Combination of results from similar experiments

Results may be combined in a natural way making an interactive use of Bayesian inference. As a first case we assume several experiments having the same efficiency and exposure time.

• Prior knowledge:
  $$f_\circ(\lambda\,\vert\,I_\circ)\,;$$

• Experiment 1 provides Data$_1$:
  $$f_1(\lambda\,\vert\,I_\circ,$$   Data$_1$$$) \propto f($$Data$_1$$$\,\vert\, \lambda,I_\circ)\cdot f_\circ(\lambda\,\vert\,I_\circ)\,;$$

• Experiment 2 provides Data$_2$:
  $$f_2(\lambda\,\vert\,I_\circ,$$   Data$_1$$$\ldots) \propto f($$Data$_2$$$\,\vert\, \lambda,I_\circ) \cdot f_1(\lambda\,\vert\,\ldots)\,;$$
  $$\Rightarrow f_2(\lambda\,\vert\,I_\circ,$$   Data$_1$$$,$$   Data$_2$$$)\,.$$

• Combining $n$ similar independent experiments we get
  

  $$f(\lambda\,\vert\,\underline{x}) \propto \Pi_{i=1}^n f(x_i\,\vert\,\lambda)\cdot f_\circ(\lambda)$$

  $$\propto f(\underline{x}\,\vert\,\lambda) \cdot f_\circ(\lambda)$$

  $$\propto \Pi_{i=1}^n \frac{e^{-\lambda}\lambda^{x_i}} {x_i!}\cdot f_\circ(\lambda)$$

  $$\propto e^{-n\,\lambda}\lambda^{\sum_{i=1}^nx_i} f_\circ(\lambda)\,.$$ (9.6)

  
  Then it is possible to evaluate expected value, standard deviation, and probability intervals.

As an exercise, let us analyse the two extreme cases, starting from a uniform prior:

$\boxed{\sum_ix_i=0}$

if none of the $n$ similar experiments has observed events we have

$$f(\lambda\,\vert\,n , 0 ) = n e^{-n\,\lambda}$$

$$F(\lambda\,\vert\,n , 0 ) = 1-e^{-n\,\lambda}$$

$$\lambda_u = -\frac{\ln(1-P_u)}{n} P_u\,.$$

$\boxed{\sum_ix_i\mbox{ \lq\lq large''}}$

If the number of observed events is large (and the prior flat), the result will be normally distributed:

$$f(\lambda) \sim {\cal N}(\mu_\lambda, \sigma_\lambda)\,.$$

Then, in this case it is more practical to use maximum likelihood methods than to make integrals (see Section ). From the maximum of $f(\lambda)$, in correspondence of $\lambda=\lambda_m$, we easily get:

$$\mu_\lambda =$$   E$$(\lambda) \approx \lambda_{m}= \frac{\sum_{i=1}^n x_i}{n}\,,$$

and from the second derivative of $\ln f(\lambda)$ around the maximum:

$$\left.\frac{\partial ^2 \ln f(\lambda)}{\partial \lambda^2}\right \vert _{\lambda_{m}} = \frac{-n^2}{\sum_{i=1}^n x_i}$$

$$\sigma_\lambda^2 \approx -\left(\left.\frac{\partial ^2 \ln f(\lambda)}{\partial \lambda}\right \vert _{\lambda_{m}}\right)^{-1} = \frac{1}{n}\frac{\sum_{i=1}^n x_i}{n}$$

$$\sigma_\lambda \approx \frac{\sqrt{\mu_\lambda}}{\sqrt{n}}\, .$$