Form of $\chi^2$ test This article uses the following form of the chi-squared test:
$\chi^2=\sum\limits_{i}\frac{(m_i-n_i)^2}{m_i+n_i}$
where $m_i$ is the number of counts of bin $i$ in a modeled histogram and $n_i$ the equivalent for the observed histogram.


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*How does one obtain this form of the $\chi^2$ test?

*Is this a non-parametric statistic or am I assuming a Gaussian distribution of the data?



Edit: so, based on @Michael's answer, am I to assume that the statistics used in that article are in fact wrong?
 A: 1) The authors of the article have written that both $m_i$ and $n_i$ are distributed Poisson, as it's a comparison of simulated to actual data; hence the difference between the two has variance $m_i + n_i$ instead of the usual case, where $m_i$ is the variance (assuming correctness of the model). Consequently, $m_i+n_i$ is the appropriate divisor. 
2) The test is indeed nonparametric.  No Gaussianity assumed!
Edit: Michael Chernick makes a valuable point in comments, namely that the approximation will be poor if the cells are sparse.  The authors do mention the importance of bin size selection, but it's certainly not as though you can take this statistic, apply it without care, and expect to get good results.
Edit the Second:  Michael Chernick makes another valuable point in comments, which is that there are better ways of validating their model against the star data.  Even if they take the "binning" approach, they'd be better off using model-calculated expected values for bin occupancy counts and doing a more typical $\chi^2$ test than comparing simulated data with actual data; the use of simulated data vs. expected values just adds randomness to the results and thereby reduces the power of the test.  It may be, however, that w/o access to the software tool's code, they can't actually do the needed calculations to get bin occupancy rates, in which case this may be about as well as they can do (I could be wrong about that, though.)  
A: The usual chi square test would only have mi in the denominator not the sum.  Chi square is an asymptotic distribution for the test statisitc and doesn't require any Gaussian assumption.
I don't think the test statistic you are using is the right one and since the one with only mi in the denominator is asymptotically chi square I don't think this one is.
