Tarea 15 - Axel

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Universidad Nacional Autónoma de México 
Facultad de Ingeniería 
 
 
 
 
 
 
 
 
 
 
 
 
 
Estadística 
 
Grupo: 09 Semestre 2021-1 
 
Tarea 15: Obtener beta cero y beta 1 
 
Fecha: 13/01/2021 
 
 
Nombre de los alumnos: 
Ibarra Moreno Sofía 
Axel Sánchez Nazario 
 
 
 
 
 
 
Se tiene que: 
SEC= ∑ 𝑒2
𝑛
𝑖=1
 = ∑ (𝑦𝑖 − ŷ𝑖)
2𝑛
𝑖=1
 = ∑ (𝑦𝑖 − ŷ𝑖)
2𝑛
𝑖=1
 
Y 
𝑦 = β0 + β1x𝑖 + 𝜀 ŷ = β̂0 + β̂1𝑥 
 
Simplificando 
 𝑆𝐸𝐶 = ∑ (𝑦𝑖 − β0 − β1x𝑖)
2𝑛
𝑖=1
 = ∑ 𝑦𝑖
2 − 2𝑦𝑖β0 − 2𝑦𝑖β1x𝑖 + β0
2 +
𝑛
𝑖=1
2β0β1x𝑖 + β1
2x𝑖
2 
 
𝑆𝐸𝐶 = ∑ 𝑦𝑖
2
𝑛
𝑖=1
− 2β0 ∑ 𝑦𝑖
𝑛
𝑖=1
− 2β1 ∑ 𝑦𝑖x𝑖
𝑛
𝑖=1
+ ∑ β0
2
𝑛
𝑖=1
+ 2β0β1 ∑ x𝑖
𝑛
𝑖=1
+ β1
2 ∑ x1
2
𝑛
𝑖=1
 
 
Derivada parcial respecto a 𝛃𝟎 
 
 1) 
𝜕 
𝜕β0
 (∑ 𝑦𝑖
2
𝑛
𝑖=1
− 2β0 ∑ 𝑦𝑖
𝑛
𝑖=1
− 2β1 ∑ 𝑦𝑖x𝑖
𝑛
𝑖=1
+ ∑ β0
2
𝑛
𝑖=1
+ 2β0β1 ∑ x𝑖
𝑛
𝑖=1
+ β1
2 ∑ x1
2
𝑛
𝑖=1
)
= 0 
 
2) 
𝜕 
𝜕β0
(∑ 𝑦𝑖
2
𝑛
𝑖=1
) − 2
𝜕 
𝜕β0
(β0 ∑ 𝑦𝑖
𝑛
𝑖=1
) − 2
𝜕 
𝜕β0
(β1 ∑ 𝑦𝑖x𝑖
𝑛
𝑖=1
) +
𝜕 
𝜕β0
(∑ β0
2
𝑛
𝑖=1
)
+ 2
𝜕 
𝜕β0
(β0β1 ∑ x𝑖
𝑛
𝑖=1
) + 
𝜕 
𝜕β0
(β1
2 ∑ x1
2
𝑛
𝑖=1
) = 0 
 
3) 0 − 2 ∑ 𝑦𝑖
𝑛
𝑖=1
+ 0 + 2nβ0 + 2β1 ∑ x𝑖
𝑛
𝑖=1
+ 0 = 0 
 
 
 
Despejando a β0 
β0 =
2 ∑ 𝑦𝑖
𝑛
𝑖=1 + 2β1 ∑ x𝑖
𝑛
𝑖=1
2𝑛
 = 
1
𝑛
∑ 𝑦𝑖
𝑛
𝑖=1
 − 
β1
𝑛
∑ x𝑖
𝑛
𝑖=1
 
