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<BLOCKQUOTE
>
----- Original Message -----
<DIV
>From:
DIMITRIS
TSOKAKIS
To: <A title=amibroker@xxxxxxxxxxxxxxx
href="">amibroker@xxxxxxxxxxxxxxx
Sent: Sunday, May 16, 2004 10:41 AM
Subject: [amibroker] Re: EFA vs FFT
Stephane,a. What should I do after downloading and save this
the minimum total error.The only way to express this quantity is to sum
the abs(y-C1) for all bars and then minimize it.[In linear reg, for
example, we minimise the sum of (y-C1)^2]b1. What is your criterion,
if not the minimum of LastValue(Cum(abs(y-C1)));b2. Since you already
have the equation of your blue line, check if it gives smaller error to
see if it fits better than my white line.
Dimitri, That's the only thing I do today, was to
compare the afl and the dll
the only difference between the Afl and the dll is
the line
error=LastValue<FONT
size=1>(Cum(<FONT
color=#0000ff size=1>abs(y-new1))); in Afl
and becomes
error=(Cum<FONT
size=1>(abs(y-new1))); in
the dll
the error is smallest in afl
<IMG alt="" hspace=0 src="jpg00063.jpg"
align=baseline border=0>
But, I must think again about lastvalue
in a loop, there is something that's I don't understand
correctly
a simple example like <FONT
color=#800000 size=1>
for(i=<FONT color=#ff00ff
size=1>10;i<<FONT color=#ff00ff
size=1>12;i++)
{
Plot(MFI<FONT
size=1>(i),""<FONT
size=1>,i,1<FONT
size=1>);
Plot(<FONT
face="Times New Roman">LastValue<FONT
size=1>(<FONT face="Courier New" color=#0000ff
size=1>MFI(i)<FONT
face="Times New Roman" size=1>)<FONT face="Courier New"
size=1>,""<FONT
face="Courier New" size=1>,i<FONT
size=1>+1<FONT
face="Courier New" size=1>,<FONT face="Courier New" color=#ff00ff
size=1>1);
}
shows that for all bars we have the MFI value of barcount
-1
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>
c. LastValue is used many times in my code, before the error
calculation[ in lastx, Daysback, aa, bb, Hor. I did not understand how
do you solve the LastValue in all these cases and where is the problem in
error=LastValue(...) line.d. In any case, it is reasonable to ask the
total error to be minimum. The error per bar would not make sense. We may
have a function that minimizes the error for the first, say, 500 bars but
not for all bars.d. I will agree that LastValue(array) is known
"before" the last bar.There is no problem here, I use the results of all
the LastValue()s exactly at the last bar, not before. This way of use does
not introduce any "looking into the future" problem.[A problem, for
example, would be to use a LastValue() result in the middle of an
array and DECIDE for tomorrows movements]TIADimitris
TsokakisPS:Stephane's attachements are at <A
href="">http://www.purebytes.com/archives/amibroker/2004/msg07780.html
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