Line fits and the use of V_rab
chozo
Tue, 05/10/2022 - 05:58 am
I realized that Line fits do not output a covariance matrix even with /M=2, and a correlation coefficient V_rab is given instead. I have a bit trouble wrapping my head around using this correlation coefficient in error propagation. Is the following correct when calculating the error from a multiplication or division of slope and intercept (e.g., a/b):
NVAR coefCorr = V_rab
Wave cw = W_coef
Wave sw = W_sigma
Variable err = sqrt( (sw[0]/cw[0])^2 + (sw[1]/cw[1])^2 + 2*(coefCorr*sw[0]*sw[1])/(cw[0]*cw[1]) )
Wave cw = W_coef
Wave sw = W_sigma
Variable err = sqrt( (sw[0]/cw[0])^2 + (sw[1]/cw[1])^2 + 2*(coefCorr*sw[0]*sw[1])/(cw[0]*cw[1]) )
Here I 'convert' the correlation coefficient into a covariance-like factor by multiplying with sigma. I am not sure if I make a mistake on the order of sqrt() or x^2.
Why I need this? To generate derived values such as the x-intercept of a line in my Super Quick Fit package, of course. ;)
I would need to think hard about your equations for error propagation, but if it helps you can force generation of the covariance matrix by using a user defined function:
Display wData
ModifyGraph mode=3
CurveFit/M=2/NTHR=0 line wData /D
// fit_wData= W_coef[0]+W_coef[1]*x
// W_coef={1.619,0.88571}
// V_chisq= 1.10476;V_npnts= 6;V_numNaNs= 0;V_numINFs= 0;
// V_startRow= 0;V_endRow= 5;V_q= 1;V_Rab= -0.825723;V_Pr= 0.96204;
// V_r2= 0.925522;
// W_sigma={0.38,0.126}
// Coefficient values ± one standard deviation
// a =1.619 ± 0.38
// b =0.88571 ± 0.126
// ** M_Covar not created **
Function myLineFun(w,x) : FitFunc
Wave w
Variable x
//CurveFitDialog/ These comments were created by the Curve Fitting dialog. Altering them will
//CurveFitDialog/ make the function less convenient to work with in the Curve Fitting dialog.
//CurveFitDialog/ Equation:
//CurveFitDialog/ f(x) = a + b * x
//CurveFitDialog/ End of Equation
//CurveFitDialog/ Independent Variables 1
//CurveFitDialog/ x
//CurveFitDialog/ Coefficients 2
//CurveFitDialog/ w[0] = a
//CurveFitDialog/ w[1] = b
return w[0] + w[1] * x
End
Make/D/N=2/O W_coef
W_coef[0] = {1,0}
FuncFit/M=2/NTHR=0 myLineFun W_coef wData /D
//Fit converged properly
// fit_wData= myLineFun(W_coef,x)
// W_coef={1.6191,0.88571}
// V_chisq= 1.10476;V_npnts= 6;V_numNaNs= 0;V_numINFs= 0;
// V_startRow= 0;V_endRow= 5;
// W_sigma={0.38,0.126}
// Coefficient values ± one standard deviation
// a =1.6191 ± 0.38
// b =0.88571 ± 0.126
print M_Covar
// M_Covar[0][0]= {0.144671,-0.0394558}
// M_Covar[0][1]= {-0.0394558,0.0157823}
May 10, 2022 at 06:37 am - Permalink
Hi Kurt, thanks a lot for the hint. While I do not want to replicate the Line function (this would introduce some clutter in my package https://www.wavemetrics.com/node/21866 ), I was able to test that my approach is correct:
Display wData
ModifyGraph mode=3
CurveFit/M=2/NTHR=0 line wData /D
print V_rab*W_sigma[0]*W_sigma[1]
-0.0394558
Which is (almost) equal to:
print M_Covar[0][1]
-0.0394559
If this is not just a strange coincidence with this particular data set, then I guess my approach using coefCorr*sw[0]*sw[1] as a replacement for the coefficient matrix is OK. Great.
May 10, 2022 at 07:43 am - Permalink