Finding Best Unbiased Estimator of Uniform Distribution
$begingroup$
Let $X_i$, $i=1,...,n$ be iid with $f(x,theta) = frac{1}{2theta}$ for $-theta<x<theta$. Find the best unbiased estimator of $theta$ if one exists.
So I first tried $T(X)=X_{(n)}$, which gave me $$mathbb{E}[T]=frac{n-1}{n+1}theta.$$ I thus concluded that $T(X)=frac{n+1}{n-1}X_{(n)}$ should be an unbiased estimator.
Next, I wanted to check whether it is the best unbiased estimator using the Cramer Rao Lower Bound. It states
$$operatorname{Var}(T)geq frac{1}{I_X(theta)},$$
so I first calculated the Fisher information for a single random variable. The calculation yields
$$I_X(theta)=mathbb{E}_theta[V(theta,x)^2] = int_{-theta}^thetaleft(partial_thetalogfrac{1}{2theta}right)^{!2}frac{1}{2theta},dx = frac{1}{theta^2}.$$
Thus the lower bound is $dfrac{theta^2}{n}$. Next, I computed the variance of the estimator
$$operatorname{Var}[T] = int_{-theta}^theta left(frac{n+1}{n-1}right)^{!2}x^2nfrac{(x+theta)^{n-1}}{2^ntheta^n},dx-theta^2 = ... = theta^2frac{4n}{(n-1)^2(n-2)},$$
which is obviously lower than the lower bound. However, I am really unsure where I made a mistake. Any help is greatly appreciated.
statistical-inference estimation parameter-estimation
edited Dec 7 '18 at 17:03
Adrian Keister
5,28571933
asked Dec 7 '18 at 16:51
Analysis801Analysis801
1267
$endgroup$
add a comment |
$begingroup$
Let $X_i$, $i=1,...,n$ be iid with $f(x,theta) = frac{1}{2theta}$ for $-theta<x<theta$. Find the best unbiased estimator of $theta$ if one exists.
So I first tried $T(X)=X_{(n)}$, which gave me $$mathbb{E}[T]=frac{n-1}{n+1}theta.$$ I thus concluded that $T(X)=frac{n+1}{n-1}X_{(n)}$ should be an unbiased estimator.
Next, I wanted to check whether it is the best unbiased estimator using the Cramer Rao Lower Bound. It states
$$operatorname{Var}(T)geq frac{1}{I_X(theta)},$$
so I first calculated the Fisher information for a single random variable. The calculation yields
$$I_X(theta)=mathbb{E}_theta[V(theta,x)^2] = int_{-theta}^thetaleft(partial_thetalogfrac{1}{2theta}right)^{!2}frac{1}{2theta},dx = frac{1}{theta^2}.$$
Thus the lower bound is $dfrac{theta^2}{n}$. Next, I computed the variance of the estimator
$$operatorname{Var}[T] = int_{-theta}^theta left(frac{n+1}{n-1}right)^{!2}x^2nfrac{(x+theta)^{n-1}}{2^ntheta^n},dx-theta^2 = ... = theta^2frac{4n}{(n-1)^2(n-2)},$$
which is obviously lower than the lower bound. However, I am really unsure where I made a mistake. Any help is greatly appreciated.
statistical-inference estimation parameter-estimation
edited Dec 7 '18 at 17:03
Adrian Keister
5,28571933
asked Dec 7 '18 at 16:51
Analysis801Analysis801
1267
$endgroup$
$begingroup$
First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:37
$begingroup$
I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:43
$begingroup$
Thank you, but how would one go about proving that it is the UMVUE without this theorem?
$endgroup$
– Analysis801
Dec 7 '18 at 17:44
$begingroup$
What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:47
add a comment |
$begingroup$
Let $X_i$, $i=1,...,n$ be iid with $f(x,theta) = frac{1}{2theta}$ for $-theta<x<theta$. Find the best unbiased estimator of $theta$ if one exists.
So I first tried $T(X)=X_{(n)}$, which gave me $$mathbb{E}[T]=frac{n-1}{n+1}theta.$$ I thus concluded that $T(X)=frac{n+1}{n-1}X_{(n)}$ should be an unbiased estimator.
