Converting a Uniform Distribution to a Normal Distribution

Question

How can I convert a uniform distribution  as most random number generators produce  e g  between 0 0 and 1 0  into a normal distribution  What if I want a mean and standard deviation of my choosing

User · Answer

It seems incredible that I could add something to this after eight years, but for the case of Java I would like to point readers to the Random.nextGaussian() method, which generates a Gaussian distribution with mean 0.0 and standard deviation 1.0 for you.

A simple addition and/or multiplication will change the mean and standard deviation to your needs.

User · Answer

There are plenty of methods    Do not use Box Muller  Especially if you draw many gaussian numbers  Box Muller yields a result which is clamped between -6 and 6  assuming double precision  Things worsen with floats    And it is really less efficient than other available methods  Ziggurat is fine  but needs a table lookup  and some platform-specific tweaking due to cache size issues  Ratio-of-uniforms is my favorite  only a few addition multiplications and a log 1 50th of the time  eg  look there   Inverting the CDF is efficient  and overlooked  why     you have fast implementations of it available if you search google  It is mandatory for Quasi-Random numbers

User · Answer

It seems incredible that I could add something to this after eight years, but for the case of Java I would like to point readers to the Random.nextGaussian() method, which generates a Gaussian distribution with mean 0.0 and standard deviation 1.0 for you.

A simple addition and/or multiplication will change the mean and standard deviation to your needs.

User · Answer

Changing the distribution of any function to another involves using the inverse of the function you want   In other words  if you aim for a specific probability function p x  you get the distribution by integrating over it -  d x    integral p x   and use its inverse  Inv d x     Now use the random probability function  which have uniform distribution  and cast the result value through the function Inv d x     You should get random values cast with distribution according to the function you chose   This is the generic math approach - by using it you can now choose any probability or distribution function you have as long as it have inverse or good inverse approximation   Hope this helped and thanks for the small remark about using the distribution and not the probability itself

User · Answer

Use the central limit theorem wikipedia entry mathworld entry to your advantage   Generate n of the uniformly distributed numbers  sum them  subtract n 0 5 and you have the output of an approximately normal distribution with mean equal to 0 and variance equal to  1 12     1 sqrt N    see wikipedia on uniform distributions for that last one    n 10 gives you something half decent fast  If you want something more than half decent go for tylers solution  as noted in the wikipedia entry on normal distributions

User · Answer

Use the central limit theorem wikipedia entry mathworld entry to your advantage   Generate n of the uniformly distributed numbers  sum them  subtract n 0 5 and you have the output of an approximately normal distribution with mean equal to 0 and variance equal to  1 12     1 sqrt N    see wikipedia on uniform distributions for that last one    n 10 gives you something half decent fast  If you want something more than half decent go for tylers solution  as noted in the wikipedia entry on normal distributions

User · Answer

Changing the distribution of any function to another involves using the inverse of the function you want   In other words  if you aim for a specific probability function p x  you get the distribution by integrating over it -  d x    integral p x   and use its inverse  Inv d x     Now use the random probability function  which have uniform distribution  and cast the result value through the function Inv d x     You should get random values cast with distribution according to the function you chose   This is the generic math approach - by using it you can now choose any probability or distribution function you have as long as it have inverse or good inverse approximation   Hope this helped and thanks for the small remark about using the distribution and not the probability itself

User · Answer

Changing the distribution of any function to another involves using the inverse of the function you want   In other words  if you aim for a specific probability function p x  you get the distribution by integrating over it -  d x    integral p x   and use its inverse  Inv d x     Now use the random probability function  which have uniform distribution  and cast the result value through the function Inv d x     You should get random values cast with distribution according to the function you chose   This is the generic math approach - by using it you can now choose any probability or distribution function you have as long as it have inverse or good inverse approximation   Hope this helped and thanks for the small remark about using the distribution and not the probability itself

User · Answer

This is a Matlab implementation using the polar form of the Box-Muller transformation   Function randn box muller m   function  values    randn box muller n  mean  std dev      if nargin    1        mean   0         std dev   1      end      r   gaussRandomN n       values   r  std dev - mean  end  function  values    gaussRandomN n       u  v  r    gaussRandomNValid n        c   sqrt -2 log r   r       values   u  c  end  function  u  v  r    gaussRandomNValid n      r   zeros n  1       u   zeros n  1       v   zeros n  1        filter   r  0   r gt  1         if outside interval  0 1  start over     while n    0         u filter    2 rand n  1 -1          v filter    2 rand n  1 -1          r filter    u filter   u filter    v filter   v filter            filter   r  0   r gt  1          n   size r filter  1       end end   And invoking histfit randn box muller 10000000  100   this is the result    Obviously it is really inefficient compared with the Matlab built-in randn

