Outside the sensitivity region

With the Bayesian method it is possible to trace the point in which an unstated condition has been introduced, and how to remove it, or how to take it into account. With the form of the likelihood used in () it was implicit that $m$ should not exceed $E_b$. A more physically motivated likelihood should be:

$$f(x=0\,\vert\,m) = \left\{ \begin{array}{lcl} \exp{\left[-k\,\lef... ...ht]} & \mbox{if} & 0\le m\le E_b \\ 1 & \mbox{if} & m > E_b \end{array} \right.$$ (9.24)

Taking a uniform prior, we get the following posterior:

$$f(m\,\vert\,x=0) = \left\{ \begin{array}{lcl} \frac{\exp{\left[-k... ...\footnotesize d}m \,+\, (m_{max}-E_b)} & \mbox{if} & m > E_b \end{array}\right.$$ (9.25)

where $(m_{max}-E_b)$ comes from the integral $\int_{E_b}^{m_{max}} 1\cdot {\rm d}m$. So, we get our solution () for $m_{max}=E_b$. In general, the probability that $m\le E_b$ is smaller than 1 and decreases for increasing $m_{max}$. For the parameters of experiment $A$ the integral in the denominator is equal to 0.0058. Therefore, if, for example, $m_{max}=3\,E_b$

$$P(m < E_b \,\vert\, x=0, m_{max}=3\,E_b) = 2.7\%$$

$$P(m < 0.078 \,\vert\, x=0, m_{max}=3\,E_b) = 0.13\% .$$

There is another reasoning which leads to the same conclusion. At $m = E_b$ the detector has zero sensitivity. For this reason, in case of null observation, this values gets the maximum degree of belief. As far as larger values are concerned, the odds ratios with respect to $m = E_b$ must be invariant, since they are not influenced by the experimental observations, i.e.

$$\frac{f(m\,\vert\,x=0)}{f(m=E_b\,\vert\,x=0)} = \frac{f_\circ(m)}{f_\circ(m=E_b)}\ \ \ \ \ (m>E_b)\,.$$ (9.26)

Since we are using, for the moment, a uniform distribution, the condition gives:

$$f(m\,\vert\,x=0) = f(m=E_b\,\vert\,x=0) \ \ \ (m>E_b)\,.$$ (9.27)

Figure: Result of the inference from experiment $A$, taking into account values of mass above the beam energy as well. These all have the same degree of belief and the normalization constant depends on the maximum value of $m$ considered. Therefore the distribution is usually improper.

We easily get the result shown in Fig. by this piece of Mathematica code:

(********************************************************)
famax=fa/.m->eba
f2a=If[m<eba, fa, famax]

(* f2a(m) represents instead the (improper) distribution
   extended also for values larger that eba, in the light 
   of a flat prior and of the Experiment A *)

Plot[f2a, {m,0.06,0.15}]
(********************************************************)

The curve is extended on the right side up to a limit which cannot be determined by this experiment, it could virtually go to infinity. For this reason the ratio of probabilities

$$\frac{P(m < E_b)}{P(m\ge E_b)}$$

decreases (i.e. we tend to believe more strongly large mass values) but its exact value is not well defined. For this reason we leave the function `open' on the right side and unnormalized. The normalization will be done when we can include other data which can provide an upper limit.