Mixing
Minimum variance unbiased estimator of exponential distribution
$begingroup$
The given model is $text{Exp}(mu,sigma),;muinBbb{R},sigmagt0$ whose pdf is
$f(xtext{;}theta)={1over sigma}e^{-{{(x-mu)}over sigma}}I_{(mu,infty)}(x)$
I easily found $(X_{(1)},bar{X}-X_{(1)})'$ is CSS for $theta=(mu,sigma)'$ with the sample size $n$
The problem is, the parameter to be estimated is $eta=P_{theta}(X_{1}gt a);(ainBbb{R}text{ : given})$, not $theta$
I'm trying to solve it with Beta distribution as an ancillary statistic, applying Lehmann-Scheffe, but it doesn't work well
$1);;$I think ${X_{1}-X_{(1)}over bar{X}-X_{(1)}}sim B(1,n-2)$ is an ancillary statistic for $theta$, is it right?
$2);;$If my guess is wrong(or too difficult to calculate an ancillary statistic), what is the key of this problem?
parameter-estimation exponential-distribution
asked Jan 13 at 17:48
GiyookGiyook
83
$endgroup$
add a comment |
$begingroup$
The given model is $text{Exp}(mu,sigma),;muinBbb{R},sigmagt0$ whose pdf is
$f(xtext{;}theta)={1over sigma}e^{-{{(x-mu)}over sigma}}I_{(mu,infty)}(x)$
I easily found $(X_{(1)},bar{X}-X_{(1)})'$ is CSS for $theta=(mu,sigma)'$ with the sample size $n$
The problem is, the parameter to be estimated is $eta=P_{theta}(X_{1}gt a);(ainBbb{R}text{ : given})$, not $theta$
I'm trying to solve it with Beta distribution as an ancillary statistic, applying Lehmann-Scheffe, but it doesn't work well
$1);;$I think ${X_{1}-X_{(1)}over bar{X}-X_{(1)}}sim B(1,n-2)$ is an ancillary statistic for $theta$, is it right?
$2);;$If my guess is wrong(or too difficult to calculate an ancillary statistic), what is the key of this problem?
parameter-estimation exponential-distribution
asked Jan 13 at 17:48
GiyookGiyook
83
$endgroup$
$begingroup$
What distribution? The usual exponential distribution has only one parameter, not two.
$endgroup$
– Robert Israel
Jan 13 at 17:57
$begingroup$
It's shifted exponential. I'm sorry. I added its pdf above.
$endgroup$
– Giyook
Jan 13 at 18:06
$begingroup$
You should mention that you are working with a sample of size $n$. Why searching for ancillary statistic? You have to find an unbiased estimator of $eta$ based on $left(X_{(1)},sum_{i=1}^n (X_i-X_{(1)})right)$, a complete sufficient statistic (yours is not correct I think). So you could use the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Jan 13 at 18:21
$begingroup$
@StubbornAtom I missed the detail of question. Yours is what I found and applying Lehmann-Scheffe with that was not quite simple. So I thought it is easier if I use ancillary statistic to calculate it.
$endgroup$
– Giyook
Jan 13 at 23:21
add a comment |
$begingroup$
The given model is $text{Exp}(mu,sigma),;muinBbb{R},sigmagt0$ whose pdf is
$f(xtext{;}theta)={1over sigma}e^{-{{(x-mu)}over sigma}}I_{(mu,infty)}(x)$
I easily found $(X_{(1)},bar{X}-X_{(1)})'$ is CSS for $theta=(mu,sigma)'$ with the sample size $n$
The problem is, the parameter to be estimated is $eta=P_{theta}(X_{1}gt a);(ainBbb{R}text{ : given})$, not $theta$
I'm trying to solve it with Beta distribution as an ancillary statistic, applying Lehmann-Scheffe, but it doesn't work well
$1);;$I think ${X_{1}-X_{(1)}over bar{X}-X_{(1)}}sim B(1,n-2)$ is an ancillary statistic for $theta$, is it right?
