3
$\begingroup$

The following is an exercise (4.5.2) from Chung's book, "A Course in Probability Theory".

If $\{X_n\}$ is dominated by some $Z$ in $L^p$ and converges in distribution to $X$, then prove that $$\lim\limits_{n\rightarrow\infty}E[|X_n|^p]=E[|X|^p]$$

I applied the Skorohod's Representation theorem. Since $X_n$ converges in distribution to $X$, there exists a sequence of random variables $\{Y_n\}$ with the same distribution as $\{X_n\}$ such that $Y_n \rightarrow Y$ a.s and $Y$ has the same distribution as $X$. By the continuous mapping theorem,

$$ \lim\limits_{n\rightarrow\infty}|Y_n|^p = |Y|^p $$

almost surely. Then,

$$ \lim\limits_{n\rightarrow\infty}E[|X_n|^p] = \lim\limits_{n\rightarrow\infty}E[|Y_n|^p] = E[|Y|^p] = E[|X|^p]$$

where I applied the Dominated Convergence Theorem in the second step. Am I allowed to apply the Dominated Convergence Theorem here? We know that $\{X_n\}$ are dominated by $Z$, but does that mean the $\{Y_n\}$ are as well?

$\endgroup$
3
$\begingroup$

No, your argument is not correct because $\{Y_n\}$ need not be dominated in $L^{p}$. Note that $$\int_{|X_n|^{p}>\Delta } |X_n|^{p} \leq \int_{Z^{p}>\Delta } Z^{p} \to 0$$ uniformly in $n$ as $\Delta \to \infty$. Now use Skhrohod Representation and get $\int_{|Y_n|^{p}>\Delta } |Y_n|^{p} \to 0$ uniformly in $n$ as $\Delta \to \infty$. Since $Y_n-Y \to 0$ almost surely and $\{|Y_n-Y|^{p}\}$ is uniformly integrable we get $E|Y_n-Y|^{p} \to 0$. From this we get $$\limsup E|Y_n|^{p} \leq E|Y|^{p}$$ so $$\limsup E|X_n|^{p} \leq E|X|^{p}$$. On the other hand $$\liminf E|Y_n|^{p} \geq E|Y|^{p}$$ by Fatou's Lemma which gives $$\liminf E|X_n|^{p} \geq E|X|^{p}$$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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