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Unbiased results

In the Bayesian approach there is a natural way of giving results in an unbiased way, so that everyone may draw his own scientific conclusion depending on his prior beliefs. One can simply present likelihoods or, for convenience, ratios of likelihoods (Bayes' factors, see Sections [*] and [*]). Some remarks are needed in order not to give the impression that, at the end of this long story, we have not just ended up at likelihood methods. Having clarified these points, let us look at two typical cases.
Classifying hypotheses.
In the case of a discrete number of hypotheses, the proper quantities to report are the likelihoods of the data for each hypothesis

$\displaystyle P($data$\displaystyle \,\vert\,H_i)\,,$

or Bayes' factor for any of the couples

$\displaystyle \frac{P(\mbox{data}\,\vert\,H_i)}{P(\mbox{data}\,\vert\,H_j)}\,.$

On the other hand, the likelihood for a given hypothesis alone, e.g. $ P($data$ \,\vert\,H_\circ)$, does not help the reader to form his idea on the hypothesis, nor on alternatives (see also Section [*]). Therefore, if a collaboration publishes experimental evidence against the Standard Model, suggesting some kind of explanation in terms of a new effect, it should report the likelihoods for both hypotheses. (See also 5th bullet of Section 8.8 in the case of Gaussian likelihood).
Values of quantities.
In this case the likelihoods are summarized by the likelihood function $ f($data$ \,\vert\,\mu)$. In this case one may also calculate Bayes' factors between any pair of values

$\displaystyle \frac{f(\mbox{data}\,\vert\,\mu_i)}{f(\mbox{data}\,\vert\,\mu_j)}\,.$

This can be interesting if only a discrete number of solutions are admissible.

When one publishes a likelihood function this should be clearly stated. Otherwise the temptation to turn $ f($data$ \,\vert\,\mu)$ into $ f(\mu\,\vert\,$data$ )$ is really strong. In fact, taking the example of the neutrino mass of Section [*], the formula

$\displaystyle \frac{1}{\sqrt{2\,\pi}\,2}\,e^{-\frac{(-4-\mu)^2}
{8}}$

(with mass in eV) can easily be considered as if it were a result for $ \mu$:

$\displaystyle \frac{1}{\sqrt{2\,\pi}\,2}\,e^{-\frac{[\mu-(-4)]^2} {8}}$

and conclude that $ \mu_\nu=-4\pm 2\,$eV.
After having criticized this way of publishing the data for the second time, I try in the next section to encourage this way of presenting the result, on condition that one is well aware of what one is writing.

Subsections
next up previous contents
Next: Uniform prior and fictitious Up: Further HEP applications Previous: Signal and background: a   Contents
Giulio D'Agostini 2003-05-15