Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 $\sup_{n\geq n_0}|E|X_n|^{\alpha}-E|X|^{\alpha}|\leq 3\varepsilon$.

share|cite|improve this answer
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
@DavideGiraudo How do you show the convergence of moments without the absolute values? – Calculon Apr 25 at 8:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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