics/geomath/intnb/int2.gif

The area under the curve, i.e. the area between the curve and the X-axis, is the sum of the areas of these columns.

Now look at the top of these columns. The curve is not quite a straight line along each top. If it were, we could calculate the area of each column very easily, since each column would be a triangle on top of a rectangle. Here's one particular column:

http://www.ucl.ac.uk/Mathematics/geomath/intnb/int3.gif


The height of the left-hand side is f(X1) and the height of the right-hand side is f(X1+dX). If the top were a straight line, then the area would be the area of the rectangle, that's f(X1)dX, added to the area of the triangle, that's 1/2dX(f(X1+dX)-f(X1)) using the "half-base-times-height" formula. Adding these two together we'd get the total area of the column,
area=1/2 dX (f(X1)+f(X1+dX)).

Now here's the crunch. If we imagine, as we always do, that dX is getting smaller and smaller, so we're going to need more and more columns which are thinner and thinner to fill the space between X=0 and X=10, then the top of the column gets closer and closer to being a straight line and our expression for the area therefore gets closer and closer to being exactly right.

All we need to do is look at what our expression for the area is approaching as dX gets smaller, and we'll know the area of the column. (OK it's going to be infinitessimally thin, but there again there's going to be infinitely many of them so when we add them all up we'll end up with an ordinary-sized number!)

As dX tends to zero, f(X1+dX) gets closer to f(X1), so we end up with 1/2dX(f(X1)+f(X1)) so that's just dX f(X1). We just need to multiply the value of f at point X1 by dX. We do the same for all points on the curve between X=0 and X=10 and add them together, so we get the result that the total area under the curve is the sum of f(X)dX for all possible values of X, in the limit as dX tends to zero and the number of columns tends to infinity. This sum is actually the integral of f(X), so the area under the curve of f(X) is the integral of f(X).

http://www.ucl.ac.uk/Mathematics/geo...b/intlnk3.html

The integral as a sum

We can think of integration as the result of taking a limit.

This diagram might make it clearer:

http://www.ucl.ac.uk/Mathematics/geomath/intnb/int4.gif

Before, when we were differentiating, we were trying to find the slope of the curve of Y(X), which was dY/dX. Now we know the slope (the derivative) and we want to know the original function (that's Y) that has this slope.

So instead of using slope=dY/dX, we rearrange the equation to find

dY = slope * dX.

Adding all these dY's together, using a different slope from every single point of the curve (that's infinitely many), we can build up the function Y that we're looking for. Since the slope at any point is the derivative dY/dX, we get

Y = sigma (I can't put this signhttp://www.ucl.ac.uk/Mathematics/geomath/intnb/sum1.gif everywhere because of the 5 pictures per post limit, sigma means sum, but almost everyone knows this) dY/dX dX.


That expression that we're adding looks very much like the one we had in the previous section for the area under the curve between X=0 and X=10. There we found that

area = sigma f(X)dX, when dX is tending to zero and there's infinitely many columns each giving a different f(X) value.

Comparing these two sums then,

Y=sigma dY/dXdX and

area = sigma f(X) dX,

we see that if we think of f(X) as a derivative dY/dX, then the area is equal to Y(X).

But if f(X) is the derivative dY/dX, then Y(X) is the integral of f(X). In other words, the area is the integral of f(X).

To sum up then, sigma f(X) dX is the integral of f(X) and so the area under the curve of f(X) is also the integral of f(X).

This shows that we can think of an integral of a function as a sum of all values of that function over the given range, multiplied by dX.

When we take the limit of dX tending to zero, this sum f(X) dX becomes the integral: http://www.ucl.ac.uk/Mathematics/geomath/intnb/int.giff(X)dX.

We can extend the notation we used earlier of http://www.ucl.ac.uk/Mathematics/geomath/intnb/int.gifcos(X)dX=sin(X) to show the range over which we want to integrate the function. We write

http://www.ucl.ac.uk/Mathematics/geo...ntnb/sum21.gif.

That shows the "sum of all possible f(X) between X=0 and X=10, multiplied by dx".

Whether all of the explanation above is clear to you or not (it does takes some getting used to), the idea you need to take away from this section is that the integral of a function is a sum of values of that function over the given range and that this also means that the area under the curve of f(X) is the integral of f(X).

http://www.ucl.ac.uk/Mathematics/geo...b/intlnk4.html

Now, what next?

Calculating the area under the curve for a length of 100 pips (or up to 10 lots)?

Finding the local maximum and inflection point for that segment?

Am I even close?

These two spreadsheets will help you to visualize and calculate area under the curve.