This is a follow-up question from my earlier one - context and motivation can be found there:

If we have a sequence of random variables $Z_n:=\sqrt n X_n$ converging in distribution to X (all defined on the same probability space), i.e. for the all continuous points of the distribution Function of $F$ $$ \lim_{n\to\infty}F_{Z_n}(x\leq c)=F_X(x\leq c)\iff\lim_{n\to\infty}P(Z_n \leq c)=\lim_{n\to\infty}P(\sqrt X_n \leq c)=P(X\leq c) $$ can we then say actually draw the following conclusion (is this implication even correct) for $c_{+}>0,c_{-}<0$: $$ P(X_n \leq c_{+})=P(\sqrt nX_n \leq \sqrt n c_{+})\implies\lim_{n\to\infty} P(\sqrt nX_n \leq \sqrt n c_{+})=P(X\leq \infty)=1 $$ and $$ P(X_n \leq c_{-})=P(\sqrt nX_n \leq \sqrt n c_{-})\implies\lim_{n\to\infty} P(\sqrt nX_n \leq \sqrt n c_{-})=P(X\leq -\infty)=0 $$ I don't see a reason why this shouldn't hold, nevertheless the implication feels somehow not quite right (in the definition of convergence in distribution the right hand side is always fixed, here we let it grow itself). On the other hand, if $\sqrt n X_n$ really converges in some way, $X_n$ would somehow need to compensate the extra growth of $\sqrt n$ and therefore need to go to zero itself in probability. (which would be the implication of the above). Anyone has a reference or any other insight?

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    $\begingroup$ Yes, if $Z_n\to X$ in distribution and if $P(|X|<\infty)=1$ then $Z_n/\sqrt{n}\to0$ in distribution. $\endgroup$ – Did May 30 '17 at 9:38
  • $\begingroup$ Thanks again :)! $\endgroup$ – user190080 May 30 '17 at 22:11

Formalizing Did's comment:

Suppose that $Z_n \overset{L}{\rightarrow} X$. Then, by Slutsky's Theorem,

$(\frac{1}{\sqrt{n}})(Z_n)\overset{L}{\rightarrow} (0)*(X)=0$.

Finally, we can extend this result by using the fact that convergence in law to a constant implies convergence in probability (see proof here). We can then conclude the following:



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