Author's dialogue. Alexander Smirnov. - page 16

 
Prival:

Z.I. Yurixx and Mathemat

I have an idea after reading your posts, I'm writing it down so as not to forget. To make an adaptive indicator based on FFT not redrawing + triangular window with peak at t=0, adaptation by threshold that removes ADC noise. It is necessary to think about variation of window width.



What is "triangular window", in particular "with a peak at t=0" ? And how to make FFT not redrawable ? None of the methods I know of for solving the boundary value problem allow for this, which is equivalent to looking ahead.
 
I don't get it.
Mathemat and Prival have just provided people with a properly computable first-order derivative,
as a result of which all previously written intuitive indicators can be happily discarded
as unfounded, i.e. glitchy.
(A serious trader-educational publication could be made out of this).
It doesn't matter that the professors already know this somewhere,
important that the audience of traders in this mathematical place is dark and uneducated (see articles and Code Base).
There's nothing wrong with the fact that the derivative is geometrically a tangent,
and geez, this conveniently calculated tangent is exactly in the middle of the computational interval.
Congratulations to both Mathemat and Prival for clearing up the erroneous-intuitive rubble on the path to a Bright Future.
 
Yurixx писал (а): And how do you make the FFT not redrawable ?
There are averages, stochastics and other indicators based on FFT. And they do not redraw......
 
Yurixx:
Prival:

Z.I. Yurixx and Mathemat

I have an idea after reading your posts, I'm writing it down so as not to forget. To make an adaptive indicator based on FFT not redrawing + triangular window with peak at t=0, adaptation by threshold that removes ADC noise. I should think over variation of window width.



What is a "triangular window", in particular "with a peak at t=0" ? And how to make FFT not redrawable ? None of the methods I know of for solving the boundary value problem allow for this, which is equivalent to looking ahead.

Redraw it as here, you may redraw straight line y(x)=a*x+b with arriving new data and you may do the same as suggested by mathematician (it is not redrawn). The window is from the field of Hemming, Henning, Butterworth, etc. It's just that they are all built with respect to the middle of the window and the triangular window is known to peak at (N-1)/2. If as you say from physics then it is more logical to move the peak to t =0, i.e. to give more weight to last values. About the adaptation... I will try to open a new branch as soon as I have enough time and explain in pictures. I think A. Smirnov will have a hard time finding questions that were asked here.

 

My two cents. For mathematicians who are not steeped in practice :), the average always refers to the centre of the interval. This is correct, because the error variance in the centre of the interval will be the smallest. At the edges, the variance will be equal to 1, i.e. commensurate with the variability of the data. If the data is random, then the predictive value is also zero. This goes to the point about the meaning of conventional MAs. On the other hand, with random data, the best estimate of the future value is the mean.

 

When there is confusion between the algorithm developer and the programmer as to who is right and what to do about it, the most reliable way out is a test case.

A control example of calculating CCC with m'=4.

C1=1.1 C2=1.3 C3=1.2 C4=1.4

Q4=1.1630 Q3=1.2889 Q2=1.2667 Q1=1.4

Q5=1.1630 Q6=1.2469 Q7=1.2601 Q8=1.3534

Calculations were performed on a CASIO fx-7400G programmable calculator, rounded to 4 decimal places. The value of alpha is 0.6667. The CCC value at bar 4 is Q8=1.3534. Now substitute C1 to C4 in your programs and if you get a result close to 1.3534, you are correct. If not, you need to look for an error in your program. I do not need to teach you how to do it. Find m'=8 with the same values of "claws" yourself. And everything will fall into place!

For comparison, the classical EMA at bar 4 with an alpha of 0.4 gives 1.2728. Do you see how it lags?

Perhaps it's time to put a full stop to our dialogue. Your dialogue is not going on the constructive way and I am not interested to continue it. To my main question: What is the essence of Juric's algorithm? (Or at least imagine the value of "clowzes" of 50-100 bars and reactions to them of the true Djuric algorithm, I have not received). I have explained my algorithm of CCC formation to you in detail.

Thank you all for your attention. The Earth is round - perhaps we will talk again sometime. Good luck to you and to "roughnecks" too.

P.S. Since the "Sun" has ordered a long life, all my following articles will be published in the USA.

 
ASmirnoff:

P.S. Since the Sun has died, all my future articles will be published in the USA.


And the whole "WS" is in the USA, which is probably why the subscription index of the journal promising a long life is not found on the website?
 
Rosh:
There are no questions about the alpha coefficient. Do I correctly understand that C1, C2, C3 and C4 are closing prices? C1 is the current bar (the freshest), C2 is the second bar that was formed before the C1 bar, and so on.
At every four consecutive bars, eight values are calculated from Q1 to Q8, and the last eight is exactly the value of the average.
C1 is the first bar of the first analysis window, C4 is the last bar where we calculate the average value, i.e. Q8. Then, like in MA, we discard С1 and add С5. This is when we calculate CCC at m'=4. By sequentially assigning values of the lower branch to the corresponding "clowes" we realize CCC with multiples of 4 after back-and-forth passes. Notice that the delay of the CCS with respect to the price chart at any m' does not exceed 1 bar. In addition, studies have shown that the fluctuation of the SSS compared to the EMA of the same order would win many times over (up to 10 times for m'=24).
 
ASmirnoff:

I think it's time to put a full stop to our dialogue. The dialogue on your part is not on a constructive path and I am not interested in pursuing it further. To my main question: What is the essence of Djuric's algorithm? (Or at least imagine the value of "clowzes" of 50-100 bars and their response to the true Djuric's algorithm, I have not received).


Pay for the work, I'll gut the whole JMA code and decompose the algorithm for you.

 
Integer:

Pay for the work, and I'll gut all the JMA code and break down the algorithm for you.

I gave you my CCC algorithm for free, so at least out of patriotism you should not ask for payment for Jurik's algorithm. In general, I do not need it. But I want to "shoe a flea" as the Russian craftsman Lefty did in his time. And no more than that.
Reason: