Generate random number with given probability matlab

>> c = cumsum([0.5, 0.1, 0.4]);
>> r = rand(1e5, 1);
>> x = arrayfun(@(x) find(x <= c, 1, 'first'), r);
>> h = hist(x, 1:3)

h =

       49953       10047       40000

x distributed as desired.

The simple solution is to generate a number with a uniform distribution (using rand), and manipulate it a bit:

r = rand;
prob = [0.5, 0.1, 0.4];
x = sum(r >= cumsum([0, prob]));

or in a one-liner:

x = sum(rand >= cumsum([0, 0.5, 0.1, 0.4]));

Explanation

Here r is a uniformly distributed random number between 0 and 1. To generate an integer number between 1 and 3, the trick is to divide the [0, 1] range into 3 segments, where the length of each segment is proportional to its corresponding probability. In your case, you would have:

• Segment [0, 0.5), corresponding to number 1.
• Segment [0.5, 0.6), corresponding to number 2.
• Segment [0.6, 1], corresponding to number 3.

The probability of r falling within any of the segments is proportional to the probabilities you want for each number. sum(r >= cumsum([0, prob])) is just a fancy way of mapping an integer number to one of the segments.

Extension

If you're interested in creating a vector/matrix of random numbers, you can use a loop or arrayfun:

r = rand(3); % # Any size you want
x = arrayfun(@(z)sum(z >= cumsum([0, prob])), r);

Of course, there's also a vectorized solution, I'm just too lazy to write it.

The answers so far are correct, but slow for large inputs: O(m*n) where n is the number of values and m is the number of random samples. Here is a O(m*log(n)) version that takes advantage of monotonicity of the cumsum result and the binary search used in histc:

% assume n = numel(prob) is large and sum(prob) == 1
r = rand(m,1);
[~,x] = histc(r,cumsum([0,prob]));