Minimal sufficient statistics for beta distribution
$begingroup$
Let $X_1, ldots, X_n$ be a simple random sample. Each $X_i, i in {1, ldots, n }$ has beta distribution $Be(alpha, alpha), alpha >0$.
My task is to find the minimal sufficient statistics. I did it in the following way:
$$f(x_1, ldots, x_n | alpha) = frac{1}{mbox{B}(alpha, alpha)}big(x_1 ldots x_n (1-x_1) ldots (1-x_n) big)^{alpha-1} tag{1}.$$
Using the form $(1)$ I can conclude (using factorization theorem) that my statistics $T(X)$ looks like this:
$$T(X) = prod_{i=1}^{n}x_i(1-x_i).$$
Now I need to show that $T$ is minimal.
$$U = frac{f(x_1, ldots, x_n | alpha)}{f(y_1, ldots, y_n | alpha)} = bigg( frac{x_1 ldots x_n (1-x_1) ldots (1-x_n)}{y_1 ldots y_n (1-y_1) ldots (1-y_n)} bigg) ^{alpha - 1}.$$
$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ thus $T(X)$ is minimal.
Is my reasoning correct?
statistics
asked Nov 28 '18 at 13:23
HendrraHendrra
1,137516
$endgroup$
add a comment |
$begingroup$
Let $X_1, ldots, X_n$ be a simple random sample. Each $X_i, i in {1, ldots, n }$ has beta distribution $Be(alpha, alpha), alpha >0$.
My task is to find the minimal sufficient statistics. I did it in the following way:
$$f(x_1, ldots, x_n | alpha) = frac{1}{mbox{B}(alpha, alpha)}big(x_1 ldots x_n (1-x_1) ldots (1-x_n) big)^{alpha-1} tag{1}.$$
Using the form $(1)$ I can conclude (using factorization theorem) that my statistics $T(X)$ looks like this:
$$T(X) = prod_{i=1}^{n}x_i(1-x_i).$$
Now I need to show that $T$ is minimal.
$$U = frac{f(x_1, ldots, x_n | alpha)}{f(y_1, ldots, y_n | alpha)} = bigg( frac{x_1 ldots x_n (1-x_1) ldots (1-x_n)}{y_1 ldots y_n (1-y_1) ldots (1-y_n)} bigg) ^{alpha - 1}.$$
$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ thus $T(X)$ is minimal.
Is my reasoning correct?
statistics
asked Nov 28 '18 at 13:23
HendrraHendrra
1,137516
$endgroup$
$begingroup$
Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
$endgroup$
– Henry
Nov 28 '18 at 13:32
$begingroup$
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
$endgroup$
– Hendrra
Nov 28 '18 at 13:35
add a comment |
$begingroup$
Let $X_1, ldots, X_n$ be a simple random sample. Each $X_i, i in {1, ldots, n }$ has beta distribution $Be(alpha, alpha), alpha >0$.
My task is to find the minimal sufficient statistics. I did it in the following way:
$$f(x_1, ldots, x_n | alpha) = frac{1}{mbox{B}(alpha, alpha)}big(x_1 ldots x_n (1-x_1) ldots (1-x_n) big)^{alpha-1} tag{1}.$$
Using the form $(1)$ I can conclude (using factorization theorem) that my statistics $T(X)$ looks like this:
$$T(X) = prod_{i=1}^{n}x_i(1-x_i).$$
Now I need to show that $T$ is minimal.
$$U = frac{f(x_1, ldots, x_n | alpha)}{f(y_1, ldots, y_n | alpha)} = bigg( frac{x_1 ldots x_n (1-x_1) ldots (1-x_n)}{y_1 ldots y_n (1-y_1) ldots (1-y_n)} bigg) ^{alpha - 1}.$$
$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ thus $T(X)$ is minimal.
Is my reasoning correct?
statistics
asked Nov 28 '18 at 13:23
HendrraHendrra
1,137516
$endgroup$
Let $X_1, ldots, X_n$ be a simple random sample. Each $X_i, i in {1, ldots, n }$ has beta distribution $Be(alpha, alpha), alpha >0$.
My task is to find the minimal sufficient statistics. I did it in the following way:
$$f(x_1, ldots, x_n | alpha) = frac{1}{mbox{B}(alpha, alpha)}big(x_1 ldots x_n (1-x_1) ldots (1-x_n) big)^{alpha-1} tag{1}.$$
Using the form $(1)$ I can conclude (using factorization theorem) that my statistics $T(X)$ looks like this:
$$T(X) = prod_{i=1}^{n}x_i(1-x_i).$$
Now I need to show that $T$ is minimal.
$$U = frac{f(x_1, ldots, x_n | alpha)}{f(y_1, ldots, y_n | alpha)} = bigg( frac{x_1 ldots x_n (1-x_1) ldots (1-x_n)}{y_1 ldots y_n (1-y_1) ldots (1-y_n)} bigg) ^{alpha - 1}.$$
$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ thus $T(X)$ is minimal.
Is my reasoning correct?
statistics
statistics
asked Nov 28 '18 at 13:23
HendrraHendrra
1,137516
asked Nov 28 '18 at 13:23
HendrraHendrra
1,137516
asked Nov 28 '18 at 13:23
HendrraHendrra
1,137516
asked Nov 28 '18 at 13:23
HendrraHendrra
1,137516
asked Nov 28 '18 at 13:23
HendrraHendrra
1,137516
1,137516
$begingroup$
Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
$endgroup$
– Henry
Nov 28 '18 at 13:32
$begingroup$
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
$endgroup$
– Hendrra
Nov 28 '18 at 13:35
add a comment |
$begingroup$
Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
$endgroup$
– Henry
Nov 28 '18 at 13:32
$begingroup$
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
$endgroup$
– Hendrra
Nov 28 '18 at 13:35
$begingroup$
Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
$endgroup$
– Henry
Nov 28 '18 at 13:32
$begingroup$
Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
$endgroup$
– Henry
Nov 28 '18 at 13:32
$begingroup$
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
$endgroup$
– Hendrra
Nov 28 '18 at 13:35
$begingroup$
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
$endgroup$
– Hendrra
Nov 28 '18 at 13:35
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%2f3017143%2fminimal-sufficient-statistics-for-beta-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%2f3017143%2fminimal-sufficient-statistics-for-beta-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
$begingroup$
Essentially yes, though I am not sure that the general statement is "$U$ is independent from $alpha$ if and only if the fraction is equal to $1$. And that happens when $T(x) = T(y)$ ..." I might prefer "$U$ is independent from $alpha$ if and only $T(x) = T(y)$ (or one of then is zero) ..."
$endgroup$
– Henry
Nov 28 '18 at 13:32
$begingroup$
@Henry I think that the general statement would be a bit different. However in my case it is the only possibility that's why I wrote it this way.
$endgroup$
– Hendrra
Nov 28 '18 at 13:35