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Combination of results: general case

The previous case is rather artificial and can be used, at most, to combine several measurements of the same experiment repeated $ n$ times, each with the same running time. In general, experiments differ in size, efficiency, and running time. A result on $ \lambda$ is no longer meaningful. The quantity which is independent from these contingent factors is the rate, related to $ \lambda$ by

$\displaystyle r=\frac{\lambda}{\epsilon\, S\,\Delta T} = \frac{\lambda}{\cal L}\, ,$

where $ \epsilon$ indicates the efficiency, $ S$ the generic `size' (either area or volume, depending on whatever is relevant for the kind of detection) and $ \Delta T$ the running time: all the factors have been grouped into a generic `integrated luminosity' $ {\cal L}$ which quantify the effective exposure of the experiment.

As seen in the previous case, the combined result can be achieved using Bayes' theorem iteratively, but now one has to pay attention to the fact that:

Starting from a prior on $ r$ (e.g. a monopole flux) and going from experiment 1 to $ n$ we have Lets us see in detail the case of null observation in all experiments ( $ \underline{x}=\underline{0}$) , starting from a uniform distribution.
Experiment 1:

$\displaystyle f_1(\lambda\,\vert\,x_1=0)$ $\displaystyle =$ $\displaystyle e^{-\lambda}$  
$\displaystyle f_1(r\,\vert\,x_1=0)$ $\displaystyle =$ $\displaystyle {\cal L}_1e^{-{\cal L}_1 r}$ (9.7)
$\displaystyle r_{u_1}$ $\displaystyle =$ $\displaystyle \frac{-\ln 0.05}{{\cal L}_1}$ at 95% probability$\displaystyle \,.$ (9.8)

Experiment 2:

$\displaystyle f_{\circ_2}$ $\displaystyle =$ $\displaystyle \frac{{\cal L}_1}{{\cal L}_2}e^{-\frac{{\cal L}_1}{{\cal L}_2}
\lambda}$  
$\displaystyle f_2(\lambda\,\vert\,x_2=0)$ $\displaystyle \propto$ $\displaystyle e^{-\lambda}
\frac{{\cal L}_1}{{\cal L}_2}e^{-\frac{{\cal L}_1}{{\cal L}_2}\lambda}$  
  $\displaystyle \propto$ $\displaystyle e^{-\left(1+\frac{{\cal L}_1}{{\cal L}_2}
\right)\,\lambda}$  
$\displaystyle f_2(r\,\vert\,x_1=x_2=0)$ $\displaystyle =$ $\displaystyle ({\cal L}_1+{\cal L}_2)
\,e^{-({\cal L}_1+{\cal L}_2)\,r}.$  

Experiment $ n$:

$\displaystyle f_n(r\,\vert\,\underline{x}=\underline{0}, f_\circ(r)=k) = \sum_i{\cal L}_i e^{-\sum_i{\cal L}_i r}\,.$ (9.9)

The final result is insensitive to the data grouping. As the intuition suggests, many experiments give the same result of a single experiment with equivalent luminosity. To get the upper limit, we calculate, as usual, the cumulative distribution and require a certain probability $ P_u$ for $ r$ to be below $ r_u$ [i.e. $ P_u=P(r\le r_u)$]:
$\displaystyle F_n(r\,\vert\,\underline{x}=\underline{0}, f_\circ(r)=k)$ $\displaystyle =$ $\displaystyle 1- e^{-\sum_i{\cal L}_i r}$  
$\displaystyle r_u$ $\displaystyle =$ $\displaystyle \frac{-\ln(1-P_u)}{\sum_i{\cal L}_i}$  
$\displaystyle \frac{1}{r_u}$ $\displaystyle =$ $\displaystyle \frac{-\sum_i{\cal L}_i}{\ln(1-P_u)}$  
  $\displaystyle =$ $\displaystyle \sum_i\frac{-{\cal L}_i}{\ln(1-P_u)}$  
  $\displaystyle =$ $\displaystyle \sum_i \frac{1}{r_{u_i}}\,,$  

obtaining the following rule for the combination of upper limits on rates:

$\displaystyle \frac{1}{r_u} = \sum_i\frac{1}{r_{u_i}}\,.$ (9.10)

We have considered here only the case in which no background is expected, but it is not difficult to take background into account, following what has been said in Section [*].
next up previous contents
Next: Including systematic effects Up: Poisson model: dependence on Previous: Combination of results from   Contents
Giulio D'Agostini 2003-05-15