Suppose $X_n \to_{p} X$, if $\limsup_n E|X_n|^r \leq E|X|^r$, how can I show that $X_n \to_r X$?

Question

If I have that $X_n \to_p X$ (convergence in probability), and if $\limsup_n E|X_n|^r \leq E|X|^r$ for all $r \geq 1$, how can I show that $X_n \to_r X$ (this means $L^{r}$ convergence)? My goal is to show that $\limsup_{n} E|X_n|^r \leq E|X|^r$ implies uniform integrability, then I can use a standard result to show the $L^r$ convergence. However, I am really not sure how to do this. Does anyone have any hints? thanks.

Answer 1

We have to assume $\mathbb{E}(|X|^r)<\infty$; otherwise the expession $\|X_n-X\|_{L^r}$ might not even be finite.

Note that

$$\|X_n-X\|_{L^r}^r = \int_{|X_n-X| \leq \epsilon} |X_n-X|^r \, d\mathbb{P} + \int_{|X_n-X| > \epsilon} |X_n-X|^r \, d\mathbb{P} \tag{1}$$

for any $\epsilon>0$ and $n \in \mathbb{N}$. Obviously,

$$\int_{|X_n-X| \leq \epsilon} |X_n-X|^r \, d\mathbb{P} \leq \epsilon^r.$$

In order to estimate the second term at the right-hand side of $(1)$, note that, by the Cauchy Schwarz inequality,

$$\int_{|X_n-X| > \epsilon} |X_n-X|^r \, d\mathbb{P} \leq \sqrt{\mathbb{P}(|X_n-X| >\epsilon)} \cdot \sqrt{\int |X_n-X|^{2r} \, d\mathbb{P}}.$$

Now use the inequality

$$|X_n-X|^{2r} \leq 2^{2r} (|X_n|^{2r}+|X|^{2r}),$$

the fact that $X_n$ converges to $X$ in probability and the estimate for $\mathbb{E}(|X_n|^{2r})$.