I want to use a random number generator that creates random numbers in a gaussian range where I can define the median by myself. I already asked a similar question here and now I’m using this code:
class RandomGaussian
{
private static Random random = new Random();
private static bool haveNextNextGaussian;
private static double nextNextGaussian;
public static double gaussianInRange(double from, double mean, double to)
{
if (!(from < mean && mean < to))
throw new ArgumentOutOfRangeException();
int p = Convert.ToInt32(random.NextDouble() * 100);
double retval;
if (p < (mean * Math.Abs(from - to)))
{
double interval1 = (NextGaussian() * (mean - from));
retval = from + (float)(interval1);
}
else
{
double interval2 = (NextGaussian() * (to - mean));
retval = mean + (float)(interval2);
}
while (retval < from || retval > to)
{
if (retval < from)
retval = (from - retval) + from;
if (retval > to)
retval = to - (retval - to);
}
return retval;
}
private static double NextGaussian()
{
if (haveNextNextGaussian)
{
haveNextNextGaussian = false;
return nextNextGaussian;
}
else
{
double v1, v2, s;
do
{
v1 = 2 * random.NextDouble() - 1;
v2 = 2 * random.NextDouble() - 1;
s = v1 * v1 + v2 * v2;
} while (s >= 1 || s == 0);
double multiplier = Math.Sqrt(-2 * Math.Log(s) / s);
nextNextGaussian = v2 * multiplier;
haveNextNextGaussian = true;
return v1 * multiplier;
}
}
}
Then to verify the results I plotted them with gaussianInRange(0, 0.5, 1) for n=100000000

As one can see the median is really at 0.5 but there isn’t really a curve visible. So what I’m doing wrong?
EDIT
What i want is something like this where I can set the highest probability by myself by passing a value.

The simplest way to draw normal deviates conditional on them being in a particular range is with rejection sampling:
The same sort of thing is used when you draw coordinates (v1, v2) in a circle in your unconditional normal generator.
Simply folding in values outside the range doesn’t produce the same distribution.
Also, if you have a good implementation of the error function and its inverse, you can calculate the values directly using an inverse CDF. The CDF of a normal distribution is
The CDF of a censored distribution is
To draw a random number using a CDF, you draw
vfrom a uniform distribution on [0, 1] and solveC(retval) = v. This givesYou can precalculate
t1andt2for specific parameters. The advantage of this approach is that there is no rejection sampling, so you only need a singleNextDouble()per draw. If the [from, to] interval is small this will be faster.However, it sounds like you might want the binomial distribution instead.