Central limit theorem (CLT) writing












5












$begingroup$


Is there a reason why we are used to write the CLT as $sqrt{n}(overline{X}_n-mu)stackrel{d}{rightarrow}N(0,sigma^2)$ and not as $overline{X}_nstackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?






distributions convergence central-limit-theorem







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
    $endgroup$
    – Neeraj
    2 days ago








  • 1




    $begingroup$
    Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing
    $endgroup$
    – therealcode
    2 days ago








  • 1




    $begingroup$
    Opinions will differ on this, but I would consider the latter statement a legitimate abuse of notation. It is not formally correct, but it is a more useful intuitive statement of the CLT, and the underlying meaning of the statement seems to me to be obvious. Others will object to this statement, but I find it useful.
    $endgroup$
    – Ben
    2 days ago






  • 3




    $begingroup$
    Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.
    $endgroup$
    – Xi'an
    2 days ago
















5












$begingroup$


Is there a reason why we are used to write the CLT as $sqrt{n}(overline{X}_n-mu)stackrel{d}{rightarrow}N(0,sigma^2)$ and not as $overline{X}_nstackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?






distributions convergence central-limit-theorem







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
    $endgroup$
    – Neeraj
    2 days ago








  • 1




    $begingroup$
    Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing
    $endgroup$
    – therealcode
    2 days ago








  • 1




    $begingroup$
    Opinions will differ on this, but I would consider the latter statement a legitimate abuse of notation. It is not formally correct, but it is a more useful intuitive statement of the CLT, and the underlying meaning of the statement seems to me to be obvious. Others will object to this statement, but I find it useful.
    $endgroup$
    – Ben
    2 days ago






  • 3




    $begingroup$
    Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.
    $endgroup$
    – Xi'an
    2 days ago














5












5








5


1



$begingroup$


Is there a reason why we are used to write the CLT as $sqrt{n}(overline{X}_n-mu)stackrel{d}{rightarrow}N(0,sigma^2)$ and not as $overline{X}_nstackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?






distributions convergence central-limit-theorem







share|cite|improve this question











$endgroup$




Is there a reason why we are used to write the CLT as $sqrt{n}(overline{X}_n-mu)stackrel{d}{rightarrow}N(0,sigma^2)$ and not as $overline{X}_nstackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?






distributions convergence central-limit-theorem




distributions convergence central-limit-theorem






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Peter Mortensen

19718




19718










asked 2 days ago









therealcodetherealcode

333




333








  • 1




    $begingroup$
    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
    $endgroup$
    – Neeraj
    2 days ago








  • 1




    $begingroup$
    Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing
    $endgroup$
    – therealcode
    2 days ago








  • 1




    $begingroup$
    Opinions will differ on this, but I would consider the latter statement a legitimate abuse of notation. It is not formally correct, but it is a more useful intuitive statement of the CLT, and the underlying meaning of the statement seems to me to be obvious. Others will object to this statement, but I find it useful.
    $endgroup$
    – Ben
    2 days ago






  • 3




    $begingroup$
    Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.
    $endgroup$
    – Xi'an
    2 days ago














  • 1




    $begingroup$
    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
    $endgroup$
    – Neeraj
    2 days ago








  • 1




    $begingroup$
    Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing
    $endgroup$
    – therealcode
    2 days ago








  • 1




    $begingroup$
    Opinions will differ on this, but I would consider the latter statement a legitimate abuse of notation. It is not formally correct, but it is a more useful intuitive statement of the CLT, and the underlying meaning of the statement seems to me to be obvious. Others will object to this statement, but I find it useful.
    $endgroup$
    – Ben
    2 days ago






  • 3




    $begingroup$
    Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.
    $endgroup$
    – Xi'an
    2 days ago








1




1




$begingroup$
Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
$endgroup$
– Neeraj
2 days ago






$begingroup$
Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
$endgroup$
– Neeraj
2 days ago






1




1




$begingroup$
Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing
$endgroup$
– therealcode
2 days ago






$begingroup$
Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing
$endgroup$
– therealcode
2 days ago






1




1




$begingroup$
Opinions will differ on this, but I would consider the latter statement a legitimate abuse of notation. It is not formally correct, but it is a more useful intuitive statement of the CLT, and the underlying meaning of the statement seems to me to be obvious. Others will object to this statement, but I find it useful.
$endgroup$
– Ben
2 days ago




$begingroup$
Opinions will differ on this, but I would consider the latter statement a legitimate abuse of notation. It is not formally correct, but it is a more useful intuitive statement of the CLT, and the underlying meaning of the statement seems to me to be obvious. Others will object to this statement, but I find it useful.
$endgroup$
– Ben
2 days ago




3




3




$begingroup$
Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.
$endgroup$
– Xi'an
2 days ago




$begingroup$
Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.
$endgroup$
– Xi'an
2 days ago










1 Answer
1






active

oldest

votes


















7












$begingroup$

The notation $stackrel{d}{rightarrow}$ in the CLT is shorthand for the formal limit statement:



$$lim_{n rightarrow infty} mathbb{P} Big( sqrt{n} (bar{X}_n - mu) leqslant t Big) = Phi(t | 0, sigma^2).$$



