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Let $X$ and $(X_n)_{n\geq 1}$ be random variables such that $X_n\to X$ in distribution. Assume that $\sup_n E[|X_n|^r]<\infty$ for some $r>0$. Then how do I show that $E[|X|^r]<\infty$ and that $E[X_n^\alpha]\to E[X^\alpha]$ and $E[|X_n|^\alpha]\to E[|X|^\alpha]$ for all $0<\alpha < r$. I'm told to show and use that $\sup_n E[|X_n|^\alpha - |X_n|^\alpha \wedge m]\to 0$ for $m\to\infty$ for all $\alpha<r$, but I don't see how this helps me. Any thoughts?

My thoughts on the first is the following: $$ E[|X|^r]=\sup_{k\in\mathbb{N}}E[|X|^r\wedge k]=\sup_{k\in\mathbb{N}}\lim_{n\to\infty}E[|X_n|^r\wedge k]\leq \sup_{k\in\mathbb{N}}\sup_{n\in\mathbb{N}}E[|X_n|^r\wedge k]\leq \sup_{n\in\mathbb{N}}E[|X_n|^r]<\infty $$ Is this correct?

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What you did to show that $E|X|^r$ seems fine to me. How do you define $X^{\alpha}$ for $\omega$ such that $X(\omega)<0$ and $\alpha$ irrational for example? –  Davide Giraudo Feb 27 '12 at 10:59
Alright thanks. I guess it's the same way as defining $x^\alpha$ for $x\in (-\infty,0)$. Is that causing any trouble here? –  Stefan Hansen Feb 27 '12 at 11:10
In fact no. Is this exercise form a book? –  Davide Giraudo Feb 27 '12 at 11:53
It's from some lecture notes, but I don't know if it is "stolen" from some book. –  Stefan Hansen Feb 27 '12 at 12:47

1 Answer 1

up vote 1 down vote accepted

We have using Hölder's inequality and denoting $M:=\sup_{n\geq 1}E|X_n|^r$ \begin{align*} \int |X_n|^{\alpha}\mathbf 1_{|X_n|^{\alpha}> m}&\leq \left(\int |X_n|^r\right)^{\alpha/r}\left(P(|X_n|^\alpha> m)\right)^{\frac {r-\alpha}r}\\ &\leq M^{\alpha /r}\left(P(|X_n|^r> m^{\frac r{\alpha}})\right)^{\frac {r-\alpha}r}\\ &\leq M^{\alpha/r}\left(\frac 1{m^{\frac r{\alpha}}}E |X_n|^r\right)^{\frac {r-\alpha}r}\\ &=M\frac 1{m^{\frac{r-\alpha}{\alpha}}}, \end{align*} and since $\frac{r-\alpha}{\alpha}>0$ we have the wanted convergence.

Now, fix $\varepsilon>0$ and pick an integer $m_0$ such that $E|X|^{\alpha}\wedge m_0\leq \varepsilon$ and $\sup_n E|X_n|^{\alpha}\wedge m_0\leq \varepsilon$.

Then for all $n_0$ $$\sup_{n\geq n_0}|E|X_n|^{\alpha}-E|X|^{\alpha}|\leq \varepsilon+\sup_{n\geq n_0}|E|X_n|^{\alpha}\wedge m_0-E|X|^{\alpha}|,$$ and using the fact that the map $x\mapsto |x|^{\alpha}\wedge m$ is continuous and bounded we take $n_0$ such that $\sup_{n\geq n_0} |E|X_n|^{\alpha}\wedge m_0-E|X|^{\alpha}\wedge m_0|\leq \varepsilon$ and we can conclude that $\sup_{n\geq n_0}|E|X_n|^{\alpha}-E|X|^{\alpha}|\leq 3\varepsilon$.

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How can you use Hölder, when $\tfrac{\alpha}{r}+\tfrac{r}{r-\alpha}\neq 1$? –  Stefan Hansen Feb 27 '12 at 10:49
Sorry, I made a mistake computing the conjugate exponent. Fixed now, I hope. –  Davide Giraudo Feb 27 '12 at 10:50
Excellent, thanks alot for your help! –  Stefan Hansen Feb 27 '12 at 11:34
You're welcome. –  Davide Giraudo Feb 27 '12 at 11:52

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