Let $\mu_n,n\in \mathbb N$ be a random probability measures and let $\mu$ be a deterministic probability measure on $\mathbb R$. That is to say, that the $\mu_n$ are measurable maps from a probability space $(\Omega,\mathcal{T},\mathbf{P})$ to the space of $M_1(\mathbb R)$ equipped with the Borel-$\sigma$-algebra generated by the topology of weak convergence. Assume that the expected measures $\nu_n:=\mathbf E\mu_n$, defined via duality as $\int f~d\nu_n:=\mathbf E\int f~d\mu_n$ for all $f\in C_b(\mathbb R)$, converge weakly to $\mu$, i.e. that for all $f\in C_b(\mathbb R)$ we have the convergence of $\mathbf E\int f d\mu_n$ to $\int f d\mu$. By Levy's continuity Theorem this is equivalent to the statement that the characteristic function $\phi_{\nu_n}(t):=\int e^{itx}~d\nu_n(x)$ converges pointwise to the characteristic function $\phi_\mu$ of $\mu$. Is a similar statement also true for weak convergence in probability the random measures, i.e. is there a connection between the following statements?

  • For all $\epsilon>0$ the sequence $\mathbf{P}(d_{BL}(\mu_n,\mu)>\epsilon)$ converges to zero, where $d_{BL}$ is the bounded Lipschitz metric $$d_{BL}(\mu,\nu)=\sup\left\{\left\lvert\int f~d\mu-\int f~d\nu\right\lvert~:~f\colon\mathbb{R}\to\mathbb{R}\text{ is 1-Lipschitz and }\lVert f\lVert_\infty\leq1\right\}$$ (which completely metrizes the topology of weak convergence)

  • For all $f\in C_b(\mathbb R)$ it holds that $\int f~d\mu_n$ converges in probability to $\int f~d\mu$.

  • For all $t\in \mathbb R$, $\phi_{\mu_n}(t)$ converges to $\phi_\mu(t)$ in probability.

  • On some small interval $[0,\epsilon]$, the sequence $\sup_{0\leq t\leq \epsilon}\lvert \phi_{\mu_n}(t)-\phi_\mu(t)\lvert$ converges in probability to zero.

  • For all $k\in\mathbb N$, the sequence of moments $\int x^k~d\mu_n$ converges in probability to $\int x^k d\mu$.

  • For all $k\in\mathbb N$, the sequence of expected moments $\mathbf E\int x^k~d\mu_n$ converges to $\int x^k d\mu$ and the variances $\mathbf{Var}\int x^k~d\mu_n$ converge to zero.

Clearly, for the last two statements to be equivalent to any of the above, we would need some condition on the determinacy of $\mu$ via its moments. Therefore, for simplicity assume that $\mu$ is subgaussian, and that the $\mu_n$ are uniformly subgaussian in the sense that $\mathbf E\mu_n([-R,R]^c)\leq C e^{-C R^2}$ for some $C$ and all $n,R$.

  • $\begingroup$ You defined the expected measure $\nu_{n}:=\mathbf{E}\mu_{n}$ via duality as $\int f~d\nu_n:=\mathbf E\int f~d\mu_n$ for all $f\in C_b(\mathbb R)$. I assume this is by the Riesz-Markov representation theorem. But for this to apply don't you need the space of test functions which are continuous bounded and compactly supported, that is $C_{c}(\mathbb{R})$? Or why is it sufficient to take test functions from the larger space $C_{b}(\mathbb{R})$? $\endgroup$ Jul 17, 2015 at 12:38

2 Answers 2


$\newcommand{\bR}{\mathbb{R}}$ For any Polish space $S$ (separable complete metric space) we denote by $M_1(S)$ the space of Borel probability measures on $S$.

The space $\newcommand{\eP}{\mathscr{P}}$ $\eP:=M_1(\bR)$ is a topological space with respect to the weak convergence. In fact, $\eP$ with this topology is a Polish space.

