ASE3: TD 2

Exercice 1

Soit

$$X=\left(\begin{array}{ccc}16& 2& 0\\ 8& 12& 10\\ 12& 16& 14\\ 20& 8& 14\\ 16& 4& 10\\ 0& 6& 12\end{array}\right)$$

On donne le meme poids a tous les individus:

$p_i=\frac{1}{6}$

$\mathrm{\forall }i$ et

$M=I_3$

1. Calculer la moyenne des variables et le centre de gravite
2. Donner la matrice
   $Y$
3. Calculer la matrice de var-covariance
   $V$
4. Diagonaliser sur
   $MV=V$
5. Calculer le
   $\mathrm{\%}$ d'inertie
6. Facteurs principaux
7. Determiner les composantes principales et calculer les coefficients de correlation

Solution

$$\begin{array}{rl}& \begin{array}{ccc}{X}^{(1)}& X^{(2)}& X^{(3)}\end{array}\\ X=& \left(\begin{array}{ccc}16& 2& 0\\ 8& 12& 10\\ 12& 16& 14\\ 20& 8& 14\\ 16& 4& 10\\ 0& 6& 12\end{array}\right)\end{array}$$

$p_i=\frac{1}{6}$

$\mathrm{\forall }i=1,2,4,5,6$ poids de chaque individu et

$M=I_3$ metrique

La moyenne des variables:

$$\begin{gathered} {\overline{X}}^{(1)}=\sum _{i=1}^6{p}_i{X}_i^{(1)}=\frac{1}{6}\sum _{i=1}^6{X}_i^{(1)}=\frac{72}{6}=12 \\ {\overline{X}}^{(2)}=\frac{1}{6}\sum _{i=1}^6{X}_i^{(2)}=\frac{1}{6}\bullet 48=8 \\ {\overline{X}}^{(3)}=\frac{1}{6}\sum _{i=1}^6{X}_i^{(3)}=\frac{60}{6}=10 \end{gathered}$$

Donc

${\overline{X}}^{(1)}=12$, X^{(2)}=8

$,X^{(3)}=10$.

Le centre de gravite du nuage forme par les 3 individus:

$$g^T=(12,8,10)$$

Tableau des donnees centrees

$Y$

$$\begin{gathered} y_i^{(j)}=X_i^{(j)}={\overline{X}}^{(j)} \\ Y=\left(\begin{array}{ccc}4& -6& -10\\ -4& 4& 0\\ 0& 8& 4\\ 8& 0& 4\\ 4& -4& 0\\ -12& -2& 2\end{array}\right) \end{gathered}$$

Matrice de var-covariance

$V=Y^{T}DY$ avec

$D=\frac{1}{6}{I}_6$

$$\Rightarrow V=\frac{1}{6}{Y}^{T}Y=\left(\begin{array}{ccc}\frac{128}{3}& -\frac{16}{3}& -\frac{16}{3}\\ -\frac{16}{3}& \frac{68}{3}& \frac{44}{3}\\ -\frac{16}{3}& \frac{44}{3}& \frac{68}{3}\end{array}\right)$$

Diagonalisation de

$MV=V$

$M=I_3$: metriques de l'espaces des individus

$P_V(\lambda )=det(V-\lambda I_3)$ polynome caracteristiques de

$V$

$$\begin{gathered} \begin{array}{rl}{P}_v(\lambda )& =\left|\begin{array}{ccc}\frac{128}{3}-\lambda & -\frac{16}{3}& -\frac{16}{3}\\ -\frac{16}{3}& \frac{68}{3}-\lambda & \frac{44}{3}\\ -\frac{16}{3}& \frac{44}{3}& \frac{68}{3}-\lambda \end{array}\right|\\ C_1\to C_1& +C_2+C_3\\ P_v(\lambda )& =(32-\lambda )\left|\begin{array}{ccc}1& -\frac{16}{3}& -\frac{16}{3}\\ 1& \frac{68}{3}-\lambda & \frac{44}{3}\\ 1& \frac{44}{3}& \frac{68}{3}-\lambda \end{array}\right|\text{par linearite}\\ L_2\to L_2-L_1& \text{ et }L3\to L_3-L_1\\ P_v(\lambda )& =(32-\lambda )\left|\begin{array}{ccc}1& -\frac{16}{3}& -\frac{16}{3}\\ 0& 28-\lambda & 20\\ 0& 20& 28-\lambda \end{array}\right|\\ & =(32-\lambda )((28-\lambda {)}^2-(20{)}^2)\\ & =(32-\lambda )(28\lambda -20)(28-\lambda +20)\end{array} \\ \text{color}red\boxed{{\displaystyle P_V(\lambda )=(32-\lambda )(8-\lambda )(48-\lambda )}} \end{gathered}$$

