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[img]https://c.mql5.com/mql4/forum/2006/07/%D1%F2%FC%FE%E4%E5%ED%F21.gif [/img]
And Solandra (thanks to him for his participation in this thread) has killed 630 lines in this case ..... with the same accuracy...
By the way - if someone calculates other quantiles I will help with formulas.
I'm very ashamed of course, but I'll ask anyway. How did you get this formula D(E) = D(Y) - a^2*D(X). If you can do it for the same price, please sell the deduction logic.) Or give me a hint where to read it.
No problem to your email. I gave mine in this thread.
Searched and searched, couldn't find it. Here's mine ********
a thought occurred to me that maybe this notorious minimum of the functional is inherent to the channel whose parabola coefficient A->0. I.e. the field sources on top of the regression line and on the bottom counterbalance each other.
And here is the situation, the channels I have chosen have the following RMS
2006.07.17 20:11:30 VGGopII EURUSD,H4: N= 160 CKO2/3= 0.00707805 CKO= 0.00739682
2006.07.17 20:11:30 VGGopII EURUSD,H4: N= 307 CKO2/3= 0.00863145 CKO= 0.00967016
We see that these channels have CKO>CO2/3. You can argue about the smaller one, but the bigger one can be seen even by eye as a channel.
Correction. The sign of parameter A for a parabola doesn't matter, you have to relate it to the A sign of the linear regression (from general considerations). This is not because of my secret knowledge, but for the love of art :)
ZS. Did you teach the parabola to draw outside the sample boundaries? Congratulations, I can't get my hands on it.
ZS. Did you teach the parabola to draw beyond the sampling boundaries? Congratulations, I can't get my hands on it.
I just looked at how ANG3110 did it in his script.
By the way - if someone calculates other quantiles I will help with formulas.
But this is a private solution, and I like general ones. Or am I missing something?
I did not see the expression of probability through the random deviation from the regression centre.