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Correct procedure

In order to solve the problem consistently with our beliefs, we have to avoid the intermediate inference9.5 on $ \lambda$, and write prior and likelihood directly in terms of $ m$:

$\displaystyle f(m\,\vert\,x=0) \propto \exp{\left[-k\,\left(1-\frac{m^2}{E_b^2}\right)^{\frac{3}{2}}\right]} \cdot f_\circ(m)\,,$ (9.18)

with $ f_\circ(m)=$   constant. Let us do it again with Mathematica:
(********************************************************)
(* Now let's do it right: *)

lik=Exp[-lambda]
norm=NIntegrate[lik, {m, 0, eba}]

(* fa(m) is the final distribution from experiment A, 
   under the condition that m < eba *)

fa=lik/norm
Plot[fa, {m, 0.06, eba}, AxesLabel -> {m, f}]
(********************************************************)
Figure: Inference on $ m$ obtained from a direct inference on $ m$, starting from a uniform prior in this quantity.
\begin{figure}\centering\epsfig{file=higgs2.eps,clip=}\end{figure}
The final distribution is shown in Fig. [*]. It is now reasonable and consistent with the expectations: The values of mass which are less believable are those which could have been produced easier, given the kinematics. From $ f(m\,\vert\,x=0)$ we can calculate several results, for example a 95% upper limit, the average and the standard deviation:
(********************************************************)
NIntegrate[fa, {m, 0, 0.0782}]
ava = NIntegrate[m*fa, {m, 0, eba}]
stda = Sqrt[NIntegrate[m*fa, {m, 0, eba}] - ava^2]
(********************************************************)
We get:
$\displaystyle m$ $\displaystyle >$ $\displaystyle 0.0782\ $   with 95% probability (9.19)
E$\displaystyle (m)$ $\displaystyle =$ $\displaystyle 0.0856$ (9.20)
$\displaystyle \sigma(m)$ $\displaystyle =$ $\displaystyle 0.0038\,.$ (9.21)


next up previous contents
Next: Interpretation of the results Up: Constraining the mass of Previous: Naïve procedure   Contents
Giulio D'Agostini 2003-05-15