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Question about forced algorithm. Certainly I want to believe that you have thought up how in one and the same cycle except for coefficients a and b to find also RMS, but to it I have not thought up yet (in general I assume that I have guessed that way which you count, i.e.Really according to this algorithm in principle we count only one biggest channel and others are received vicariously, but it is impossible to change RMS in the same cycle as it should be counted for each channel running all its number of bars, and it will again increase time and I think it will be about 100-300 ms for 3000 bars.
I would like you to reassure me that you have no error here and that there is still a way to fit RMS into this cycle.
I can encourage you and sell the secret for a small price: D(E) = D(Y) - a^2*D(X)
Here X and Y are random variables for which the regression Y = a*X + b is calculated
E - regression error, i.e. the deviation of Y from the regression line
D(E), D(Y) and D(X) are dispersions of the corresponding quantities. By the way, the RMS of an error = square root of D(E).
So you don't have to build a series of errors and calculate the RMS by head-on summing. You have to be more lazy.
Just don't tell anyone else about it ! :-)
Good luck.
:-D Well I won't. Thank you very much.
:))
Let's denote by S[N] the sum of squares of deviations Si, where i=1,...N , then D[N]=S[N]/N.
RMS2/3[N]=({D[N]-D[2N/3]}/{N-2N/3})^0.5
All ratios (for linear regression, for parabola, RMS, kinetic and potential energies, RMS from parabola, sum of gradients from parabola and other yet undesigned channel characteristics) are calculated for any bar (read channel of given length) by simple analytical formulae.
This whole bunch of parameters is calculated in one go.
I was only too lazy to calculate the Hearst indicators using accelerated algorithms reasoning that they are calculated for certain channels.
True, again a mistake somewhere, the results are very big so far.
Well, there's kind of a little turnaround here.
.
tempBar=k_bar-n*2.0/3.0;
lastBar2=MathRound(tempBar);//here you recalculate the first and last bar for 2/3
StDev23=GetStDevFromArraysZ(k_bar,lastBar2,a_CH,b_CH);//and substitute A and B found for the entire sample into the function
Maybe this is how Vladislav's statement should be interpreted
.
But I understood that we build a 2/3rds channel and check how the data of the last third fit in it, and if it does fit in, then the channel is built over its entire length.
Hmm, I didn't know that formula. But pen and paper and without it help :)