What is J in while calculating SST in multiple regression?
$begingroup$
I am little confused what actually is the J in the formula of the SST and SSR for multiple regression
SST= $Y^Tleft[ 1-frac{1}{n}Jright]Y$
SSR=$Y^Tleft[ H-frac{1}{n}Jright]Y$
matrices statistics regression linear-regression
edited Dec 8 '18 at 16:27
V. Vancak
11k2926
asked Dec 8 '18 at 4:26
surbhi groversurbhi grover
61
$endgroup$
add a comment |
$begingroup$
I am little confused what actually is the J in the formula of the SST and SSR for multiple regression
SST= $Y^Tleft[ 1-frac{1}{n}Jright]Y$
SSR=$Y^Tleft[ H-frac{1}{n}Jright]Y$
matrices statistics regression linear-regression
edited Dec 8 '18 at 16:27
V. Vancak
11k2926
asked Dec 8 '18 at 4:26
surbhi groversurbhi grover
61
$endgroup$
add a comment |
$begingroup$
I am little confused what actually is the J in the formula of the SST and SSR for multiple regression
SST= $Y^Tleft[ 1-frac{1}{n}Jright]Y$
SSR=$Y^Tleft[ H-frac{1}{n}Jright]Y$
matrices statistics regression linear-regression
edited Dec 8 '18 at 16:27
V. Vancak
11k2926
asked Dec 8 '18 at 4:26
surbhi groversurbhi grover
61
$endgroup$
I am little confused what actually is the J in the formula of the SST and SSR for multiple regression
SST= $Y^Tleft[ 1-frac{1}{n}Jright]Y$
SSR=$Y^Tleft[ H-frac{1}{n}Jright]Y$
matrices statistics regression linear-regression
matrices statistics regression linear-regression
edited Dec 8 '18 at 16:27
V. Vancak
11k2926
asked Dec 8 '18 at 4:26
surbhi groversurbhi grover
61
edited Dec 8 '18 at 16:27
V. Vancak
11k2926
asked Dec 8 '18 at 4:26
surbhi groversurbhi grover
61
edited Dec 8 '18 at 16:27
V. Vancak
11k2926
edited Dec 8 '18 at 16:27
V. Vancak
11k2926
edited Dec 8 '18 at 16:27
V. Vancak
11k2926
11k2926
asked Dec 8 '18 at 4:26
surbhi groversurbhi grover
61
asked Dec 8 '18 at 4:26
surbhi groversurbhi grover
61
asked Dec 8 '18 at 4:26
surbhi groversurbhi grover
61
61
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$J$ is the matrix of all $1$s. i.e., let
$$
mathbf{1}=(1,1,...,1)^Tin mathbb{R}^n,
$$
then
$$
J = mathbf{1}mathbf{1}^T.
$$
While $frac{1}{n}J$ can be called "means generating matrix", namely, for some $y=(y_1, y_2,...,y_n)^T in mathbb{R}^n$, then
$$
frac{1}{n}Jy= (bar{y}_n, bar{y}_n, ..., bar{y}_n)^T,
$$
which is an essential part of any sum of squares.
answered Dec 8 '18 at 16:24
V. VancakV. Vancak
11k2926
$endgroup$
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
$begingroup$
$J$ is the matrix of all $1$s. i.e., let
$$
mathbf{1}=(1,1,...,1)^Tin mathbb{R}^n,
$$
then
$$
J = mathbf{1}mathbf{1}^T.
$$
While $frac{1}{n}J$ can be called "means generating matrix", namely, for some $y=(y_1, y_2,...,y_n)^T in mathbb{R}^n$, then
$$
frac{1}{n}Jy= (bar{y}_n, bar{y}_n, ..., bar{y}_n)^T,
$$
which is an essential part of any sum of squares.
answered Dec 8 '18 at 16:24
V. VancakV. Vancak
11k2926
$endgroup$
add a comment |
$begingroup$
$J$ is the matrix of all $1$s. i.e., let
$$
mathbf{1}=(1,1,...,1)^Tin mathbb{R}^n,
$$
then
$$
J = mathbf{1}mathbf{1}^T.
$$
While $frac{1}{n}J$ can be called "means generating matrix", namely, for some $y=(y_1, y_2,...,y_n)^T in mathbb{R}^n$, then
$$
frac{1}{n}Jy= (bar{y}_n, bar{y}_n, ..., bar{y}_n)^T,
$$
which is an essential part of any sum of squares.
answered Dec 8 '18 at 16:24
V. VancakV. Vancak
11k2926
$endgroup$
add a comment |
$begingroup$
$J$ is the matrix of all $1$s. i.e., let
$$
mathbf{1}=(1,1,...,1)^Tin mathbb{R}^n,
$$
then
$$
J = mathbf{1}mathbf{1}^T.
$$
While $frac{1}{n}J$ can be called "means generating matrix", namely, for some $y=(y_1, y_2,...,y_n)^T in mathbb{R}^n$, then
$$
frac{1}{n}Jy= (bar{y}_n, bar{y}_n, ..., bar{y}_n)^T,
$$
which is an essential part of any sum of squares.
answered Dec 8 '18 at 16:24
V. VancakV. Vancak
11k2926
$endgroup$
$J$ is the matrix of all $1$s. i.e., let
$$
mathbf{1}=(1,1,...,1)^Tin mathbb{R}^n,
$$
then
$$
J = mathbf{1}mathbf{1}^T.
$$
While $frac{1}{n}J$ can be called "means generating matrix", namely, for some $y=(y_1, y_2,...,y_n)^T in mathbb{R}^n$, then
$$
frac{1}{n}Jy= (bar{y}_n, bar{y}_n, ..., bar{y}_n)^T,
$$
which is an essential part of any sum of squares.
answered Dec 8 '18 at 16:24
V. VancakV. Vancak
11k2926
answered Dec 8 '18 at 16:24
V. VancakV. Vancak
11k2926
answered Dec 8 '18 at 16:24
V. VancakV. Vancak
11k2926
answered Dec 8 '18 at 16:24
V. VancakV. Vancak
11k2926
11k2926
add a comment |
add a comment |
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