Showing that estimator is minimax
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I have the following question. Let $Xsim textBin(n,p)$ and consider estimating $pin(0,1)$ with loss function,
$$
L(p,hatp)=left(1-frachatppright)^2.
$$
I need to show that the estimator $hatp=0$ is minimax.
I have an idea on how to do this since I have a theorem that says that if an estimator is admissible and has constant risk then it is minimax. Constant risk is easy to check is this case,
beginalign
mathbbEleft[L(p,0)right] &= mathbbEleft[left(1-frac0pright)^2right]\
&= mathbbEleft[1right]\
&= 1.
endalign
I am having a harder time showing that $hatp=0$ is admissible. My idea is to assume that it is not admissible; then there exists an estimator $tildep$ with the property,
$$
mathbbEleft[L(p,tildep)right] leq 1
$$
for all $p$ and with strict inequality for at least one $pin(0,1)$. The left-hand side can be expanded,
beginalign
mathbbEleft[L(p,tildep)right] &= mathbbEleft[fractilde p^2p^2 - 2fractilde pp + 1right] \
&= fracmathbbEleft[tilde p^2right]p^2 - 2fracmathbbEleft[tilde pright]p + 1.
endalign
Combining this with the definition of inadmissibility above we have,
$$
fracmathbbEleft[tilde p^2right]p^2 leq 2fracmathbbEleft[tilde pright]p,
$$
which can be written,
$$
mathbbEleft[tilde p^2right] leq 2pmathbbEleft[tilde pright].
$$
Unfortunately this is as far as I can get without some further guidance. Can someone point me in the right direction? I haven't used the fact that $X$ is Binomial...
statistics statistical-inference binomial-distribution
asked Aug 17 at 22:34
user3064222
82
add a comment |Â
up vote
1
down vote
favorite
I have the following question. Let $Xsim textBin(n,p)$ and consider estimating $pin(0,1)$ with loss function,
$$
L(p,hatp)=left(1-frachatppright)^2.
$$
I need to show that the estimator $hatp=0$ is minimax.
I have an idea on how to do this since I have a theorem that says that if an estimator is admissible and has constant risk then it is minimax. Constant risk is easy to check is this case,
beginalign
mathbbEleft[L(p,0)right] &= mathbbEleft[left(1-frac0pright)^2right]\
&= mathbbEleft[1right]\
&= 1.
endalign
I am having a harder time showing that $hatp=0$ is admissible. My idea is to assume that it is not admissible; then there exists an estimator $tildep$ with the property,
$$
mathbbEleft[L(p,tildep)right] leq 1
$$
for all $p$ and with strict inequality for at least one $pin(0,1)$. The left-hand side can be expanded,
beginalign
mathbbEleft[L(p,tildep)right] &= mathbbEleft[fractilde p^2p^2 - 2fractilde pp + 1right] \
&= fracmathbbEleft[tilde p^2right]p^2 - 2fracmathbbEleft[tilde pright]p + 1.
endalign
Combining this with the definition of inadmissibility above we have,
$$
fracmathbbEleft[tilde p^2right]p^2 leq 2fracmathbbEleft[tilde pright]p,
$$
which can be written,
$$
mathbbEleft[tilde p^2right] leq 2pmathbbEleft[tilde pright].
$$
Unfortunately this is as far as I can get without some further guidance. Can someone point me in the right direction? I haven't used the fact that $X$ is Binomial...
statistics statistical-inference binomial-distribution
asked Aug 17 at 22:34
user3064222
82
This is an exercise from Wasserman's book right?
– cdipaolo
Aug 17 at 22:40
1
Yes, but Wasserman's question also asks the reader to prove that the estimator is unique in addition to being minimax. The question is apparently based on this paper projecteuclid.org/euclid.aos/1176346248 but it is too advanced for me to read.
– user3064222
Aug 17 at 22:47
I must say Wasserman has great exercises. Fun question.
