The probability that the size of a set is N after observing M unique elements from a sample of X (with...
$begingroup$
Problem
Given a set of unknown size, X elements are sampled with replacement. Of these sampled elements, M elements are unique, meaning (X - M) elements were chosen more than once. Knowing this, what is the probability that the set has size N?
Related question on SE
If a set has n elements and x elements are selected with replacement, what is the probability that m unique elements are selected? answer
probability combinatorics
asked Dec 6 '18 at 14:15
Beamer159Beamer159
31
$endgroup$
add a comment |
$begingroup$
Problem
Given a set of unknown size, X elements are sampled with replacement. Of these sampled elements, M elements are unique, meaning (X - M) elements were chosen more than once. Knowing this, what is the probability that the set has size N?
Related question on SE
If a set has n elements and x elements are selected with replacement, what is the probability that m unique elements are selected? answer
probability combinatorics
asked Dec 6 '18 at 14:15
Beamer159Beamer159
31
$endgroup$
add a comment |
$begingroup$
Problem
Given a set of unknown size, X elements are sampled with replacement. Of these sampled elements, M elements are unique, meaning (X - M) elements were chosen more than once. Knowing this, what is the probability that the set has size N?
Related question on SE
If a set has n elements and x elements are selected with replacement, what is the probability that m unique elements are selected? answer
probability combinatorics
asked Dec 6 '18 at 14:15
Beamer159Beamer159
31
$endgroup$
Problem
Given a set of unknown size, X elements are sampled with replacement. Of these sampled elements, M elements are unique, meaning (X - M) elements were chosen more than once. Knowing this, what is the probability that the set has size N?
Related question on SE
If a set has n elements and x elements are selected with replacement, what is the probability that m unique elements are selected? answer
probability combinatorics
probability combinatorics
asked Dec 6 '18 at 14:15
Beamer159Beamer159
31
asked Dec 6 '18 at 14:15
Beamer159Beamer159
31
asked Dec 6 '18 at 14:15
Beamer159Beamer159
31
asked Dec 6 '18 at 14:15
Beamer159Beamer159
31
asked Dec 6 '18 at 14:15
Beamer159Beamer159
31
31
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We require the probability $operatorname{Pr}left(N = nmiddle| M = mright)$ after observing $m$ unique elements sampled from a sample size of $X$. By Bayes' theorem (where $X$ has been suppressed in the notation):
$$ operatorname{Pr}left(N = nmiddle| M = mright) = dfrac{operatorname{Pr}left( M = mmiddle|N = nright)operatorname{Pr}left(N = nright)}{operatorname{Pr}left( M = mright)}$$
From this it appears that the problem is ill-defined without knowledge of the prior distribution $operatorname{Pr}left(N = nright)$. However if we proceed with the mindset that '$N$ is equally likely to be anything' (in the positive integers), then we could choose the improper prior that $operatorname{Pr}left(N = nright) propto 1$. This allows us to take
$$ operatorname{Pr}left(N = nmiddle| M = mright) propto operatorname{Pr}left( M = mmiddle|N = nright) $$
where the right-hand side is the same probability as from the related question. Hence this expresses that the desired probability is proportional to the likelihood, and we can obtain a valid figure for the probability through normalisation:
$$ operatorname{Pr}left(N = nmiddle| M = mright) = dfrac{operatorname{Pr}left( M = mmiddle|N = nright)}{sum_{i = m}^{infty}operatorname{Pr}left( M = mmiddle|N = iright)} $$
answered Dec 6 '18 at 17:14
rzchrzch
1365
$endgroup$
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%2f3028539%2fthe-probability-that-the-size-of-a-set-is-n-after-observing-m-unique-elements-fr%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
$begingroup$
We require the probability $operatorname{Pr}left(N = nmiddle| M = mright)$ after observing $m$ unique elements sampled from a sample size of $X$. By Bayes' theorem (where $X$ has been suppressed in the notation):
$$ operatorname{Pr}left(N = nmiddle| M = mright) = dfrac{operatorname{Pr}left( M = mmiddle|N = nright)operatorname{Pr}left(N = nright)}{operatorname{Pr}left( M = mright)}$$
From this it appears that the problem is ill-defined without knowledge of the prior distribution $operatorname{Pr}left(N = nright)$. However if we proceed with the mindset that '$N$ is equally likely to be anything' (in the positive integers), then we could choose the improper prior that $operatorname{Pr}left(N = nright) propto 1$. This allows us to take
$$ operatorname{Pr}left(N = nmiddle| M = mright) propto operatorname{Pr}left( M = mmiddle|N = nright) $$
