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Assume that I have a discrete set $L$ and a transformation $\phi: L \rightarrow [0,1]$ that normalizes set $L$ such that now values belonging $L$ are uniformly distributed among the unit interval.

How do I mathematically state that $\phi(L)$ adheres to uniform distribution?

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    $\begingroup$ I'm not sure I understand what you're discussing, since a set $L$ isn't a random variable. If I have $L=\{1,2,4,17,190\}$ what makes $L$ uniformly distributed - rather than (say) merely equispaced? $\endgroup$ – Glen_b -Reinstate Monica Jun 17 '14 at 9:16
  • $\begingroup$ @Glen_b I'm sorry, my bad. Maybe the correct term is uqispaced. So what I've in mind is a transformation $\psi$ where the min element of $L$ is mapped to 0, max element of $L$ is mapped to 1 and the rest are mapped to some value in the unit interval s.t. $x_i - x_{i-1} = x_{i+1} - x_i$. For example wrt. your definition of $L$, $\psi(4) = 0.5$ $\endgroup$ – alexT Jun 17 '14 at 9:46
  • $\begingroup$ Are multisets (like $\{1,1,1,1,9\}$) as might be found with data excluded? $\endgroup$ – Glen_b -Reinstate Monica Jun 17 '14 at 10:06
  • $\begingroup$ @Glen_b for the sake of simplicity let's just assume sets a.k.a. all values in the input set being unique. $\endgroup$ – alexT Jun 17 '14 at 10:09
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If repeated values can't occur in $L$, I think the word you're probably after is a form of probability points (used in probability plotting, something like a sample percentile rank).

If the values in $L$ are $\mathbf{x}=(x_1,x_2,...,x_n)$ and $r(L)=\text{rank}(\mathbf{x})$ (which for distinct values are equivalently the indices of the order statistics), then one definition, $p_i=\frac{r_i-\alpha}{n+1-2\alpha}$ would seem to give the kind of thing you seem to be after. Specifically, with $\alpha=1$, you would reproduce the example we discussed in comments:

This is available as a standard function in a number of stats packages. So for example, R has such a probability points function, ppoints:

(L=sort(rnorm(4)))
[1] -2.5736722 -0.1084315  0.1672061  1.4856733

ppoints(L,a=1)
[1] 0.0000000 0.3333333 0.6666667 1.0000000

ppoints(c(1,2,4,17,190),a=1)
[1] 0.00 0.25 0.50 0.75 1.00

The probability points are equispaced by construction; they're a linear transformation of the ranks, $1,2, ..., n$.

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