Not the Grail, just a regular one - Bablokos!!! - page 11
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Warm greetings to the high assembly!
As promised to the author of the thread, I am posting a mathematical proof of the possibility of profitable Forex trading.
However, since the last post, the idea that such a proof has existed for a long time has come to mind. It's martingale! The system of the game, proved mathematically strictly a long time ago, and it is not the business of mathematics to delve into the fact that the dealer or the owner of the casino limits the bets from above and below, depriving players of the opportunity to use the martingale to the fullest. Even if they have enough money to play martingale...
But since I promised, I will have to, especially since the system still takes into account the peculiarities of Forex.
To begin with, consider the nature of the movement of the exchange rate within the hour. For the order to work, it is necessary that the maximum deviation value is not less than the set order. Therefore, we are interested in the probability distribution of the maximum hourly value of the exchange rate. It is easy to obtain such a distribution in the form of a histogram if we take hourly exchange rate bars for a sufficiently long period, count all bars of the same height, and arrange the resulting dropout frequencies according to the value of the bar. Such a histogram is shown in Fig.1. The abscissa shows the size of the bar (High – Open), and the ordinate shows the number of such bars for the period under study. Unfortunately, I don't remember which currency the histogram was calculated for and for what period. Most likely for EUR for the period from December 16, 1998 to, approximately, April of this year. Although, in the end, this is not important for the proof, since the nature of this distribution is almost the same for all currency pairs and differs only in specific numerical parameters.
Picture 1.
If you look closely at the histogram, you will notice that the distribution is very similar to the binomial distribution as N tends to infinity. The limiting case of the binomial distribution of a discrete random variable with N equal to infinity is the exponential distribution of a continuous random variable. Since we do not know what maximum value the size of an hourly bar can take in principle, we have the right to assume that this value is not limited by anything and use the exponential distribution law. Such a replacement is quite justified, because. the formulas describing the binomial and exponential distributions differ in complexity like "a locomotive from a bicycle". The exponential distribution -
p(x) = λ*exp(-λ*x)
it's just an exponent, which, after integration and after differentiation, remains the same exponent. Handy little thing.
In addition, both laws are derived from the assumption that the random variable is independent of history. In other words, they characterize absolutely unpredictable processes. And, if we approximate the existing statistical distribution - exponential, then, thereby, we will already consider the process, the forecast on which is impossible, i.e. Markovsky.
Figure 2 shows: the normalized statistical distribution of the currency pair (presumably EUR/USD) in brown, and the exponential distribution approximating it in blue.
Figure 2.
The figure shows that the maximum deviation of the statistical distribution from the exponential one is concentrated in the region of small values, up to about 13 points. In the region of larger values, the coincidence is almost complete, and in the region of “very large values”, the distribution densities again diverge, because the statistical one simply ends, and the exponential lasts “forever”.
Since the degree and area of deviation of the statistical distribution from the “unpredictable” exponential one characterizes the degree of exchange rate predictability, it can be concluded that the exchange rate predictability, regardless of the forecast method, is very, very low, almost none. Except for very small values (to the delight of pipsers) and very large values. Those. we can confidently predict that a stop order placed at a distance of, say, eight figures from the current price, within the next hour, the price will not reach ...
And where should the “poor” trader go? The forecast is impossible, but I want a denyushka!
Let's consider the equation of the mathematical expectation of the profitability of the trading system:
M(sys) = M(T) – M(L),
where M(T) – profit expectation;
M(L) – loss expectation.
It is known that the mathematical expectation of a random variable can be calculated as the product of this value and its probability, i.e.
M(x) = x * p(x), then
M(sys) = (T - S) * p(T) - (L + S) * p(L),
where T is the value of the profit order;
L is the size of the stop order;
S – spread value;
p(T) – probability of triggering a take profit order;
p(L) – probability of triggering a side-loss order.
Slightly transform the original equation:
M(sys) = T* p(T) – L * p(L) – S * (p(T) + p(L))
and taking into account the fact that p(T) + p(L) is a complete group of events, i.e. equals 1, because we will stand “until blue in the face” until either the stop or the profit works. Finally:
M(sys) = T* p(T) – L * p(L) – S or
M(sys) = T* p(T) – L * (1 - p(T)) – S(1)
It remains only to calculate p(T) and, we have a win-win system in our pocket ...
Now it's time to look at the exponential distribution again.
Figure 3
Figure 3 shows orders: profit - point A, and stop - point B. The projections of these points on the abscissa axis are equal to the value of the placed order, and on the ordinate axis - the probability of its triggering. In accordance with the formula for calculating the mathematical expectation, the area of the formed rectangles is equal to the mathematical expectation of the corresponding order. Red - profit, blue - stop, green - spread. It remains only to decide whether there is a maximum for these rectangles and bubble take profit there.
