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?

Answer 1

Create and array with the distribution you require

e.g. 1,1,1,2,3,3,6,6,7,8,9,9,9,9,9,9,10

use a standard random number generator to select indexes into this array.

Answer 2

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

Answer 3

Java's Random class has the nextGaussian() method for this.

Answer 4

I would use Box-Muller. Two things about this:

1. 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.
2. 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.

Answer 5

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

Answer 6

function distRandom(){
  do{
    x=random(DISTRIBUTION_DOMAIN);
  }while(random(DISTRIBUTION_RANGE)>=distributionFunction(x));
  return x;
}

Answer 7

Changing the distribution of any function to another involves using the inverse of the probability function you want.

In other words, if you know the probability function p(x) and it has an inverse: Inv(p(x)) then by using the random probability function (uniform distribution) and casting the result value through the function Inv(p(x)) you should get random values cast with distribution according to the function you wanted, so now it's only a matter of choosing your desired probability function and its inverse.

Hope this helped and that I didn't mixed my math :)

Answer 8

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)

Answer 9

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
 */
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 > 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 */
}

Answer 10

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.