$2);;$If my guess is wrong(or too difficult to calculate an ancillary statistic), what is the key of this problem?
parameter-estimation exponential-distribution
asked Jan 13 at 17:48
GiyookGiyook
83
$endgroup$
The given model is $text{Exp}(mu,sigma),;muinBbb{R},sigmagt0$ whose pdf is
$f(xtext{;}theta)={1over sigma}e^{-{{(x-mu)}over sigma}}I_{(mu,infty)}(x)$
I easily found $(X_{(1)},bar{X}-X_{(1)})'$ is CSS for $theta=(mu,sigma)'$ with the sample size $n$
The problem is, the parameter to be estimated is $eta=P_{theta}(X_{1}gt a);(ainBbb{R}text{ : given})$, not $theta$
I'm trying to solve it with Beta distribution as an ancillary statistic, applying Lehmann-Scheffe, but it doesn't work well
$1);;$I think ${X_{1}-X_{(1)}over bar{X}-X_{(1)}}sim B(1,n-2)$ is an ancillary statistic for $theta$, is it right?
$2);;$If my guess is wrong(or too difficult to calculate an ancillary statistic), what is the key of this problem?
parameter-estimation exponential-distribution
parameter-estimation exponential-distribution
asked Jan 13 at 17:48
GiyookGiyook
83
asked Jan 13 at 17:48
GiyookGiyook
83
edited Jan 13 at 23:46
Giyook
asked Jan 13 at 17:48
GiyookGiyook
83
asked Jan 13 at 17:48
GiyookGiyook
83
asked Jan 13 at 17:48
GiyookGiyook
83
83
$begingroup$
What distribution? The usual exponential distribution has only one parameter, not two.
$endgroup$
– Robert Israel
Jan 13 at 17:57
$begingroup$
It's shifted exponential. I'm sorry. I added its pdf above.
$endgroup$
– Giyook
Jan 13 at 18:06
$begingroup$
You should mention that you are working with a sample of size $n$. Why searching for ancillary statistic? You have to find an unbiased estimator of $eta$ based on $left(X_{(1)},sum_{i=1}^n (X_i-X_{(1)})right)$, a complete sufficient statistic (yours is not correct I think). So you could use the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Jan 13 at 18:21
$begingroup$
@StubbornAtom I missed the detail of question. Yours is what I found and applying Lehmann-Scheffe with that was not quite simple. So I thought it is easier if I use ancillary statistic to calculate it.
$endgroup$
– Giyook
Jan 13 at 23:21
add a comment |
$begingroup$
What distribution? The usual exponential distribution has only one parameter, not two.
$endgroup$
– Robert Israel
Jan 13 at 17:57
$begingroup$
It's shifted exponential. I'm sorry. I added its pdf above.
$endgroup$
– Giyook
Jan 13 at 18:06
$begingroup$
You should mention that you are working with a sample of size $n$. Why searching for ancillary statistic? You have to find an unbiased estimator of $eta$ based on $left(X_{(1)},sum_{i=1}^n (X_i-X_{(1)})right)$, a complete sufficient statistic (yours is not correct I think). So you could use the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Jan 13 at 18:21
$begingroup$
@StubbornAtom I missed the detail of question. Yours is what I found and applying Lehmann-Scheffe with that was not quite simple. So I thought it is easier if I use ancillary statistic to calculate it.
$endgroup$
– Giyook
Jan 13 at 23:21
$begingroup$
What distribution? The usual exponential distribution has only one parameter, not two.
$endgroup$
– Robert Israel
Jan 13 at 17:57
$begingroup$
What distribution? The usual exponential distribution has only one parameter, not two.
$endgroup$
– Robert Israel
Jan 13 at 17:57
$begingroup$
It's shifted exponential. I'm sorry. I added its pdf above.
$endgroup$
– Giyook
Jan 13 at 18:06
$begingroup$
It's shifted exponential. I'm sorry. I added its pdf above.
