Mixing
$X_i$ follows Bernoulli distribution find UMVUE of $theta(1-theta)$
$begingroup$
Let $X_1,X_2,X_3 ...X_n$ be a random sample from Bernoulli
distribution with parameter $theta$.Find UMVUE of $theta(1-theta)$.
I know that $T=sum_{i=1}^{n}X_i$ is complete sufficient statistic for our paramenter $theta$.
I am trying to find out function of T which is a unique unbiased estimator of $theta(1-theta)$.
Now $Tsim Bin(n,theta)$
$E(T^2)-E(T)^2=V(T)$
$ntheta(1-theta)+n^2theta^2-n^2theta^2=ntheta(1-theta)$
$impliesdfrac{1}{n}(T^2-bar{T}^2)$ is umvue of $theta(1-theta)$
If I have sample size $n=10$ with observations $1,1,1,1,1,0,0,0,0,0$
obtain the value of this estimator.
Now I am stuck at this point that is $T^2$ is $T^2=sum_{i=1}^{n}X_i^2$ or $T^2=(sum_{i=1}^{n}X_i)^2$. Can someone help me and tell at what point I am doing things wrongly?
statistics statistical-inference estimation
asked Jan 25 at 9:44
Daman deepDaman deep
756419
$endgroup$
add a comment |
$begingroup$
Let $X_1,X_2,X_3 ...X_n$ be a random sample from Bernoulli
distribution with parameter $theta$.Find UMVUE of $theta(1-theta)$.
I know that $T=sum_{i=1}^{n}X_i$ is complete sufficient statistic for our paramenter $theta$.
I am trying to find out function of T which is a unique unbiased estimator of $theta(1-theta)$.
Now $Tsim Bin(n,theta)$
$E(T^2)-E(T)^2=V(T)$
$ntheta(1-theta)+n^2theta^2-n^2theta^2=ntheta(1-theta)$
$impliesdfrac{1}{n}(T^2-bar{T}^2)$ is umvue of $theta(1-theta)$
If I have sample size $n=10$ with observations $1,1,1,1,1,0,0,0,0,0$
obtain the value of this estimator.
Now I am stuck at this point that is $T^2$ is $T^2=sum_{i=1}^{n}X_i^2$ or $T^2=(sum_{i=1}^{n}X_i)^2$. Can someone help me and tell at what point I am doing things wrongly?
statistics statistical-inference estimation
asked Jan 25 at 9:44
Daman deepDaman deep
756419
$endgroup$
add a comment |
$begingroup$
Let $X_1,X_2,X_3 ...X_n$ be a random sample from Bernoulli
distribution with parameter $theta$.Find UMVUE of $theta(1-theta)$.
I know that $T=sum_{i=1}^{n}X_i$ is complete sufficient statistic for our paramenter $theta$.
I am trying to find out function of T which is a unique unbiased estimator of $theta(1-theta)$.
Now $Tsim Bin(n,theta)$
$E(T^2)-E(T)^2=V(T)$
$ntheta(1-theta)+n^2theta^2-n^2theta^2=ntheta(1-theta)$
$impliesdfrac{1}{n}(T^2-bar{T}^2)$ is umvue of $theta(1-theta)$
If I have sample size $n=10$ with observations $1,1,1,1,1,0,0,0,0,0$
obtain the value of this estimator.
Now I am stuck at this point that is $T^2$ is $T^2=sum_{i=1}^{n}X_i^2$ or $T^2=(sum_{i=1}^{n}X_i)^2$. Can someone help me and tell at what point I am doing things wrongly?
statistics statistical-inference estimation
asked Jan 25 at 9:44
Daman deepDaman deep
756419
$endgroup$
Let $X_1,X_2,X_3 ...X_n$ be a random sample from Bernoulli
distribution with parameter $theta$.Find UMVUE of $theta(1-theta)$.
I know that $T=sum_{i=1}^{n}X_i$ is complete sufficient statistic for our paramenter $theta$.
