Optimal strategy under statistical uncertainty - unsteady markets - page 5
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HideYourRichess, well, try it out with Bernoulli... I won't scare you too much... maybe it'll work out...
PS. If Mathemat hasn't become a millionaire yet, then it's not that simple...
Shall I?! Do you want me to study with Bernoulli?! I'm embarrassed to ask, is there nothing wrong with your perception of reality? (it's a rhetorical question, you don't have to answer)
I may not know everything... But if you're already an expert on Bernoulli, then what are you asking about?
I'm embarrassed to ask... Why does a skewed system have to be non-bernullian? How can you be so sure?
_____________
Imho, it's time to call Mathemat.
I may not know everything... but if you're already an expert on Bernoulli, then what are you asking?
No offence, comrade, have a snack or a drink. Or speak clearly.
I'm embarrassed to ask... Why does a skewed system have to be non-bernullian? How can you be so sure?
_____________
Imho, it's time to call Mathemat.
yeah that's the question right there... that's the point, it doesn't have to be non-bernoullium... as far as I'm concerned, the Bernoullianness can only be estimated... with a reasonable degree of approximation...
Comrade, no offence - have a snack! Or sleep it off. >> or write it down.
What's the problem, comrade? I don't get it.
So the difficulty is about the same as finding a profitable strategy in general.
Not really, although it's not all that trivial.
I found it much easier, i.e. I experimented with the code of ready-made TS and by mistake did not remove one of the conditions. I ran the test. The balance is growing. Profit is not large, but more or less stable. I have tested it using deeper history. It still grows. On other symbols and timeframes. Growth again.
The first thought was that it was another grail on the tester glitches (I have discovered similar ones before). I started to re-check it on separate trades. I was not able to detect any differences. I went to the code. Something there is not what was intended by the algorithm. I began to sort it out. It turned out to be a Shannon algorithm. I remembered that I'd read about it somewhere before.
In short, some trading strategies have properties of a wrong coin, i.e. they change from one stationary state to another and these stationary states have a decent duration. As a result, the TS itself turns out to be non-stationary. But the point is that it abruptly sinks in one state and profits in the other one. Since it is almost impossible to calculate the exact moment of switching from one state to another (as well as to determine the moments of switching from a flat condition to a trend and back), we can only profit from Shannon's algorithm. It's not much, but it makes money.
.... then you can only make money on the Shannon algorithm. It's not much, but it's a lot of money.
I wonder how it is possible to make money on an information compression algorithm. https://ru.wikipedia.org/wiki/%D0%90%D0%BB%D0%B3%D0%BE%D1%80%D0%B8%D1%82%D0%BC_%D0%A8%D0%B5%D0%BD%D0%BD%D0%BE%D0%BD%D0%B0_%E2%80%94_%D0%A4%D0%B0%D0%BD%D0%BE
Unless, of course, you're trading it.
Let's make the problem even simpler, let us have a wrong coin (wrong means that one side is struck more often than the other). We don't know in advance which side is more frequent and with what exact probability, but we know for sure that the coin is wrong.
Under the terms, it is necessary to create a profitable betting system, which does not allow to calculate statistically the advantage of one of the sides of the coin, and therefore its algorithm should be built on the knowledge of only two parameters:
1. The number of the next flip.
2. The side of the coin, which was struck in the previous flip.
It is possible to bet on either side of a coin before the next flip. It is possible to skip a particular coin toss, i.e., not bet, i.e., the bet amount is 0. It is possible to increase or decrease bets.
We know for a fact that it is a sandwich toss. The probability of one side falling out is p, the other q = 1 - p. Bernoulli's scheme.
I have this strong intuitive feeling that skipping deals in Bernoulli's scheme does not statistically change it in any way. It will still be the same Bernoulli scheme with the same probabilities. The reason is that deals are independent of history.
Expectation of a deal when its reward is equal to its loss and the value of the deal is constant, is in any case not equal to zero:
| p * M + ( 1 - p ) * (- M ) | = | ( 2 * p - 1 ) * M | # 0
So whether or not we know p > 0.5 or vice versa, it's still not a martingale. Varying the size of the bets... I don't know what it can do yet - but it's unlikely to change anything in terms of m.o. sign either.
2 PapaYozh:
Here's a simple sequence of heads and tails for you: ORORORORORORORORORORORORORO
That is, we have: 20 events of which 9 are heads and 11 are tails
I hope you will not deny that there is a statistical advantage of "tails" over "heads".
No statistical advantage of 11 over 9 in a series of only 20 trials is out of the question. It is just a very small deviation of frequency from probability - even if the coin is correct.
I wonder how it is possible to make money on an information compression algorithm.
Unless, of course, you're trading it.
Mm-hmm. You might also ask how other C. Shannon algorithms, like diffusion and confusion for cryptography or his algorithm for a computer game of chess, can be used to make dough.