Dependence on priors

The results of Sections and were obtained using a uniform prior. One may worry how much the result changes if different priors are used in the analysis. Bearing in mind the rule of coherence, we are clearly interested only in reasonable^9.1 priors.

In frontier physics the choice of $f_\circ(\lambda)=k$ is often not reasonable. For example, searching for monopoles, one does not believe that $\lambda=10^6$ and $\lambda=1$ are equally possible. Realistically, one would expect to observe, with the planned experiment and running time, ${\cal O}(10)$ monopoles, if they exist at all. We follow the same arguments of Section (negative neutrino mass), modelling the prior beliefs of a community of rational people who have planned and run the experiment. For reasons of mathematical convenience, we model $f_\circ(\lambda)$ with an exponential, but, extrapolating the results of Section , it is easy to understand that the exact function is not really crucial for the final result.

The function

$$f_\circ(\lambda) = \frac{1}{10}e^{-\lambda/10}\, ,$$ (9.1)

with

$$_\circ[\lambda] = 10$$

$$\sigma_\circ(\lambda) = 10$$

may be well suited to the case: the highest beliefs are for small values of $\lambda$, but also values up to 30 or 50 would not be really surprising. We obtain the following results:

$$f(\lambda\,\vert\,x=0) = \frac{e^{-\lambda}\frac{1}{10}\,e^{-\lambda/10}} {\int_0^\infty (\ldots)\mbox{d}\lambda}$$ (9.2)

$$= \frac{11}{10}\,e^{-\frac{11}{10}\lambda}$$ (9.3)

$$[\lambda] = 0.91$$

$$P(\lambda \le 2.7) = 95\%$$

$$\lambda_u = 2.7 \,.$$ (9.4)

The result is very stable. Changing E$_\circ[\lambda]$ from `$\infty$' to 10 has only a 10% effect on the upper limit. As far as the scientific conclusions are concerned, the two limit are identical. For this reason one should not worry about using a uniform prior, and complicate one's life to model a more realistic prior.

As an exercise, we can extend this result to a generic expected value of events, still sticking to the exponential:

$$f_\circ(\lambda) = \frac{1}{\lambda_\circ}e^{-\lambda/\lambda_\circ}\,,$$

which has an expected value $\lambda_\circ$. The uniform distribution is recovered for $\lambda_\circ\rightarrow\infty$. We get:

$$f(\lambda\,\vert\,x=0,\lambda_\circ) \propto e^{-\lambda}\frac{1}{\lambda_\circ} e^{-\lambda/\lambda_\circ}$$

$$f(\lambda\,\vert\,x=0,\lambda_\circ) = (1+\lambda_\circ) e^{-\lambda\,(1+\lambda_\circ)/\lambda_\circ}$$

$$= \frac{1}{\lambda_1}e^{-\lambda/\lambda_1}$$

$$\frac{1}{\lambda_1} = \frac{1}{1} + \frac{1}{\lambda_\circ}$$

$$F(\lambda\,\vert\,x=0,\lambda_\circ) = 1-e^{-\lambda/\lambda_\circ}\, .$$

The upper limit, at a probability level $P_u$, becomes:

$$\lambda_u = - \lambda_1\,\ln(1-P_u)\,.$$ (9.5)