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A sequence $X_n$ of random variables is said to converge weakly (i.e. in distribution) to $X$ if for every $f \in C_b$ we have $$ Ef(X_n) \to Ef(X). $$

Now, I was wondering if there was a class of test functions such that for every $g \in Class$, we would have an equivalent definition for convergence in probability. That is $X_n$ converges in probability to $X$ if for every $g \in Class$, we have $$ Eg(X_n) \to E g(X). $$

  • The class of simple functions is somewhat obvious. –  Sep 02 '21 at 14:12
  • I doubt there is such a characterization because convergence $Eg(X_n)\to Eg(X)$ only deals with the distribution of $X_n$ towards that of $X$. Convergence in probability gives closeness between $X$ and $X_n$ as random variables and thus provides more information. In particular, $X_n$ and $X$ must be defined on the same probability space, which is something a statement like “$Eg(X_n)\to Eg(X)$” does not assert. – nejimban Sep 02 '21 at 14:16
  • Would this be true for the class of all measurable functions? – Davi Barreira Sep 02 '21 at 15:14

2 Answers2

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Convergence in probability (or convergence in measure or convergence in $L_0$) happens at the level of functions; furthermore,

  • A measure space $(\Omega,\mathscr{F},\mu)$, $\mu$ finite, is fixed.
  • Define $L_0(\Omega,\mathscr{F},\mu)$ as the space of finite $\mu$-measurable functions (measurable w.r.t completion of $\mu$)
  • Define $\|f\|_0=\inf\{\varepsilon>0:\mu(|f|>\varepsilon)\leq \varepsilon\}$. $\|\;\|_0$ defines a topology $\tau_c$ on $L_0$ which makes it a topological linear space. Moreover, a sequence $f_n\xrightarrow{n\rightarrow\infty}f$ in $L_0$ iff for any $\varepsilon>0$ $$\lim_{n\rightarrow\infty}\mu(|f-f_n|>\varepsilon)=0$$
  • Weak convergence of random variables happens at the level of measures (the laws of the random variables). In fact the Radom variables need not be defined in a common probability space. Weak convergence of measures is defined by a weak topology in the sense of functional analysis.
  • The topology $\tau_c$ is not a weak topology (in the sense of functional analysis).
  • Determining classes are are natural int context of weak topologies.

Here is the outline of weak topologies and weak convergence.

  • Weak convergence refers to a weak topology in a topological vector space.

  • Recall that if $X$ is a topological vector space and $X'$ is its dual space (the space of linear functionals that of continuous with respect to the topology in $X$). The weak topology on $X$ $\sigma(X,X')$ is the smallest topology for which all functionals in $X'$ are continuous. A local basis for this topology is generated by sets of the form $$ V=\{x\in X: |\lambda_j(x)|<\varepsilon,\,j=1,\ldots,n\}$$ where $\varepsilon>0$, $n\in\mathbb{N}$, and $\lambda_j\in X'$ for $j=1,\ldots,n$.

  • Conversely, if $X$ is a linear space and $W$ is a linear space of linear functionals (algebraic linear functionals) that separates points of $X$ (i.e. for any $x\in X$ there is $\lambda\in W$ with $\lambda(x)\neq0$) then the weak topology $\sigma(X,W)$ is the smallest topology that makes any $\lambda\in W$ continuous. A local basis is as above by with functionals restricted to $W$.

  • It turns out that that a net $\{x_\alpha:\alpha\in D\}\subset X$ converges to $x\in X$ in the topology $\sigma(X,W)$ iff for any $\lambda\in W$, $$\lim_\alpha\lambda(x_\alpha)=\lambda(x)$$

  • In the case of weak convergence of measures, consider for simplicity the space $\mathcal{M}(\mathbb{R},\mathscr{B}(\mathbb{R}))$ of Borel complex measure (one can consider finite real measures instead) on $\mathbb{R}$. This is a complete normed space with the total variation norm. The Riesz representation theorem states that its dual space is $\mathcal{C}_0(\mathbb{R})$, the space of continuous functions that vanish at infinity. The weak topology $\sigma(\mathcal{M}(\mathbb{R}),\mathcal{C}_0(\mathbb{R}))$ is an abject of interest in its own right. However, in applications there are other weak topologies on $\mathcal{M}(\mathbb{R})$ that are of interest:

  1. Vague topology: the weak topology $\sigma(\mathcal{M}(\mathbb{R}),\mathcal{C}_{00}(\mathbb{R}))$, where $\mathcal{C}_{00}(\mathbb{R})$ is the space of continuous functions with compact support. In this topology, it is enough to consider sequences instead of nets.
  2. Weak convergence in measure topology: $W=\mathcal{C}_b(\mathbb{R})$ is the space of bounded continuous functions on $\mathbb{R}$.
  • Example (2) happened to be very useful in Probability theory. Notice that the space of probability measures on $\mathbb{R}$ is a convex subset of $\mathcal{M}(\mathbb{R})$. One can show that it is also a closed subspace in the weak topology $\sigma(\mathcal{M}(\mathbb{R},\mathcal{C}_b(\mathbb{R}))$.
  • It turns out that instead of $\mathcal{C}_b(\mathbb{R})$, one can consider $\mathcal{U}_b(\mathbb{R})$, the space if uniformly continuous in $\mathbb{R}$.
  • Moreover, the following result holds: a net $\{\mu_\alpha:\alpha\in D\}$ of positive finite measures on $\mathscr{B}(\mathbb{R})$ converges to $\mu$ in $\sigma(\mathcal{M}(\mathbb{R},\mathcal{C}_b(\mathbb{R})$ iff for any bounded below lower semicontinuous function $f$ $$\liminf_\alpha\int_{\mathbb{R}}f\,d\mu_\alpha\geq\int_\mathbb{R} f\,d\mu$$
  • There are other criteria summarized by the Portmanteau theorem.

I hope this clarifies things.

Mittens
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If $X$ is a constant, you may use $\mathcal{C}_b$. Otherwise, if $X\overset{d}{=}-X$, such a condition would imply that $X=-X$ a.s.

  • Do you mean, the class of all measurable functions? I.e., if $f$ is measurable, then $Ef(X_n) \to E f(X)$? – Davi Barreira Sep 02 '21 at 15:15
  • @DaviBarreira Which class do you mean? $\mathcal{C}_b$? –  Sep 02 '21 at 15:28
  • I mean, are you saying that for any measurable function $g$, if $X_n \to_p X$, then $Eg(X_n) \to E g(X)$? – Davi Barreira Sep 02 '21 at 15:36
  • No. Suppose that there is such a class $\mathcal{G}$ and for each $g\in\mathcal{G}$, $\mathsf{E}g(X_n)\to \mathsf{E}g(X)$, which implies that $X_n\xrightarrow{p}X$. Since $\mathsf{E}g(X)=\mathsf{E}g(Y)$ with $Y\equiv-X\overset{d}{=}X$, this also implies that $X_n\xrightarrow{p}Y$ and so $X=Y$ a.s. –  Sep 02 '21 at 15:46
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    So are you saying that there is not such class? – Davi Barreira Sep 02 '21 at 15:49