X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/book_chic.git/blobdiff_plain/a6cad0acff5edb59be08de7a97cbf8b1d78a837a..bc528e875a70eb7d9dd810f644ba646bf6a4e6f6:/chapter2.tex

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+++ b/chapter2.tex
@@ -253,3 +253,47 @@ Assumptions:
   interval $[t,~ t+T[$ depends only on T;
 \item h3: two such events cannot occur simultaneously
 \end{itemize}
+
+It is then demonstrated (for example in~\cite{Saporta}) that the
+number of events occurring during a period of fixed duration $n$
+follows a Poisson's law of parameter $c.n$ where $c$ is called the
+rate of the apparitions process during the unit of time.
+
+
+However, for each transaction assumed to be random, the event $[a=1]$
+has the probability of the frequency $\frac{n_a}{n}$, the event[b=0]
+has as probability the frequency, therefore the joint event $[a=1~
+  and~ b=0]$ has for probability estimated by the frequency
+$\frac{n_a}{n}. \frac{n_{\overline{b}}}{b}$ in the hypothesis of absence of an a priori link between a and b (independence).
+
+We can then estimate the rate $c$ of this event by $\frac{n_a}{n}. \frac{n_{\overline{b}}}{b}$.
+
+Thus for a duration of time $n$, the occurrences of the event $[a~ and~ not~b]$ follow a Poisson's law of parameter : 
+$$\lambda = \frac{n_a.n_{\overline{b}}}{n}$$
+
+As a result, $Pr[Card(X\cap \overline{Y})= s]= e^{-\lambda}\frac{\lambda^s}{s!}$
+
+Consequently, the probability that the hazard will lead, under the
+assumption of the absence of an a priori link between $a$ and $b$, to
+more counter-examples than those observed is:
+
+$$Pr[Card(X\cap \overline{Y})\leq card(A\cap \overline{B})] =
+\sum^{card(A\cap \overline{B})}_{s=0}  e^{-\lambda}\frac{\lambda^s}{s!} $$
+
+ But other legitimate drawing processes lead to a binomial law, or
+ even a hypergeometric law (itself not semantically adapted to the
+ situation because of its symmetry). Under suitable convergence
+ conditions, these two laws are finally reduced to the Poisson Law
+ above (see Annex to this chapter).
+ 
+If $n_{\overline{b}}\neq 0$, we reduce and center this Poison variable
+into the variable:
+
+$$Q(a,\overline{b})= \frac{card(X \cap \overline{Y})) -  \frac{n_a.n_{\overline{b}}}{n}}{\sqrt{\frac{n_a.n_{\overline{b}}}{n}}}  $$
+
+In the experimental realization, the observed value of
+$Q(a,\overline{b})$ is $q(a,\overline{b})$.
+It estimates a gap between the contingency $(card(A\cap
+\overline{B}))$ and the value it would have taken if there had been
+independence between $a$ and $b$.
+