What is the reason that Student-t Distribution is used when the number of samples is small












0












$begingroup$


Let $bar{X}$ be the distribution of sample mean for $n$ identical and independent distributed as Normal distributions $N(mu, sigma^2)$.



The random variable
$$ frac{bar{X} - mu}{frac{sigma}{sqrt{n}}} $$
has standard normal distribution. Now let
$$ S^2 =frac{1}{n-1} sum_{i=1}^{n} (X_{i} - bar{X})^2, $$
then random variable
$$ frac{bar{X} - mu}{frac{S}{sqrt{n}}} $$
has student-t distribution with $n-1$ degrees of freedom.
From this, we can conclude then when $n$ is large, the random variable above will converge to the standard normal distribution,
$$ lim_{n rightarrow infty} S^2 = lim_{n rightarrow infty} frac{1}{n-1} sum_{i=1}^{n} (X_{i} - bar{X})^2 = lim_{n rightarrow infty} frac{1}{n} sum_{i=1}^{n} (X_{i} - mu)^2 = sigma^{2} $$



But why we should choose student-t distribution when the sample size is small..? What is the mathematical explanation..? thanks.










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$endgroup$












  • $begingroup$
    You provided the mathematical explanation already, namely $(bar{X}-mu)/frac{S}{sqrt{n}}sim t_{n-1}$.
    $endgroup$
    – user10354138
    Nov 10 '18 at 7:51










  • $begingroup$
    @user10354138 Yes, but that does not explain why we use student-t when $n$ is small..
    $endgroup$
    – Arief Anbiya
    Nov 10 '18 at 7:55










  • $begingroup$
    Why not? The question should be why we could avoid it when $n$ is large, not why we use it when $n$ is small.
    $endgroup$
    – user10354138
    Nov 10 '18 at 7:57










  • $begingroup$
    @user10354138 Is it because we have no other choices? and student-t resembles Normal distribution which makes it an approximation to the Normal distribution..?
    $endgroup$
    – Arief Anbiya
    Nov 10 '18 at 7:59
















0












$begingroup$


Let $bar{X}$ be the distribution of sample mean for $n$ identical and independent distributed as Normal distributions $N(mu, sigma^2)$.



The random variable
$$ frac{bar{X} - mu}{frac{sigma}{sqrt{n}}} $$
has standard normal distribution. Now let
$$ S^2 =frac{1}{n-1} sum_{i=1}^{n} (X_{i} - bar{X})^2, $$
then random variable
$$ frac{bar{X} - mu}{frac{S}{sqrt{n}}} $$
has student-t distribution with $n-1$ degrees of freedom.
From this, we can conclude then when $n$ is large, the random variable above will converge to the standard normal distribution,
$$ lim_{n rightarrow infty} S^2 = lim_{n rightarrow infty} frac{1}{n-1} sum_{i=1}^{n} (X_{i} - bar{X})^2 = lim_{n rightarrow infty} frac{1}{n} sum_{i=1}^{n} (X_{i} - mu)^2 = sigma^{2} $$



But why we should choose student-t distribution when the sample size is small..? What is the mathematical explanation..? thanks.










share|cite|improve this question









$endgroup$












  • $begingroup$
    You provided the mathematical explanation already, namely $(bar{X}-mu)/frac{S}{sqrt{n}}sim t_{n-1}$.
    $endgroup$
    – user10354138
    Nov 10 '18 at 7:51










  • $begingroup$
    @user10354138 Yes, but that does not explain why we use student-t when $n$ is small..
    $endgroup$
    – Arief Anbiya
    Nov 10 '18 at 7:55










  • $begingroup$
    Why not? The question should be why we could avoid it when $n$ is large, not why we use it when $n$ is small.
    $endgroup$
    – user10354138
    Nov 10 '18 at 7:57










  • $begingroup$
    @user10354138 Is it because we have no other choices? and student-t resembles Normal distribution which makes it an approximation to the Normal distribution..?
    $endgroup$
    – Arief Anbiya
    Nov 10 '18 at 7:59














0












0








0


2



$begingroup$


Let $bar{X}$ be the distribution of sample mean for $n$ identical and independent distributed as Normal distributions $N(mu, sigma^2)$.



