Converting a Uniform Distribution to a Normal Dist

问题:

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?

回答1:

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

回答2:

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.

回答3:

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.

回答4:

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 */
}

回答5:

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)

回答6:

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.

回答7:

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

回答8:

Where R1, R2 are random uniform numbers:

NORMAL DISTRIBUTION, with SD of 1: sqrt(-2*log(R1))*cos(2*pi*R2)

This is exact... no need to do all those slow loops!

回答9:

It seems incredible that I could add something to this after eight years, but for the case of Java I would like to point readers to the Random.nextGaussian() method, which generates a Gaussian distribution with mean 0.0 and standard deviation 1.0 for you.

A simple addition and/or multiplication will change the mean and standard deviation to your needs.

回答10:

I thing you should try this in EXCEL: =norminv(rand();0;1). This will product the random numbers which should be normally distributed with the zero mean and unite variance. \"0\" can be supplied with any value, so that the numbers will be of desired mean, and by changing \"1\", you will get the variance equal to the square of your input.

For example: =norminv(rand();50;3) will yield to the normally distributed numbers with MEAN = 50 VARIANCE = 9.

回答11:

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)

回答12:

This is a Matlab implementation using the polar form of the Box-Muller transformation:

Function randn_box_muller.m:

function [values] = randn_box_muller(n, mean, std_dev)
    if nargin == 1
       mean = 0;
       std_dev = 1;
    end

    r = gaussRandomN(n);
    values = r.*std_dev - mean;
end

function [values] = gaussRandomN(n)
    [u, v, r] = gaussRandomNValid(n);

    c = sqrt(-2*log(r)./r);
    values = u.*c;
end

function [u, v, r] = gaussRandomNValid(n)
    r = zeros(n, 1);
    u = zeros(n, 1);
    v = zeros(n, 1);

    filter = r==0 | r>=1;

    % if outside interval [0,1] start over
    while n ~= 0
        u(filter) = 2*rand(n, 1)-1;
        v(filter) = 2*rand(n, 1)-1;
        r(filter) = u(filter).*u(filter) + v(filter).*v(filter);

        filter = r==0 | r>=1;
        n = size(r(filter),1);
    end
end

And invoking histfit(randn_box_muller(10000000),100); this is the result:

Obviously it is really inefficient compared with the Matlab built-in randn.

回答13:

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)]

回答14:

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

回答15:

Approximation:

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

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