Non-uniform random distribution: How do I get a random between 100 and 180 that is on average close to 120? (like in a Gaussian distribution)

Question

Presume I have the following case:

• int value a
• int value b, for which a < b
• int value i, for which a <= i < b
• I need an int value x, for which a <= x < b, randomly chosen, according to a non-uniform distribution: the distribution should follow a bell curve which is centered around i.

In Java, I have the following methods available on java.util.Random:

• nextInt(int m): int between 0 and m
• nextDouble(): double between 0.0 and 1.0
• nextGaussian(): double between -Infinity and +Infinity, usually close to 0.0.

How do I build such a non-uniform distribution from these building blocks? How do I reliably and efficiently transform nextGaussian() into nextGaussian(a, b, i)? The part I am struggling with is to enforce that x is selected between a and b (without doing a trial-and-error).

Answer 1

I would generate a beta distribution, and then expand to the desired range and round to the nearest integer.

A beta distribution on the 0-1 scale has two parameters: $Beta(a,b)$. $Beta(1,1)$ is simply a flat uniform. You want $a$ and $b$ to be > 1 or it won't have a central peak. The mean is $a/(a+b)$, and the mode (peak) is $(a-1)/(a+b-2)$.

The larger $a$ and $b$ are, the more "peaked" the distribution is.

Here's how to generate it from uniform random numbers if $a$ and $b$ are integers:

First, here's how you generate a Gamma variate $Gamma(1,a)$.

double Gamma1(int n){
  double sum = 0;
  for (int i = 0; i < n; i++){
    sum += -log(getUniform());
  }
  return sum;
}

Then

double Beta(int a, int b){
  double x = Gamma1(a);
  double y = Gamma1(b);
  return x/(x+y);
}

Then for your problem, you have a span from A to B with mode at I. Then the distance from A to I is fraction $(I-A)/(B-A)$, so that is the mode of the distribution. Then decide how large you want $a+b$ to be, the more the sharper. You might start with 10, call it $n$. Then calculate $$a = (I-A)/(B-A)*(n-2)+1$$ and use the nearest integer value for $a$, and of course $b=n-a$.

Then all you gotta do is calculate

A + (B-A)*Beta(a,b)

and round to the nearest integer.

It sounds like your requirements are pretty flexible, so this should be good enough.

Anyway, look up Beta distribution.

EDIT: Another way that might be simpler: Use order statistics. Generate $n$ uniforms, sort them, and then choose the $a$th one, and scale that.

Answer 2

Your problem is rather vague: "bell curve" and "centered around i" are not well defined.

One possible recipe is to generate a truncated gaussian: define s= min(b-i,i-a)/c where c is a free parameter (suggested values: [1,3] range), generate a gaussian with mean i and standard deviation s - and round it to the nearest integer. If the value falls outside the [a,b] range, try again.

Related.

Answer 3

Use the alias method. There's a fairly elaborate description by Keith Schwarz, which there is no reason to repeat here. Instead, I'll explain what the final algorithm looks like.

Before describing the algorithm, let me explain the data you need to come up with. You need to decide on a probability distribution over the set $\{100,\ldots,180\}$. That is, you need to decide on numbers $p_{100},\ldots,p_{180}$ such that the probability that your random number is $i \in \{100,\ldots,180\}$ equals $p_i$.

Given the $p_i$, you can construct another array $q_{100},\ldots,q_{180} \in [0,1]$ and two integer arrays $a_{100},b_{100},\ldots,a_{180},b_{180} \in \{100,\ldots,180\}$ such that the following function samples the distribution $p_i$:

1. Sample uniformly a number $t \in \{100,\ldots,180\}$ and a real $\theta \in [0,1]$.
2. If $\theta < q_t$ then return $a_t$, otherwise return $b_t$.

Keith Schwarz's essay explains how to generate the arrays $q_i,a_i,b_i$ (he uses different letters; look under the section The Alias Method). The essay gives several algorithms for generating these arrays; in your case the range $\{100,\ldots,180\}$ is very small, so you can use whichever you prefer.