User · Answer

I thing you should try this in EXCEL   norminv rand   0 1   This will product the random numbers which should be normally distributed with the zero mean and unite variance   0  can be supplied with any value  so that the numbers will be of desired mean  and by changing  1   you will get the variance equal to the square of your input   For example   norminv rand   50 3  will yield to the normally distributed numbers with MEAN   50 VARIANCE   9

User · Answer

function distRandom      do      x random DISTRIBUTION DOMAIN      while random DISTRIBUTION RANGE  gt  distributionFunction x      return x

User · Answer

This is my JavaScript implementation of Algorithm P  Polar method for normal deviates  from Section 3 4 1 of Donald Knuth s book The Art of Computer Programming   function normal random mean stddev        var V1     var V2     var S     do          var U1   Math random      return uniform distributed in  0 1          var U2   Math random           V1   2 U1-1         V2   2 U2-1         S   V1 V1 V2 V2      while S  gt   1      if S   0  return 0     return mean stddev  V1 Math sqrt -2 Math log S  S

User · Answer

The standard Python library module random has what you want      normalvariate mu  sigma    Normal distribution  mu is the mean  and sigma is the standard deviation    For the algorithm itself  take a look at the function in random py in the Python library   The manual entry is here

User · Answer

function distRandom      do      x random DISTRIBUTION DOMAIN      while random DISTRIBUTION RANGE  gt  distributionFunction x      return x

User · Answer

Where R1  R2 are random uniform numbers  NORMAL DISTRIBUTION  with SD of 1  sqrt -2 log R1   cos 2 pi R2   This is exact    no need to do all those slow loops  Reference  dspguide com ch2 6 htm

User · Answer

The Ziggurat algorithm is pretty efficient for this  although the Box-Muller transform is easier to implement from scratch  and not crazy slow

User · Answer

I thing you should try this in EXCEL   norminv rand   0 1   This will product the random numbers which should be normally distributed with the zero mean and unite variance   0  can be supplied with any value  so that the numbers will be of desired mean  and by changing  1   you will get the variance equal to the square of your input   For example   norminv rand   50 3  will yield to the normally distributed numbers with MEAN   50 VARIANCE   9

User · Answer

Here is a javascript implementation using the polar form of the Box-Muller transformation         Returns member of set with a given mean and standard deviation    mean  mean    standard deviation  std dev      function createMemberInNormalDistribution mean std dev       return mean    gaussRandom   std dev            Returns random number in normal distribution centering on 0      95  of numbers returned should fall between -2 and 2    ie within two standard deviations     function gaussRandom         var u   2 Math random  -1      var v   2 Math random  -1      var r   u u   v v        if outside interval  0 1  start over       if r    0    r  gt   1  return gaussRandom         var c   Math sqrt -2 Math log r  r       return u c          todo  optimize this algorithm by caching  v c          and returning next time gaussRandom   is called         left out for simplicity

User · Answer

The standard Python library module random has what you want      normalvariate mu  sigma    Normal distribution  mu is the mean  and sigma is the standard deviation    For the algorithm itself  take a look at the function in random py in the Python library   The manual entry is here

User · Answer

The Ziggurat algorithm is pretty efficient for this  although the Box-Muller transform is easier to implement from scratch  and not crazy slow

User · Answer

It is also easier to use the implemented function rnorm   since it is faster than writing a random number generator for the normal distribution  See the following code as prove  n  lt - length z  t0  lt - Sys time   z  lt - rnorm n  t1  lt - Sys time   t1-t0

User · Answer

I would use Box-Muller  Two things about this    You end up with two values per iteration Typically  you cache one value and return the other  On the next call for a sample  you return the cached value  Box-Muller gives a Z-score You have to then scale the Z-score by the standard deviation and add the mean to get the full value in the normal distribution

User · Answer

Here is a javascript implementation using the polar form of the Box-Muller transformation         Returns member of set with a given mean and standard deviation    mean  mean    standard deviation  std dev      function createMemberInNormalDistribution mean std dev       return mean    gaussRandom   std dev            Returns random number in normal distribution centering on 0      95  of numbers returned should fall between -2 and 2    ie within two standard deviations     function gaussRandom         var u   2 Math random  -1      var v   2 Math random  -1      var r   u u   v v        if outside interval  0 1  start over       if r    0    r  gt   1  return gaussRandom         var c   Math sqrt -2 Math log r  r       return u c          todo  optimize this algorithm by caching  v c          and returning next time gaussRandom   is called         left out for simplicity

User · Answer

I have the following code which maybe could help   set seed 123  n  lt - 1000 u  lt - runif n   creates U x  lt - -log u  y  lt - runif n  max u sqrt  2 exp 1   pi    create Y z  lt - ifelse  y  lt  dnorm x  2  -x  NA  z  lt - ifelse   y  gt  dnorm x  2   amp   y  lt  dnorm x    x  z  z  lt - z  is na z