$endgroup$
– Giyook
Jan 13 at 18:06
$begingroup$
You should mention that you are working with a sample of size $n$. Why searching for ancillary statistic? You have to find an unbiased estimator of $eta$ based on $left(X_{(1)},sum_{i=1}^n (X_i-X_{(1)})right)$, a complete sufficient statistic (yours is not correct I think). So you could use the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Jan 13 at 18:21
$begingroup$
You should mention that you are working with a sample of size $n$. Why searching for ancillary statistic? You have to find an unbiased estimator of $eta$ based on $left(X_{(1)},sum_{i=1}^n (X_i-X_{(1)})right)$, a complete sufficient statistic (yours is not correct I think). So you could use the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Jan 13 at 18:21
$begingroup$
@StubbornAtom I missed the detail of question. Yours is what I found and applying Lehmann-Scheffe with that was not quite simple. So I thought it is easier if I use ancillary statistic to calculate it.
$endgroup$
– Giyook
Jan 13 at 23:21
$begingroup$
@StubbornAtom I missed the detail of question. Yours is what I found and applying Lehmann-Scheffe with that was not quite simple. So I thought it is easier if I use ancillary statistic to calculate it.
$endgroup$
– Giyook
Jan 13 at 23:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I will use the more common notations, i.e., $1/sigma = lambda$ and $mu = gamma$, hence
$$
mathbb{P}(X>a)= exp{-lambda(a-gamma)},
$$
hence the MLE is
$$
hat{P}=exp{-frac{1}{bar{X}_n}(a-X_{(1)})}.
$$
This is a biased estimator, so you can find its expectation using the joint probability function of $bar{X_n}$ and $X_{(1)}$ and then correcting the bias (this is basically an application of the Lehmann-Scehffe theorem. I'm not sure that this is an easy exercise. However, finding UMVU estimators is an old-fashion problem in parametric statistics. You can find here
https://projecteuclid.org/download/pdf_1/euclid.aoms/1177706256 in eq. (7.9) a UMVUE of the tail probability for $lambda = 1$ or use Thoerem~3 in order to construct an UMVUE for any functional of an exponential shifted distribution (in exponential distribution, truncation is equivalent to shifting, therefore you can apply all the result from this paper).
answered Jan 19 at 14:30
V. VancakV. Vancak
11.1k2926
$endgroup$
add a comment |
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%2f3072303%2fminimum-variance-unbiased-estimator-of-exponential-distribution%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
$begingroup$
I will use the more common notations, i.e., $1/sigma = lambda$ and $mu = gamma$, hence
$$
mathbb{P}(X>a)= exp{-lambda(a-gamma)},
$$
hence the MLE is
$$
hat{P}=exp{-frac{1}{bar{X}_n}(a-X_{(1)})}.
$$
This is a biased estimator, so you can find its expectation using the joint probability function of $bar{X_n}$ and $X_{(1)}$ and then correcting the bias (this is basically an application of the Lehmann-Scehffe theorem. I'm not sure that this is an easy exercise. However, finding UMVU estimators is an old-fashion problem in parametric statistics. You can find here
https://projecteuclid.org/download/pdf_1/euclid.aoms/1177706256 in eq. (7.9) a UMVUE of the tail probability for $lambda = 1$ or use Thoerem~3 in order to construct an UMVUE for any functional of an exponential shifted distribution (in exponential distribution, truncation is equivalent to shifting, therefore you can apply all the result from this paper).
answered Jan 19 at 14:30
V. VancakV. Vancak
11.1k2926
$endgroup$
add a comment |
$begingroup$
I will use the more common notations, i.e., $1/sigma = lambda$ and $mu = gamma$, hence
$$
mathbb{P}(X>a)= exp{-lambda(a-gamma)},
$$
hence the MLE is
$$
hat{P}=exp{-frac{1}{bar{X}_n}(a-X_{(1)})}.
