#Ejercicio 1 (datos del ejemplo)
#Ingresando matriz X
options(scipen = 99999999)
matriz_xx<-matrix(data = c(3.92,3.61,3.32,3.07,3.06,3.11,3.21,3.26,3.42,3.42,3.45,3.58,3.66,3.78,3.82,3.97,4.07,4.25,4.41,4.49,4.7,4.58,4.69,4.71,4.78,7298,6855,6636,6506,6450,6402,6368,6340,6349,6352,6361,6369,6546,6672,6890,7115,7327,7546,7931,8097,8468,8717,8991,9179,9318),nrow = 25,ncol = 2,byrow = FALSE)
print(matriz_xx)## [,1] [,2]
## [1,] 3.92 7298
## [2,] 3.61 6855
## [3,] 3.32 6636
## [4,] 3.07 6506
## [5,] 3.06 6450
## [6,] 3.11 6402
## [7,] 3.21 6368
## [8,] 3.26 6340
## [9,] 3.42 6349
## [10,] 3.42 6352
## [11,] 3.45 6361
## [12,] 3.58 6369
## [13,] 3.66 6546
## [14,] 3.78 6672
## [15,] 3.82 6890
## [16,] 3.97 7115
## [17,] 4.07 7327
## [18,] 4.25 7546
## [19,] 4.41 7931
## [20,] 4.49 8097
## [21,] 4.70 8468
## [22,] 4.58 8717
## [23,] 4.69 8991
## [24,] 4.71 9179
## [25,] 4.78 9318#Generando X’X
XtrX<- t(matriz_xx) %*% matriz_xx
print(XtrX)## [,1] [,2]
## [1,] 379.2928 710932.3
## [2,] 710932.3200 1335796275.0#Ingresando matriz Y
options(scipen = 99999999)
matriz_Y<-matrix(data = c(0.75,0.71,0.66,0.61,0.7,0.72,0.77,0.74,0.9,0.82,0.75,0.77,0.78,0.84,0.79,0.7,0.68,0.72,0.55,0.63,0.56,0.41,0.51,0.47,0.32),nrow = 25,ncol = 1,byrow = FALSE)
print(matriz_Y)## [,1]
## [1,] 0.75
## [2,] 0.71
## [3,] 0.66
## [4,] 0.61
## [5,] 0.70
## [6,] 0.72
## [7,] 0.77
## [8,] 0.74
## [9,] 0.90
## [10,] 0.82
## [11,] 0.75
## [12,] 0.77
## [13,] 0.78
## [14,] 0.84
## [15,] 0.79
## [16,] 0.70
## [17,] 0.68
## [18,] 0.72
## [19,] 0.55
## [20,] 0.63
## [21,] 0.56
## [22,] 0.41
## [23,] 0.51
## [24,] 0.47
## [25,] 0.32#Generando X’Y
XtrY<- t(matriz_xx) %*% matriz_Y
print(XtrY)## [,1]
## [1,] 63.6124
## [2,] 119215.9400#Calculando la Inversa de X′X, es decir (X′X)−1 A través de Gauss -Jordan, u otro método (usando 10 decimales para el redondedo) se obtiene:
inv_matriz_xx<-solve(XtrX)
print(inv_matriz_xx)## [,1] [,2]
## [1,] 1.0832428129 -0.0005765192945
## [2,] -0.0005765193 0.0000003075815#Calculando el estimador de parametros β
beta<-inv_matriz_xx%*%XtrY
print(beta)## [,1]
## [1,] 0.177385485833
## [2,] -0.000005160319#También es posible a través de la solución del sistema de ecuaciones normales X′X⋅β=X′Y
beta<-solve(XtrX,XtrY)
print(beta)## [,1]
## [1,] 0.177385485833
## [2,] -0.000005160319#Ejercicio 2 (datos de ejercicio)
options(scipen = 99999999)
matriz_A<-matrix(data = c(50,53,60,63,69,82,100,104,113,130,150,181,202,217,229,240,243,247,249,254,
7.4,5.1,4.2,3.9,1.4,2.2,7.0,5.7,13.1,16.4,5.1,2.9,4.5,6.2,3.2,2.4,4.9,8.8,10.1,6.1),nrow = 20,ncol = 2,byrow = FALSE)
print(matriz_A)## [,1] [,2]
## [1,] 50 7.4
## [2,] 53 5.1
## [3,] 60 4.2
## [4,] 63 3.9
## [5,] 69 1.4
## [6,] 82 2.2
## [7,] 100 7.0
## [8,] 104 5.7
## [9,] 113 13.1
## [10,] 130 16.4
## [11,] 150 5.1
## [12,] 181 2.9
## [13,] 202 4.5
## [14,] 217 6.2
## [15,] 229 3.2
## [16,] 240 2.4
## [17,] 243 4.9
## [18,] 247 8.8
## [19,] 249 10.1
## [20,] 254 6.1#Generando A´A
AtrA<-t(matriz_A)%*%matriz_A
print(AtrA)## [,1] [,2]
## [1,] 574618.0 18601.80
## [2,] 18601.8 992.26#Ingresando matriz B
options(scipen = 99999999)
matriz_B<-matrix(data = c(320,450,370,470,420,500,570,640,670,780,690,700,910,930,940,1070,1160,1210,1450,1220),nrow = 20,ncol = 1,byrow = FALSE)
print(matriz_B)## [,1]
## [1,] 320
## [2,] 450
## [3,] 370
## [4,] 470
## [5,] 420
## [6,] 500
## [7,] 570
## [8,] 640
## [9,] 670
## [10,] 780
## [11,] 690
## [12,] 700
## [13,] 910
## [14,] 930
## [15,] 940
## [16,] 1070
## [17,] 1160
## [18,] 1210
## [19,] 1450
## [20,] 1220#Generando A´B
AtrB<- t(matriz_A) %*% matriz_B
print(AtrB)## [,1]
## [1,] 2801880
## [2,] 98350#Calculando la Inversa de X′X, es decir (X′X)−1 A través de Gauss -Jordan, u otro método (usando 10 decimales para el redondedo) se obtiene:
inv_matriz_A<-solve(AtrA)
print(inv_matriz_A)## [,1] [,2]
## [1,] 0.000004426896 -0.00008299059
## [2,] -0.000082990587 0.00256361669#Calculando el estimador de parametros β
betaAB<-inv_matriz_A%*%AtrB
print(betaAB)## [,1]
## [1,] 4.241508
## [2,] 19.602036##También es posible a través de la solución del sistema de ecuaciones normales X′X⋅β=X′Y
beta<-solve(AtrA,AtrB)
print(beta)## [,1]
## [1,] 4.241508
## [2,] 19.602036