User · Answer

function distRandom      do      x random DISTRIBUTION DOMAIN      while random DISTRIBUTION RANGE  gt  distributionFunction x      return x

User · Answer

Where R1  R2 are random uniform numbers  NORMAL DISTRIBUTION  with SD of 1  sqrt -2 log R1   cos 2 pi R2   This is exact    no need to do all those slow loops  Reference  dspguide com ch2 6 htm

User · Answer

Here is a javascript implementation using the polar form of the Box-Muller transformation         Returns member of set with a given mean and standard deviation    mean  mean    standard deviation  std dev      function createMemberInNormalDistribution mean std dev       return mean    gaussRandom   std dev            Returns random number in normal distribution centering on 0      95  of numbers returned should fall between -2 and 2    ie within two standard deviations     function gaussRandom         var u   2 Math random  -1      var v   2 Math random  -1      var r   u u   v v        if outside interval  0 1  start over       if r    0    r  gt   1  return gaussRandom         var c   Math sqrt -2 Math log r  r       return u c          todo  optimize this algorithm by caching  v c          and returning next time gaussRandom   is called         left out for simplicity

User · Answer

function distRandom      do      x random DISTRIBUTION DOMAIN      while random DISTRIBUTION RANGE  gt  distributionFunction x      return x

User · Answer

It is also easier to use the implemented function rnorm   since it is faster than writing a random number generator for the normal distribution  See the following code as prove  n  lt - length z  t0  lt - Sys time   z  lt - rnorm n  t1  lt - Sys time   t1-t0

User · Answer

This is my JavaScript implementation of Algorithm P  Polar method for normal deviates  from Section 3 4 1 of Donald Knuth s book The Art of Computer Programming   function normal random mean stddev        var V1     var V2     var S     do          var U1   Math random      return uniform distributed in  0 1          var U2   Math random           V1   2 U1-1         V2   2 U2-1         S   V1 V1 V2 V2      while S  gt   1      if S   0  return 0     return mean stddev  V1 Math sqrt -2 Math log S  S

User · Answer

I have the following code which maybe could help   set seed 123  n  lt - 1000 u  lt - runif n   creates U x  lt - -log u  y  lt - runif n  max u sqrt  2 exp 1   pi    create Y z  lt - ifelse  y  lt  dnorm x  2  -x  NA  z  lt - ifelse   y  gt  dnorm x  2   amp   y  lt  dnorm x    x  z  z  lt - z  is na z

User · Answer

Q How can I convert a uniform distribution  as most random number generators produce  e g  between 0 0 and 1 0  into a normal distribution    For software implementation I know couple random generator names which give you a pseudo uniform random sequence in  0 1   Mersenne Twister  Linear Congruate Generator   Let s call it U x  It is exist mathematical area which called probibility theory   First thing  If you want to model r v  with integral distribution F then you can try just to evaluate F -1 U x    In pr theory it was proved that such r v  will have integral distribution F  Step 2 can be appliable to generate r v  F without usage of any counting methods when F -1 can be derived analytically without problems   e g  exp distribution  To model normal distribution you can cacculate y1 cos y2   where y1 is uniform in 0 2pi   and y2 is the relei distribution    Q  What if I want a mean and standard deviation of my choosing   You can calculate sigma N 0 1  m   It can be shown that such shifting and scaling lead to N m sigma

User · Answer

Changing the distribution of any function to another involves using the inverse of the function you want   In other words  if you aim for a specific probability function p x  you get the distribution by integrating over it -  d x    integral p x   and use its inverse  Inv d x     Now use the random probability function  which have uniform distribution  and cast the result value through the function Inv d x     You should get random values cast with distribution according to the function you chose   This is the generic math approach - by using it you can now choose any probability or distribution function you have as long as it have inverse or good inverse approximation   Hope this helped and thanks for the small remark about using the distribution and not the probability itself

User · Answer

The standard Python library module random has what you want      normalvariate mu  sigma    Normal distribution  mu is the mean  and sigma is the standard deviation    For the algorithm itself  take a look at the function in random py in the Python library   The manual entry is here

User · Answer

The standard Python library module random has what you want      normalvariate mu  sigma    Normal distribution  mu is the mean  and sigma is the standard deviation    For the algorithm itself  take a look at the function in random py in the Python library   The manual entry is here

User · Answer

I would use Box-Muller  Two things about this    You end up with two values per iteration Typically  you cache one value and return the other  On the next call for a sample  you return the cached value  Box-Muller gives a Z-score You have to then scale the Z-score by the standard deviation and add the mean to get the full value in the normal distribution

User · Answer

I would use Box-Muller  Two things about this    You end up with two values per iteration Typically  you cache one value and return the other  On the next call for a sample  you return the cached value  Box-Muller gives a Z-score You have to then scale the Z-score by the standard deviation and add the mean to get the full value in the normal distribution

