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Feuille 3 Exercice 1

La variable aleatoire

$X$ suit une loi

$N(0,{\sigma }^2)$ avec

$\sigma >0$.
Nous etudierons le test

$H_0:{\sigma }^2={\sigma }_0^2$ contre

$H_1:{\sigma }^2={\sigma }_2^1$ avec

$0<{\sigma }_0<{\sigma }_1$.

1. Determiner la statistique de Neyman-Pearson que nous noterons
   $T_n$.
2. Determiner
   $\alpha$ en fonction du seuil du test.
3. Determiner
   $\beta$ en fonction du seuil du test.
4. Determiner les courbes COR associees a ce test.

Solution

$$\begin{gathered} \begin{array}{rl}T& =\frac{{\mathrm{\Pi }}_{i=1}^{n}f(X_i,{\sigma }_1)}{{\mathrm{\Pi }}_{i=1}^{n}f(X_i,{\sigma }_0)}\\ & =\frac{{\mathrm{\Pi }}_{i=1}^n\frac{1}{{\sigma }_1\sqrt{2\pi }}\exp (\frac{-X_i^2}{2{\sigma }_1^2})}{{\mathrm{\Pi }}_{i=1}^n\frac{1}{{\sigma }_0\sqrt{2\pi }}\exp (\frac{-X_i^2}{2{\sigma }_0^2})}\\ & =(\frac{{\sigma }_0}{{\sigma }_1}{)}^n\exp (-\frac{1}{2}\sum _{i=1}^n{X}_i^2(\frac{1}{{\sigma }_1^2}-\frac{1}{{\sigma }_0^2}))\end{array} \\ \begin{array}{rl}\ln (T)& =\underset{\text{color}greena}{\underbrace{n\ln (\frac{{\sigma }_0}{{\sigma }_1})}}-\frac{1}{2}\sum _{i=1}^n{X}_i^2\underset{\text{color}greenb}{\underbrace{(\frac{1}{{\sigma }_1^2}-\frac{1}{{\sigma }_0^2})}}\\ & =a-\frac{b}{2}\sum _{i=1}^n{X}_i^2\end{array} \end{gathered}$$

D'apres le lemme de Neyman-Pearson:

L'hypothese

$H_0$ est rejetee lorsque:

$$\begin{gathered} \begin{array}{rl}T& >C_{\alpha }\\ \ln (T)& >\ln (C_{\alpha })\\ a-\frac{b}{2}\sum _{i=1}^n{X}_i^2& >\ln (C_{\alpha })\\ \sum _{i=1}^n{X}_i^2& >-\frac{2}{b}(\ln (C_{\alpha })-a)\end{array} \\ \text{color}red\boxed{{\displaystyle \sum _{i=1}^n{X}_i^2>S_{\alpha }}} \\ {T}_n=\sum _{i=1}^n{X}_i^2 \end{gathered}$$

On cherche

$\alpha$ et

$\beta$:

$$\begin{array}{rl}\alpha & =P(\text{rejeter }{H}_0|H_0\text{ vraie})\\ & =P(\sum X_i^2>S_{\alpha }|\sigma ={\sigma }_0)\\ & =P(\frac{T_n}{{\sigma }_0^2}>\frac{S_{\alpha }}{{\sigma }_0^2})\end{array}$$

Les variables aleatoires sont normales centrees et independantes donc

$$\begin{gathered} W_n:=\frac{T_n}{{\sigma }_0^2}\sim {\chi }_2(n) \\ \alpha =P(W_n>\frac{S_{\alpha }}{{\sigma }_0^2}) \end{gathered}$$

