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$\left \vert\varepsilon _K \right \vert$

Since the third constraint depends on two strongly correlated parameters, we need to consider a joint bivariate Gaussian distribution:
$\displaystyle f(c,d)$ $\textstyle =$ $\displaystyle \frac{1}{2\,\pi\,\sigma_c\,\sigma_d\,\sqrt{1-\rho_{cd}^2}}\cdot
\...
...rac{1}{2\,(1-\rho_{cd}^2)}
\left[ \frac{(x-\mu_c)^2}{\sigma_c^2} \right.\right.$  
    $\displaystyle \left.\left. - 2\,\rho_{cd}\,\frac{(c-\mu_c)(d-\mu_d)}{\sigma_c\,\sigma_d}
+ \frac{(d-\mu_d)^2}{\sigma_d^2}
\right]
\right\} \,,$ (14)

with $\mu_c=\mbox{E}[c]=3.4$, $\sigma_c=0.9$, $\mu_d=\mbox{E}[d]=1.23$, $\sigma_d=0.33$, and $\rho_{cd}=\rho(c,d)= 0.76$. The integral
\begin{displaymath}
f(\bar {\rho},\bar{\eta}\,\vert\,C_3) = \int\!\!\!\!\int_{-\...
...+c\,(1-\bar {\rho})]-d\right) \,
f(c,d)\,\mbox{d}c\,\mbox{d}c
\end{displaymath} (15)

can be still evaluated analytically but we omit here the final formula, just giving the 3-D plot of the resulting p.d.f. normalized in the region of interest (Fig. 5).
Figure: Probability density function of $\bar {\rho}$ and $\bar{\eta}$ obtained by the constraint given by $\left \vert\varepsilon _K \right \vert$.
\begin{figure}\begin{center}
\epsfig{file=fC3.eps,clip=,width=12.0cm}\end{center}\end{figure}



Giulio D'Agostini 2004-01-20