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What are the potential distributions for a continuous variable *X* on [0, 1], if |2*X* - 1| is known to have a uniform distribution on [0, 1]? Will the distribution of INT(2*X*) 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(2*X*) and obviously being equal to 0 or 1, are uniformly distributed (50% of 0's and 50% of 1's.)

**Update**

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(2*X*) 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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