Machine learning in trading: theory, models, practice and algo-trading - page 2569
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Who knows if there is a test for trend determinism...
Tests for different kinds of trends out of the box
R best!!!!
There is a good trend package in R. For a linear trend, sens.slope() is good from there.
What is there to understand in financial mathematics and IR, you need to know the mechanics of the market and its players
The crowd is bound to lose in most cases, because its counter-agent is a "major player".
1) it is necessary to see the imbalance of retail buyers and sellers, for example if there are many sellers, then on the other side of the deal there is a "major player" (the buyer)
For example now on the Jew, there are a lot of sellers
2) There is also trading in the moment against the crowd - this is a market maker
You can see that the price always moves against the crowd (inverse correlation).
While the crowd is buying and believes in growth, the price will fall and vice versa...
That's the whole market.
p.s. And I'll watch the video for sure.
There are problems with it too, and not a small one. We have only a small slice of the market, which is not known how representative and reliable. The second picture (the total range) is useless. First, reversals regularly occur at extremums where it would seem that the trend "against the crowd" should have continued. Secondly, the range simply repeats the price (upside down and not trending) - invert any oscillator and you get a very similar picture. Well, if the indicator repeats the price, then it is no easier to forecast it than the price chart itself, right? The first picture (the glass and open positions) and the value of profit ratio are more valuable, but it's also so... ...is highly questionable.
It does not make much difference (in terms of combinatorics) how exactly it is encoded. The essence is the same - each line has as signs which rules are executed and which are not. This is always 2^N choices, where N is the number of rules. We then choose whether each of these rules is included or not in the final set, which is 2^(2^N) variants. It is clear that it is simply unrealistic to formally enumerate such a set of variants. That's why it makes sense to arrange them in a reasonable way. For example, first we take all variants that are described by only one rule, then by only two, etc. Or something like that.
Maybe I'm not explaining it well, but I want to emphasize again that we're working with an initial sample, which within the CatBoost learning algorithm is quantized once, i.e. let's say we have 1000 predictors, we quantized them each into 10 segments, as a result we have 10000 segments - further the CatBoost algorithm randomly pulls predictors and running through the boundaries/quanta does a split/split, to select the quantum which is in the middle it needs to build 3 inequalities or create 8 leaves, with 7 leaves talking about garbage and one leaf about valuable information. Whether the algorithm will build such a tree is also not a fact, since there is a love of randomness and the neighboring quantum can get into the leaf. And let it be only one tree. In my method one inequality will be enough - 2 leaves. Even if all predictors are used and each of them has 1 useful segment, the number of combinations when creating the model will be much less. But importantly, there will also be less noise, as there will not be so many erroneous partitions of samples to stabilize which you need to build a new tree (boosting principle).
Sooner or later many other players will find them, for example.
Then maybe we should somehow test this theory? According to which there is a minimum frequency of recurrence of advantage, after the increase of which there is a higher probability of its disappearance than continuation of existence?
For example we measure every half year, there is an advantage of 1,5 years - we believe that half a year will still work, and if it is already 2,5 years, there is no sense and take it - the probability is higher that it will stop working.
I'm not strong in scientific formulas, can you tell me how to make such a calculation and to check the hypothesis?
R has a nice trend package. For a linear trend, sens.slope() is good from there.
thanks
There are problems with that too, and quite a few. We only have a small slice of the market available to us, which we don't know how indicative and reliable it is. The second picture (overall sentiment) is not useful at all. First, reversals regularly occur at extremums where it would seem that the trend "against the crowd" should have continued. Secondly, the range simply repeats the price (upside down and not trending) - invert any oscillator and you get a very similar picture. And if the indicator repeats the price, it means that predicting it is not easier than predicting the price, right? The first picture (the glass and open positions) and the value of profit ratio are more valuable, but it's also so... is highly questionable.
You're getting lost in many things, you perceive the information too superficially and thoughtlessly.