– cdipaolo
Aug 17 at 22:58
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have the following question. Let $Xsim textBin(n,p)$ and consider estimating $pin(0,1)$ with loss function,
$$
L(p,hatp)=left(1-frachatppright)^2.
$$
I need to show that the estimator $hatp=0$ is minimax.
I have an idea on how to do this since I have a theorem that says that if an estimator is admissible and has constant risk then it is minimax. Constant risk is easy to check is this case,
beginalign
mathbbEleft[L(p,0)right] &= mathbbEleft[left(1-frac0pright)^2right]\
&= mathbbEleft[1right]\
&= 1.
endalign
I am having a harder time showing that $hatp=0$ is admissible. My idea is to assume that it is not admissible; then there exists an estimator $tildep$ with the property,
$$
mathbbEleft[L(p,tildep)right] leq 1
$$
for all $p$ and with strict inequality for at least one $pin(0,1)$. The left-hand side can be expanded,
beginalign
mathbbEleft[L(p,tildep)right] &= mathbbEleft[fractilde p^2p^2 - 2fractilde pp + 1right] \
&= fracmathbbEleft[tilde p^2right]p^2 - 2fracmathbbEleft[tilde pright]p + 1.
endalign
Combining this with the definition of inadmissibility above we have,
$$
fracmathbbEleft[tilde p^2right]p^2 leq 2fracmathbbEleft[tilde pright]p,
$$
which can be written,
$$
mathbbEleft[tilde p^2right] leq 2pmathbbEleft[tilde pright].
$$
Unfortunately this is as far as I can get without some further guidance. Can someone point me in the right direction? I haven't used the fact that $X$ is Binomial...
statistics statistical-inference binomial-distribution
asked Aug 17 at 22:34
user3064222
82
I have the following question. Let $Xsim textBin(n,p)$ and consider estimating $pin(0,1)$ with loss function,
$$
L(p,hatp)=left(1-frachatppright)^2.
$$
I need to show that the estimator $hatp=0$ is minimax.
I have an idea on how to do this since I have a theorem that says that if an estimator is admissible and has constant risk then it is minimax. Constant risk is easy to check is this case,
beginalign
mathbbEleft[L(p,0)right] &= mathbbEleft[left(1-frac0pright)^2right]\
&= mathbbEleft[1right]\
&= 1.
endalign
I am having a harder time showing that $hatp=0$ is admissible. My idea is to assume that it is not admissible; then there exists an estimator $tildep$ with the property,
$$
mathbbEleft[L(p,tildep)right] leq 1
$$
for all $p$ and with strict inequality for at least one $pin(0,1)$. The left-hand side can be expanded,
beginalign
mathbbEleft[L(p,tildep)right] &= mathbbEleft[fractilde p^2p^2 - 2fractilde pp + 1right] \
&= fracmathbbEleft[tilde p^2right]p^2 - 2fracmathbbEleft[tilde pright]p + 1.
endalign
Combining this with the definition of inadmissibility above we have,
$$
fracmathbbEleft[tilde p^2right]p^2 leq 2fracmathbbEleft[tilde pright]p,
$$
which can be written,
$$
mathbbEleft[tilde p^2right] leq 2pmathbbEleft[tilde pright].
$$
Unfortunately this is as far as I can get without some further guidance. Can someone point me in the right direction? I haven't used the fact that $X$ is Binomial...
statistics statistical-inference binomial-distribution
asked Aug 17 at 22:34
user3064222
82
asked Aug 17 at 22:34
user3064222
82
asked Aug 17 at 22:34
user3064222
82
asked Aug 17 at 22:34
user3064222
82
82
This is an exercise from Wasserman's book right?
– cdipaolo
Aug 17 at 22:40
1
Yes, but Wasserman's question also asks the reader to prove that the estimator is unique in addition to being minimax. The question is apparently based on this paper projecteuclid.org/euclid.aos/1176346248 but it is too advanced for me to read.
– user3064222
Aug 17 at 22:47
I must say Wasserman has great exercises. Fun question.
