Subsystem "Asset Management" - page 7
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Seryoga, what's the matter with you?
Yeah, all right, all right!
I'm not saying you can't do it, it's just that I don't feel a deep sense of inner satisfaction when I'm told, for example, "cow squared...". What's that?
You know what I was thinking...
Suppose I have 2000 daily bars (their amplitudes X). I write a SLU of equations of the form:
а[1000]*x[1000]+а[999]*x[999]+...+а[1]*x[1]=x[0]
а1001*x1001+а1000*x1000+...+а2*x2=x1
.
.
а2000*x2000+а1999*x1999+...+а1001*x1001=x1000.
And I find coefficients A0,A1...A999. Then I substitute the current daily bar x[0] into the equation
a999*x999+a998*x998...+a0*x0=x[-1] (increment for tomorrow)
and I get the forecast for tomorrow's afternoon bar.
You haven't fiddled around in this direction. It would seem, what could be easier?
Yeah, all right, all right!
I'm not saying you can't do that, it's just that I don't get a deep sense of inner satisfaction when I'm told, for example, "cow squared...". What's that?
You know what I was thinking...
Suppose I have 2000 daily bars (their amplitudes X). I write a SLU of equations of the form:
а[1000]*x[1000]+а[999]*x[999]+...+а[1]*x[1]=x[0]
а1001*x1001+а1000*x1000+...+а2*x2=x1
.
.
а2000*x2000+а1999*x1999+...+а1001*x1001=x1000.
And I find coefficients A0,A1...A999. Then I substitute the current daily bar x[0] into the equation
a999*x999+a998*x998...+a0*x0=x[-1] (increment for tomorrow)
and I get the forecast for tomorrow's afternoon bar.
You haven't fiddled around in this direction. It would seem, what could be easier?
I tried it a long time ago - it turns out very rarely.) Besides it's the same problem of identification - which sample length to take...
Sergei, why are you duplicating my entire message in your post? Is it a habit? It's a bit cumbersome.
What about sample length, you can run in the tester by this parameter and choose the best :-)
Here's what caught my attention:
If the rank of the matrix is noticeably large (like 1000 or more), then when predicting one step ahead, the impact of the "new" member of the equation should be minimal and the system should give a stable solution (no chattering +/- 100000). So it seems to me. This differs from the case where the system consists of, say, 10 equations and the coincidence on 10 examples from the sample is 100%, and on the next term the difference is 2000000 times...
Yeah, all right, all right!
I'm not saying you can't do that, it's just that I don't get a deep sense of inner satisfaction when I'm told, for example, "cow squared...". What's that?
You know what I was thinking...
Suppose I have 2000 daily bars (their amplitudes X). I write a SLU of equations of the form:
а[1000]*x[1000]+а[999]*x[999]+...+а[1]*x[1]=x[0]
а1001*x1001+а1000*x1000+...+а2*x2=x1
.
.
а2000*x2000+а1999*x1999+...+а1001*x1001=x1000.
And I find coefficients A0,A1...A999. Then I substitute the current daily bar x[0] into the equation
a999*x999+a998*x998...+a0*x0=x[-1] (increment for tomorrow)
and I get the forecast for tomorrow's afternoon bar.
You haven't fiddled around in this direction. It would seem, what could be easier?
It actually works out pretty good. Only it's not just the bars you need. Daybars are best. Sampling distance is 8 years.
Sergei, why are you duplicating my entire message in your post? Is that a habit? It's a bit cumbersome, isn't it?
What about sample length, you can run in the tester by this parameter and choose the best :-)
Here's what caught my attention:
If the rank of the matrix is noticeably large (like 1000 or more), then when predicting one step ahead, the impact of the "new" member of the equation should be minimal and the system should give a stable solution (no chattering +/- 100000). So it seems to me. This differs from the case where the system consists of, say, 10 equations and the coincidence on 10 examples from the sample is 100%, and on the next term the difference is 2000000 times...
1. If the rank of the matrix is noticeably large -- then even the chattering of the last term will have an impact at the level of error. What kind of calculation accuracy are we talking about then?
