2 Tight families of measures
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 \(\mu \in S\), \(\mu (K^c) \le \varepsilon \).
Let \(S\) be a set of measures on a finite-dimensional normed space \(E\). Then \(S\) is tight iff \(\lim _{r \to +\infty }\sup _{\mu \in S} \mu \{ \Vert x \Vert {\gt} r\} = 0\).
Let \(S\) be a set of measures on a finite-dimensional inner product space \(E\). Let \(b_1, \ldots , b_k\) be an orthonormal basis of \(E\). The \(S\) is tight iff for all \(i \in \{ 1, \ldots , k\} \), \(\lim _{r \to +\infty }\sup _{\mu \in S} \mu \{ \vert \langle b_i, x\rangle \vert {\gt} r\} = 0\).
Let \((\mu _n)_{n \in \mathbb {N}}\) be a sequence of measures on a finite-dimensional inner product space \(E\). Let \(b_1, \ldots , b_k\) be an orthonormal basis of \(E\). Then the set \(\{ \mu _n \mid n \in \mathbb {N}\} \) is tight iff for all \(i \in \{ 1, \ldots , k\} \), \(\lim _{r \to +\infty }\limsup _{n \to \infty } \mu _n\{ \vert \langle b_i, x\rangle \vert {\gt} r\} = 0\).
If \(\mu _1, \mu _2, \ldots \) converge weakly to \(\mu \), then \(\{ \mu _n \mid n \in \mathbb {N}\} \) is tight.
Fix \(r {\gt} 0\) and let \(f(x) = (1 - (r - \vert x \vert )_+)_+\) . \(f\) is a bounded continuous function. Then
As \(r \to +\infty \), the last quantity tends to 0.
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 \(c_n \to 0\), \(c_n X_n \xrightarrow {p} 0\).
The closure of a tight set of measures is tight.
2.1 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.
Let \(E\) be a metric space and let \(S \subseteq \mathcal P(E)\) be tight. Then the closure of \(S\) is compact.