I don't understand the following as I read along a proof in a paper (Page 66, "Asymptotic Behaviour of some interacting systems", by Sylvie Meleard):

We denote by $\mathcal{P}({M})$ the space of probability measures on a metric space $M$, equipped with the weak topology.

Let $E$ be a metric space. Let $\{ \mu_n \}$ be a sequence of random measures on $E$, i.e. for each $n$, $\mu_n$ is a $\mathcal{P} (E)$-valued random variable. Also, let $\mathbb{Q}$ be a deterministic probability measure on the same probability space. We can then treat $\delta_{\mathbb{Q}}$ be a constant $\mathcal{P}(E)$-valued random variable.

Since both $\text{Law} ( \mu_n) $ and $\text{Law} ( \delta_{\mathbb{Q}})$ are measures on $\mathcal{P}(E)$, the paper defines that $\mu_n$ converges in law to $\mathbb{Q}$ if $$\text{Law} ( \mu_n) \implies \text{Law} ( \delta_{\mathbb{Q}}). \quad \quad \quad \quad \, \, \, \, (*)$$ However, in the paper, the fact that $\mu_n$ converges in law to $\mathbb{Q}$ is concluded by establishing that $$\mathbb{E} \bigg[ \bigg| \int_E f \,d \mu_n - \int_E f \,d \mathbb{Q} \bigg| \bigg] \rightarrow 0, \quad \quad \quad (**)$$ for all continuous bounded functions $f$ on $E$.

By definition of $(*)$, this is equivalent to saying that $$ \int_{\mathcal{P}(E)} f \, d\text{Law} ( \mu_n) \rightarrow \int_{\mathcal{P}(E)} f \, d \text{Law} ( \delta_{\mathbb{Q}}),$$ for all continuous bounded functions $f$ on $\mathcal{P}(E)$. How does this follow from $(**)$? Any ideas?


1 Answer 1


Let $F : \mathcal{P}(E) \to \mathbb{R}$ be bounded and continuous, and let $M := \sup |F|$. In adjusted notation, we wish to show $\mathbb{E} F(\mu_n) \to \mathbb{E} F(\delta_{\mathbb{Q}}) = F(\mathbb{Q})$. To save me some typing, let's suppose without loss of generality that $F(\mathbb{Q}) = 0$.

Fix $\epsilon > 0$. By continuity of $F$, there is a weakly open set $U \subset \mathcal{P}(E)$, containing $\mathbb{Q}$, such that if $\nu \in U$ then $|F(\nu)| < \epsilon$. By definition of the weak topology, there are finitely many bounded continuous $f_1, \dots, f_k : E \to \mathbb{R}$ and $\delta > 0$ such that if $\left| \int f_i\,d\nu - \int f_i \,d\mathbb{Q}\right| < \delta$ for $i = 1,\dots,k$ then $\nu \in U$.

Now for each $i$ we have from (**) that $\mathbb{E} \left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \to 0$. By Chebyshev's inequality we therefore have $\mathbb{P}\left(\left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \ge \delta\right) \to 0$. So we may find a large enough $N$ such that for all $n \ge N$ and all $i = 1,\dots,k$ we have $\mathbb{P}\left(\left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \ge \delta\right) < \epsilon/k$. For such $n$, a union bound gives $\mathbb{P}\left(\exists i : \left| \int f_i\,d\mu_n - \int f_i\,d\mathbb{Q}\right| \ge \delta\right) < \epsilon$. In particular, for $n \ge N$ we have $\mathbb{P}(\mu_n \notin U) < \epsilon$.

Now we can write $$|\mathbb{E} F(\mu_n)| \le |\mathbb{E}[F(\mu_n) ; \mu_n \in U]| + |\mathbb{E} [F(\mu_n); \mu_n \notin U]|.$$ The first term is bounded by $\epsilon$, since we have $|F| < \epsilon$ on $U$. And for $n \ge N$ the second term is bounded by $M \mathbb{P}(\mu_n \notin U) < M \epsilon$. Thus we conclude $$\limsup_{n \to \infty} |\mathbb{E} F(\mu_n)| \le \epsilon + \epsilon M.$$ Letting $\epsilon \to 0$ we have the desired conclusion.

Note this used the fact that the limiting distribution $\delta_\mathbb{Q}$ was degenerate. I'm not sure how to do it without that. We might need more assumptions on $E$ in that case.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.