0.1 Tightness
Tightness (or being bounded in probability i.e. stochastically bounded) is a property of sequences of distribution functions whereby the limsup of the probability of the tails of the sequence eventually vanish to zero.
A sequence of distribution functions is tight if \[ \forall \varepsilon > 0,~\exists M_{\epsilon}>0: \limsup_{n \to \infty}\{ 1-F_{n}(M_{\varepsilon})+F_{n}(-M_{\varepsilon}) \}\leq \varepsilon. \]
Notice that this is taking the limsup of \(\mathbb{P}(X_{n}\leq -M_{\varepsilon})+\mathbb{P}(X_{n}\geq M_{\epsilon})\) or equivalently \(\mathbb{P}(X_{n}\not\in[-M_{\varepsilon}, M_{\varepsilon}])\) so proving that a sequence of distribution functions is tight means to show that in the limit supremum, for some small \(\varepsilon\) there is some interval \([-M_{\varepsilon}, M_{\varepsilon}]\) such that the probability of \(X_{n}\) falling within this interval is less than \(\varepsilon\).
A sequence of distribution functions \(\{ F_{n} \}\) is tight if for all \(\varepsilon >0\) there exists \(M=M_\varepsilon>0\) such that \[ \begin{align}& \forall n,~F_{n}([-M_{\varepsilon}, M_{\varepsilon}])>1-\varepsilon \\ &\iff \inf_{n}F_{n}([-M_{\varepsilon}, M_{\varepsilon}])>1-\varepsilon \\ & \iff \sup_{n}F_{n}([-M_{\varepsilon},M_{\varepsilon}]^c)\leq \varepsilon.\end{align} \]
0.2 Tightness Results
A sequence of distribution functions is tight if and only if every sub-sequential limit is a distribution function.
\begin{proof}
\end{proof} Roughly, we have that weak convergence equals vague convergence plus tightness!
0.3 Sufficient Conditions for Tightness
Let \(\{ X_{n} \}\) be r.v.s with d.f.s \(\{ F_{n} \}\). If there exists \(\phi \geq 0\) such that \(\phi(x)\uparrow \infty\) as \(\lvert x \rvert\uparrow \infty\) and \(C=\sup_{n}\mathbb{E}[\phi(X_{n})]<\infty\), then \(\{ F_{n} \}\) is tight.
Note that the most common criterion for tightness is for \(\phi(x)=\lvert x \rvert^r\), \(r >0\).
If there exists \(r>0\) such that \(\limsup_{n}\mathbb{E}\lvert X_{n} \rvert^r<\infty\) then \(\{ F_{n} \}\) is tight.
Note that it is therefore sufficient to show \(L_{p}\) bounded.