I am trying to find out function of T which is a unique unbiased estimator of $theta(1-theta)$.
Now $Tsim Bin(n,theta)$
$E(T^2)-E(T)^2=V(T)$
$ntheta(1-theta)+n^2theta^2-n^2theta^2=ntheta(1-theta)$
$impliesdfrac{1}{n}(T^2-bar{T}^2)$ is umvue of $theta(1-theta)$
If I have sample size $n=10$ with observations $1,1,1,1,1,0,0,0,0,0$
obtain the value of this estimator.
Now I am stuck at this point that is $T^2$ is $T^2=sum_{i=1}^{n}X_i^2$ or $T^2=(sum_{i=1}^{n}X_i)^2$. Can someone help me and tell at what point I am doing things wrongly?
statistics statistical-inference estimation
statistics statistical-inference estimation
asked Jan 25 at 9:44
Daman deepDaman deep
756419
asked Jan 25 at 9:44
Daman deepDaman deep
756419
asked Jan 25 at 9:44
Daman deepDaman deep
756419
asked Jan 25 at 9:44
Daman deepDaman deep
756419
asked Jan 25 at 9:44
Daman deepDaman deep
756419
756419
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An unbiased estimator of $theta$ is $frac{T}{n} $ where $T=sumlimits_{k=1}^n X_k$.
From your approach that $operatorname{Var}_{theta}(T)=mathrm E_{theta}(T^2)-(mathrm E_{theta}(T))^2=ntheta(1-theta)$, it follows that an unbiased estimator of $theta^2$ is $frac{T(T-1)}{n(n-1)}$. You do not get unbiased estimator of $theta(1-theta)$ directly from this step.
An unbiased estimator of $theta(1-theta)$ is therefore $frac{T}{n}-frac{T(T-1)}{n(n-1)}$, which is also the UMVUE.
answered Jan 25 at 10:14
StubbornAtomStubbornAtom
6,22811339
$endgroup$
$begingroup$
If $T sim Bin(n, theta)$ then T is random variable right ?
$endgroup$
– Daman deep
Jan 25 at 10:24
$begingroup$
Of course it is.
$endgroup$
– StubbornAtom
Jan 25 at 10:26
$begingroup$
estimator I used $dfrac{1}{n}(T^2-bar{T}^2)$ if I take expectation I get same result how come mine is not umvue ? $dfrac{E(T^2)-E(E(T)^2)}{n}=dfrac{ntheta(1-theta)+n^2theta^2-n^2theta^2}{n}=theta(1-theta)$ but umvue is unique. Can you tell me whats wrong ? I didn't get it .
$endgroup$
– Daman deep
Jan 25 at 10:32
$begingroup$
$bar T=frac{sum_{i=1}^{n}X_i}{n}$ sorry I didn't specify that .
$endgroup$
– Daman deep
Jan 25 at 10:36
$begingroup$
No need to define things like $bar T$; your notation is confusing. You appear to be doing something like $E(T^2)-(E(T))^2=ntheta(1-theta)impliesleft[E(frac{1}{n}(T^2-(E(T))^2))right]=theta(1-theta)$, and hence concluding that $frac{1}{n}(T^2-(E(T))^2)$ is UMVUE. This is not correct because $E(T)=ntheta$, so that your proposed estimator is not even a statistic (it depends on the parameter).
$endgroup$
– StubbornAtom
Jan 25 at 10:47
|
show 2 more comments
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%2f3086919%2fx-i-follows-bernoulli-distribution-find-umvue-of-theta1-theta%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An unbiased estimator of $theta$ is $frac{T}{n} $ where $T=sumlimits_{k=1}^n X_k$.
From your approach that $operatorname{Var}_{theta}(T)=mathrm E_{theta}(T^2)-(mathrm E_{theta}(T))^2=ntheta(1-theta)$, it follows that an unbiased estimator of $theta^2$ is $frac{T(T-1)}{n(n-1)}$. You do not get unbiased estimator of $theta(1-theta)$ directly from this step.
