Convergence of Gaussian measures - MathOverflow

Convergence of Gaussian measures

2010-02-25T17:38:17Z

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.

Question: How do I show that $\mathbb P_n \to \mathbb P$ weakly?

Surely this is a theorem or an exercise somewhere; e.g. in Talagrand and Ledoux's Probability in Banach Spaces or Vakhania, Tarieladze and Chobanyan's Probability Distributions in Banach Spaces.

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 flatly concentrated. 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.

Answer by Mark Meckes for Convergence of Gaussian measures

2010-02-25T18:48:42Z

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 tight 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 Prokhorov's theorem. The first condition is in particular implied by convergence of characteristic functions.

Flat concentration is not actually related to concentration of measure. It's actually an alternative way of characterizing tightness. Quoting L&T:

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.

Answer by Mark Meckes for Convergence of Gaussian measures

2010-02-26T15:21:39Z

Somehow I didn't register how strong the assumptions Tom was making were, hence the fact that my other answer missed the point.

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.