– cdipaolo
Aug 17 at 22:58
add a comment |Â
This is an exercise from Wasserman's book right?
– cdipaolo
Aug 17 at 22:40
1
Yes, but Wasserman's question also asks the reader to prove that the estimator is unique in addition to being minimax. The question is apparently based on this paper projecteuclid.org/euclid.aos/1176346248 but it is too advanced for me to read.
– user3064222
Aug 17 at 22:47
I must say Wasserman has great exercises. Fun question.
– cdipaolo
Aug 17 at 22:58
This is an exercise from Wasserman's book right?
– cdipaolo
Aug 17 at 22:40
This is an exercise from Wasserman's book right?
– cdipaolo
Aug 17 at 22:40
1
1
Yes, but Wasserman's question also asks the reader to prove that the estimator is unique in addition to being minimax. The question is apparently based on this paper projecteuclid.org/euclid.aos/1176346248 but it is too advanced for me to read.
– user3064222
Aug 17 at 22:47
Yes, but Wasserman's question also asks the reader to prove that the estimator is unique in addition to being minimax. The question is apparently based on this paper projecteuclid.org/euclid.aos/1176346248 but it is too advanced for me to read.
– user3064222
Aug 17 at 22:47
I must say Wasserman has great exercises. Fun question.
– cdipaolo
Aug 17 at 22:58
I must say Wasserman has great exercises. Fun question.
– cdipaolo
Aug 17 at 22:58
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
$newcommandEmathbbE$From your last expression, Jensen's inequality tells us that for an estimator to have risk uniformly less than one
$$E[tilde p]^2 leq E[tilde p^2] leq 2pE[tilde p]$$
so $E[tilde p]$ satisfies $x(x-2p) leq x^2 - 2px leq 0$ or equivalently $E[tilde p]in[0,2p]$. But since $p$ is could be arbitrarily close to zero this implies $E[tilde p]=0$ is necessary for any estimator to have risk uniformly at most one for all $p$.
Plugging this back into the risk expansion we have
$$E[L(p,tilde p)] = 1 + fracE[tilde p^2]p^2geq 1$$ so the only estimator which could possibly have risk uniformly less than one must have $E[tilde p^2] = 0$ or $tilde p = 0$ almost surely. $blacksquare$
answered Aug 17 at 22:58
cdipaolo
677311
After this solution was posted, I thought of an alternative approach as follows. After we see that $mathbbEleft[tilde pright] = 0$, we can use the completeness of $X$ ($X$ is a complete statistic for a Binomial distribution) to say that if $mathbbEleft[tilde p(X)right] = 0$ then we must have that $textPrleft[tilde p(X) = 0right] = 1$.
– user3064222
Aug 18 at 15:44
@user3064222 how do we know the estimator is complete?
– cdipaolo
Aug 18 at 16:46
The sample $X$ is complete so any function $g(X)$ with expectation zero must equal zero with probability one. There is a proof for the Binomial distribution on the Wikipedia page: en.wikipedia.org/wiki/Completeness_(statistics)
– user3064222
Aug 18 at 20:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$newcommandEmathbbE$From your last expression, Jensen's inequality tells us that for an estimator to have risk uniformly less than one
$$E[tilde p]^2 leq E[tilde p^2] leq 2pE[tilde p]$$
so $E[tilde p]$ satisfies $x(x-2p) leq x^2 - 2px leq 0$ or equivalently $E[tilde p]in[0,2p]$. But since $p$ is could be arbitrarily close to zero this implies $E[tilde p]=0$ is necessary for any estimator to have risk uniformly at most one for all $p$.
Plugging this back into the risk expansion we have
$$E[L(p,tilde p)] = 1 + fracE[tilde p^2]p^2geq 1$$ so the only estimator which could possibly have risk uniformly less than one must have $E[tilde p^2] = 0$ or $tilde p = 0$ almost surely. $blacksquare$
answered Aug 17 at 22:58
cdipaolo
677311
After this solution was posted, I thought of an alternative approach as follows. After we see that $mathbbEleft[tilde pright] = 0$, we can use the completeness of $X$ ($X$ is a complete statistic for a Binomial distribution) to say that if $mathbbEleft[tilde p(X)right] = 0$ then we must have that $textPrleft[tilde p(X) = 0right] = 1$.
