Optimized Risk vs Reward Equation

bigvlada,

I am glad that I have given you a new goal in life to pursue..

It took me 4~8 hours a day, for 2 years.. So the progress you are making trumps mine by a lot.. Keep it up.

Well, It's a combination of interesting problem (just like the Rubik cube and 22 moves solution; btw. - have you tried the Rubik clock?), mixed with something that is extremely useful, and not just in currency trading.

(quoted from The Limit of Delta Y Over Delta X)

She whispered. ``He scares me," and if Jimmy heard, he gave no clue.

He said, ``While we are waiting, I will explain how the derivative of a function works."

Sally taunted, ``a2+b2equals ______ all?"

Jim ignored her. ``Let's say two people together make a couple. We'll call the male y and the female x. And let's say that in this couple, the way they function together is such that the male is dominant by a power of two. That means y=x2. It's a curve. You can plot it on a graph," He drew the parabola in the air for them. ``So, assuming x =10 for the initial value of the female, this fixes y = 100 as the initial value of the male. Got it so far? Okay, suppose x increases to x = 12, what's the change in x and y?"

Magda said, ``Delta x is 2, delta y is 44."

``Very good. Now if you plug in different values, you see that as x diminishes, so does y. As the influence of the female diminishes, so does the male's This to me is the interesting part. They work together as a co-dependent unit. They're tied to each other. The thing we want to find out is, when you compare the change in x with the change in y, what's the limit of the ratio, delta y over delta x, as delta x approaches zero?"

``Now you've lost me," said Stephen.

The derivative of a function y = f(x) is the limit of delta y over delta x as delta x approaches zero.

The problem is (twoblink mentioned this before) that we determine this man/woman ratio (stop loss and lot size) before we enter the transaction.

We can then use the equation to get the best result possible but we do not know if our first ratio assumption was correct.

I'll use the Formula 1 example. You know how to drive in the most efficient way in order to win the Monte Carlo GP but you do not know what is the best rate of change between thickness of the tire and amount of gasoline at the start. You may win regardless of the starting conditions because you know how, but you do not know if you could done it easier if you knew the best starting conditions.

hmm, we'll need another roof equation that will test all possible rates of change of the rates of change. It also has to allow us the best starting conditions for the variable risk appetite (for example, what's the best starting condition for 1% risk? or 2,4...)

A lot to think about.

btw. I'm reading Godel, Echer, Bach, in hope I'll get a new perspective to look from when reviewing the problem at hand.

Well, here's the first attempt at the first problem.

Given the initial balance and risk size, there exists a finite number of lot and stop loss sizes.

If we punch balance, risk, preferred stop loss and lot size, we'll get something like attached table. Here's how to use it: You either use the combination of data from stop loss and recommended trade volume cells or recommended stop loss and trade volume cells.

If everything is OK, the last cell will be green. If it's red, you violated the starting conditions.

Bear in mind that recommended stop loss and recommended trade volume are the limits, the boundaries of initial terms.

The equation (first problem) must provide the way to test all possible stop loss/trade volume combinations within these boundaries. But, you'll have to know the optimized risk vs. reward equation (second problem) in order to do that (you'll need it to calculate profit for first n add up positions with different initial terms).

The third problem could be summed up in sentence: When is it enough?

Reasoning: Because the lot size, distance between subsequent entry points and lot size all increase as price moves in our direction, after a while we'll have a situation where we have 700 lots with 1000 pips SL and the next add-on point 3500 pips away.

The good thing is that the transactions need babysitting mostly at the beginning, but we may get that next add-on point in a few years or never.

The problem is how to determine when the adding should stop, when it becomes a burden? Or unrealistic? Swap should play a role in this (if it's positive, you could just leave it be, but what if it isn't?). I can't even determine all the variables yet.


manual: start the mt4 and then the excel file. If you want full lots, delete little m's in all cells (just renaming EUR/USDm in EUR/USD won't work)

Let me go ahead and share how I do my testing; not with MT4..

I start off with ramdisk software; About 12 gigs of ramdisk. This is because most of the I/O kills the HD's..

My usual setup: 2 Gig ramdisk; sqlite (don't bother with mysql, it's crap). this allows for fast data import.. just export MT4 data to csv, and import it to sqlite. Make sure the flatfile is on the ramdisk. I then use python for Rapid Applications Development. With dictionaries, lists, arrays etc.. python is the recommended programming language.. easy to debug, and you can still OO with overloading.

Hope everybody had a great xmas and wish you and yours a happy new years.

Here's a few thoughts on risk graph. The risk is a function of length and stop loss. We have our entry point, a stop loss point and a point where the risk is 0%. Now, I do not know the ideal rate of change for the risk, but I do know where I should add my next position if my only concern is my risk size.