The random variable
$$ frac{bar{X} - mu}{frac{sigma}{sqrt{n}}} $$
has standard normal distribution. Now let
$$ S^2 =frac{1}{n-1} sum_{i=1}^{n} (X_{i} - bar{X})^2, $$
then random variable
$$ frac{bar{X} - mu}{frac{S}{sqrt{n}}} $$
has student-t distribution with $n-1$ degrees of freedom.
From this, we can conclude then when $n$ is large, the random variable above will converge to the standard normal distribution,
$$ lim_{n rightarrow infty} S^2 = lim_{n rightarrow infty} frac{1}{n-1} sum_{i=1}^{n} (X_{i} - bar{X})^2 = lim_{n rightarrow infty} frac{1}{n} sum_{i=1}^{n} (X_{i} - mu)^2 = sigma^{2} $$



But why we should choose student-t distribution when the sample size is small..? What is the mathematical explanation..? thanks.










share|cite|improve this question









$endgroup$




Let $bar{X}$ be the distribution of sample mean for $n$ identical and independent distributed as Normal distributions $N(mu, sigma^2)$.



The random variable
$$ frac{bar{X} - mu}{frac{sigma}{sqrt{n}}} $$
has standard normal distribution. Now let
$$ S^2 =frac{1}{n-1} sum_{i=1}^{n} (X_{i} - bar{X})^2, $$
then random variable
$$ frac{bar{X} - mu}{frac{S}{sqrt{n}}} $$
has student-t distribution with $n-1$ degrees of freedom.
From this, we can conclude then when $n$ is large, the random variable above will converge to the standard normal distribution,
$$ lim_{n rightarrow infty} S^2 = lim_{n rightarrow infty} frac{1}{n-1} sum_{i=1}^{n} (X_{i} - bar{X})^2 = lim_{n rightarrow infty} frac{1}{n} sum_{i=1}^{n} (X_{i} - mu)^2 = sigma^{2} $$



But why we should choose student-t distribution when the sample size is small..? What is the mathematical explanation..? thanks.







normal-distribution statistical-inference






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asked Nov 10 '18 at 7:25









Arief AnbiyaArief Anbiya

1,3601622




1,3601622












  • $begingroup$
    You provided the mathematical explanation already, namely $(bar{X}-mu)/frac{S}{sqrt{n}}sim t_{n-1}$.
    $endgroup$
    – user10354138
    Nov 10 '18 at 7:51










  • $begingroup$
    @user10354138 Yes, but that does not explain why we use student-t when $n$ is small..
    $endgroup$
    – Arief Anbiya
    Nov 10 '18 at 7:55










  • $begingroup$
    Why not? The question should be why we could avoid it when $n$ is large, not why we use it when $n$ is small.
    $endgroup$
    – user10354138
    Nov 10 '18 at 7:57










  • $begingroup$
    @user10354138 Is it because we have no other choices? and student-t resembles Normal distribution which makes it an approximation to the Normal distribution..?
    $endgroup$
    – Arief Anbiya
    Nov 10 '18 at 7:59


















  • $begingroup$
    You provided the mathematical explanation already, namely $(bar{X}-mu)/frac{S}{sqrt{n}}sim t_{n-1}$.
    $endgroup$
    – user10354138
    Nov 10 '18 at 7:51










  • $begingroup$
    @user10354138 Yes, but that does not explain why we use student-t when $n$ is small..
    $endgroup$
    – Arief Anbiya
    Nov 10 '18 at 7:55










  • $begingroup$
    Why not? The question should be why we could avoid it when $n$ is large, not why we use it when $n$ is small.
    $endgroup$
    – user10354138
    Nov 10 '18 at 7:57










  • $begingroup$
    @user10354138 Is it because we have no other choices? and student-t resembles Normal distribution which makes it an approximation to the Normal distribution..?
    $endgroup$
    – Arief Anbiya
    Nov 10 '18 at 7:59
















$begingroup$
You provided the mathematical explanation already, namely $(bar{X}-mu)/frac{S}{sqrt{n}}sim t_{n-1}$.
$endgroup$
– user10354138
Nov 10 '18 at 7:51




$begingroup$
You provided the mathematical explanation already, namely $(bar{X}-mu)/frac{S}{sqrt{n}}sim t_{n-1}$.
$endgroup$
– user10354138
Nov 10 '18 at 7:51












$begingroup$
@user10354138 Yes, but that does not explain why we use student-t when $n$ is small..
$endgroup$
– Arief Anbiya
Nov 10 '18 at 7:55




$begingroup$
@user10354138 Yes, but that does not explain why we use student-t when $n$ is small..
$endgroup$
– Arief Anbiya
Nov 10 '18 at 7:55












