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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ Cts mapping + skorohod + uniform integrability convergence theorem $\endgroup$
    – Taylor
    Commented May 3, 2023 at 23:22
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    $\begingroup$ Use Fatou's lemma and Skorohod theorem for part (a). $\endgroup$
    – Zhanxiong
    Commented May 4, 2023 at 0:20

1 Answer 1

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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}

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

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