Application to $\rho_\lambda=\lambda_1/\lambda_2$ and to $\rho=r_1/r_2$

In the case of ratio of $\lambda$'s, and starting from uniform prior, as done in Sec. , we get, applying Eq. () and writing the conditions in terms of the Gamma parameters,

$$f(\rho_\lambda\!=\!\frac{\lambda_1}{\lambda_2}\,\vert\,x_1+1,1,x_2+1,1)\!\! = \frac{\Gamma(x_1+x_2+2)}{\Gamma(x_1+1)\cdot \Gamma(x_2+1)}\cdot \rho_\lambda^{\,x_1}\cdot (1+\rho_\lambda)^{\,-(x_1+\,x_2+2)}$$

$$= \frac{(x_1+x_2+1)!}{x_1!\,x_2!} \cdot \rho_\lambda^{\,x_1}\cdot (1+\rho_\lambda)^{\,-(x_1+\,x_2+2)}\,,$$ (61)

re-obtaining exactly Eq. ().

As far as the ratio of rates, starting again from a uniform prior, implying then $\alpha_i=x_i+1$ and $\beta_i=T_i$, we get, writing, as in Eq. (), the conditions in terms of the Gamma parameters,

$$f(\rho\!=\!\frac{r_1}{r_2}\,\vert\,x_1\!+\!1,T_1,x_2\!+\!1,T_2)\!\! = \frac{(x_1+x_2+1)!}{x_1!\,x_2!} \!\cdot\! T_1^{\,x_1+1}\!\cdot\! ... ...,x_2+1}\!\cdot\! \rho^{\,x_1}\!\cdot\! (T_2+ T_1\cdot \rho)^{\,-(x_1+\,x_2+2)},$$

that is, without redundant details,²⁵

$$f(\rho\,\vert\,x_1,T_1,x_2,T_2)\!\! = \frac{(x_1+x_2+1)!}{x_1!\,x_2!} \!\cdot\! T_1^{\,x_1+1}\!\cdot\! ... ...\! \rho^{\,x_1}\!\cdot\! (T_2+ T_1\cdot \rho)^{\,-(x_1+\,x_2+2)}\!,\ \ \ \ \ \$$ (62)

from which we re-obtain Eqs. () and () in the special case $T_1=T_2$, as it has to be. Mode, expected value and standard deviation can be obtained quite easily from Eqs. ()-(), just noting that

$$\rho = \frac{r_1}{r_2} = \frac{\lambda_1/T_1}{\lambda_2/T_2} = \frac{\lambda_1}{\lambda_2}\cdot \frac{T_2}{T_1} = \frac{T_2}{T_1} \cdot \rho_\lambda\,,$$

and then

$$(\rho) = \frac{T_2}{T_1} \cdot (\rho_\lambda) = \frac{T_2}{T_1} \cdot \frac{x_1}{x_2+2} = \frac{x_1/T_1}{(x_2+2)/T_2}$$ (63)

$$(\rho) = \frac{T_2}{T_1} \cdot (\rho_\lambda) = \frac{T_2}{T_1} \cdot \frac{x_1+1}{x_2} = \frac{(x_1+1)/T_1}{x_2/T_2} \hspace{2.1cm}(\mathbf{x_2>0})$$ (64)

$$\sigma(\rho) = \frac{T_2}{T_1}\cdot \sigma(\rho_\lambda) = \frac{T_2}{T_1}\cdot ... ...{x_1+2}{x_2-1} - \frac{x_1+1}{x_2}\right)} \hspace{0.6cm}(\mathbf{x_2>1})\ \ \$$ (65)

that we can rewrite in a more compact form, in terms of $\mu_\rho\equiv$   E$(\rho)$, as

$$\sigma(\rho) = \sqrt{\mu_\rho\cdot \left(\frac{T_2}{T_1}\cdot\frac{x_1+2}{x_2-1} - \mu_\rho\right)}\,, \hspace{2.1cm} (\mathbf{x_2>1})$$ (66)

Some examples are provided in Figs.  and ,

Figure: Ratios of rates, given counts and times.

Figure: As Fig.  for larger values of counts, observed in proportionally larger times.

for low and relatively high numbers of counts, respectively. Each plot shows both the curve of the pdf, calculated with the closed formulae just derived, and the histogram of Monte Carlo simulation (the script to reproduce these plots is given in Appendix B.3). The value of mode, expected value and standard deviation calculated from exact formulae are reported too, together with `mean' and `std' (`empirical standard deviation') evaluated from the sampling. The excellent agreement can be considered a cross check of the exact formulae, derived above for the purpose.

The counts and the measuring times have been chosen such that $(x_1/T_1)/(x_2/T_2)$ are equal to one in all cases. Therefore the plots are comparable to those of Figs.  and reporting $\rho_\lambda$ for several values of $\lambda_1=\lambda_2$ (but in that case all summaries were evaluated from sampling, having, at that stage of the work, not yet derived the closed formulae of interest). As we can again see, for small numbers of counts the distribution of the ratio of rates is strongly asymmetric, with mode and expected value systematically below and above, respectively, the ratio calculated naively as $(x_1/T_1)/(x_2/T_2)$. This value is reached asymptotically, as we see in Fig. , and as expected by the fact that for high numbers of counts we get

$$(\rho) = \frac{x_1/T_1}{(x_2+2)/T_2}\, \xrightarrow[\ x_2\gg 2\ ]{} \frac{x_1/T_1}{x_2/T_2}$$ (67)

$$(\rho) = \frac{(x_1+1)/T_1}{x_2/T_2}\, \xrightarrow[\ x_1\gg 1\ ]{} \frac{x_1/T_1}{x_2/T_2}$$ (68)

$$\sigma(\rho) = \frac{T_2}{T_1}\cdot \sqrt{ \frac{x_1+1}{x_2}\cdot\left(\frac{x_1+2}{x_2-1} - \frac{x_1+1}{x_2}\right)} \xrightarrow[\ x_1,x_2\rightarrow\infty \ ]{} 0\,.$$ (69)