On prend

$\alpha =0.05$ et

$n=33$

$$\frac{S_{\alpha }}{{\sigma }_0^2}\simeq 47,40$$

chi2.ppf(0.95, 33)
chi2.isf(0.05, 33)

s comme survie

$$\begin{gathered} \alpha =1-F_N(\frac{S_{\alpha }}{{\sigma }_0^2}) \\ {F}_N(\frac{S_{\alpha }}{{\sigma }_0^2})=1-\alpha \\ \frac{S_{\alpha }}{{\sigma }_0^2}=F_N^{-1}(1-\alpha ) \\ \text{color}red\boxed{{\displaystyle S_{\alpha }={\sigma }_0^2{F}_N^{-1}(1-\alpha )}} \end{gathered}$$

$$\begin{array}{rl}\beta & =P(\text{Accepte H\_0}|H_1\text{ vraie})\\ & =P(T_n\le S_{\alpha }|{\sigma }^2={\sigma }_1^2)\end{array}$$

Sous l'hypothese

$H_1$:

$W_n^{\prime }:=\frac{T_n}{{\sigma }_1^2}\sim {\chi }^2(n)$

$$\begin{gathered} \begin{array}{rl}\beta & =P(W_n^{\prime }\le \frac{S_{\alpha }}{{\sigma }_1^2})\\ & =F_n(\frac{S_{\alpha }}{{\sigma }_1^2})\end{array} \\ \text{color}red\boxed{{\displaystyle \beta =F_n(\frac{{\sigma }_0^2{F}_n^{-1}(1-\alpha )}{{\sigma }_1^2})}} \end{gathered}$$

Feuille 3 Exercice 3

La variable aleatoire

$X$ suit une loi geometrique de parametre

$p$. A l’aide du theoreme de Wilks, ecrire la zone de rejet du test

$H_0:p=0,25$ contre

$H_1:p=0,5$.

Solution

D'apres le theoreme de Wilks,

$$\begin{gathered} R_n=2\log (T_n)\sim {\chi }^2(1) \\ \{R_n>3,84\} \end{gathered}$$

Il suffit d'expliciter en

$R_n$

Il suffit d'expliciter

$R_n$

$$\begin{array}{rl}{T}_n& =\frac{L(X_1,\dots ,X_n,0.5)}{L(X_1,\dots ,X_n,0.25)}\\ & =\frac{\prod _{i=1}^{n}0.5\times (1-0.5{)}^{X_i-1}}{\prod _{i=1}^{n}0.25\times (1-0.25{)}^{X_i-1}}\\ & =2^n\times \prod _{i=1}^n(\frac{0.5}{0.75}{)}^{X_i-1}\\ & =2^n\times \prod _{i=1}^n(\frac{2}{3})\\ & =2^n\times (\frac{2}{3}{)}^{\sum _{i=1}^n(X_i-1)}\end{array}$$

Passons au logarithme neperien:

$$\begin{gathered} \ln (T_n)=n\ln (2)+\ln (\frac{2}{3})\sum _{i=1}^n(X_i-1) \\ \begin{array}{rl}{R}_n& =2\ln (T_n)\\ & =2(n\ln (2)+\ln (\frac{2}{3})\sum _{i=1}^n(X_i-1))\end{array} \\ \alpha =5\mathrm{\%} \end{gathered}$$

Rappel:

$R_n$ suit asymptotiquement

${\chi }^2(1)$

Zone de rejet:

$$\{R_n>3,84\}$$

On veut resoudre l'equation pour isoler

$\sum _{i=1}^n{X}_i$

$$\begin{array}{rl}2[n\ln (2)+\ln (\frac{2}{3})\sum _{i=1}^n(X_i-1)]& >3.84\\ \ln (\frac{2}{3})\sum _{i=1}^n(X_i-1)& >\frac{3.84}{2}-\ln (2)\\ \sum _{i=1}^n{X}_i-n& <\frac{\frac{3.84}{2}-n\ln (2)}{ln(\frac{2}{3})}\text{car }\ln (\frac{2}{3})<0\\ \frac{\sum _{i=1}^n{X}_i}{n}& <\frac{1.92-n\ln (2)}{n\ln (\frac{2}{3})}+1\\ {\overline{X}}_n& <\frac{1.92-n\ln (2)}{n\ln (\frac{2}{3})}+1\end{array}$$

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