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@wiso
Last active March 22, 2016 10:47
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chi2
The data have been binned in such a way that every bin contains more than 10 events. For every bin the integral of the S+B postfit pdf has been computed ($E_i$).
In the table the value of the Pearson-$\chi^2 = \sum_i (E_i - O_i)^2 / E_i$ is reported with the number of bins of $m_{\gamma\gamma}$.
Note that the $\chi^2$ is taking into account only the physical pdf, and not the product of constraints. In fact it is difficult to compute the number of degrees of freedom. We have 5 NPs for the background (4 "$\alpha$" + normalization) plus all the NPs for the statistical fluctuations (100+). Since these parameters are constrained they don't count -1 in the sum of the degree of freedom (something between 0 and -1). To try to evaluate their contribution we can imagine to add the constraint pdf to the computation of the $\chi^2$. This means to add 1 "bin", to subtract 1 dof, and to add a contribution to the $\chi^2$. This can (?) be evaluated as
$$-2\log (pdf(x | x_{true})) + 2\log(pdf(x_{true}|x_{true}))$$
for example for a Gaussian constraint it is equal to $((x - x_{true}) / \sigma_{x})^2$. This (summed over all the statistical constraints) is the last column in the table. Unfortunately this is not able to justify the difference between the $\chi^2$ computed with all the NPs free and the one with the NPs of the MC statistical error fixed to the prefit value (1).
| sample | all NPs free | stat NP fixed | stat contribution |
| -----|---------|----|
| 8 TeV graviton | 73 / 116 bins | 87 / 116 bins | 6.6 |
| 13 TeV EKEI | 80 / 99 bins| 121 / 99 bins | 5.71 |
| 13 TeV EKHI | 68 / 85 bins| 102 / 85 bins | 6.47 |
| 13 TeV EKHI func-bkg | 70 / 86 bins | |
| 13 TeV scalar (L. ws) | 65 / 71 bins | |
***To conclude: $\chi^2$ is small, even if we neglect the systematics due to the statisticsl uncertainty of the MC.***
Caveat: I have used S+ fit. As usual I assume the post-fit to be the "true".
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