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so the XAvg solution is something like this:
Inputs:Price(numeric),Max(numericsimple);
Vars:len(1),Min(1000),Save(10);
For Len=5 to Max begin
Min=XAverage(price,Max)-XAverage(Price,len);
If AbsValue(Min)<Save then begin
Cyclelen=len;
Save=AbsValue(Min);
end;
end;
.. but there is an error somewhere since this gives a result of 5 .. always
I did work with my close-close[1] but there is way too much - noise as you
anticipated
----- Original Message -----
From: "Gary Funck" <gary@xxxxxxxxxxxx>
To: <omega-list@xxxxxxxxxx>
Sent: Thursday, January 08, 2004 12:25 PM
Subject: RE: Cycle Length
>
>
>
> > From: Gary Fritz
> > Sent: Thursday, January 08, 2004 10:35 AM
> [...]
> >
> > BTW I should have said the xavg-xavg signal was most similar to
> > Chris's summation(C-C[1]) signal. It still has some trend
> > component, just like his signal did. Running it through LRS just
> > detrends it further. And in fact LRS acts as a bandpass filter,
> > removing both noise AND trend. LRS actually does a better job of
> > filtering out the trend component than the highpass xaverage.
> > But I still think LRS(xavg-xavg) does the best job of extracting
> > the pure cycle component.
>
> For this application (finding the dominant cycle in the range 5...50,
say),
> I would question the use of a bypass filter of (15,20). This will chop out
> a lot of signal at the Len=5 and len=50 time periods. And further
detrending
> seems unecessary.
>
> As you note, LinRegSlope() amounts to an improved versrion of Summation(),
> because
> it will tend to elimnate noise. I'd think the following is likely more
> adequate and accurate for the purpose of determining dominate cycle:
> LRS = LinRegSlope(Price - Xaverage(Price, 50), 5);
> Here, we assume 5 is the min. cycle of interest and 50 is the max.
>
>
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