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Interpretation of the results

At this point we need to discuss the results achieved so far, to understand what they mean, and what they don't. In particular, the first statement gives the impression that we can say something about the mass even if $ H$ was too heavy to be produced. In general, a statement like
``A 95% confidence level lower bound of $ 77.5\,$$ \mbox \rm {{GeV}/\it {c}^2}$ is obtained for the mass of the Standard Model Higgs boson.''[85]
may be misleading, because it transmits information which is inconsistent with the experimental observation. The interpretation of the result ([*]) is limited to
$\displaystyle P(m > 0.0782\,\vert\, \underline{0\le m\le 0.09})$ $\displaystyle =$ $\displaystyle 0.95\,,$ (9.22)

as may be understood intuitively and will be shown in a while (there is also the condition of the uniform prior, but at this level it is irrelevant). So, given this status of information, I could bet 1:19 that the $ m$ is below 0.0782, but only on condition that the bet is invalidated if $ m$ turns out to be greater than the beam energy (see Section [*]). Otherwise, I would choose the other direction (19:1 on `$ m>0.0782$') without hesitation (and wish fervently that somebody accepts my bet...).

What are our rational beliefs on $ m$, on the basis of experiment A, releasing the condition $ \le m\le E_b$? The data cannot help much because there is no experimental sensitivity, and the conclusions depend essentially on the priors.

To summarize, the result of the inference is:

$ \mathbf{m < E_b}$:
$ P(m > 0.0782) = 0.95$; $ m=0.0856\pm 0.038$, etc. ;
$ \mathbf{m \ge E_b}$:

As a final remark on the presentation of the result, I would like to comment on the three significative digits with which the result on the `conditional lower bound' has been given. For the sake of the exercise the mass bound has been evaluated from the condition ([*]). But does it really matter if the limit is 0.0782, rather than 0.0780, or 0.0800? As stated in Sections [*] and [*], the limits have to be considered in the same way as the uncertainty. Nobody cares if the uncertainty of the uncertainty is 10 or 20%, and nobody would redo a MACRO-like experiment to lower the monopole limit by 20%. Simply translating this argument to the case under study, it may give the impression that one significant digit would be enough (0.08), but this is not true, if we stick to presenting the result under the condition that $ m$ is smaller than $ E_b$. In fact, what really matters, is not the absolute mass, but the mass difference with respect to the kinematical limit. If the experiment ran with infinite statistics and found `nothing', there is no interest in providing a detailed study for the limit: it will be exactly the same as the kinematical limit. Therefore, the interesting piece of information that the experimenter should provide is how far the lower bound is from the kinematical limit, i.e. what really matters is not the absolute mass scale, but rather the mass difference. In our case we have

$\displaystyle \Delta m = E_b -$   lower bound$\displaystyle = 0.0118 \rightarrow 0.012\,.$ (9.23)

Two digits for this number are enough (or even only one, if the first were greater that 5) and the value of the lower bound becomes9.7

$\displaystyle m > 0.078 \ $   at 95%, if $\displaystyle 0\le m \le 0.09\,.$

next up previous contents
Next: Outside the sensitivity region Up: Constraining the mass of Previous: Correct procedure   Contents
Giulio D'Agostini 2003-05-15