Hi Troels,

You should be very careful with your interpretation here.  The
curvature of the chi-squared space does not correlate with the
parameter errors!  Well, it most cases it doesn't.  You will see this
if you map the space for different Monte Carlo simulations.  Some
extreme edge cases might help in understanding the problem.  Lets say
you have a kex value of 100 with a real error of 1000.  In this case,
you could still have a small, perfectly quadratic minimum.  But this
minimum will jump all over the place with the simulations.  Another
extreme example might be kex of 100 with a real error of 0.00000001.
In this case, the chi-squared space could look similar to the
screenshot you attached to the task ( http://gna.org/task/?7882).
However Monte Carlo simulations may hardly perturb the chi-squared
space.  I have observed scenarios similar to these hypothetical cases
with the Lipari and Szabo model-free protein dynamics analysis.

There is one case where the chi-squared space and error space match,
and that is at the limit of the minimum when the chi-squared space
becomes quadratic.  This happens when you zoom right into the minimum.
The correlation matrix approach makes this assumption.  Monte Carlo
simulations do not.  In fact, Monte Carlo simulations are the gold
standard.  There is no technique which is better than Monte Carlo
simulations, if you use enough simulations.  You can only match it by
deriving exact symbolic error equations.

Therefore you really should investigate how your optimisation space is
perturbed by Monte Carlo simulations to understand the correlation -
or non-correlation - of the chi-squared curvature and the parameter
errors.  Try mapping the minimum for the simulations and see if the
distribution of minima matches the chi-squared curvature
(http://gna.org/task/download.php?file_id=23527).

Regards,

Edward


On 16 January 2015 at 17:14, Troels E. Linnet
<no-reply.invalid-addr...@gna.org> wrote:
> URL:
>   <http://gna.org/task/?7882>
>
>                  Summary: Implement Monte-Carlo simulation, where errors are
> generated with width of standard deviation or residuals
>                  Project: relax
>             Submitted by: tlinnet
>             Submitted on: Fri 16 Jan 2015 04:14:30 PM UTC
>          Should Start On: Fri 16 Jan 2015 12:00:00 AM UTC
>    Should be Finished on: Fri 16 Jan 2015 12:00:00 AM UTC
>                 Category: relax's source code
>                 Priority: 5 - Normal
>                   Status: In Progress
>         Percent Complete: 0%
>              Assigned to: tlinnet
>              Open/Closed: Open
>          Discussion Lock: Any
>                   Effort: 0.00
>
>     _______________________________________________________
>
> Details:
>
> This is implemented due to strange results.
>
> A relaxation dispersion on data with 61 spins, and a monte carlo simulation
> with 500 steps, showed un-expected low errors.
>
> -------
> results.read(file=fname_results, dir=dir_results)
>
> # Number of MC
> mc_nr = 500
>
> monte_carlo.setup(number=mc_nr)
> monte_carlo.create_data()
> monte_carlo.initial_values()
> minimise.execute(min_algor='simplex', func_tol=1e-25, max_iter=int(1e7),
> constraints=True)
> monte_carlo.error_analysis()
> --------
>
> The kex was 2111 and with error 16.6.
>
> When performing a dx.map, some weird results was found:
>
> i_sort    dw_sort    pA_sort    kex_sort      chi2_sort
> 471       4.50000    0.99375    2125.00000    4664.31083
> 470       4.50000    0.99375    1750.00000    4665.23872
>
> So, even a small change with chi2, should reflect a larger
> deviation with kex.
>
> It seems, that change of R2eff values according to their errors, is not
> "enough".
>
> According to the regression book of Graphpad
> http://www.graphpad.com/faq/file/Prism4RegressionBook.pdf
>
> Page 33, and 104.
> Standard deviation of residuals is:
>
> Sxy = sqrt(SS/(N-p))
>
> where SS is sum of squares. N - p, is the number of degrees of freedom.
> In relax, SS is spin.chi2, and is weighted.
>
> The random scatter to each R2eff point should be drawn from a gaussian
> distribution with a mean of Zero and SD equal to Sxy.
>
> Additional, find the 2.5 and 97.5 percentile for each parameter.
> The range between these values is the confidence interval.
>
>
>
>
>     _______________________________________________________
>
> File Attachments:
>
>
> -------------------------------------------------------
> Date: Fri 16 Jan 2015 04:14:30 PM UTC  Name: Screenshot-1.png  Size: 161kB
> By: tlinnet
>
> <http://gna.org/task/download.php?file_id=23527>
>
>     _______________________________________________________
>
> Reply to this item at:
>
>   <http://gna.org/task/?7882>
>
> _______________________________________________
>   Message sent via/by Gna!
>   http://gna.org/
>
>
> _______________________________________________
> relax (http://www.nmr-relax.com)
>
> This is the relax-devel mailing list
> relax-devel@gna.org
>
> To unsubscribe from this list, get a password
> reminder, or change your subscription options,
> visit the list information page at
> https://mail.gna.org/listinfo/relax-devel

_______________________________________________
relax (http://www.nmr-relax.com)

This is the relax-devel mailing list
relax-devel@gna.org

To unsubscribe from this list, get a password
reminder, or change your subscription options,
visit the list information page at
https://mail.gna.org/listinfo/relax-devel

Reply via email to