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I'm trying to formulate a way to generate a random set of 4 numbers that sum to 1 (a discrete probability distribution) such that each is uniform, ~$U(0,1)$. They obviously can't be iid, and normalized uniforms have a mean 0.25 which is way too low for our purposes. The way I've been trying to tackle it is by giving each one a 25% chance of coming from a "bigger" distribution.

Let $X_1,X_2,X_3$ iid ~$U(0,1)$ and $Y$ with support $(0,\infty)$. Let $A={X_1\over X_1+X_2+X_3+Y}, B={Y\over X_1+X_2+X_3+Y}$. Find the $F_Y$ such that ${1\over 4}f_B(z)+{3\over 4}f_A(z)=1.$

jrclimer
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The mean of each of them will have to be $0.25$ if they come from the same distribution because means add. The easiest approach is to pull four numbers from $U(0,1)$ and normalize them by dividing by the sum. This will give each a distribution which ranges all the way up to $1$, but there will be little density there.

Ross Millikan
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  • Yeah the means adding makes total sense. I tried the U(0,1) thing first, and there's just too little density up there for our purposes. I also tried to scale things by "bigger" distributions and didn't like how dramatic the dropoff in the middle was. I'll keep messing around. Thanks. – jrclimer Apr 07 '18 at 23:58
  • The point is you can't have four big numbers sum to $1$. They need to be small. That is what the means being additive means. You are asking the impossible. – Ross Millikan Apr 08 '18 at 00:05
  • You're right and I'm not disputing that - you can do better than normalizing uniform randoms by the sum. The Kullback-Leibler divergence for that distribution: – jrclimer Apr 11 '18 at 01:51
  • you are correct and I don't dispute that U(0,1) is impossible, and that whatever function I use to generate the values must have a mean of 0.25. However, you can do better than normalizing uniform randoms by their sum. The Kullback-Leibler divergence for that distribution is 1.83. If we instead use a procedure where we pick one of the four numbers to be drawn from a uniform, and the other three to have a sum of 1 minus that number, the divergence drops to 0.445. I am curious about optimizing this, but this question is poorly posed. Thanks for your help. – jrclimer Apr 11 '18 at 02:03