# Does this seems like a reasonable definition of a uniform distribution?

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}

• 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. – Pohoua May 18 at 16:09
• This definition even doesn't have a probability measure $P$ in it so it is invalid. – Zhanxiong May 18 at 16:20
• Cross-posted at math.stackexchange.com/q/3680737/321264. – 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 $$X.$$

• 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? – jld May 18 at 18:48
• @jld That's an interesting question. It exposes the role of a group operation on the codomain of $X.$ – whuber May 18 at 20:12
• Ah interesting, thank you – jld May 18 at 20:36