$begingroup$
Why not? The question should be why we could avoid it when $n$ is large, not why we use it when $n$ is small.
$endgroup$
– user10354138
Nov 10 '18 at 7:57




$begingroup$
Why not? The question should be why we could avoid it when $n$ is large, not why we use it when $n$ is small.
$endgroup$
– user10354138
Nov 10 '18 at 7:57












$begingroup$
@user10354138 Is it because we have no other choices? and student-t resembles Normal distribution which makes it an approximation to the Normal distribution..?
$endgroup$
– Arief Anbiya
Nov 10 '18 at 7:59




$begingroup$
@user10354138 Is it because we have no other choices? and student-t resembles Normal distribution which makes it an approximation to the Normal distribution..?
$endgroup$
– Arief Anbiya
Nov 10 '18 at 7:59










3 Answers
3






active

oldest

votes


















1












$begingroup$

The t-test is based on a Student's t-distribution which is sensitive to the number of observations. Furthermore, a t-statistic is calculated for small sample sizes where you do not know the population standard deviation. Even in the case of a large sample, we likely do not "know" the population standard deviation, but there are some nice results that help here.



The law of large numbers (LLN) gives us the result that sample averages converge in probability to the population average (formally, $bar{x} to mu$ as $n to infty$). That is, that as the sample size increases, the sample average gets closer and closer to the population average.



The central limit theorem (CLT) describes how the distribution of the difference in the sample and population averages ($bar{x}-mu$) changes with respect to the sample size $n$. Ultimately, what it tells us is that for sufficiently large $n$, this distribution approximates the normal distribution, $N(0,sigma^2 / n)$. Manipulating this expression, we can show the following
$$ bar{x}-mu sim N(0,sigma^2 / n) implies Z = frac{bar{x}-mu}{sigma/sqrt{n}} sim N(0,1) $$



So, in summary, the reason we can use a normal distribution for large samples is due to these results regarding convergence in distribution. This convergence happens fairly quickly (typically $nge30$ is sufficient), but for smaller samples these results do not hold and the Student's t-distribution is more appropriate.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The $t$ distribution is exact for all $n$. Therefore, we must use it or an adequate approximation. The Normal distribution is only a valid approximation for large $n$. The exact pdf is $frac{Gamma (frac{nu+1}{2})}{sqrt{nupi}Gamma(frac{nu}{2})}(1+frac{x^2}{nu})^{-frac{nu+1}{2}}$ with $nu=n-1$. For large $n$, this approximates $sqrt{frac{Gamma (frac{nu}{2}+1)}{Gamma (frac{nu}{2})nupi}}exp-frac{x^2}{2}=frac{1}{sqrt{2pi}}exp-frac{x^2}{2}$, the $N(0,,1)$ pdf.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      I would like to post my view relating to @Chris answer also.



      By the Central Limit Theorem, for large $m$, random var $X_{i}$ with mean $mu$ and st.dev $sigma$ from any distribution ($X_{i}$s have same dist) will have $$ sqrt{m} frac{ bar{X} - mu }{ sigma } $$ converge to standard normal. Also, the random.var $bar{X} - mu$ will be normal with mean $0$.



      Now.., if $X_{i}$s are normal, then distribution
      $$ T = sqrt{m} frac{bar{X} - mu}{S} $$ has student-t distribution with $m-1$ degrees of freedom. But we can only approximate the CLT using student-t when $X_{i}$s are normal.



      Now I have an argument:
      If $X_{i}$s are from the same distribution (any dist) with mean $mu$, then $bar{X} - mu$ will converge to normal with mean $0$ as $m$ gets large. So that
      $$ sqrt{m}frac{frac{sum (bar{X}_{i} - mu)}{m} - 0}{S} $$
      will have student-t distribution, and we can rewrite it as
      $$ sqrt{m}frac{frac{sum (bar{X}_{i} )}{m} - mu}{S} $$
      and for large $m$, $frac{sum (bar{X}_{i} )}{m}$ will converger to $bar{X}$. So we can still use $T$ as approximation even when $X_{i}$s are not normal.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
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        active

        oldest

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        3 Answers
        3






        active

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        active

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        active

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        1












        $begingroup$

        The t-test is based on a Student's t-distribution which is sensitive to the number of observations. Furthermore, a t-statistic is calculated for small sample sizes where you do not know the population standard deviation. Even in the case of a large sample, we likely do not "know" the population standard deviation, but there are some nice results that help here.