An unbiased estimator of $theta(1-theta)$ is therefore $frac{T}{n}-frac{T(T-1)}{n(n-1)}$, which is also the UMVUE.
answered Jan 25 at 10:14
StubbornAtomStubbornAtom
6,22811339
$endgroup$
$begingroup$
If $T sim Bin(n, theta)$ then T is random variable right ?
$endgroup$
– Daman deep
Jan 25 at 10:24
$begingroup$
Of course it is.
$endgroup$
– StubbornAtom
Jan 25 at 10:26
$begingroup$
estimator I used $dfrac{1}{n}(T^2-bar{T}^2)$ if I take expectation I get same result how come mine is not umvue ? $dfrac{E(T^2)-E(E(T)^2)}{n}=dfrac{ntheta(1-theta)+n^2theta^2-n^2theta^2}{n}=theta(1-theta)$ but umvue is unique. Can you tell me whats wrong ? I didn't get it .
$endgroup$
– Daman deep
Jan 25 at 10:32
$begingroup$
$bar T=frac{sum_{i=1}^{n}X_i}{n}$ sorry I didn't specify that .
$endgroup$
– Daman deep
Jan 25 at 10:36
$begingroup$
No need to define things like $bar T$; your notation is confusing. You appear to be doing something like $E(T^2)-(E(T))^2=ntheta(1-theta)impliesleft[E(frac{1}{n}(T^2-(E(T))^2))right]=theta(1-theta)$, and hence concluding that $frac{1}{n}(T^2-(E(T))^2)$ is UMVUE. This is not correct because $E(T)=ntheta$, so that your proposed estimator is not even a statistic (it depends on the parameter).
$endgroup$
– StubbornAtom
Jan 25 at 10:47
|
show 2 more comments
$begingroup$
An unbiased estimator of $theta$ is $frac{T}{n} $ where $T=sumlimits_{k=1}^n X_k$.
From your approach that $operatorname{Var}_{theta}(T)=mathrm E_{theta}(T^2)-(mathrm E_{theta}(T))^2=ntheta(1-theta)$, it follows that an unbiased estimator of $theta^2$ is $frac{T(T-1)}{n(n-1)}$. You do not get unbiased estimator of $theta(1-theta)$ directly from this step.
An unbiased estimator of $theta(1-theta)$ is therefore $frac{T}{n}-frac{T(T-1)}{n(n-1)}$, which is also the UMVUE.
answered Jan 25 at 10:14
StubbornAtomStubbornAtom
6,22811339
$endgroup$
$begingroup$
If $T sim Bin(n, theta)$ then T is random variable right ?
$endgroup$
– Daman deep
Jan 25 at 10:24
$begingroup$
Of course it is.
$endgroup$
– StubbornAtom
Jan 25 at 10:26
$begingroup$
estimator I used $dfrac{1}{n}(T^2-bar{T}^2)$ if I take expectation I get same result how come mine is not umvue ? $dfrac{E(T^2)-E(E(T)^2)}{n}=dfrac{ntheta(1-theta)+n^2theta^2-n^2theta^2}{n}=theta(1-theta)$ but umvue is unique. Can you tell me whats wrong ? I didn't get it .
$endgroup$
– Daman deep
Jan 25 at 10:32
$begingroup$
$bar T=frac{sum_{i=1}^{n}X_i}{n}$ sorry I didn't specify that .
$endgroup$
– Daman deep
Jan 25 at 10:36
$begingroup$
No need to define things like $bar T$; your notation is confusing. You appear to be doing something like $E(T^2)-(E(T))^2=ntheta(1-theta)impliesleft[E(frac{1}{n}(T^2-(E(T))^2))right]=theta(1-theta)$, and hence concluding that $frac{1}{n}(T^2-(E(T))^2)$ is UMVUE. This is not correct because $E(T)=ntheta$, so that your proposed estimator is not even a statistic (it depends on the parameter).
