✅ Lo Bueno (The Good):
⚠️ Lo Tedioso (The Bad):
❌ Lo Feo (The Ugly):
El sistema es:
[ \begincases 4\beta_0 + 10\beta_1 + 7\beta_2 + 10\beta_3 = 55 \ 10\beta_0 + 30\beta_1 + 21\beta_2 + 30\beta_3 = 151 \ 7\beta_0 + 21\beta_1 + 18\beta_2 + 21\beta_3 = 113 \ 10\beta_0 + 30\beta_1 + 21\beta_2 + 30\beta_3 = 151 \endcases ]
Vemos que las ecuaciones 2 y 4 son iguales, por lo que tenemos infinitas soluciones (multicolinealidad). Elegimos una solución particular: hacemos (\beta_3 = 0).
Con (\beta_3=0), el sistema se reduce a:
(1) (4\beta_0 + 10\beta_1 + 7\beta_2 = 55)
(2) (10\beta_0 + 30\beta_1 + 21\beta_2 = 151)
(3) (7\beta_0 + 21\beta_1 + 18\beta_2 = 113)
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)
Multiplicamos (1) por 1.75: (7\beta_0 + 17.5\beta_1 + 12.25\beta_2 = 96.25)
Restamos de (3): ((7-7)\beta_0 + (21-17.5)\beta_1 + (18-12.25)\beta_2 = 113 - 96.25) ⇒ (3.5\beta_1 + 5.75\beta_2 = 16.75) (II)
Resolvemos (I) y (II):
(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)
Restamos: ((28.75-12.25)\beta_2 = 83.75 - 47.25) ⇒ (16.5\beta_2 = 36.5) ⇒ (\beta_2 = 2.2121)
Luego (5\beta_1 + 3.5(2.2121)=13.5) ⇒ (5\beta_1 = 13.5 - 7.7424 = 5.7576) ⇒ (\beta_1 = 1.1515) regresion lineal multiple ejercicios resueltos a mano
De (1): (4\beta_0 + 10(1.1515) + 7(2.2121) = 55)
(4\beta_0 + 11.515 + 15.4847 = 55) ⇒ (4\beta_0 + 27 = 55) ⇒ (4\beta_0 = 28) ⇒ (\beta_0 = 7)
Modelo elegido (con (\beta_3=0)): [ \hatY = 7 + 1.1515 X_1 + 2.2121 X_2 ]
Nota: La multicolinealidad revela que (X_3) no aporta información adicional si ya tenemos (X_1) y (X_4)(??)… En este caso, (X_3) es combinación lineal.
Para un modelo con dos variables independientes ($Y$ vs $X_1$ y $X_2$), el sistema de ecuaciones es:
Aquí es donde se resuelve el sistema (usualmente por Reducción de Gauss o Regla de Cramer si es $2x2$).
Simplificar (1): Dividimos entre 1 (no simplifica mucho). Lo dejamos como está.
Vamos a eliminar (\hat\beta_0) de (2) y (3) usando (1).
Multiplicamos (1) por (24/5) y restamos de (2), pero es más ordenado hacer:
De (1): (5\hat\beta_0 = 150 - 24\hat\beta_1 - 25\hat\beta_2) ⇒ (\hat\beta_0 = 30 - 4.8\hat\beta_1 - 5\hat\beta_2)
Sustituir en (2):
(24(30 - 4.8\hat\beta_1 - 5\hat\beta_2) + 138\hat\beta_1 + 135\hat\beta_2 = 774)
(720 - 115.2\hat\beta_1 - 120\hat\beta_2 + 138\hat\beta_1 + 135\hat\beta_2 = 774)
(720 + (22.8)\hat\beta_1 + 15\hat\beta_2 = 774)
(22.8\hat\beta_1 + 15\hat\beta_2 = 54) (Ecuación A)
Sustituir en (3):
(25(30 - 4.8\hat\beta_1 - 5\hat\beta_2) + 135\hat\beta_1 + 135\hat\beta_2 = 786)
(750 - 120\hat\beta_1 - 125\hat\beta_2 + 135\hat\beta_1 + 135\hat\beta_2 = 786)