I have already said that there is a common opinion that it does not matter what the size of the stop and profit orders, because. the larger the order size, the less likely it is to trigger and vice versa, and as a result, we get neither gain nor loss from varying the order size.
Even the author of the thread in one place said this:
Quote: Message from M. Jobbaryannik
Indeed, if the profit is shorter than the stop, then it starts to work more often, but at the same time it is necessary that the position be oriented towards the highest probability of movement, otherwise a large stop will appear behind a series of small profits, which will destroy all profits...
, and in the other, like this:
Quote: Message from M. Jobbaryannik
It seems to me that the statement about the presence of goals larger than the loss is not enough.
You can check it in the following way - test the system with random entries where the size of the expected profit is 2-3 times greater than the size of the expected loss.
However, tests of such a system show a sure minus, because if the loss is shorter than the profit, then, according to statistics, it will work more often than the profit.
You would decide, finally, what is better "yesterday, five - but large, or today three - but small." (c) M. Zhvanetsky
The reality, however, is not as terrible as they think, because if the area of the inscribed rectangle (Fig. 3) is constant
x * y = Const - then this is the equation of a hyperbola.
And there is no hyperbolic distribution, because the graph of the probability density of a random variable, although it can have any shape, as fate wills, there is one indispensable condition: the integral of this graph must be equal to one. A hyperbola has an integral equal to infinity. Moreover, all smooth curves with a curvature greater than a hyperbola have a minimum area of the inscribed rectangle in the middle with an increase in its edges, and with a smaller curvature - a maximum in the center and a decrease in the edges.
Well, actually, the proof can be considered almost complete. It remains only to differentiate the distribution density of the exponential law, equate it to zero, solve the equation and obtain the naturally expected value:
T(opt) = 1/ λ .
But, this decision does not suit us, because. we agreed to hold the orders “until blue in the face” until they work, and we calculate the probability for a work within an hour. This will not work! To get the correct solution, you need to go to the probabilities of triggering orders without taking into account the time - until they work.
In my workbook, the derivation of these formulas takes more than three pages of "juggling with hieroglyphs", so I will not give the derivation here. But, I will tell you the way, for those who want to do it on their own. It is necessary to make a recursive expression for the probability of the order triggering, assuming that it did not work during the previous hour. As a result, we get a geometric progression, the sum of which is calculated. After calculating this amount, the following order trigger probability formulas should be obtained:
p(T) = (p(t) * q(l))/(1 - q(t)*q(l) – p(t)*p(l));
where
q(t) = 1 – p(t),
q(l) = 1 – p(l);
and finally
p(t) = exp(-λ*T), p(l) = exp(-λ*L).
Now we can substitute the obtained formulas into the formula (1) of the expectation of the system and, to find the solution, take partial derivatives with respect to T and with respect to L. Equating both equations obtained to zero, we find that the resulting system of equations has no solution in an analytical form. She has no solution at all! And this is natural, because. with an exponential distribution, the most profitable solution, from the point of view of the maximum profit of the system, lies in the stop-loss area equal to infinity. But we don't need it!
We then know that the real, statistical distribution is limited and does not extend to infinity - hence the solution exists, but it must be sought by numerical methods. Well, now we can consider the proof complete. I do not present the resulting chart, according to the refined formulas, because the nature of the probability curve for the order triggering has not changed, but only the specific digital expression of the curve has changed, which we do not need, since the solution still needs to be sought by numerical methods. Yes, and this image does not look so beautiful, as it should be depicted by a surface in space.
M(sys) = f(T, S).
Findings:
1. The possibility of profitable Forex trading without the use of predictive methods has been proven. To do this, it is necessary to set the take profit approximately in the area of the mathematical expectation of the probabilistic law of distribution of the used currency pair and stop loss or in the area of sufficiently large values, where the statistical distribution of the currency pair ends, or in the area of small values. In this case, the direction of the opened position does not matter. The second version of the system (with a short stop) is perhaps more interesting, because. the variance of the system is very high and I don't think that anyone will have enough deposit to survive her turbulence. However, for those “who are not interested in profit” this is not important ...
2. Analysis of Fig. 3 in the area of small take profit values shows that pipsing systems have a “strongly negative” profit expectation (on the mountain for pipsers). Indeed, if we look at the red rectangle and mentally direct point A to the origin, we will see that the difference between the areas of the red and green rectangles tends to zero, i.e. profit tends to zero. But the loss, no matter how small we make the stop loss, does not tend to zero, because. it is equal to the sum of the areas of the blue and green rectangles. Now it is clear what the myth about the high profitability of pipsing is based on: the predictability of the exchange rate in the area of small values. But in summary, we can say that a pipser needs: a powerful mind (for forecasting), nimble hands (to quickly enter and exit even faster), and a VERY friendly dealer, because. even by accidentally sneezing at the monitor, the dealer can brush off a whole flock of pipsers from the market...
