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Show that $hatbeta $ and $hatsigma^2$ are unbiased in special case of linear regression model [closed]
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Let's consider we have OLS model $Y= Xbeta+ epsilon$, where rows of matrix $X$ are multivariate normal independent vectors with expected value $0$ and variance $Sigma$. Vector $epsilon$ is independent of X and its mean value is zero ($E(epsilon) =0$) and $Var(epsilon)=sigma^2 I$.
I need to show that $hatbeta = (X^T X)^-1 X^T Y$ and $hatsigma^2= frac1rank(I-H) sum_i=1^n hatepsilon_i^2$, where $H=X(X^T X)^-1X^T$, are still unbiased estimators.
I would really appreciate any help.
statistics regression linear-regression
edited Jul 12 at 15:21
V. Vancak
9,8202926
asked Jul 11 at 17:17
User1999
549
closed as off-topic by Davide Giraudo, amWhy, Nils Matthes, callculus, Parcly Taxel Jul 19 at 1:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, amWhy, Nils Matthes, callculus, Parcly Taxel
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up vote
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Let's consider we have OLS model $Y= Xbeta+ epsilon$, where rows of matrix $X$ are multivariate normal independent vectors with expected value $0$ and variance $Sigma$. Vector $epsilon$ is independent of X and its mean value is zero ($E(epsilon) =0$) and $Var(epsilon)=sigma^2 I$.
I need to show that $hatbeta = (X^T X)^-1 X^T Y$ and $hatsigma^2= frac1rank(I-H) sum_i=1^n hatepsilon_i^2$, where $H=X(X^T X)^-1X^T$, are still unbiased estimators.
I would really appreciate any help.
statistics regression linear-regression
edited Jul 12 at 15:21
V. Vancak
9,8202926
asked Jul 11 at 17:17
User1999
549
closed as off-topic by Davide Giraudo, amWhy, Nils Matthes, callculus, Parcly Taxel Jul 19 at 1:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, amWhy, Nils Matthes, callculus, Parcly Taxel
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let's consider we have OLS model $Y= Xbeta+ epsilon$, where rows of matrix $X$ are multivariate normal independent vectors with expected value $0$ and variance $Sigma$. Vector $epsilon$ is independent of X and its mean value is zero ($E(epsilon) =0$) and $Var(epsilon)=sigma^2 I$.
I need to show that $hatbeta = (X^T X)^-1 X^T Y$ and $hatsigma^2= frac1rank(I-H) sum_i=1^n hatepsilon_i^2$, where $H=X(X^T X)^-1X^T$, are still unbiased estimators.
I would really appreciate any help.
statistics regression linear-regression
edited Jul 12 at 15:21
V. Vancak
9,8202926
asked Jul 11 at 17:17
User1999
549
Let's consider we have OLS model $Y= Xbeta+ epsilon$, where rows of matrix $X$ are multivariate normal independent vectors with expected value $0$ and variance $Sigma$. Vector $epsilon$ is independent of X and its mean value is zero ($E(epsilon) =0$) and $Var(epsilon)=sigma^2 I$.
I need to show that $hatbeta = (X^T X)^-1 X^T Y$ and $hatsigma^2= frac1rank(I-H) sum_i=1^n hatepsilon_i^2$, where $H=X(X^T X)^-1X^T$, are still unbiased estimators.
I would really appreciate any help.
statistics regression linear-regression
edited Jul 12 at 15:21
V. Vancak
9,8202926
asked Jul 11 at 17:17
User1999
549
edited Jul 12 at 15:21
V. Vancak
9,8202926
edited Jul 12 at 15:21
V. Vancak
9,8202926
edited Jul 12 at 15:21
V. Vancak
9,8202926
9,8202926
asked Jul 11 at 17:17
User1999
549
asked Jul 11 at 17:17
User1999
549
asked Jul 11 at 17:17
User1999
549
549
closed as off-topic by Davide Giraudo, amWhy, Nils Matthes, callculus, Parcly Taxel Jul 19 at 1:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, amWhy, Nils Matthes, callculus, Parcly Taxel
closed as off-topic by Davide Giraudo, amWhy, Nils Matthes, callculus, Parcly Taxel Jul 19 at 1:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Davide Giraudo, amWhy, Nils Matthes, callculus, Parcly Taxel
add a comment |Â
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1 Answer
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$$
mathbbE[hatbeta|X]= mathbbE[(X'X)^-1X'y|X] = (X'X)^-1X'(Xbeta + mathbbE[epsilon|X])= (X'X)^-1(X'X)beta = beta.
