$\left \vert \frac{V_{ub}}{V_{cb}} \right \vert$

Applying our reasoning to the constraint given by $\left \vert \frac{V_{ub}}{V_{cb}} \right \vert$, we obtain

$$f(\bar {\rho},\bar{\eta}\,\vert\,C_1) = \int_{-\infty}^{+\infty}\! \delta(\bar {\rho}^2+\bar{\eta}^2-a)\,... ...\pi}\,0.031}\, \exp{\left[-\frac{(a-0.145)^2}{2\,(0.031)^2}\right]}\, \mbox{d}a$$ (9)

$$= \frac{1}{\sqrt{2\,\pi}\,0.031}\, \exp{\left[-\frac{(\bar {\rho}^2+\bar{\eta}^2-0.145)^2}{2\,(0.031)^2}\right]}\,.$$ (10)

The contour plot shown in Fig. 1 for $\bar{\eta}\ge 0$. 3-D plots are given in Fig. 2 (in all figures ``rho'' and ``eta'' stand for $\bar {\rho}$ and $\bar{\eta}$).

Figure: Contour plot of the p.d.f. of Fig. 2 for $\bar{\eta}\ge 0$. Note that contours are simply iso-p.d.f. levels obtained by 12 contour lines. In order to assign them a probabilistic meaning, one needs to evaluate the p.d.f. integrals inside the contours.

Figure: Probability density function of $\bar {\rho}$ and $\bar{\eta}$ obtained by the constraint given by $\left \vert \frac{V_{ub}}{V_{cb}} \right \vert$.

Note that the plot of Fig. 1 is obtained slicing $f(\bar {\rho},\bar{\eta}\,\vert\,C_1)$ into 12 iso-p.d.f. contours, and hence the regions shown there have no straightforward probabilistic interpretation, since one should make the integrals of the p.d.f. inside the region. We leave it as an exercise for the interested readers.