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I see that without this last requirement $n^r(X_n-X)$ would blow up in probability, how to prove this formally? By contradiction maybe? Like assuming it doesn't converge in probability and hence finding a subsequence such that $\mathbb{P}(|X_n-X|>\varepsilon)>\Delta$ for all $k$, and then showing that this implies $n^r(X_n-X)$ to go to infinity in probability?

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    $\begingroup$ What happened to this question of yours? Seems that they are kind of related, aren't they? Moreover, I have the impression that you are confusing "convergence in probability" and "convergence in distribution"... $\endgroup$ – saz Oct 30 '18 at 14:57
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It's overkill, but it follows trivially from the Skorokhod representation theorem. Otherwise you can use a uniform integrability argument to show the characteristic function of $X_n$ converges to that of $X$, where the uniformity comes from the uniform tightness implied by your convergence in distribution to $Z$.

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It is true for $r\gt 0$, but not necessarily for $r\gt 0$. Indeed, for each positive $\varepsilon$, $$ \Pr\{\left\lvert X_n-X\right\rvert \gt \varepsilon\}=\Pr\{n^r\left\lvert X_n-X\right\rvert \gt n^r\varepsilon\}. $$ A real number $R$ which is a continuity point of the cumulative distribution function (c.d.f.) of $Z$. Then for all $n$ such that $n^r\varepsilon\geqslant R$, the following inequality holds: $$ \Pr\{\left\lvert X_n-X\right\rvert \gt \varepsilon\} \leqslant \Pr\{n^r\left\lvert X_n-X\right\rvert \gt R\} $$ and using the convergence in distribution of $n^r\left\lvert X_n-X\right\rvert$ to $Z$ combined with the fact that $R$ is a continuity point of the c.d.f. of $Z$, we derive that for each $R$ being a continuity point of the c.d.f. of $Z$, $$ \limsup_{n\to +\infty}\Pr\{\left\lvert X_n-X\right\rvert \gt \varepsilon\} \leqslant \Pr\{Z \gt R\}. $$ As $R$ can be chosen as large as wished, we deduce that $X_n\to X$ in probability. Since convergence in probability implies weak convergence, we are done.

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