– user3064222
Aug 18 at 15:44
@user3064222 how do we know the estimator is complete?
– cdipaolo
Aug 18 at 16:46
The sample $X$ is complete so any function $g(X)$ with expectation zero must equal zero with probability one. There is a proof for the Binomial distribution on the Wikipedia page: en.wikipedia.org/wiki/Completeness_(statistics)
– user3064222
Aug 18 at 20:09
add a comment |Â
up vote
0
down vote
accepted
$newcommandEmathbbE$From your last expression, Jensen's inequality tells us that for an estimator to have risk uniformly less than one
$$E[tilde p]^2 leq E[tilde p^2] leq 2pE[tilde p]$$
so $E[tilde p]$ satisfies $x(x-2p) leq x^2 - 2px leq 0$ or equivalently $E[tilde p]in[0,2p]$. But since $p$ is could be arbitrarily close to zero this implies $E[tilde p]=0$ is necessary for any estimator to have risk uniformly at most one for all $p$.
Plugging this back into the risk expansion we have
$$E[L(p,tilde p)] = 1 + fracE[tilde p^2]p^2geq 1$$ so the only estimator which could possibly have risk uniformly less than one must have $E[tilde p^2] = 0$ or $tilde p = 0$ almost surely. $blacksquare$
answered Aug 17 at 22:58
cdipaolo
677311
After this solution was posted, I thought of an alternative approach as follows. After we see that $mathbbEleft[tilde pright] = 0$, we can use the completeness of $X$ ($X$ is a complete statistic for a Binomial distribution) to say that if $mathbbEleft[tilde p(X)right] = 0$ then we must have that $textPrleft[tilde p(X) = 0right] = 1$.
– user3064222
Aug 18 at 15:44
@user3064222 how do we know the estimator is complete?
– cdipaolo
Aug 18 at 16:46
The sample $X$ is complete so any function $g(X)$ with expectation zero must equal zero with probability one. There is a proof for the Binomial distribution on the Wikipedia page: en.wikipedia.org/wiki/Completeness_(statistics)
– user3064222
Aug 18 at 20:09
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$newcommandEmathbbE$From your last expression, Jensen's inequality tells us that for an estimator to have risk uniformly less than one
$$E[tilde p]^2 leq E[tilde p^2] leq 2pE[tilde p]$$
so $E[tilde p]$ satisfies $x(x-2p) leq x^2 - 2px leq 0$ or equivalently $E[tilde p]in[0,2p]$. But since $p$ is could be arbitrarily close to zero this implies $E[tilde p]=0$ is necessary for any estimator to have risk uniformly at most one for all $p$.
Plugging this back into the risk expansion we have
$$E[L(p,tilde p)] = 1 + fracE[tilde p^2]p^2geq 1$$ so the only estimator which could possibly have risk uniformly less than one must have $E[tilde p^2] = 0$ or $tilde p = 0$ almost surely. $blacksquare$
answered Aug 17 at 22:58
cdipaolo
677311
$newcommandEmathbbE$From your last expression, Jensen's inequality tells us that for an estimator to have risk uniformly less than one
$$E[tilde p]^2 leq E[tilde p^2] leq 2pE[tilde p]$$
so $E[tilde p]$ satisfies $x(x-2p) leq x^2 - 2px leq 0$ or equivalently $E[tilde p]in[0,2p]$. But since $p$ is could be arbitrarily close to zero this implies $E[tilde p]=0$ is necessary for any estimator to have risk uniformly at most one for all $p$.
