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Roy,
Thanks!! You never cease to amaze me. I appreciate the explanation as
well.
Preston
--- In equismetastock@xxxxxxxxxxxxxxx, "Roy" <rlarsen@xxxx> wrote:
> Preston
>
> It seems that the formula below for predicting the closing price
> needed for the signal line to cross the MACD is not as accurate as
> it could be. The reason, at least with my version of MS (7.03), is
> that the MACD in MetaStock is based on EMA's that are slightly off
> the assumed 12/26 periods. The MACD EMA's are calculated using
> proportions of 0.075 and 0.15 instead of 2/13 and 2/27. The latter
> is what the code you posted uses for the 'top' variable, hence the
> small errors that occur when tested against the MS canned MACD.
>
> I've taken the liberty of revamping the code so that the predictive
> prices work for my version of MS.
>
> Roy
>
> {MACD MS Look Ahead}
> x:=0.15; y:=0.075;
> a:=If(Cum(1)=1,C,PREV*(1-x)+C*x);
> b:=If(Cum(1)=1,C,PREV*(1-y)+C*y);
> a:=a*(1-x); b:=b*(1-y); top:=b-a;
> {C required for signal/MACD crossover}
> (Mov(MACD(),9,E)+top)/(x-y);
>
>
>
> --- In equismetastock@xxxxxxxxxxxxxxx, pumrysh <no_reply@xxxx>
wrote:
> > This is from another group. See if it gives you any ideas.
> >
> > Preston
> >
> > a user asked if the tommorows close could calculated which would
> > cause the MACD to cross the 0 line, or the trigger line. we
> worked
> > it out and got:
> >
> > a:=12;b:=26;x:=2/(1+a);y:=2/(1+b);
> > top:=Mov(C,b,E)*(1-y)-Mov(C,a,E)*(1-x);
> >
> > {C required for trigger line crossover}
> > (Mov(MACD(),9,E)+top)/(x-y);
> >
> > {C required for 0 value MACD}
> > top/(x-y)
> >
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