Simulating rnorm() using runif()












3












$begingroup$


I am trying to 'simulate' rnorm() using only runif().



I don't know if I should do:



sqrt(-2*log(U1))*cos(U2)


or



sqrt(-2*log(U1))*sin(U2)


Where U1 is a runif(0,1) and U2 runif(0,6.28)



I do not know if I should do it using cos or sin, or is it that I need to sample from one and the other consecutively? What is the mathematical logic behind it?






r normalization







share|cite|improve this question











$endgroup$












  • $begingroup$
    You will want to investigate the Probability Integral Transform.
    $endgroup$
    – StatsStudent
    Mar 14 at 15:28






  • 1




    $begingroup$
    Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
    $endgroup$
    – whuber♦
    Mar 14 at 18:44
















3












$begingroup$


I am trying to 'simulate' rnorm() using only runif().



I don't know if I should do:



sqrt(-2*log(U1))*cos(U2)


or



sqrt(-2*log(U1))*sin(U2)


Where U1 is a runif(0,1) and U2 runif(0,6.28)



I do not know if I should do it using cos or sin, or is it that I need to sample from one and the other consecutively? What is the mathematical logic behind it?






r normalization







share|cite|improve this question











$endgroup$












  • $begingroup$
    You will want to investigate the Probability Integral Transform.
    $endgroup$
    – StatsStudent
    Mar 14 at 15:28






  • 1




    $begingroup$
    Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
    $endgroup$
    – whuber♦
    Mar 14 at 18:44














3












3








3





$begingroup$


I am trying to 'simulate' rnorm() using only runif().



I don't know if I should do:



sqrt(-2*log(U1))*cos(U2)


or



sqrt(-2*log(U1))*sin(U2)


Where U1 is a runif(0,1) and U2 runif(0,6.28)



I do not know if I should do it using cos or sin, or is it that I need to sample from one and the other consecutively? What is the mathematical logic behind it?






r normalization







share|cite|improve this question











$endgroup$




I am trying to 'simulate' rnorm() using only runif().



I don't know if I should do:



sqrt(-2*log(U1))*cos(U2)


or



sqrt(-2*log(U1))*sin(U2)


Where U1 is a runif(0,1) and U2 runif(0,6.28)



I do not know if I should do it using cos or sin, or is it that I need to sample from one and the other consecutively? What is the mathematical logic behind it?






r normalization




r normalization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 14 at 16:51









masoud

675




675










asked Mar 14 at 14:13









Chicago1988Chicago1988

162




162












  • $begingroup$
    You will want to investigate the Probability Integral Transform.
    $endgroup$
    – StatsStudent
    Mar 14 at 15:28






  • 1




    $begingroup$
    Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
    $endgroup$
    – whuber♦
    Mar 14 at 18:44


















  • $begingroup$
    You will want to investigate the Probability Integral Transform.
    $endgroup$
    – StatsStudent
    Mar 14 at 15:28






  • 1




    $begingroup$
    Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
    $endgroup$
    – whuber♦
    Mar 14 at 18:44
















$begingroup$
You will want to investigate the Probability Integral Transform.
$endgroup$
– StatsStudent
Mar 14 at 15:28




$begingroup$
You will want to investigate the Probability Integral Transform.
$endgroup$
– StatsStudent
Mar 14 at 15:28




1




1




$begingroup$
Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
$endgroup$
– whuber♦
Mar 14 at 18:44




$begingroup$
Because $cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $cos(U_2)$ and $sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2pi$).
$endgroup$
– whuber♦
Mar 14 at 18:44










2 Answers
2






active

oldest

votes


















5












$begingroup$

It looks like you are trying to use the Box-Muller transform.



The method is to use $U_1, U_2$ which are independently distribution Uniform(0,1). Then



$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2) $



and



$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2) $



are a pair of independent $N(0,1)$.



The derivation comes from using polar coordinates, but to be honest the exact steps escape me at the moment (see link if you are interested).






