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Here is a question from a past exam from probability theory that I try to tackle:

Let $X,X_1,X_2,\ldots$ be real random variables. We know that:

(a) $X_n^2$ converges in distribution to $X^2$

(b) $X_n^3$ converges in distribution to $X^3$

Does this imply that $X_n$ converges in distribution to $X$?

To be honest, I don't know how to even start with this so I would appreciate any hints.

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    $\begingroup$ (a) isn't even necessary here. Notice that function $x \to x^3$ has inverse on the whole real line, so you can just use $f: \mathbb R \to \mathbb R$, $f(x) = x^{\frac{1}{3}}$ and apply continuous mapping theorem to $f(X_n^3)$ getting $X_n = f(X_n^3)$ converges weakly to $f(X^3) = X$. $\endgroup$ – Dominik Kutek Dec 5 '19 at 20:35
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Yes. You can use the Continuous Mapping Theorem (CMT) to realize this. One of the consequences of the CMT is that if $g$ is a univariate continuous function and $X_n$ converges in distribution to $X$, then $g(X_n)$ converges in distribution to $g(X)$. Now, (b) and the CMT with $g(x)=x^{1/3}$ gives that $g(X_n^3)=X_n$ converges in distribution to $g(X^3)=X$.

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