Market model: constant throughput - page 24
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You can compress the dubbles too, not the point. For the experiment in question it didn't matter, the result when compressing the difference in integer form would be pretty much the same. (In relative form, not absolute.)
From here:
Над этим пришлось поэкспериментировать. Сначала брал относительные изменения цен. Архиватор сжимал на 30%.
Then I tried absolute price changes (in pips) - the archiver started compressing by 60%.
Further it turned out that if I locate data on financial instruments not by lines, but sequentially (first one, then the second one, etc.), then the compression is increased by another 5%. I didn't experiment any further.
It was important to get not only the minimum representation of data, but also the maximum compression.
Hence:
It was important to get not only the minimum representation of the data, but also the maximum compression.
It's like that joke, "I can do a hundred and twenty experiments an hour, but it's rubbish..."
What did knowing the maximum compression rate do for you?
You rearrange the numbers and the compression got better, you know, like less information...
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Crosses would be worse compressed because they have a larger spread and are not as filtered as majors. You would think there would be more information in the crosses than in the majors.
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You ponderous fucking pilots...
I read it.
Read it standing up.
Now searching for the inner Pinocchio.
What did knowing the maximum compression rate do for you?
I rearranged the numbers and the compression got better, like there was less information...
"Information" is not made less or more by compression. The better the compression, the more accurately you can estimate the amount of "information" contained in the data. But this is still a very rough estimate.
Crosses will be worse compressed because they have a larger spread and are not as filtered as majors. According to your ideas it would appear that there is more information in crosses than in majors.
I did not compress the crosses, because it would not make sense, because all "information" about the crosses is already contained in the "majors". Only the "majors" were compressed.
"Information" is not made less or more by compression. The better the compression, the more accurately the amount of information contained in the data can be estimated. But this is still a very rough estimate.
If shuffling changes the estimate, what are you estimating? Obviously you are evaluating the arrangement itself.
The best practical conclusion from your experiments is what order you should compress the pairs with the archiver.
If shuffling changes the assessment, what are you assessing? Obviously you're evaluating the formation itself. It doesn't matter what the fuck you're evaluating and what you're taking it for, but why are you rubbing it in that you're evaluating the amount of information? Where in your experiments is even a hint of that?
The smaller the compressed data takes up, the closer it is to the minimum set. Is there something unclear here? It is impossible to outrun the minimal set. It can only be approached in a more accurate estimation.
The smaller the compressed data takes up, the closer it is to a minimum set.
It's pretty clear here, it's just plain stupid.
It's pretty clear, it's just plain stupid.
What's this:
Если от перестановки меняется оценка, то что ты оцениваешь? Очевидно ты оцениваешь саму расстановку.
???
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P.S. As it is difficult to absorb, I am writing directly: this is the clarification(!) You are evaluating anything but information.