Optimal strategy under statistical uncertainty - unsteady markets - page 2
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Yes, if there is an edge, then the right side will fall out more often.
Hence we have a similar algorithm for trading systems of which we do not know in advance if they are losing or profitable, but we know in advance that they will either lose more of the spread or
they will profit
- If the previous trade signal resulted in a loss, the next position must be opened against the previous interpretation of the trade signal
- If previous trade signal resulted in profit, the next position shall be opened against the previous trade signal interpretation.
Hence we have a similar algorithm for trading systems of which we do not know in advance whether they are losing or profitable, but we know in advance that they will either lose more of the spread or take a profit.
- If the previous trade signal resulted in a loss, the next position should be opened against the previous trade signal interpretation
- If the previous trade signal resulted in a profit, the next position must be opened in the same way as in the previous interpretation of the trade signal
There are nuances, namely, the size of the spread. The point is that the smaller the spread, the smaller the profit on such a strategy will be. Therefore, even the spread may be insufficient.
So better still to test on history and identify the side :)
The yield is very low. But it is positive, yes.
There are nuances -- namely the size of the spread.
it's got nothing to do with spread size... he'll make a bet on the whole deposit and he'll get out of the market... because it's easy to fall 2 or 3 times in a row on the "not heavy" side... depending on how un-heavy it is...
There are nuances -- namely the size of the spread. The point is that profits from such a strategy will be smaller the smaller the spread is. That is why we may not have enough energy even for the spread.
So it's better to test it on history and define the side :)
I mentioned about spread.
And what about testing on history, we've seen many times that the test gives one thing, and then the market turns and the opposite happens. For example, TS for a rebound from channels. We have tested it on history, obtained an amazing profit. When we set it on the real account, the sideways loop finishes, the trends start reversing and we need to trade on the opposite direction, i.e. channel breakthrough because the system fucks up.
Thus, we do not know in advance how to interpret signals unambiguously in order to maximize the probability of response.
As for the fact that expectation of profitability will be lower than that of a fit, it is understandable: any hedge will always give much less profitability in exchange for more reliability. Therefore, it is better to make less, but in profit (and sometimes even not to get into a drawdown, but to hold on to zero and that is the result).
This results in a similar algorithm for trading systems of which we do not know in advance whether they are losing or profitable, but we know in advance that they will either lose more than the spread, or
they will profit
- If the previous trade signal resulted in a loss, the next position must be opened against the previous interpretation of the trade signal
- If previous trade signal gave profit, next position should be opened against the previous trade signal interpretation
Here is a simple sequence of "heads" and "tails": ORORORORORORORORORORORO
That is, we have: 20 events, 9 of which are heads and 11 are tails.
I hope you will not deny the existing statistical advantage of "tails" over "heads". All that remains is to bet according to the suggested system and make sure of its unprofitability.
The guessing probability in this system, when the next bet is made on the previous one, is p^2+q^2, for example, if we have a coin bent 2/3, i.e. probability 0.66, then the resulting probability is 0.55.
You can calculate the expectation from there.
Model experiments have shown results close to this.
And by the way, it doesn't matter which way the coin is bent. The important thing is that the odds are the same.
And by the way, it doesn't matter which way the coin is bent.
The main thing is that it must necessarily be bent, i.e. wrong. The curvier it is, the higher the expectation.
The main thing is that it must necessarily be bent, i.e. wrong. The more crooked it is, the higher the expectation.
Yes, at least according to the formulas this is what it turns out to be.
Yes, at least according to the formulas that's how it works.
p^2 + q^2 = p ^ 2 + (1 - p)^2 = p^2 + 1 - 2*p + p^2 = 1 + 2 * p * (p - 1) = 1 - 2 * p * (1 - p)
I.e. at p equal to 1 or 0 we get the probability of winning 1 - absolutely winning no matter which side will fall with 100% guarantee. Lowest probability is 0.5 when p = q = 0.5, i.e. if the coin is perfectly right, the game turns into martingale and expectation is 0.
In all cases the strategy is win-win, because in the worst single case we simply have nothing and lose nothing, while in all others we gain profit.