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 $0<r\le 1$ case.

  • 1
    $\begingroup$ Are you interested only in part b? It seems like it, but you should make that clear either way. $\endgroup$
    – jbowman
    Commented May 3, 2023 at 21:38
  • 1
    $\begingroup$ For (a), see stats.stackexchange.com/questions/244202. $\endgroup$
    – whuber
    Commented May 3, 2023 at 21:39
  • $\begingroup$ @jbowman yes, I'll place an edit $\endgroup$ Commented May 3, 2023 at 21:53
  • 1
    $\begingroup$ Cts mapping + skorohod + uniform integrability convergence theorem $\endgroup$
    – Taylor
    Commented May 3, 2023 at 23:22
  • 1
    $\begingroup$ Use Fatou's lemma and Skorohod theorem for part (a). $\endgroup$
    – Zhanxiong
    Commented May 4, 2023 at 0:20

1 Answer 1


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.


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}

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.