$$
$$
frac1sigma ^ 2sum_i=1^n hatepsilon^2_i sim chi^2_n-p
$$
and
$$
rank(I-H)=n-p,
$$
hence,
$$
mathbbE[hatsigma^2|X] = fracsigma^2n-p mathbbE[ chi^2_n-p |X] = fracsigma^2(n-p)n-p = sigma^2.
$$
answered Jul 12 at 15:20
V. Vancak
9,8202926
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$$
mathbbE[hatbeta|X]= mathbbE[(X'X)^-1X'y|X] = (X'X)^-1X'(Xbeta + mathbbE[epsilon|X])= (X'X)^-1(X'X)beta = beta.
$$
$$
frac1sigma ^ 2sum_i=1^n hatepsilon^2_i sim chi^2_n-p
$$
and
$$
rank(I-H)=n-p,
$$
hence,
$$
mathbbE[hatsigma^2|X] = fracsigma^2n-p mathbbE[ chi^2_n-p |X] = fracsigma^2(n-p)n-p = sigma^2.
$$
answered Jul 12 at 15:20
V. Vancak
9,8202926
add a comment |Â
up vote
1
down vote
$$
mathbbE[hatbeta|X]= mathbbE[(X'X)^-1X'y|X] = (X'X)^-1X'(Xbeta + mathbbE[epsilon|X])= (X'X)^-1(X'X)beta = beta.
$$
$$
frac1sigma ^ 2sum_i=1^n hatepsilon^2_i sim chi^2_n-p
$$
and
$$
rank(I-H)=n-p,
$$
hence,
$$
mathbbE[hatsigma^2|X] = fracsigma^2n-p mathbbE[ chi^2_n-p |X] = fracsigma^2(n-p)n-p = sigma^2.
$$
answered Jul 12 at 15:20
V. Vancak
9,8202926
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$
mathbbE[hatbeta|X]= mathbbE[(X'X)^-1X'y|X] = (X'X)^-1X'(Xbeta + mathbbE[epsilon|X])= (X'X)^-1(X'X)beta = beta.
$$
$$
frac1sigma ^ 2sum_i=1^n hatepsilon^2_i sim chi^2_n-p
$$
and
$$
rank(I-H)=n-p,
$$
hence,
$$
mathbbE[hatsigma^2|X] = fracsigma^2n-p mathbbE[ chi^2_n-p |X] = fracsigma^2(n-p)n-p = sigma^2.
$$
answered Jul 12 at 15:20
V. Vancak
9,8202926
$$
mathbbE[hatbeta|X]= mathbbE[(X'X)^-1X'y|X] = (X'X)^-1X'(Xbeta + mathbbE[epsilon|X])= (X'X)^-1(X'X)beta = beta.
$$
$$
frac1sigma ^ 2sum_i=1^n hatepsilon^2_i sim chi^2_n-p
$$
and
$$
rank(I-H)=n-p,
$$
hence,
$$
mathbbE[hatsigma^2|X] = fracsigma^2n-p mathbbE[ chi^2_n-p |X] = fracsigma^2(n-p)n-p = sigma^2.
$$
answered Jul 12 at 15:20
V. Vancak
9,8202926
answered Jul 12 at 15:20
V. Vancak
9,8202926
answered Jul 12 at 15:20
V. Vancak
9,8202926
answered Jul 12 at 15:20
V. Vancak
9,8202926
9,8202926
add a comment |Â
add a comment |Â