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Adam,
I've come across this before with comparing the ATR to a moving average of
the ATR and found them to give different values. See attached chart which
gives an example.
In the data window shown (in red type)the top value is mov(atr(1),10,e) and
the value underneath is mov(atr(1),10,s) and the bottom value is
atr(10). I think the "average" in the ATR calculation is a different method
at least in comparison to an EMA and SMA.
regards...Ian
>From: "Adam Hefner" <vonhef@xxxxxxxxxxxx>
>Reply-To: metastock@xxxxxxxxxxxxx
>To: <metastock@xxxxxxxxxxxxx>
>Subject: Re: atr
>Date: Fri, 14 Jul 2000 08:00:57 -0500
>
>Al,
> If you used a ATR(1) and then plotted a 10 day simple moving average
>of this ATR(1)..... it should calculate the same value as an ATR(10).
>Now if you needed an "Exponential ATR" you could plot a 10 exponential
>moving average of ATR(1).
> I believe this is what the previous ( Mike) e-mail was trying to show.
>
> Adam
>
>
>----- Original Message -----
>From: "Al Taglavore" <altag@xxxxxxxxxx>
>To: <metastock@xxxxxxxxxxxxx>
>Sent: Friday, July 14, 2000 3:10 AM
>Subject: Re: atr
>
>
> > ATR(1) would simply be the true range for one day. I fail to see the
>value
> > of taking an "n" day moving average of one day. I am looking for the
> > average true range of price over "x" period of days....what is the
>average
> > price movement for the past 10, 50 day period. If, as is the case for
>WMT,
> > the 10 day ATR is 2 2/16, and the 50 day ATR is 2 7/16, after price
>has
> > moved, during the trading day, 2 points, I would anticipate little
>reward
> > to buy the stock as I could only presume a futhur movement of 2-7
> > sixteenths. If however, the stock fell 2 1/2 points, I have a low risk
> > entry point for a countertrend trade. If I owned the stock from a lower
> > price point, after the 10/50 day ATR is reached I have a good exit point
> > for my day trade.
> >
> > Al Taglavore
> >
> > ----------
> > > From: Bob Jagow <bjagow@xxxxxxx>
> > > To: metastock@xxxxxxxxxxxxx
> > > Subject: RE: atr
> > > Date: Thursday, July 13, 2000 8:28 PM
> > >
> > > Right. The Equis ATR(period) matches Wilder's original version and the
>TR
> > > isn't a builtin.
> > > ATR(1) is actually the TR so taking its ma will give SMA or EMA
> > versions
> > > of ATR -- Chande uses the SMA for stops.
> > >
> > > Bob
> > >
> > > -----Original Message-----
> > > From: owner-metastock@xxxxxxxxxxxxx
> > > [mailto:owner-metastock@xxxxxxxxxxxxx]On Behalf Of Mike Campbell
> > > Sent: Thursday, July 13, 2000 2:18 PM
> > > To: metastock@xxxxxxxxxxxxx
> > > Subject: Re: atr
> > >
> > >
> > > Al Taglavore writes:
> > >
> > > > Neither. As Welles Wilder developed it, a moving average was not
>used.
> > > > MetaStock has it programmed. Simply pull up the indicator and type
>in
> > the
> > > > number of days. Today's ATR is the distance from today's low to
> > today's
> > > > high OR from yesterdays close to today's high.....whichever is
>greater.
> > > > This accounts for any gaps from the previous close to the low of the
> > > > current day.
> > >
> > > I believe you are mistaken there. What you described is the "true
> > > range" calcuation. ATR is some moving average of THOSE values.
> > >
> > > Otherwise, what would the "number of days" have to do with it?
> >
>
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