a trading strategy based on Elliott Wave Theory - page 54
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Well, I've already written above. Just calculating quantiles to build confidence intervals. How else can it be used? Bulashev wrote how to calculate these very quantiles in Excle. In general I have the same file that you posted above, but only for Student's distribution. Here is the difference. Just think how can you apply the normal probability distribution to a sample of 30 bars for example if there are only a few bars? Just compare quantiles of Student's distribution at different degrees of freedom and everything will become clear at once.
"If there is no difference - why pay more?" :)
Solandr, compare the behaviour of the diagram at different values of N and the problem disappears.
https://c.mql5.com/mql4/forum/2006/06/Student.zip
By the way, I wanted to ask you - didn't you try to use Chi square? Maybe Vladislav tried? On the one hand - redundancy, when we are not sure that the approximation should obey a normal distribution, on the other hand - also a criterion for sample selection.
Frankly speaking, it's the difference at different values of N that I'm interested in. The Student's distribution remaining qualitatively the same in form changes its quantitative parameters at different degrees of freedom. When the number of degrees of freedom is very large, it should coincide with the normal distribution. When the number of degrees of freedom is small, it differs from the normal distribution.
The Student's distribution at different degrees of freedom:
For probability of 99%:
q(30 bars)=2.750
q(100 bars)=2.626
q(300 bars)=2.593
q(1000 bars)=2.581
If you think that a difference of 6% between the quantile value for 30 bars and 1000 bars is not worth the additional suffering, it is your personal choice. I hold a slightly different opinion.
Hee square hasn't tried it. Although, after reading Bulashev, I had such thoughts in the beginning. But it will not give us anything since you will not see the normal distribution in the real sample anyway. We only use irrefutable evidence that a BIG sample will have a normal distribution. Our 30-1000 bars can only be called BIG by a stretch.
In principle, I don't really think there's any difference, but out of an innate pest, I've made a diagram and that's the end of the Steudent. I can assume that if you calculate ranges through probabilities, the error can reach 6 percent, but if you put the horse before the cart (calculating probability through the deviation from the centre of the regression), it will not exceed 0.6
Rosh, how do you plan to trade with this strategy? Without setting of stoplosses, but only using current probabilities? Judging by what you said to prove that there is no difference between the two distributions, it looks like without any stoplosses. Actually 6% difference means 5-10 pips difference in stoploss setting. That is all! Whether it is much or little, I cannot judge since I have not checked it myself. Perhaps you are completely right.
Bulashev uses Student's method in his example of statistical methods in chapter 10, page 147. And he does this with 816 points in the sample!
As I am not a big expert in the field of statistics I just did as Bulashev does and settled down. If you have a desire to prove something else, you will be happy to see the final result, which you get based on a normal distribution.
find linear regression coefficients Y=A*X+B. Further, we set a confidence interval, for example 95% (P=0.95), and try to find the limits of this interval (i.e. such limits that prices in 95% of cases lie at a distance +- delta Y from the central regression line).
Using the properties of the normal distribution, I would do a simple thing - set off two sigmas each from the centre of the linear regression (2 sigmas from the centre is also ~95%). As long as the number_in_intrevalue/total_number <=95% - the channel has a right to life.
Give me your methodology in the form of formulas, and I'll put it into Excel to compare it to a normal distribution.
Thank you for the reference to this section of Bulashev, otherwise no one knows when I would have got there :)
Stop-loss is a different song, so far I only have its tune, and it sounds in such a way that you can only hear it an octave lower :)
I am doing the same thing - I am putting aside sigmas. Only how many sigmas to defer from the centre line is determined by Student's (using the example in Bulashev). I'm not trying to determine the viability of a channel for trading, i.e. I consider a channel valid if prices are not out of the 99% interval. I am simply trying to repeat Vladislav's methodology (at least the way I understand it) - find the turning zone first and not determine how long the channel will exist - it is a thankless task in general and the Hurst indicator itself will not give you the exact time of its disappearance until that time is very close (or already passed), which is not very convenient in practical application. Vladislav said at the very beginning that market entry is in the reversal zone, and then if the pose is successful, the question is raised about the holding time of an already opened position, and not about being added in the middle of the confidence interval, when the channel is formed and stable. So you're going to enter (be added) in a stable channel, which in itself is very risky. Although, on the other hand, the very concept of a channel is very flexible in time. Maybe you are going to enter (be added) only in long term channels (several weeks long), in this case I completely agree with you, if on the border of this large channel there are all signs of reversal zone, I act the same way as you do. Well, if you are going to catch channels of 1-2 days and enter these channels, then of course there is an increased risk.
A pivot zone can be just a superposition of several probability zones of several channels and it can contribute with MareiMaz (approach to the MM level according to Vladislav's system means making a stand and till that moment the advisor can easily doze off).
Second, imposing of criterion on the regression channel (RMS convergence) introduces some serious contradictions: we have a linear channel, and according to the criterion R/S expression will grow linearly in time, as well as N/2. As long as we are in the channel, the Hurst criterion will not change, and when it changes, we are no longer in the channel, isn't that funny :)
This can be solved in two ways:
1) we build the channel at hovels, then go down to 15 minutes, R remains the same, the RMS also doesn't change much but N/2 increases twofold, so we artificially halved the Hearst index in the channel - it's not ~0.6 but ~0.3
2) we rebuild the channel breaking the confidence interval and at some moment it becomes flatter (the lines will move apart and the channel becomes longer) and there Hirst shows a possible reversal. But I looked more attentively and came to a conclusion - H<0.5 rather means a bounce from the channel border than a trend (channel) reversal.
On the contrary, entering in a stable channel is less risky. That is what the confidence intervals are for, so you get a stable channel and wait for the correction to the 5% area and enter there with minimal risk. I thought we had the same interpretation of risk :(
Vladislav did not say anything about tuning to different timeframes. I don't see the point of setting up on different timeframes either. As far as I understand, you are inventing your own approach to the problem. Well, maybe it will also be very successful. We are waiting for the first results of the Expert Advisor using your system. Also I do not quite understand your assumption about the Hearst index changing by a factor of 2, if the TF differs by a factor of 4. As far as I imagine for one channel built on different timeframes, the Hearst value should differ ONLY by the magnitude of error arising due to different number of degrees of freedom if you like(Student's distribution probably plays some role here by the way), and not the way you say in this phrase.