Empirical distribution of sorted Gaussian numbers












1












$begingroup$


I wrote a small program that does the following :




  1. Pick $N$ independent standard Gaussian numbers (expected value : 0, standard deviation : 1). Call that list $L={y_1, ldots, y_N}$.

  2. Sort that list in increasing order : $tilde{L}=mathrm{sort}(L)$.

  3. Plot that list on the $[-1,1]$ interval using a regularly distributed grid $x_i=-1+frac{2i}{N-1}$, with $i=0,ldots,N-1$.


I found that the plot was similar to that of the inverse error function, only differing by a multiplicative factor $a>0$.



Inverse error function & sorted Gaussian numbers plot, with N=10^5



I made a linear regression to find an approximate value of $1.42104$ for $a$. The two functions are very close for $N=10^5$ :
Fitted inverse error function and sorted Gaussian numbers for N=10^5



I have two questions :




  1. What is the exact value of $a$ ?

  2. How to prove that the limit function is indeed $a*mathrm{inverf}$ as $Nto infty$ ?






probability normal-distribution probability-limit-theorems sorting







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$endgroup$








  • 1




    $begingroup$
    What about $sqrt 2$ ?
    $endgroup$
    – Claude Leibovici
    Jan 4 at 9:07
















1












$begingroup$


I wrote a small program that does the following :




  1. Pick $N$ independent standard Gaussian numbers (expected value : 0, standard deviation : 1). Call that list $L={y_1, ldots, y_N}$.

  2. Sort that list in increasing order : $tilde{L}=mathrm{sort}(L)$.

  3. Plot that list on the $[-1,1]$ interval using a regularly distributed grid $x_i=-1+frac{2i}{N-1}$, with $i=0,ldots,N-1$.


I found that the plot was similar to that of the inverse error function, only differing by a multiplicative factor $a>0$.



Inverse error function & sorted Gaussian numbers plot, with N=10^5



I made a linear regression to find an approximate value of $1.42104$ for $a$. The two functions are very close for $N=10^5$ :
Fitted inverse error function and sorted Gaussian numbers for N=10^5



I have two questions :




  1. What is the exact value of $a$ ?

  2. How to prove that the limit function is indeed $a*mathrm{inverf}$ as $Nto infty$ ?






probability normal-distribution probability-limit-theorems sorting







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What about $sqrt 2$ ?
    $endgroup$
    – Claude Leibovici
    Jan 4 at 9:07














1












1








1





$begingroup$


I wrote a small program that does the following :




  1. Pick $N$ independent standard Gaussian numbers (expected value : 0, standard deviation : 1). Call that list $L={y_1, ldots, y_N}$.

  2. Sort that list in increasing order : $tilde{L}=mathrm{sort}(L)$.

  3. Plot that list on the $[-1,1]$ interval using a regularly distributed grid $x_i=-1+frac{2i}{N-1}$, with $i=0,ldots,N-1$.


I found that the plot was similar to that of the inverse error function, only differing by a multiplicative factor $a>0$.



Inverse error function & sorted Gaussian numbers plot, with N=10^5



I made a linear regression to find an approximate value of $1.42104$ for $a$. The two functions are very close for $N=10^5$ :
Fitted inverse error function and sorted Gaussian numbers for N=10^5



I have two questions :




  1. What is the exact value of $a$ ?

  2. How to prove that the limit function is indeed $a*mathrm{inverf}$ as $Nto infty$ ?






probability normal-distribution probability-limit-theorems sorting







share|cite|improve this question











$endgroup$




I wrote a small program that does the following :




  1. Pick $N$ independent standard Gaussian numbers (expected value : 0, standard deviation : 1). Call that list $L={y_1, ldots, y_N}$.

  2. Sort that list in increasing order : $tilde{L}=mathrm{sort}(L)$.

  3. Plot that list on the $[-1,1]$ interval using a regularly distributed grid $x_i=-1+frac{2i}{N-1}$, with $i=0,ldots,N-1$.


I found that the plot was similar to that of the inverse error function, only differing by a multiplicative factor $a>0$.



