Probability, how do you turn it into a pattern ...? - page 28
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конечно дема. Там вопрос задавал, сам же отвечу- да, 27 пар, без экзотики. Может быть такое... эти валютные пары друг друга неким образом хеджируют (количетсвом, стоимостью пункта), при этом не исключены колебательные движения в стороны от точки равновесия, это происходит от того, что некотыре резывые участики группы 27 (G 27) время от времени уходят в отрыв (падают или растут), остальная же часть их догоняет и так постоянно. Но самое главное так это то, что есть точка равновесия и есть возврат к ней в любом случае.
Вот я купил все 27, посмотрим каков будет макс. + и - и будет ли стремление к 0.
There is also a point of no return, called a margin call. There's also a possibility of a call.
http://www.fx4u.ru/neveteran-m10133.html&tab=topics
if the volatility is oversized in pips and the result is positive or underprovided and the result is negative - the lock plus again a bit
and where there is an underweight of pips with a positive result or an overweight of pips with a negative result - there is a share
in general it may be so ))
Turka, я тока читаю инструкцию в клозете, летать не собираюсь :)
)))1. Assume that market price movements are random.
2. from the first assumption, we begin to consider the price movement of all markets as an independent set of random increments of the base value by one of two numbers equal in modulo but different in sign. Or to put it simply:
[int]$count+=Get-Random -InputObject -1, 1
In this case, a thousand throws are made, each of which either adds 1 or (-1) to the current value, i.e. one is taken away from the current value.
3. The delta between the initial value of the $count variable and the maximum and minimum values of the increments will increase as the tests increase. In other words, we get a potentially infinite divergence of the current value from the original value.
4. This is the most interesting part. The topicstarter argues that yes, indeed, this process is random and it is impossible to predict where we will be in the future relative to the original value: in the plus zone or in the minus zone. However, if we take a series of such martingales, we find that about half of them will be in the plus zone and the other half will be in the minus zone. While some of these non-independent trials will move incredibly far into the plus zone, others (and there will be enough of them to compensate for a positive move) will move into the minus zone. In other words positive tests will be compensated by negative ones and their total effect will tend to zero.
Looking for confirmation of these words I went to https://ru.wikipedia.org/wiki/Случайное_блуждание and came across this picture:
It seemed to confirm this argument. As can be seen, although the tests diverge from the initial value (zero) by significant distances with increasing number of throws, their median is still zero.
5. Let us try to simulate this process by improvised means:
#-----------------------------------------------------------------------------
# Name: Random-Stabilisation.ps1
# Lung: PoSH 1.0
# Date: 03-24-2010
# Author: Vasily Sokolov (C-4)
# Intro: ---------------
#-----------------------------------------------------------------------------
for($j=1;$j -le 30;$j++){
for($i=1;$i -le 1000;$i++){
[int]$count+=Get-Random -InputObject -1, 1
}
$count_array += @($count)
$count=0;
}
ForEach($elements in $count_array){
$general_variable+=$elements
}
$general_variable
Remove-Variable count_array
Remove-Variable general_variable
This code generates 30 independent trials of 1000 coin tosses each. If you call this script enough times (say, 100) and change the number of trials in it, you can calculate the cumulative value of the deviation from zero. From statement #4, it follows that it should stay about the same. I have not seen this in reality. The first time, when the number of trials was 10, the cumulative deviation was -454 units. The second time with 20 trials the average deviation was +1748 units, the third time with 30 trials the deviation was +204 units.
6. Assuming that the topstarter is right, this means that one can calculate the current cumulative deviation of the instruments from their median and thus get into the market at the point where the deviation peaks or fails, in the hope that it will return to normal.
And the number of trajectories averaged affects only the variance, but not the nature of the curve
20 trajectories
200:
So there is no way to play on this.
6. Если предположить что топикстартер прав, то это означает что можно рассчитать текущее совокупное отклонение инструментов от их медианы и таким образом залезть в рынок в месте пика или провала этого отклонения, в надежде на то, что оно вернется в норму.
It doesn't work in rendom. In forex... - Unless you have a known buy-in..... :-))
This code generates 30 independent tests of 1000 coin tosses each. If you call this script enough times (say 100) and change the number of trials in it, you can calculate the cumulative deviation value from zero. From statement #4, it follows that it should stay about the same. I have not seen this in reality. The first time, when the number of trials was 10, the cumulative deviation was -454 units. The second time with 20 trials the average deviation was +1748 units, the third time with 30 trials the deviation was +204 units.
6. Assuming that the topstarter is right, it means that one can calculate the current cumulative deviation of instruments from their median and thus get into the market at the point of peak or failure of this deviation, in the hope that it will return to normal.
Let me add a couple of words.
1) ... If you change the number of trials in it, then you can calculate the cumulative value of the deviation from zero - reduce the number of trials. My method of elimination of positive results may be primitive, but it does not matter for the general idea, the goal is achieved, I systematically eliminate tools from series of tests)
2) . ..It is possible to calculate the current cumulative deviation of the instruments from their median - the ratio of the instrumentsto be deviated (from the median) hasto be taken into account, I calculate it as a percentage of the instruments which have gone to (+) to (-)
3) .. .To get into the market at the place of the peak or failure of this deviation - the time of thefirst cycle, in order to achieve a given deviation. I have been repeating throughout this thread that this is the most important value, absolutely indicative.
Looks like you don't need me anymore.
Thank you
...можно рассчитать текущее совокупное отклонение инструментов от их медианы и таким образом залезть в рынок в месте пика или провала этого отклонения, в надежде на то, что оно вернется в норму.
the rest seems to be true, but this...
the author states: "each subsequent position is hedged by the previous one ........., by a certain %, I set this % of hedge in the setup, determining it by the degree of aggression I intend to apply to the given depo".
source http://www.fx4u.ru/rinki-forex-commodities-cfd-futures-f14/obschenie-f7/graal-v-forekse-t7914.html&st=40
остальное похоже на правду, но это...
автор утверждает: "каждая последующая позиция захеджированна предидущей ........., на определенный %, я этот % хэджа задаю в настройке определяя его по степени агрессии, кот. намерен применить к данному дэпо."
источник http://www.fx4u.ru/rinki-forex-commodities-cfd-futures-f14/obschenie-f7/graal-v-forekse-t7914.html&st=40
This is from another thread ......mr. Sherlock Holmes