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Tomasz,
thank you for the reply, I know the definition of the linear
regression model.
The point is that the sum of the square residuals method does not
minimize the cumulative absolute error, as it comes from the codes.
Dimitris Tsokakis
--- In amibroker@xxxxxxxxxxxxxxx, "Tomasz Janeczko" <amibroker@xxxx>
wrote:
> Dimitris,
>
> The linear (Least-Squares ) regression model postulates that
>
> Y= a+bX+e
> Where the "residual" e is a random variable with mean zero. The
coefficients a and b are determined by
> the condition that the sum of the square residuals is as small as
possible.
>
> So error is SUM OF THE SQUARE of residuals.
>
> You are not using sum of squares of residuals as error estimate.
>
> For more info see:
> http://www.stat.yale.edu/Courses/1997-98/101/linreg.htm
>
> Best regards,
> Tomasz Janeczko
> amibroker.com
> ----- Original Message -----
> From: Dimitris Tsokakis
> To: amibroker@xxxxxxxxxxxxxxx
> Sent: Monday, March 15, 2004 12:17 PM
> Subject: [amibroker] The linear trend calculations
>
>
> I tried to duplicate the LinRegIntercept( ARRAY, periods )
example from the 4.50 users guide.
> I used as example the C1=MA(RSI(50),200).
> The error was calculated from Cum(abs(y-C1)).
>
> C1=MA(RSI(50),200);Plot(C1,"C1",1,1);
> x = Cum(1);
> lastx = LastValue( x );
> Daysback = LASTX-ValueWhen(Cum(IsTrue(C1))==1,X);
> aa = LastValue( LinRegIntercept( C1, Daysback) );
> bb = LastValue( LinRegSlope( C1, Daysback ) );
> y = Aa + bb * ( x - (Lastx - DaysBack) );
> Plot( IIf( x >= (lastx - Daysback), y, -1e10 ), "LinReg",
colorRed,8 );
> Title=Name()+"\nLine="+WriteVal(aa,1.1)+"+"+WriteVal(bb,1.3)+"*(t-
t1)"+"\nError="+WriteVal(Cum(abs(y-C1)),1.0);
> start=Cum(IsTrue(C1))==1;
> PlotShapes(shapeUpTriangle*start,colorWhite);
>
> My analytic code
> t=Cum(1)-1;tmax=LastValue(t);
> C1=MA(RSI(50),200);Plot(C1,"C1",1,1);
> C1max=1.5*LastValue(ceil(Highest(C1)));C1min=0.5*LastValue(floor
(Lowest(C1)));
> start=Cum(IsTrue(C1))==1;
> t1=ValueWhen(start,t);
> PlotShapes(shapeUpTriangle*start,colorWhite);
> C10=ValueWhen(start,C1);
> d0=Param("d0",-0.03,-0.03,-0.005,0.005);
> Line0=C10+d0*(t-t1);
> d1=-d0;Line1=C10+d1*(t-t1);
> Error0=LastValue(Cum(abs(C1-Line0)));
> Error1=LastValue(Cum(abs(C1-Line1)));
> Err=Max(Error0,Error1);
> Title=Name()+"\nInitial Error="+WriteVal(err,1.0);
> stepi=0.1;stepj=0.001;
> j1=d1;j0=d0;
> for(i=C1min;i<=C1max;i=i+stepi)
> {
> for(j=j0;j<=j1;j=j+stepj)
> {
> Line=i+j*(t-t1);
> err1=LastValue(Cum(abs(C1-Line)));
> if(err1<err)
> err=Min(err,err1);
> }}
> Title=Title+"\nMinError="+WriteVal(err,1.0);iA=0;jA=0;
> for(i=C1min;i<=C1max;i=i+stepi)
> {
> for(j=j0;j<=j1;j=j+stepj)
> {
> Line=i+j*(t-t1);
> err1=LastValue(Cum(abs(C1-Line)));
> if(err1==err)
> {Plot(Line,"\n"+WriteVal(j,1.4),colorRed,8);
> Title=Title+"\n\nLine="+WriteVal(i,1.1)+"+"+WriteVal(j,1.3)+"*(t-
t1)";
> iA=i;jA=j;
> }}}
> Title=Title+"\nFirst Approx [iA,jA]=["+WriteVal(iA,1.1)
+","+WriteVal(jA,1.3)+"]";
>
> gave, in many cases, smaller error.
> Is there any explanation ?
> Dimitris Tsokakis
>
>
>
>
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