(750 + 15\hat\beta_1 + 10\hat\beta_2 = 786)
(15\hat\beta_1 + 10\hat\beta_2 = 36) (Ecuación B) ✅ Lo Bueno (The Good):
Ahora resolvemos A y B:
A: (22.8\beta_1 + 15\beta_2 = 54)
B: (15\beta_1 + 10\beta_2 = 36)
Multiplicamos B por 1.5: (22.5\beta_1 + 15\beta_2 = 54)
Restamos A - (B×1.5):
((22.8 - 22.5)\beta_1 + (15-15)\beta_2 = 54 - 54)
(0.3\beta_1 = 0) ⇒ (\beta_1 = 0)
Sustituir (\beta_1=0) en B: (15(0) + 10\beta_2 = 36) ⇒ (\beta_2 = 3.6)
Ahora hallamos (\beta_0):
(\beta_0 = 30 - 4.8(0) - 5(3.6) = 30 - 18 = 12)
Eliminate (b_0):
From (1): (5b_0 = 375 - 20b_1 - 32b_2 \Rightarrow b_0 = 75 - 4b_1 - 6.4b_2)
Substitute into (2): [ 1550 = 20(75 - 4b_1 - 6.4b_2) + 90b_1 + 123b_2 ] [ 1550 = 1500 - 80b_1 - 128b_2 + 90b_1 + 123b_2 ] [ 1550 - 1500 = 10b_1 - 5b_2 ] [ 50 = 10b_1 - 5b_2 \quad \text(Divide by 5) \Rightarrow 10 = 2b_1 - b_2 \quad (A) ]
Substitute into (3): [ 2375 = 32(75 - 4b_1 - 6.4b_2) + 123b_1 + 210b_2 ] [ 2375 = 2400 - 128b_1 - 204.8b_2 + 123b_1 + 210b_2 ] [ 2375 - 2400 = -5b_1 + 5.2b_2 ] [ -25 = -5b_1 + 5.2b_2 \quad \text(Multiply by -1) \Rightarrow 25 = 5b_1 - 5.2b_2 \quad (B) ]
Now solve (A) and (B): From (A): (b_2 = 2b_1 - 10)
Sub into (B): [ 25 = 5b_1 - 5.2(2b_1 - 10) ] [ 25 = 5b_1 - 10.4b_1 + 52 ] [ 25 - 52 = -5.4b_1 ] [ -27 = -5.4b_1 \Rightarrow b_1 = 5 ]
Then (b_2 = 2(5) - 10 = 0)
Finally (b_0 = 75 - 4(5) - 6.4(0) = 75 - 20 = 55)
Problem: A teacher wants to predict students' final exam scores ($Y$) based on:
Data from 5 students:
| Student | $X_1$ (Hours) | $X_2$ (Quizzes) | $Y$ (Exam Score) | |---------|--------------|----------------|------------------| | 1 | 2 | 1 | 70 | | 2 | 3 | 2 | 75 | | 3 | 5 | 3 | 85 | | 4 | 7 | 4 | 95 | | 5 | 8 | 5 | 100 |
Goal: Find the regression equation: $\hatY = \beta_0 + \beta_1 X_1 + \beta_2 X_2$
From (1): (4b_0 = 38 - 10b_1 - 14b_2 \Rightarrow b_0 = 9.5 - 2.5b_1 - 3.5b_2)
Sub into (2): [ 110 = 10(9.5 - 2.5b_1 - 3.5b_2) + 30b_1 + 40b_2 ] [ 110 = 95 - 25b_1 - 35b_2 + 30b_1 + 40b_2 ] [ 110 - 95 = 5b_1 + 5b_2 ] [ 15 = 5(b_1 + b_2) \Rightarrow b_1 + b_2 = 3 \quad (A) ]
Sub into (3): [ 148 = 14(9.5 - 2.5b_1 - 3.5b_2) + 40b_1 + 54b_2 ] [ 148 = 133 - 35b_1 - 49b_2 + 40b_1 + 54b_2 ] [ 148 - 133 = 5b_1 + 5b_2 ] [ 15 = 5(b_1 + b_2) \Rightarrow b_1 + b_2 = 3 \quad (B) ]
(A) and (B) are identical → infinite solutions? Let's check raw data:
Notice (Y = 3X_1 + 2)? No: (Y = 3X_1 + 2) gives Week 1: 5 (yes), Week 2: 8 (yes), Week 3: 11 (yes), Week 4: 14 (yes).
Also (X_2 = X_1 + 1) always. So perfect multicollinearity: (X_2) provides no unique info.
Thus, we can set (b_2 = 0): Then (b_1 = 3), (b_0 = 9.5 - 2.5(3) = 9.5 - 7.5 = 2).
Equation: (\hatY = 2 + 3X_1 + 0X_2).