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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@xxxxxxxxx]
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);
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > > Check AmiBroker web page at:
> > > > > > http://www.amibroker.com/
> > > > > >
> > > > > > Check group FAQ at:
> > > > > >
> > http://groups.yahoo.com/group/amibroker/files/groupfaq.html
> > > > > >
> > > > > >
> > > > > > Yahoo! Groups Sponsor
> > > > > > ADVERTISEMENT
> > > > > >
> > > > > >
> > > > > >
> > > > > >
> > > > > >
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> > to:
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Check AmiBroker web page at:
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[Non-text portions of this message have been removed]
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