        The law of large numbers (LLN) gives us the result that sample averages converge in probability to the population average (formally, $bar{x} to mu$ as $n to infty$). That is, that as the sample size increases, the sample average gets closer and closer to the population average.



        The central limit theorem (CLT) describes how the distribution of the difference in the sample and population averages ($bar{x}-mu$) changes with respect to the sample size $n$. Ultimately, what it tells us is that for sufficiently large $n$, this distribution approximates the normal distribution, $N(0,sigma^2 / n)$. Manipulating this expression, we can show the following
        $$ bar{x}-mu sim N(0,sigma^2 / n) implies Z = frac{bar{x}-mu}{sigma/sqrt{n}} sim N(0,1) $$



        So, in summary, the reason we can use a normal distribution for large samples is due to these results regarding convergence in distribution. This convergence happens fairly quickly (typically $nge30$ is sufficient), but for smaller samples these results do not hold and the Student's t-distribution is more appropriate.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          The t-test is based on a Student's t-distribution which is sensitive to the number of observations. Furthermore, a t-statistic is calculated for small sample sizes where you do not know the population standard deviation. Even in the case of a large sample, we likely do not "know" the population standard deviation, but there are some nice results that help here.



          The law of large numbers (LLN) gives us the result that sample averages converge in probability to the population average (formally, $bar{x} to mu$ as $n to infty$). That is, that as the sample size increases, the sample average gets closer and closer to the population average.



          The central limit theorem (CLT) describes how the distribution of the difference in the sample and population averages ($bar{x}-mu$) changes with respect to the sample size $n$. Ultimately, what it tells us is that for sufficiently large $n$, this distribution approximates the normal distribution, $N(0,sigma^2 / n)$. Manipulating this expression, we can show the following
          $$ bar{x}-mu sim N(0,sigma^2 / n) implies Z = frac{bar{x}-mu}{sigma/sqrt{n}} sim N(0,1) $$



          So, in summary, the reason we can use a normal distribution for large samples is due to these results regarding convergence in distribution. This convergence happens fairly quickly (typically $nge30$ is sufficient), but for smaller samples these results do not hold and the Student's t-distribution is more appropriate.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            The t-test is based on a Student's t-distribution which is sensitive to the number of observations. Furthermore, a t-statistic is calculated for small sample sizes where you do not know the population standard deviation. Even in the case of a large sample, we likely do not "know" the population standard deviation, but there are some nice results that help here.



            The law of large numbers (LLN) gives us the result that sample averages converge in probability to the population average (formally, $bar{x} to mu$ as $n to infty$). That is, that as the sample size increases, the sample average gets closer and closer to the population average.



            The central limit theorem (CLT) describes how the distribution of the difference in the sample and population averages ($bar{x}-mu$) changes with respect to the sample size $n$. Ultimately, what it tells us is that for sufficiently large $n$, this distribution approximates the normal distribution, $N(0,sigma^2 / n)$. Manipulating this expression, we can show the following
            $$ bar{x}-mu sim N(0,sigma^2 / n) implies Z = frac{bar{x}-mu}{sigma/sqrt{n}} sim N(0,1) $$



            So, in summary, the reason we can use a normal distribution for large samples is due to these results regarding convergence in distribution. This convergence happens fairly quickly (typically $nge30$ is sufficient), but for smaller samples these results do not hold and the Student's t-distribution is more appropriate.






            share|cite|improve this answer









            $endgroup$



            The t-test is based on a Student's t-distribution which is sensitive to the number of observations. Furthermore, a t-statistic is calculated for small sample sizes where you do not know the population standard deviation. Even in the case of a large sample, we likely do not "know" the population standard deviation, but there are some nice results that help here.



            The law of large numbers (LLN) gives us the result that sample averages converge in probability to the population average (formally, $bar{x} to mu$ as $n to infty$). That is, that as the sample size increases, the sample average gets closer and closer to the population average.



