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I don't understand the following as I read along a proof in a paper:

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?

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    $\begingroup$ You forgot to cite the paper itself for us to look at. $\endgroup$ – MathematicalPhysicist Jan 24 '16 at 20:16
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Maybe my answer is not the simplest, but what we have is: for all $f\in \mathcal C_b$, let $\Phi_f : \nu \in \mathcal P(E) \mapsto \int_E f d\nu$ (is continuous and bounded),

$$ E[ \Phi_f(\mu_n)] \to E[ \Phi_f(\mu)] $$

In fact, there is some stuff I never really understand that are Convergent determining function... You can found that in the book of Dawson if I remember. The set $\{ \Phi_f \, : \, f\in \mathcal C_b \} \subset \mathcal C_b(\mathcal P(E),\mathbb R)$ if it separates the points, etc (I imagine your $E$ is a Polish). It is a direction I think...

** EDIT: Ethier & Kurtz Chap. 3 sec 4 on convergent determining set.

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  • $\begingroup$ I still not have a real answer but in Measure valued Markov process, b Dawson, around Lemma 2.1.2 he construct an algebra.It is in the case of compact space, but... $\endgroup$ – Airlast Jan 25 '16 at 3:45

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