Central Limit Theorem different versions












0












$begingroup$


Which one is correct about the Central Limit Theorem:




  1. If $X_{1}, ... , X_{n}$ are i.i.d continuous random variables with mean $mu$ and variance $sigma^2$, then as $n rightarrow infty$,
    $$ sqrt{n} frac{bar{X} - mu}{sigma} $$ will have the standard normal distribution


or




  1. If $X_{1}, ... , X_{n}$ are i.i.d continuous random variables with mean $mu$ and variance $sigma^2$, then as $n rightarrow infty$,
    $$ sqrt{n} frac{bar{X} - mu}{sigma} $$ will have normal distribution $N(0, sigma^2)$


According to a book: "Introduction to Mathematical Statistics", by Hogg and Craig, the first one is true. But I also see from other sources that the second one is true.



Thanks.






statistics normal-distribution statistical-inference







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Work out the variance.
    $endgroup$
    – J.G.
    Dec 2 '18 at 7:49












  • $begingroup$
    @J.G. but it should be just one version that is true..
    $endgroup$
    – Arief Anbiya
    Dec 2 '18 at 7:57










  • $begingroup$
    Both versions are incorrect (for example, if the distribution of every $X_i$ is discrete, then none of the random variables $sqrt n(bar X_n-mu)/sigma$ is normally distributed).
    $endgroup$
    – Did
    Dec 2 '18 at 9:25












  • $begingroup$
    The above statement is itself incorrect. The conditions of this version of CLT is simply that $mathrm{var}(X)<infty$. CLT claims nothing about finite $sqrt{n}(bar{X}_n-mu)/sigma$ (you need the Berry-Esseen theorem for that), it simply says this sequence converges in distribution to a normal.
    $endgroup$
    – obscurans
    Dec 2 '18 at 23:59
















0












$begingroup$


Which one is correct about the Central Limit Theorem:




  1. If $X_{1}, ... , X_{n}$ are i.i.d continuous random variables with mean $mu$ and variance $sigma^2$, then as $n rightarrow infty$,
    $$ sqrt{n} frac{bar{X} - mu}{sigma} $$ will have the standard normal distribution


or




  1. If $X_{1}, ... , X_{n}$ are i.i.d continuous random variables with mean $mu$ and variance $sigma^2$, then as $n rightarrow infty$,
    $$ sqrt{n} frac{bar{X} - mu}{sigma} $$ will have normal distribution $N(0, sigma^2)$


According to a book: "Introduction to Mathematical Statistics", by Hogg and Craig, the first one is true. But I also see from other sources that the second one is true.



Thanks.






statistics normal-distribution statistical-inference







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Work out the variance.
    $endgroup$
    – J.G.
    Dec 2 '18 at 7:49












  • $begingroup$
    @J.G. but it should be just one version that is true..
    $endgroup$
    – Arief Anbiya
    Dec 2 '18 at 7:57










  • $begingroup$
    Both versions are incorrect (for example, if the distribution of every $X_i$ is discrete, then none of the random variables $sqrt n(bar X_n-mu)/sigma$ is normally distributed).
    $endgroup$
    – Did
    Dec 2 '18 at 9:25












  • $begingroup$
    The above statement is itself incorrect. The conditions of this version of CLT is simply that $mathrm{var}(X)<infty$. CLT claims nothing about finite $sqrt{n}(bar{X}_n-mu)/sigma$ (you need the Berry-Esseen theorem for that), it simply says this sequence converges in distribution to a normal.
    $endgroup$
    – obscurans
    Dec 2 '18 at 23:59














0












0








0





$begingroup$


Which one is correct about the Central Limit Theorem:




  1. If $X_{1}, ... , X_{n}$ are i.i.d continuous random variables with mean $mu$ and variance $sigma^2$, then as $n rightarrow infty$,
    $$ sqrt{n} frac{bar{X} - mu}{sigma} $$ will have the standard normal distribution


or




  1. If $X_{1}, ... , X_{n}$ are i.i.d continuous random variables with mean $mu$ and variance $sigma^2$, then as $n rightarrow infty$,
    $$ sqrt{n} frac{bar{X} - mu}{sigma} $$ will have normal distribution $N(0, sigma^2)$


According to a book: "Introduction to Mathematical Statistics", by Hogg and Craig, the first one is true. But I also see from other sources that the second one is true.



