Central limit theorem (CLT) writing












5












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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 $overline{X}_nstackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?










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$endgroup$








  • 1




    $begingroup$
    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
    $endgroup$
    – Neeraj
    Jan 12 at 9:21








  • 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
    Jan 12 at 9:30








  • 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
    Jan 12 at 9:34






  • 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
    Jan 12 at 10:32
















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 $overline{X}_nstackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
    $endgroup$
    – Neeraj
    Jan 12 at 9:21








  • 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
    Jan 12 at 9:30








  • 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
    Jan 12 at 9:34






  • 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
    Jan 12 at 10:32














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 $overline{X}_nstackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?










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






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edited Jan 12 at 21:57









Peter Mortensen

19718




19718










asked Jan 12 at 9:16









therealcodetherealcode

333




333








  • 1




    $begingroup$
    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
    $endgroup$
    – Neeraj
    Jan 12 at 9:21








  • 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
    Jan 12 at 9:30








  • 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
    Jan 12 at 9:34






  • 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
    Jan 12 at 10:32














  • 1




    $begingroup$
    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
    $endgroup$
    – Neeraj
    Jan 12 at 9:21








  • 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
    Jan 12 at 9:30








  • 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
    Jan 12 at 9:34






  • 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
    Jan 12 at 10:32








1




1




$begingroup$
Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
$endgroup$
– Neeraj
Jan 12 at 9:21






$begingroup$
Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
$endgroup$
– Neeraj
Jan 12 at 9:21






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
Jan 12 at 9:30






$begingroup$
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
Jan 12 at 9:30






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
Jan 12 at 9:34




$begingroup$
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
Jan 12 at 9:34




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
Jan 12 at 10:32




$begingroup$
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
Jan 12 at 10:32










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
    Jan 12 at 9:49








  • 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
    Jan 12 at 9:51











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1 Answer
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1 Answer
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active

oldest

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active

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active

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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
    Jan 12 at 9:49








  • 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
    Jan 12 at 9:51
















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
    Jan 12 at 9:49








  • 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
    Jan 12 at 9:51














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 Jan 12 at 9:25









BenBen

22.9k224110




22.9k224110












  • $begingroup$
    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
    Jan 12 at 9:49








  • 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
    Jan 12 at 9:51


















  • $begingroup$
    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
    Jan 12 at 9:49








  • 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
    Jan 12 at 9:51
















$begingroup$
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
Jan 12 at 9:49






$begingroup$
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
Jan 12 at 9:49






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
Jan 12 at 9:51




$begingroup$
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
Jan 12 at 9:51


















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