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Yes. If the hypothesis
if(d<f)
is wrong for all n, then n1 can not have a value and remains
undefined.
With the initial n1=na, if the hypothesis is wrong, n1 will have a
final value equal to na.
It happens when the MA(C,10) is the closest MA.
Dimitris Tsokakis
--- In amibroker@xxxxxxxxxxxxxxx, "Steve Almond" <steve2@xxxx> wrote:
> Dimitris,
>
> Yes, that works perfectly. Do you understand why the previous
version worked
> some times, but not others?
>
> Steve
>
> ----- Original Message -----
> From: "DIMITRIS TSOKAKIS" <TSOKAKIS@xxxx>
> To: <amibroker@xxxxxxxxxxxxxxx>
> Sent: Saturday, April 10, 2004 11:37 AM
> Subject: [amibroker] Re: Mathematical question.
>
>
> > I hope it will come full now.[the n1=na; was missing...]
> >
> > na=10;nb=100;step=1;n1=na;
> > f=LastValue(abs(C-MA(C,na)));
> > for(n=na;n<=nb;n=n+step)
> > {
> > //Plot(MA(C,n),"",colorLightGrey,1);
> > d=LastValue(abs(C-MA(C,n)));
> > if(d<f)
> > {
> > n1=n;
> > f=d;
> > }
> > }
> > Plot(C,"\nC",1,8);
> > Plot(MA(C,n1),"MA(C,"+WriteVal(n1,1.0)+")",colorDarkRed,8);
> >
> >
> > --- In amibroker@xxxxxxxxxxxxxxx, "DIMITRIS TSOKAKIS"
<TSOKAKIS@xxxx>
> > wrote:
> > > If you have 5 different MAs [17, 35, 49, 112 and 238 bars] you
> > should
> > > compare the distances with
> > > min(d1,min(d2,min(d3,min(d4,d5)))).
> > > The second question is more interesting, you can use loops to
find
> > > the closest MA.
> > > If you search various MAs from na to nb, then the closest MA
makes
> > > the abs(C-MA(C,n)) minimum.
> > >
> > > na=10;nb=100;step=1;
> > > f=LastValue(abs(C-MA(C,na)));
> > > for(n=na;n<=nb;n=n+step)
> > > {
> > > //Plot(MA(C,n),"",colorLightGrey,1);
> > > d=LastValue(abs(C-MA(C,n)));
> > > if(d<f)
> > > {
> > > n1=n;
> > > f=d;
> > > }
> > > }
> > > Plot(C,"\nC",1,8);
> > > Plot(MA(C,n1),"MA(C,"+WriteVal(n1,1.0)+")",colorDarkRed,8);
> > >
> > > Uncomment the //Plot... line to see the rest MAs.
> > > Calibrate the na, nb, step according to your research field.
> > > Dimitris Tsokakis
> > > --- In amibroker@xxxxxxxxxxxxxxx, "mmqp" <mmqp@xxxx> wrote:
> > > > Hi, I have 5 different moving average. I'd like to
know/measure
> > > the
> > > > least distance of last close to these average. Another word
is
> > > > which moving average is closest to today close. I understand
> > that
> > > > this can be done in a brute force way however I would like to
> > know
> > > > if there is a better mathematical way to do this in case of
more
> > > > than 5 moving average. TIA.
> >
> >
> >
> > Send BUG REPORTS to bugs@xxxx
> > Send SUGGESTIONS to suggest@xxxx
> > -----------------------------------------
> > Post AmiQuote-related messages ONLY to: amiquote@xxxxxxxxxxxxxxx
> > (Web page: http://groups.yahoo.com/group/amiquote/messages/)
> > --------------------------------------------
> > Check group FAQ at:
> http://groups.yahoo.com/group/amibroker/files/groupfaq.html
> > Yahoo! Groups Links
> >
> >
> >
> >
> >
> >
> >
Send BUG REPORTS to bugs@xxxxxxxxxxxxx
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