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I've read in a working paper (bottom of page 9) that the following is a "standard result":

Let $A$ be a compact metric space and $T$ be a Polish space. Let $\rho$ be a Borel probability measure on $T$. Let $\mathcal{M}^\rho(T\times A)$ be the set of Borel probability measures on $T\times A$ such that the marginal on $T$ is equal to $\rho$. Then $\mathcal{M}^\rho(T\times A)$ is a compact set in the narrow topology on the space of probability measures.

Could anyone tell me how to show this or give me a reference? The narrow topology is the same as the weak* topology or the topology of weak convergence of measures.

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As $T$ is separable and complete, by Ulam's theorem, for each $j\geq 1$, we can find a compact subset of $T$, say $K_j$, such that $\rho(K_j)\geq 1-j^{-1}$. If $P\in \mathcal M^{\rho}(T\times A)$, then $$P(K_j\times A)=\rho(K_j)\geq 1-j^{-1}.$$ As $K_j\times A$ is a compact subset of $T\times A$, the set $\mathcal M^{\rho}(T\times A)$ is uniformly tight. By of Prokhorov theorem, as $T\times A$ is separable and complete, $\mathcal M^{\rho}(T\times A)$ has a compact closure in the narrow topology. As this one is metrizable, to see that $\mathcal M^{\rho}(T\times A)$ is closed, we just have to check sequential closeness. Let $\{ P_n\}\subset \mathcal M^{\rho}(T\times A)$ converging in law to $P$. We just have to show that for each open set $O$ of $S$, $P(O\times A)=\rho(O)$. We have by portmanteau theorem, $$P(O\times A)\leq \liminf_n P_n(O\times A)=\rho(O).$$ As a closed set in a metric space is a countable intersection of open sets, we have $P(O^c\times A)\leq \rho(O^c)$, so $\rho(O)=P(O\times A)$ for all open set $O$.


Theorem.(Ulam) Let $(S,d)$ a complete separable metric space. Then each Borel probability measure on $S$ is tight.

Proof: Let $\{x_n\}$ a sequence dense in $S$, and $\varepsilon>0$. For each $j$, let $N_j$ integer such that $P\left(\bigcup_{l=1}^{N_j}B(x_l,j^{-1})\right)\geq 1-2^{-1}\varepsilon$. Then $K:=\bigcap_{j\geq 1}\overline{\bigcup_{l=1}^{N_j}B(x_l,j^{-1})}$ is pre-compact and closed, hence compact, and $P(K)\geq 1-\varepsilon$.

Theorem.(Prokhorov) Let $(S,d)$ a separable metric space, and $\mathcal P$ a family of Borel probability measures. If $\mathcal P$ is uniformly tight, that is, for all $\varepsilon>0$, we can find a compact subset $K$ of $S$ such that for each element $P$ of $\mathcal P$, $P(K)\geq 1-\varepsilon$, then $\mathcal P$ is relatively compact for the narrow topology.

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Thank you very much! –  Michael Greinecker Nov 2 '12 at 12:13
    
You are welcome! –  Davide Giraudo Nov 2 '12 at 12:14
    
In the first version, I forget to check that $\mathcal M^{\rho}$ is closed for the narrow topology. –  Davide Giraudo Nov 2 '12 at 18:53
    
The Prokhorov theorem as stated is valid without the assumption that $S$ is separable. This assumption is needed for the converse statement, namely that a relatively compact set is tight. –  Ahriman Nov 2 '12 at 19:40
    
@Ahriman Yes, the fact that $T\times A$ is Polish gives us that the narrow topology is metrizable (that what is needed here). And you are right, this direction of Prokhorov theorem doesn't need separability. –  Davide Giraudo Nov 2 '12 at 19:43
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