# Do these conditions imply weak convergence of the random variable?

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.

• (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 $f(X^3) = X$. – Dominik Kutek Dec 5 '19 at 20:35

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 $$g(X^3)=X$$.