0
$\begingroup$

In probability theory, there are a number of equivalent ways to characterize a distribution on $\mathbb R^n$. For example, the distribution of a random vector $X\in\mathbb R^n$ may be characterized by:

  1. $F(x)\equiv \mathrm{Pr}[X\leq x]$ for $x\in\mathbb R^n$.
  2. $\mathbb E[g(X)]$ for all continuous functions $g$.
  3. $\mathbb E[g(X)]$ for all bounded continuous functions $g$.

Notably, characterization (3) weakens characterization (2) quite a bit in the sense that one only needs to know the expected value for a smaller set of functions $g(\cdot)$. My question is as follows: are there any results characterizing just how small the set of functions $g(\cdot)$ for which we need to know the expectations in order to characterize the distribution of $X$? Something like the Stone-Wierstrass approximation theorem seems to be related, but I am somewhat at a loss in terms of what to search for in connecting the dots.

$\endgroup$
1
  • 3
    $\begingroup$ Every family of functions that is total in $\mathcal{C}_0(\mathbb{R}^n)$ works, by linearity and density. $\endgroup$ Commented Jan 17, 2023 at 21:36

2 Answers 2

1
$\begingroup$

Let us considering indicator functions of sets. Patrick Billingsley in his book "convergence of probability measures" defines a convergence-determining class of sets as a class $\cal A$ such that $P_n(A) \rightarrow P(A)$ for all $A \in {\cal A}$ implies weak convergence of $P_n$ to $P$.

He proves that the following conditions implies that $\cal A$ is a convergence-determining class.

-- $\cal A$ is closed under finite intersection,

-- For each $x$ and $\varepsilon>0$, either there is no set $A \in {\cal A}$ satisfying $x \in int(A) \subset A \subset B(x,\varepsilon)$ or there is a countable family of such sets with disjoint boundaries.

$\endgroup$
0
$\begingroup$

[A]re there any results characterizing just how small the set of functions $g(\cdot)$ for which we need to know the expectations in order to characterize the distribution of $X$?

This is a very basic answer, if an additional condition on $X$ is imposed. If the random variable $X$ is bounded, then it is well known that the powers of $X$ form a sufficient set of functions. That is, if $\Pr[X<a]=\Pr[X>b]=0$, then the distribution of $X$ is completely characterized by the moments $$m_{n}=\mathbb{E}[X^{n}],$$ for all $n\in \mathbb{Z}^{+}$. The reverse problem, of determining which moments actually correspond to a random variable with support on $[a,b]$, is the classic Hausdorff Moment Problem.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .