Inferred distribution of $\rho$

Let us start with pdf of $\rho$. Our starting point is

$$f(\rho\,\vert\,x_1,T_1,x_2,T_2)\!\! \propto \int_0^\infty\!\!\!\int_0^\infty\! \tilde f(\ldots) \, r_1\, r_2\,,% \\$$ (82)

from which it follows

$$f(\rho\,\vert\,x_1,T_1,x_2,T_2)\!\! \propto \int_0^\infty\!\!\!\int_0^\infty\! r_2^{x_2} e^{-T_2\,r_2} \cdot ... ...,r_1}\cdot \delta(r_1\!-\!\rho\cdot r_2)\!\cdot\! f_0(r_2)\!\cdot\! f_0(\rho)\, r_1\, r_2$$ (83)

$$\propto \left[ \int_0^\infty\! r_2^{x_2} e^{-T_2\,r_2} \cdot (\rho\cdot r... ...dot\! e^{-T_1\,\rho\,r_2}\cdot f_0(r_2)\, \mbox{d}r_2 \right] \cdot\! f_0(\rho)$$ (84)

$$\propto \left[\, \rho^{x_1}\cdot \int_0^\infty\! r_2^{x_1+x_2}\cdot e^{-(... ...1\cdot\rho)\cdot r_2}\cdot f_0(r_2) \, \mbox{d}r_2 \right] \cdot\! f_0(\rho)\,,$$ (85)

in which we have explicitly factorized $f_0(\rho)$. Again, besides $f_0(r_2)$, we recognize in the integrand something proportional to a Gamma pdf. If we then model also $f_0(r_2)$ by a Gamma of parameters $\alpha_0$ and $\beta_0$, and again neglect irrelevant factors, we get

$$f(\rho\,\vert\,x_1,T_1,x_2,T_2)\!\! \propto \left[\, \rho^{x_1}\cdot \int_0^\infty\! r_2^{x_1+x_2}\cdot e^{-(... ...pha_0-1} \cdot e^{-\beta_0\,r_2} \, \mbox{d}r_2 \right] \cdot\! f_0(\rho)\ \ \$$ (86)

$$ \propto \left[\, \rho^{x_1}\cdot \int_0^\infty\! r_2^{\alpha_0+x_1+x_2-1}... ...eta_0+T_2+T_1\cdot\rho)\cdot r_2} \, \mbox{d}r_2\, \right] \cdot\! f_0(\rho)\,.$$ (87)

Indicating, in analogy to what done to obtain Eq. (), the power of $r_2$ as $\alpha_* - 1 = \alpha_0+ x_1+x_2-1$, and the factor multiplying $r_2$ at the exponent as $\beta_*=\beta_0+T_2+T_1\cdot\rho$, we get

$$f(\rho\,\vert\,x_1,T_1,x_2,T_2)\!\! \propto \left[\, \rho^{x_1}\cdot \int_0^\infty\! r_2^{\alpha_*-1}\cdot e^{-\beta_*\cdot r_2} \, \mbox{d}r_2 \right] \cdot\! f_0(\rho)$$ (88)

$$ \propto \left[\, \rho^{x_1}\cdot \frac{\Gamma(\alpha_*)}{\beta_*^{\alpha_*}} \right] \cdot\! f_0(\rho)$$ (89)

$$ \propto \bigg[\, \rho^{x_1}\!\cdot\! (\beta_0\!+\!T_2\!+\!T_1\!\cdot\!\rho)^{-(\alpha_0+x_1+x_2)} \bigg] \cdot\! f_0(\rho)\,.\ \ \ \ \ \$$ (90)

What is interesting with this result is that we can consider the term inside the square brackets as an effective likelihood (remember that multiplicative factors are irrelevant), and therefore we can rewrite Eq. () as

$$f(\rho\,\vert\,x_1,T_1,x_2,T_2) \propto {\cal L}(\rho\,;x_1,T_1,x_2,T_2,\alpha_0,\beta_0)\cdot f_0(\rho)\,.$$ (91)