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Let $N_1,N_2$ be independent $N(0,1)$ random variates. $(N_1,N_2)$ defines a point in Cartesian coordinates. We transform to polar by



    $$ N_1 = R cos theta$$
    $$ N_2 = R sin theta$$



    This transformation is one to one and has continuous derivatives. So we can derive the joint distribution $f_{R,theta}(r,theta)$ by using



    $$f_{R,theta}(r,theta) = f_{N_1,N_2}(n_1,n_2)|J^{-1}|$$



    where $J$ is the Jacobian of the transformation and



    $$ J^{-1}(r,theta )=
    {begin{bmatrix}{dfrac {partial n_1}{partial r}}{dfrac {partial n_1}{partial theta }}\{dfrac {partial n_2}{partial r}}{dfrac {partial n_2}{partial theta }}end{bmatrix}}
    = begin{bmatrix}cos theta & -rsin theta \sin theta & rcos theta end{bmatrix}$$



    Hence $$ |J^{-1}| = r cos^2 theta + r sin^2 theta = r $$



    and



    $$ f_{R,theta}(r,theta) = r frac{1}{2pi} e^{-frac{1}{2}(n_1^2+n_2^2)} = boxed{frac{r}{2 pi}e^{-frac{1}{2}(r^2)}} qquad 0 leq theta leq 2 pi, 0 leq r leq infty $$



    And we see $theta sim unif[0,2pi]$ (since $f_{Theta}(theta) = frac{1}{2pi}$ ) and $R^2 sim exp(1/2)$



    Hence, to simulate $Theta$ we simply take $2 pi U_2$ where $$U_2 sim unif[0,1]$$ and to simulate $R$ we can take $$-2ln U_1$$ where $U_1 sim unif[0,1]$. ( to see why find the density of $X=-ln U, U sim unif[0,1]$).



    So, Box and Muller simply inverted $N_1= R cos theta$, $N_2= R sin theta$ and moved from $(R,Theta)$ to $(N_1,N_2)$ by simulating $Theta$ from $2 pi U_2$, and an independent $R$ from $ sqrt{- 2 ln U_1}$



    explicitly



    $$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2)$$



    $$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2)$$



    And that is that is the mathematical logic behind it.




    p.s. As I don't know whether it is clear through the mathematical justification (I believe it is, but you may have been lost in details), note you need a pair of drawings from the uniform to get a pair of normals






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      It looks like you are trying to use the Box-Muller transform.



      The method is to use $U_1, U_2$ which are independently distribution Uniform(0,1). Then



      $Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2) $



      and



      $Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2) $



      are a pair of independent $N(0,1)$.



      The derivation comes from using polar coordinates, but to be honest the exact steps escape me at the moment (see link if you are interested).






      share|cite|improve this answer











      $endgroup$


















        5












        $begingroup$

        It looks like you are trying to use the Box-Muller transform.



        The method is to use $U_1, U_2$ which are independently distribution Uniform(0,1). Then



        $Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2) $



        and



        $Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2) $



        are a pair of independent $N(0,1)$.



        The derivation comes from using polar coordinates, but to be honest the exact steps escape me at the moment (see link if you are interested).






        share|cite|improve this answer











        $endgroup$
















          5












          5








          5





          $begingroup$

          It looks like you are trying to use the Box-Muller transform.



          The method is to use $U_1, U_2$ which are independently distribution Uniform(0,1). Then



          $Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2) $



          and



          $Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2) $



          are a pair of independent $N(0,1)$.



          The derivation comes from using polar coordinates, but to be honest the exact steps escape me at the moment (see link if you are interested).






          share|cite|improve this answer











          $endgroup$



          It looks like you are trying to use the Box-Muller transform.



          The method is to use $U_1, U_2$ which are independently distribution Uniform(0,1). Then



          $Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2) $



          and



          $Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2) $



          are a pair of independent $N(0,1)$.



          The derivation comes from using polar coordinates, but to be honest the exact steps escape me at the moment (see link if you are interested).







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 14 at 18:15

























          answered Mar 14 at 14:58









          Cliff ABCliff AB

          13.6k12567




          13.6k12567

























              2












              $begingroup$

              Let $N_1,N_2$ be independent $N(0,1)$ random variates. $(N_1,N_2)$ defines a point in Cartesian coordinates. We transform to polar by



              $$ N_1 = R cos theta$$
              $$ N_2 = R sin theta$$



              This transformation is one to one and has continuous derivatives. So we can derive the joint distribution $f_{R,theta}(r,theta)$ by using



              $$f_{R,theta}(r,theta) = f_{N_1,N_2}(n_1,n_2)|J^{-1}|$$



              where $J$ is the Jacobian of the transformation and



              $$ J^{-1}(r,theta )=
              {begin{bmatrix}{dfrac {partial n_1}{partial r}}{dfrac {partial n_1}{partial theta }}\{dfrac {partial n_2}{partial r}}{dfrac {partial n_2}{partial theta }}end{bmatrix}}
              = begin{bmatrix}cos theta & -rsin theta \sin theta & rcos theta end{bmatrix}$$



