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Suppose that I can generate some random variable $X$ that is distributed according to the CDF $F$. If $F$ is continuous then $F(X)$ is uniform $[0,1]$ (can anyone explain this to me).

My question is if $F$ is a CDF of some discrete R.V. $X$, is there some method for generating a uniform $[0,1]$ random variable from $F$ in this case (references or sketch of proofs appreciated)?

This is purely from a probability theoretic perspective (the application is information theoretic cryptography) so I am not interesting in pseudo generation. I am looking for a way to transform $X$ to be a uniform R.V. on $[0,1]$. Thanks in advance.

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2 Answers 2

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No. There is no way to find any function F that maps a discrete random variable to a random variable on [0,1]. One way to think about this is cardinality. Since X is discrete, the range of X is at most countable. But you need the image of F(X) to be uncountable. That just ain't going to happen, with any function F.

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  • $\begingroup$ I get that. What if I can take an infinite number of draws from $F$? Then transform that random variable (the infinite sequence) to a uniform random? $\endgroup$
    – h4nusGT
    Commented Aug 6, 2014 at 15:56
  • $\begingroup$ We can generate the binary expansion of a uniform on $[0,1]$ by sampling infinitely often. $\endgroup$ Commented Aug 6, 2014 at 16:10
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(1) The distribution of a discrete random variable can be approximated by a continuous distribution function -- a good example is the normal approximation to the binomial distribution.

(2) If you mean that you have a finite sample $x_1, x_2, ..., x_n$ of a continuous random variable $X$ with CDF $F$. Then $F(X)$ is uniformly distributed in $[0,1]$.

Since $F$ is non-decreasing and right-continuous then for $z \in [0,1]$

$$P[F(X) \leq z] = P[X \leq F^{-1}(z)]=F(F^{-1}(z)) = z.$$

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