Regardless of starting conditions (risk size, stop loss size, rate of change) the point where the risk is zero is always the one where the length is twice the size of the stop loss.

Example: If the stop loss is 50 pips and the risk is 2%, the risk is zero at a point 50 pips in profit. If we want to add to our position, but wish to leave risk at 0%, we must use the same position size and also add the worst possible scenario (the position activates, the price drops and activates the stop loss). In worst case scenario, the risk will remain at 0% if we add to our position at a point twice the size of the stop loss. We can calculate another add-on position, but although in worst case scenario we are at break even, this is just one view at the problem, purely from risk perspective.

I can't use the same logic to determine when the profit reaches the maximum potential and then declines. I even tried to experiment with the equations from physics used to determine the mass and thrust ratio for multiple stage rockets in order to reach space... The problem the rockets face is backward to our own - they lose mass and thrust but keep the ideal trajectory (function) to reach space.

Twoblink, can you recommend some books on risk and profit profile? I do not think that I've grasped both terms fully and correctly.

I'm an idiot. This could be calculated easier with the ellipse.

The area under the curve is always half the area of the ellipse.

http://upload.wikimedia.org/wikipedi...erties.svg.png

And of course, there is only one point where the profit is at it's peak (or risk at it's bottom).

ps. I know there is only left half of the ellipse in the pdf (probably that's why the pdf generation at wolfram's site is classified as experimental).


hmm, the ideal ellipse is circle, but somehow I do not think that's the curve we are looking for.

You should read this, http://epchan.blogspot.com/2010/01/d...g-in-work.html.
It's an argument about scaling in which doesn't require pages of irrelevant math.

hmm, it's about averaging down, we are talking about averaging up.
Don't like? Don't read.
Posting irrelevant math helps me to better understand the problem by attacking it from different angles. The advantage of forum over plain paper is that I can discuss it with others.

Agreed. Don't confuse "averaging in" with the problem Twoblink has posed: how to determine the optimal point (defined as a tuple of price,risk,lot size) at which to add to your position when you're *already* profitable. The "averaging in" issue has been effectively deposed by Twoblink in many previous posts. I'm glad to see that Chan has debunked it yet again.

Happy New Years to everybody.. I found I had a typo up top, it should say "1 ~ 2 Gigs" not 12 Gigs..

Optimization: Not just of profit, but of profit risk profile.

I have been working on Averaging Up for the last almost 3 years of my life; and now I might have found a slight mathematical wiggle against Averaging Down as well.. So I'm excited!!

I'm not a math major, I'm no even an engineering major; and so I'm curious why I don't read about most of these mathematical "findings" online. The conspiracy factualist in me wants to say it's because those who found out, don't want to share.. because I gotta believe I'm not the only person figuring this stuff out..

BTW.. I did read that article, and it has almost nothing to do with what I'm talking about. It talks about optimized reward only, which is a lie because that's only half the story..

Scenerio:

A) If you bet $10, you can win $20.
B) If you bet $1, you can win $3.

Given the same expectancy, A would be the optimized winner in terms of $$$, but given a risk/reward optimization profile, B would be the better bet.

quote=Craig;3359464]You should read this, http://epchan.blogspot.com/2010/01/d...g-in-work.html.
It's an argument about scaling in which doesn't require pages of irrelevant math.[/quote]

It isn't simple one bit. I'm not a mathematican...but I look at the market
pretty much in the same vein as bigvlada where math can give you a
possible prespective where literal symbols can be allowed to take on any
given value. I'm working on such a routine...which I have named as the
"Cartesian Cross." It's very simple and has been laying around for the
past four hundred years. Why hadn't someone come up with it before,
is beyond me. It's nothing but a rectangular coordinate. After all...the
chart you look at is such one. If you want to get cute about it...you
can go into oblique, homogeneous, spherical coordinates. You can do
all kinds of thing mathematically with and in the market...but why not
try to keep it simple. There is a very simple one that has been laying
around for years, and has been in Charles Drummond's "Point and Line
Methodology". It's pretty much the same as Joe Ross's Swing trading...
I think he refers to it...being P1, P2, P3, P4. The "active market"...
would be from P1 (low point) to P2 (last high point)...then you have
a "reaction" (retracement)....down to a possible Fibbo (has to, if it's a
trend run) which becomes the P3. The market has to go up from there
if it's a trend. A established P3 to P4 (TP..target price) is figured by the
distance from P1 to P2. The EP (Entry Price) is 1/4 the distance from
P1 to P2. Or EP= P3 + 1/4(P2-P1). SP (Stop Price) 1 or 2 ticks below
P3. Once the market gets 1/2 the distance (or 1/2 distance) from
P3 to P4 (target) you bring up your SL (stop loss) to your EP (Entry
Price). The half way point is called the CP (Critical Price). Many traders
take their profits here,,,( a bird on hand). However...if the market
goes past the P2 level...you go into the TS (Trailing Stop). The TS
is figured by 1/4 (P2-P3). Now this TS is taken from the highest High
that the market has moved. You are assured of a profit, regardless
what happens. The Ratio of Profit to Risk is always 3 to 1. Most of
the time it will be something like 2.95 to 1. This method can also
stop you from making a trade....or rather "prevent you from making
a trade". Is it perfect?.....of course not. Just be careful about trying
to mathematically force the market to become a 'mathematical servant'.
I assure you that the market will blow you out of the box and if it's
in a good nature...."hang you up to dry". You're better off, in most
instance, to 'sit on your hands and wait'....watch the 'chicken with
false teeth do it's soft shoe dance....with the rooster looking on. The
moment the rooster mounts the chicken....hop on the train and take
off".
ciao,
Tonyj

