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I must consider plots like these enter image description here

where the bounds on $x$ (in this case $\sin^2\theta_{12}$ or $\delta/\pi$) are shown in terms of the number of standard deviations $N\sigma$ from the best-fit value.

How can I obtain the PDF of the $x$ values from this information?

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  • $\begingroup$ It will be a Gaussian. So just read off the mean and standard deviation from your plot... $\endgroup$
    – lemon
    May 27, 2016 at 9:30
  • $\begingroup$ Yeah in this case it's a gaussian, but I have much worse plots with non-linear, non-symmetric curves.... $\endgroup$
    – mrf1g12
    May 27, 2016 at 9:46
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    $\begingroup$ Oh okay. Either way, you should probably post this question in cross validated since it's more stats than physics. $\endgroup$
    – lemon
    May 27, 2016 at 10:17
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    $\begingroup$ This kind of data scraping is common in physics, so the question should stay here imho $\endgroup$
    – innisfree
    May 27, 2016 at 10:33

1 Answer 1

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I assume the plots follow common frequentist methodology. The numbers of sigma are found by

  1. Building a likelihood function $$\mathcal{L}(x) \equiv \max_\text{other params} p(D|x)$$ where $D$ is the data. NB that this is not a PDF of the variable $x$. The PDF $p(x|D)$ is never calculated in frequentist methods.
  2. Finding a test-statistic $\lambda = -2\ln \frac{\mathcal{L}(x)}{\max \mathcal{L}(x)}$
  3. Assuming that the test statistic is chi-squared distributed by Wilks' theorem, such that the p-value $$ p= p(\lambda≥\lambda_\text{obs}) = 2(1 - \Phi(\sqrt\lambda_\text{obs})) $$ where $\Phi$ is a standard normal CDF.
  4. Lastly converting the p-value into a one-tailed $z$-score, $$ z=\Phi^{-1}(1-p) $$ This is the $N\sigma$.

This whole sequence could be inverted. However, you'll never recover the maximum likelihood, but hopefully you only required the shape of the likelihood function.

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  • $\begingroup$ However, it's not absolutely clear what they mean by $N\sigma$. E.g. whether it's one- or two-tailed $\endgroup$
    – innisfree
    May 27, 2016 at 10:43
  • $\begingroup$ Thank you! I was attempting at reconstructing sort of a "distribution" of $x$, from the experimental results, to employ as a bayesian prior. $\endgroup$
    – mrf1g12
    May 27, 2016 at 12:54
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    $\begingroup$ I'm guessing the model has more than one parameter? It won't be possible as information about correlations was lost by profiling the likelihood. You could say $p(x|D) \sim L(x)$. it'd be approximate but possibly reasonable for a noninformative prior p(x) and strong, informative data $\endgroup$
    – innisfree
    May 27, 2016 at 13:01
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    $\begingroup$ Oh of course, i could have written it with a chisquared CDF with 1 dof and argument $\lambda$, $p=1-F_{\chi^2}(\lambda, n=1)$. That should be exactly equal to the expression I've given $\endgroup$
    – innisfree
    May 27, 2016 at 13:22
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    $\begingroup$ But the standard normal CDF is a more well known function (in numerical libraries etc) $\endgroup$
    – innisfree
    May 27, 2016 at 13:23

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