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Hi MG Ferreira:
Thank everyone, and specially thank to MG, who makes a very
valuable contribution on Kalman Filter. I will keeping exploring more
on this managing subject.
Thank you very much
Eric :>
--- In equismetastock@xxxxxxxxxxxxxxx, mgf_za_1999 <no_reply@xxxx>
wrote:
> Thanks TecloGeo and Eric,
>
> This is half of what I am looking for, although the
>
> > sum(C*(Cum(1)=1))
>
> example I included in the post does the same. Since an array, such
as
> C, can have practically unlimited values, this could be used as a
way
> to bridge the MSFL 20 variable limit. You could do something like
>
> Variables := 0; {Unlimited number of variables, all = 0}
>
> Now, to get say 50 variables in a formula, you could do something
like
>
> Sum( Variables*(Cum(1)=1) + {Returns first variable}
> Variables*(Cum(1)=2) + {Returns second variable}
> Variables*(Cum(1)=3) + {Returns third variable}
> ...
> Variables*(Cum(1)=50) ) {Returns 50th variable}
>
> The problem with this part of the solution is the complexity. It
> would have been a lot easier if we could just write say
>
> Sum( Variables[1] +
> Variables[2] +
> Variables[3] +
> ...
> Variables[50] );
>
> But we can at least 'read' the variables.
>
> The problem is that we need to also assign values to these - we need
> to be able to 'write' to these variables. This is the missing
half.
> So if we could do something like say
>
> Variables[1] := 20;
> Variables[2] := Variables[3]+0.5*Variables[4];
> Variables[3] := xyz;
>
> then we have a really neat way in which to keep all the say weights
> together, manipulate them, and even plot them. But I don't think
this
> is possible in MSFL.
>
> Anyhow, Eric, I think using MSFL the best is to first stick to only
> one PC and one MS. To optimise each one you would do something like
> what you propose, although that assumes that the various indicators
> are completely independant of one another. This is sometimes
assumed
> to simplify things, but is bound to give suboptimal answers.
>
> You could ensure that each indicator has a minimum of say 10% in the
> final function, via something like
>
> TT := ( (W1+0.1)*T1 + (W2+0.1)*T2 + (W3+0.1)*T3 ) /
> (W1+0.1+W2+0.1+W3+0.1);
>
> or you can just give say W3 a very slowly adjusting smooth function,
> which calls for individual smooth functions!
>
> Regards
> MG Ferreira
> TsaTsa EOD Programmer and trading model builder
> http://www.ferra4models.com
> http://fun.ferra4models.com
>
>
>
>
>
>
>
> --- In equismetastock@xxxxxxxxxxxxxxx, "TecloGeo" <teclogeo@xxxx>
wrote:
> > MG,
> >
> >
> >
> > Is this what you're looking for?
> >
> >
> >
> > Valuewhen(1,cum(1)=1,C)
> >
> >
> >
> > _____
> >
> > From: equismetastock@xxxxxxxxxxxxxxx
> [mailto:equismetastock@xxxxxxxxxxxxxxx]
> > On Behalf Of mgf_za_1999
> > Sent: Monday, May 16, 2005 7:49 AM
> > To: equismetastock@xxxxxxxxxxxxxxx
> > Subject: [EquisMetaStock Group] Absolute reference
> >
> >
> >
> > Hi Eric,
> >
> > I am working on your question, just a quick question to the group,
> > that relates to our discussion.
> >
> > Is it possible to get an absolute reference? I know
> >
> > Ref(C,-1)
> >
> > gives the close one bar ago. Also, something like
> >
> > sum(C*(Cum(1)=1))
> >
> > gives the close of the first bar. But is there a function,
say 'X',
> > so that
> >
> > X(C,1)
> >
> > gives the close of the first bar?
> >
> > Thanks in advance
> > MG Ferreira
> > TsaTsa EOD Programmer and trading model builder
> > http://www.ferra4models.com
> > http://fun.ferra4models.com
> >
> >
> >
> >
> >
> >
> > _____
> >
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> >
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