Plugging this back into the risk expansion we have
$$E[L(p,tilde p)] = 1 + fracE[tilde p^2]p^2geq 1$$ so the only estimator which could possibly have risk uniformly less than one must have $E[tilde p^2] = 0$ or $tilde p = 0$ almost surely. $blacksquare$
answered Aug 17 at 22:58
cdipaolo
677311
answered Aug 17 at 22:58
cdipaolo
677311
answered Aug 17 at 22:58
cdipaolo
677311
answered Aug 17 at 22:58
cdipaolo
677311
677311
After this solution was posted, I thought of an alternative approach as follows. After we see that $mathbbEleft[tilde pright] = 0$, we can use the completeness of $X$ ($X$ is a complete statistic for a Binomial distribution) to say that if $mathbbEleft[tilde p(X)right] = 0$ then we must have that $textPrleft[tilde p(X) = 0right] = 1$.
– user3064222
Aug 18 at 15:44
@user3064222 how do we know the estimator is complete?
– cdipaolo
Aug 18 at 16:46
The sample $X$ is complete so any function $g(X)$ with expectation zero must equal zero with probability one. There is a proof for the Binomial distribution on the Wikipedia page: en.wikipedia.org/wiki/Completeness_(statistics)
– user3064222
Aug 18 at 20:09
add a comment |Â
After this solution was posted, I thought of an alternative approach as follows. After we see that $mathbbEleft[tilde pright] = 0$, we can use the completeness of $X$ ($X$ is a complete statistic for a Binomial distribution) to say that if $mathbbEleft[tilde p(X)right] = 0$ then we must have that $textPrleft[tilde p(X) = 0right] = 1$.
– user3064222
Aug 18 at 15:44
@user3064222 how do we know the estimator is complete?
– cdipaolo
Aug 18 at 16:46
The sample $X$ is complete so any function $g(X)$ with expectation zero must equal zero with probability one. There is a proof for the Binomial distribution on the Wikipedia page: en.wikipedia.org/wiki/Completeness_(statistics)
– user3064222
Aug 18 at 20:09
After this solution was posted, I thought of an alternative approach as follows. After we see that $mathbbEleft[tilde pright] = 0$, we can use the completeness of $X$ ($X$ is a complete statistic for a Binomial distribution) to say that if $mathbbEleft[tilde p(X)right] = 0$ then we must have that $textPrleft[tilde p(X) = 0right] = 1$.
– user3064222
Aug 18 at 15:44
After this solution was posted, I thought of an alternative approach as follows. After we see that $mathbbEleft[tilde pright] = 0$, we can use the completeness of $X$ ($X$ is a complete statistic for a Binomial distribution) to say that if $mathbbEleft[tilde p(X)right] = 0$ then we must have that $textPrleft[tilde p(X) = 0right] = 1$.
– user3064222
Aug 18 at 15:44
@user3064222 how do we know the estimator is complete?
– cdipaolo
Aug 18 at 16:46
@user3064222 how do we know the estimator is complete?
– cdipaolo
Aug 18 at 16:46
The sample $X$ is complete so any function $g(X)$ with expectation zero must equal zero with probability one. There is a proof for the Binomial distribution on the Wikipedia page: en.wikipedia.org/wiki/Completeness_(statistics)
– user3064222
Aug 18 at 20:09
The sample $X$ is complete so any function $g(X)$ with expectation zero must equal zero with probability one. There is a proof for the Binomial distribution on the Wikipedia page: en.wikipedia.org/wiki/Completeness_(statistics)
– user3064222
Aug 18 at 20:09
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2886255%2fshowing-that-estimator-is-minimax%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
This is an exercise from Wasserman's book right?
– cdipaolo
Aug 17 at 22:40
1
Yes, but Wasserman's question also asks the reader to prove that the estimator is unique in addition to being minimax. The question is apparently based on this paper projecteuclid.org/euclid.aos/1176346248 but it is too advanced for me to read.
– user3064222
Aug 17 at 22:47
I must say Wasserman has great exercises. Fun question.
– cdipaolo
Aug 17 at 22:58