Ejercicio Parcial Econometría I
Utilizaremos la siguiente información para calcular una regresión sobre el salario.
En este examen se debe hallar lo siguiente.
El coeficiente de determinación R 2 R^2 R 2
La descomposición de la varianza.
La Matriz X ′ X X'X X ′ X X ′ Y X'Y X ′ Y
La matriz de varianza y covarianza.
Estimaremos la ecuación ln w i = β 0 + e d u c ⋅ β 1 + e x p e r ⋅ β 2 \ln w_i = \beta_0 + educ \cdot \beta_1 + exper \cdot \beta_2 ln w i = β 0 + e d u c ⋅ β 1 + e x p er ⋅ β 2
Y por último hallaremos el error estándar y el estadístico t.
Vamos a ello.
Estadísticos descriptivos
Variable Promedio ln.wi 1.623268 educ 12.56274 exper 17.01711
Matriz de varianzas y covarianzas
ln.wi educ exper ln.wi 0.2825329 0.6344412 0.8034574 educ 0.6344412 7.6674851 -11.257266 exper 0.8034574 -11.257266 184.2035162
Descomposición de la varianza
S T C = S E C + S R C STC = SEC + SRC STC = SEC + SRC
148.329762 = 36.9850408 + S R C 148.329762 = 36.9850408 + SRC 148.329762 = 36.9850408 + SRC
S R C = 111.3447212 SRC = 111.3447212 SRC = 111.3447212
Coeficiente de determinación
R 2 = S R C S T C = 111.3447212 148.329762 = 0.75066 R^2 = \frac{SRC}{STC} = \frac{111.3447212}{148.329762} = 0.75066 R 2 = STC SRC = 148.329762 111.3447212 = 0.75066
σ ^ 2 = S R C n − k = 111.3447212 523 = 0.212896 \hat{\sigma}^2 = \frac{SRC}{n-k} = \frac{111.3447212}{523} = 0.212896 σ ^ 2 = n − k SRC = 523 111.3447212 = 0.212896
Matrices X'X y X'Y
Además, se sabe que:
X ′ X = [ n ∑ i = 1 n x 1 i ∑ i = 1 n x 2 i ∑ i = 1 n x 1 i ∑ i = 1 n x 1 i 2 ∑ i = 1 n x 1 i x 2 i ∑ i = 1 n x 2 i ∑ i = 1 n x 1 i x 2 i ∑ i = 1 n x 2 i 2 ] X ′ Y = [ ∑ i = 1 n y i ∑ i = 1 n x 1 i y i ∑ i = 1 n x 2 i y i ] X'X = \begin{bmatrix} n & \sum_{i=1}^{n} x_{1i} & \sum_{i=1}^{n} x_{2i} \\ \sum_{i=1}^{n} x_{1i} & \sum_{i=1}^{n} x_{1i}^2 & \sum_{i=1}^{n} x_{1i}x_{2i} \\ \sum_{i=1}^{n} x_{2i} & \sum_{i=1}^{n} x_{1i}x_{2i} & \sum_{i=1}^{n} x_{2i}^2 \end{bmatrix} \quad X'Y = \begin{bmatrix} \sum_{i=1}^{n} y_i \\ \sum_{i=1}^{n} x_{1i}y_i \\ \sum_{i=1}^{n} x_{2i}y_i \end{bmatrix} X ′ X = n ∑ i = 1 n x 1 i ∑ i = 1 n x 2 i ∑ i = 1 n x 1 i ∑ i = 1 n x 1 i 2 ∑ i = 1 n x 1 i x 2 i ∑ i = 1 n x 2 i ∑ i = 1 n x 1 i x 2 i ∑ i = 1 n x 2 i 2 X ′ Y = ∑ i = 1 n y i ∑ i = 1 n x 1 i y i ∑ i = 1 n x 2 i y i
Recordando que:
∑ i = 1 n x i 2 = var ( x ) ⋅ ( n − 1 ) + n X ˉ 2 \sum_{i=1}^{n} x_i^2 = \text{var}(x) \cdot (n-1) + n\bar{X}^2 ∑ i = 1 n x i 2 = var ( x ) ⋅ ( n − 1 ) + n X ˉ 2
∑ i = 1 n x 1 i x 2 i = cov ( x 1 , x 2 ) ⋅ ( n − 1 ) + n X ˉ 1 X ˉ 2 \sum_{i=1}^{n} x_{1i}x_{2i} = \text{cov}(x_1, x_2) \cdot (n-1) + n\bar{X}_1\bar{X}_2 ∑ i = 1 n x 1 i x 2 i = cov ( x 1 , x 2 ) ⋅ ( n − 1 ) + n X ˉ 1 X ˉ 2
Con esta información se pueden completar las matrices X ′ X X'X X ′ X X ′ Y X'Y X ′ Y
X ′ X = [ 526 6608.00124 8950.99986 6608.00124 87040.0311755 106539.016331 8950.99986 106539.016331 249026.995233 ] X'X = \begin{bmatrix} 526 & 6608.00124 & 8950.99986 \\ 6608.00124 & 87040.0311755 & 106539.016331 \\ 8950.99986 & 106539.016331 & 249026.995233 \end{bmatrix} X ′ X = 526 6608.00124 8950.99986 6608.00124 87040.0311755 106539.016331 8950.99986 106539.016331 249026.995233
Donde:
