central limit theorem writing












2















Is there any 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 $overline{X}stackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?






distributions convergence central-limit-theorem







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  • 1





    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473

    – Neeraj
    2 hours ago








  • 1





    Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing

    – therealcode
    1 hour ago








  • 1





    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.

    – Ben
    1 hour ago






  • 1





    Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.

    – Xi'an
    52 mins ago
















2















Is there any 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 $overline{X}stackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?






distributions convergence central-limit-theorem







share|cite|improve this question


















  • 1





    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473

    – Neeraj
    2 hours ago








  • 1





    Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing

    – therealcode
    1 hour ago








  • 1





    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.

    – Ben
    1 hour ago






  • 1





    Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.

    – Xi'an
    52 mins ago














2












2








2








Is there any 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 $overline{X}stackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?






distributions convergence central-limit-theorem







share|cite|improve this question














Is there any 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 $overline{X}stackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?






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










asked 2 hours ago









therealcodetherealcode

183




183








  • 1





    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473

    – Neeraj
    2 hours ago








  • 1





    Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing

    – therealcode
    1 hour ago








  • 1





    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.

    – Ben
    1 hour ago






  • 1





    Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.

    – Xi'an
    52 mins ago














  • 1





    Because second is not true. Read this: stats.stackexchange.com/a/194331/90473

    – Neeraj
    2 hours ago








  • 1





    Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing

    – therealcode
    1 hour ago








  • 1





    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.

    – Ben
    1 hour ago






  • 1





    Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.

    – Xi'an
    52 mins ago








1




1





Because second is not true. Read this: stats.stackexchange.com/a/194331/90473

– Neeraj
2 hours ago







Because second is not true. Read this: stats.stackexchange.com/a/194331/90473

– Neeraj
2 hours ago






1




1





Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing

– therealcode
1 hour ago







Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing

– therealcode
1 hour ago






1




1





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.

– Ben
1 hour ago





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.

– Ben
1 hour ago




1




1





Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.

– Xi'an
52 mins ago





Simply because the term on the right hand side is the limit as $n$ goes to $infty$ and therefore cannot depend on $n$.

– Xi'an
52 mins ago










1 Answer
1






active

oldest

votes


















3














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
























  • 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 $stackrel{d}{rightarrow}N(0,sigma^2)$ ?

    – therealcode
    1 hour ago








  • 1





    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.

    – Ben
    1 hour ago











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1 Answer
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1 Answer
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3














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
























  • 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 $stackrel{d}{rightarrow}N(0,sigma^2)$ ?

    – therealcode
    1 hour ago








  • 1





    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.

    – Ben
    1 hour ago
















3














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
























  • 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 $stackrel{d}{rightarrow}N(0,sigma^2)$ ?

    – therealcode
    1 hour ago








  • 1





    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.

    – Ben
    1 hour ago














3












3








3







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













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 1 hour ago









BenBen

22.2k224106




22.2k224106













  • 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 $stackrel{d}{rightarrow}N(0,sigma^2)$ ?

    – therealcode
    1 hour ago








  • 1





    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.

    – Ben
    1 hour ago



















  • 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 $stackrel{d}{rightarrow}N(0,sigma^2)$ ?

    – therealcode
    1 hour ago








  • 1





    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.

    – Ben
    1 hour ago

















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 $stackrel{d}{rightarrow}N(0,sigma^2)$ ?

– therealcode
1 hour ago







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 $stackrel{d}{rightarrow}N(0,sigma^2)$ ?

– therealcode
1 hour ago






1




1





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.

– Ben
1 hour ago





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.

– Ben
1 hour ago


















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