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how to change uniformly distributed random variable to exponential or normal distribution

Posted on 2004-11-30
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Last Modified: 2012-05-05
hi,
    function rand() in C++ generates a random variable which is uniformly distributed but i want to generate a random variable that is exponentialy distributed or normally distributed. can any one tell me the code for doing this.

Saira
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Question by:sairaseven
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8 Comments
 
LVL 30

Expert Comment

by:Axter
ID: 12713593
use time to seed the random generator
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LVL 30

Expert Comment

by:Axter
ID: 12713603
Example:

#include <time.h>

int main( void )
{
   int i;

   /* Seed the random-number generator with current time so that
    * the numbers will be different every time we run.
    */
   srand( (unsigned)time( NULL ) );

   /* Display 10 numbers. */
   for( i = 0;   i < 10;i++ )
      printf( "  %6d\n", rand() );
  return 0;
}
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Accepted Solution

by:
Axter earned 136 total points
ID: 12713608
Make sure you only call srand one time throughout the life of your program.
This should be done in the main() when the program first starts up.
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LVL 4

Expert Comment

by:skypalae
ID: 12713647
Axter,

the original question was not about same rand() values all the time (as usually is) but about the distribution.

The answer is NO. You may find some algorithms on internet, but you have to implement other distributions by yourself. If I want to use some approximation to normal distribution I usually make mean of several uniformly distributed values:

randVal = (rand () + rand () + rand () + rand ()) / 4 ;

This is enough for me (though it is 4x slower than just rand()). If you need more precise solution I'll try to find something else.

S.
0
 

Author Comment

by:sairaseven
ID: 12713715
i want to generate a normally distributed random variable with given mean and variance. moreover, for exponential distribution above solution will not work. if no such function exists then please tell me the algorithm to do that.

saira
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Assisted Solution

by:skypalae
skypalae earned 132 total points
ID: 12713828
well ... yes, the above trick doesn't use mean and variance (same as original rand() doesn't). you have to recalculate the result to specific mean and variance by yourself. for example:

randVal -= RAND_MAX / 2 ; // this makes mean 0
randVal *= X ; // this sets the variance to X

-----

anyway. have you tried simple rewriting of mathematical formulas (http://astronomy.swin.edu.au/~pbourke/analysis/distributions/) ?

normal distribution:

x = rand () ;
m = mean ;
s = variance ;
pi = 3.1415 ;
y = 1 / (s * sqrt (2*pi)) * exp (-0.5 * pow ((x-m)/s, 2)) ;


exponential:

x = rand () ;
l = lambda ;
y = (x>=0 && l>0)? l * exp (-l * x) : 0 ;


S.
0
 

Author Comment

by:sairaseven
ID: 12723565
well the solution given above don't generate normally distributed random number it just apply a transformation to a uniformly distributed random variable. same is the case with exponential. it did not work
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Assisted Solution

by:guitaristx
guitaristx earned 132 total points
ID: 12822809
rand() gives you a values between a given range (0 - 1), all you've got to do is take that value and apply a function to it.  For instance, if you want a gaussian distribution, simply do this:

double value = my_gaussian_function(rand());

The fun part is implementing those functions.  If it were me implementing it, I would use a lookup table of about 256 elements, and do a second-order polynomial extrapolation taking the nearest four values.  Good luck!
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