              Hence $$ |J^{-1}| = r cos^2 theta + r sin^2 theta = r $$



              and



              $$ f_{R,theta}(r,theta) = r frac{1}{2pi} e^{-frac{1}{2}(n_1^2+n_2^2)} = boxed{frac{r}{2 pi}e^{-frac{1}{2}(r^2)}} qquad 0 leq theta leq 2 pi, 0 leq r leq infty $$



              And we see $theta sim unif[0,2pi]$ (since $f_{Theta}(theta) = frac{1}{2pi}$ ) and $R^2 sim exp(1/2)$



              Hence, to simulate $Theta$ we simply take $2 pi U_2$ where $$U_2 sim unif[0,1]$$ and to simulate $R$ we can take $$-2ln U_1$$ where $U_1 sim unif[0,1]$. ( to see why find the density of $X=-ln U, U sim unif[0,1]$).



              So, Box and Muller simply inverted $N_1= R cos theta$, $N_2= R sin theta$ and moved from $(R,Theta)$ to $(N_1,N_2)$ by simulating $Theta$ from $2 pi U_2$, and an independent $R$ from $ sqrt{- 2 ln U_1}$



              explicitly



              $$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2)$$



              $$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2)$$



              And that is that is the mathematical logic behind it.




              p.s. As I don't know whether it is clear through the mathematical justification (I believe it is, but you may have been lost in details), note you need a pair of drawings from the uniform to get a pair of normals






              share|cite|improve this answer











              $endgroup$


















                2












                $begingroup$

                Let $N_1,N_2$ be independent $N(0,1)$ random variates. $(N_1,N_2)$ defines a point in Cartesian coordinates. We transform to polar by



                $$ N_1 = R cos theta$$
                $$ N_2 = R sin theta$$



                This transformation is one to one and has continuous derivatives. So we can derive the joint distribution $f_{R,theta}(r,theta)$ by using



                $$f_{R,theta}(r,theta) = f_{N_1,N_2}(n_1,n_2)|J^{-1}|$$



                where $J$ is the Jacobian of the transformation and



                $$ J^{-1}(r,theta )=
                {begin{bmatrix}{dfrac {partial n_1}{partial r}}{dfrac {partial n_1}{partial theta }}\{dfrac {partial n_2}{partial r}}{dfrac {partial n_2}{partial theta }}end{bmatrix}}
                = begin{bmatrix}cos theta & -rsin theta \sin theta & rcos theta end{bmatrix}$$



                Hence $$ |J^{-1}| = r cos^2 theta + r sin^2 theta = r $$



                and



                $$ f_{R,theta}(r,theta) = r frac{1}{2pi} e^{-frac{1}{2}(n_1^2+n_2^2)} = boxed{frac{r}{2 pi}e^{-frac{1}{2}(r^2)}} qquad 0 leq theta leq 2 pi, 0 leq r leq infty $$



                And we see $theta sim unif[0,2pi]$ (since $f_{Theta}(theta) = frac{1}{2pi}$ ) and $R^2 sim exp(1/2)$



                Hence, to simulate $Theta$ we simply take $2 pi U_2$ where $$U_2 sim unif[0,1]$$ and to simulate $R$ we can take $$-2ln U_1$$ where $U_1 sim unif[0,1]$. ( to see why find the density of $X=-ln U, U sim unif[0,1]$).



                So, Box and Muller simply inverted $N_1= R cos theta$, $N_2= R sin theta$ and moved from $(R,Theta)$ to $(N_1,N_2)$ by simulating $Theta$ from $2 pi U_2$, and an independent $R$ from $ sqrt{- 2 ln U_1}$



                explicitly



                $$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2)$$



                $$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2)$$



                And that is that is the mathematical logic behind it.




                p.s. As I don't know whether it is clear through the mathematical justification (I believe it is, but you may have been lost in details), note you need a pair of drawings from the uniform to get a pair of normals






                share|cite|improve this answer











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Let $N_1,N_2$ be independent $N(0,1)$ random variates. $(N_1,N_2)$ defines a point in Cartesian coordinates. We transform to polar by



                  $$ N_1 = R cos theta$$
                  $$ N_2 = R sin theta$$



                  This transformation is one to one and has continuous derivatives. So we can derive the joint distribution $f_{R,theta}(r,theta)$ by using



                  $$f_{R,theta}(r,theta) = f_{N_1,N_2}(n_1,n_2)|J^{-1}|$$



                  where $J$ is the Jacobian of the transformation and



                  $$ J^{-1}(r,theta )=
                  {begin{bmatrix}{dfrac {partial n_1}{partial r}}{dfrac {partial n_1}{partial theta }}\{dfrac {partial n_2}{partial r}}{dfrac {partial n_2}{partial theta }}end{bmatrix}}
                  = begin{bmatrix}cos theta & -rsin theta \sin theta & rcos theta end{bmatrix}$$