Thanks.






statistics normal-distribution statistical-inference







share|cite|improve this question











$endgroup$




Which one is correct about the Central Limit Theorem:




  1. If $X_{1}, ... , X_{n}$ are i.i.d continuous random variables with mean $mu$ and variance $sigma^2$, then as $n rightarrow infty$,
    $$ sqrt{n} frac{bar{X} - mu}{sigma} $$ will have the standard normal distribution


or




  1. If $X_{1}, ... , X_{n}$ are i.i.d continuous random variables with mean $mu$ and variance $sigma^2$, then as $n rightarrow infty$,
    $$ sqrt{n} frac{bar{X} - mu}{sigma} $$ will have normal distribution $N(0, sigma^2)$


According to a book: "Introduction to Mathematical Statistics", by Hogg and Craig, the first one is true. But I also see from other sources that the second one is true.



Thanks.






statistics normal-distribution statistical-inference




statistics normal-distribution statistical-inference






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 9:29







Arief Anbiya

















asked Dec 2 '18 at 7:43









Arief AnbiyaArief Anbiya

1,3601622




1,3601622








  • 1




    $begingroup$
    Work out the variance.
    $endgroup$
    – J.G.
    Dec 2 '18 at 7:49












  • $begingroup$
    @J.G. but it should be just one version that is true..
    $endgroup$
    – Arief Anbiya
    Dec 2 '18 at 7:57










  • $begingroup$
    Both versions are incorrect (for example, if the distribution of every $X_i$ is discrete, then none of the random variables $sqrt n(bar X_n-mu)/sigma$ is normally distributed).
    $endgroup$
    – Did
    Dec 2 '18 at 9:25












  • $begingroup$
    The above statement is itself incorrect. The conditions of this version of CLT is simply that $mathrm{var}(X)<infty$. CLT claims nothing about finite $sqrt{n}(bar{X}_n-mu)/sigma$ (you need the Berry-Esseen theorem for that), it simply says this sequence converges in distribution to a normal.
    $endgroup$
    – obscurans
    Dec 2 '18 at 23:59














  • 1




    $begingroup$
    Work out the variance.
    $endgroup$
    – J.G.
    Dec 2 '18 at 7:49












  • $begingroup$
    @J.G. but it should be just one version that is true..
    $endgroup$
    – Arief Anbiya
    Dec 2 '18 at 7:57










  • $begingroup$
    Both versions are incorrect (for example, if the distribution of every $X_i$ is discrete, then none of the random variables $sqrt n(bar X_n-mu)/sigma$ is normally distributed).
    $endgroup$
    – Did
    Dec 2 '18 at 9:25












  • $begingroup$
    The above statement is itself incorrect. The conditions of this version of CLT is simply that $mathrm{var}(X)<infty$. CLT claims nothing about finite $sqrt{n}(bar{X}_n-mu)/sigma$ (you need the Berry-Esseen theorem for that), it simply says this sequence converges in distribution to a normal.
    $endgroup$
    – obscurans
    Dec 2 '18 at 23:59








1




1




$begingroup$
Work out the variance.
$endgroup$
– J.G.
Dec 2 '18 at 7:49






$begingroup$
Work out the variance.
$endgroup$
– J.G.
Dec 2 '18 at 7:49














$begingroup$
@J.G. but it should be just one version that is true..
$endgroup$
– Arief Anbiya
Dec 2 '18 at 7:57




