Re: Is there a bug in FIT?

• To: mathgroup at smc.vnet.net
• Subject: [mg4402] Re: Is there a bug in FIT?
• From: rubin at msu.edu (Paul A. Rubin)
• Date: Fri, 19 Jul 1996 03:26:46 -0400
• Organization: Michigan State University
• Sender: owner-wri-mathgroup at wolfram.com

```In article <4rq9ek\$ld5 at dragonfly.wolfram.com>,
welter at .ti.com (Jason Welter) wrote:
->I am trying to fit some data and have discovered that the
->FIT function does'nt work with the following data:
->
->data1 = {{1.42 10^13, -1.26}, {1.5 10^13, -1.3}, {1.63 10^13, -1.37}};
->
->When I do:
->
->funcdata1 = Fit[data1,{1,x},x];
->
->Show[{Plot[data1,{x,1.42 10^13,1.63 10^13}],ListPlot[data1]},
->Prolog->AbsolutePointSize[16]]
->
->I get terrible results, but when I take the exponents out:
->
->data2 = {{1.42, -1.26}, {1.5, -1.3}, {1.63, -1.37}};
->
->an I do:
->
->funcdata2 = Fit[data2,{1,x},x];
->
->Show[{Plot[data2,{x,1.42,1.63}],ListPlot[data2]},
->Prolog->AbsolutePointSize[16]]
->
->It fits it just fine.  What is wrong?

I suspect it's a problem with mathematical precision.  The formula for the
coefficient estimates is

Inverse[ Transpose[ x ] . x ] . Transpose[ x ] . y,

where x is the data matrix and y the vector of dependent values.  In your
case, the data matrix is

{{1, 1.42 10^13}, {1, 1.5 10^13}, {1, 1.63 10^13}},

the first column (of 1's) corresponding to the constant term in your model.
When you go to compute the inverse, the disparity in scaling of the

Now I computed the coefficients explicitly (as written above), using the
formula above, with both versions of your data, and got correct answers
both times.  That's because Inverse[] coped pretty well with the
conditioning problem.  Fit, on the other hand, may not be using Inverse,
but rather some other solution method which is more challenged by ill
conditioning.

Paul

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* Paul A. Rubin                                  Phone: (517) 432-3509   *
* Department of Management                       Fax:   (517) 432-1111   *
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