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Let $X_n$ have as a distribution function $F_n$.

Let's assume that exists an $r>0$ such that $\limsup_n E(|X_n|^r)<\infty$. Why does it imply that $\{F_n\}$ is tight?

My try: $\limsup_n E(|X_n|^r)<\infty \implies \forall_n \forall_{\epsilon>0}\exists_{\lambda_{\epsilon,n} >0} P(|X_n|\geq\lambda_{\epsilon,n})\leq \frac{E(|X_n|^r)}{\lambda_{\epsilon,n}^r}=\epsilon$ (by Markov inequality). This means that for all $\epsilon>0$, exists a finite interval $I$ such that $F_n(I^c)\leq \epsilon$ for all $F_n$.

Resnick's textbook is one text giving this criteria.

But why do we have the $\limsup$? why not just $\forall_n E|X_n|^r<\infty$?

Any help would be appreciated.

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    $\begingroup$ Asking that every $E(|X_n|^r)$ is finite does not suffice to imply uniform integrability, a property that should hold simultaneously for every $n$. The relevant condition is that $$\sup_nE(|X_n|^r)$$ is finite but the limsup may help if finitely many random variables $X_n$ are not in $L^r$. $\endgroup$ – Did Oct 24 '17 at 20:16
  • $\begingroup$ @Did, thanks for the comment. But isn't uniform integrability( of a family of r.v.) definition not equivalent to tightness (of a family of dist.functions)? Isn't u.i. stronger than tightness? $\endgroup$ – An old man in the sea. Oct 24 '17 at 20:34
  • $\begingroup$ Also, it seems that even sup is not enough. If I read the following link correctly math.stackexchange.com/questions/2247130/… $\endgroup$ – An old man in the sea. Oct 24 '17 at 20:41
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    $\begingroup$ Except that one usually considers $$r>1$$ $\endgroup$ – Did Oct 24 '17 at 20:50
  • $\begingroup$ Oh... I didn't know that. In the book, the author just writes $r>0$. $\endgroup$ – An old man in the sea. Oct 24 '17 at 20:56

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