An estimator whose variance attains Cramer-Rao lower bound is consistent
$begingroup$
If the variance of an estimator attains the Cramer-Rao lower bound
then the estimator is
$(A)$ most efficient
$(B)$ sufficient
$(C)$ consistent
$(D)$ admissible
I can show that the the variance is most efficient. But I'm unable to test whether it is consistent or NOT.
Please help me.
probability statistics statistical-inference
edited Aug 16 '16 at 12:40
Michael Hardy
1
asked Aug 14 '16 at 13:59
TopoTopo
311214
$endgroup$
add a comment |
$begingroup$
If the variance of an estimator attains the Cramer-Rao lower bound
then the estimator is
$(A)$ most efficient
$(B)$ sufficient
$(C)$ consistent
$(D)$ admissible
I can show that the the variance is most efficient. But I'm unable to test whether it is consistent or NOT.
Please help me.
probability statistics statistical-inference
edited Aug 16 '16 at 12:40
Michael Hardy
1
asked Aug 14 '16 at 13:59
TopoTopo
311214
$endgroup$
add a comment |
$begingroup$
If the variance of an estimator attains the Cramer-Rao lower bound
then the estimator is
$(A)$ most efficient
$(B)$ sufficient
$(C)$ consistent
$(D)$ admissible
I can show that the the variance is most efficient. But I'm unable to test whether it is consistent or NOT.
Please help me.
probability statistics statistical-inference
edited Aug 16 '16 at 12:40
Michael Hardy
1
asked Aug 14 '16 at 13:59
TopoTopo
311214
$endgroup$
If the variance of an estimator attains the Cramer-Rao lower bound
then the estimator is
$(A)$ most efficient
$(B)$ sufficient
$(C)$ consistent
$(D)$ admissible
I can show that the the variance is most efficient. But I'm unable to test whether it is consistent or NOT.
Please help me.
probability statistics statistical-inference
probability statistics statistical-inference
edited Aug 16 '16 at 12:40
Michael Hardy
1
asked Aug 14 '16 at 13:59
TopoTopo
311214
edited Aug 16 '16 at 12:40
Michael Hardy
1
asked Aug 14 '16 at 13:59
TopoTopo
311214
edited Aug 16 '16 at 12:40
Michael Hardy
1
edited Aug 16 '16 at 12:40
Michael Hardy
1
edited Aug 16 '16 at 12:40
Michael Hardy
1
1
asked Aug 14 '16 at 13:59
TopoTopo
311214
asked Aug 14 '16 at 13:59
TopoTopo
311214
asked Aug 14 '16 at 13:59
TopoTopo
311214
311214
add a comment |
add a comment |
1 Answer
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Given an ubiased estimator $hattheta$ of the parameter $thetainTheta$, this estimator is efficient if its var-cov matrix equals the Cramér-Rao lower bound. In other words, the Cramér-Rao inequality provides a lower bound for the var-cov matrix of unbiased estimators. As you know, unbiasedness is different from consistency.
To show the consistency of $hattheta$, one must show that $plimhattheta$ = $theta_0$. If the set $Theta$ is compact; the objective function $mathbb{Q}_0(theta)$ is continuous and has a unique maximum in $theta_0$, and; $widehat{mathbb{Q}_n}(theta)$ converges (uniformily) in probability to $mathbb{Q}_0(theta)$; then $plimhattheta=hattheta_0$.
Lets consider the case of a linear model $Y_i=X_itheta$ + $epsilon_i$. The OLS estimator or MLE is $hattheta=theta_0+(X'X)^{-1}X'epsilon$. If one assumes that plim $n^{-1}(X'epsilon)=0$ and that $lim_{nto infty}n^{-1}X'X=Q$ where $Q$ is not singular, then $hatthetarightarrow_ptheta_0$.
answered Aug 14 '16 at 18:26
jgastaizajgastaiza
314
$endgroup$
$begingroup$
Okk...So both options (A) and (C) are correct. Is it ?
$endgroup$
– Topo
Aug 15 '16 at 4:22
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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active
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$begingroup$
Given an ubiased estimator $hattheta$ of the parameter $thetainTheta$, this estimator is efficient if its var-cov matrix equals the Cramér-Rao lower bound. In other words, the Cramér-Rao inequality provides a lower bound for the var-cov matrix of unbiased estimators. As you know, unbiasedness is different from consistency.
To show the consistency of $hattheta$, one must show that $plimhattheta$ = $theta_0$. If the set $Theta$ is compact; the objective function $mathbb{Q}_0(theta)$ is continuous and has a unique maximum in $theta_0$, and; $widehat{mathbb{Q}_n}(theta)$ converges (uniformily) in probability to $mathbb{Q}_0(theta)$; then $plimhattheta=hattheta_0$.