Les valeurs propres de

$V$:

${\lambda }_1=48$,

${\lambda }_2=32$,

${\lambda }_3=8$ (ordre decroissant)

Le

$\mathrm{\%}$ d'inertie

• Le
  $1^{er}$ axe:
  $\frac{{\lambda }_1}{{\lambda }_1+{\lambda }_2+{\lambda }_3}=\frac{48}{88}=0,54=54\mathrm{\%}$
• Le
  $2^e$ axe:
  $\frac{{\lambda }_2}{{\lambda }_1+{\lambda }_2+{\lambda }_3}=\frac{32}{88}=0,36=36\mathrm{\%}$
• Le
  $3^e$ axe:
  $\frac{{\lambda }_3}{{\lambda }_1+{\lambda }_2+{\lambda }_3}=\frac{8}{88}=0,09=9\mathrm{\%}$

Le plan factoriel:

$\frac{{\lambda }_1+{\lambda }_2}{{\lambda }_1+{\lambda }_2+{\lambda }_3}=\frac{80}{88}=90\mathrm{\%}$

Les facteurs principaux sont les deux vecteurs propres associes aux valeurs propres

${\lambda }_1=48$ et

${\lambda }_2=32$.

$$\begin{gathered} E_{48}=Ker(V-48I_3) \\ \mathrm{\forall }u=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)\in E_{48}\iff (V-48I_3)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\overrightarrow{0} \\ \{\begin{array}{l}-\frac{16}{3}x-\frac{16}{3}y-\frac{16}{3}z=0\\ -\frac{16}{3}x-\frac{76}{3}y+\frac{44}{3}z=0\\ -\frac{16}{3}x+\frac{44}{3}-\frac{76}{3}z=0\end{array}\iff \{\begin{array}{l}x+y+z=0(1)\\ -16x-76y+44z=0(2)\\ -16x+44y-76z=0(3)\end{array} \\ (2)-(3)\Rightarrow -120y+120z=0\Rightarrow \text{color}green\boxed{{\displaystyle y=z}} \\ (1)\Rightarrow \text{color}green\boxed{{\displaystyle x=-2z}} \\ {E}_{48}=Vect(\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right))\text{Droite vectorielle} \\ {u}^{(1)}=\frac{1}{\sqrt{4+1+1}}\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)=\frac{1}{\sqrt{6}}\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right) \\ \text{color}red{u}^{(1)}\text{ est norme} \\ \Vert \left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)\Vert =\sqrt{4+1+1}=\sqrt{6} \\ {E}_{32}=Ker(V-32I_3) \\ \mathrm{\forall }u=\left(\begin{array}{c}x\\ y\\ z\end{array}\right)\in E_{32}\iff (V-32I+3)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\overrightarrow{0} \\ \{\begin{array}{l}32x-16y-16z=0(1)\\ -16x-28y+44z=0(2)\\ -16x+44y-28z=0(3)\end{array} \\ \{\begin{array}{ll}(2)-(3)& \Rightarrow \text{color}green\boxed{{\displaystyle y=z}}\\ (1)& \Rightarrow \text{color}green\boxed{{\displaystyle y=x}}\end{array} \\ \text{color}red\boxed{{\displaystyle E_{32}=Vect\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)\text{Droite}}} \\ {u}^{(2)}=\frac{1}{\sqrt{3}}\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right)\text{norme} \end{gathered}$$

