Does this seem like a reasonable definition of a uniform distribution?

Let \begin{align} &S \text{ be a sample space}; \tag{1} \\ &X:S\rightarrow \mathbb{R} \; \text{ be a random variable on } S. \tag{2} \\ \end{align} Then \begin{align} & X(S) \text{ is distributed uniformly } \\ \Leftrightarrow &\text{ every realisation of } X \text{ has pre-image of equal size}, \tag{3.1}\\ \Leftrightarrow &\forall \; x_1,x_2 \in X(S): \; |X^{-1}(\{x_1\})|=|X^{-1}(\{x_2\})|. \tag{3.2} \end{align}

  • $\begingroup$ I think this is correct if $X$ can only take a finite number of values. For a continuous random variable, the preimage of any value is of measure equal to zero, even if the random variable is not uniform. $\endgroup$ – Pohoua May 18 at 16:09
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    $\begingroup$ This definition even doesn't have a probability measure $P$ in it so it is invalid. $\endgroup$ – Zhanxiong May 18 at 16:20
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    $\begingroup$ Cross-posted at math.stackexchange.com/q/3680737/321264. $\endgroup$ – StubbornAtom May 18 at 17:33

Let me answer the implicit question: what is a uniform distribution?

Because $X$ is a random variable, $S$ really is the underlying set in a probability space $(S,\mathfrak F, \mathbb P).$

We say $X$ has a continuous uniform distribution when there exists a subset $A\subset \mathbb R$ such that, for every interval $(a,b]\subset \mathbb{R},$ $$\mathbb{P}(X\in (a,b]) \ \propto\ \lambda((a,b]\cap A)$$ where $\lambda$ is Lebesgue measure.

$X$ has a discrete uniform distribution when there exists $A\subset \mathbb{R}$ such that for every interval $(a,b],$ $$\mathbb{P}(X\in (a,b])\ \propto\ |(a,b] \cap A|$$ where $|\cdot |$ is the cardinality of a set.

The implicit normalizing constants (denominators) in these equations are $\lambda(A)$ in the first case and $|A|$ in the second, both of which (therefore) must be finite and nonzero. In particular,

  • in the continuous case, $X$ has a probability density function equal to $$f_X(x) = \frac{1}{\lambda(A)} \mathcal{I}_A(x)$$ (where $\mathcal{I}_A$ is the indicator function of $A$);

  • in the discrete case we may write $|A|=n$ (a nonzero natural number) and see that for any number $x\in \mathbb R,$ $\Pr(X=x) = 1/n$ when $x\in A$ and otherwise $\Pr(X=x)=0.$ This defines the probability mass function of $X.$

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  • $\begingroup$ Are there some fundamental properties of $\lambda$ and $|\cdot|$ as measures that make them suitable for uniformity, like translation invariance plus $\sigma$-finiteness, in a way that doesn't require a discrete/continuous distinction? $\endgroup$ – jld May 18 at 18:48
  • $\begingroup$ @jld That's an interesting question. It exposes the role of a group operation on the codomain of $X.$ $\endgroup$ – whuber May 18 at 20:12
  • $\begingroup$ Ah interesting, thank you $\endgroup$ – jld May 18 at 20:36

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