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Herman,
As I wrote in previous post, Ct is a function of Ti3Ct [and vice
versa]
In your example, Ti3C and Ti3H should be calculated separately, to
avoid any confusion.
The expected next bar Close Ct would be
Plot(C,"C",colorBlack,8);
periods=5;s=0.7;f=2/(periods+1);R=1-f;
//the Ti3H
price=H;
e1H=EMA(price,periods);
e2H=EMA(e1H,Periods);
e3H=EMA(e2H,Periods);
e4H=EMA(e3H,Periods);
e5H=EMA(e4H,Periods);
e6H=EMA(e5H,Periods);
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;
Ti3H=c1*e6H+c2*e5H+c3*e4H+c4*e3H;
//Plot(Ti3H,"Ti3H",1,1);
//the Ti3C
price=C;
e1C=EMA(price,periods);
e2C=EMA(e1C,Periods);
e3C=EMA(e2C,Periods);
e4C=EMA(e3C,Periods);
e5C=EMA(e4C,Periods);
e6C=EMA(e5C,Periods);
Ti3C=c1*e6C+c2*e5C+c3*e4C+c4*e3C;
//Plot(Ti3C,"Ti3C",2,1);
Ti3Ct=Ti3H;//your condition:the next bar Ti3C is equal to todays Ti3H
//the Ct as a function of Ti3Ct
Ct=(Ti3Ct-R*(c1*(f^5*e1C+f^4*e2C+f^3*e3C+f^2*e4C+f*e5C+e6C)+c2*
(f^4*e1C+f^3*e2C+f^2*e3C+f*e4C+e5C)+c3*(f^3*e1C+f^2*e2C+f*e3C+e4C)+c4*
(f^2*e1C+f*e2C+e3C)))/(C1*f^6+C2*f^5+C3*f^4+C4*f^3) ;
Plot(Ct,"Ct",colorRed,8);
For periods>10 the condition is quite rare.
Dimitris
--- In amibroker@xxxxxxxxxxxxxxx, "Herman van den Bergen"
<psytek@xxxx> wrote:
> 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@x...]
> 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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