$(u^{(1)},u^{(2)})$ base orthonormee

Composantes principales

$$C^{(i)}=Y{u}^{(i)}i=1,2$$

La

$1^{ere}$ composante:

$$\begin{array}{rl}{C}^{(1)}& =Y{u}^{(1)}=\left(\begin{array}{ccc}4& -6& -10\\ -4& 4& 0\\ 0& 8& 4\\ 8& 0& 4\\ 4& -4& 0\\ -12& -2& 2\end{array}\right)\bullet \frac{1}{\sqrt{6}}\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)\\ & =\left(\begin{array}{c}4\sqrt{6}\\ -2\sqrt{6}\\ -2\sqrt{6}\\ 2\sqrt{6}\\ 2\sqrt{6}\\ -4\sqrt{6}\end{array}\right)\text{variable centree}\end{array}$$

$$C^{(2)}=Y{u}^{(2)}=\left(\begin{array}{c}-4\sqrt{3}\\ 0\\ 4\sqrt{3}\\ 4\sqrt{3}\\ 0\\ -4\sqrt{3}\end{array}\right)\text{variable centree}$$

Remarque
Ces composantes principales contiennent les projections des individus sur les 2 axes factoriels.

Calcul des coefficients de correlation:

$$\begin{gathered} \rho (X_1^{(1)},C^{(1)})=\frac{Cov(X^{(1)},C^{(1)})}{{\sigma }_{X^{(1)}}{\sigma }_{C^{(1)}}} \\ Cov(X^{(1)},C^{(1)})=<y^{(1)},C^{(1)}>={y^{(1)}}^{T}D.C^{(1)}\text{produit scalaire de l'espace des variables} \\ D=\frac{1}{6}{I}_6\text{Metriques dans l'espace des variables} \\ \text{color}green\boxed{{\displaystyle Cov(X^{(1)},C^{(1)})=\frac{1}{6}{y^{(1)}}^T{C}^{(1)}}} \end{gathered}$$

$$\begin{gathered} \begin{array}{rl}Cov(X^{(1)},C^{(1)})& =\frac{1}{6}(16\sqrt{6}+8\sqrt{6}+16\sqrt{6}+8\sqrt{6}+48\sqrt{6})\\ & =\frac{96\sqrt{6}}{6}=\text{color}green\boxed{{\displaystyle 16\sqrt{6}}}\end{array} \\ {\sigma }_{X^{(1)}}=\sqrt{V(X^{(1)})}=\sqrt{\frac{128}{3}} \\ \begin{array}{rl}{\sigma }_{C^{(1)}}=\Vert C^{(1)}\Vert & =\sqrt{<C^{(1)},C^{(1)}>}\\ & =\sqrt{\frac{1}{6}(96+24+24+24+24+96)}\\ & =\text{color}green\boxed{{\displaystyle 4\sqrt{3}}}\end{array} \\ \rho (X^{(1)},C^{(1)})=\frac{16\sqrt{6}}{\sqrt{\frac{128}{3}}\bullet 4\sqrt{3}}=\text{color}green\boxed{{\displaystyle \frac{\sqrt{3}}{2}}} \end{gathered}$$

Tableau des correlations:

	$C^{(1)}$	$C^{(2)}$
$X^{(1)}$	$\frac{\sqrt{3}}{2}=0,87$	$\frac{1}{2}=0,5$
$X^{(2)}$	$-\sqrt{\frac{6}{17}}=-0,59$	$\frac{2\sqrt{34}}{17}=0,69$
$X^{(3)}$	$-0,59$	$0,69$

:::

Exercice 2

$$X=\left(\begin{array}{ccc}2& 2& 3\\ 3& 1& 2\\ 1& 0& 3\\ 2& 1& 4\\ 2& 1& 3\end{array}\right)$$