2. Don't forget the accuracy of the calculation. Which will start to limp with this amount of calculation.
3. Correct me if I'm wrong, but the system is not correct! The last equation has 2 unknowns. a[0] and x[-1]
Day trippers are best. The sampling distance is eight years.
Well, well - here come the Experts!
Well, the fact that the girls are better, it's already clear and the 8 years difference is just right :-)
But what about the rank of the matrix where the optimum. Is it worth it? I will now try to put some code in MQL to solve an SLU.
1. If the matrix rank is considerably high, even a jitter of the last term will have an impact at the level of error. How precise are your calculations going to be then?
2. Let us not forget about calculation accuracy. Which will start to lag behind in such a large volume of calculations.
3. Correct me if I'm wrong, but the system is wrong! The last equation has 2 unknowns. a[0] and x[-1].
1. I have a hope that with a large number of coefficients in the equation, the role of the new term is smoothed out. Why is this not the case?
2. I agree.
3. it is not an equation, it is not a part of SLU, it is an equation which allows you to make predictions that the coefficients are changing slower than the price.
x[0] is the current value. We wait for the formation of today's day candle and then get a forecast for the next bar - x[-1]
Sergei, why are you duplicating my entire message in your post? Is it a habit? It's a bit cumbersome.
What about sample length, you can run it in the tester by this parameter and choose the best :-)
I don't always write this way (if you notice), only when I use a PDA, it makes it a lot easier, so sorry - I'll "use" it that way sometimes
If matrix rank is noticeably big (like 1000 and more), then while predicting one step ahead, the influence of the "new" term should be minimal and the system should give a stable solution (without +/- 100000 chattering). So it seems to me. This differs from the case where the system consists of, say, 10 equations and the coincidence on 10 examples from the sample is 100%, and on the next term the difference is 2000000 times...
then you have to watch, but I was not at all happy with the results of this particular method. But the idea itself to predict the next bar is excellent and "statistically secured": Do you remember my strategy described at the forum where I predicted the expected payoff and SCO of the next bar as an example. Let me remind you of a sample of trade modelling, x-axis is hours, y-axis is points gained, EURUSD, modelling result (and profit taking) is accurate though in MathCAD in the sense that if expectation and +/- RMS levels lie entirely within a bar, then profit is taken, if not, SL shows a loss:
Jokes are jokes - but it actually works. Maximum 24 trades - one every hour, but it is actually less. And I have set for each transaction "minus 9 points, for unforeseen expenses", i.e. all profitable and loss-making trades "took away" my 9 points. Well, you never know...
1. I have a hope that with a large number of coefficients in the equation, the role of the new term is smoothed out. Why is this not the case?
3. it is not an equation, it is not part of SLU, it is an equation that allows us to predict that the coefficients are changing more slowly than the price.
x[0] is the current value. We wait for the formation of today's daily candlestick and then get a forecast for the next bar - x[-1]
1. OK, here is an example. Let's take a 500 wave. But if you try to make a forecast based on a waveform, you will get a huge divergence because of the same small error.
3. What is the difference between an equality and an equation? :) a[0] cannot be seen, it is unknown.
http://forex.sunstation.com/pics/gbpjpydance.gif
The quality isn't great, but it's good enough for me.
1. OK, example. Let's take a Mach 500. Predicting is not a problem, but if you try to predict based on the forecast of the mashka -- because of that very small margin of error, you end up with a huge spread.
3. What's the difference between an equality and an equation? :) a[0] I don't see, it's unknown.
That's it. I see a clerical error. It should read like this:
а[999]*x[1000]+а[998]*x[999]+...+а[0]*x[1]=x[0]
а1000*x1001+а999*x1000+...+а1*x2=x1
.
.
а1998*x1999+а1997*x1998+...+а999*x1000=x999.
And I find coefficients A0,A1...A999. Then I substitute the current daily bar x[0] into the equation
a999*x999+a998*x998... +a0*x0=x[-1] (increment for tomorrow)
As for the stability of the solution, you are probably right. By the way, advantage of NS is exactly in that they solve essentially overpredetermined systems, i.e. those in which number of unknowns is much less than number of equations (dimension of NS inputs is less than training sample length). Exactly for this reason, solution given by NS is stable and its fraction is smaller than training sample!
http://forex.sunstation.com/pics/gbpjpydance.gif