R-squared and variance relation
$begingroup$
According to Wiki:
https://en.wikipedia.org/wiki/Fraction_of_variance_unexplained
$1 - R^2 = VAR_{err}/VAR_{tot}$
Where $VAR_{err} = sum_{i = 1}^N (y_i - hat{y}_i)^2$
is the variance of the residuals.
I don't see how this is correct, unless the residuals have mean zero. To see this, define the residuals:
$e_i = y_i - hat{y}_i$
The variance of the residuals is:
$sum_{i = 1}^N (e_i - bar{e}_i)^2 =
sum_{i = 1}^N (y_i - hat{y}_i - bar{e}_i)^2 $
Do you disagree or is Wikipedia wrong?
statistics regression
edited Dec 9 '18 at 20:05
Picaud Vincent
1,49439
asked Dec 9 '18 at 19:57
hanshans
11
$endgroup$
add a comment |
$begingroup$
According to Wiki:
https://en.wikipedia.org/wiki/Fraction_of_variance_unexplained
$1 - R^2 = VAR_{err}/VAR_{tot}$
Where $VAR_{err} = sum_{i = 1}^N (y_i - hat{y}_i)^2$
is the variance of the residuals.
I don't see how this is correct, unless the residuals have mean zero. To see this, define the residuals:
$e_i = y_i - hat{y}_i$
The variance of the residuals is:
$sum_{i = 1}^N (e_i - bar{e}_i)^2 =
sum_{i = 1}^N (y_i - hat{y}_i - bar{e}_i)^2 $
Do you disagree or is Wikipedia wrong?
statistics regression
edited Dec 9 '18 at 20:05
Picaud Vincent
1,49439
asked Dec 9 '18 at 19:57
hanshans
11
$endgroup$
add a comment |
$begingroup$
According to Wiki:
https://en.wikipedia.org/wiki/Fraction_of_variance_unexplained
$1 - R^2 = VAR_{err}/VAR_{tot}$
Where $VAR_{err} = sum_{i = 1}^N (y_i - hat{y}_i)^2$
is the variance of the residuals.
I don't see how this is correct, unless the residuals have mean zero. To see this, define the residuals:
$e_i = y_i - hat{y}_i$
The variance of the residuals is:
$sum_{i = 1}^N (e_i - bar{e}_i)^2 =
sum_{i = 1}^N (y_i - hat{y}_i - bar{e}_i)^2 $
Do you disagree or is Wikipedia wrong?
statistics regression
edited Dec 9 '18 at 20:05
Picaud Vincent
1,49439
asked Dec 9 '18 at 19:57
hanshans
11
$endgroup$
According to Wiki:
https://en.wikipedia.org/wiki/Fraction_of_variance_unexplained
$1 - R^2 = VAR_{err}/VAR_{tot}$
Where $VAR_{err} = sum_{i = 1}^N (y_i - hat{y}_i)^2$
is the variance of the residuals.
I don't see how this is correct, unless the residuals have mean zero. To see this, define the residuals:
$e_i = y_i - hat{y}_i$
The variance of the residuals is:
$sum_{i = 1}^N (e_i - bar{e}_i)^2 =
sum_{i = 1}^N (y_i - hat{y}_i - bar{e}_i)^2 $
Do you disagree or is Wikipedia wrong?
statistics regression
statistics regression
edited Dec 9 '18 at 20:05
Picaud Vincent
1,49439
asked Dec 9 '18 at 19:57
hanshans
11
edited Dec 9 '18 at 20:05
Picaud Vincent
1,49439
asked Dec 9 '18 at 19:57
hanshans
11
edited Dec 9 '18 at 20:05
Picaud Vincent
1,49439
edited Dec 9 '18 at 20:05
Picaud Vincent
1,49439
edited Dec 9 '18 at 20:05
Picaud Vincent
1,49439
1,49439
asked Dec 9 '18 at 19:57
hanshans
11
asked Dec 9 '18 at 19:57
hanshans
11
asked Dec 9 '18 at 19:57
hanshans
11
11
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that in OLS regression with an intercept term, due to the "first order condition", when you take derivative w.r.t. $beta_0$ you have
$$
-2sum_{i=1}^n (y_i - hat{beta}_0 - hat{beta}_1x_{1i} - cdots hat{beta}_px_{pi}) = -2sum e_i =0,
$$
hence
$$
bar{e}_n = 0,
$$
thus
$$
hat{sigma}^2 = frac{1}{n}sum(e_i - bar{e})^2 = frac{sum e_i ^ 2}{n}.
$$
answered Dec 10 '18 at 21:23
V. VancakV. Vancak
11k2926
$endgroup$
$begingroup$
Thanks for the answer, but this not only hold for linear models? I think Wiki is wrong since it treats the more general case, but still imposes this assumption.
$endgroup$
– hans
Dec 25 '18 at 9:13
$begingroup$
@hans The residuals have zero mean for any model with an intercept term and least-squares estimation.