1. Calculer
   $\overline{X^{(1)}}$,
   $\overline{X^{(2)}}$,
   $\overline{X^{(3)}}$ et le centre de gravite
2. Calculer la matrice
   $Y$
3. Calculer
   $V$
4. Diagonaliser
   $V$ et calculer le
   $\mathrm{\%}$ d'inertie
5. Facteurs principaux et composantes principales

Solution

$$\begin{gathered} P_i=\frac{1}{5}\mathrm{\forall }i \\ \overline{X^{(1)}}=2,\overline{X^{(2)}}=1,\overline{X^{(2)}}=3 \\ {g}^T=(2,1,3) \end{gathered}$$

$$Y=\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& -1\\ -1& -1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right)$$

$$\begin{gathered} V=Y^{T}DY \\ D=\frac{1}{5}{I}_5 \\ \begin{array}{rl}V& =\frac{1}{5}{Y}^{T}Y\\ & =\frac{1}{5}\left(\begin{array}{ccc}2& 1& -1\\ 1& 2& 0\\ -1& 0& 2\end{array}\right)\end{array} \\ \begin{array}{rl}{P}_{Y^{T}Y}(\lambda )& =\left|\begin{array}{ccc}2-\lambda & 1& -1\\ 1& 2-\lambda & 0\\ -1& 0& 2-\lambda \end{array}\right|C_2\to C_2+C_3\\ & =\left|\begin{array}{ccc}2-\lambda & 0& -1\\ 1& 2-\lambda & 0\\ -1& 2-\lambda & 2-\lambda \end{array}\right|L_2\to L_3-L_2\\ & =\left|\begin{array}{ccc}2-\lambda & 0& -1\\ 1& 2-\lambda & 0\\ -2& 0& 2-\lambda \end{array}\right|\\ & =(2-\lambda )((2-\lambda {)}^2-2)\\ & =(2-\lambda )(2-\lambda -\sqrt{2})(2-\lambda +\sqrt{2})\end{array} \\ \{\begin{array}{l}{\mathrm{\Gamma }}_1=2+\sqrt{2}\\ {\mathrm{\Gamma }}_2=2\\ {\mathrm{\Gamma }}_3=2-\sqrt{2}\end{array}\Rightarrow \{\begin{array}{l}{\lambda }_1=\frac{2+\sqrt{2}}{5}\\ {\lambda }_2=\frac{2}{5}\\ {\lambda }_3=\frac{2-\sqrt{2}}{5}\end{array} \end{gathered}$$

$$\mathrm{\%}\frac{{\lambda }_1+{\lambda }_2}{{\lambda }_1+{\lambda }_2+{\lambda }_3}=?$$

Facteurs principaux:

$$\begin{gathered} E_{2+\sqrt{2}}=Ker(Y^{T}Y-(2+\sqrt{2})I)=Vect\left(\begin{array}{c}-\sqrt{2}\\ -1\\ 1\end{array}\right) \\ \text{color}green\boxed{{\displaystyle u^{(1)}=\frac{1}{\sqrt{4}}\left(\begin{array}{c}-\sqrt{2}\\ -1\\ 1\end{array}\right)=\frac{1}{2}\left(\begin{array}{c}-\sqrt{2}\\ -1\\ 1\end{array}\right)}} \\ \mathrm{\%}\frac{{\lambda }_1+{\lambda }_2}{{\lambda }_1+{\lambda }_2+{\lambda }_3}=90\mathrm{\%}\text{(inertie maximale)} \\ {E}_2=Ker(Y^{T}Y-2I_3)=Vect\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right) \\ \text{color}green\boxed{{\displaystyle u^{(2)}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)}} \end{gathered}$$

Composantes principales:

$$\begin{gathered} C^{(1)}=Y{u}^{(1)}=\left(\begin{array}{c}-\frac{1}{2}\\ -\frac{\sqrt{2}}{2}-\frac{1}{2}\\ \frac{\sqrt{2}}{2}+\frac{1}{2}\\ \frac{1}{2}\\ 0\end{array}\right)\text{centrees} \\ {C}^{(1)}=Y{u}^{(1)}=\left(\begin{array}{c}\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}0\end{array}\right) \end{gathered}$$