 
➔ �̂�𝟎 = ŷ − �̂�𝟏𝒙 
 
Derivada parcial respecto a 𝛃𝟏 
 
 1) 
𝜕 
𝜕β1
 (∑ 𝑦𝑖
2
𝑛
𝑖=1
− 2β0 ∑ 𝑦𝑖
𝑛
𝑖=1
− 2β1 ∑ 𝑦𝑖x𝑖
𝑛
𝑖=1
+ ∑ β0
2
𝑛
𝑖=1
+ 2β0β1 ∑ x𝑖
𝑛
𝑖=1
+ β1
2 ∑ x1
2
𝑛
𝑖=1
)
= 0 
 
2) 
𝜕 
𝜕β1
(∑ 𝑦𝑖
2
𝑛
𝑖=1
) − 2
𝜕 
𝜕β1
(β0 ∑ 𝑦𝑖
𝑛
𝑖=1
) − 2
𝜕 
𝜕β1
(β1 ∑ 𝑦𝑖x𝑖
𝑛
𝑖=1
) +
𝜕 
𝜕β1
(∑ β0
2
𝑛
𝑖=1
)
+ 2
𝜕 
𝜕β1
(β0β1 ∑ x𝑖
𝑛
𝑖=1
) + 
𝜕 
𝜕β1
(β1
2 ∑ x1
2
𝑛
𝑖=1
) = 0 
 
3) 0 + 0 − 2 ∑ 𝑦𝑖
𝑛
𝑖=1
𝑥𝑖 + 0 + 2β0 ∑ x𝑖
𝑛
𝑖=1
+ 2β1 ∑ x1
2
𝑛
𝑖=1
= 0 
 
Sust a β0 𝑒𝑛 3) 
−2 ∑ 𝑦𝑖
𝑛
𝑖=1
𝑥𝑖 + 2 ( 
1
𝑛
∑ 𝑦𝑖
𝑛
𝑖=1
 − 
β1
𝑛
∑ x𝑖
𝑛
𝑖=1
) ∑ x𝑖
𝑛
𝑖=1
+ 2β1 ∑ x1
2
𝑛
𝑖=1
= 0 
 
Simplificando 
−2 ∑ 𝑦𝑖
𝑛
𝑖=1
𝑥𝑖 +
2
𝑛
∑ 𝑦𝑖
𝑛
𝑖=1
∑ x𝑖
𝑛
𝑖=1
 − 
2β1
𝑛
∑ x𝑖
𝑛
𝑖=1
∑ x𝑖
𝑛
𝑖=1
+ 2β1 ∑ x1
2
𝑛
𝑖=1
= 0 
 
Despejando a β1 
2 ∑ 𝑦𝑖
𝑛
𝑖=1
𝑥𝑖 −
2
𝑛
∑ 𝑦𝑖
𝑛
𝑖=1
∑ x𝑖
𝑛
𝑖=1
= 2β1(∑ x1
2
𝑛
𝑖=1
−
1
𝑛
∑ x𝑖
𝑛
𝑖=1
∑ 𝑥𝑖
𝑛
𝑖=1
) 
 
β1 =
2 ∑ 𝑦𝑖
𝑛
𝑖=1 𝑥𝑖 −
2
𝑛
∑ 𝑦𝑖
𝑛
𝑖=1 ∑ x𝑖
𝑛
𝑖=1
2 (∑ x12
𝑛
𝑖=1 −
1
𝑛
∑ x𝑖
𝑛
𝑖=1 ∑ 𝑥𝑖
𝑛
𝑖=1 )
 = 
∑ 𝑦𝑖
𝑛
𝑖=1 𝑥𝑖 −
1
𝑛
∑ 𝑦𝑖
𝑛
𝑖=1 ∑ x𝑖
𝑛
𝑖=1
∑ x12
𝑛
𝑖=1 −
1
𝑛
(∑ x𝑖
𝑛
𝑖=1 )
2
= 
𝑆𝑆𝑥𝑦
𝑆𝑆𝑥𝑥
 
 
 
➔ 𝛃𝟏 =
𝑺𝑺𝒙𝒚
𝑺𝑺𝒙𝒙