$endgroup$
– V. Vancak
Dec 25 '18 at 9:58
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that in OLS regression with an intercept term, due to the "first order condition", when you take derivative w.r.t. $beta_0$ you have
$$
-2sum_{i=1}^n (y_i - hat{beta}_0 - hat{beta}_1x_{1i} - cdots hat{beta}_px_{pi}) = -2sum e_i =0,
$$
hence
$$
bar{e}_n = 0,
$$
thus
$$
hat{sigma}^2 = frac{1}{n}sum(e_i - bar{e})^2 = frac{sum e_i ^ 2}{n}.
$$
answered Dec 10 '18 at 21:23
V. VancakV. Vancak
11k2926
$endgroup$
$begingroup$
Thanks for the answer, but this not only hold for linear models? I think Wiki is wrong since it treats the more general case, but still imposes this assumption.
$endgroup$
– hans
Dec 25 '18 at 9:13
$begingroup$
@hans The residuals have zero mean for any model with an intercept term and least-squares estimation.
$endgroup$
– V. Vancak
Dec 25 '18 at 9:58
add a comment |
$begingroup$
Note that in OLS regression with an intercept term, due to the "first order condition", when you take derivative w.r.t. $beta_0$ you have
$$
-2sum_{i=1}^n (y_i - hat{beta}_0 - hat{beta}_1x_{1i} - cdots hat{beta}_px_{pi}) = -2sum e_i =0,
$$
hence
$$
bar{e}_n = 0,
$$
thus
$$
hat{sigma}^2 = frac{1}{n}sum(e_i - bar{e})^2 = frac{sum e_i ^ 2}{n}.
$$
answered Dec 10 '18 at 21:23
V. VancakV. Vancak
11k2926
$endgroup$
$begingroup$
Thanks for the answer, but this not only hold for linear models? I think Wiki is wrong since it treats the more general case, but still imposes this assumption.
$endgroup$
– hans
Dec 25 '18 at 9:13
$begingroup$
@hans The residuals have zero mean for any model with an intercept term and least-squares estimation.
$endgroup$
– V. Vancak
Dec 25 '18 at 9:58
add a comment |
$begingroup$
Note that in OLS regression with an intercept term, due to the "first order condition", when you take derivative w.r.t. $beta_0$ you have
$$
-2sum_{i=1}^n (y_i - hat{beta}_0 - hat{beta}_1x_{1i} - cdots hat{beta}_px_{pi}) = -2sum e_i =0,
$$
hence
$$
bar{e}_n = 0,
$$
thus
$$
hat{sigma}^2 = frac{1}{n}sum(e_i - bar{e})^2 = frac{sum e_i ^ 2}{n}.
$$
answered Dec 10 '18 at 21:23
V. VancakV. Vancak
11k2926
$endgroup$
Note that in OLS regression with an intercept term, due to the "first order condition", when you take derivative w.r.t. $beta_0$ you have
$$
-2sum_{i=1}^n (y_i - hat{beta}_0 - hat{beta}_1x_{1i} - cdots hat{beta}_px_{pi}) = -2sum e_i =0,
$$
hence
$$
bar{e}_n = 0,
$$
thus
$$
hat{sigma}^2 = frac{1}{n}sum(e_i - bar{e})^2 = frac{sum e_i ^ 2}{n}.
$$
answered Dec 10 '18 at 21:23
V. VancakV. Vancak
11k2926
answered Dec 10 '18 at 21:23
V. VancakV. Vancak
11k2926
answered Dec 10 '18 at 21:23
V. VancakV. Vancak
11k2926
answered Dec 10 '18 at 21:23
V. VancakV. Vancak
11k2926
11k2926
$begingroup$
Thanks for the answer, but this not only hold for linear models? I think Wiki is wrong since it treats the more general case, but still imposes this assumption.
$endgroup$
– hans
Dec 25 '18 at 9:13
$begingroup$
@hans The residuals have zero mean for any model with an intercept term and least-squares estimation.
$endgroup$
– V. Vancak
Dec 25 '18 at 9:58
add a comment |
$begingroup$
Thanks for the answer, but this not only hold for linear models? I think Wiki is wrong since it treats the more general case, but still imposes this assumption.
$endgroup$
– hans
Dec 25 '18 at 9:13
$begingroup$
@hans The residuals have zero mean for any model with an intercept term and least-squares estimation.
$endgroup$
– V. Vancak
Dec 25 '18 at 9:58
$begingroup$
Thanks for the answer, but this not only hold for linear models? I think Wiki is wrong since it treats the more general case, but still imposes this assumption.
$endgroup$
– hans
Dec 25 '18 at 9:13
$begingroup$
Thanks for the answer, but this not only hold for linear models? I think Wiki is wrong since it treats the more general case, but still imposes this assumption.
$endgroup$
– hans
Dec 25 '18 at 9:13
$begingroup$
@hans The residuals have zero mean for any model with an intercept term and least-squares estimation.
$endgroup$
– V. Vancak
Dec 25 '18 at 9:58
$begingroup$
@hans The residuals have zero mean for any model with an intercept term and least-squares estimation.
$endgroup$
– V. Vancak
Dec 25 '18 at 9:58
add a comment |
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