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I got confused by a silly question. Suppose $\sqrt nX_n\rightarrow_dN(0,1)$, so $\sqrt nX_n=O_p(1)$, which implies $X_n=O_p(n^{-1/2})=o_p(1)$. Does this mean convergence in distribution implies convergence in probability?

wsong
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  • In general, convergence "in distribution" does not imply "in probability," since you may have ${Z_n}{n=1}^{\infty}$ i.i.d. uniform over [0,1] (they have the same distribution as $Z_1$ but do not converge to $Z_1$ in probability). But in this case, if $\sqrt{n}X_n \rightarrow_d N(0,1)$ you can indeed conclude $X_n\rightarrow 0$ in probability. Specifically, fix $\epsilon>0$, can you give a rigorous computation of $\lim{n\rightarrow\infty}P[|X_n|>\epsilon]$ ? – Michael Jul 05 '17 at 01:48
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    PS: Feel free to answer your own question with the computation of $\lim_{n\rightarrow\infty} P[|X_n|>\epsilon]$. That is standard practice on this site when you can answer based on hints. – Michael Jul 05 '17 at 01:57

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