Generate random numbers according to distributions

问题

I want to generate random numbers according some distributions. How can I do this?

回答1:

The standard random number generator you've got (rand() in C after a simple transformation, equivalents in many languages) is a fairly good approximation to a uniform distribution over the range [0,1]. If that's what you need, you're done. It's also trivial to convert that to a random number generated over a somewhat larger integer range.

Conversion of a Uniform distribution to a Normal distribution has already been covered on SO, as has going to the Exponential distribution.

[EDIT]: For the triangular distribution, converting a uniform variable is relatively simple (in something C-like):

double triangular(double a,double b,double c) {
   double U = rand() / (double) RAND_MAX;
   double F = (c - a) / (b - a);
   if (U <= F)
      return a + sqrt(U * (b - a) * (c - a));
   else
      return b - sqrt((1 - U) * (b - a) * (b - c));
}

That's just converting the formula given on the Wikipedia page. If you want others, that's the place to start looking; in general, you use the uniform variable to pick a point on the vertical axis of the cumulative density function of the distribution you want (assuming it's continuous), and invert the CDF to get the random value with the desired distribution.

回答2:

The right way to do this is to decompose the distribution into n-1 binary distributions. That is if you have a distribution like this:

A: 0.05
B: 0.10
C: 0.10
D: 0.20
E: 0.55

You transform it into 4 binary distributions:

1. A/E: 0.20/0.80
2. B/E: 0.40/0.60
3. C/E: 0.40/0.60
4. D/E: 0.80/0.20

Select uniformly from the n-1 distributions, and then select the first or second symbol based on the probability if each in the binary distribution.

Code for this is here

回答3:

It actually depends on distribution. The most general way is the following. Let P(X) be the probability that random number generated according to your distribution is less than X.

You start with generating uniform random X between zero and one. After that you find Y such that P(Y) = X and output Y. You could find such Y using binary search (since P(X) is an increasing function of X).

This is not very efficient, but works for distributions where P(X) could be efficiently computed.

回答4:

You can look up inverse transform sampling, rejection sampling as well as the book by Devroye "Nonuniform random variate generation"/Springer Verlag 1986

回答5:

You can convert from discrete bins to float/double with interpolation. Simple linear works well. If your table memory is constrained other interpolation methods can be used. -jlp

回答6:

It's a standard textbook matter. See here for some code, or here at Section 3.2 for some reference mathematical background (actually very quick and simple to read).

来源：https://stackoverflow.com/questions/3510475/generate-random-numbers-according-to-distributions

标签

random

distribution