# $\mathbb E[|X_n|^r]<\infty$ and $\mathbb E[|X_n|^r]\to \mathbb E[|X|^r]$ as $n\to \infty$

Let $$\{X_n\}\xrightarrow{d}X$$ and for some $$p>0$$, we have $$\sup_{n\ge 1} \mathbb E[|X_n|^p]<\infty$$ Show that for any $$r\in (0,p)$$, we have

a. $$\mathbb E[|X|^r]<\infty$$

b. $$\mathbb E[|X_n|^r]\to \mathbb E[|X|^r]$$ as $$n\to \infty$$

[Note: You must not use (b) to prove (a)]

I am pretty sure we need to use Skorohod Representation Theorem. Maybe, we also need to use the fact that $$\{X_n\}\xrightarrow{d}X \iff \mathbb E[f(X_n)]\to \mathbb E[f(X)]\;\;\forall f\in \mathcal C_B(\mathbb R)$$ but I can't figure out how to do that.

The actual question had $$\mathbb E[|X|^r]<\infty$$ instead of $$\mathbb E[|X_n|^r]<\infty$$ which was wrongly written in the first question. So, now I have doubts in part (a) as well. The $$1\le r\le p$$ case can be tackled using some theorems done in class, but I can't find any argument for the $$0 case.

• Are you interested only in part b? It seems like it, but you should make that clear either way. Commented May 3, 2023 at 21:38
• For (a), see stats.stackexchange.com/questions/244202.
– whuber
Commented May 3, 2023 at 21:39
• @jbowman yes, I'll place an edit Commented May 3, 2023 at 21:53
• Cts mapping + skorohod + uniform integrability convergence theorem Commented May 3, 2023 at 23:22
• Use Fatou's lemma and Skorohod theorem for part (a). Commented May 4, 2023 at 0:20

#### Part (a)

I assume you already know how to prove $$\sup_n E[|X_n|^r] < \infty$$.

By continuous mapping theorem, $$X_n \overset{d}{\to} X$$ implies $$|X_n|^r \overset{d}{\to} |X|^r$$. It then follows by Skorohod's representation theorem that there exist $$\{Y_n\}$$ and $$Y$$ such that $$Y_n \overset{d}{=} |X_n|^r$$, $$Y \overset{d}{=} |X|^r$$, and $$Y_n \to Y$$ with probability $$1$$. Therefore, by Fatou's lemma, \begin{align} E[|X|^r] = E[Y] = E[\liminf_n Y_n] \leq \liminf_n E[Y_n] = \liminf_n E[|X_n|^r] \leq \sup_n E[|X_n|^r] < \infty. \end{align}

#### Part (b)

By continuous mapping theorem, $$X_n \overset{d}{\to} X$$ implies $$|X_n|^r \overset{d}{\to} |X|^r$$. In view of Theorem 25.12$$^\dagger$$ (the proof to this theorem indeed uses Skorohod's theorem) in Probability and Measure by Patrick Billingsley, to show $$E[|X_n|^r] \to E[|X|^r]$$, it suffices to prove $$\{|X_n|^r\}$$ is uniformly integrable. Indeed, suppose by condition $$\sup_n E[|X_n|^p] = M < \infty$$, then for any $$\alpha > 0$$, we have \begin{align} & E[|X_n|^rI_{[|X_n|^r \geq \alpha]}] \\ =& E\left[|X_n|^p\frac{1}{|X_n|^{p - r}}I_{[|X_n| \geq \alpha^{1/r}]}\right] \\ \leq & \frac{1}{\alpha^{(p - r)/r}}E[|X_n|^p] \leq \frac{1}{\alpha^{(p - r)/r}}M \to 0 \end{align} as $$\alpha \to \infty$$. This completes the proof.

$$\dagger$$

Theorem 25.12. If $$X_n \overset{d}{\to} X$$ and the $$X_n$$ are uniformly integrable, then $$X$$ is integrable and \begin{align} E[X_n] \to E[X]. \end{align}

• Excellent answer (+1)! Commented May 4, 2023 at 0:57