            The central limit theorem (CLT) describes how the distribution of the difference in the sample and population averages ($bar{x}-mu$) changes with respect to the sample size $n$. Ultimately, what it tells us is that for sufficiently large $n$, this distribution approximates the normal distribution, $N(0,sigma^2 / n)$. Manipulating this expression, we can show the following
            $$ bar{x}-mu sim N(0,sigma^2 / n) implies Z = frac{bar{x}-mu}{sigma/sqrt{n}} sim N(0,1) $$



            So, in summary, the reason we can use a normal distribution for large samples is due to these results regarding convergence in distribution. This convergence happens fairly quickly (typically $nge30$ is sufficient), but for smaller samples these results do not hold and the Student's t-distribution is more appropriate.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 10 '18 at 8:49









            ChrisChris

            362




            362























                1












                $begingroup$

                The $t$ distribution is exact for all $n$. Therefore, we must use it or an adequate approximation. The Normal distribution is only a valid approximation for large $n$. The exact pdf is $frac{Gamma (frac{nu+1}{2})}{sqrt{nupi}Gamma(frac{nu}{2})}(1+frac{x^2}{nu})^{-frac{nu+1}{2}}$ with $nu=n-1$. For large $n$, this approximates $sqrt{frac{Gamma (frac{nu}{2}+1)}{Gamma (frac{nu}{2})nupi}}exp-frac{x^2}{2}=frac{1}{sqrt{2pi}}exp-frac{x^2}{2}$, the $N(0,,1)$ pdf.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  The $t$ distribution is exact for all $n$. Therefore, we must use it or an adequate approximation. The Normal distribution is only a valid approximation for large $n$. The exact pdf is $frac{Gamma (frac{nu+1}{2})}{sqrt{nupi}Gamma(frac{nu}{2})}(1+frac{x^2}{nu})^{-frac{nu+1}{2}}$ with $nu=n-1$. For large $n$, this approximates $sqrt{frac{Gamma (frac{nu}{2}+1)}{Gamma (frac{nu}{2})nupi}}exp-frac{x^2}{2}=frac{1}{sqrt{2pi}}exp-frac{x^2}{2}$, the $N(0,,1)$ pdf.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    The $t$ distribution is exact for all $n$. Therefore, we must use it or an adequate approximation. The Normal distribution is only a valid approximation for large $n$. The exact pdf is $frac{Gamma (frac{nu+1}{2})}{sqrt{nupi}Gamma(frac{nu}{2})}(1+frac{x^2}{nu})^{-frac{nu+1}{2}}$ with $nu=n-1$. For large $n$, this approximates $sqrt{frac{Gamma (frac{nu}{2}+1)}{Gamma (frac{nu}{2})nupi}}exp-frac{x^2}{2}=frac{1}{sqrt{2pi}}exp-frac{x^2}{2}$, the $N(0,,1)$ pdf.






                    share|cite|improve this answer









                    $endgroup$



                    The $t$ distribution is exact for all $n$. Therefore, we must use it or an adequate approximation. The Normal distribution is only a valid approximation for large $n$. The exact pdf is $frac{Gamma (frac{nu+1}{2})}{sqrt{nupi}Gamma(frac{nu}{2})}(1+frac{x^2}{nu})^{-frac{nu+1}{2}}$ with $nu=n-1$. For large $n$, this approximates $sqrt{frac{Gamma (frac{nu}{2}+1)}{Gamma (frac{nu}{2})nupi}}exp-frac{x^2}{2}=frac{1}{sqrt{2pi}}exp-frac{x^2}{2}$, the $N(0,,1)$ pdf.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 10 '18 at 9:21









                    J.G.J.G.

                    26.1k22539




                    26.1k22539























                        0












                        $begingroup$

                        I would like to post my view relating to @Chris answer also.



                        By the Central Limit Theorem, for large $m$, random var $X_{i}$ with mean $mu$ and st.dev $sigma$ from any distribution ($X_{i}$s have same dist) will have $$ sqrt{m} frac{ bar{X} - mu }{ sigma } $$ converge to standard normal. Also, the random.var $bar{X} - mu$ will be normal with mean $0$.



                        Now.., if $X_{i}$s are normal, then distribution
                        $$ T = sqrt{m} frac{bar{X} - mu}{S} $$ has student-t distribution with $m-1$ degrees of freedom. But we can only approximate the CLT using student-t when $X_{i}$s are normal.



                        Now I have an argument:
                        If $X_{i}$s are from the same distribution (any dist) with mean $mu$, then $bar{X} - mu$ will converge to normal with mean $0$ as $m$ gets large. So that
                        $$ sqrt{m}frac{frac{sum (bar{X}_{i} - mu)}{m} - 0}{S} $$
                        will have student-t distribution, and we can rewrite it as
                        $$ sqrt{m}frac{frac{sum (bar{X}_{i} )}{m} - mu}{S} $$
                        and for large $m$, $frac{sum (bar{X}_{i} )}{m}$ will converger to $bar{X}$. So we can still use $T$ as approximation even when $X_{i}$s are not normal.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          I would like to post my view relating to @Chris answer also.