For this reason we can serenely proceed assuming a flat prior about $\rho$, because we can reshape in a second step the result (see Ref. [1] for details). So, assuming $f_0(\rho)=k$ and comparing the expression inside the square bracket of Eq. () with Eq. () we get the normalization just by analogy, thus getting

$$f(\rho\,\vert\,x_1,T_1,x_2,T_2)\!\! = \frac{\Gamma(\alpha_0+x_1+x_2)}{\Gamma(x_1+1)\cdot\Gamma(\alpha_0+x_2-1)} \cdot T_1^{x_1+1}\cdot (\beta_0+T_2)^{\alpha_0+x_2-1} \cdot$$

$$\ \ \rho^{x_1} \cdot (\beta_0 + T_2+T_1\cdot\rho)^{-(\alpha_0+x_1+x_2)} \,,$$ (92)

or

$$f(\rho\,\vert\,x_1,T_1,x_2,T_2)\!\! = \frac{ T_1^{\,x_1+1}\!\cdot\! (\beta_0+T_2)^{\alpha_0+x_2-1}} {\m... ...!1)} \cdot \rho^{x_1} \cdot (\beta_0 + T_2+T_1\cdot\rho)^{-(\alpha_0+x_1+x_2)},$$ (93)

that, for a flat prior about $r_2$, i.e. $\alpha_0=1$ and $\beta_0=0$, becomes

$$f(\rho\,\vert\,x_1,T_1,x_2,T_2)\!\! = \frac{\Gamma(x_1+x_2+1)}{\Gamma(x_1+1)\cdot\Gamma(x_2)} \cdot T_1... ...1+1}\cdot T_2^{\,x_2} \cdot \rho^{x_1} \cdot ( T_2+T_1\cdot\rho)^{-(x_1+x_2+1)}$$ (94)

$$ = \frac{(x_1+x_2)!}{x_1!\cdot (x_2-1)!} \cdot T_1^{\,x_1+1}\cdot T_2^{\,x_2} \cdot \rho^{x_1} \cdot ( T_2+T_1\cdot\rho)^{-(x_1+x_2+1)}$$ (95)

The comparison of this result with Eq. (), obtained using flat priors for $r_1$ and $r_2$, is at least surprising: the structures of the pdf's are the same, but $x_2$ in Eq. () is replaced by $x_2\!-\!1$ in Eq. (). Obviously, the two results will coincide for large $x_2$, and also for small $x_2$ there is not a dramatic difference, as we can see from Fig. .

Figure: Dependence of the inference of $\rho$ from the priors. Solid lines: flat priors on $r_1$ and $r_2$, as in Figs.  and , following from the causal model depicted in Fig. . Dashed lines: flat priors on $r_2$ and $\rho$ (causal model of Fig. .)

As far as the summaries of the distribution are concerned, we get

$$(\rho) = \frac{x_1/T_1}{(x_2+1)/T_2}$$ (96)

$$(\rho) = \mu_\rho = \frac{(x_1+1)/T_1}{(x_2-1)/T_2} \hspace{4.05cm}(\mathbf{x_2>1})$$ (97)

$$\sigma(\rho) = \sqrt{\mu_\rho\cdot \left(\frac{T_2}{T_1}\cdot\frac{x_1+2}{x_2-2} - \mu_\rho\right)} \hspace{1.5cm}(\mathbf{x_2>2})\,.$$ (98)

At this point, instead of taking comfort for the fact that the differences are irrelevant in practical cases, or tout court `rejecting Bayesian methods because of their dependence of priors', it is interesting to try to understand the origin of this effect, certainly related to the priors.

But, before proceeding, let us not forget that Eq. () was obtained assuming a flat prior about $\rho$ and that in that model this prior can be factorized. Therefore the more general pdf of the rate ratio for the model of Fig.  is

$$f(\rho\,\vert\,x_1,T_1,x_2,T_2)\!\! = \frac{1}{\mbox{B}(x_1+1,\alpha_0+x_2-1)} \cdot T_1^{\,x_1+1}\cdot (\beta_0+T_2)^{\alpha_0+x_2-1} \cdot$$

$$\rho^{x_1} \cdot (\beta_0 + T_2+T_1\cdot\rho)^{-(\alpha_0+x_1+x_2)}\cdot f_0(\rho) \,,$$ (99)

having only the limitation (but in reality almost irrelevant, given the flexibility of the Gamma distribution) of depending on the chosen parametrization for $f_0(r_2)$.