Machine learning in trading: theory, models, practice and algo-trading - page 3197

 
Maxim Dmitrievsky #:

Randomly shuffle the symbol n times, see the average of the quatrains (which are the best).

As I understand, you need to take the increments (OHLC) and randomly shuffle the values? Or take, say, and divide the increments into 100 windows and mix the windows? I've heard somewhere that time series can only be mixed with windows.....

I haven't dealt with custom characters - do you have a script for this purpose?

Maxim Dmitrievsky #:

If the average is not the best, then there is nothing to lose there.

Very rambling - I don't understand what "average" means and not better than what? It can't be better than the original - it limits the search area, so if the average value is close to the original, then "there is nothing to lose there", in other words, it would mean that quantum segments could be randomly selected, i.e. the criterion for their selection is not good enough?

 
Aleksey Vyazmikin #:

As I understand it, should we take the increments (OHLC) and randomise the values? Or to take, say, and divide the increments into 100 windows and mix the windows? I've heard somewhere that time series can only be interleaved with windows....

I have not dealt with custom characters - do you have a script for this purpose?

Very rambling - I don't understand what it means in "average" and not better than what? It can't be better than the original - it limits the search area, so if the average is close to the original then "nothing to lose there", in other words, it would mean that quantum segments could have been selected randomly, i.e. the criterion for their selection is not good enough?

You no longer have the original. And a shuffled symbol. The increments, yes.

on each such symbol, you determine the best quadrangle, then look at the average of the best quadrangle across all simulations.

if I understand correctly what a quatrain is. If it's a range of chip values, then everything fits.

I don't do anything in the terminal, I only compile bots in it
 
Maxim Dmitrievsky #:

You no longer have the original. It's a shuffled symbol. The increments, yes.

on each such symbol, you determine the best quadratic, look at the average of all simulations.

I don't do anything in the terminal, I just compile bots into it.

It's not clear - one will obviously be better than the others purely by chance will fall out more often (even if only twice) - what is the point?

For simplification we can take one quantum table.

As a result, we will get that on predictor 1 more often falls out by 5 quantum pieces, for predictor 2 by 3, and so on.

It should be realised that the cutoffs for each predictor are different.

And, how do you want to summarise all this into one averaging for any conclusion?

 
Aleksey Vyazmikin #:

It's not clear - one will obviously be better than the others purely by chance will fall out more often (even if only twice) - what's the point?

For simplification we can take one quantum table.

As a result, we will get that on predictor 1 quantum more often falls out by 5 quantum pieces, for predictor 2 by 3, and so on.

It should be realised that the cutoffs for each predictor are different.

And, how do you want to summarise all this into one averaging for any conclusion?

Each feature has its own average. How many of these segments are there? There should be more multiples of simulations.

It's all solvable and it's a technical issue, not a matter of principle.

Purely by chance 10k times it will fall out, if the graph is really random. If not, the segment that contains regularities will be found.

 
Maxim Dmitrievsky #:

each chip has its own average. How many of these segments are there in total? There should be more simulations in multiples.

This is all solvable and is already a technical issue, not a matter of principle.

The number of segments for each predictor will be different - I will take a summary based on the results of selection on the original sample - in fact, it is not important for randomisation. In general, I have 900 tables, but it will be excessive to do table accounting here.

 
Aleksey Vyazmikin #:

The number of segments for each predictor will be different - I will take a summary based on the results of selection on the original sample - in fact, it does not matter for randomisation. In general, I have 900 tables, but it will be redundant to do table accounting here.

I just wrote how you can estimate your segments with monte-carla.

I don't know any other way.
 
Maxim Dmitrievsky #:

I just wrote how you can estimate your sections with monte carla.

Well, if you're off already, then again wasting time and getting an answer on results in Alexey's style is not motivating enough.

 
Aleksey Vyazmikin #:

Well, if you're off already, then wasting time again and getting an answer on results in Alexey's style is not motivating enough.

because if you don't understand what you're doing, it's better not to do it at all and not to torture others )

and there is nothing to do there, everything is simple

 
mytarmailS #:

And what are the advantages of this package vs. other options?

I haven't met any other options to open R session and send requests from MT5 to it.

 
Maxim Dmitrievsky #:

because if you don't know what you're doing, it's better not to do it at all, rather than torturing others )

and there's nothing to do, it's simple.

Where do I not understand? I understand that not everything is not so simple and I write about the details, for coordination.

In general, when I help a person in anything, it is important to me that he would have a positive result - I thought that I was not unique before, but now I have doubts....

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