# Generate pair of random numbers with respect to a sum constraint?

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Michael Ziedalski il 21 Feb 2018
Commentato: César il 26 Mar 2019
Hello, all. So I am still gaining experience with Matlab and am currently trying to generate sequences of two numbers, x and y, such that their sum is <= 1. The naive way I immediately thought to do this would be by 1) generating x within the range [0, .5], and 2) keep generating y until it is < x, which would guarantee my condition, but introduce some significant statistical bias.
Is there some standard, statistically robust way to do this, guys? I would be very grateful for any of your input on this matter.
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### Risposta accettata

Roger Stafford il 22 Feb 2018
Modificato: Roger Stafford il 22 Feb 2018
(Corrected) Assuming you restrict x and y to non-negative values, the set of x and y values for which x+y<=1 would be a triangular area in the xy plane. You can obtain an area-wise uniform distribution of x and y with the following.
For generating a single pair:
x = 1-sqrt(rand);
y = (1-x)*rand;
To get row vectors with n elements each:
x = 1-sqrt(rand(1,n));
y = (1-x).*rand(1,n);
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### Più risposte (1)

Jeff Miller il 21 Feb 2018
It isn't entirely clear what joint distribution you want for (x,y), but here is one possibility:
x = rand; % uniform 0 to 1
y = (1-x)*rand; % uniform 0 to 1-x
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Roger Stafford il 22 Feb 2018
Modificato: Roger Stafford il 22 Feb 2018
Michael: I will attempt to answer your question. Draw a vertical line at a value x and consider the area of the triangle to the right of the line. It is proportional to (1-x)^2, so the probability of choosing an x to the right should be proportional to (1-x)^2:
p = k*(1-x)^2
and setting x = 0 it is clear that k equals 1. Hence
1-x = sqrt(p)
We replace p by Matlab's 'rand' which then plays to role of p to get:
1-x = sqrt(rand)
x = 1-sqrt(rand)
That is, let r1 < r2 be two possible values of rand. Then the probability of rand lying between them is r2-r1. The corresponding values of x are x1 = 1-sqrt(r1) and x2 = 1-sqrt(r2) and we have r1 = (1-x1)^2 and r2 = (1-x2)^2. Hence the probability of lying between x1 and x2 is
r2-r1 = (1-x2)^2-(1-x1)^2
which is what we wish to achieve since that is proportional to the area between x1 and x2 under the line x+y=1.
[For higher n dimensional "triangles", otherwise known as 'simplexes', the answer will of course be different and involve n-th roots.]
César il 26 Mar 2019
Dear @Roger Stafford,
Could you please give also some explanations of how do y = (1-x).*rand(N,1) give the correcrt distribution please?

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