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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                      Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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                      up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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