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I've seen this result in several places, however, I've yet to find a proof for it and I'm struggling to come up with one on my own. So far I know that I want to show that for all $\epsilon > 0$ there exists a finite $M$ and a finite $N$ such that for all $n > N$

\begin{equation} P(|X_n| > M) < \epsilon \end{equation}

So far I know that

\begin{equation} P(|X_n| > M) = F_{X_n}(-M) + (1-F_{X_n}(M)) \end{equation}

Additionally, I know that since $X_n \xrightarrow{D} X$ that $F_{X_n} \rightarrow F_X$. So that for some $n > N$ I can make the distance between $F_{X_n}$ and $F_X$ arbitrarily small. However, I'm not seeing how I can leverage this to prove the result. Any help is much appreciated!

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Start by showing that $X=O_p(1)$, then you can take advantage of the distribution of $X_n$ being close to that of $X$ to show that for large $n$, $X_n$ can't be much more likely than $X$ to exceed any specified bound $M$

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    $\begingroup$ I was able to use the portmanteau lemma and your hint to form a solution. Thanks for the help! $\endgroup$ Mar 27, 2021 at 2:11

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