You will notice that the formal statement is a limit statement in $n$ and so all of the values of $n$ are on the left-hand-side of the equation. On the right-hand-side this value does not appear, since the limit operation removes it. The latter statement in your question is generally considered an acceptable shorthand, but it is a slight abuse of notation with respect this formal limit statement, since $n$ appears in the right-hand-side of the limiting symbol. So the reason it is not commonly used is that, formally speaking, the limit of a function taken with respect to $n$ cannot itself be a function of $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would it be correct then to state that the latter statement is wrong as statement of CLT but nevertheless a true property of $overline{X}_n$, for example using the continuos mapping theorem and the first CLT statement $stackrel{d}{rightarrow}N(0,sigma^2)$ ?
    $endgroup$
    – therealcode
    2 days ago








  • 1




    $begingroup$
    No. It is formally a false statement, and therefore not a true property of $bar{X}_n$. Arguably it may be interpreted as an abuse of notation for the above formal property.
    $endgroup$
    – Ben
    2 days ago











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: "65"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f386809%2fcentral-limit-theorem-clt-writing%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









7












$begingroup$

The notation $stackrel{d}{rightarrow}$ in the CLT is shorthand for the formal limit statement:



$$lim_{n rightarrow infty} mathbb{P} Big( sqrt{n} (bar{X}_n - mu) leqslant t Big) = Phi(t | 0, sigma^2).$$



You will notice that the formal statement is a limit statement in $n$ and so all of the values of $n$ are on the left-hand-side of the equation. On the right-hand-side this value does not appear, since the limit operation removes it. The latter statement in your question is generally considered an acceptable shorthand, but it is a slight abuse of notation with respect this formal limit statement, since $n$ appears in the right-hand-side of the limiting symbol. So the reason it is not commonly used is that, formally speaking, the limit of a function taken with respect to $n$ cannot itself be a function of $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would it be correct then to state that the latter statement is wrong as statement of CLT but nevertheless a true property of $overline{X}_n$, for example using the continuos mapping theorem and the first CLT statement $stackrel{d}{rightarrow}N(0,sigma^2)$ ?
    $endgroup$
    – therealcode
    2 days ago








  • 1




    $begingroup$
    No. It is formally a false statement, and therefore not a true property of $bar{X}_n$. Arguably it may be interpreted as an abuse of notation for the above formal property.
    $endgroup$
    – Ben
    2 days ago
















7












$begingroup$

The notation $stackrel{d}{rightarrow}$ in the CLT is shorthand for the formal limit statement:



$$lim_{n rightarrow infty} mathbb{P} Big( sqrt{n} (bar{X}_n - mu) leqslant t Big) = Phi(t | 0, sigma^2).$$



You will notice that the formal statement is a limit statement in $n$ and so all of the values of $n$ are on the left-hand-side of the equation. On the right-hand-side this value does not appear, since the limit operation removes it. The latter statement in your question is generally considered an acceptable shorthand, but it is a slight abuse of notation with respect this formal limit statement, since $n$ appears in the right-hand-side of the limiting symbol. So the reason it is not commonly used is that, formally speaking, the limit of a function taken with respect to $n$ cannot itself be a function of $n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would it be correct then to state that the latter statement is wrong as statement of CLT but nevertheless a true property of $overline{X}_n$, for example using the continuos mapping theorem and the first CLT statement $stackrel{d}{rightarrow}N(0,sigma^2)$ ?
    $endgroup$
    – therealcode
    2 days ago








  • 1




    $begingroup$
    No. It is formally a false statement, and therefore not a true property of $bar{X}_n$. Arguably it may be interpreted as an abuse of notation for the above formal property.
    $endgroup$
    – Ben
    2 days ago














7












7








7





$begingroup$

The notation $stackrel{d}{rightarrow}$ in the CLT is shorthand for the formal limit statement:



$$lim_{n rightarrow infty} mathbb{P} Big( sqrt{n} (bar{X}_n - mu) leqslant t Big) = Phi(t | 0, sigma^2).$$



You will notice that the formal statement is a limit statement in $n$ and so all of the values of $n$ are on the left-hand-side of the equation. On the right-hand-side this value does not appear, since the limit operation removes it. The latter statement in your question is generally considered an acceptable shorthand, but it is a slight abuse of notation with respect this formal limit statement, since $n$ appears in the right-hand-side of the limiting symbol. So the reason it is not commonly used is that, formally speaking, the limit of a function taken with respect to $n$ cannot itself be a function of $n$.






share|cite|improve this answer









$endgroup$



The notation $stackrel{d}{rightarrow}$ in the CLT is shorthand for the formal limit statement:



$$lim_{n rightarrow infty} mathbb{P} Big( sqrt{n} (bar{X}_n - mu) leqslant t Big) = Phi(t | 0, sigma^2).$$