Lets consider the case of a linear model $Y_i=X_itheta$ + $epsilon_i$. The OLS estimator or MLE is $hattheta=theta_0+(X'X)^{-1}X'epsilon$. If one assumes that plim $n^{-1}(X'epsilon)=0$ and that $lim_{nto infty}n^{-1}X'X=Q$ where $Q$ is not singular, then $hatthetarightarrow_ptheta_0$.
answered Aug 14 '16 at 18:26
jgastaizajgastaiza
314
$endgroup$
$begingroup$
Okk...So both options (A) and (C) are correct. Is it ?
$endgroup$
– Topo
Aug 15 '16 at 4:22
add a comment |
$begingroup$
Given an ubiased estimator $hattheta$ of the parameter $thetainTheta$, this estimator is efficient if its var-cov matrix equals the Cramér-Rao lower bound. In other words, the Cramér-Rao inequality provides a lower bound for the var-cov matrix of unbiased estimators. As you know, unbiasedness is different from consistency.
To show the consistency of $hattheta$, one must show that $plimhattheta$ = $theta_0$. If the set $Theta$ is compact; the objective function $mathbb{Q}_0(theta)$ is continuous and has a unique maximum in $theta_0$, and; $widehat{mathbb{Q}_n}(theta)$ converges (uniformily) in probability to $mathbb{Q}_0(theta)$; then $plimhattheta=hattheta_0$.
Lets consider the case of a linear model $Y_i=X_itheta$ + $epsilon_i$. The OLS estimator or MLE is $hattheta=theta_0+(X'X)^{-1}X'epsilon$. If one assumes that plim $n^{-1}(X'epsilon)=0$ and that $lim_{nto infty}n^{-1}X'X=Q$ where $Q$ is not singular, then $hatthetarightarrow_ptheta_0$.
answered Aug 14 '16 at 18:26
jgastaizajgastaiza
314
$endgroup$
$begingroup$
Okk...So both options (A) and (C) are correct. Is it ?
$endgroup$
– Topo
Aug 15 '16 at 4:22
add a comment |
$begingroup$
Given an ubiased estimator $hattheta$ of the parameter $thetainTheta$, this estimator is efficient if its var-cov matrix equals the Cramér-Rao lower bound. In other words, the Cramér-Rao inequality provides a lower bound for the var-cov matrix of unbiased estimators. As you know, unbiasedness is different from consistency.
To show the consistency of $hattheta$, one must show that $plimhattheta$ = $theta_0$. If the set $Theta$ is compact; the objective function $mathbb{Q}_0(theta)$ is continuous and has a unique maximum in $theta_0$, and; $widehat{mathbb{Q}_n}(theta)$ converges (uniformily) in probability to $mathbb{Q}_0(theta)$; then $plimhattheta=hattheta_0$.
Lets consider the case of a linear model $Y_i=X_itheta$ + $epsilon_i$. The OLS estimator or MLE is $hattheta=theta_0+(X'X)^{-1}X'epsilon$. If one assumes that plim $n^{-1}(X'epsilon)=0$ and that $lim_{nto infty}n^{-1}X'X=Q$ where $Q$ is not singular, then $hatthetarightarrow_ptheta_0$.
answered Aug 14 '16 at 18:26
jgastaizajgastaiza
314
$endgroup$
Given an ubiased estimator $hattheta$ of the parameter $thetainTheta$, this estimator is efficient if its var-cov matrix equals the Cramér-Rao lower bound. In other words, the Cramér-Rao inequality provides a lower bound for the var-cov matrix of unbiased estimators. As you know, unbiasedness is different from consistency.
To show the consistency of $hattheta$, one must show that $plimhattheta$ = $theta_0$. If the set $Theta$ is compact; the objective function $mathbb{Q}_0(theta)$ is continuous and has a unique maximum in $theta_0$, and; $widehat{mathbb{Q}_n}(theta)$ converges (uniformily) in probability to $mathbb{Q}_0(theta)$; then $plimhattheta=hattheta_0$.
Lets consider the case of a linear model $Y_i=X_itheta$ + $epsilon_i$. The OLS estimator or MLE is $hattheta=theta_0+(X'X)^{-1}X'epsilon$. If one assumes that plim $n^{-1}(X'epsilon)=0$ and that $lim_{nto infty}n^{-1}X'X=Q$ where $Q$ is not singular, then $hatthetarightarrow_ptheta_0$.
answered Aug 14 '16 at 18:26
jgastaizajgastaiza
314
answered Aug 14 '16 at 18:26
jgastaizajgastaiza
314
answered Aug 14 '16 at 18:26
jgastaizajgastaiza
314
answered Aug 14 '16 at 18:26
jgastaizajgastaiza
314
314
$begingroup$
Okk...So both options (A) and (C) are correct. Is it ?
$endgroup$
– Topo
Aug 15 '16 at 4:22
add a comment |
$begingroup$
Okk...So both options (A) and (C) are correct. Is it ?
$endgroup$
– Topo
Aug 15 '16 at 4:22
$begingroup$
Okk...So both options (A) and (C) are correct. Is it ?
$endgroup$
– Topo
Aug 15 '16 at 4:22
$begingroup$
Okk...So both options (A) and (C) are correct. Is it ?
$endgroup$
– Topo
Aug 15 '16 at 4:22
add a comment |
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