LES ORIGINES DE LA LOI NORMALE

Ci dessous le long et douloureux calcul qui montre le lien entre loi binomiale et loi normale:

Point de départ, la formule de Stirling: n!\sim(\dfrac{n}{e})^n\sqrt{2\pi n}.

Dans une suite de n expériences, la probabilité d’obtenir \alpha et \beta fois les évènement A et B est donné par: P=p^{\alpha}q^{\beta}\dfrac{n!}{\alpha!\beta!}

Le terme le plus grand du développement de (p+q)^n correspond aux valeurs de \alpha et \beta les plus voisinnes de np et nq , donc on pose: a=np+x et b=nq+y .

On a alors

\ln n! \sim n\ln n-n+ \frac{1}{2}\ln (2\pi n)

et donc: \ln P\sim a \ln p+b \ln q+n \ln n-a \ln a-b \ln b+\frac{1}{2}\ln 2\pi n -\frac{1}{2}\ln 2\pi a-\frac{1}{2}\ln 2\pi \b -n+a+b

Comme n=a+b on obtient: n \ln n-a \ln a-b \ln b=a \ln{\frac{n}{a}}+b \ln{\frac{n}{b}}

et donc

\ln P\sim -a \ln{\frac{a}{np}}-b \ln{\frac{b}{nq}}+\ln n-\frac{1}{2}\ln a-\frac{1}{2}\ln b-\frac{1}{2}\ln 2\pi n

Soit

\ln P\sim -(a+\frac{1}{2}) \ln{\frac{a}{np}}-(b+\frac{1}{2}) \ln{\frac{b}{nq}}-\frac{1}{2}\ln pq-\frac{1}{2}\ln2\pi n

En posant: \ln P'= -(a+\frac{1}{2}) \ln{\frac{a}{np}}-(b+\frac{1}{2}) \ln{\frac{b}{nq}} on obtient:

P\sim \frac{1}{\sqrt{pq}}\frac{1}{\sqrt{2\pi n}}P'.

Occupons nous maintenant de P’:

\ln{\frac{a}{np}}=\ln{(1+\frac{x}{np})} et donc : \ln{\frac{a}{np}}=\frac{x}{np}-\frac{x^2}{2n^2 p^2}+\frac{x^3}{3n^3 p^3}

donc :(a+\frac{1}{2}) \ln{\frac{a}{np}}=(np+x+\frac{1}{2})(\frac{x}{np}-\frac{x^2}{2n^2 p^2}+\frac{x^3}{3n^3 p^3})=x+\frac{x}{2np}+\frac{x^2}{2np}-\frac{x^2}{4n^2 p^2}-\frac{x^3}{6n^2 p^2}

En négligeant le terme de degré 3, on obtient:

\ln P'=-\frac{1}{2n}(\frac{x^2}{p}+\frac{y^2}{q}+\frac{x}{p}+\frac{y}{q})+\frac{1}{6n^2}(\frac{x^3}{p^2}+\frac{y^3}{q^2})

y=-x et l’on arrive à:

P= \frac{1}{\sqrt{2\pi npq}}e^{\frac{-x^2}{2npq}+\frac{(q-p)x^3}{6n^2p^2q^2}-\frac{(q-p)x}{2npq}}

Le terme impair est négligeable si x est de l’ordre de \sqrt{n}, c’est à dire au voisinnage du max de probabilité.

Posons alors x=t\sqrt{npq} il vient alors:

P= \frac{1}{\sqrt{2\pi npq}}e^{\frac{-t^2}{2}+\frac{(q-p)}{\sqrt{npq}}(\frac{t^3}{3}-t)}

n étant très grand, on négligera le terme impair et \Delta x=\sqrt{npq}\Delta t permet d’obtenir que la probabilité pour que t soit compris entre t et t+dt est égale à:

\dfrac{1}{\sqrt{2\pi}}e^{\frac{-t^2}{2}} dt!

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