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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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