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In my requirements there is only one unknown, the next day Ct as the other
Ti3 was using the previous bar's value. The equation is solvable by
iteration however it is too slow for Real Time data, where in a backtests i
want to process 100,000 bars. The problem could also be stated, since both
Ti3s have the same value when crossing, and assigning arbitrary familiar
price arrays to Array1 and Array2, as:
Ti3( ref(H,-1), period, sensitivity ) = Ti3( C, period, sensitivity )
and solving for C. In this equation C is the only unknown because H is
yesterday's known value. Your earlier solutions are close to this but i
haven't been able to modify them to this requirement. Note that periods and
sensitivities are the same on both sides of the equation.
Best regards,
herman
-----Original Message-----
From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@xxxxxxxxx]
Sent: Monday, October 25, 2004 6:58 AM
To: amibroker@xxxxxxxxxxxxxxx
Subject: [amibroker] Re: DT's Ct prediction for the Ti3
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/
> > >
> > > Check group FAQ at:
> > > http://groups.yahoo.com/group/amibroker/files/groupfaq.html
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