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To be more specific:
Ti3t, the next bar Ti3 of an array, is ALWAYS a [known] function of
the next bar array value arrayt.
Arrayt is NOT ALWAYS a [known] function of the next bar Close Ct.
Dimitris
PS : If you could be more specific about array1, array2 we could
probably come to some result...
--- In amibroker@xxxxxxxxxxxxxxx, "Herman van den Bergen"
<psytek@xxxx> wrote:
> Thank you Dt.
>
> herman
> -----Original Message-----
> From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@x...]
> Sent: Monday, October 25, 2004 6:41 AM
> To: amibroker@xxxxxxxxxxxxxxx
> Subject: [amibroker] Re: DT's Ct prediction for the Ti3
>
>
>
> Herman,
> There is no answer for this general question.
> Some arrays may be RevEnged [RSI, Ti3], some others not [StochD].
> The next bar RSI is a function of the next bar Close Ct, the next
bar
> StochD is [unfortunately] a function of Ht, Lt and Ct.
> Dimitris
> --- In amibroker@xxxxxxxxxxxxxxx, "Herman van den Bergen"
> <psytek@xxxx> wrote:
> > Thank you DT, I tried this code but it requires that pb is
greater
> than pa
> > and also it uses C in both Ti3s. I am looking for a sulution
where
> pa==pb
> > and we use different price arrays.
> >
> > best regards,
> > herman
> >
> >
> >
> >
> > -----Original Message-----
> > From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@x...]
> > Sent: Monday, October 25, 2004 1:57 AM
> > To: amibroker@xxxxxxxxxxxxxxx
> > Subject: [amibroker] Re: DT's Ct prediction for the Ti3
> >
> >
> >
> > --- In amibroker@xxxxxxxxxxxxxxx, "Herman van den Bergen"
> > <psytek@xxxx> wrote:
> > > DT, I am not sure i understand the Ti3t... what i am trying
to
> find
> > out is
> > > what tomorrows closing price would be to cause the crossing
of
> two
> > Ti3
> > > functions.
> >
> > Herman,
> > this specific question is already in
> > http://finance.groups.yahoo.com/group/amibroker-ts/files/A%
20Ti3%
> > 20application/
> > "Cross(Ti3a,Ti3b) predictions.txt"
> > Dimitris
> > For example:
> > >
> > > F1 = T3( Array1, 3, 0.8 );
> > > F2 = T3( ref(Array2,-1), 3, 0.8 );
> > >
> > > Array1 and Array2 are two different arrays and could be any
> type of
> > > indicator array. For development you can plug in any of the
OHLC
> > arrays, but
> > > you cannot use the same price array for both function
calles.
> > > The Periods and Sensitivities are the same for both T3
function
> > calls.
> > > F2 is based on yesterdays values, F1 is based on todays
values.
> > > I would like to calculate tomorrows value for Array1 that
would
> > cause the
> > > two functions to cross: cross(F1, F2).
> > >
> > > Do you think this is possible?
> > >
> > > thanks for you help DT!
> > > herman
> > > -----Original Message-----
> > > From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@x...]
> > > Sent: Saturday, October 23, 2004 5:34 AM
> > > To: amibroker@xxxxxxxxxxxxxxx
> > > Subject: [amibroker] Re: DT's Ct prediction for the Ti3
> > >
> > >
> > >
> > > Herman,
> > > Ti3t (the next bar Ti3) is a direct function of the next
bar
> > Close Ct.
> > > Ti3t=
> > > c1*(f*(f*(f*(f*(f*(f*Ct+(1-f)*e1)+
> > > (1-f)*e2)+(1-f)*e3)+(1-f)*e4)+(1-f)*e5)+(1-f)*e6)+
> > > c2*(f*(f*(f*(f*(f*Ct+(1-f)*e1)+(1-f)*e2)+(1-f)*e3)+
> > > (1-f)*e4)+(1-f)*e5)+
> > > c3*(f*(f*(f*(f*Ct+(1-f)*e1)+(1-f)*e2)+(1-f)*e3)+(1-f)*e4)+
> > > c4*(f*(f*(f*Ct+(1-f)*e1)+(1-f)*e2)+(1-f)*e3);
> > > If I understand your question, you want to solve Ct for
any
> given
> > > Ti3t.
> > > Let me know and I will do it.
> > > Dimitris
> > > --- In amibroker@xxxxxxxxxxxxxxx, "Herman van den Bergen"
> > > <psytek@xxxx> wrote:
> > > > Hello,
> > > >
> > > > 'been looking at DT's Ct formula (nice work!) to predict
> where
> > > tomorrow's
> > > > Close will touch the Ti3 - see code below. Can anybody
see a
> > way to
> > > use this
> > > > formula to predict the Close of tomorrow needed to have
the
> Ti3
> > > touch any
> > > > arbitrary point? For example a point on another
indicator.
> > > >
> > > > My math is not up to this, any help would be
appreciated!
> > > >
> > > > herman.
> > > >
> > > > p=3;s=0.84;f=2/(p+1);
> > > > // Ti3
> > > > e1=EMA(C,p);
> > > > e2=EMA(e1,p);
> > > > e3=EMA(e2,p);
> > > > e4=EMA(e3,p);
> > > > e5=EMA(e4,p);
> > > > e6=EMA(e5,p);
> > > > c1=-s*s*s;
> > > > c2=3*s*s+3*s*s*s;
> > > > c3=-6*s*s-3*s-3*s*s*s;
> > > > c4=1+3*s+s*s*s+3*s*s;
> > > > Ti3=c1*e6+c2*e5+c3*e4+c4*e3;
> > > > //The value of tomorrow´s Close Ct that touches
tomorrow´s
> Ti3
> > is
> > > > Ct=
> > > > (1-f)*((c1*f^5+c2*f^4+c3*f^3+c4*f^2)*e1+
> > (c1*f^4+c2*f^3+c3*f^2+c4*f)
> > > *e2+(c1*f
> > > > ^3+c2*f^2+c3*f+c4*1)*e3+(c1*f^2+c2*f+c3*1)*e4+
(c1*f+c2*1)
> > *e5+c1*e6)/
> > > (1-(C1*f
> > > > ^6+C2*f^5+C3*f^4+C4*f^3)) ;//relation III
> > > > Plot(C,"Close",1,128);
> > > > Plot(Ti3,"Ti3",4,1);
> > > > Plot(Ct,"Ct",2,1);
> > >
> > >
> > >
> > >
> > >
> > > Check AmiBroker web page at:
> > > http://www.amibroker.com/
> > >
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> > > http://groups.yahoo.com/group/amibroker/files/groupfaq.html
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