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Alexander, in the machine learning thread they gave a link to https://smart-lab.ru/blog/499678.php where they write that "Somehow it is not too strict and almost without formulas to "prove" that price increments on large timeframes are non-stationary normal". I remember you were just looking for normality.
Alexander is missing something, he's not getting in touch. Probably disappointed...... or maybe he is working on a new idea.
Alexander is missing something, he does not get in touch. Apparently disappointed ...... or maybe he is working on a new idea.
Ain moment... Working, of course. I promised the Grail to the sufferers - I'm used to keeping my promises.
Eugene, do you know anything about the MT tester?
Because I am tired of checking everything manually...
The algorithm is as follows (modification of Koldun's algorithm):
1. count the sum of increments in sliding window = 24 hours (1440 values of returnees CLOSE M1, see attached files)
2. we count variance = 2.9814*(SUM(ABS(return))/SQRT(1440))
3. calculate simple MA of this sum of increments in the sliding window.
4. enter the trade not immediately upon crossing the upper limit of the channel by the sum of increments, but as soon as the MA>0
5. exit when the sum of increments <0
6. entry into buy trade not immediately after crossing the lower boundary of the channel by the sum of increments, but with MA<0
7. exit at the increment sum >0
I have the wildest trash in the form of profits and gold on the history.
The results can be found here.
It should go like this:
where:
black is the sum of the increments in the sliding window = 24 hours.
blue - lower limit of dispersion
red - upper limit of dispersion
green - moving MA(1440)
Ain moment... Working, of course. I promised the Grail to the sufferers - I'm used to keeping my promises.
Eugene, do you know anything about the MT tester?
Because I am tired of checking everything manually...
The algorithm is as follows (modification of Koldun's algorithm):
1. Consider the sum of increments in the sliding window = 24 hours (1440 values of CLOSE M1 returnees, see attached files)
...
Window either less than 24 hours or more, e.g. 20/28(30)
The window is either less than 24 hours or more, e.g. 20/28(30)
Why?
To be honest, I still can't give a 100% recommendation on window size...
I only have 2 arguments in defence of exactly 24 hours:
1. it has a pseudo-Poisson flow of tick quotes
2. it was used successfully by old Gunn.
That's it.
2. Consider variance = 2.9814*(SUM(ABS(return)/SQRT(1440))
I don't understand a bit about the brackets. Should I divide the sum of the increments by the root of t or should I calculate the sum of the sums of increments/t ?
2. Consider variance = 2.9814*(SUM(ABS(return)/SQRT(1440))
I do not understand a bit about brackets. I need to divide sum of increments by root of t or to calculate sum from sum of increments/t ?
:))) I will correct it now.
Everything is as before, nothing new - only the condition of crossing МА(1440) zero when entering a trade is added.
It is now necessary to explain why the quantile = 2.9814 should be used.
Let's look at the distribution formed by the sum of the increments in the sliding window = 24 hours for EURUSD for 2017:
Its statistical characteristics:
We see that it is a VERY normal distribution. But.... Because of the correlation between the values, Lyapunov's TSP is not fulfilled... Well, let's just say it's a little bit off.
So what?!
We have the Petunin-Vysokovsky inequality for unimodal distributions, which states that 95% of all distribution values lie in the interval +-2.9814*sigma.
Given that, on the average, with a large number of measurements, we have a negative correlation for any pair in the sliding window = 24 hours, which guarantees a return to the expectation - conclude deals when going beyond the specified range and go, Vasya...
That's how algorithms should be written, children! Not to suffer from infirmity and emptiness in their pockets.
What is the distribution in the market? Maybe it's not open yet, one modal or two or the other.....
Maybe I'm being silly now, but it seems to me that there should be two mat expectations in this distribution.