6608.00124 = ( 526 ) ( 12.56275 ) 6608.00124 = (526)(12.56275) 6608.00124 = ( 526 ) ( 12.56275 ) 8950.99986 = ( 526 ) ( 17.01711 ) 8950.99986 = (526)(17.01711) 8950.99986 = ( 526 ) ( 17.01711 ) 87040.0311755 = ( 525 ) ( 7.6674851 ) + ( 526 ) ( 12.56274 ) 2 87040.0311755 = (525)(7.6674851) + (526)(12.56274)^2 87040.0311755 = ( 525 ) ( 7.6674851 ) + ( 526 ) ( 12.56274 ) 2 249026.995233 = ( 525 ) ( 184.2035162 ) + ( 526 ) ( 17.01711 ) 2 249026.995233 = (525)(184.2035162) + (526)(17.01711)^2 249026.995233 = ( 525 ) ( 184.2035162 ) + ( 526 ) ( 17.01711 ) 2 106539.019331 = ( 525 ) ( − 11.257266 ) + ( 526 ) ( 17.01711 ) ( 12.56274 ) 106539.019331 = (525)(-11.257266) + (526)(17.01711)(12.56274) 106539.019331 = ( 525 ) ( − 11.257266 ) + ( 526 ) ( 17.01711 ) ( 12.56274 )
X ′ Y = [ 853.838968 11059.6232569 14951.6867757 ] X'Y = \begin{bmatrix} 853.838968 \\ 11059.6232569 \\ 14951.6867757 \end{bmatrix} X ′ Y = 853.838968 11059.6232569 14951.6867757
Donde:
853.838968 = 1.623268 × 526 853.838968 = 1.623268 \times 526 853.838968 = 1.623268 × 526 11059.6232569 = 0.634412 × 525 + 526 × 1.623268 × 12.56274 11059.6232569 = 0.634412 \times 525 + 526 \times 1.623268 \times 12.56274 11059.6232569 = 0.634412 × 525 + 526 × 1.623268 × 12.56274 14951.6867757 = 0.8034574 × 525 + 526 × 1.623268 × 17.01711 14951.6867757 = 0.8034574 \times 525 + 526 \times 1.623268 \times 17.01711 14951.6867757 = 0.8034574 × 525 + 526 × 1.623268 × 17.01711
Determinante e inversa
det ( X ′ X ) = 186392173300 \det(X'X) = 186392173300 det ( X ′ X ) = 186392173300
Cofactores
Cof ( X ′ X ) = [ 10324754788.6 − 691939946.174 − 75085335.018 − 691939946.174 50867800.9993 3108694.006 − 75085335.018 3108694.006 2117376.0105 ] \text{Cof}(X'X) = \begin{bmatrix} 10324754788.6 & -691939946.174 & -75085335.018 \\ -691939946.174 & 50867800.9993 & 3108694.006 \\ -75085335.018 & 3108694.006 & 2117376.0105 \end{bmatrix} Cof ( X ′ X ) = 10324754788.6 − 691939946.174 − 75085335.018 − 691939946.174 50867800.9993 3108694.006 − 75085335.018 3108694.006 2117376.0105
Inversa
( X ′ X ) − 1 = [ 0.05539264 − 0.00371228 − 0.00040284 − 0.00371228 0.00027291 0.00001668 − 0.00040284 0.00001668 0.00001136 ] (X'X)^{-1} = \begin{bmatrix} 0.05539264 & -0.00371228 & -0.00040284 \\ -0.00371228 & 0.00027291 & 0.00001668 \\ -0.00040284 & 0.00001668 & 0.00001136 \end{bmatrix} ( X ′ X ) − 1 = 0.05539264 − 0.00371228 − 0.00040284 − 0.00371228 0.00027291 0.00001668 − 0.00040284 0.00001668 0.00001136
Coeficientes estimados
( X ′ X ) − 1 X ′ Y = [ 0.21691 0.09793 0.01035 ] = [ β 0 β 1 β 2 ] (X'X)^{-1}X'Y = \begin{bmatrix} 0.21691 \\ 0.09793 \\ 0.01035 \end{bmatrix} = \begin{bmatrix} \beta_0 \\ \beta_1 \\ \beta_2 \end{bmatrix} ( X ′ X ) − 1 X ′ Y = 0.21691 0.09793 0.01035 = β 0 β 1 β 2
Varianza de los estimadores
var ( β ^ ) = σ ^ 2 ( X ′ X ) − 1 = [ 0.01179287 − 0.00079033 − 0.00008576 − 0.00079033 0.0000581 0.000003551 − 0.00008576 0.000003551 0.000002418 ] \text{var}(\hat{\beta}) = \hat{\sigma}^2 (X'X)^{-1} = \begin{bmatrix} 0.01179287 & -0.00079033 & -0.00008576 \\ -0.00079033 & 0.0000581 & 0.000003551 \\ -0.00008576 & 0.000003551 & 0.000002418 \end{bmatrix} var ( β ^ ) = σ ^ 2 ( X ′ X ) − 1 = 0.01179287 − 0.00079033 − 0.00008576 − 0.00079033 0.0000581 0.000003551 − 0.00008576 0.000003551 0.000002418
Resultados finales
Coeficiente Valor Error estándar Estadístico t Constante 0.2169 0.10859498 1.99742766 Educación 0.097931 0.00762234 12.8479503 Experiencia 0.103467 0.00155499 6.65385582
El estadístico t al 95% de significancia y 523 grados de libertad es equivalente a 1.96, por lo cual todos los coeficientes son estadísticamente diferentes de cero.
La educación y la experiencia laboral son determinantes para el salario.
