Volumes, volatility and Hearst index - page 16
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Yurixx, according to your observations the ratio of mean spread to mean increment (in your terms R/M) converges to 2 as N increases? Or is it just a lack of data that gives this impression?
The impression is correct. I wrote about it to Nikolai in our private correspondence: this ratio for SB converges to 2, as well as Hurst index converges to 0.5.
The impression is correct. I wrote to Nikolai about this in our private correspondence: this ratio for SB converges to 2, just as Hurst's ratio converges to 0.5.
Well, then Hearst is not so bad))), if you calculate it on a sufficiently large range of elementary increments (ticks in our case).
Candid has given the formula R/S = k * (N^h) - now it remains to clarify how these letters are calculated, an example would be better. Suppose it will be a series of 0, 1, 2 ...,29,30,29 ...2,1,0.
Calculate and show everything on it. He who says things wrongly. On the same line, give the formula and show how it is right.
PZY You'll erase all the keyboard here, but the truth will not come to me so it seems for some reason ...
R - the average spread. Range is equal to the difference between the maximum and minimum values of the series on the interval.
N - number of samples in the interval.
S - RMS of increments of a series.
k - constant coefficient.
h - Hurst index.
It means that the entire series is divided into equal intervals of N counts. For each interval, the increment and the spread are calculated. On the basis of these data, the RMS of the increments and the average spread are determined. The Hurst index must be selected so that the formula is satisfied. :-)))
If Hurst was right and the average spread did satisfy this equation, then it would have a solution with respect to h. This solution would be determined by two points
R1/S1 = k * (N1^h) and R2/S2 = k * (N2^h)
The series can be broken in two ways: into intervals of magnitude N1 and magnitude N2. Correspondingly, we obtain ranges R1 and R2, and RMS S1 and S2. Coefficient k is constant. Thus we get a system of two equations. Excluding the coefficient k we get the expression for the Hurst ratio calculation:
h = [ Log(R1/S1) - Log(R2/S2)]/[Log(N1) - Log(N2)]
Geometrically, it is the tangent of the slope of the straight line drawn through the two points [Log(R1/S1),Log(N1)] and [Log(R2/S2),Log(N2)]. A curve expressing the dependence of R/S on N in logarithmic coordinates has been plotted. Its graph is shown. It shows that the angle of slope changes, i.e. depends on N. This implies that the coefficient k in Hurst's formula is not a constant, that it depends on N, and that Hurst's formula is only asymptotically true for large N. Since the object of the study was SB, there were no problems with the amount of data, unlike the series of quotes.
Well, then Hurst is not so bad))), if we calculate it over a sufficiently large range of elementary increments (ticks in our case).
Yeah ... :-)
I was counting on ticks. Naturally model ones. I could investigate any range - both in terms of the size of the interval and the necessary statistics. With limitations, of course, on the capabilities of the computer. But I have reached this ceiling.
The scissors here are simple: the larger the interval size you choose, the smaller your statistics will be. After all, a series of quotes is finite. In the relative sense it's even worse, because as the interval increases you need more intervals, so that averages become closer to their real values.
However, I've already written about this on page 5.
I have run out of arguments.
I can only recommend that you remember some basics. If k is k1 for N1 and k2 for N2, this is called the dependence of k on N. It is synonymous with the formulation: k is a function of N. Formally it is written as k = k(N). So I just translated Vita's phrase into stricter language.
I simply did not understand the passage about problems with calculation of the Hurst exponent for series other than SB. For a moment I had a wild idea whether the author thinks that for any series the Hearst exponent must be 1/2, but I immediately dismissed it.
For High - Low = k * (N^3) series the Hearst exponent will be equal to 3.
For example Vita 0, 1, 8, 27, 64, 125, ..., 1000*1000*1000 let's take for certainty the points with N=2 and N=3 (numbering from 0).
So, h=(ln(8)-ln(27))/(ln(2)-ln(3)) = 3*(ln(2)-ln(3))/(ln(2)-ln(3)) = 3.
h = 3 denotes that the formula is rubbish, the author is ignorant.
I see that the substitution of the average mileage is repulsive to you. Forget about it.
I suggest you substitute 1 old pips = 10 new pips. Q=10R.
Compare the results of the formula for both cases. I'm sure the results will be different. This means that by measuring with a different ruler we get different fractal dimensions for the same series. For this it is of course necessary to know that H complements the fractal dimension to 2 and that the choice of the ruler does not change the fractal dimension. But one has to know that before one passes off any rubbish as Hearst.
Hurst was doing R/S analysis, so his exponent does not depend on ruler choice. The topikaster's result is dependent, no matter how many times he spells the letters R and S. The result of the topikcaster does not complement the fractal dimension to 2, and therefore is not in any way meaningful to Hurst. Topikcaster's result shows for his fictional row 1/2, and for all other rows it is simply a number that has nothing to do with Hearst. If this were not the case, the topikmaster would have long ago posted the results for the various rows and shown how they converge to the theory. This is not the case as his formula is completely wrong. And he has nothing to show.
Question for everyone here. Has anyone seen the file attached by Vita ? I don't see anything, but maybe I missed something ?
p. 10
What about the three simple questions ?
Probably everyone. Candid has given the formula R / S = k * (N ^ h) - now it remains to clarify how to calculate these letters, the example will be better. Suppose it will be a number 0, 1, 2...,29,30,29...2,1,0.
On it calculate and show everything. And the appointee is the one who says the wrong thing. He will show you the right way on the same row by giving you a formula.
Z.I. You'll erase all the keyboard here, but the truth does not come to me so it seems to me for some reason ...
There is no need to prove it. This formula was postulated by Hurst, at least this is how it is written in Peters' book. That is why it is the actual definition of the Hurst index. Only not in this form, but in this one:
R/S = k * (N^h)
The entry (High-Low) is generally delusional from my point of view (sorry Nikolai, I understand that you're just following Wit's designations). The High and Low values are used everywhere as purely local. And R in Hearst's formula is the average spread.
Amazing logic, I appreciate it /:o) I'll take it on board, because I'm afraid I won't be able to cope next time.
As for the formula, it's absolutely correct, except that historically I don't quite remember what the primary was. But it is still one way of calculating it, not the definition of the indicator. To be fair - this indicator has been rediscovered several times. However - it doesn't matter anymore.
Yeah ... :-)
I was counting on ticks. Naturally model ones. I could investigate any range - both in terms of the size of the interval and the necessary statistics. With limitations, of course, on the capabilities of the computer. But I have reached this ceiling.
The scissors here are simple: the larger the interval size you choose, the smaller your statistics will be. After all, a series of quotes is finite. In the relative sense it's even worse, because as the interval increases you need more intervals, so that averages become closer to their real values.
However, I've already written about this on page 5.
The idea is that if we calculate Hirst on some range of data, and then divide this range into a sufficiently large number of intervals and Hirst is calculated on each of them, then their average value must converge to the Hirst coefficient calculated for the entire range. If this is so, the only limitation when you calculate Hirst is that N must be large enough. Judging by your studies, the accuracy at N=15 is already quite high. Therefore, perhaps this is an acceptable number of ticks on which it makes sense to calculate Hirst. And it is not necessary to average N ticks by segments - it will be more exact Hirst calculated for the whole range.
P.S. On second thought, I have decided that 15 is not enough. What I need is a sequence of K intervals of at least 15 ticks (or one time to calculate the Hurst in the range K*15 ticks). I do not know how many such intervals must be at least for the acceptable accuracy. It seems to depend on the dispersion of the spread - how it decreases when increasing K. But it is probably easier, just as an experimental estimate for SB.