# Is this moment inequality valid?

$X$ is a positive continuous random variable. $E[X^p]$ is the $p$-th moment of $X$, $p\ge2$. Is the following moment inequality valid? $E[X^p]\le (p-1)^{p/2}(E[X^2])^{p/2}$

If so, What is the name of this inequality, and how to prove it?

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Try $p=1$. (extra characters) – Byron Schmuland Mar 20 '12 at 15:36
Have you tried playing around with $E(X^p)=\int_0^\infty py^{p-1}P(X>y)dy$, which holds for $p>0$? As Byron mentions, it's not true for $p=1$, but maybe try looking at $1<p<2$ and that $P(X>y)dy/E(X)$ is a probability measure along with Jenson's Inequality. – Alex R. Mar 20 '12 at 17:25
Let me up the ante:) For $1<p<2$ consider the degenerate random variable $X\equiv 1$. – Byron Schmuland Mar 20 '12 at 20:35
Sam, could you please explain your answer in more detail? I just modify p>=2 and p is integer. I really appreciate your answers! – huoshanzhx Mar 24 '12 at 2:59

No inequality $\mathrm E(X^p)^2\leqslant c(p)\cdot\mathrm E(X^2)^p$ $(\ast)$ may hold for a finite $c(p)$ and for every nonnegative random variable $X$, as soon as $p\gt2$.
To see why, note that the LHS of $(\ast)$ may be infinite while its RHS is finite. This happens, for example, when the tail distribution is $\mathrm P(X\geqslant x)\sim c/x^q$ when $x\to+\infty$, for some $c\gt0$ and $2\lt q\leqslant p$.
The same argument also shows that, even when restricted to bounded random variables (and in particular, with every moment finite), $(\ast)$ cannot hold for $p\gt2$, for any finite $c(p)$. To wit, consider a random variable $X$ as above and, for every $u\gt0$, $X_u=\min\{X,u\}$. Then every moment of every $X_u$ is finite and, when $u\to+\infty$, $\mathrm E(X_u^2)\to\mathrm E(X^2)$, which is finite, and $\mathrm E(X_u^p)\to+\infty$. Hence, for every finite $c$, one sees that $\mathrm E(X_u^p)^2\gt c\cdot\mathrm E(X_u^2)^p$ for some finite $u\gt0$, and in particular for some bounded random variable $X_u$.
If $1\leqslant p\lt2$, any nonzero $X$ which is almost surely constant disproves the proposed inequality. Finally, if $p=2$, the proposed inequality reduces to a tautology.