2 Tight families of measures

Definition 2.1

✓

A set \(S\) of measures on \(\Omega \) is tight if for all \(\varepsilon {\gt} 0\) there exists a compact set \(K\) such that for all \(\mu \in S\), \(\mu (K^c) \le \varepsilon \).

Lemma 2.2

If \(\mu _1, \mu _2, \ldots \) converge weakly to \(\mu \), then \(\{ \mu _n \mid n \in \mathbb {N}\} \) is tight.

Proof ▶

Fix \(r {\gt} 0\) and let \(f(x) = (1 - (r - \vert x \vert )_+)_+\) . \(f\) is a bounded continuous function. Then

\begin{align*} \limsup _{n \to +\infty } \mu _n\{ \vert X \vert {\gt} r\} \le \lim _{n \to +\infty } \mu _n\left[ f \right] = \mu [f] \le \mu \{ \vert X \vert {\gt} r - 1\} \: . \end{align*}

As \(r \to +\infty \), the last quantity tends to 0.

Lemma 2.3

For random variables \((X_i)_{i \in \mathbb {N}}\) (TODO in which kind of space?), the two following conditions are equivalent:

\((\mathcal L(X_n))\) is tight,
For all \((c_n) \ge 0\) with \(c_n \to 0\), \(c_n X_n \xrightarrow {p} 0\).

Proof ▶

Lemma 2.4

The closure of a tight set of measures is tight.

Proof ▶

2.1 Prokhorov’s theorem

Lemma 2.5

Let \((U_n)_{n \in \mathbb {N}}\) be open sets in a complete separable metric space \(E\) such that \(\bigcup _{n=1}^{+ \infty } U_n = E\). Let \(\Gamma \) be a relatively compact set of probability measures on \(E\) for the topology of weak convergence of measures. Then for all \(\varepsilon {\gt} 0\) there exists a finite set \(S \subseteq \mathbb {N}\) such that for all \(\mu \in \Gamma \), \(\mu (\bigcup _{n \in S} U_n) {\gt} 1 - \varepsilon \).

Proof ▶

Lemma 2.6

Let \(E\) be a complete separable metric space and let \(S \subseteq \mathcal P(E)\). If the closure of \(S\) is compact, then \(S\) is tight.

Proof ▶

Theorem 2.7 Riesz-Markov-Kakutani representation theorem, for compact spaces

Let \(E\) be a compact T2 space and let \(\psi \) be a positive linear functional on \(C(E, \mathbb {C})\). There exists a regular measure \(\mu \) finite on compacts such that

\begin{align*} \forall f \in C(E, \mathbb {C}), \ \psi (f) = \int _{x \in E} f(x) \partial \mu (x) \: . \end{align*}

Proof ▶

Lemma 2.8

If \(E\) is a compact metric space, then \(\mathcal P(E)\) with the Prokhorov metric is a compact metric space.

Proof ▶

Theorem 2.9 Prokhorov’s theorem

Let \(E\) be a complete separable metric space and let \(S \subseteq \mathcal P(E)\). Then the following are equivalent:

\(S\) is tight.
the closure of \(S\) is compact.

Proof ▶

Lemma 2.6 provides one direction.

TODO: proof of the other one.

Lemma 2.10

Let \(E\) be a metric space and let \(S \subseteq \mathcal P(E)\) be tight. Then the closure of \(S\) is compact.

Proof ▶