Mixing
For a general linear regression , are Y and Y hat independent?
$begingroup$
For a general linear regression , are $Y$ and $hat{Y}$ independent?
$Y$=XB+e
$hat{Y}$=X$hat{B}$
I think they are dependent, because if they rely on the same data, they should have some sort of relationship.
that is to say , $hat{Y}$=HY , so they are dependent right?
regression
asked Nov 6 '14 at 21:06
user1919987user1919987
621414
$endgroup$
add a comment |
$begingroup$
For a general linear regression , are $Y$ and $hat{Y}$ independent?
$Y$=XB+e
$hat{Y}$=X$hat{B}$
I think they are dependent, because if they rely on the same data, they should have some sort of relationship.
that is to say , $hat{Y}$=HY , so they are dependent right?
regression
asked Nov 6 '14 at 21:06
user1919987user1919987
621414
$endgroup$
$begingroup$
"...because if they rely on the same data..." Could you explain what you mean by that?
$endgroup$
– daOnlyBG
Nov 6 '14 at 21:09
$begingroup$
well, isn't Y= xb +e , and Y hat = xBhat . they all rely on x and y data?
$endgroup$
– user1919987
Nov 6 '14 at 21:20
$begingroup$
Perhaps define $Y$ and $hat{Y}$.
$endgroup$
– parsiad
Nov 6 '14 at 21:28
add a comment |
$begingroup$
For a general linear regression , are $Y$ and $hat{Y}$ independent?
$Y$=XB+e
$hat{Y}$=X$hat{B}$
I think they are dependent, because if they rely on the same data, they should have some sort of relationship.
that is to say , $hat{Y}$=HY , so they are dependent right?
regression
asked Nov 6 '14 at 21:06
user1919987user1919987
621414
$endgroup$
For a general linear regression , are $Y$ and $hat{Y}$ independent?
$Y$=XB+e
$hat{Y}$=X$hat{B}$
I think they are dependent, because if they rely on the same data, they should have some sort of relationship.
that is to say , $hat{Y}$=HY , so they are dependent right?
regression
regression
asked Nov 6 '14 at 21:06
user1919987user1919987
621414
asked Nov 6 '14 at 21:06
user1919987user1919987
621414
edited Nov 6 '14 at 21:51
user1919987
asked Nov 6 '14 at 21:06
user1919987user1919987
621414
asked Nov 6 '14 at 21:06
user1919987user1919987
621414
asked Nov 6 '14 at 21:06
user1919987user1919987
621414
621414
$begingroup$
"...because if they rely on the same data..." Could you explain what you mean by that?
$endgroup$
– daOnlyBG
Nov 6 '14 at 21:09
$begingroup$
well, isn't Y= xb +e , and Y hat = xBhat . they all rely on x and y data?
$endgroup$
– user1919987
Nov 6 '14 at 21:20
$begingroup$
Perhaps define $Y$ and $hat{Y}$.
$endgroup$
– parsiad
Nov 6 '14 at 21:28
add a comment |
$begingroup$
"...because if they rely on the same data..." Could you explain what you mean by that?
$endgroup$
– daOnlyBG
Nov 6 '14 at 21:09
$begingroup$
well, isn't Y= xb +e , and Y hat = xBhat . they all rely on x and y data?
$endgroup$
– user1919987
Nov 6 '14 at 21:20
$begingroup$
Perhaps define $Y$ and $hat{Y}$.
$endgroup$
– parsiad
Nov 6 '14 at 21:28
$begingroup$
"...because if they rely on the same data..." Could you explain what you mean by that?
$endgroup$
– daOnlyBG
Nov 6 '14 at 21:09
$begingroup$
"...because if they rely on the same data..." Could you explain what you mean by that?
$endgroup$
– daOnlyBG
Nov 6 '14 at 21:09
$begingroup$
well, isn't Y= xb +e , and Y hat = xBhat . they all rely on x and y data?
$endgroup$
– user1919987
Nov 6 '14 at 21:20
$begingroup$
well, isn't Y= xb +e , and Y hat = xBhat . they all rely on x and y data?
$endgroup$
– user1919987
Nov 6 '14 at 21:20
$begingroup$
Perhaps define $Y$ and $hat{Y}$.
$endgroup$
– parsiad
Nov 6 '14 at 21:28
$begingroup$
Perhaps define $Y$ and $hat{Y}$.
$endgroup$
– parsiad
Nov 6 '14 at 21:28
add a comment |
1 Answer
1
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$begingroup$
As long as B is not equal to 0, the answer is no. If B is not 0, then both terms will depend on X and so they cannot be independent.
$Cov(Y,hat{Y})=Cov(XB+e,Ŷ) = B*cov(X,Xhat{b})$
Edit: I suppose if your model is misspecified, you could get an incorrect estimate of $hat{b}=0$, and thus no correlation, but I don't think thats what you had in mind.
answered Nov 7 '14 at 23:01
GregGreg
527311
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1 Answer
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active
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1 Answer
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active
oldest
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active
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votes
$begingroup$
As long as B is not equal to 0, the answer is no. If B is not 0, then both terms will depend on X and so they cannot be independent.
$Cov(Y,hat{Y})=Cov(XB+e,Ŷ) = B*cov(X,Xhat{b})$
Edit: I suppose if your model is misspecified, you could get an incorrect estimate of $hat{b}=0$, and thus no correlation, but I don't think thats what you had in mind.
answered Nov 7 '14 at 23:01
GregGreg
527311
$endgroup$
add a comment |
$begingroup$
As long as B is not equal to 0, the answer is no. If B is not 0, then both terms will depend on X and so they cannot be independent.
$Cov(Y,hat{Y})=Cov(XB+e,Ŷ) = B*cov(X,Xhat{b})$
Edit: I suppose if your model is misspecified, you could get an incorrect estimate of $hat{b}=0$, and thus no correlation, but I don't think thats what you had in mind.
answered Nov 7 '14 at 23:01
GregGreg
527311
$endgroup$
add a comment |
$begingroup$
As long as B is not equal to 0, the answer is no. If B is not 0, then both terms will depend on X and so they cannot be independent.
$Cov(Y,hat{Y})=Cov(XB+e,Ŷ) = B*cov(X,Xhat{b})$
Edit: I suppose if your model is misspecified, you could get an incorrect estimate of $hat{b}=0$, and thus no correlation, but I don't think thats what you had in mind.
answered Nov 7 '14 at 23:01
GregGreg
527311
$endgroup$
As long as B is not equal to 0, the answer is no. If B is not 0, then both terms will depend on X and so they cannot be independent.
$Cov(Y,hat{Y})=Cov(XB+e,Ŷ) = B*cov(X,Xhat{b})$
Edit: I suppose if your model is misspecified, you could get an incorrect estimate of $hat{b}=0$, and thus no correlation, but I don't think thats what you had in mind.
answered Nov 7 '14 at 23:01
GregGreg
527311
answered Nov 7 '14 at 23:01
GregGreg
527311
answered Nov 7 '14 at 23:01
GregGreg
527311
answered Nov 7 '14 at 23:01
GregGreg
527311
527311
add a comment |
add a comment |
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$begingroup$
"...because if they rely on the same data..." Could you explain what you mean by that?
$endgroup$
– daOnlyBG
Nov 6 '14 at 21:09
$begingroup$
well, isn't Y= xb +e , and Y hat = xBhat . they all rely on x and y data?
$endgroup$
– user1919987
Nov 6 '14 at 21:20
$begingroup$
Perhaps define $Y$ and $hat{Y}$.
$endgroup$
– parsiad
Nov 6 '14 at 21:28