Re: Difference Fit vs. Correlation

*To*: mathgroup at smc.vnet.net*Subject*: [mg96866] Re: Difference Fit vs. Correlation*From*: Claus <clausenator at gmail.com>*Date*: Thu, 26 Feb 2009 07:57:35 -0500 (EST)*Organization*: Comp.Center (RUS), U of Stuttgart, FRG*References*: <go0j8u$n53$1@smc.vnet.net>

Gentlemen, thanks for your comments, I was writing my original post in a hurry, and I must have been in a hurry when I was writing the original code. My problem was that I had calculated the slope of the regression line by multiplying Correlation[GaltonX,GaltonY] with StandardDeviation[GaltonX]/StandardDeviation[GaltonY] However, the correct slope of the regression line is Correlation[GaltonX,GaltonY] * StandardDeviation[GaltonY]/StandardDeviation[GaltonX] Below is some code that should clarify things (use GaltonX and GaltonY from original post). Thank you, Claus In[98]:= lm = LinearModelFit[Transpose@{GaltonX, GaltonY}, {1, x}, x] lm["BestFit"] Out[98]= FittedModel[\!\(\* PanelBox[ TagBox[ RowBox[{"33.88660435407788`", " ", "+", RowBox[{"0.5140930386233082`", " ", "x"}]}], Short], FrameMargins->5]\)] Out[99]= 33.8866 + 0.514093 x should be the same if I use the original pairs In[100]:= lm2 = LinearModelFit[GaltonDat, {1, x}, x] lm2["BestFit"] Out[100]= FittedModel[\!\(\* PanelBox[ TagBox[ RowBox[{"33.88660435407788`", " ", "+", RowBox[{"0.5140930386233082`", " ", "x"}]}], Short], FrameMargins->5]\)] Out[101]= 33.8866 + 0.514093 x In[102]:= CorrelXY = Correlation[GaltonX, GaltonY] Out[102]= 0.501338 In[103]:= SDevX = StandardDeviation[GaltonX] SDevY = StandardDeviation[GaltonY] MY = Mean[GaltonY] MX = Mean[GaltonX] CovXY = Covariance[GaltonX, GaltonY] Out[103]= 2.74487 Out[104]= 2.8147 Out[105]= 68.6841 Out[106]= 67.6871 Out[107]= 3.87333 CorrelXY2 should be the same as CorrelXY In[108]:= CorrelXY2 = CovXY/(SDevX*SDevY) Out[108]= 0.501338 Slope of the regression line In[109]:= slopeRegrL = CorrelXY*SDevY/SDevX Out[109]= 0.514093 In[110]:= interceptRegrL = MY - slopeRegrL*MX Out[110]= 33.8866 In[111]:= regrLine = interceptRegrL + slopeRegrL*x Out[111]= 33.8866 + 0.514093 x Now this is the same as the result from LinearModelFit In[112]:= lm["BestFit"] Out[112]= 33.8866 + 0.514093 x