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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.

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up vote 3 down vote accepted

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.

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I knew it had to be easy. Sometimes, the lure of advanced technology keeps one from seeing the simple, direct argument. Thanks a lot, Mark. – Tom LaGatta Feb 26 '10 at 15:47
I'm actually making a project of reading Ledoux and Talagrand right now. There's more than enough advanced technology in this subject to make all the simple things really hard to see. – Mark Meckes Feb 26 '10 at 16:25

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.

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Mark, thanks for the response. What I mean is that in this particular setting (Gaussian measures with the same covariance and convergent means), my intuition tells me it's the isoperimetry or concentration property of Gaussian measures that gives flat concentration. – Tom LaGatta Feb 25 '10 at 18:51
Ah, I see. My intuition certainly agrees, but I don't see anything explicit along these lines in L&T. I'll think about it some more and let you know if I come up with something. – Mark Meckes Feb 25 '10 at 18:56
Thanks a lot, Mark! – Tom LaGatta Feb 25 '10 at 19:02

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