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Tightness

Yuri Vasilevich Prokhorov (1929 - 2013)
Yuri V. Prokhorov (1929 – 2013)

This post is about a basic fact which is not very well known. Let \( {{(\mu_i)}_{i\in I}} \) be a family of probability measures on a topological space \( {E} \) equipped with its Borel sigma-field. The following two properties are equivalent, and when they hold we say that \( {{(\mu_i)}_{i\in I}} \) is tight.

  1. for any \( {\varepsilon>0} \) there exists a compact set \( {K_\varepsilon\subset E} \) such that

    \[ \sup_{i\in I}\mu_i(K_\varepsilon^c)\leq\varepsilon; \]

  2. there exists a measurable \( {f:E\rightarrow[0,\infty]} \) with compact sub-level-sets such that

    \[ \sup_{i\in I}\int\!f\,d\mu_i<\infty. \]

Recall that the sub-level-sets of \( {f} \) are the sets \( {\{x\in E:f(x)\leq r\}} \), \( {r\geq0} \). If \( {E} \) is not compact then it cannot be a sub-level-set of \( {f} \) and thus \( {f} \) cannot be bounded, while if \( {E} \) is compact then the property holds trivially with \( {f} \) constant.

To deduce 1. from 2. we write, using the Markov inequality, for any \( {r>0} \),

\[ \sup_{i\in I}\mu_i(\{x\in E:f(x)\leq r\}^c) \leq \frac{1}{r}\sup_{i\in I}\int\!f\,d\mu_i=\frac{C}{r}, \]

which leads to take \( {r=r_\varepsilon=C/\varepsilon} \) and \( {K_\varepsilon=\{x\in E:f(x)\leq r_\varepsilon\}} \), for any \( {\varepsilon>0} \).

To deduce 2. from 1. we first extract from 1. a sequence of compact subsets \( {{(K_{1/n^2})}_{n\geq1}} \). We can assume without loss of generality that it grows: \( {K_{1/n^2}\subset K_{1/m^2}} \) if \( {n\leq m} \). Now, for any \( {x\in F=\cup_{n}K_n} \), there exists \( {n_x} \) such that \( {x\in K_{1/m^2}} \) for any \( {m\geq n_x} \), and thus \( {\sum_n \mathbf{1}_{K_{1/n^2}^c}(x)=n_x-1<\infty} \). As a consequence, if one defines

\[ f:=\sum_n\mathbf{1}_{K_{1/n^2}^c} \]

then \( {f<\infty} \) on \( {F} \) while \( {f=\infty} \) on \( {F^c} \), and \( {f} \) has compact sub-level-sets since \( {\{x\in E:f(x)\leq n-1\}=K_{1/n^2}} \) for any \( {n\geq1} \). On the other hand, by definition of \( {K_\varepsilon} \),

\[ \sup_{i\in I}\int\!f\,d\mu_i \leq\sum_n\frac{1}{n^2}<\infty. \]

Tightness is an important concept of probability theory. A famous theorem of Prokhorov states that a family of probability measures is tight if and only if it is relatively compact for the topology of narrow convergence.

Taking \( {X_i\sim\mu_i} \) for every \( {i\in I} \), the second property reads

\[ \sup_{i\in I}\mathbb{E}(f(X_i))<\infty. \]

It plays for tightness the role played by the famous de la Vallée Poussin criterion for uniform integrability.

One Comment

  1. student 2017-08-10

    Nice characterization. Having a close look at the proof it seems sufficient to require the level sets of $f$ to be only relatively compact. The inclusion ${\{x\in E:f(x)\leq n-1\}\subset K_{1/n^2}}$ shows this (besides, I could not see why the other inclusion would hold).

    Regards

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