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







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  • $begingroup$
    You will want to investigate the Probability Integral Transform.
    $endgroup$
    – StatsStudent
    4 hours ago










  • $begingroup$
    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♦
    1 hour ago
















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









New contributor




Chicago1988 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    You will want to investigate the Probability Integral Transform.
    $endgroup$
    – StatsStudent
    4 hours ago










  • $begingroup$
    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♦
    1 hour ago














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









New contributor




Chicago1988 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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









New contributor




Chicago1988 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Chicago1988 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









masoud

74




74






New contributor




Chicago1988 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 6 hours ago









Chicago1988Chicago1988

162




162




New contributor




Chicago1988 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Chicago1988 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Chicago1988 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    You will want to investigate the Probability Integral Transform.
    $endgroup$
    – StatsStudent
    4 hours ago










  • $begingroup$
    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♦
    1 hour ago


















  • $begingroup$
    You will want to investigate the Probability Integral Transform.
    $endgroup$
    – StatsStudent
    4 hours ago










  • $begingroup$
    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♦
    1 hour ago
















$begingroup$
You will want to investigate the Probability Integral Transform.
$endgroup$
– StatsStudent
4 hours ago




$begingroup$
You will want to investigate the Probability Integral Transform.
$endgroup$
– StatsStudent
4 hours ago












$begingroup$
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♦
1 hour ago




$begingroup$
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♦
1 hour ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

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}$



    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









      4












      $begingroup$

      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











      $endgroup$


















        4












        $begingroup$

        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











        $endgroup$
















          4












          4








          4





          $begingroup$

          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 1 hour ago

























          answered 5 hours ago









          Cliff ABCliff AB

          13.4k12567




          13.4k12567

























              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}$



              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}$



                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}$



                  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}$



                  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 3 hours ago

























                  answered 3 hours ago









                  StatsStats

                  44117




                  44117






















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                      Tabula Rosettana

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                      Tabula multilinguis Rosettana in Museo Britannico ostenditur. Tabula Rosettana, [1] etiam titulo OGIS 90 agnita, est stela decreto de rebus sacris in Aegypto anno 196 a.C.n. lato inscripta. Tabula iuxta Rosettam Aegypti, urbem in delta Nili et ad oram maris Mediterranei iacentem, anno 1799 a milite Francico reperta est. Inventio stelae, linguis duabus scripturisque tribus inscriptae, eruditis Instituti Aegypti statim nuntiata est; ibi enim iussu imperatoris Napoleonis eruditi omnium scientiarum (sub aegide Commissionis Scientiarum et Artium) properaverant cum expeditione Francica. Qua a Britannis mox debellata, tabula Rosettana Londinium missa hodie apud Museum Britannicum iacet. Textus Graecus cito lectus interpretationi textuum Aegyptiorum (in formis hieroglyphica et demotica expressorum) gradatim adiuvit. Denique textum plene interpretatus est Ioannes Franciscus Champollion. Ab opere eruditorum cumulativo coepit hodiernus scripturae hieroglyphicae linguaeque Aegyptiae a...
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                      How to label and detect the document text images

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                      1 $begingroup$ This is what I mean as document text image: I want to label the texts in image as separate blocks and my model should detect these labels as classes. NOTE: This is how the end result should be like: The labels like Block 1, Block 2, Block 3,.. should be Logo, Title, Date,.. Others, etc. Work done: First approach : I tried to implement this method via Object Detection, it didn't work. It didn't even detect any text. Second approach : Then I tried it using PixelLink. As this model is build for scene text detection, it detected each and every text in the image. But this method can detect multiple lines of text if the threshold values are increased. But I have no idea how do I add labels to the text blocks. PIXEL_CLS_WEIGHT_all_ones = 'PIXEL_CLS_WEIGHT_all_ones' PIXEL_C...
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