Let $X$ be a random variable and let $p\in(0,\infty)$ such that $\mathbf{E}(|X|^p)<\infty$. Prove that for all $q\in(0,p)$, we have $\mathbf{E}(|X|^q)\leqslant [\mathbf{E}(|X|^p)]^{\frac{q}{p}}$. You can use that $|x|^r$ is convex for all $r\geqslant 1$.

My try. Consider the random variable $X^p$. From $p>0$ and $0<q<p$, we have $0<\frac{q}{p}<1$. This means that $f(x)= |x|^{q/p}$ is concave, but then $g(x)= -|x|^{q/p}$ is convex. From Jensen's Inequality we get $\mathbf{E}[-f(X^p)]=-\mathbf{E}[f(X^p)]\geqslant -f[\mathbf{E}(X^p)] \iff \mathbf{E}[f(X^p)]\leqslant f[\mathbf{E}(X^p)]$. This gives $$\mathbf{E}[|X^p|^{\frac{q}{p}}]\leqslant [\mathbf{E}(|X^p|)]^{\frac{q}{p}} \iff \mathbf{E}(|X|^q)\leqslant [\mathbf{E}(|X|^p)]^{\frac{q}{p}}$$ as asked.

My thoughts

I would like to know how I can use the fact that $\mathbf{E}(|X|^p)<\infty$. Also, when I look at the graph of $g$, it is not convex for all values of $x$, only on the right and left branch separately. Do I need to split cases in my proof for that? I would also like to know if there is a cleaner or better proof.


You may look at $g(x)=-x^{q/p}$ on $x\geq 0$ side, and consider the random variable $|X|^{p}$. The assumption that $E(|X|^{p})<\infty$ is for the purpose when you put the real number $E(|X|^{p})$ into $g(\cdot)$.

Another way:

For the convex function $\varphi(x)=x^{p/q}$ on $x\geq 0$, then $[E(|X|^{q})]^{p/q}\leq E[\varphi(|X|^{q})]=E(|X|^{p})$, then $E(|X|^{q})\leq [E(|X|^{p})]^{q/p}$.

  • $\begingroup$ What confuses me is, why was given that $|x|^r$ is convex for all $r\geqslant 1$, if I need to look at $-x^{q/p}$? $\endgroup$ – Heinz Doofenschmirtz Jan 18 '18 at 8:50
  • $\begingroup$ Details added, please take a look. $\endgroup$ – user284331 Jan 18 '18 at 8:54
  • $\begingroup$ Thanks, but it does not answer my question. Why the absolute values? $\endgroup$ – Heinz Doofenschmirtz Jan 18 '18 at 9:00
  • $\begingroup$ I think that is a little strange, if we use $|x|^{r}$ ($r=p/q>1$), we still need to put $|X|^{q}$, not only $X^{q}$, because negative number does not have meaning in real for taking exponent. $\endgroup$ – user284331 Jan 18 '18 at 9:05
  • $\begingroup$ Anyway, $|x|^{r}$ and my writing that $x^{r}$ for $x\geq 0$ do not make any difference, but we must put the random variable $|X|^{q}$ (or $|X|^{p}$ for $g$). $\endgroup$ – user284331 Jan 18 '18 at 9:08

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