Stochastic resonance - page 16
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Interesting article http://elementy.ru/lib/164581
So, what do we have? There is noise - quite strong: volatility. There is a weak regular signal (hardly periodic, but it is definitely there). Weakness of the regular signal is confirmed by a very low return value in comparison to the volatility itself even on strong trends. I already gave these values somewhere by the example: EUR goes ascending trend on daily data for 6 years that is about 1600 daily bars. During this time period EUR has gone through 6000 points. So, the expectation is less than 4 pips (regular low impact). At the same time the volatility on the daily bars is about tens of points (noise).
Steady states are flat on tops during reversals or corrections. Trends are unstable states of transition from one flat to the next. Before a trend, a regular signal is amplified by flat noise and appears as sharp, often momentary jumps from level to level.
How can we learn something practical from this?
P.S. For example, how can we extract only the random component (pure noise) from volatility to get a regular signal? Volatility is known to be an antipersistent process. Simply subtracting a constant from it will not work, as the signal is getting stronger during the trend. Detrend? And what, I wonder, is the amplification coefficient equal to?
Hi.
I have been using resonances in my systems for a long time now. Without revealing any particularly interesting options I can say the following - take any trend indicator.
You make another one with 2 periods of the same indicator and get the resonances on market peaks and troughs.
The only thing you have to learn how to identify the moment of entering a position. I am attaching a screenshot of one such indicator.
I think that since one copy of the indicator has longer period than the other one, it needs tuning on pairs and TF to get clear resonances.
I would like to wish you a good trend and more profits.
Suppose there is a normally distributed sequence of values X. The number of members of the sequence is N=1000000, the average value is A, and the ska is S. Obviously, the set of X element values is bounded from above, i.e. all X's belong to the interval [0,Xmax]. We take a sample of M=100 members of the sequence and calculate its average XM. We form a new sequence Y = {XM} from all sequential samples containing M elements of the original sequence. It is clear that the set of Y values is bounded too.
How to find its upper and lower bounds, i.e. the interval of [Ymin,Ymax] values ?
I'm naturally interested in the analytical evaluation by means of mathematical statistics (in which I, alas, am not strong). To calculate in head-on is not difficult, but it is not interesting. Interesting to get a dependence of the limits of the interval on the ratio of N and M and statistical properties of the initial sequence.
If X is a random variable, then Y is the sum of M independent random variables with the same distribution as X. Thus if X is normal, then Y will also be normal, with variance S/sqrt(M). The question of maximum and minimum values can only be posed for a particular realisation of the series (i.e., counting head-on), for an arbitrary realisation we can only talk about probabilities.
P.S. The above does not mean that I consider myself a specialist in mathematical statistics :)
I don't pretend to be an expert either, but the variance of the sum of Nsum=the sum of variances. So Dsum=M*D => Ssum=sqrt(M)*S (Ssum-sigma of Y distribution, S-sigma of X distribution).
The mathematical expectation of the sum of random variables is equal to the sum of mathematical expectations Asum=M*A
The probability of SV Y in any intervals can be found using Laplace function value tables. For example, as a consequence in 3 sigmas will be with probability 0.9973. It means that this probability will be in the range: -3*Ssum+Asum<Y<3*Ssum+Asum => -3*S*sqrt(M)+A*M<Y<3*S*sqrt(M)+A*M
For example. If the distribution function is known, then for any X0 we know the probability P of having an element with value >=X0 in the sequence. If the sequence contains N elements, the total number of elements in the sequence that satisfy the condition X>=X0 is P*N. If this value is less than 1, i.e. 0, then statistically Xmax<X0. But of course, it does not mean that in fact no element >=X0 can appear in such a sequence.
... If X>=X0, the mathematical expectation of the number of sequence elements satisfying the condition is P*N. This value is always less than 1 (unless the distribution function is artificially cut of course). The probability that there will be no number >= X0 in the sequence of length N is (1-P)^N.
P. S. The words "This quantity is always less than 1 (unless the distribution function is artificially truncated)" refer to P, i.e. they do not provide essentially new information and are redundant in this phrase :)
I don't pretend to be an expert either, but the variance of the sum of NSV=the sum of variances. Therefore Dsum=M*D => Ssum=sqrt(M)*S (Ssum-sigma of Y distribution, S-sigma of X distribution).
I don't pretend to be an expert either, but the variance of the sum of the NSV=the sum of the variances. Therefore Dsum=M*D => Ssum=sqrt(M)*S (Ssum-sigma of Y distribution, S-sigma of X distribution).
Where is the condition: divided by M?
Where did the condition come from: divided by M?
Yurixx wrote (a):
... We take a sample of M=100 members of the sequence and calculate its average XM. Form a new sequence Y = {XM} ...
We feed one neuron - an oscillator with a small averaging period to the input, and another neuron - an oscillator with a large period to the input. Add another neuron with an oscillator of a very long period.
The outputs of these neurons are fed to the input of the fourth neuron that already outputs data on resonance: if the number is around zero, there is no resonance; if it is above zero and growing, an upward impulse and an upward trend enter into resonance; and vice versa: if it is below zero and falling, a downward impulse and a downward trend enter into resonance.
Where does the condition come from: divided by M?
Yurixx wrote (a):
.
..We take a sample of M=100 terms of the sequence and calculate its average XM. Form a new sequence Y = {XM} .
..Then I'm sorry, I didn't understand the conditions.
If a series of averages is considered, and even on overlapping sections, they are dependent. You have to consider the increment (it will be independent).
XMi - XMi-1=(Xi - Xi-M)/M
It seems to suggest that the SV has the mathematical expectation=0, D=2*D1/M, RMS=sqrt(2*D1/M)
If this is correct, then continue with the table of Laplace function values.
If a series of averages is considered, and even on overlapping plots, they are dependent.
Yurixx wrote (a):
Form a new sequence Y = {XM} from all consecutive samples containing M elements of the original sequence.
So they will be just independent
If a series of averages is considered, and even on overlapping plots, they are dependent.
Yurixx wrote (a):
Form a new sequence Y = {XM} from all consecutive samples containing M elements of the original sequence.
So they will be just independent
Then it won't work:
Yurixx wrote (a):
No, we are just talking about a sliding window of length M samples. Therefore the number of elements in sequence Y is N-M+1.