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



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Herman,
Do not add this "NOT when the Ct crosses anything", it makes things 
complicated without reason. No Ct touch was used in the
http://finance.groups.yahoo.com/group/amibroker/message/72276
solution.
Ct is the [unconditional] next bar Close, as already explained.
Ti3Ct is the [unconditional] next bar Ti3C.
The formulas

Ti3Ct=(C1*f^6+C2*f^5+C3*f^4+C4*f^3)*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);
and
Ct=(Ti3Ct-R*(c1*(f^5*e1+f^4*e2+f^3*e3+f^2*e4+f*e5+e6)+c2*
(f^4*e1+f^3*e2+f^2*e3+f*e4+e5)+c3*(f^3*e1+f^2*e2+f*e3+e4)+c4*
(f^2*e1+f*e2+e3)))/(C1*f^6+C2*f^5+C3*f^4+C4*f^3) ;

are nothing but mathematical definitions.
The first gives the Ti3Ct as a function of Ct.Give a value to Ct and 
you will get the Ti3Ct.
The second gives the Ct as a function of Ti3Ct. Give a value to Ti3Ct 
and you will get Ct.
See also the verification in
http://finance.groups.yahoo.com/group/amibroker/message/72290
Dimitris
--- In amibroker@xxxxxxxxxxxxxxx, "Herman van den Bergen" 
<psytek@xxxx> wrote:
> Thank you DT,
> 
> The problem is different. We need to know the value of tomorrow's 
Ct when
> two OTHER arrays cross, NOT when the Ct crosses anything: Ct is not 
required
> to touch any line. I do not know how to explain it better: the 
objective is
> to know at what value of Ct the other two functions cross, i.e. 
find Ct when
> a F1(Ct) touches F2(Ref(Array,-1)). This is probably a more useful
> application than having the raw C touch any line because the C is 
subject to
> a lot of noise and would give many false signals.  I think your 
innovative
> math functions provide the solution for this problem but i can't 
figure out
> how, I hoped my code would illustrated the problem more clearly... 
Forgive
> me if i am just plain dumb and the solution is staring at me.
> 
> Anyway, thanks very much for your time DT,
> best regards,
> herman.
> 
> 
> 
>   -----Original Message-----
>   From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@x...]
>   Sent: Tuesday, October 26, 2004 2:08 AM
>   To: amibroker@xxxxxxxxxxxxxxx
>   Subject: [amibroker] Re: DT's Ct prediction for the Ti3
> 
> 
> 
>   Herman,
>   The formula is correct and gives the next bar close Ct as a 
function
>   of tomorrow´s Ti3Ct.
>   When Ti3Ct value is known [or equal to another function] then Ct 
is
>   calculated.
>   For a simple verification, set
>   Ti3Ct=Ref(Ti3C,1);
>   and then Ct will be EXACTLY the Ref(C,1).
> 
>   Plot(C,"C",colorBlack,1);
>   periods=5;s=0.7;f=2/(periods+1);R=1-f;
>   //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);
>   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;
>   Ti3C=c1*e6C+c2*e5C+c3*e4C+c4*e3C;
>   Ti3Ct=Ref(Ti3C,1);// condition
>   //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(Ref(Ct,-1),"Ct",colorRed,8);
> 
>   Dimitris
> 
> 
>   --- In amibroker@xxxxxxxxxxxxxxx, "Herman van den Bergen"
>   <psytek@xxxx> wrote:
>   > Hello DT,
>   >
>   > I really appreciate your help with this however is that your
>   solution looks
>   > somewhat the same but isn't accurate...perhaps you misunderstood
>   and are
>   > calculating something else...
>   >
>   > The best way for me to illustrate my need is with a generic
>   iterative
>   > solution. Sorry to impose on you DT but to see how the result
>   differs you
>   > have to display both your solution and mine. When you have 
loaded
>   the demo
>   > code below, click on any bar and open the Param() window. Then
>   apply an
>   > offset to the selected close price until you see the two T3s 
(white
>   and
>   > black) cross, at that point the C price is exactly on the 
Predicted
>   Close
>   > price (Red). This is not the case with your formula. Note that I
