2
$\begingroup$

I stumbled upon this exercise:

Prove the following : A sequence of measures $(P_n = P_{X_n})_{n \in \mathbb{N}}$ is tight if and only if the corresponding distribution functions $(F_n = F_{X_n})_{n \in \mathbb{N}}$ fulfill the uniform convergences in $n$ :

$\lim\limits_{x\to\infty} \inf\limits_{n\in\mathbb{N}}$ $F_n(x)$ = $1$,

$\lim\limits_{x\to-\infty} \sup\limits_{n\in\mathbb{N}}$ $F_n(x) = 0.$

I fail to see any link between tightness and this kind of convergence. It is also not clear to me why the requirements for the distribution functions need to be stronger than in the usual definition. I appreciate any piece of advice and tip.

$\endgroup$
2
$\begingroup$

Suppose $\{X_n\}$ is tight. Let $\epsilon >0$. Then there exists $T$ such that $P\{|X_n|>T\} <\epsilon$ for all $n$. Now $1-F_n(x)=P\{X_n >x\} \leq P\{|X_n|>T\} <\epsilon$ for all $n$ if $x >T$. Hence $F_n(x) >1-\epsilon$ for all $n$ if $x >T$. This gives $\inf_n F_n(x) \geq 1-\epsilon$ whenever $x>T$. Since $\inf_n F_n(x) \leq 1$ always we have proved that $\inf_n F_n(x) \to 1$ as $x \to \infty$. The proof of the second assertion is very similar: $X_n <-x$ implies $|X_n| >T$ provided $x>T$. Converse part: there exist $T_1$ and $T_2$ such that $F_n(x) >1-\epsilon /2$ for all$n$ if $x>T_1$ and $F_n(x) <\epsilon$ for all $n$ if $x<-T_2$. Take $T=\max \{T_1,T_2\}$ and verify that $P\{X_n| >x\} < \epsilon$ for all $n$ if $x>T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.