Regresion Lineal Multiple Ejercicios Resueltos A Mano Direct

Modelo: (Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \beta_3 X_3 + \varepsilon) [ \mathbfY = \mathbfX\boldsymbol\beta + \boldsymbol\varepsilon ] [ \mathbfY = \beginbmatrix 10 \ 12 \ 15 \ 18 \endbmatrix, \quad \mathbfX = \beginbmatrix 1 & 1 & 2 & 1 \ 1 & 2 & 1 & 2 \ 1 & 3 & 3 & 3 \ 1 & 4 & 2 & 4 \endbmatrix, \quad \boldsymbol\beta = \beginbmatrix \beta_0 \ \beta_1 \ \beta_2 \ \beta_3 \endbmatrix ] Paso 2: Estimar por Mínimos Cuadrados Ordinarios La solución es: [ \hat\boldsymbol\beta = (\mathbfX'\mathbfX)^-1 \mathbfX'\mathbfY ] Paso 3: Calcular (\mathbfX'\mathbfX) [ \mathbfX'\mathbfX = \beginbmatrix 1 & 1 & 1 & 1 \ 1 & 2 & 3 & 4 \ 2 & 1 & 3 & 2 \ 1 & 2 & 3 & 4 \endbmatrix \beginbmatrix 1 & 1 & 2 & 1 \ 1 & 2 & 1 & 2 \ 1 & 3 & 3 & 3 \ 1 & 4 & 2 & 4 \endbmatrix ]

Sustituir en (2):

(I) × 3.5: (17.5\beta_1 + 12.25\beta_2 = 47.25) (II) × 5: (17.5\beta_1 + 28.75\beta_2 = 83.75) regresion lineal multiple ejercicios resueltos a mano

Resolvemos: Multiplicamos (1) por 2.5: (10\beta_0 + 25\beta_1 + 17.5\beta_2 = 137.5) Restamos de (2): ((10-10)\beta_0 + (30-25)\beta_1 + (21-17.5)\beta_2 = 151 - 137.5) ⇒ (5\beta_1 + 3.5\beta_2 = 13.5) (I) Modelo: (Y = \beta_0 + \beta_1 X_1 +