Mixing
Trouble with Gauss-Markov Theorem and with finding a Best “Non-Linear” Unbiased Estimator
$begingroup$
Let us consider a simple model.
$y_i = beta + epsilon_i$
If we assume that $epsilon_i$ has 0 mean, constant variance and is uncorrelated. Then via Gauss-Markov theorem we know that $hat{beta} = bar{y}$ is the BLUE.
However, when I assume that $epsilon_i sim^{iid} U(-sigma,sigma)$ then I am getting $hat{beta}_{UMVUE}=frac{y_{(1)}+y_{(n)}}{2}$.
Isn't $hat{beta}_{UMVUE}$ Linear? If so isn't it a violation of the Gauss-Markov Theorem? Where am I going wrong?
Also can anyone suggest me a distribution of $epsilon$ where I can get a Non-Linear unbiased estimator (in closed form) which is better than OLS.
statistical-inference regression
asked Dec 18 '18 at 13:01
Vishaal SudarsanVishaal Sudarsan
995
$endgroup$
add a comment |
$begingroup$
Let us consider a simple model.
$y_i = beta + epsilon_i$
If we assume that $epsilon_i$ has 0 mean, constant variance and is uncorrelated. Then via Gauss-Markov theorem we know that $hat{beta} = bar{y}$ is the BLUE.
However, when I assume that $epsilon_i sim^{iid} U(-sigma,sigma)$ then I am getting $hat{beta}_{UMVUE}=frac{y_{(1)}+y_{(n)}}{2}$.
Isn't $hat{beta}_{UMVUE}$ Linear? If so isn't it a violation of the Gauss-Markov Theorem? Where am I going wrong?
Also can anyone suggest me a distribution of $epsilon$ where I can get a Non-Linear unbiased estimator (in closed form) which is better than OLS.
statistical-inference regression
asked Dec 18 '18 at 13:01
Vishaal SudarsanVishaal Sudarsan
995
$endgroup$
add a comment |
$begingroup$
Let us consider a simple model.
$y_i = beta + epsilon_i$
If we assume that $epsilon_i$ has 0 mean, constant variance and is uncorrelated. Then via Gauss-Markov theorem we know that $hat{beta} = bar{y}$ is the BLUE.
However, when I assume that $epsilon_i sim^{iid} U(-sigma,sigma)$ then I am getting $hat{beta}_{UMVUE}=frac{y_{(1)}+y_{(n)}}{2}$.
Isn't $hat{beta}_{UMVUE}$ Linear? If so isn't it a violation of the Gauss-Markov Theorem? Where am I going wrong?
Also can anyone suggest me a distribution of $epsilon$ where I can get a Non-Linear unbiased estimator (in closed form) which is better than OLS.
statistical-inference regression
asked Dec 18 '18 at 13:01
Vishaal SudarsanVishaal Sudarsan
995
$endgroup$
Let us consider a simple model.
$y_i = beta + epsilon_i$
If we assume that $epsilon_i$ has 0 mean, constant variance and is uncorrelated. Then via Gauss-Markov theorem we know that $hat{beta} = bar{y}$ is the BLUE.
However, when I assume that $epsilon_i sim^{iid} U(-sigma,sigma)$ then I am getting $hat{beta}_{UMVUE}=frac{y_{(1)}+y_{(n)}}{2}$.
Isn't $hat{beta}_{UMVUE}$ Linear? If so isn't it a violation of the Gauss-Markov Theorem? Where am I going wrong?