>   moved the
>   > Red predicted value forward one bar to allow easier comparison 
to
>   the Close
>   > and show a little square when the target is hit. The key
>   requirement of
>   > course it that C falls exactly on the predicted value when the 
T3s
>   cross -
>   > this doesn't happen with your indicator.
>   >
>   > With EOD data and restricting the calculation to the display 
area
>   only, my
>   > solutions works OK, but i need a continuous indicator for the
>   entire data RT
>   > history, this means about 100K bars. If your formula worked it
>   would enable
>   > me to do that!!!
>   >
>   > Any change of making your solution match mine?
>   >
>   > thanks again DT,
>   > herman.
>   >
>   >
>   > ----------------------------------------------------------------
----
>   --------
>   > ----
>   >
>   > SetBarsRequired(1000000,1000000);
>   >
>   > function Ti3(array,p,s)
>   >  {
>   >  f=2/(p+1);
>   >  e1=EMA(array,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;
>   >  T3=c1*e6+c2*e5+c3*e4+c4*e3;
>   >  return T3;
>   >  }
>   >
>   > function ReferenceFunction( ReferenceArray )
>   >  {
>   >  global T3Periods, T3Sensitivity, ResetReference;
>   >  R = Ti3( ReferenceArray, T3Periods ,T3Sensitivity);
>   >  return R;
>   >  }
>   >
>   > function TestFunction( TestArray )
>   >  {
>   >  global T3Periods, T3Sensitivity, ResetReference;
>   >  R = Ti3( TestArray, T3Periods ,T3Sensitivity);
>   >  return R;
>   >  }
>   >
>   > function GetTriggerPrice( ReferenceArray, TestArray, TestBarNum,
>   T3Period,
>   > T3Sensitivity )
>   >  {
>   >  Precision  = 0.0001;
>   >
>   >  RefArray  = ReferenceFunction( ReferenceArray );
>   >  RefValue  = RefArray[TestBarNum ];
>   >
>   >  TestValue = TestArray[TestBarNum];
>   >  TestIncr  = TestValue/3;
>   >
>   >  do {
>   >   TestArray[TestBarNum ]  = TestValue;
>   >   T3t     = TestFunction( TestArray);
>   >   TodaysT3   = T3t[TestBarNum ];
>   >   if( abs(TodaysT3-RefValue) < Precision );
>   >   else if(TodaysT3< RefValue) TestValue= TestValue+ TestIncr ;
>   >   else TestValue= TestValue- TestIncr ;
>   >   TestIncr = TestIncr /2;
>   >   Error = abs(TodaysT3- RefValue);
>   >   } while ( Error > Precision );
>   >
>   >  return TestArray[TestBarNum ];
>   >  }
>   >
>   > PriceOffSet = Param("Price offset",0,-2,2,0.001);
>   > BarNum = SelectedValue(BarIndex());
>   > CursorBar = BarNum == BarIndex();
>   > ParamPrice = C + PriceOffSet;
>   > C = IIf(CursorBar, ParamPrice, C);
>   > ReferenceArray = Ref(H,-1);
>   > TestArray = C;
>   > T3Periods = 3;
>   > T3Sensitivity = 0.7;
>   >
>   > FirstVisibleBar = Status( "FirstVisibleBar");
>   > Lastvisiblebar = Status("LastVisibleBar");
>   > TP=Null;
>   > for( b = Firstvisiblebar; b < Lastvisiblebar AND b < BarCount; 
b++)
>   >  {
>   >  TP[b] = GetTriggerPrice( ReferenceArray, TestArray, b, 
T3Periods,
>   > T3Sensitivity );
>   >  }
>   >
>   > TargetHit = IIf(abs(C-TP) < 0.01,1,Null);
>   > Plot(C,"C",1,128);
>   > Plot(Ti3( ReferenceArray,
>   T3Periods ,T3Sensitivity),"ReferenceArray",1,1);
>   > Plot(Ti3( TestArray, T3Periods ,T3Sensitivity),"TestArray",2,1);
>   > PlotShapes(TargetHit*shapeHollowSquare,9,0,TP,0);
>   > Plot(TP,"TP",4,1);
>   >
>   > ----------------------------------------------------------------
----
>   --------
>   > ----
>   >
>   >
>   >
>   >
>   >
>   >   -----Original Message-----
>   >   From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@x...]
>   >   Sent: Monday, October 25, 2004 3:04 PM
>   >   To: amibroker@xxxxxxxxxxxxxxx
>   >   Subject: [amibroker] Re: DT's Ct prediction for the Ti3
>   >
>   >
>   >
>   >   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);
>   >   >   >   >   >
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