Converting a Uniform Distribution to a Normal Distribution

Question

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

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.

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.

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 >= 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 */
}

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)