Hi, Do the R2eff errors look reasonable? Another issue is in clustered analysis, certain parameters can be over-constrained by being shared between multiple data sets. This is the biased introduced by an under-fitted problem. This can artificially decrease the errors. Anyway, you should plot the Monte Carlo simulations, a bit like I did in figure 4 of my paper:
d'Auvergne, E. J. and Gooley, P. R. (2006). Model-free model elimination: A new step in the model-free dynamic analysis of NMR relaxation data. J. Biomol. NMR, 35(2), 117-135. (http://dx.doi.org/10.1007/s10858-006-9007-z) That might indicate if something is wrong - i.e. if optimisation of certain simulations have failed. However this problem only causes errors to be bigger than they should be (unless all simulations have failed). I don't know how Monte Carlo simulations could fail otherwise. Monte Carlo simulations are the gold standard for error analysis. All other error analysis techniques are judged based on how close the approach this gold standard. Saying that the Monte Carlo simulations technique failed is about equivalent to claiming the Earth is flat! I challenge you to test the statement on a statistics professor at your Uni ;) Anyway, if Monte Carlo failed, using residuals will not save you as the failure point will be present in both techniques. What could have failed is the model or the input data. Under-fitting due to too much R2eff data variability in the spins of the cluster would be my guess. Do you see similarly small errors in the non-clustered analysis of the same data? Regards, Edward On 16 January 2015 at 17:48, Troels Emtekær Linnet <tlin...@nmr-relax.com> wrote: > Hi Edward. > > At the moment, I am fairly confident that I should investigate the > distribution from which the errors are drawn. > > The method in relax draws from a Gauss distribution of the R2eff errors, but > I should try to draw errors from the > overall residual instead. > > It is two different methods. > > My PI, has earlier has before analysed the data with the aforementioned > method, and got errors in the hundreds. > Errors are 5-10% of the fitted global parameters. > > Having 0.5-1 percent error is way to small, and I see this for 4 of my > datasets. > > So, something is fishy. > > Best > Troels > > 2015-01-16 17:30 GMT+01:00 Edward d'Auvergne <edw...@nmr-relax.com>: >> >> 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
