Generating N numbers that sum to 1

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Given an array of size n I want to generate random probabilities for each index such that Sigma(a[0]..a[n-1])=1

One possible result might be:

0     1     2     3     4
0.15  0.2   0.18  0.22  0.25

Another perfectly legal result can be:

0     1     2     3     4
0.01  0.01  0.96  0.01  0.01

How can I generate these easily and quickly? Answers in any language are fine, Java preferred.

6 Answers

Get n random numbers, calculate their sum and normalize the sum to 1 by dividing each number with the sum.

The task you are trying to accomplish is tantamount to drawing a random point from the N-dimensional unit simplex.

http://en.wikipedia.org/wiki/Simplex#Random_sampling might help you.

A naive solution might go as following:

public static double[] getArray(int n)
    {
        double a[] = new double[n];
        double s = 0.0d;
        Random random = new Random();
        for (int i = 0; i < n; i++)
        {
           a [i] = 1.0d - random.nextDouble();
           a [i] = -1 * Math.log(a[i]);
           s += a[i];
        }
        for (int i = 0; i < n; i++)
        {
           a [i] /= s;
        }
        return a;
    }

To draw a point uniformly from the N-dimensional unit simplex, we must take a vector of exponentially distributed random variables, then normalize it by the sum of those variables. To get an exponentially distributed value, we take a negative log of uniformly distributed value.

public static double[] array(int n){

    double[] a = new double[n];
    double flag = 0;

    for(int i=0;i<n;i++){
        a[i] = Math.random();
        flag += a[i];
    }

    for(int i=0;i<n;i++) a[i] /= flag;

    return a;
}

Here, at first a stores random numbers. And the flag will keep the sum all the numbers generated so that at the next for loop the numbers generated will be divided by the flag, which at the end the array will have random numbers in probability distribution.

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