Complete statistics
$begingroup$
Let $(x_{1}, ldots, x_{n})$ be a sample from some known distribution $F_{theta}$ with unknown parameter $theta$ and let $$mathbb{E}_{theta}{X_{i}} = theta$$
Let $T$ be an identity function of a sample e.g. for the given sample $T$ returns the sample itself. How to prove that $T$ is not a complete statistics?
By definition, the complete statistics is a function $T$ such that for any measurable $g$ such that $$mathbb{E}(g(T)) = 0$$
for any $theta$
$$mathbb{P}_{theta}(g(T) = 0) = 1$$
for any $theta$.
I guess that the problem is quite easy but i cannot find an easy way to attack it. Are there any hints that might help?
statistics statistical-inference
asked Dec 17 '18 at 18:51
hyperkahlerhyperkahler
1,487714
$endgroup$
add a comment |
$begingroup$
Let $(x_{1}, ldots, x_{n})$ be a sample from some known distribution $F_{theta}$ with unknown parameter $theta$ and let $$mathbb{E}_{theta}{X_{i}} = theta$$
Let $T$ be an identity function of a sample e.g. for the given sample $T$ returns the sample itself. How to prove that $T$ is not a complete statistics?
By definition, the complete statistics is a function $T$ such that for any measurable $g$ such that $$mathbb{E}(g(T)) = 0$$
for any $theta$
$$mathbb{P}_{theta}(g(T) = 0) = 1$$
for any $theta$.
I guess that the problem is quite easy but i cannot find an easy way to attack it. Are there any hints that might help?
statistics statistical-inference
asked Dec 17 '18 at 18:51
hyperkahlerhyperkahler
1,487714
$endgroup$
add a comment |
$begingroup$
Let $(x_{1}, ldots, x_{n})$ be a sample from some known distribution $F_{theta}$ with unknown parameter $theta$ and let $$mathbb{E}_{theta}{X_{i}} = theta$$
Let $T$ be an identity function of a sample e.g. for the given sample $T$ returns the sample itself. How to prove that $T$ is not a complete statistics?
By definition, the complete statistics is a function $T$ such that for any measurable $g$ such that $$mathbb{E}(g(T)) = 0$$
for any $theta$
$$mathbb{P}_{theta}(g(T) = 0) = 1$$
for any $theta$.
I guess that the problem is quite easy but i cannot find an easy way to attack it. Are there any hints that might help?
statistics statistical-inference
asked Dec 17 '18 at 18:51
hyperkahlerhyperkahler
1,487714
$endgroup$
Let $(x_{1}, ldots, x_{n})$ be a sample from some known distribution $F_{theta}$ with unknown parameter $theta$ and let $$mathbb{E}_{theta}{X_{i}} = theta$$
Let $T$ be an identity function of a sample e.g. for the given sample $T$ returns the sample itself. How to prove that $T$ is not a complete statistics?
By definition, the complete statistics is a function $T$ such that for any measurable $g$ such that $$mathbb{E}(g(T)) = 0$$
for any $theta$
$$mathbb{P}_{theta}(g(T) = 0) = 1$$
for any $theta$.
I guess that the problem is quite easy but i cannot find an easy way to attack it. Are there any hints that might help?
statistics statistical-inference
statistics statistical-inference
asked Dec 17 '18 at 18:51
hyperkahlerhyperkahler
1,487714
asked Dec 17 '18 at 18:51
hyperkahlerhyperkahler
1,487714
asked Dec 17 '18 at 18:51
hyperkahlerhyperkahler
1,487714
asked Dec 17 '18 at 18:51
hyperkahlerhyperkahler
1,487714
asked Dec 17 '18 at 18:51
hyperkahlerhyperkahler
1,487714
1,487714
add a comment |
add a comment |
1 Answer
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$begingroup$
Take $g(T)=X_1-X_2$. Then $E(g(T))=0$ but $g$ is not identically $0$ with probability $1$.
answered Dec 17 '18 at 18:58
pulpfictionalpulpfictional
1986
$endgroup$
add a comment |
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$begingroup$
Take $g(T)=X_1-X_2$. Then $E(g(T))=0$ but $g$ is not identically $0$ with probability $1$.
answered Dec 17 '18 at 18:58
pulpfictionalpulpfictional
1986
$endgroup$
add a comment |
$begingroup$
Take $g(T)=X_1-X_2$. Then $E(g(T))=0$ but $g$ is not identically $0$ with probability $1$.
answered Dec 17 '18 at 18:58
pulpfictionalpulpfictional
1986
$endgroup$
add a comment |
$begingroup$
Take $g(T)=X_1-X_2$. Then $E(g(T))=0$ but $g$ is not identically $0$ with probability $1$.
answered Dec 17 '18 at 18:58
pulpfictionalpulpfictional
1986
$endgroup$
Take $g(T)=X_1-X_2$. Then $E(g(T))=0$ but $g$ is not identically $0$ with probability $1$.
answered Dec 17 '18 at 18:58
pulpfictionalpulpfictional
1986
answered Dec 17 '18 at 18:58
pulpfictionalpulpfictional
1986
answered Dec 17 '18 at 18:58
pulpfictionalpulpfictional
1986
answered Dec 17 '18 at 18:58
pulpfictionalpulpfictional
1986
1986
add a comment |
add a comment |
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