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Scenario III
If the next bar Ti3 should be equal to the Close 3 bars ago, then
calculate the next bar Close Ct.
We shall work again the scenario II formula, since we know Ti3t and
we ask for Ct.
Plot(C,Date()+", C",colorBlack,8);
p=5;s=0.7;f=2/(p+1);R=1-f;
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^3;
C2=3*s^2*(1+s);
C3=-3*s*(s+1)^2;
C4=(1+s)^3;
Ti3=c1*e6+c2*e5+c3*e4+c4*e3;
Plot(Ti3,"Ti3",colorWhite,1);
Ti3t=Ref(C,-3);//the condition
Plot(Ti3t,"Ti3t...",colorRed,8);
//the calculation
Ct=(Ti3t-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) ;
Plot(Ct,"Ct...",5,8);
PlotShapes(shapeCircle*(abs(Ct-Ref(C,1))<0.02),colorYellow);
In CSCO chart I see a yellow circle on Feb3, 2004.
For this date we read Ct=24.06, Ti3t=25.96.It means : If the next bar
Close is 24.06, then the next bar Ti3 will be equal to 25.96.
The next bar close was indeed 24.08 and the next bar Ti3 was 25.96
I hope it is clear, I have nothing more to add.
Dimitris
--- In amibroker@xxxxxxxxxxxxxxx, "DIMITRIS TSOKAKIS" <TSOKAKIS@xxxx>
wrote:
>
> Until then, let us see how it works.
>
> Scenario I:
> If the next bar CSCO close is 19, calculate the next bar Ti3.
>
> Plot(C,Date()+", C",colorBlack,8);
> p=5;s=0.7;f=2/(p+1);R=1-f;
> 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^3;
> C2=3*s^2*(1+s);
> C3=-3*s*(s+1)^2;
> C4=(1+s)^3;
> Ti3=c1*e6+c2*e5+c3*e4+c4*e3;
> Plot(Ti3,"Ti3",colorWhite,1);
> Ct=19;//the next bar Close condition
> Plot(Ct,"Ct",colorBrightGreen,8);
> //the next bar Ti3 calculation
> Ti3t=(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);
> Plot(Ti3t,"Ti3t...",colorRed,8);
>
> On Aug17, for example, we read Ti3t=18.26
> On the next bar, Aug18, the actual close is C=18.99 and the Ti3 is
> indeed 18.26
>
> Scenario II
> If the next CSCO bar Ti3 is 19, then calculate the next bar close.
>
> Plot(C,Date()+", C",colorBlack,8);
> p=5;s=0.7;f=2/(p+1);R=1-f;
> 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^3;
> C2=3*s^2*(1+s);
> C3=-3*s*(s+1)^2;
> C4=(1+s)^3;
> Ti3=c1*e6+c2*e5+c3*e4+c4*e3;
> Plot(Ti3,"Ti3",colorWhite,1);
> Ti3t=19;//the next bar Ti3 condition
> Plot(Ti3t,"Ti3t...",colorRed,8);
> //the next bar Ct calculation
> Ct=(Ti3t-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) ;
> Plot(Ct,"Ct...",colorbrightgreen,8);
>
> On Oct07 we see that we need a next bar Ct=18.75, in order to have
a
> next bar Ti3t=19.
> On the next bar, Oct8, C=18.78 and Ti3=19.00
>
> Dimitris
>
> --- In amibroker@xxxxxxxxxxxxxxx, "DIMITRIS TSOKAKIS"
<TSOKAKIS@xxxx>
> wrote:
> >
> > Herman,
> > Athens would not be a bad idea.
> > 28 Celsius now [20min before 12:00], it will be around 30 some h
> > later.
> > The menu is broiled fresh fish and will be served 4h later.
> > Since the fisher and the broiler is the same person, the
> word "fresh"
> > is guaranteed.
> > 10-year red wine will be also available [unlimited quantity and
> free
> > of charge].
> > It would be, perhaps, the ideal condition to discuss Ti3 details.
> > Although this message looks [strongly] into the future, it is not
> > dangerous as the zig functions.
> > Dimitris
> > --- In amibroker@xxxxxxxxxxxxxxx, "Herman van den Bergen"
> > <psytek@xxxx> wrote:
> > > DT, your patience is much appreciated, it is 4:00 am here so
I'll
> > get some
> > > sleep first and will study your formulas in the (later)
morning,
> > thanks for
> > > the extra explanations, I think they will help. i have a strong
> gut
> > feeling
> > > your formulas really give me what i want - I may just be a bit
> > overwhelmed
> > > by all the terms, they look a bit intimidating :-)
> > >
> > > Thanks again DT, I sure will enjoy working on it later.
> > > herman
> > > -----Original Message-----
> > > From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@x...]
> > > Sent: Wednesday, October 27, 2004 3:13 AM
> > > To: amibroker@xxxxxxxxxxxxxxx
> > > Subject: [amibroker] Re: DT's Ct prediction for the Ti3
> > >
> > >
> > >
> > > 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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