$endgroup$
– StubbornAtom
Jan 25 at 10:47
|
show 2 more comments
$begingroup$
An unbiased estimator of $theta$ is $frac{T}{n} $ where $T=sumlimits_{k=1}^n X_k$.
From your approach that $operatorname{Var}_{theta}(T)=mathrm E_{theta}(T^2)-(mathrm E_{theta}(T))^2=ntheta(1-theta)$, it follows that an unbiased estimator of $theta^2$ is $frac{T(T-1)}{n(n-1)}$. You do not get unbiased estimator of $theta(1-theta)$ directly from this step.
An unbiased estimator of $theta(1-theta)$ is therefore $frac{T}{n}-frac{T(T-1)}{n(n-1)}$, which is also the UMVUE.
answered Jan 25 at 10:14
StubbornAtomStubbornAtom
6,22811339
$endgroup$
An unbiased estimator of $theta$ is $frac{T}{n} $ where $T=sumlimits_{k=1}^n X_k$.
From your approach that $operatorname{Var}_{theta}(T)=mathrm E_{theta}(T^2)-(mathrm E_{theta}(T))^2=ntheta(1-theta)$, it follows that an unbiased estimator of $theta^2$ is $frac{T(T-1)}{n(n-1)}$. You do not get unbiased estimator of $theta(1-theta)$ directly from this step.
An unbiased estimator of $theta(1-theta)$ is therefore $frac{T}{n}-frac{T(T-1)}{n(n-1)}$, which is also the UMVUE.
answered Jan 25 at 10:14
StubbornAtomStubbornAtom
6,22811339
edited Jan 25 at 10:23
answered Jan 25 at 10:14
StubbornAtomStubbornAtom
6,22811339
answered Jan 25 at 10:14
StubbornAtomStubbornAtom
6,22811339
answered Jan 25 at 10:14
StubbornAtomStubbornAtom
6,22811339
6,22811339
$begingroup$
If $T sim Bin(n, theta)$ then T is random variable right ?
$endgroup$
– Daman deep
Jan 25 at 10:24
$begingroup$
Of course it is.
$endgroup$
– StubbornAtom
Jan 25 at 10:26
$begingroup$
estimator I used $dfrac{1}{n}(T^2-bar{T}^2)$ if I take expectation I get same result how come mine is not umvue ? $dfrac{E(T^2)-E(E(T)^2)}{n}=dfrac{ntheta(1-theta)+n^2theta^2-n^2theta^2}{n}=theta(1-theta)$ but umvue is unique. Can you tell me whats wrong ? I didn't get it .
$endgroup$
– Daman deep
Jan 25 at 10:32
$begingroup$
$bar T=frac{sum_{i=1}^{n}X_i}{n}$ sorry I didn't specify that .
$endgroup$
– Daman deep
Jan 25 at 10:36
$begingroup$
No need to define things like $bar T$; your notation is confusing. You appear to be doing something like $E(T^2)-(E(T))^2=ntheta(1-theta)impliesleft[E(frac{1}{n}(T^2-(E(T))^2))right]=theta(1-theta)$, and hence concluding that $frac{1}{n}(T^2-(E(T))^2)$ is UMVUE. This is not correct because $E(T)=ntheta$, so that your proposed estimator is not even a statistic (it depends on the parameter).
$endgroup$
– StubbornAtom
Jan 25 at 10:47
|
show 2 more comments
$begingroup$
If $T sim Bin(n, theta)$ then T is random variable right ?
$endgroup$
– Daman deep
Jan 25 at 10:24
$begingroup$
Of course it is.
$endgroup$
– StubbornAtom
Jan 25 at 10:26
$begingroup$
estimator I used $dfrac{1}{n}(T^2-bar{T}^2)$ if I take expectation I get same result how come mine is not umvue ? $dfrac{E(T^2)-E(E(T)^2)}{n}=dfrac{ntheta(1-theta)+n^2theta^2-n^2theta^2}{n}=theta(1-theta)$ but umvue is unique. Can you tell me whats wrong ? I didn't get it .