$begingroup$
@J.G. but it should be just one version that is true..
$endgroup$
– Arief Anbiya
Dec 2 '18 at 7:57












$begingroup$
Both versions are incorrect (for example, if the distribution of every $X_i$ is discrete, then none of the random variables $sqrt n(bar X_n-mu)/sigma$ is normally distributed).
$endgroup$
– Did
Dec 2 '18 at 9:25






$begingroup$
Both versions are incorrect (for example, if the distribution of every $X_i$ is discrete, then none of the random variables $sqrt n(bar X_n-mu)/sigma$ is normally distributed).
$endgroup$
– Did
Dec 2 '18 at 9:25














$begingroup$
The above statement is itself incorrect. The conditions of this version of CLT is simply that $mathrm{var}(X)<infty$. CLT claims nothing about finite $sqrt{n}(bar{X}_n-mu)/sigma$ (you need the Berry-Esseen theorem for that), it simply says this sequence converges in distribution to a normal.
$endgroup$
– obscurans
Dec 2 '18 at 23:59




$begingroup$
The above statement is itself incorrect. The conditions of this version of CLT is simply that $mathrm{var}(X)<infty$. CLT claims nothing about finite $sqrt{n}(bar{X}_n-mu)/sigma$ (you need the Berry-Esseen theorem for that), it simply says this sequence converges in distribution to a normal.
$endgroup$
– obscurans
Dec 2 '18 at 23:59










1 Answer
1






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oldest

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1












$begingroup$

Since $X_i$ are iid, we know that
$$mathrm{var}left(sqrt{n}frac{bar{X}-mu}{sigma}right)=frac{n}{sigma^2}mathrm{var}left(bar{X}-muright)=frac{n}{sigma^2}mathrm{var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{n}{sigma^2}frac{1}{n^2}(nsigma^2)=1$$
so the limit is a standard normal.



Version 1 is correct, version 2 is wrong.






share|cite|improve this answer











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    $begingroup$

    Since $X_i$ are iid, we know that
    $$mathrm{var}left(sqrt{n}frac{bar{X}-mu}{sigma}right)=frac{n}{sigma^2}mathrm{var}left(bar{X}-muright)=frac{n}{sigma^2}mathrm{var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{n}{sigma^2}frac{1}{n^2}(nsigma^2)=1$$
    so the limit is a standard normal.



    Version 1 is correct, version 2 is wrong.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Since $X_i$ are iid, we know that
      $$mathrm{var}left(sqrt{n}frac{bar{X}-mu}{sigma}right)=frac{n}{sigma^2}mathrm{var}left(bar{X}-muright)=frac{n}{sigma^2}mathrm{var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{n}{sigma^2}frac{1}{n^2}(nsigma^2)=1$$
      so the limit is a standard normal.



      Version 1 is correct, version 2 is wrong.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Since $X_i$ are iid, we know that
        $$mathrm{var}left(sqrt{n}frac{bar{X}-mu}{sigma}right)=frac{n}{sigma^2}mathrm{var}left(bar{X}-muright)=frac{n}{sigma^2}mathrm{var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{n}{sigma^2}frac{1}{n^2}(nsigma^2)=1$$
        so the limit is a standard normal.



        Version 1 is correct, version 2 is wrong.






        share|cite|improve this answer











        $endgroup$



        Since $X_i$ are iid, we know that
        $$mathrm{var}left(sqrt{n}frac{bar{X}-mu}{sigma}right)=frac{n}{sigma^2}mathrm{var}left(bar{X}-muright)=frac{n}{sigma^2}mathrm{var}left(frac{1}{n}sum_{i=1}^nX_iright)=frac{n}{sigma^2}frac{1}{n^2}(nsigma^2)=1$$
        so the limit is a standard normal.



        Version 1 is correct, version 2 is wrong.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 2 '18 at 7:58

























        answered Dec 2 '18 at 7:52









        obscuransobscurans

        1,027311




        1,027311






























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