Can these two regression models be compared based on the coefficient of determination?
As part of a statistical project I'm trying to deal with, I've ran into the question with the following models:
(1) $y=beta_0+beta_1x_1+beta_2x_2+u$
(2) $y=beta_0+beta_1x_1+(1-beta_1)x_2+u$
The variables are the same but (2) is linearly constrained, and cannot calculate its $R^2$(?).
Can the two models be compared based on the coefficient of determination?
statistics regression
asked Dec 29 '18 at 9:10
O B.O B.
1
add a comment |
As part of a statistical project I'm trying to deal with, I've ran into the question with the following models:
(1) $y=beta_0+beta_1x_1+beta_2x_2+u$
(2) $y=beta_0+beta_1x_1+(1-beta_1)x_2+u$
The variables are the same but (2) is linearly constrained, and cannot calculate its $R^2$(?).
Can the two models be compared based on the coefficient of determination?
statistics regression
asked Dec 29 '18 at 9:10
O B.O B.
1
Why can't $R^2$ be calculated for model (2)? I'm not sure I completely understand the question, but it might be helpful to see that if we define $x_3 = x_1 - x_2$ then (2) can be written as $y = beta_0 + beta_1(x_1 - x_2) + 1x_2 + u = beta_0 + beta_1x_3 + 1x_2 + u$.
– Alex
Dec 29 '18 at 11:13
add a comment |
As part of a statistical project I'm trying to deal with, I've ran into the question with the following models:
(1) $y=beta_0+beta_1x_1+beta_2x_2+u$
(2) $y=beta_0+beta_1x_1+(1-beta_1)x_2+u$
The variables are the same but (2) is linearly constrained, and cannot calculate its $R^2$(?).
Can the two models be compared based on the coefficient of determination?
statistics regression
asked Dec 29 '18 at 9:10
O B.O B.
1
As part of a statistical project I'm trying to deal with, I've ran into the question with the following models:
(1) $y=beta_0+beta_1x_1+beta_2x_2+u$
(2) $y=beta_0+beta_1x_1+(1-beta_1)x_2+u$
The variables are the same but (2) is linearly constrained, and cannot calculate its $R^2$(?).
Can the two models be compared based on the coefficient of determination?
statistics regression
statistics regression
asked Dec 29 '18 at 9:10
O B.O B.
1
asked Dec 29 '18 at 9:10
O B.O B.
1
asked Dec 29 '18 at 9:10
O B.O B.
1
asked Dec 29 '18 at 9:10
O B.O B.
1
asked Dec 29 '18 at 9:10
O B.O B.
1
1
Why can't $R^2$ be calculated for model (2)? I'm not sure I completely understand the question, but it might be helpful to see that if we define $x_3 = x_1 - x_2$ then (2) can be written as $y = beta_0 + beta_1(x_1 - x_2) + 1x_2 + u = beta_0 + beta_1x_3 + 1x_2 + u$.
– Alex
Dec 29 '18 at 11:13
add a comment |
Why can't $R^2$ be calculated for model (2)? I'm not sure I completely understand the question, but it might be helpful to see that if we define $x_3 = x_1 - x_2$ then (2) can be written as $y = beta_0 + beta_1(x_1 - x_2) + 1x_2 + u = beta_0 + beta_1x_3 + 1x_2 + u$.
– Alex
Dec 29 '18 at 11:13
Why can't $R^2$ be calculated for model (2)? I'm not sure I completely understand the question, but it might be helpful to see that if we define $x_3 = x_1 - x_2$ then (2) can be written as $y = beta_0 + beta_1(x_1 - x_2) + 1x_2 + u = beta_0 + beta_1x_3 + 1x_2 + u$.
– Alex
Dec 29 '18 at 11:13
Why can't $R^2$ be calculated for model (2)? I'm not sure I completely understand the question, but it might be helpful to see that if we define $x_3 = x_1 - x_2$ then (2) can be written as $y = beta_0 + beta_1(x_1 - x_2) + 1x_2 + u = beta_0 + beta_1x_3 + 1x_2 + u$.
– Alex
Dec 29 '18 at 11:13
add a comment |
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Why can't $R^2$ be calculated for model (2)? I'm not sure I completely understand the question, but it might be helpful to see that if we define $x_3 = x_1 - x_2$ then (2) can be written as $y = beta_0 + beta_1(x_1 - x_2) + 1x_2 + u = beta_0 + beta_1x_3 + 1x_2 + u$.
– Alex
Dec 29 '18 at 11:13