Inverse error function & sorted Gaussian numbers plot, with N=10^5



I made a linear regression to find an approximate value of $1.42104$ for $a$. The two functions are very close for $N=10^5$ :
Fitted inverse error function and sorted Gaussian numbers for N=10^5



I have two questions :




  1. What is the exact value of $a$ ?

  2. How to prove that the limit function is indeed $a*mathrm{inverf}$ as $Nto infty$ ?






probability normal-distribution probability-limit-theorems sorting




probability normal-distribution probability-limit-theorems sorting






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Jan 3 at 13:22







Florian Omnès

















asked Jan 3 at 10:32









Florian OmnèsFlorian Omnès

266




266








  • 1




    $begingroup$
    What about $sqrt 2$ ?
    $endgroup$
    – Claude Leibovici
    Jan 4 at 9:07














  • 1




    $begingroup$
    What about $sqrt 2$ ?
    $endgroup$
    – Claude Leibovici
    Jan 4 at 9:07








1




1




$begingroup$
What about $sqrt 2$ ?
$endgroup$
– Claude Leibovici
Jan 4 at 9:07




$begingroup$
What about $sqrt 2$ ?
$endgroup$
– Claude Leibovici
Jan 4 at 9:07










1 Answer
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$begingroup$

What you observe is the convergence of the empirical distribution function to cumulative distribution function of the sample - more accurately, of the empirical quantiles to theoretical quantiles (= values of inverse cumulative distribution function).



Specifically, for a continuous increasing cdf, theoretical quantiles are given by
$$
x_q = F^{-1}(q) = sup{xinmathbb R: F(x)<q}, qin(0,1).
$$
The definition of empirical quantiles varies. For a sample $X_1,dots,X_n$ of iid variables they can e.g. be defined by
$$
hat x_q = X_{(lfloor nqrfloor +1)}, qin (0,1),
$$
where $X_{(1)}le dotsle X_{(n)}$ is the sorted sample. It is known that whenever the cdf $F$ is strictly increasing, $hat x_q to x_q$ for all $qin (0,1)$ with probability $1$ as the sample size $ntoinfty$.



In your case, $F(x) = Phi(x)$ is the standart normal cdf, which is related to the error function by $$Phi(x) = frac{1+mathrm{Erf}(x/sqrt{2})}{2},$$
so
$$
Phi^{-1}(y) = sqrt{2}operatorname{Erf}^{-1}(2y-1), yin (0,1).
$$
What you are doing is applying a similar transformation to the empirical quantiles, so the convergence to $sqrt{2}operatorname{Erf}^{-1} approx 1.4142 operatorname{Erf}^{-1}$ is not surprising.






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    $begingroup$

    What you observe is the convergence of the empirical distribution function to cumulative distribution function of the sample - more accurately, of the empirical quantiles to theoretical quantiles (= values of inverse cumulative distribution function).



    Specifically, for a continuous increasing cdf, theoretical quantiles are given by
    $$
    x_q = F^{-1}(q) = sup{xinmathbb R: F(x)<q}, qin(0,1).
    $$
    The definition of empirical quantiles varies. For a sample $X_1,dots,X_n$ of iid variables they can e.g. be defined by
    $$
    hat x_q = X_{(lfloor nqrfloor +1)}, qin (0,1),
    $$
    where $X_{(1)}le dotsle X_{(n)}$ is the sorted sample. It is known that whenever the cdf $F$ is strictly increasing, $hat x_q to x_q$ for all $qin (0,1)$ with probability $1$ as the sample size $ntoinfty$.



    In your case, $F(x) = Phi(x)$ is the standart normal cdf, which is related to the error function by $$Phi(x) = frac{1+mathrm{Erf}(x/sqrt{2})}{2},$$
    so
    $$
    Phi^{-1}(y) = sqrt{2}operatorname{Erf}^{-1}(2y-1), yin (0,1).
    $$
    What you are doing is applying a similar transformation to the empirical quantiles, so the convergence to $sqrt{2}operatorname{Erf}^{-1} approx 1.4142 operatorname{Erf}^{-1}$ is not surprising.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      What you observe is the convergence of the empirical distribution function to cumulative distribution function of the sample - more accurately, of the empirical quantiles to theoretical quantiles (= values of inverse cumulative distribution function).



