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The short-term (shorter of the two) MA contains not only the cycle of
interest but also all longer-term cycles and trends. The purpose behind the
MA with period equal to the cycle of interest IS to remove that cycle, but
retain all others of greater length. When you take the difference,
therefore, what remains is only the cycle of interest (within the inherent
accuracy of the filter, of course).
On the subject of "one-day MAs" - you should always try to have as many data
included in a MA as possible. If you need to consider a shorter cycle, you
should perhaps use hourly data. On the other hand, if your approach is
based solidly on short-term daily data, then MAs are probably not
appropriate.
Regards,
Carroll S.
----- Original Message -----
From: "Gary Funck" <gary@xxxxxxxxxxxx>
To: "Carroll Slemaker" <cslemaker1@xxxxxxxx>; "Monte C. Smith"
<mcs@xxxxxxxxxxx>; "Gary Funck" <gary@xxxxxxxxxxxx>; <omega-list@xxxxxxxxxx>
Sent: Saturday, September 30, 2000 5:54 PM
Subject: Re: Exponential moving average
> > What you want is the shortest-period MA
> > which produces the desired degree of smoothing (removal of short-period
> > variations) withOUT significantly reducing the size of the desired
cycle.
> >
> > You can take this one step further by also calculating a MA with span
equal
> > to the cycle of interest
> > and then displaying the difference - short period
> > MA minus long period MA. This gives you a rough band-pass filter.
>
> As Bob Fulks mentioned, the problem with using a simple average
> whose length is that of the cycle is that it will average to zero (note
> in a rising market, with a cycle imposed on top of it, the average
> would work out to the upward bias). EMA's won't do that, but they do
> attenuate the signal, and they're late.
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