3. I want to immediately warn those who like to criticize indicators and TA, so that they do not refer to me for allegedly proving the unpredictability of the exchange rate. The exchange rate is really unpredictable, not by any means, even with neural networks, even with digital filters, even with the Caterpillar, even with astrology, but (!) Only in the area of 15-150 points from the current price. In the area of more than 100-150 points, the statistical distribution and the exponential distribution diverge again and the rate predictability increases. If we take the statistical distribution of non-hourly, let's say, daily and more bars, then the distribution there is not at all similar to the exponential one and is much more accurately approximated by the Cauchy distribution. And show me a competent analyst who would draw trends within the day? If "someone" is looking for a divergence of three to five hourly bars; advises to exit on ten-minute MACD; Yes, at the same time, he also recommends not to put stops when working out gaps (!), And when he is hinted at a resemblance to Vasya Pupkin, he does not understand the comparison point-blank; it’s not surprising that then branches appear with names like: “so-and-so is a scammer!”.
Name the formula (19)
Name the formula (19)
Please clean up after yourself comrade. It wasn't my post, no need to take it out of context, it was a response to my post in which I provided a link to the author. just copied that entire post here for convenience.
Not to make someone believe. It is more fun to discuss from different angles.
Everything you've written here is cheap trolling - I haven't seen a single word from you that says your system can generate such profits - just an incoherent stream of words and boorishness. And if you want someone to believe in your fantasies - provide them with the password to the account or connect the account to monitor the same
You're out of luck :-) Over 350 posts with nothing but trolling have gone down the drain... at the whim of the local moderators... by deleting 300 pages of threads... so... you won't get anything...
I will speak in terms of intelligible information about TC :-) but I will speak in gibberish (with grains of information) :-) for that is MY WILL :-)
nope - it's simpler here... We make a synthetic tool (total equity) of the form we need - where we can easily apply simple lot management tricks...
Well, I'll figure it out, too,
And you can put it in an EA, let it pick pairs, lots, for the right form + maximum variation + all pairs (not just majors) ;)
Alekzundera, have you tried more than 2 pairs in synthetics?
hmmm... i seem to be showing synthetics of 4 pairs in this thread :-) - there are 8 pairs in analysis, 4 of them going into bidding....
yeah... by the way... today's text block :-) almost forgot :-)
---
xxxxxx7902 d6 t6 buy 0.15 p1 1.3613 0.0000 0.0000 d6 t6.1 1.3581 0.00 0.00 0.00 -48.00
xxxxxx7903 d6 t6 sell 0.43 p3 0.9891 0.0000 0.0000 d6 t6.1 0.9852 0.00 0.00 0.00 167.70
xxxxx7905 d6 t6 sell 0.51 p4 0.9847 0.0000 0.0000 d6 t6.1 0.9850 0.00 0.00 0.00 -15.53
xxxxx7907 d6 t6 sell 0.38 p2 1.6086 0.0000 0.0000 d6 t6.1 1.6058 0.00 0.00 0.00 106.40
---
trade closed the same day, about 6 hours after the opening that's how the close signal went off...
No :) I have to look: I pressed a combination... now it's crossed out... - i may be able to get a few things at different times :-) p1 p2 p3 p4 and so on
the link to the post where UP on forexclub forum publishes his mathematical proof that profitable trading is possible in forex. And (!) not as a result of violation of the Markov process, but just based on the assumption that it is a perfectly random, i.e. Markov process.
The actual link is http://forum.fxclub.org/showthread.php?t=22097&page=3
I wrote, too, by the way. Shit... I can't be mad enough, they erased everything.... now it's all gone.....
so, i.e. a baltic was created near zero with gradual widening of borders in both directions from zero. i.e. long trends were crushing into more frequent changes of small trends. And that hnach a flan martini would be enough for the depo to become voomat good. But that's the thing, you have to check on the quotes,
And how do you do it in 48 moves, it's too fast, I can hardly believe it, if you go out on it steadily every time, then hundreds and not one move to do probably, or is it purely by chance you so it?
In hashby I dialed about 20 in a short time.
Then I decided to test the theory that random behaviour is the most effective. In coinflep threw a coin, and in hashby depending on the eagle / hashby put + -. By move 400 I got to 24. Was always on the plus side before that, then quickly went to 0 and then minus. Heehee.