I was doing some HD Benchmarks and they indeed confirm what I already know to be true:

Average transfer rates in Mb/Sec.

Ramdisk ~ 4,484Mb/Sec (That's not a typo)
RAID 0 ~ 150Mb/Sec (Dual SATA2 drives)
Single Disk ~ 100Mb/Sec

The Ramdisk is in nanoseconds, not milliseconds, order of magnitude.. of a few orders. If you are running data crunching on a regular disk, I urge you to go ahead and do as I recommend, go ramdisk. 45x the performance.. but it's much more than that, because since it's ramdrive, there's no bussing of data; it's in memory already, so not only is there 45x the speed, but no I/O performance hit. From my personal number crunching experience, I get about a total of 600x (again, not a typo) performance.

if you go ahead and use sqlite, the only thing that's slow about it is I/O, which you've eliminated in ramdisk. And if you go ahead and use a persistent connection against the file, it will open, and remain open, requiring no additional file openings and closings.

My old partner in crime programs in Ruby, and I do it in Python. Both are OO script languages and we picked them for a reason. They have mature data types like lists, arrays, dictionaries etc.. Both allow for RAD, and direct connect to sqlite is fast and easy in about 2~4 lines.

I'm reprogramming a lot of stuff again, so once again, I'd like to go ahead and make the same recommendation I did before; this time with benchmarks..

1~2 Gigs Ramdisk
Sqlite file on the ramdisk
python or ruby against the sqlite db

Happy number crunching people!

I love sqlite and use it for my tick data.

Here are a couple of twoblinks' past posts that may help clear some of this up......
Pascals' Triangle
Here is the blue and orange hockey stick post

Another view is from the turtles trading system .... how they entered and how they added to their position by pyramiding can be found on page 19 of this PDF The Turtle Trading Rules.
.
I would think this would be a good foundation to start from.
.
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You better start with this:

http://en.wikipedia.org/wiki/Minimax


Minimax theorem

The minimax theorem states:
For every two-person, zero-sum game with finite strategies, there exists a value V and a mixed strategy for each player, such that (a) Given player 2's strategy, the best payoff possible for player 1 is V, and (b) Given player 1's strategy, the best payoff possible for player 2 is -V.

Equivalently, Player 1's strategy guarantees him a payoff of V regardless of Player 2's strategy, and similarly Player 2 can guarantee himself a payoff of -V. The name minimax arises because each player minimizes the maximum payoff possible for the other—since the game is zero-sum, he also maximizes his own minimum payoff.

Maybe like this

"A differentiable manifold is a topological space on which we can do calculus"

http://everything2.com/title/differential+manifold


saddle point

In the context of undergraduate multivariate calculus courses, a saddle point is a point on a two-dimensional surface in three dimensions that is a local minimum in one direction and a local maximum in another. More generally, it's a point on a differential manifold where the Gauss curvature is negative. Without going into the mathematical details, one can think of it as a point where the surface is both concave and convex, like a saddle: you can sit in it and on it at the same time.


Many results in game theory are linked to topological notions; a saddle point is an example. This writeup applies to a game in two player, zero-sum strategic form X,Y,A.
An element aij of a matrix A is described as a Saddle point if it equals both the minimum of row i and the maximum of column j.

A payoff matrix needn't have a saddle point, but if it does, then the usually technical minimax theorem is easily shown to hold true:

(Minimax Theorem for a game with a Saddle point)

Let aij be a saddle point for a game in strategic form given by X,Y,A. Then the game has value aij , achieved when Player 1 plays the pure strategy xi and Player 2 the pure strategy yj .

http://everything2.com/title/Saddle+Point