                          By the Central Limit Theorem, for large $m$, random var $X_{i}$ with mean $mu$ and st.dev $sigma$ from any distribution ($X_{i}$s have same dist) will have $$ sqrt{m} frac{ bar{X} - mu }{ sigma } $$ converge to standard normal. Also, the random.var $bar{X} - mu$ will be normal with mean $0$.



                          Now.., if $X_{i}$s are normal, then distribution
                          $$ T = sqrt{m} frac{bar{X} - mu}{S} $$ has student-t distribution with $m-1$ degrees of freedom. But we can only approximate the CLT using student-t when $X_{i}$s are normal.



                          Now I have an argument:
                          If $X_{i}$s are from the same distribution (any dist) with mean $mu$, then $bar{X} - mu$ will converge to normal with mean $0$ as $m$ gets large. So that
                          $$ sqrt{m}frac{frac{sum (bar{X}_{i} - mu)}{m} - 0}{S} $$
                          will have student-t distribution, and we can rewrite it as
                          $$ sqrt{m}frac{frac{sum (bar{X}_{i} )}{m} - mu}{S} $$
                          and for large $m$, $frac{sum (bar{X}_{i} )}{m}$ will converger to $bar{X}$. So we can still use $T$ as approximation even when $X_{i}$s are not normal.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            I would like to post my view relating to @Chris answer also.



                            By the Central Limit Theorem, for large $m$, random var $X_{i}$ with mean $mu$ and st.dev $sigma$ from any distribution ($X_{i}$s have same dist) will have $$ sqrt{m} frac{ bar{X} - mu }{ sigma } $$ converge to standard normal. Also, the random.var $bar{X} - mu$ will be normal with mean $0$.



                            Now.., if $X_{i}$s are normal, then distribution
                            $$ T = sqrt{m} frac{bar{X} - mu}{S} $$ has student-t distribution with $m-1$ degrees of freedom. But we can only approximate the CLT using student-t when $X_{i}$s are normal.



                            Now I have an argument:
                            If $X_{i}$s are from the same distribution (any dist) with mean $mu$, then $bar{X} - mu$ will converge to normal with mean $0$ as $m$ gets large. So that
                            $$ sqrt{m}frac{frac{sum (bar{X}_{i} - mu)}{m} - 0}{S} $$
                            will have student-t distribution, and we can rewrite it as
                            $$ sqrt{m}frac{frac{sum (bar{X}_{i} )}{m} - mu}{S} $$
                            and for large $m$, $frac{sum (bar{X}_{i} )}{m}$ will converger to $bar{X}$. So we can still use $T$ as approximation even when $X_{i}$s are not normal.






                            share|cite|improve this answer









                            $endgroup$



                            I would like to post my view relating to @Chris answer also.



                            By the Central Limit Theorem, for large $m$, random var $X_{i}$ with mean $mu$ and st.dev $sigma$ from any distribution ($X_{i}$s have same dist) will have $$ sqrt{m} frac{ bar{X} - mu }{ sigma } $$ converge to standard normal. Also, the random.var $bar{X} - mu$ will be normal with mean $0$.



                            Now.., if $X_{i}$s are normal, then distribution
                            $$ T = sqrt{m} frac{bar{X} - mu}{S} $$ has student-t distribution with $m-1$ degrees of freedom. But we can only approximate the CLT using student-t when $X_{i}$s are normal.



                            Now I have an argument:
                            If $X_{i}$s are from the same distribution (any dist) with mean $mu$, then $bar{X} - mu$ will converge to normal with mean $0$ as $m$ gets large. So that
                            $$ sqrt{m}frac{frac{sum (bar{X}_{i} - mu)}{m} - 0}{S} $$
                            will have student-t distribution, and we can rewrite it as
                            $$ sqrt{m}frac{frac{sum (bar{X}_{i} )}{m} - mu}{S} $$
                            and for large $m$, $frac{sum (bar{X}_{i} )}{m}$ will converger to $bar{X}$. So we can still use $T$ as approximation even when $X_{i}$s are not normal.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 16 '18 at 4:23









                            Arief AnbiyaArief Anbiya

                            1,3601622




                            1,3601622






























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