Convergence of Gaussian measures - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T23:13:01Z http://mathoverflow.net/feeds/question/16422 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16422/convergence-of-gaussian-measures Convergence of Gaussian measures Tom LaGatta 2010-02-25T17:38:17Z 2010-02-26T15:21:39Z <p>Let $X$ be a separable Banach space with its Borel $\sigma$-algebra $\mathcal F$. Let $x_n \to x$ in $X$. Fix a Gaussian covariance operator $K$, and let $\mathbb P_n$ and $\mathbb P$ be Gaussian measures on $X$ with covariance $K$ and means $x_n$ and $x$, respectively. </p> <p><b>Question:</b> How do I show that $\mathbb P_n \to \mathbb P$ weakly?</p> <p>Surely this is a theorem or an exercise somewhere; e.g. in Talagrand and Ledoux's <i>Probability in Banach Spaces</i> or Vakhania, Tarieladze and Chobanyan's <i>Probability Distributions in Banach Spaces</i>. </p> <p>The characteristic functions of $\mathbb P_n$ converge to those of $\mathbb P$ (simple exercise). By de Acosta's theorem, this implies that $\mathbb P_n \to \mathbb P$, provided that the family ${\mathbb P_n}$ is <i>flatly concentrated</i>. I'm not so familiar with the concept (hence this question), but I'm guessing this is related to the concentration of measure property of Gaussian measures.</p> http://mathoverflow.net/questions/16422/convergence-of-gaussian-measures/16431#16431 Answer by Mark Meckes for Convergence of Gaussian measures Mark Meckes 2010-02-25T18:48:42Z 2010-02-25T18:48:42Z <p>In general, a sequence of Banach space-valued random variables $Y_n$ converges weakly to $Y$ if $f(Y_n)\to f(Y)$ for every $f\in X^*$, and $Y_n$ is <em>tight</em> in the sense that for each $\varepsilon > 0$ there is a compact set $K\subset X$ such that $\mathbb{P}(Y_n \in K) \ge 1-\varepsilon$ for every $n$ (Ledoux and Talagrand, p. 41). This is a consequence of <a href="http://en.wikipedia.org/wiki/Prokhorov%27s_theorem" rel="nofollow">Prokhorov's theorem</a>. The first condition is in particular implied by convergence of characteristic functions.</p> <p>Flat concentration is not actually related to concentration of measure. It's actually an alternative way of characterizing tightness. Quoting L&amp;T:</p> <blockquote> <p>The idea is simply that bounded sets in finite dimension are relatively compact and, therefore, if a set of measurse is concentrated near a finite dimensional subspace, then it should be close to be relatively compact.</p> </blockquote> http://mathoverflow.net/questions/16422/convergence-of-gaussian-measures/16518#16518 Answer by Mark Meckes for Convergence of Gaussian measures Mark Meckes 2010-02-26T15:21:39Z 2010-02-26T15:21:39Z <p>Somehow I didn't register how strong the assumptions Tom was making were, hence the fact that my other answer missed the point.</p> <p>Unless I'm still missing something, this is very easy. Say $Z$ is a Gaussian random vector in $X$ with covariance $K$ and mean $0$. You want to show that $Z+x_n \to Z+x$ weakly, i.e. $\mathbb{E} f(Z+x_n) \to \mathbb{E} f(Z+x)$ for every bounded continuous $f:X\to \mathbb{R}$. Since $f$ is both bounded and continuous, this follows immediately from dominated convergence.</p>