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

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

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:

• the observable is Poisson distributed, and the each experiment can infer a $\lambda$ parameter;
• the result on $\lambda$ must be translated^9.2into a result on $r$.

Starting from a prior on $r$ (e.g. a monopole flux) and going from experiment 1 to $n$ we have

• from $f_\circ(r)$ and ${\cal L}_1$ we get $f_\circ(\lambda)$; then, from the data we perform the inference on $\lambda$ and then on $r$:
  

  $$f_\circ(r) \,\&\, {\cal L}_1 \rightarrow f_{\circ_1}(\lambda)$$

  $$_1 \rightarrow f_1(\lambda\,\vert\, _1, f_{\circ_1}(\lambda))$$

  $$\rightarrow f_1(r\,\vert\, _1, {\cal L}_1, f_\circ(r))\,.$$

  
  
• The process is iterated for the second experiment:
  

  $$f_1(r) \,\&\, {\cal L}_2 \rightarrow f_{\circ_2}(\lambda)$$

  $$_2 \rightarrow f_2(\lambda\,\vert\, _2, f_{\circ_2}(\lambda))$$

  $$\rightarrow f_2(r\,\vert\, _2, {\cal L}_2, f_1(r))$$

  $$\rightarrow f_2(r\,\vert\,( _1,{\cal L}_1), ( _2,{\cal L}_2), f_\circ(r))\,,$$

  
  
• and so on for all the experiments.

Lets us see in detail the case of null observation in all experiments ( $\underline{x}=\underline{0}$) , starting from a uniform distribution.

Experiment 1:

$$f_1(\lambda\,\vert\,x_1=0) = e^{-\lambda}$$

$$f_1(r\,\vert\,x_1=0) = {\cal L}_1e^{-{\cal L}_1 r}$$ (9.7)

$$r_{u_1} = \frac{-\ln 0.05}{{\cal L}_1} \,.$$ (9.8)

Experiment 2:

$$f_{\circ_2} = \frac{{\cal L}_1}{{\cal L}_2}e^{-\frac{{\cal L}_1}{{\cal L}_2} \lambda}$$

$$f_2(\lambda\,\vert\,x_2=0) \propto e^{-\lambda} \frac{{\cal L}_1}{{\cal L}_2}e^{-\frac{{\cal L}_1}{{\cal L}_2}\lambda}$$

$$\propto e^{-\left(1+\frac{{\cal L}_1}{{\cal L}_2} \right)\,\lambda}$$

$$f_2(r\,\vert\,x_1=x_2=0) = ({\cal L}_1+{\cal L}_2) \,e^{-({\cal L}_1+{\cal L}_2)\,r}.$$

Experiment $n$:

$$f_n(r\,\vert\,\underline{x}=\underline{0}, f_\circ(r)=k) = \sum_i{\cal L}_i e^{-\sum_i{\cal L}_i r}\,.$$ (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. $P_u=P(r\le r_u)$]:

$$F_n(r\,\vert\,\underline{x}=\underline{0}, f_\circ(r)=k) = 1- e^{-\sum_i{\cal L}_i r}$$

$$r_u = \frac{-\ln(1-P_u)}{\sum_i{\cal L}_i}$$

$$\frac{1}{r_u} = \frac{-\sum_i{\cal L}_i}{\ln(1-P_u)}$$

$$= \sum_i\frac{-{\cal L}_i}{\ln(1-P_u)}$$

$$= \sum_i \frac{1}{r_{u_i}}\,,$$

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

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