PDF from number of standard deviations plot

I must consider plots like these  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?

• It will be a Gaussian. So just read off the mean and standard deviation from your plot... – lemon May 27 '16 at 9:30
• Yeah in this case it's a gaussian, but I have much worse plots with non-linear, non-symmetric curves.... – mrf1g12 May 27 '16 at 9:46
• Oh okay. Either way, you should probably post this question in cross validated since it's more stats than physics. – lemon May 27 '16 at 10:17
• This kind of data scraping is common in physics, so the question should stay here imho – innisfree May 27 '16 at 10:33

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.

• However, it's not absolutely clear what they mean by $N\sigma$. E.g. whether it's one- or two-tailed – innisfree May 27 '16 at 10:43
• Thank you! I was attempting at reconstructing sort of a "distribution" of $x$, from the experimental results, to employ as a bayesian prior. – mrf1g12 May 27 '16 at 12:54
• 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 – innisfree May 27 '16 at 13:01
• 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 – innisfree May 27 '16 at 13:22
• But the standard normal CDF is a more well known function (in numerical libraries etc) – innisfree May 27 '16 at 13:23