$$
This is a biased estimator, so you can find its expectation using the joint probability function of $bar{X_n}$ and $X_{(1)}$ and then correcting the bias (this is basically an application of the Lehmann-Scehffe theorem. I'm not sure that this is an easy exercise. However, finding UMVU estimators is an old-fashion problem in parametric statistics. You can find here
https://projecteuclid.org/download/pdf_1/euclid.aoms/1177706256 in eq. (7.9) a UMVUE of the tail probability for $lambda = 1$ or use Thoerem~3 in order to construct an UMVUE for any functional of an exponential shifted distribution (in exponential distribution, truncation is equivalent to shifting, therefore you can apply all the result from this paper).
answered Jan 19 at 14:30
V. VancakV. Vancak
11.1k2926
$endgroup$
add a comment |
$begingroup$
I will use the more common notations, i.e., $1/sigma = lambda$ and $mu = gamma$, hence
$$
mathbb{P}(X>a)= exp{-lambda(a-gamma)},
$$
hence the MLE is
$$
hat{P}=exp{-frac{1}{bar{X}_n}(a-X_{(1)})}.
$$
This is a biased estimator, so you can find its expectation using the joint probability function of $bar{X_n}$ and $X_{(1)}$ and then correcting the bias (this is basically an application of the Lehmann-Scehffe theorem. I'm not sure that this is an easy exercise. However, finding UMVU estimators is an old-fashion problem in parametric statistics. You can find here
https://projecteuclid.org/download/pdf_1/euclid.aoms/1177706256 in eq. (7.9) a UMVUE of the tail probability for $lambda = 1$ or use Thoerem~3 in order to construct an UMVUE for any functional of an exponential shifted distribution (in exponential distribution, truncation is equivalent to shifting, therefore you can apply all the result from this paper).
answered Jan 19 at 14:30
V. VancakV. Vancak
11.1k2926
$endgroup$
I will use the more common notations, i.e., $1/sigma = lambda$ and $mu = gamma$, hence
$$
mathbb{P}(X>a)= exp{-lambda(a-gamma)},
$$
hence the MLE is
$$
hat{P}=exp{-frac{1}{bar{X}_n}(a-X_{(1)})}.
$$
This is a biased estimator, so you can find its expectation using the joint probability function of $bar{X_n}$ and $X_{(1)}$ and then correcting the bias (this is basically an application of the Lehmann-Scehffe theorem. I'm not sure that this is an easy exercise. However, finding UMVU estimators is an old-fashion problem in parametric statistics. You can find here
https://projecteuclid.org/download/pdf_1/euclid.aoms/1177706256 in eq. (7.9) a UMVUE of the tail probability for $lambda = 1$ or use Thoerem~3 in order to construct an UMVUE for any functional of an exponential shifted distribution (in exponential distribution, truncation is equivalent to shifting, therefore you can apply all the result from this paper).
answered Jan 19 at 14:30
V. VancakV. Vancak
11.1k2926
answered Jan 19 at 14:30
V. VancakV. Vancak
11.1k2926
answered Jan 19 at 14:30
V. VancakV. Vancak
11.1k2926
answered Jan 19 at 14:30
V. VancakV. Vancak
11.1k2926
11.1k2926
add a comment |
add a comment |
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%2f3072303%2fminimum-variance-unbiased-estimator-of-exponential-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
$begingroup$
What distribution? The usual exponential distribution has only one parameter, not two.
$endgroup$
– Robert Israel
Jan 13 at 17:57
$begingroup$
It's shifted exponential. I'm sorry. I added its pdf above.
$endgroup$
– Giyook
Jan 13 at 18:06
$begingroup$
You should mention that you are working with a sample of size $n$. Why searching for ancillary statistic? You have to find an unbiased estimator of $eta$ based on $left(X_{(1)},sum_{i=1}^n (X_i-X_{(1)})right)$, a complete sufficient statistic (yours is not correct I think). So you could use the Lehmann-Scheffe theorem.
$endgroup$
– StubbornAtom
Jan 13 at 18:21
$begingroup$
@StubbornAtom I missed the detail of question. Yours is what I found and applying Lehmann-Scheffe with that was not quite simple. So I thought it is easier if I use ancillary statistic to calculate it.
$endgroup$
– Giyook
Jan 13 at 23:21