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What are the potential distributions for a continuous variable X on [0, 1], if |2X - 1| is known to have a uniform distribution on [0, 1]? Will the distribution of INT(2X) always be uniform on {0, 1} ?

This question arises in a potential proof that the digits of the number Pi in base 2 (see exercise 7 in this article), distributed as INT(2X) and obviously being equal to 0 or 1, are uniformly distributed (50% of 0's and 50% of 1's.) 


I spent more time on this problem, and it is not an easy one. There are actually infinitely many solutions, as many as there are real numbers on [0, 1]. The vast majority of these distributions are nowhere continuous -- they don't have a density. To understand this, do the following simulation:

  • Simulate n random deviates u(n) uniformly distributed on [0, 1].
  • Generate n numbers d(n) distributed on {-1, +1}. They don't need to be uniformly distributed: they can all be -1 or +1 or any combination of both. For instance d(n) can be -1 if the n-th digit of Pi in base 2, is zero, and +1 if the n-th digit of Pi in base 2, is one. You can use any other number instead of Pi, for instance 7/13, and then the final result will be different.
  • For each n, compute v(n) = d(n) * u(n).
  • For each n, compute x(n) = (1 + v(n)) / 2.

The limiting random variable X attached to the x(n)'s, as n tends to infinity, is solution to the problem. However, there are as many solutions as there are ways to generate the d(n)'s, and the distribution of INT(2X) will be discrete on {0, 1}, but usually not uniform: it will depend on the proportions of +1 and -1 in the d(n)'s. If you use the number Pi to compute the d(n)'s, it will be uniform.

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