Convergence of Moments (2)

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Let $\{X_n\}$ and $X$ be random variables.  If $X_n$ converges in distribution to $X$, and for some $p>0$, $\sup_n\mathscr{E}|X_n|^p=M<\infty$, then for each $r<p$, $$\underset{n\rightarrow\infty}{\lim}\mathscr{E}|X_n|^r=\mathscr{E}|X|^r<\infty.$$

$\bullet$ Proof.

For simplifying the notations, WLOG, we suppose that $\{X_n\}$ and $X$ are positive random variables.  Let $F_n$ and $F$ be the distribution functions of $X_n$ and $X$ respectively, then $X_n\overset{d}{\longrightarrow}X\implies F_n\longrightarrow F$.  For $A>0$, define $$f_A(x)=\begin{cases}x^r, &\mbox{if }|x|\leq A;\\ A^r, &\mbox{if }x>A;\\ (-A)^r, &\mbox{if }x<-A.\end{cases}$$Then $f_A\in C_B=\{\mbox{bounded continuous functions}\}$, hence by Converge in Distribution and Vague Convergence (2): Equivalence for p.m.'s, we have $$\int_\mathbb{R}f_A(x)\,dF_n(x)\rightarrow\int_\mathbb{R}f_A(x)\,dF(x).$$Next, consider the distance between $f_A(x)$ and $x^r$, $$\begin{array}{rl}\int_\mathbb{R}\left|f_A(x)-x^r\right|\,dF_n(x)&\leq\int_{|x|>a}|x|^r\,dF_n(x)=\int_{|X_n|>A}|X_n|^r\,d\mathscr{P}\\ &\leq\int_{|X_n|>A}\frac{|X_n|^p}{A^{p-r}}\,d\mathscr{P}\qquad\qquad\left(\because\,|X_n|^{-(p-r)}<A^{-(p-r)}\right)\\ &\leq\frac{1}{A^{p-r}}\int_\Omega|X_n|^p\,d\mathscr{P}\qquad\qquad\left(\because\,\mbox{release int. region}\right)\\ &\leq\frac{M}{A^{p-r}}.\qquad\qquad\qquad\qquad\left(\because\,\sup_n\mathscr{E}|X_n|^p=M<\infty\right)\end{array}$$The last term $M/A^{p-r}$ does not depend on $n$, and converges to $0$ as $A\rightarrow\infty$.  That is, $f_A(x)$ converges uniformly to $x^r$ w.r.t. $n$, hence by interchanging the $\lim$ signs, we have $$\begin{array}{rl}\int_\mathbb{R}x^r\,dF &=\underset{A\rightarrow\infty}{\lim}\int_\mathbb{R}f_A\,dF=\underset{A\rightarrow\infty}{\lim}\underset{n\rightarrow\infty}{\lim}\int_\mathbb{R}f_A\,dF_n\\ &=\underset{n\rightarrow\infty}{\lim}\underset{A\rightarrow\infty}{\lim}\int_\mathbb{R}f_A\,dF_n=\underset{n\rightarrow\infty}{\lim}\int_\mathbb{R}x^r\,dF_n. \end{array}$$That is, $$\underset{n\rightarrow\infty}{\lim}\mathscr{E}|X_n|^r=\mathscr{E}|X|^r<\infty.$$

$\Box$