      Specifically, for a continuous increasing cdf, theoretical quantiles are given by
      $$
      x_q = F^{-1}(q) = sup{xinmathbb R: F(x)<q}, qin(0,1).
      $$
      The definition of empirical quantiles varies. For a sample $X_1,dots,X_n$ of iid variables they can e.g. be defined by
      $$
      hat x_q = X_{(lfloor nqrfloor +1)}, qin (0,1),
      $$
      where $X_{(1)}le dotsle X_{(n)}$ is the sorted sample. It is known that whenever the cdf $F$ is strictly increasing, $hat x_q to x_q$ for all $qin (0,1)$ with probability $1$ as the sample size $ntoinfty$.



      In your case, $F(x) = Phi(x)$ is the standart normal cdf, which is related to the error function by $$Phi(x) = frac{1+mathrm{Erf}(x/sqrt{2})}{2},$$
      so
      $$
      Phi^{-1}(y) = sqrt{2}operatorname{Erf}^{-1}(2y-1), yin (0,1).
      $$
      What you are doing is applying a similar transformation to the empirical quantiles, so the convergence to $sqrt{2}operatorname{Erf}^{-1} approx 1.4142 operatorname{Erf}^{-1}$ is not surprising.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        What you observe is the convergence of the empirical distribution function to cumulative distribution function of the sample - more accurately, of the empirical quantiles to theoretical quantiles (= values of inverse cumulative distribution function).



        Specifically, for a continuous increasing cdf, theoretical quantiles are given by
        $$
        x_q = F^{-1}(q) = sup{xinmathbb R: F(x)<q}, qin(0,1).
        $$
        The definition of empirical quantiles varies. For a sample $X_1,dots,X_n$ of iid variables they can e.g. be defined by
        $$
        hat x_q = X_{(lfloor nqrfloor +1)}, qin (0,1),
        $$
        where $X_{(1)}le dotsle X_{(n)}$ is the sorted sample. It is known that whenever the cdf $F$ is strictly increasing, $hat x_q to x_q$ for all $qin (0,1)$ with probability $1$ as the sample size $ntoinfty$.



        In your case, $F(x) = Phi(x)$ is the standart normal cdf, which is related to the error function by $$Phi(x) = frac{1+mathrm{Erf}(x/sqrt{2})}{2},$$
        so
        $$
        Phi^{-1}(y) = sqrt{2}operatorname{Erf}^{-1}(2y-1), yin (0,1).
        $$
        What you are doing is applying a similar transformation to the empirical quantiles, so the convergence to $sqrt{2}operatorname{Erf}^{-1} approx 1.4142 operatorname{Erf}^{-1}$ is not surprising.






        share|cite|improve this answer









        $endgroup$



        What you observe is the convergence of the empirical distribution function to cumulative distribution function of the sample - more accurately, of the empirical quantiles to theoretical quantiles (= values of inverse cumulative distribution function).



        Specifically, for a continuous increasing cdf, theoretical quantiles are given by
        $$
        x_q = F^{-1}(q) = sup{xinmathbb R: F(x)<q}, qin(0,1).
        $$
        The definition of empirical quantiles varies. For a sample $X_1,dots,X_n$ of iid variables they can e.g. be defined by
        $$
        hat x_q = X_{(lfloor nqrfloor +1)}, qin (0,1),
        $$
        where $X_{(1)}le dotsle X_{(n)}$ is the sorted sample. It is known that whenever the cdf $F$ is strictly increasing, $hat x_q to x_q$ for all $qin (0,1)$ with probability $1$ as the sample size $ntoinfty$.



        In your case, $F(x) = Phi(x)$ is the standart normal cdf, which is related to the error function by $$Phi(x) = frac{1+mathrm{Erf}(x/sqrt{2})}{2},$$
        so
        $$
        Phi^{-1}(y) = sqrt{2}operatorname{Erf}^{-1}(2y-1), yin (0,1).
        $$
        What you are doing is applying a similar transformation to the empirical quantiles, so the convergence to $sqrt{2}operatorname{Erf}^{-1} approx 1.4142 operatorname{Erf}^{-1}$ is not surprising.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 9:22









        zhorasterzhoraster

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