A random measure $\mu$ on $\bR$ is by definition a measure on $\eP$, i.e.,

$$\mu \in M_1\bigl(\;\eP\;\bigr).$$

The space $C_b(\bR)$ embeds in $C_b\bigl(\eP\bigr)$ $\newcommand{\bsE}{\boldsymbol{E}}$

$$ f\in C_b(\bR)\mapsto \widehat{f}\in C_b\bigl(\eP\bigr), \;\;\widehat{f}(\pi):=\bsE_\pi(f)=\int_{\bR} f(t) \pi(dt),\;\;\forall \pi\in \eP. $$

wher $\bsE_\pi$ denotes the expectation of a random variable with respect to the probability distribution $\pi$. Denote by $\newcommand{\eX}{\mathscr{X}}$ $\eX\subset C_b\bigl(M_1(\bR)\bigr)$ the image of this embedding.

In your question you are given $\newcommand{\bmu}{\boldsymbol{\mu}}$ a sequence $\bmu_n\in M_1(\eP)$ and another measure $\bmu$ with the property that

$$\lim_n\int_{\eP}F(\pi) \bmu_n(d\pi)=\int_{\eP}F(\pi)\bmu(d\pi),\;\;\forall F\in \eX $$

and you ask if the above holds for all $F\in C_b(\eP)$. Clearly this happens iff $\eX$ is dense in $C_b(\eP)$. It is not hard to see that is not the case. Fix a nontrivial, compactly supported continuous function $\rho:\bR\to[0,1]$. Then the bounded, continuous function

$$ Q_\rho:\eP\to\bR,\;\;Q_\rho(\pi)=\left(\int_{\bR}\rho(t) \pi(dt)\,\right)^2, $$

is not in the closure of $\eX$.

  • $\begingroup$ I am not entirely sure how your answer relates to my question. Firstly, I'd rather say that a random measure $\mu$ is, by definition, a measurable map $\omega\mapsto\mu(\omega)$ from some probability space $(\Omega,\mathcal{T},\mathbf P)$ to $(M_1(\mathbb R),\mathcal{B})$ where $\mathcal{B}$ is the Borel $\sigma$-algebra generated by the topology of weak convergence. What you defined, is in my mind, the distribution of such a random measure and I don't think that a concept like convergence in probability can be phrased in terms of the distribution. $\endgroup$
    – whz
    Feb 16, 2015 at 17:19
  • $\begingroup$ Secondly, in your answer you only assumed the convergence of $\int_{\mathcal{P}}\int_{\mathbb R} f(t)~d\pi(t)~d\mu_n(\pi)$ towards $\int_{\mathcal{P}}\int_{\mathbb R} f(t)~d\pi(t)~d\mu(\pi)$ for all $f\in C_b(\mathbb R)$, which I would call 'convergence in expectation'. It is clear that convergence in expectation is not sufficient on its own. The assumptions in my question are much stronger than convergence in expectation. $\endgroup$
    – whz
    Feb 16, 2015 at 17:28
  • $\begingroup$ @whz You are right. $\endgroup$ Feb 17, 2015 at 19:59

I found a way to prove the implication I was mostly interested in (the last statement implies the second): Below my argument, for the sake of completeness.

Fix a function $f$. A sequence of random variables $X_n$ converges to a random variable $X$ in probability if and only if any subsequence $X_{n_m}$ has a further subsequence $X_{n_{m_l}}$ which converges to $X$ almost surely. Apply this to the sequence in question and pick the subsequence $n_{m_l}$ in such a way that $\mathbf{Var}\int x~d\mu_{n_{m_l}}$ is summable. By Borel-Cantelli this implies that $\int x~d\mu_{n_{m_l}}$ converges almost surely to $\int x~d\mu$. Continue inductively and choose a diagonal sequence (for convenience still named the same) such that the sequences $\int x^k~d\mu_{n_{m_l}}$ converge to $\int x^k~d\mu$ almost surely for all $k\in\mathbb N$. If the limiting distribution is uniquely determined by its moment this implies that with full probability $\mu_{n_{m_l}}$ converges weakly to $\mu$, which, in particular, implies that the integral against $f$ converges almost surely.


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