Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a set of dihedral angle values that I have fitted using a polygaussian function via the Levenberg-Marquardt algorithm http://en.wikipedia.org/wiki/Levenberg-Marquardt. Specifically, the function $$f(x) = G_1(x) + G_2(x) + ... + G_n(x)$$ where $$Gi(x) = \frac{\omega_i}{\sqrt(2\pi)\sigma_i}e^{-\frac{(x - \mu_i)^2}{2\sigma_i²}}$$

was defined. The values of $\mu_i,\sigma_i,\omega_i$ are declared as the respective mean, standard deviation and weight constants for that particular subgaussian. It is also true that $\sum^n_{i=1}{\omega_i} = 1$

Given that the original values were a dihedral angles vector, I plotted a histogram from them, using a particular bin size. Then, I extracted the $(x, y)$ pair values from the final bars. $x$ are the x-axis values in the middle of the bin, and $y$ are the frequencies for that bin, within the range $(x - binsize/2,x + binsize/2)$.

I am trying to figure out a way to assess the statistical significance of the fitting. I know that from this analysis a goodness of fit value (Q) can be obtained. However, I am aware that the values that I plugged in where modified to create the $(x, y)$ pairs, and there are no estimated errors for them.

I can currently measure the fitting quality via the area difference of the final function and the area from an averaged shifted histogram of the same dataset (http://sfb649.wiwi.hu-berlin.de/fedc_homepage/xplore/ebooks/html/spm/spmhtmlnode11.html). A low area difference value shows that the fittings are good, but not that they are statistically significant.

I would also like to ask if there exist a Normality Test available for multiple Gaussians (http://en.wikipedia.org/wiki/Normality_test)

Thanks for any advice, Ignacio

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.