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4 months or so it's been running... ( just as predicted)
To give you an example ... EA chart for 7 years... ( profit 10 p) stop 300 but profit floats with the price even if in loss.... Profit to drawdown ratio is about 25 for seven years... it's not much in principle... but it's possible to make about 200 annual profits.
Realistically, however, this is not the case. It's just a beautiful mirage. And there are a lot of them. This should always be borne in mind.
What is the average profit in pips? And over what period of time? Thanks for the answer :)
To give you an example ... EA chart for 7 years... ( profit 10 p) stop 300 but profit floats with the price even if in loss.... Profit to drawdown ratio is about 25 for 7 years... it's not much in principle... but it's about 200 p.a.
Realistically, however, this is not the case. It's just a beautiful mirage. And there are a lot of them. This should always be borne in mind.
What is the average profit in pips? And over what period of time? Thanks for the reply :)
4 months or so running... ( also as predicted)
No, I can tell you about it... write to clin-p@inbox.ru... In my opinion, the trader is separated from the big money not only by the absence of a good algorithm but by something else :))
At the moment I have the best results only for those optimization parameters with upward yield curve (it filters itself out if you turn off useless results) and linear correlation coefficient of this curve is closer to 1 in absolute value. I.e. the program takes one by one variants given by the optimizer, runs a test on each of them and analyzing profits and losses finds the one having the graph most similar to a straight line. Obviously, such a graph will never give the best balance of the optimization result, in most cases it is a middle one. But those yield curves that are more linear have a very small drawdown and practically do not have sharp upward moves either.
And it's not just about reducing risk. I think that the main aspect is the functionality of the TS. The absence of significant drawdowns shows the adequacy of the system, i.e. that it uses real market properties. And this in turn means that the parameters on which it is achieved are not the result of stupid fitting to history, but reflect these very properties.
By the way, alas, I don't know what the "linear correlation coefficient" is and how it differs from a simple correlation coefficient.
I would use RMS to assess the quality of the approximation of the yield curve by a linear regression line. So maybe you can explain what it is and how the linear correlation coefficient option is better than the RMS estimate ?
By the way, alas, I don't know what a "linear correlation coefficient" is and how it differs from a simple correlation coefficient.
They are the same thing. Just a play on words meaning the same concept.
PS: The statement is true of course for this case, as the participants in the conversation mean the same correlation.
By the way, alas, I don't know what a "linear correlation coefficient" is and how it differs from a simple correlation coefficient.
It's the same thing. Just a play on words meaning the same concept.
I currently have the best results only for those optimization parameters with upward yield curve (it filters itself out if you turn off useless results) and linear correlation coefficient of this curve is more close to 1 in absolute value. I.e. the program takes one by one variants given by the optimizer, runs a test on each of them and analyzing profits and losses finds the one having the graph most similar to a straight line. Obviously, such a graph will never give the best balance of the optimization result, in most cases it is a middle one. But those yield curves that are more linear have a very small drawdown and practically no excessive upward spikes as well.
And it's not just about reducing risk. I think that the main aspect is the functionality of the TS. The absence of significant drawdowns shows the adequacy of the system, i.e. that it uses real market properties. And this in turn means that the parameters on which it is achieved are not the result of stupid fitting to history, but reflect these very properties.
By the way, alas, I don't know what the "linear correlation coefficient" is and how it differs from a simple correlation coefficient.
I would use RMS to assess the quality of the approximation of the yield curve by a linear regression line. So maybe you can explain what it is and how the linear correlation coefficient option is better than the RMS estimate ?
"The degree of correlation between two quantities x and y (values of coordinates of points in the plane) can be measured by the linear correlation coefficient - r. If the value of r is close to 0, then the claim that there is a linear relationship between the quantities x and y can be rejected. If r is close to (+/-)1, then we must assume that the points lie around some line y = A*x + B. If the quantities are uncorrelated, we can calculate the probability that a random sample correlation coefficient modulo exceeds some value of r0 with sample size N. If the number of measurements is small, the probability to get a large value of correlation coefficient |r| > 0.5 may be high for uncorrelated variables."
The last two sentences say that if the yield curve was drawn with a small amount of transactions (small sample volume), it is possible that the correlation coefficient will exceed 0.5, i.e. a fitting will take place.
In fact, you can also calculate the RMS, but you should consider that the RMS for a straight line, but not for a horizontal one, will always be greater than 0. And the linear correlation coefficient for any straight line, regardless of its slope, will be 1.
In fact, you can calculate the RMS, but note that the RMS for a straight line, but not for a horizontal one, will always be greater than 0. And the linear correlation coefficient for any straight line, regardless of its slope, will be 1.
I was referring to the RMS error of a linear regression approximating the yield curve.
In this case the RMS will be 0 only if the yield line is a straight line. In this case, the result does not depend on the angle of slope of the line. In all other cases the RMS>0. In general I think the error RMS is related to the correlation coefficient (since it is the same as the linear correlation coefficient) and this relationship is not difficult to express analytically. Therefore the variants are to be assumed to be equivalent. The only difference is that the error RMS allows to estimate the drawdown at a given risk level. What do you think?
Reshetov, I am very pleased that you address me as "you". Thank you very much.
But the phrase
If you can't open a maths book and read it...
is a masterpiece! I had a lot of fun. Don't you think there's a mix of styles in it? :-))Essentially you can calculate the RMS as well, but note that the RMS for a straight but not horizontal line will always be greater than 0. And the linear correlation coefficient for any straight line, regardless of its slope, will be 1.
I was referring to the RMS of the error of the linear regression approximating the yield curve.
In this case, the RMS will be 0 only if the yield line is a straight line. In this case, the result does not depend on the angle of slope of the line. In all other cases the RMS>0. In general I think the error RMS is related to the correlation coefficient (since it is the same as the linear correlation coefficient) and this relationship is not difficult to express analytically. Therefore the variants are to be assumed to be equivalent. The only difference is that the error RMS allows to estimate the drawdown at a given risk level. What do you think?
Reshetov, I am very pleased that you call me "you". Thank you very much.
But the phrase
If you don't have the heart to open a maths handbook and read ...
is a masterpiece! I had a lot of fun. Don't you think there's a bit of a mix of styles in it? :-))I would like to know how you approximate the yield curve without calculating the linear correlation coefficient?