                  Hence $$ |J^{-1}| = r cos^2 theta + r sin^2 theta = r $$



                  and



                  $$ f_{R,theta}(r,theta) = r frac{1}{2pi} e^{-frac{1}{2}(n_1^2+n_2^2)} = boxed{frac{r}{2 pi}e^{-frac{1}{2}(r^2)}} qquad 0 leq theta leq 2 pi, 0 leq r leq infty $$



                  And we see $theta sim unif[0,2pi]$ (since $f_{Theta}(theta) = frac{1}{2pi}$ ) and $R^2 sim exp(1/2)$



                  Hence, to simulate $Theta$ we simply take $2 pi U_2$ where $$U_2 sim unif[0,1]$$ and to simulate $R$ we can take $$-2ln U_1$$ where $U_1 sim unif[0,1]$. ( to see why find the density of $X=-ln U, U sim unif[0,1]$).



                  So, Box and Muller simply inverted $N_1= R cos theta$, $N_2= R sin theta$ and moved from $(R,Theta)$ to $(N_1,N_2)$ by simulating $Theta$ from $2 pi U_2$, and an independent $R$ from $ sqrt{- 2 ln U_1}$



                  explicitly



                  $$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2)$$



                  $$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2)$$



                  And that is that is the mathematical logic behind it.




                  p.s. As I don't know whether it is clear through the mathematical justification (I believe it is, but you may have been lost in details), note you need a pair of drawings from the uniform to get a pair of normals






                  share|cite|improve this answer











                  $endgroup$



                  Let $N_1,N_2$ be independent $N(0,1)$ random variates. $(N_1,N_2)$ defines a point in Cartesian coordinates. We transform to polar by



                  $$ N_1 = R cos theta$$
                  $$ N_2 = R sin theta$$



                  This transformation is one to one and has continuous derivatives. So we can derive the joint distribution $f_{R,theta}(r,theta)$ by using



                  $$f_{R,theta}(r,theta) = f_{N_1,N_2}(n_1,n_2)|J^{-1}|$$



                  where $J$ is the Jacobian of the transformation and



                  $$ J^{-1}(r,theta )=
                  {begin{bmatrix}{dfrac {partial n_1}{partial r}}{dfrac {partial n_1}{partial theta }}\{dfrac {partial n_2}{partial r}}{dfrac {partial n_2}{partial theta }}end{bmatrix}}
                  = begin{bmatrix}cos theta & -rsin theta \sin theta & rcos theta end{bmatrix}$$



                  Hence $$ |J^{-1}| = r cos^2 theta + r sin^2 theta = r $$



                  and



                  $$ f_{R,theta}(r,theta) = r frac{1}{2pi} e^{-frac{1}{2}(n_1^2+n_2^2)} = boxed{frac{r}{2 pi}e^{-frac{1}{2}(r^2)}} qquad 0 leq theta leq 2 pi, 0 leq r leq infty $$



                  And we see $theta sim unif[0,2pi]$ (since $f_{Theta}(theta) = frac{1}{2pi}$ ) and $R^2 sim exp(1/2)$



                  Hence, to simulate $Theta$ we simply take $2 pi U_2$ where $$U_2 sim unif[0,1]$$ and to simulate $R$ we can take $$-2ln U_1$$ where $U_1 sim unif[0,1]$. ( to see why find the density of $X=-ln U, U sim unif[0,1]$).



                  So, Box and Muller simply inverted $N_1= R cos theta$, $N_2= R sin theta$ and moved from $(R,Theta)$ to $(N_1,N_2)$ by simulating $Theta$ from $2 pi U_2$, and an independent $R$ from $ sqrt{- 2 ln U_1}$



                  explicitly



                  $$Z_0 = sqrt{-2 ln(U_1)}cos(2 pi U_2)$$



                  $$Z_1 = sqrt{-2 ln(U_1)} sin(2 pi U_2)$$



                  And that is that is the mathematical logic behind it.




                  p.s. As I don't know whether it is clear through the mathematical justification (I believe it is, but you may have been lost in details), note you need a pair of drawings from the uniform to get a pair of normals







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 14 at 21:11

























                  answered Mar 14 at 16:14









                  StatsStats

                  668210




                  668210






























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                      Can I use Tabulator js library in my java Spring + Thymeleaf project?

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                      0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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