The usual definition of CDFs is explicitly only for random variables (as opposed to random elements), i.e., only for $\mathbb{R}$.

I’ve seen definitions to extend CDFs to random vectors, i.e., $\mathbb R^n$.

But can we go crazier? Can CDFs work for random sets? Random functions? Random stochastic processes? (I’m pulling examples from this Wikipedia list).

What exactly are the core axioms CDFs need?

If CDFs don’t work for many exotic random elements, that’s a good reason why probability needs measure theory. However, I have been fooled about needing measure theory where CDFs would have sufficed before (e.g., for $\mathbb R$, CDFs can represent any RV just fine).

  • 3
    $\begingroup$ Since a cdf is a thing that gives $P(X\leq x)$ you would presumably need to be able to order the elements in the set in some sense. $\endgroup$ – Glen_b Mar 13 '19 at 11:32
  • 4
    $\begingroup$ If you're going to generalize "CDF," you must begin with a clear concept of what you mean by "CDF." As @Glen_b points out, if the inequality in its usual definition is essential, then (obviously) you need variables with values in an ordered space (in which the Borel sets determined by the order are measurable). But the inequality can be dispensed with: after all, the CDF is just a tool for describing distributions efficiently. Could you please, then, provide a specific and clear definition of the aspects of the usual CDF you wish to generalize and what the purpose would be? $\endgroup$ – whuber Mar 13 '19 at 15:10

The reason CDFs are so useful for random variables is because a random variable's distribution (a Borel probability measure on $\mathbb{R}$) is uniquely determined by its cumulative distribution function; if you know the CDF, then you (in principle) know everything there is to know about the distribution.

That is, if $\mu$ and $\nu$ are two Borel probability measures on $\mathbb{R}$ and $$ \mu((-\infty, x]) = \nu((-\infty, x]) $$ for every $x \in \mathbb{R}$ (equality of CDFs), then in fact $$ \mu(B) = \nu(B) $$ for every Borel subset $B \subseteq \mathbb{R}$ (equality of measures).

The general fact behind this result is the following corollary of Dynkin's $\pi$-$\lambda$ Theorem:

Theorem. Let $\mu$ and $\nu$ be measures defined on a measurable space $(\mathcal{X}, \mathcal{B})$. Let $\mathcal{E}$ be a subset of $\mathcal{B}$ such that

  1. $\mu(E) = \nu(E)$ for every $E \in \mathcal{E}$
  2. $\mathcal{E}$ is a $\pi$-system (i.e., if $E, F \in \mathcal{E}$, then $E \cap F \in \mathcal{E}$)
  3. $\mu$ is $\sigma$-finite on $\mathcal{E}$ (i.e., there exists a sequence $E_1, E_2, \ldots \in \mathcal{E}$ such that $\bigcup_{n=1}^\infty E_n = \mathcal{X}$ and $\mu(E_n) < \infty$ for all $n$)

Then $\mu(E) = \nu(E)$ for all $E \in \sigma(\mathcal{E})$.

The correspondence between Borel probability measures on $\mathbb{R}$ and CDFs comes from the fact that the Borel $\sigma$-algebra on $\mathbb{R}$ is generated by the set $$ \mathcal{E} = \big\{(-\infty, x] : x \in \mathbb{R}\big\}, $$ which is a $\pi$-system on which every probability measure trivially satisfies condition (3) of the Theorem above. It is very convenient that $\mathcal{E}$ (which is a set of subsets of $\mathbb{R}$) can be very naturally parameterized by $\mathbb{R}$ (via the map $x \mapsto (-\infty, x]$).

Following this idea, if you want to generalize CDFs to functions that determine probability distributions (AKA probability measures) on some more general measurable space $(\mathcal{X}, \mathcal{B})$, then all you need to do is find a subset $\mathcal{E}$ of $\mathcal{B}$ that is a $\pi$-system and that contains a sequence of sets whose union is $\mathcal{X}$.

For example (you already mentioned this in the question), you can consider $\mathcal{X} = \mathbb{R}^n$, $\mathcal{B} = \mathcal{B}(\mathbb{R}^n)$, and $$ \mathcal{E} = \big\{(-\infty, \mathbf{x}] : \mathbf{x} \in \mathbb{R}^n\big\} $$ (where $(-\infty, \mathbf{x}] = \prod_{i=1}^n (-\infty, x_i]$ for each $\mathbf{x} = (x_1,\ldots,x_n) \in \mathbb{R}^n$) to conclude that a Borel probability measure $\mu$ on $\mathbb{R}^n$ is uniquely determined by its "CDF": $$ \mathbf{x} \mapsto \mu((-\infty, \mathbf{x}]). $$


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.