User · Answer

I would use Box-Muller  Two things about this    You end up with two values per iteration Typically  you cache one value and return the other  On the next call for a sample  you return the cached value  Box-Muller gives a Z-score You have to then scale the Z-score by the standard deviation and add the mean to get the full value in the normal distribution

User · Answer

Use the central limit theorem wikipedia entry mathworld entry to your advantage   Generate n of the uniformly distributed numbers  sum them  subtract n 0 5 and you have the output of an approximately normal distribution with mean equal to 0 and variance equal to  1 12     1 sqrt N    see wikipedia on uniform distributions for that last one    n 10 gives you something half decent fast  If you want something more than half decent go for tylers solution  as noted in the wikipedia entry on normal distributions

User · Answer

Use the central limit theorem wikipedia entry mathworld entry to your advantage   Generate n of the uniformly distributed numbers  sum them  subtract n 0 5 and you have the output of an approximately normal distribution with mean equal to 0 and variance equal to  1 12     1 sqrt N    see wikipedia on uniform distributions for that last one    n 10 gives you something half decent fast  If you want something more than half decent go for tylers solution  as noted in the wikipedia entry on normal distributions

User · Answer

This is a Matlab implementation using the polar form of the Box-Muller transformation   Function randn box muller m   function  values    randn box muller n  mean  std dev      if nargin    1        mean   0         std dev   1      end      r   gaussRandomN n       values   r  std dev - mean  end  function  values    gaussRandomN n       u  v  r    gaussRandomNValid n        c   sqrt -2 log r   r       values   u  c  end  function  u  v  r    gaussRandomNValid n      r   zeros n  1       u   zeros n  1       v   zeros n  1        filter   r  0   r gt  1         if outside interval  0 1  start over     while n    0         u filter    2 rand n  1 -1          v filter    2 rand n  1 -1          r filter    u filter   u filter    v filter   v filter            filter   r  0   r gt  1          n   size r filter  1       end end   And invoking histfit randn box muller 10000000  100   this is the result    Obviously it is really inefficient compared with the Matlab built-in randn

User · Answer

There are plenty of methods    Do not use Box Muller  Especially if you draw many gaussian numbers  Box Muller yields a result which is clamped between -6 and 6  assuming double precision  Things worsen with floats    And it is really less efficient than other available methods  Ziggurat is fine  but needs a table lookup  and some platform-specific tweaking due to cache size issues  Ratio-of-uniforms is my favorite  only a few addition multiplications and a log 1 50th of the time  eg  look there   Inverting the CDF is efficient  and overlooked  why     you have fast implementations of it available if you search google  It is mandatory for Quasi-Random numbers

User · Answer

The Ziggurat algorithm is pretty efficient for this  although the Box-Muller transform is easier to implement from scratch  and not crazy slow

User · Answer

Here is a javascript implementation using the polar form of the Box-Muller transformation         Returns member of set with a given mean and standard deviation    mean  mean    standard deviation  std dev      function createMemberInNormalDistribution mean std dev       return mean    gaussRandom   std dev            Returns random number in normal distribution centering on 0      95  of numbers returned should fall between -2 and 2    ie within two standard deviations     function gaussRandom         var u   2 Math random  -1      var v   2 Math random  -1      var r   u u   v v        if outside interval  0 1  start over       if r    0    r  gt   1  return gaussRandom         var c   Math sqrt -2 Math log r  r       return u c          todo  optimize this algorithm by caching  v c          and returning next time gaussRandom   is called         left out for simplicity

User · Answer

The Ziggurat algorithm is pretty efficient for this  although the Box-Muller transform is easier to implement from scratch  and not crazy slow

User · Answer

Q How can I convert a uniform distribution  as most random number generators produce  e g  between 0 0 and 1 0  into a normal distribution    For software implementation I know couple random generator names which give you a pseudo uniform random sequence in  0 1   Mersenne Twister  Linear Congruate Generator   Let s call it U x  It is exist mathematical area which called probibility theory   First thing  If you want to model r v  with integral distribution F then you can try just to evaluate F -1 U x    In pr theory it was proved that such r v  will have integral distribution F  Step 2 can be appliable to generate r v  F without usage of any counting methods when F -1 can be derived analytically without problems   e g  exp distribution  To model normal distribution you can cacculate y1 cos y2   where y1 is uniform in 0 2pi   and y2 is the relei distribution    Q  What if I want a mean and standard deviation of my choosing   You can calculate sigma N 0 1  m   It can be shown that such shifting and scaling lead to N m sigma

[algorithm] Converting a Uniform Distribution to a Normal Distribution

Examples related to algorithm

Examples related to language-agnostic

Examples related to random

Examples related to normal-distribution