Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

How can I prove the following inequality:

If $r>p$, then \begin{equation} \left\|X\right\|_{p}\le\left(\frac{r}{r-p}\right)^{1/p}\left\|X\right\| _{r,\infty}, \end{equation} where \begin{equation} \left\|X\right\|_p=\left(\operatorname{E}\left|X\right|^p\right)^{1/p}\qquad\text{and}\qquad\left\|X\right\| _{r,\infty}=\left(\sup _{t>0}\ t^r\Pr\left(\left|X\right|>t\right) \right)^{1/r}. \end{equation}

I found this inequality on page 10 of the book by Ledoux and Talagrand.

Thank you for your answers!

share|cite|improve this question
up vote 4 down vote accepted

The RHS is the upper bound of $\displaystyle\|X\|_p^p=\mathrm E(|X|^p)=p\int_0^{+\infty}x^{p-1}\,\mathrm P(|X|\geqslant x)\,\mathrm dx $ that one gets using the upper bound $ \mathrm P(|X|\geqslant x)\leqslant\min\{1,\|X\|_{r,\infty}^rx^{-r}\} $ for every $x\gt0$.

share|cite|improve this answer

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.