How to generate random positive floating point numbers with normal distribution from 0 up to a given ceiling?

Question

I need to generate a number of float numbers with approximately normal distribution over a range from 0 to a specific ceiling.

I've searched on stack overflow and found similar questions for other languages, but none for .net core.

internal List<float> function(int ceiling, int repetitions)
{
    List<float> list = new List<float>();
    for (int i = 0; i<= repetitions;i++)
    {
         list.Add(Random.nextFloat() * ceiling);
    }
    return list;
}

I expect the function to return a list of random positive float numbers, in range from 0 to a given ceiling with at least approximately normal distribution.

Answer 1

If you're seeking something "at least approximately normal" with bounds at 0 and ceiling , summing three uniforms will yield a result which is symmetric, bell-shaped, and bounded, and can subsequently be rescaled to any range you wish. I'm not a C# programmer, but if you have a PRNG named prng :

(prng.NextDouble() + prng.NextDouble() + prng.NextDouble()) * ceiling / 3.0

will yield a result in the range [0, ceiling] . Here's what 100,000 observations look like with ceiling set to 3:

You can generalize this to sum k uniforms and replace the 3 by k in the divisor for the rescaling. The larger k is, the closer this will get to normality by the central limit theorem, but since you don't seem to be asking for actual normals (which don't have a bounded range anyway) that quickly gets into diminishing returns.

Note that while this approach uses multiple uniforms, it is computationally relatively efficient because it avoids transcendental functions.

Answer 2

Well, you could use truncated normal together with taking absolute values to make result positive.

Along the lines

double R = 10.0; // upper value of the truncated normal

var seed = 31234567;
Random rng = new Random( seed );

double u1 = rng.NextDouble();
double u2 = rng.NextDouble();

double phi = 2.0*Math.PI*u2;
double r   = Math.Sqrt(-2.0*Math.Log(1.0 - u1*(1.0 - Math.Exp(-R*R/2.0))));

return new Tuple<double,double>(Math.Abs(r*Math.Cos(phi)), Math.Abs(r*Math.Sin(phi)));

Code above shall return couple of sampled values in the interval from 0 to R which looks like truncated gaussian. You could compare with Box-Muller for standard gaussain sampling