The following subsets of $\mathbb{R}$ (not an exhaustive list) admit a uniform probability distribution:

  1. Finite sets
  2. Intervals (open or closed) with both endpoints finite
  3. Finite unions of the above

Is there a simpler way to characterize the set of subsets $S \subset \mathbb{R}$ that admit a uniform probability distribution?

One naive guess is that $S$ is bounded, but this requirement is neither necessary nor sufficient: e.g. the set $$\left \{ \frac{1}{n} : n \in \mathbb{Z}^+ \right \}$$ is bounded but does not admit a uniform probability distribution, while the set $$\bigcup_{n = 0}^\infty \left[ n - 2^{-n}, n + 2^{-n} \right]$$ is not bounded but does admit a uniform distribution.

My guess is that the answer is no, because by combining together sets of types 1 and 2 in the list above, we are combining "apples and oranges", as discrete and continuous probability distributions are qualitatively different. If we want to extend the support of a discrete random variable taking on a finite number of possible values to the real line, then we shouldn't think of its probability distribution as a probability mass function, but instead as a finite sum of Dirac delta functions, which are a very different beast from continuous probability density functions. Is this correct?

| cite | improve this question | | | | |
  • $\begingroup$ What is your definition of a uniform probability distribution? I'm also confused regarding your point 3: it seems to me that any finite intersection of finite sets and bounded intervals is either a finite set or a bounded interval. $\endgroup$ – Anthony Carapetis Feb 20 '18 at 3:48
  • $\begingroup$ @AnthonyCarapetis Oops, I meant to say "finite unions", fixed. $\endgroup$ – tparker Feb 20 '18 at 4:11

The essence of being a discrete distribution is concentrating all probability at point masses. I.e. we have some random variable (capital) $X$ and there is some set of objects (lower-case $X$) $x$ for which $\Pr(X=x)>0,$ and we have $$ \sum_x \Pr(X=x) = 1. $$ We can call such a distribution "uniform" if $\Pr(X=x)$ is the same for all values of $x.$ If, but only if, the aforementioned "set of objects" is finite, a uniform distribution exists.

That's for discrete distributions.

A "continuous" distribution on $\mathbb R$ assigns probabilities to intervals, and hence to Borel sets (i.e. those sets that can be constructed by starting with intervals and closing under countable unions and complements) and has a continuous c.d.f.; thus does not concentrate positive probability at any point. For every $x\in\mathbb R,$ we have $\Pr(X=x) = 0.$ This is true regardless of whether the distribution is what is conventionally called "uniform". In that case, one calls a distribution "uniform" if it assigns equal probabilities, not to every point, but to intervals of equal lengths. And one can say that the probability density function has the same value everywhere within the support of the distribution. The "support" is the set of all points $x$ for which, for every positive number $\varepsilon,$ no matter how small, $\Pr(x-\varepsilon < X<x+\varepsilon)>0.$ If you want to look at sets other than intervals, you can say that a "uniform" distribution assigns equal probabilities to sets with the same "measure", and for that one must know what "measure" is. The measure of an interval is its length; every Borel set has a measure; if you add the same constant to every member of a set to get a new set, you don't alter the measure; if a sequence of sets is pairwise mutually exclusive, then the measure of their union is the sum of their measures. That's enough to determine what the measure of every Borel set is. And the bottom line will be that there is a "uniform" distribution on a Borel set if and only if its measure is finite. For that, it need not be bounded. For example, the set $\displaystyle \bigcup_{n=1}^\infty [n, n + 1/2^n)$ has finite measure but is not bounded.

| cite | improve this answer | | | | |
  • $\begingroup$ I actually edited my question right before reading your answer in a way that addresses your last point already. I was (loosely) also considering "hybrid" distributions consisting of both intervals and discrete points, like $$p(x) = \begin{cases} 2/3 & \text{if } x \in [1, 1/2] \text{ or } x = 1 \\ 0 & \text{otherwise} \end{cases}.$$ I think the fact that such hybrid distributions are a bit contrived to formalize reflects the fact that the answer to my question is no. $\endgroup$ – tparker Feb 20 '18 at 4:24
  • $\begingroup$ @tparker : I diagree with the punctuation in your comment. I might have written something like this: $$ p(x) = \begin{cases} 2/3 & \text{if } x \in [1, 1/2] \text{ or } x = 1, \\ 0 & \text{otherwise}. \end{cases} $$ $\endgroup$ – Michael Hardy Feb 20 '18 at 15:58
  • $\begingroup$ Hmm, interesting. I'll have to think about that, but you may well be right. $\endgroup$ – tparker Feb 20 '18 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.