Converting a Uniform Distribution to a Normal Dist

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

I have the following code which maybe could help:

set.seed(123)
n <- 1000
u <- runif(n) #creates U
x <- -log(u)
y <- runif(n, max=u*sqrt((2*exp(1))/pi)) #create Y
z <- ifelse (y < dnorm(x)/2, -x, NA)
z <- ifelse ((y > dnorm(x)/2) & (y < dnorm(x)), x, z)
z <- z[!is.na(z)]

Answer 2

Q 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?

1. For software implementation I know couple random generator names which give you a pseudo uniform random sequence in [0,1] (Mersenne Twister, Linear Congruate Generator). Let's call it U(x)

2. It is exist mathematical area which called probibility theory. First thing: If you want to model r.v. with integral distribution F then you can try just to evaluate F^-1(U(x)). In pr.theory it was proved that such r.v. will have integral distribution F.

3. Step 2 can be appliable to generate r.v.~F without usage of any counting methods when F^-1 can be derived analytically without problems. (e.g. exp.distribution)

4. To model normal distribution you can cacculate y1*cos(y2), where y1~is uniform in[0,2pi]. and y2 is the relei distribution.

Q: What if I want a mean and standard deviation of my choosing?

You can calculate sigma*N(0,1)+m.

It can be shown that such shifting and scaling lead to N(m,sigma)

Answer 3

Approximation:

function rnd_snd() {
    return (Math.random()*2-1)+(Math.random()*2-1)+(Math.random()*2-1);
}

See http://www.protonfish.com/random.shtml

Answer 4

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 5

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

Answer 6

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