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I am trying to show the following statement

Let $D\subset \mathbb{R}^2$ be an open and bounded subset. $\Pi=(P^x : x \in D )$ a Family of standard Brownian Motions started at $x \in D$. Then $\Pi$ ist a tight (or relativly compact due to Prorhorvs Theorem).

I need that result to proof the existence of a certain type of Process. Unfortunately I do not have any experience working with the concept of "tightness".

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So you see each $P^x$ as the probability measure associated with the random variable which makes correspond to each $\omega$ a continuous function on $\{t\in\Bbb R, t\geq\}$. You can endow this space with the topology of uniform convergence over compact subsets, and try to find a characterization of compact subsets for this topology. –  Davide Giraudo Jun 23 '12 at 12:04
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Karatzas book's about Brownian motion contains good information about that, since Donker's invariance principle is shown for the half-line. You will find criterion of tighness. –  Davide Giraudo Jun 23 '12 at 12:32

2 Answers 2

up vote 3 down vote accepted

Using the fact that $S:=C[0,+\infty):=(f\colon [0,+\infty)\to \Bbb R, f\mbox{ continuous })$ endowed with the metric $$d(f,g):=\sum_{j=1}^{+\infty}2^{—j}\min(1,\sup_{0\leq t\leq j}|f(x)-g(x)|)$$ is polish, it's enough to show that each subsequence is tight. Using theorem 4.10 in Karatzas-Shreve book's, we know that a sequence of probability measures $(\mathbb P_n)_{n\geq 1}$ over the Borel $\sigma$-albegra is tight if and only if the two following conditions are satisfied:

  1. $\lim_{\lambda\to+\infty}\sup_{n\geq 1}\mathbb P_n(\omega,|\omega(0)|\geq \lambda)=0$ and
  2. $\lim_{\delta\to 0}\sup_{n\geq 1}\mathbb P_n(\omega,m^T(\omega,\delta)\geq \varepsilon)=0$ for each $T>0$ and $\varepsilon>0$, where $m^T(\omega,\delta)=\sup_{|t_1-t_2|\leq\delta,0\leq t_1,t_2\leq T}|\omega(t_1)-\omega(t_2)|$.

We apply this to $\mathbb P_n=\mathbb P_{x_n}$, where $(x_n)_{n\geq 1}$ is a sequence in $D$. By boundedness of $D$, the first condition is satisfied. We can control the modulus of continuity using the fact that $$\mathbb P_n(\omega,m^T(\omega,\delta)\geq \varepsilon) \leq \frac 1{\varepsilon^2}\delta^2,$$ which is a consequence of the fact that the increments of the Brownian motion $W_{t_2}-W_{t_1}$ are normally distributed, of normal distribution of mean $0$ and variance $t_2-t_1$.

In fact, with condition 1., we can see that $(\mathcal P_x,x\in D)$ is tight (where $\mathcal P_x$ is associated to a Brownian motion started at $x$) if and only if $D$ is bounded.

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this is precisely what I was looking for :D thank you very much. The first condition was somehow obvious to me too but I had trouble seeing why the second one of 4.10 is satisfied. My first guess was to use the continuity of sample paths of brownian motion. –  Boldwing Jun 23 '12 at 13:20

Here is another argument: The map $x \mapsto \mathcal{P}_x$ is a continuous map from $\mathbb{R}^2$ (the same in higher dimensions) to the probability measures on the path space with the topology of convergence in law. Therefore the family $(\mathcal{P}_x,x \in \overline{D})$ is compact, being the continuous image of the compact set $\overline{D}$. The path space is Polish and hence the family is uniformly tight (the converse of Prokhorov's theorem).

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