Generating a number belonging to N(0,1) using *m* numbers from U(0,1) using central limit theorem












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I was going through a blog which details how to generate a multivariate Gaussian vector, given a mean vector μ and co-variance matrix σ.



As a starting point, author uses generated uniform random numbers to simulate Gaussian random numbers using central limit theorem. The relevant extract from the blog is below:



Let's say you generate m uniform random numbers (each between 0 and 1) and you use the variable xi to denote each of these. The Central Limit Theorem allows us to convert these m numbers belonging to U(0,1) into a single number that belongs to the Guassian distribution N(0,1).



$$x = frac{sum_ix_i-frac{m}{2}}{sqrt{frac{m}{12}}}$$



Here, x is a one dimensional Gaussian random number - produced using the help of m uniform random variables. The $frac{m}{2}$ is derived from the term mμu (where μu is the mean of the uniform distribution - $frac{1-0}{2} = 0.5$). The denominator is derived from the term σ$sqrt{m}$ where $sigma^2$ is the variance of the uniform distribution between 0 and 1 (comes to exactly $frac{1}{12}$).



Link to the article : http://www.aishack.in/tutorials/generating-multivariate-gaussian-random/



I wanted to understand the details of the underlying math as I am not sure how central limit theorem is used to generate N(0,1) number from m U(0,1) numbers. I googled the same but was not able to find an explanation for this.






normal-distribution uniform-distribution central-limit-theorem







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    1












    $begingroup$


    I was going through a blog which details how to generate a multivariate Gaussian vector, given a mean vector μ and co-variance matrix σ.



    As a starting point, author uses generated uniform random numbers to simulate Gaussian random numbers using central limit theorem. The relevant extract from the blog is below:



    Let's say you generate m uniform random numbers (each between 0 and 1) and you use the variable xi to denote each of these. The Central Limit Theorem allows us to convert these m numbers belonging to U(0,1) into a single number that belongs to the Guassian distribution N(0,1).



    $$x = frac{sum_ix_i-frac{m}{2}}{sqrt{frac{m}{12}}}$$



    Here, x is a one dimensional Gaussian random number - produced using the help of m uniform random variables. The $frac{m}{2}$ is derived from the term mμu (where μu is the mean of the uniform distribution - $frac{1-0}{2} = 0.5$). The denominator is derived from the term σ$sqrt{m}$ where $sigma^2$ is the variance of the uniform distribution between 0 and 1 (comes to exactly $frac{1}{12}$).



    Link to the article : http://www.aishack.in/tutorials/generating-multivariate-gaussian-random/



    I wanted to understand the details of the underlying math as I am not sure how central limit theorem is used to generate N(0,1) number from m U(0,1) numbers. I googled the same but was not able to find an explanation for this.






    normal-distribution uniform-distribution central-limit-theorem







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I was going through a blog which details how to generate a multivariate Gaussian vector, given a mean vector μ and co-variance matrix σ.



      As a starting point, author uses generated uniform random numbers to simulate Gaussian random numbers using central limit theorem. The relevant extract from the blog is below:



      Let's say you generate m uniform random numbers (each between 0 and 1) and you use the variable xi to denote each of these. The Central Limit Theorem allows us to convert these m numbers belonging to U(0,1) into a single number that belongs to the Guassian distribution N(0,1).



      $$x = frac{sum_ix_i-frac{m}{2}}{sqrt{frac{m}{12}}}$$



      Here, x is a one dimensional Gaussian random number - produced using the help of m uniform random variables. The $frac{m}{2}$ is derived from the term mμu (where μu is the mean of the uniform distribution - $frac{1-0}{2} = 0.5$). The denominator is derived from the term σ$sqrt{m}$ where $sigma^2$ is the variance of the uniform distribution between 0 and 1 (comes to exactly $frac{1}{12}$).



      Link to the article : http://www.aishack.in/tutorials/generating-multivariate-gaussian-random/



      I wanted to understand the details of the underlying math as I am not sure how central limit theorem is used to generate N(0,1) number from m U(0,1) numbers. I googled the same but was not able to find an explanation for this.






      normal-distribution uniform-distribution central-limit-theorem







      share|cite|improve this question









      $endgroup$




      I was going through a blog which details how to generate a multivariate Gaussian vector, given a mean vector μ and co-variance matrix σ.



