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RE: [amibroker] Re: DT's Ct prediction for the Ti3



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Thank you Dt.

herman
  -----Original Message-----
  From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@xxxxxxxxx]
  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);
  >   >
  >   >
  >   >
  >   >
  >   >
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  >   >
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