where the right-hand side is the same probability as from the related question. Hence this expresses that the desired probability is proportional to the likelihood, and we can obtain a valid figure for the probability through normalisation:
$$ operatorname{Pr}left(N = nmiddle| M = mright) = dfrac{operatorname{Pr}left( M = mmiddle|N = nright)}{sum_{i = m}^{infty}operatorname{Pr}left( M = mmiddle|N = iright)} $$
answered Dec 6 '18 at 17:14
rzchrzch
1365
$endgroup$
add a comment |
$begingroup$
We require the probability $operatorname{Pr}left(N = nmiddle| M = mright)$ after observing $m$ unique elements sampled from a sample size of $X$. By Bayes' theorem (where $X$ has been suppressed in the notation):
$$ operatorname{Pr}left(N = nmiddle| M = mright) = dfrac{operatorname{Pr}left( M = mmiddle|N = nright)operatorname{Pr}left(N = nright)}{operatorname{Pr}left( M = mright)}$$
From this it appears that the problem is ill-defined without knowledge of the prior distribution $operatorname{Pr}left(N = nright)$. However if we proceed with the mindset that '$N$ is equally likely to be anything' (in the positive integers), then we could choose the improper prior that $operatorname{Pr}left(N = nright) propto 1$. This allows us to take
$$ operatorname{Pr}left(N = nmiddle| M = mright) propto operatorname{Pr}left( M = mmiddle|N = nright) $$
where the right-hand side is the same probability as from the related question. Hence this expresses that the desired probability is proportional to the likelihood, and we can obtain a valid figure for the probability through normalisation:
$$ operatorname{Pr}left(N = nmiddle| M = mright) = dfrac{operatorname{Pr}left( M = mmiddle|N = nright)}{sum_{i = m}^{infty}operatorname{Pr}left( M = mmiddle|N = iright)} $$
answered Dec 6 '18 at 17:14
rzchrzch
1365
$endgroup$
add a comment |
$begingroup$
We require the probability $operatorname{Pr}left(N = nmiddle| M = mright)$ after observing $m$ unique elements sampled from a sample size of $X$. By Bayes' theorem (where $X$ has been suppressed in the notation):
$$ operatorname{Pr}left(N = nmiddle| M = mright) = dfrac{operatorname{Pr}left( M = mmiddle|N = nright)operatorname{Pr}left(N = nright)}{operatorname{Pr}left( M = mright)}$$
From this it appears that the problem is ill-defined without knowledge of the prior distribution $operatorname{Pr}left(N = nright)$. However if we proceed with the mindset that '$N$ is equally likely to be anything' (in the positive integers), then we could choose the improper prior that $operatorname{Pr}left(N = nright) propto 1$. This allows us to take
$$ operatorname{Pr}left(N = nmiddle| M = mright) propto operatorname{Pr}left( M = mmiddle|N = nright) $$
where the right-hand side is the same probability as from the related question. Hence this expresses that the desired probability is proportional to the likelihood, and we can obtain a valid figure for the probability through normalisation:
$$ operatorname{Pr}left(N = nmiddle| M = mright) = dfrac{operatorname{Pr}left( M = mmiddle|N = nright)}{sum_{i = m}^{infty}operatorname{Pr}left( M = mmiddle|N = iright)} $$
answered Dec 6 '18 at 17:14
rzchrzch
1365
$endgroup$
We require the probability $operatorname{Pr}left(N = nmiddle| M = mright)$ after observing $m$ unique elements sampled from a sample size of $X$. By Bayes' theorem (where $X$ has been suppressed in the notation):
$$ operatorname{Pr}left(N = nmiddle| M = mright) = dfrac{operatorname{Pr}left( M = mmiddle|N = nright)operatorname{Pr}left(N = nright)}{operatorname{Pr}left( M = mright)}$$
From this it appears that the problem is ill-defined without knowledge of the prior distribution $operatorname{Pr}left(N = nright)$. However if we proceed with the mindset that '$N$ is equally likely to be anything' (in the positive integers), then we could choose the improper prior that $operatorname{Pr}left(N = nright) propto 1$. This allows us to take
$$ operatorname{Pr}left(N = nmiddle| M = mright) propto operatorname{Pr}left( M = mmiddle|N = nright) $$
where the right-hand side is the same probability as from the related question. Hence this expresses that the desired probability is proportional to the likelihood, and we can obtain a valid figure for the probability through normalisation:
$$ operatorname{Pr}left(N = nmiddle| M = mright) = dfrac{operatorname{Pr}left( M = mmiddle|N = nright)}{sum_{i = m}^{infty}operatorname{Pr}left( M = mmiddle|N = iright)} $$
answered Dec 6 '18 at 17:14
rzchrzch
1365
edited Dec 6 '18 at 17:23
answered Dec 6 '18 at 17:14
rzchrzch
1365
answered Dec 6 '18 at 17:14
rzchrzch
1365
answered Dec 6 '18 at 17:14
rzchrzch
1365
1365
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%2f3028539%2fthe-probability-that-the-size-of-a-set-is-n-after-observing-m-unique-elements-fr%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
