# Is it true that $n^r(X_n-X)\xrightarrow{d}Z$ implies $X_n\xrightarrow{d}X$? [duplicate]

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?

• 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"... – saz Oct 30 '18 at 14:57

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 $$Z$$.
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.