$endgroup$
– Daman deep
Jan 25 at 10:32
$begingroup$
$bar T=frac{sum_{i=1}^{n}X_i}{n}$ sorry I didn't specify that .
$endgroup$
– Daman deep
Jan 25 at 10:36
$begingroup$
No need to define things like $bar T$; your notation is confusing. You appear to be doing something like $E(T^2)-(E(T))^2=ntheta(1-theta)impliesleft[E(frac{1}{n}(T^2-(E(T))^2))right]=theta(1-theta)$, and hence concluding that $frac{1}{n}(T^2-(E(T))^2)$ is UMVUE. This is not correct because $E(T)=ntheta$, so that your proposed estimator is not even a statistic (it depends on the parameter).
$endgroup$
– StubbornAtom
Jan 25 at 10:47
$begingroup$
If $T sim Bin(n, theta)$ then T is random variable right ?
$endgroup$
– Daman deep
Jan 25 at 10:24
$begingroup$
If $T sim Bin(n, theta)$ then T is random variable right ?
$endgroup$
– Daman deep
Jan 25 at 10:24
$begingroup$
Of course it is.
$endgroup$
– StubbornAtom
Jan 25 at 10:26
$begingroup$
Of course it is.
$endgroup$
– StubbornAtom
Jan 25 at 10:26
$begingroup$
estimator I used $dfrac{1}{n}(T^2-bar{T}^2)$ if I take expectation I get same result how come mine is not umvue ? $dfrac{E(T^2)-E(E(T)^2)}{n}=dfrac{ntheta(1-theta)+n^2theta^2-n^2theta^2}{n}=theta(1-theta)$ but umvue is unique. Can you tell me whats wrong ? I didn't get it .
$endgroup$
– Daman deep
Jan 25 at 10:32
$begingroup$
estimator I used $dfrac{1}{n}(T^2-bar{T}^2)$ if I take expectation I get same result how come mine is not umvue ? $dfrac{E(T^2)-E(E(T)^2)}{n}=dfrac{ntheta(1-theta)+n^2theta^2-n^2theta^2}{n}=theta(1-theta)$ but umvue is unique. Can you tell me whats wrong ? I didn't get it .
$endgroup$
– Daman deep
Jan 25 at 10:32
$begingroup$
$bar T=frac{sum_{i=1}^{n}X_i}{n}$ sorry I didn't specify that .
$endgroup$
– Daman deep
Jan 25 at 10:36
$begingroup$
$bar T=frac{sum_{i=1}^{n}X_i}{n}$ sorry I didn't specify that .
$endgroup$
– Daman deep
Jan 25 at 10:36
$begingroup$
No need to define things like $bar T$; your notation is confusing. You appear to be doing something like $E(T^2)-(E(T))^2=ntheta(1-theta)impliesleft[E(frac{1}{n}(T^2-(E(T))^2))right]=theta(1-theta)$, and hence concluding that $frac{1}{n}(T^2-(E(T))^2)$ is UMVUE. This is not correct because $E(T)=ntheta$, so that your proposed estimator is not even a statistic (it depends on the parameter).
$endgroup$
– StubbornAtom
Jan 25 at 10:47
$begingroup$
No need to define things like $bar T$; your notation is confusing. You appear to be doing something like $E(T^2)-(E(T))^2=ntheta(1-theta)impliesleft[E(frac{1}{n}(T^2-(E(T))^2))right]=theta(1-theta)$, and hence concluding that $frac{1}{n}(T^2-(E(T))^2)$ is UMVUE. This is not correct because $E(T)=ntheta$, so that your proposed estimator is not even a statistic (it depends on the parameter).
$endgroup$
– StubbornAtom
Jan 25 at 10:47
|
show 2 more comments
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%2f3086919%2fx-i-follows-bernoulli-distribution-find-umvue-of-theta1-theta%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
