Re: [Gnuplot-info] Outputs of fit: details of how they're defined

[Dan <vi5...@ya...>]

> Am 20.11.2017 um 16:32 schrieb Daniel Hatton via gnuplot-info:
>
>> Please can anyone confirm whether I've correctly assessed what the fit
>> routine is putting in those output variables?

On Tue, 21 Nov 2017, Hans-Bernhard Bröker wrote:

> Not from those Unicode jungles I can't, sorry.

Thanks for trying, Hans-Bernhard.  I'll try and say the same things in
LaTeX, then...

Let's say that I have run

set fit noerrorscaling
set fit errorvariables
set fit covariancevariables

fit some_equation(L,M,x) 'some_data_set.dat' using ($1):($3):($2):($4)
errors x,z via L,M

where some_equation(L,M,x) is a theoretical formula, with adjustable
parameters L and M, which is supposed to predict the quantity in the
third column of the data file, when the value of the quantity in the
first column of the data file is x; the second column of the data file
contains the standard uncertainties in the measured values in the
first column of the data file 'some_data_set.dat'; and the fourth
column of the data file contains the standard uncertainties in the
measured values in the third column of the data file
'some_data_set.dat'.

Then I _think_ I expect to find:

- in the variables L and M, the values l and m of the L and M
parameters respectively, which, among all the (L,M) pairs that the
Marquardt-Levenberg algorithm has tried, produce the lowest value of
the \chi^2 statistic;
- in the variable FIT_WSSR, the value \Chi^2 of the \chi^2 statistic
that is achieved by setting L = l and M = m;
- in the variable L_err, the value (presumably calculated in a
finite-difference way) of
\sqrt{\frac{2}{\left(\partial\!^{2}\!\left(\chi^2\right)/\partial L\!^{2}\right)_{L = l,M = m}-\left(\left(\partial\!^{2}\!\left(\chi^2\right)/\left(\partial\!\!L\!\partial\!\!M\right)\right)_{L = l,M = m}\right)^2/\left(\partial\!^{2}\!\left(\chi^2\right)/\partial M\!^{2}\right)_{L = l,M = m}}}
- in the variable M_err, the value (presumably calculated in a
finite-difference way) of
\sqrt{\frac{2}{\left(\partial\!^{2}\!\left(\chi^2\right)/\partial M\!^{2}\right)_{L = l,M = m}-\left(\left(\partial\!^{2}\!\left(\chi^2\right)/\left(\partial\!\!L\!\partial\!\!M\right)\right)_{L = l,M = m}\right)^2/\left(\partial\!^{2}\!\left(\chi^2\right)/\partial L\!^{2}\right)_{L = l,M = m}}}
- in the variable FIT_COV_M_L, the value (presumably calculated in a
finite-difference way) of
\frac{2\left(\partial\!^{2}\!\left(\chi^2\right)/\left(\partial\!\!L\!\partial\!\!M\right)\right)_{L = l,M = m}}{\left(\partial\!^{2}\!\left(\chi^2\right)/\partial L\!^{2}\right)_{L = l,M = m}\left(\partial\!^{2}\!\left(\chi^2\right)/\partial M\!^{2}\right)_{L = l,M = m}-\left(\left(\partial\!^{2}\!\left(\chi^2\right)/\left(\partial\!\!L\!\partial\!\!M\right)\right)_{L = l,M = m}\right)^2}

Please can anyone confirm whether I've correctly assessed what the fit
routine is putting in those output variables?

> Had you found and considered the documentation under "help 
> statistical_overview"?

I've now had a look, but it doesn't tell me any more than the manual
did.

> You can rest assured that no Bayesian-style reasoning whatsoever was
> used in the design or implementation of this function.  This is
> strictly classical statistics and optimization theory, assuming
> normal-distributed inputs etc.

OK... but irrespective of whether the author of the relevant section
of the manual (and of "help statistical_overview") was a Bayesian or a
frequentist, his/her use of the word 'asymptotic' (in 'asymptotic
standard error') strongly suggests that s/he has seen a relevant
integral approximated using Laplace's method.  What I'm after is a
reference to a published source where that Laplace's method
approximation to a relevant integral took place.  (A Bayesian would
have had a subtly different integrand from a frequentist, but in many
ways, the real magic is that I'd expect the leading-order Laplace's
method approximation to the Bayesian's integral to be exactly the same
as the leading-order Laplace's method approximation to the
frequentist's integral.)

Thanks again.

-- 

Kind regards,

Dan