Also can anyone suggest me a distribution of $epsilon$ where I can get a Non-Linear unbiased estimator (in closed form) which is better than OLS.
statistical-inference regression
statistical-inference regression
asked Dec 18 '18 at 13:01
Vishaal SudarsanVishaal Sudarsan
995
asked Dec 18 '18 at 13:01
Vishaal SudarsanVishaal Sudarsan
995
asked Dec 18 '18 at 13:01
Vishaal SudarsanVishaal Sudarsan
995
asked Dec 18 '18 at 13:01
Vishaal SudarsanVishaal Sudarsan
995
asked Dec 18 '18 at 13:01
Vishaal SudarsanVishaal Sudarsan
995
995
add a comment |
add a comment |
1 Answer
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$begingroup$
I have found the answer to my own question. It's simple $hat{beta}_{UMVUE} = frac{y_{(1)} + y_{(n)}}{2}$ is not linear. We can check that it doesn't satisfy the property of linear transformation that $T(x+y)=T(x)+T(y)$. Therefore Gauss-Markov theorem is not violated.In this case, i.e when $epsilon sim^{iid} U(-sigma,sigma)$, the estimator $frac{y_{(1)} + y_{(n)}}{2}$ is the best unbiased estimator while $bar{y}$ is the best linear unbiased estimator.
answered Jan 1 at 10:22
Vishaal SudarsanVishaal Sudarsan
995
$endgroup$
add a comment |
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1 Answer
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$begingroup$
I have found the answer to my own question. It's simple $hat{beta}_{UMVUE} = frac{y_{(1)} + y_{(n)}}{2}$ is not linear. We can check that it doesn't satisfy the property of linear transformation that $T(x+y)=T(x)+T(y)$. Therefore Gauss-Markov theorem is not violated.In this case, i.e when $epsilon sim^{iid} U(-sigma,sigma)$, the estimator $frac{y_{(1)} + y_{(n)}}{2}$ is the best unbiased estimator while $bar{y}$ is the best linear unbiased estimator.
answered Jan 1 at 10:22
Vishaal SudarsanVishaal Sudarsan
995
$endgroup$
add a comment |
$begingroup$
I have found the answer to my own question. It's simple $hat{beta}_{UMVUE} = frac{y_{(1)} + y_{(n)}}{2}$ is not linear. We can check that it doesn't satisfy the property of linear transformation that $T(x+y)=T(x)+T(y)$. Therefore Gauss-Markov theorem is not violated.In this case, i.e when $epsilon sim^{iid} U(-sigma,sigma)$, the estimator $frac{y_{(1)} + y_{(n)}}{2}$ is the best unbiased estimator while $bar{y}$ is the best linear unbiased estimator.
answered Jan 1 at 10:22
Vishaal SudarsanVishaal Sudarsan
995
$endgroup$
add a comment |
$begingroup$
I have found the answer to my own question. It's simple $hat{beta}_{UMVUE} = frac{y_{(1)} + y_{(n)}}{2}$ is not linear. We can check that it doesn't satisfy the property of linear transformation that $T(x+y)=T(x)+T(y)$. Therefore Gauss-Markov theorem is not violated.In this case, i.e when $epsilon sim^{iid} U(-sigma,sigma)$, the estimator $frac{y_{(1)} + y_{(n)}}{2}$ is the best unbiased estimator while $bar{y}$ is the best linear unbiased estimator.
answered Jan 1 at 10:22
Vishaal SudarsanVishaal Sudarsan
995
$endgroup$
I have found the answer to my own question. It's simple $hat{beta}_{UMVUE} = frac{y_{(1)} + y_{(n)}}{2}$ is not linear. We can check that it doesn't satisfy the property of linear transformation that $T(x+y)=T(x)+T(y)$. Therefore Gauss-Markov theorem is not violated.In this case, i.e when $epsilon sim^{iid} U(-sigma,sigma)$, the estimator $frac{y_{(1)} + y_{(n)}}{2}$ is the best unbiased estimator while $bar{y}$ is the best linear unbiased estimator.
answered Jan 1 at 10:22
Vishaal SudarsanVishaal Sudarsan
995
answered Jan 1 at 10:22
Vishaal SudarsanVishaal Sudarsan
995
answered Jan 1 at 10:22
Vishaal SudarsanVishaal Sudarsan
995
answered Jan 1 at 10:22
Vishaal SudarsanVishaal Sudarsan
995
995
add a comment |
add a comment |
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