      As a starting point, author uses generated uniform random numbers to simulate Gaussian random numbers using central limit theorem. The relevant extract from the blog is below:



      Let's say you generate m uniform random numbers (each between 0 and 1) and you use the variable xi to denote each of these. The Central Limit Theorem allows us to convert these m numbers belonging to U(0,1) into a single number that belongs to the Guassian distribution N(0,1).



      $$x = frac{sum_ix_i-frac{m}{2}}{sqrt{frac{m}{12}}}$$



      Here, x is a one dimensional Gaussian random number - produced using the help of m uniform random variables. The $frac{m}{2}$ is derived from the term mμu (where μu is the mean of the uniform distribution - $frac{1-0}{2} = 0.5$). The denominator is derived from the term σ$sqrt{m}$ where $sigma^2$ is the variance of the uniform distribution between 0 and 1 (comes to exactly $frac{1}{12}$).



      Link to the article : http://www.aishack.in/tutorials/generating-multivariate-gaussian-random/



      I wanted to understand the details of the underlying math as I am not sure how central limit theorem is used to generate N(0,1) number from m U(0,1) numbers. I googled the same but was not able to find an explanation for this.






      normal-distribution uniform-distribution central-limit-theorem




      normal-distribution uniform-distribution central-limit-theorem






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      asked Jan 30 at 10:04









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          The idea is that the if there are $m$ independent random variables $X_i$ each uniformly distributed on $[0,1]$, so with mean $frac12$ and variance $frac1{12}$, then their sum $S_m$ has mean $frac{m}2$ and variance $frac{m}{12}$ while their average $frac1m S_m$ has mean $frac{1}2$ and variance $frac{1}{12m}$



          The Central Limit theorem says that $sqrt{m}left(frac1m S_m - frac{1}2right)$ converges in distribution to $mathcal Nleft(0,frac1{12}right)$, i.e. $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ converges in distribution to $mathcal N(0,1)$ as $m$ increases



          So for sufficiently large $m$, you might use $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ to generate a random variable which has a distribution close to that of a standard normal random variable






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            $begingroup$

            The idea is that the if there are $m$ independent random variables $X_i$ each uniformly distributed on $[0,1]$, so with mean $frac12$ and variance $frac1{12}$, then their sum $S_m$ has mean $frac{m}2$ and variance $frac{m}{12}$ while their average $frac1m S_m$ has mean $frac{1}2$ and variance $frac{1}{12m}$



            The Central Limit theorem says that $sqrt{m}left(frac1m S_m - frac{1}2right)$ converges in distribution to $mathcal Nleft(0,frac1{12}right)$, i.e. $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ converges in distribution to $mathcal N(0,1)$ as $m$ increases



            So for sufficiently large $m$, you might use $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ to generate a random variable which has a distribution close to that of a standard normal random variable






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              The idea is that the if there are $m$ independent random variables $X_i$ each uniformly distributed on $[0,1]$, so with mean $frac12$ and variance $frac1{12}$, then their sum $S_m$ has mean $frac{m}2$ and variance $frac{m}{12}$ while their average $frac1m S_m$ has mean $frac{1}2$ and variance $frac{1}{12m}$



              The Central Limit theorem says that $sqrt{m}left(frac1m S_m - frac{1}2right)$ converges in distribution to $mathcal Nleft(0,frac1{12}right)$, i.e. $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ converges in distribution to $mathcal N(0,1)$ as $m$ increases



              So for sufficiently large $m$, you might use $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ to generate a random variable which has a distribution close to that of a standard normal random variable






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                The idea is that the if there are $m$ independent random variables $X_i$ each uniformly distributed on $[0,1]$, so with mean $frac12$ and variance $frac1{12}$, then their sum $S_m$ has mean $frac{m}2$ and variance $frac{m}{12}$ while their average $frac1m S_m$ has mean $frac{1}2$ and variance $frac{1}{12m}$



                The Central Limit theorem says that $sqrt{m}left(frac1m S_m - frac{1}2right)$ converges in distribution to $mathcal Nleft(0,frac1{12}right)$, i.e. $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ converges in distribution to $mathcal N(0,1)$ as $m$ increases



                So for sufficiently large $m$, you might use $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ to generate a random variable which has a distribution close to that of a standard normal random variable






                share|cite|improve this answer









                $endgroup$



                The idea is that the if there are $m$ independent random variables $X_i$ each uniformly distributed on $[0,1]$, so with mean $frac12$ and variance $frac1{12}$, then their sum $S_m$ has mean $frac{m}2$ and variance $frac{m}{12}$ while their average $frac1m S_m$ has mean $frac{1}2$ and variance $frac{1}{12m}$



                The Central Limit theorem says that $sqrt{m}left(frac1m S_m - frac{1}2right)$ converges in distribution to $mathcal Nleft(0,frac1{12}right)$, i.e. $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ converges in distribution to $mathcal N(0,1)$ as $m$ increases



                So for sufficiently large $m$, you might use $dfrac{ S_m - frac{m}2}{sqrt{frac{m}{12}}}$ to generate a random variable which has a distribution close to that of a standard normal random variable







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 30 at 10:19









                HenryHenry

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