Central limit theorem (CLT) writing
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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 $overline{X}_nstackrel{d}{rightarrow}N(mu, frac{sigma^2}{n})$?
distributions convergence central-limit-theorem
edited Jan 12 at 21:57
Peter Mortensen
19718
asked Jan 12 at 9:16
therealcodetherealcode
333
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add a comment |
$begingroup$
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
edited Jan 12 at 21:57
Peter Mortensen
19718
asked Jan 12 at 9:16
therealcodetherealcode
333
$endgroup$
1
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Because second is not true. Read this: stats.stackexchange.com/a/194331/90473
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– Neeraj
Jan 12 at 9:21
1
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Have a look at the Wikipedia article on CLT>Remarks>Proof of classical CLT. At the bottom you will find exactly this writing
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– therealcode
Jan 12 at 9:30
1
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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.
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– Ben
Jan 12 at 9:34
3
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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
add a comment |
$begingroup$
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
edited Jan 12 at 21:57
Peter Mortensen
19718
asked Jan 12 at 9:16
therealcodetherealcode
333
$endgroup$
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
edited Jan 12 at 21:57
Peter Mortensen
19718
asked Jan 12 at 9:16
therealcodetherealcode
333
edited Jan 12 at 21:57
Peter Mortensen
19718
asked Jan 12 at 9:16
therealcodetherealcode
333
edited Jan 12 at 21:57
Peter Mortensen
19718
edited Jan 12 at 21:57
Peter Mortensen
19718
edited Jan 12 at 21:57
Peter Mortensen
19718
19718
asked Jan 12 at 9:16
therealcodetherealcode
333
asked Jan 12 at 9:16
therealcodetherealcode
333
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $n$.
answered Jan 12 at 9:25
BenBen
22.9k224110
$endgroup$
$begingroup$
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 $stackrel{d}{rightarrow}N(0,sigma^2)$ ?
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– 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
add a comment |
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1 Answer
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$begingroup$
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 $n$.
answered Jan 12 at 9:25
BenBen
22.9k224110
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $n$.
answered Jan 12 at 9:25
BenBen
22.9k224110
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $n$.
answered Jan 12 at 9:25
BenBen
22.9k224110
$endgroup$
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 $n$.
answered Jan 12 at 9:25
BenBen
22.9k224110
answered Jan 12 at 9:25
BenBen
22.9k224110
answered Jan 12 at 9:25
BenBen
22.9k224110
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
add a comment |
$begingroup$
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
add a comment |
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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