[Ifeffit] running ifeffit under 64-bit windows7

[Matt Newville]

Hi Kicaj,

2012/3/10 "Dr. Dariusz A. Zając" <kicaj at ifj.edu.pl>:
> Hi,
> maybe these below clarify a little bit the problem, but the problem sounds
> very intriguing
> http://millenia.cars.aps.anl.gov/pipermail/ifeffit/2004-July/005729.html
> http://millenia.cars.aps.anl.gov/pipermail/ifeffit/2005-October/006613.html
> http://cars9.uchicago.edu/ifeffit/FAQ/FeffitModeling
>
> I am waiting also for the answer from authors

I would have said these questions have been answered, but maybe I
misunderstand... What is the question you are waiting to be answered?

All of chi-square, reduced chi-square, and R factor express the sum of
squares of the residual (data-model) after a fit has finished.  The
difference between these statistics is how they are scaled.

In particular, chi-square is scaled by the estimated error in the
data. If you look at a (naive?) introduction to statistics, you will
see it stated that this should be approximately the number of degrees
of freedom in the fit.  Reduced chi-square is then defined to be
chi-squared / (the number of degrees of freedom in the fit), so that
it should be 1 (according to statistics 101).   This presupposes a
couple of things that aren't very true for us:
  a) it assumes we actually know the uncertainty in the data -- the
automated estimate in ifefit is pretty simplistic.
  b) it assumes our model of the data is much better than that data
uncertainty. Many people describe these as "systematic errors" and
include alll sorts of data processing artifacts as well as errors in
the Feff calculations.

For us, reduced chi-square is almost always >> 1, unless the data is very noisy.

R-factor scales the fit residual by the magnitude of the data itself,
for some estimate of "fractional misfit".   This gives a convenient
measure that is independent of the scale of the data (and so also
independent of data k-range and k-weight for fits in R-space), and can
more easily be made into a "rule of thumb", say "If R-factor > 0.05,
then you should  be wary of the results".

Hope that helps,

--Matt