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Exercices

Laboratoires et exercices ISLR

Dans le manuel An introduction to statistical learning [JWHT13], vous pouvez faire les exercices suivants;

Laboratoires chap#3

• 3.6.1 Libraries

• 3.6.2 Simple Linear Regression

• 3.6.3 Multiple Linear Regression

• 3.6.4 Interaction Terms

• 3.6.5 Non-linear Transformations of the Predictors

• 3.6.6 Qualitative Predictors

• 3.6.7 Writing Functions

Laboratoires chap#4

• 4.7.1 The Stock Market Data

• 4.7.2 Logistic Regression

• 4.7.7 Poisson Regression

Exercices ISLR chap#3

1-3-8-12-13-14

Exercices ISLR chap#4

6-7-13(a-b-c-d)- 14(a-b-c-f)- 16 (juste la régression logistique)

Cliquez sur le bouton pour la solution

• chap #3: Voici une proposition de solutions aux exercices.

• chap #4: Voici une proposition de solutions aux exercices.

Exercice 1

Soit les données suivantes

Period	y	x_(1)	x_(2)
1	1.3	6	4.5
2	1.5	7	4.6
3	1.8	7	4.5
4	1.6	8	4.7
5	1.7	8	4.6

En utilisant le modèle de regression suivant pour \(i=1,2, \dots 5\) ; $\( y_i=\beta_{0}+\beta_{1} x_{i1}+\beta_{2} x_{i2}+\varepsilon_{i} \)$

et \(\left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1}\)

\[\begin{split}\left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1}=\left(\begin{array}{ccc} 1522.73 & 26.87 & -374.67 \\ 26.87 & 0.93 & -7.33 \\ -374.67 & -7.33 & 93.33 \end{array}\right)\end{split}\]

Calculer \(\hat{\varepsilon}_{2}\);

Exercice 2

Vous ajuster le modèle \(y_i=\beta_{0}+\beta_{1} x_{i1}+\beta_{2} x_{i2}+\beta_{3} x_{i3}+\varepsilon_{i} \) sur les données suivantes:

\(y\)	\(x_1\)	\(x_2\)	\(x_3\)
8	1	1	0
7	0	0	1
6	1	0	0
8	1	1	1
9	0	0	0
3	0	1	1

Après avoir calcuer \(\left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1}\), vous obtenez;

\[\begin{split}\left(\mathbf{X}^{\prime} \mathbf{X}\right)^{-1}=\left(\begin{array}{ccc} 1522.73 & 26.87 & -374.67 \\ 26.87 & 0.93 & -7.33 \\ -374.67 & -7.33 & 93.33 \end{array}\right)\end{split}\]

Calculer \(\hat{\beta}_1\)

Solution 1

x_1 <- c(6,7,7,8,8)
x_2 <- c(4.5,4.6,4.5,4.7,4.6)
X <- cbind(Intercept = 1, x_1,x_2)
print(X)

     Intercept x_1 x_2
[1,]         1   6 4.5
[2,]         1   7 4.6
[3,]         1   7 4.5
[4,]         1   8 4.7
[5,]         1   8 4.6

y <- c(1.3,1.5,1.8,1.6,1.7)

sum(c(1,1,1,1,1)*y)
sum(x_1*y)
sum(x_2*y)

7.9

57.3

36.19

print(t(cbind(sum(c(1,1,1,1,1)*y),
      sum(x_1*y),
      sum(x_2*y))))

      [,1]
[1,]  7.90
[2,] 57.30
[3,] 36.19

print(t(X) %*% y)

           [,1]
Intercept  7.90
x_1       57.30
x_2       36.19

XX_1 <- matrix(c(1522.73, 26.87,-374.67,
                 26.87,0.93,-7.33,
                 -374.67,-7.33,93.33)
               ,nrow = 3, byrow =T)

print(XX_1)

        [,1]  [,2]    [,3]
[1,] 1522.73 26.87 -374.67
[2,]   26.87  0.93   -7.33
[3,] -374.67 -7.33   93.33

print(XX_1 %*% (t(X) %*% y))

        [,1]
[1,]  9.9107
[2,]  0.2893
[3,] -2.2893

Solution 2

y <- c(5,3,10,4,3,5)
x_1 <- c(0,1,0,1,0,1)
x_2 <- c(1,0,1,1,0,1)
x_3 <- c(0,1,1,0,0,0)

X <- cbind(Intercept = 1, x_1,x_2,x_3)
X
t(X) 
t(X)%*%y



XX_1 <- (1/30)*matrix(c(26,-10,-18,-12,
                       -10,20,0,0,
                       -18,0,24,6,
                       -12,0,6,24),
                     nrow = 4, byrow =T)
XX_1
beta_i <- XX_1%*% (t(X) %*% y)
beta_1 <- beta_i[2]
beta_1

print(XX_1%*% (t(X) %*% y))

donne=data.frame(cbind(y,X))
reg_model <- lm(y ~ x_1 + x_2 + x_3, data = donne)

reg_model$coefficients

A matrix: 6 × 4 of type dbl

Intercept	x_1	x_2	x_3
1	0	1	0
1	1	0	1
1	0	1	1
1	1	1	0
1	0	0	0
1	1	1	0

A matrix: 4 × 6 of type dbl

Intercept	1	1	1	1	1	1
x_1	0	1	0	1	0	1
x_2	1	0	1	1	0	1
x_3	0	1	1	0	0	0

A matrix: 4 × 1 of type dbl

Intercept	30
x_1	12
x_2	24
x_3	13

A matrix: 4 × 4 of type dbl

0.8666667	-0.3333333	-0.6	-0.4
-0.3333333	0.6666667	0.0	0.0
-0.6000000	0.0000000	0.8	0.2
-0.4000000	0.0000000	0.2	0.8

-2

     [,1]
[1,]  2.4
[2,] -2.0
[3,]  3.8
[4,]  3.2

(Intercept)

2.4

x_1

-2

x_2

3.8

x_3

3.2

JWHT13

Gareth James, Daniela Witten, Trevor Hastie, and Robert Tibshirani. An introduction to statistical learning. Volume 112. Springer, 2013.

Régression logistique Régularisation