Perhaps I'm not explaining it well, but I want to emphasize again that we are working with the original sample, which within the CatBoost learning algorithm is quantized once, i.e. let's say we have 1000 predictors, we quantized them each into 10 segments, as a result we have 10000 segments - further the CatBoost algorithm randomly pulls predictors and running through the boundaries/quanta does a split/split, to select the quantum that is in the middle it needs to build 3 inequalities or create 8 leaves, with 7 leaves talking about garbage and one leaf about valuable information. Whether the algorithm will build such a tree is also not a fact, since there is a love of randomness and the neighboring quantum can get into the leaf. And let it be only one tree. In my method one inequality will be enough - 2 leaves. Even if all the predictors are used and each has 1 useful cutoff, the number of combinations when creating the model will be much smaller. But importantly, there would also be less noise, since there wouldn't be as many erroneous sample partitions to stabilize which would require building a new tree (boosting principle).
Perhaps you should describe your approach on the example of a single decision tree. Still, by itself, the boosting already builds quite a complex structure - a refining sequence of trees. And youdo something else with it).
Then maybe we need to somehow check this theory? According to which there is a minimum frequency of recurrence of the advantage, after increasing which there is more likely its disappearance than continuation of existence?
For example, we measure every half year, there is an advantage of 1.5 years - we believe that half a year will still work, and if it is already 2.5 years, there is no sense and take it - the probability is higher that it will stop working.
I am not strong in scientific formulas, can you tell me how to make such calculation and check the hypothesis?
Something like radioactivity formula with half-life) I do not think that human activity is described by such simple formulas) Your "frequency of advantage" will change, most likely, somehow very unpredictable.
Perhaps you should describe your approach on the example of a single solver tree. After all, the boosting itself already builds quite a complex structure - a refining sequence of trees. And youstill do something else with it.)
Yes, I understand that it's not easy to immediately understand the method. I plan to describe it in more detail. And the data is processed before submitting it to CatBoost, but I want to make my own algorithm, which will take into account the changing dynamics of "decay", as you called it, when splitting.
Somewhat like a formula for radioactivity with half-life) I don't think people's activities are described by any such simple formula) Your "frequency of advantage" will probably change in some very unpredictable way.
Maybe there's nothing, but how do you check it?
thanks
You're getting lost in a lot of things, you're too superficial and unthinking about this information.
Yeah, I've just been working with this tool for years, but I've never really thought about it.
What exactly do you disagree with?
Yes, let's say so.
No, now we take the execution of the inequality rule as one and look at the average value of the target (let's say for binary classification) when the rule triggers on the sample, if the initial average value, say, 0.45 in the sample, and after evaluation only by responses became 0.51, then we believe that the predictor (its plot/quantum) has a predictive power of 0.06, i.e. 6%.
Let's gather a set of such predictors with sections, which are already independent binary predictors, and use them to build a model.
Combining all such quanta with all (with or without predictive power) is really not a quick matter, but may not be unreasonable if done with the base predictor on which the quantum with predictive power is identified.
But, even in theory, this retraining will be less, since there are fewer possible combinations than there were in the full sample.
It remains to be understood why such quantum patterns can work for 7 years and then suddenly stop...
1) We get 1 tree.
2) Each node can give up to 10 branches (in the picture we found less, let's say 10 branches), each branch is generated by 1 quantum (a quantum is a piece of predictor in 10%: either percentile or 10% in amplitude, depending on which method of quantization was used).
3) after the first split, we find 3 quanta, which subsequently lead to a successful leaf
4) subsequent splits find several more good splits/quanta leading to successful leaves
5) remember successful quanta before successful leaves
6) build a new tree that uses as predictors only the quanta that we selected
To do it by the same method as we quantized the first tree, we quantize predictors with our script, we get 1000 out of 100 predictors, they are already binary 0 or 1. If the value of the predictor is in this range, it = 1, if not, it = 0.
Since we are selecting only successful paths/quanta, all values of the selected quanta = 1. If all predictors = 1, then the new tree cannot learn. The answer is already known.
Or is there no need to build a new tree anymore? If the value of the predictor falls into the selected quantum, can we do that?
Well, yeah, it's just that I've been working with this tool for years, so I haven't really thought about it at all.
What exactly do you disagree with?
What tool exactly?
with what kind of tool?
Well, what's in the picture? The glass and the sanction.