Why this expectation inequality holds?

Question

Can someone explain me how to obtain the following result:

where $E\mid X\mid^p<\infty$, $E\mid Y\mid^q<\infty$, $p,q>1$ and $1/p+1/q<1$.

UPDATE

I just noticed that, by definition, $I(\mid Y\mid>C)^k=I(\mid Y\mid>C)$ for any real number $k>0$. Maybe it can help.

It's in Corollary A.2's proof of this book (pp. 278).

Given $(\Omega,\mathcal{F},P)$, for any $w\in\Omega$, it holds $C^{p'}I(\lvert Y\rvert>C)\leq \lvert Y\rvert^{p'}I(\lvert Y\rvert>C)$ where $p'=p/(p-1)$, by the definition of indicator/characteristic function,i.e., $\lvert Y\rvert^{p'}\geq C^{p'}$ when $\lvert Y\rvert> C$. Taking expectations and using Markov's inequality, $$E(C^{p'}I(\lvert Y\rvert>C))=C^{p'}P(\lvert Y\rvert>C)\leq C^{p'-q} E\lvert Y\rvert^q.$$ If I had $C^{p'}I(\lvert Y\rvert>C)\geq \lvert Y\rvert^{p'}I(\lvert Y\rvert>C)$, which is not true, the result would follow. Now, I'm suspecting that this result doesn't hold.

Answer 1

Let $p'=p/(p-1)$. The condition on $p$ and $q$ can be written as $q>p'$. We thus have to prove that $$ \mathbb E\left[Z^{p'}I(Z>C)\right]\leqslant C^{-q+p'}\mathbb E\left[Z^q\right], $$ where $Z=\left\lvert Y\right\rvert$. This follows by taking the expectation on both sides in the inequality $$ Z^{p'}I(Z>C)\leqslant C^{-q+p'} Z^q. $$

Answer 2

Put $p'=p/(p-1)$. By Holder's and Markov's inequalities \begin{align} E(\lvert X \rvert^{p'} I(\lvert X \rvert>C))&\leq [E\lvert X\rvert^{r}]^{p'/r} (P(\lvert X\rvert>C))^{1-p'/r}\\ &\leq [E\lvert X\rvert^{r}]^{p'/r} [E(\lvert X \rvert^r/C^r)]^{1-p'/r}\\ &=E\lvert X \rvert^r C^{p'-r}, \end{align} the desired result.