You will notice that the formal statement is a limit statement in $n$ and so all of the values of $n$ are on the left-hand-side of the equation. On the right-hand-side this value does not appear, since the limit operation removes it. The latter statement in your question is generally considered an acceptable shorthand, but it is a slight abuse of notation with respect this formal limit statement, since $n$ appears in the right-hand-side of the limiting symbol. So the reason it is not commonly used is that, formally speaking, the limit of a function taken with respect to $n$ cannot itself be a function of $n$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









BenBen

22.4k224107




22.4k224107












  • $begingroup$
    Would it be correct then to state that the latter statement is wrong as statement of CLT but nevertheless a true property of $overline{X}_n$, for example using the continuos mapping theorem and the first CLT statement $stackrel{d}{rightarrow}N(0,sigma^2)$ ?
    $endgroup$
    – therealcode
    2 days ago








  • 1




    $begingroup$
    No. It is formally a false statement, and therefore not a true property of $bar{X}_n$. Arguably it may be interpreted as an abuse of notation for the above formal property.
    $endgroup$
    – Ben
    2 days ago


















  • $begingroup$
    Would it be correct then to state that the latter statement is wrong as statement of CLT but nevertheless a true property of $overline{X}_n$, for example using the continuos mapping theorem and the first CLT statement $stackrel{d}{rightarrow}N(0,sigma^2)$ ?
    $endgroup$
    – therealcode
    2 days ago








  • 1




    $begingroup$
    No. It is formally a false statement, and therefore not a true property of $bar{X}_n$. Arguably it may be interpreted as an abuse of notation for the above formal property.
    $endgroup$
    – Ben
    2 days ago
















$begingroup$
Would it be correct then to state that the latter statement is wrong as statement of CLT but nevertheless a true property of $overline{X}_n$, for example using the continuos mapping theorem and the first CLT statement $stackrel{d}{rightarrow}N(0,sigma^2)$ ?
$endgroup$
– therealcode
2 days ago






$begingroup$
Would it be correct then to state that the latter statement is wrong as statement of CLT but nevertheless a true property of $overline{X}_n$, for example using the continuos mapping theorem and the first CLT statement $stackrel{d}{rightarrow}N(0,sigma^2)$ ?
$endgroup$
– therealcode
2 days ago






1




1




$begingroup$
No. It is formally a false statement, and therefore not a true property of $bar{X}_n$. Arguably it may be interpreted as an abuse of notation for the above formal property.
$endgroup$
– Ben
2 days ago




$begingroup$
No. It is formally a false statement, and therefore not a true property of $bar{X}_n$. Arguably it may be interpreted as an abuse of notation for the above formal property.
$endgroup$
– Ben
2 days ago


















draft saved

draft discarded




















































Thanks for contributing an answer to Cross Validated!


  • 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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f386809%2fcentral-limit-theorem-clt-writing%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Paul Cézanne

Image
Pour l’article ayant un titre homophone, voir Sézanne. Paul Cézanne.mw-parser-output .entete.artiste{background-image:url("//upload.wikimedia.org/wikipedia/commons/f/f3/Picto_infobox_artiste.png")} Autoportrait vers 1861 Naissance 19 janvier 1839 Aix-en-Provence (France) Décès 22 octobre 1906 (à 67 ans) Aix-en-Provence (France) Nationalité Française Activité Peintre Formation Académie de Charles Suisse Maître Joseph Gilbert, Antoine Guillemet, Camille Pissarro Mouvement Impressionnisme postimpressionnisme Influencé par Eugène Delacroix, Gustave Courbet, Véronèse A influencé Braque, Picasso, Matisse Conjoint Hortense Fiquet (depuis 1886) Enfant Paul Cézanne  ( d ) Œuvres principales La Montagne Sainte-Victoire, Portrait de madame Cézanne signature modifier - modifier le code - modifier Wikidata Paul Cézanne , né le 19 janvier 1839 à Aix-en-Provence et mort le 22 octobre 1906 dans la mê...
Read more

UIScrollView CustomStickyHeader Resize height generates problems when scroll is too fast

Image
0 I have been trying for a while to create a UIViewController that has a CustomStickyHeader and a UITableView. The idea I have been following is use the UITableView scrolling for changing the header view with the value of the contentOffset. The layout for the ViewController (AlbumsViewController) looks like: The UItableView Scrolling functions from the delegate are implemented like that: func scrollViewDidScroll(_ scrollView: UIScrollView) { let endReached = contentTableView.contentOffset.y >= (contentTableView.contentSize.height - contentTableView.frame.size.height) let isUP = lastContentOffset > scrollView.contentOffset.y if scrollView.contentOffset.y < 0 { headerHeightConstraint.constant += abs(scrollView.contentOffset.y) } else{ if !endReached{ ...
Read more

Angular material date-picker (MatDatepicker) auto completes the date on focus out

Image
3 3 I am using angular material/mat-datepicker in my project and whenever the user enters one digit inside of it, and goes out of focus - the date picker is auto completing with a random date. Example Just enter 1 and go out of focus, or enter 5/5/20 The date picker is adding date automatically Update : Also I am using reactive forms and I'm validating the input with that approach Can I disable this ? angular datepicker angular-material angular6 angular-material-datetimepicker share | improve this question edited Nov 20 '18 at 13:21 ...
Read more