Next, I wanted to check whether it is the best unbiased estimator using the Cramer Rao Lower Bound. It states
$$operatorname{Var}(T)geq frac{1}{I_X(theta)},$$
so I first calculated the Fisher information for a single random variable. The calculation yields
$$I_X(theta)=mathbb{E}_theta[V(theta,x)^2] = int_{-theta}^thetaleft(partial_thetalogfrac{1}{2theta}right)^{!2}frac{1}{2theta},dx = frac{1}{theta^2}.$$
Thus the lower bound is $dfrac{theta^2}{n}$. Next, I computed the variance of the estimator
$$operatorname{Var}[T] = int_{-theta}^theta left(frac{n+1}{n-1}right)^{!2}x^2nfrac{(x+theta)^{n-1}}{2^ntheta^n},dx-theta^2 = ... = theta^2frac{4n}{(n-1)^2(n-2)},$$
which is obviously lower than the lower bound. However, I am really unsure where I made a mistake. Any help is greatly appreciated.
statistical-inference estimation parameter-estimation
edited Dec 7 '18 at 17:03
Adrian Keister
5,28571933
asked Dec 7 '18 at 16:51
Analysis801Analysis801
1267
$endgroup$
Let $X_i$, $i=1,...,n$ be iid with $f(x,theta) = frac{1}{2theta}$ for $-theta<x<theta$. Find the best unbiased estimator of $theta$ if one exists.
So I first tried $T(X)=X_{(n)}$, which gave me $$mathbb{E}[T]=frac{n-1}{n+1}theta.$$ I thus concluded that $T(X)=frac{n+1}{n-1}X_{(n)}$ should be an unbiased estimator.
Next, I wanted to check whether it is the best unbiased estimator using the Cramer Rao Lower Bound. It states
$$operatorname{Var}(T)geq frac{1}{I_X(theta)},$$
so I first calculated the Fisher information for a single random variable. The calculation yields
$$I_X(theta)=mathbb{E}_theta[V(theta,x)^2] = int_{-theta}^thetaleft(partial_thetalogfrac{1}{2theta}right)^{!2}frac{1}{2theta},dx = frac{1}{theta^2}.$$
Thus the lower bound is $dfrac{theta^2}{n}$. Next, I computed the variance of the estimator
$$operatorname{Var}[T] = int_{-theta}^theta left(frac{n+1}{n-1}right)^{!2}x^2nfrac{(x+theta)^{n-1}}{2^ntheta^n},dx-theta^2 = ... = theta^2frac{4n}{(n-1)^2(n-2)},$$
which is obviously lower than the lower bound. However, I am really unsure where I made a mistake. Any help is greatly appreciated.
statistical-inference estimation parameter-estimation
statistical-inference estimation parameter-estimation
edited Dec 7 '18 at 17:03
Adrian Keister
5,28571933
asked Dec 7 '18 at 16:51
Analysis801Analysis801
1267
edited Dec 7 '18 at 17:03
Adrian Keister
5,28571933
asked Dec 7 '18 at 16:51
Analysis801Analysis801
1267
edited Dec 7 '18 at 17:03
Adrian Keister
5,28571933
edited Dec 7 '18 at 17:03
Adrian Keister
5,28571933
edited Dec 7 '18 at 17:03
Adrian Keister
5,28571933
5,28571933
asked Dec 7 '18 at 16:51
Analysis801Analysis801
1267
asked Dec 7 '18 at 16:51
Analysis801Analysis801
1267
asked Dec 7 '18 at 16:51
Analysis801Analysis801
1267
1267
$begingroup$
First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:37
$begingroup$
I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:43
$begingroup$
Thank you, but how would one go about proving that it is the UMVUE without this theorem?
$endgroup$
– Analysis801
Dec 7 '18 at 17:44
$begingroup$
What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:47
add a comment |
$begingroup$
First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:37
$begingroup$
I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:43
$begingroup$
Thank you, but how would one go about proving that it is the UMVUE without this theorem?
$endgroup$
– Analysis801
Dec 7 '18 at 17:44
$begingroup$
What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:47
$begingroup$
First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:37
$begingroup$
First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:37
$begingroup$
I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:43
$begingroup$
I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:43
$begingroup$
Thank you, but how would one go about proving that it is the UMVUE without this theorem?
$endgroup$
– Analysis801
Dec 7 '18 at 17:44
$begingroup$
Thank you, but how would one go about proving that it is the UMVUE without this theorem?
$endgroup$
– Analysis801
Dec 7 '18 at 17:44
$begingroup$
What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:47
$begingroup$
What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:47
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030117%2ffinding-best-unbiased-estimator-of-uniform-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030117%2ffinding-best-unbiased-estimator-of-uniform-distribution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
First of all, the information inequality does not apply to pdfs where support depends on parameter. And if by best unbiased estimator you mean in the sense of having minimum variance, then your proposed unbiased estimator is not the UMVUE.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:37
$begingroup$
I am not mistaken, the UMVUE of $theta$ is $frac{n+1}{n}max_{1le ile n} |X_i|$, where $max |X_i|$ is a complete sufficient statistic.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:43
$begingroup$
Thank you, but how would one go about proving that it is the UMVUE without this theorem?
$endgroup$
– Analysis801
Dec 7 '18 at 17:44
$begingroup$
What theorem have you used here? UMVUE in this case is shown by using the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Dec 7 '18 at 17:47