Convergence of moment of functional of random variable

Define $$X_n$$ a continuous random variable that converges in distribution to $$X$$. Morever, we know that $$E[|X_n|^p] \rightarrow E[|X|^p]$$ for some $$p > 0$$.

Then, could we prove that for any continuous function $$f$$

$$E[|f(X_n)|^p] \rightarrow E[|f(X)|^p]?$$

Or which additional assumptions do we need? Or if $$X_n$$ is uniformly integrable does it imply that $$f(X_n)$$ is as well U.I ?

• The limit you ask about doesn't even need to be defined, because all the expectations involved could be infinite. Consider, for instance, distributions with finite absolute $p$ moments but infinite absolute $2p$ moments and let $f(x)=x^2.$
– whuber
Commented Mar 6, 2023 at 16:52

This is not true. Set the probability space $$(\Omega, \mathscr{F}, P)$$ to be $$((0, 1], \mathscr{B}, \lambda)$$ and define $$X_n = \sqrt{n}I_{(0, n^{-1})}(x)$$, $$n = 1, 2, \ldots$$, $$X \equiv 0$$, then \begin{align} E[|X_n|] = \frac{1}{\sqrt{n}} \to 0 = E[|X|] \end{align} as $$n \to \infty$$.
Let $$f(x) = x^2$$, which is continuous everywhere in $$\mathbb{R}$$, then $$f(X_n) = nI_{(0, n^{-1})}(x)$$, $$f(X) = 0$$, hence \begin{align} E[|f(X_n)|] = 1 \not\to 0 = E[|f(X)|]. \end{align}
I know $$X_n$$ in this example may not be considered as a strict "continuous random variable", but from measure theory perspective, it is continuous almost everywhere. Besides, you can easily modify it to be strictly continuous by linear interpolating $$(n^{-1}, \sqrt{n})$$ and $$(2n^{-1}, 0)$$, say, and the result should still stand.
• +1 For a continuous example, just let every $X_n$ and $X$ have a Student $t(p-1)$ distribution with $p\gt 2$ and use the same $f.$