Cross-influences of priors

One might say that in the first case, that of Fig. , yielding Eq. () starting from $f_0(r_1)=f_0(r_2)=k$ there were no priors on $\rho$. But this is quite not true, because the flat priors on $r_1$ and $r_2$ impinge on the prior on $\rho$, due to the relation $\rho=r_1/r_2$. The easiest way to see what is going on is by Monte Carlo, that is, in R,

n = 10^7
rM = 100
r1 = runif(n, 0, rM)
r2 = runif(n, 0, rM)
rho = r1/r2
rho.h <- rho[rho<5] 
hist(rho.h, nc=200, col='blue', freq=FALSE)
abline(v=1, col='red')

where the selection of the values below $\rho\!=\!5$ is to visualize the more interesting region, shown in the top plot of Fig.  (a more complete script, which also performs the correct normalization of the histogram, is shown in Appendix B.4). The histogram is characterized by a plateau till $\rho=1$, followed by a slow decreasing. Curiously, the histogram does not depend on the maximum value rM.

Figure: Distribution of $\rho$ implied by flat priors on $r_1$ and $r_2$ in linear and log scale. The vertical line in the upper plot shows the discontinuity of the distribution at $\rho=1$.

Although it might be bizarre, this histogram shows in essence the prior on $\rho$ we have been tacitly assumed, when flat priors on $r_1$ and $r_2$ were chosen (as a cross check, the commented instructions of the script of Appendix B.4, executed one by one, plot the distribution of $r_1$ assuming a flat prior for $r_2$ and the curious distribution of the top plot of Fig.  for $\rho$).

In order to have a better insight of what is going on, the bottom plot of the same figure shows the histogram of $\log_{10}\rho$. The maximum is at $\log\rho=0$ and it decreases symmetrically, exponentially,²⁸as $\vert\log\rho\vert$ increases. This symmetry indicates that the probabilities to get a value of $\rho$ below or above 1 are the same. The same conclusion, within the uncertainties due to sampling, can be drawn from the histogram in linear scale, since $f(\rho)$ is `about $1/2$' for $0\le \rho \le 1$. Similarly, from the comparison of the two histograms we can evaluate, by symmetry arguments, that the probability that $\rho$ is between 0.1 and 10 is equal to 90% (exact value, indeed as we shall see in a while).

It is interesting to get the distribution shown in the top plot of Fig.  making a transformation of variables, as we have done in Eq. () and following equations:²⁹

$$f(\rho) = \int_0^{r_M}\!\!\int_0^{r_M}\!\delta(\rho-r_1/r_2)\cdot f(r_1)\cdot f(r_2)\, r_1\, r_2$$ (100)

$$= \int_0^{r_M}\!\!\int_0^{r_M}\!r_2\cdot \delta(r_1-\rho\cdot r_2)\cdot \frac{1}{r_M}\cdot \frac{1}{r_M} \, r_1\, r_2\,,$$ (101)

where $r_M$ is the maximum value of $r_1$ and $r_2$.³⁰

At this point, some care is needed with the limits of the integral over $r_2$, due to its `natural' upper limit at $r_M$ and to that given by the constraint $\rho\cdot r_2\le 1$, i.e. $r_2\le 1/\rho$. Therefore, after the trivial integration over $r_1$, we are left with

$$f(\rho) = \frac{1}{r_M^2}\cdot \int_0^{r_2^u}\!r_2 \, r_2\,,$$ (102)

where the upper limit $r_2^u$ depends on $\rho$ in the following way:

$$\rho \le 1 \longrightarrow r_2^u = r_M$$

$$\rho > 1 \longrightarrow r_2^u = r_M/\rho\,.$$

and therefore

$$\rho \le 1 \longrightarrow f(\rho) = \frac{1}{r_M^2}\cdot \int_0^{r_M}\!\!r_2 \, r_2 = \frac{1}{r_M^2}\cdot \frac{r_M^2}{2} = \frac{1}{2}$$

$$\rho > 1 \longrightarrow f(\rho) = \frac{1}{r_M^2}\cdot \int_0^{r_M/\rho}\!\!r_2 \, r_2 = \frac{1}{r_M^2}\cdot \frac{r_M^2}{2\,\rho^2} = \frac{1}{2\,\rho^2}\,,$$

that we summarize as³¹

$$f\left(\rho\,\,\vert\,\,f(r_1)\!\!=\!\!\frac{1}{r_M},f(r_2)\!\!=\!\!\frac{1}{r_M}\right) = \left\{ \begin{array}{ll} \frac{1}{2} & \ \ \ \ \ \ (0 \le \r... ...c{1}{2\,\rho^2} & \ \ \ \ \ \ (\rho > 1) \,, \end{array}\right.$$ (103)

which, indeed, does not depend on the the maximum values of $r_1$ and $r_2$, as we had already learned playing with Monte Carlo simulations.³²

For completeness, let also make the game of seeing how flat priors on $r_2$ and $\rho$ (up to $r_{2_M}$ and $\rho_M$, respectively) are reflected into $r_1$ in the model of Fig.:

$$f(r_1) = \int_0^{\rho_M}\!\int_0^{r_{2_M}}\!\delta(r_1-\rho\,r_2)\cdot f(\rho)\cdot f(r_2)\, \rho\, r_2$$ (104)

$$f(r_1) = \int_0^{\rho_M}\!\int_0^{r_{2_M}} \frac{\delta(\rho-r_1/r_2)}{r_2}\cdot \frac{1}{\rho_M}\cdot \frac{1}{r_{2_M}}\, \rho\, r_2$$ (105)

$$= \frac{1}{\rho_M\cdot r_{2_M}}\cdot \int_{r_{2_L}}^{r_{2_U}}\frac{1}{r_2}\, r_2$$ (106)

where the extremes of integration are $r_{2_L}=r_1/\rho_M$ and $r_{2_U}=r_{2_M}$. Here is, finally, the pdf of $r_1$, in which we have written explicitly the conditions:

$$f\left(\!r_1\,\,\vert\,\,f_0(r_2)\!\!=\!\!\frac{1}{r_{2_M}}, %non è una prior f_0(\rho)\!\!=\!\!\frac{1}{\rho_M}\!\right) \!\!\! = \frac{1}{\rho_M\cdot r_{2_M}}\!\cdot\! \log\left( \frac{r_{2_M}\!\cdot\! \rho_M}{r_1} \right) \hspace{0.5cm}(0<r_1\le r_{2_M}\cdot\rho_{M})\ \ \ \$$

$$$$ (107)

Figure: Histogram of $r_1$ implied by priors on $r_2$ and $\rho$ flat up to $\rho_M=r_{2_M}/$s$^{-1}\!=10$, compared to the exact evaluation of the pdf (solid line). Dashed lines: pdf of $r_1$ for $\rho_M=r_{2_M}/$s$^{-1}\!=30, 100, 300$ (higher to lower, that is from steeper to flatter).

An example with $\rho_M=1$ and $r_{2_M}=10\,$s$^{-1}$ is reported in Fig. , in which the exact pdf (blue solid line) is compared with the Monte Carlo result. The plot also shows the pdf's of $r_1$ for increasing maximum values ( $\rho_M=r_{2_M}/$s$^{-1}=30, 100, 300$, from higher to lower curves). We see that for $\rho_M\rightarrow \infty$ and $r_{2_M}\rightarrow \infty$ also the distribution of $r_1$ becomes flat. This is an interesting result, showing that, contrary to the model of Fig. , the model of Fig.  can accommodate in practice flat prior distributions for the three quantities of interest.³³