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Thanks, Jose. Many thanks for your reply. However, I'm not sure that I am
at one with your answer. Does not the Linear Regression Indicator equate to
the last point on a LR Trendline with the same period parameter? In fact,
the LR Trendline and the LR Indicator (the centre line if you will) do
converge at the final point. It's your variation of the standard deviation
calculation in which I'm interested.
(The 63 days makes good sense.)
Regards,
Kevin
At 16:24 11/04/2004 +0000, you wrote:
>
>
>"Please may I ask you a question about your useful Linear Regression
>Trendline Channel v2.0 indicator?"
>http://users.bigpond.com/prominex/MetaStock/LinRegTrend.txt
>
>Hmmm... Who dares to question my MS code?
>Ok, Ok... shoot. :)
>
>
>"When I plot this indicator together with your Channel indicator,
> I would expect to see the three lines converge.
> The LR lines do but the Bands/Channels don't."
>
>Your code is taking the standard deviation of the Linear Regression
>*Indicator*. My code takes the standard deviation of the Linear
>Regression *trendline*.
>
>
>"why do you favour a value of 63 as your default?"
>
>63 trading periods is 3 months' data on a complete daily chart.
>IMO, it looks like a useful number of bars to determine intermediate
>trend.
>
>jose '-)
>
>
>--- In Metastockusers@xxxxxxxxxxxxxxx, Kevin <kevin_barry@xxxx> wrote:
> > Hello Jose,
> >
> > Please may I ask you a question about your useful Linear Regression
> > Trendline Channel v2.0 indicator?
> >
> > I have created a LR Bands indicator thus:
> >
> > Pds:=Input("LR Periods",2,100,20);
> > SD:=Input("Standard Deviations",0,3,2);
> > LR:=LinearReg(C,Pds,S,1);
> > Upper:=LR + Stdev(LR,Pds)*SD;
> > Lower:=LR - Stdev(LR,Pds)*SD;
> > Upper;
> > LR;
> > Lower;
> >
> > When I plot this indicator together with your Channel indicator, I
>would
> > expect to see the three lines converge. The LR lines do but the
> > Bands/Channels don't.
> >
> > I see that in your code, you have made an adjustment to the
>calculation of
> > the Standard Deviation. Would you mind explaining the reasoning
>behind that
> > part of your formula? And, as I'm at it, why do you favour a value
> > of 63 as your default?
> >
> > Thanks in advance.
> >
> > Regards,
> > Kevin
>
>
>
>
>
>Yahoo! Groups Links
>
>
>
>
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