# Prove or disprove the following inequality

If $X$ is a non-constant random variable with $X \geq 0$ then $$[E(X^{\alpha+1})]^{\alpha}-[E(X^{\alpha})]^{\alpha+1} > 0$$ for $\alpha=1,2,3,4,...$

It is easy for $\alpha=1$, because this is same as variance. But for other values of $\alpha$, I am confused. Also I could not find any counter example for $\alpha=2,3$ etc. But how to prove then if this is correct.

Please help me. Thanks.

## 1 Answer

Adapted from another similar answer.

Let $L^p=(\Omega, {\cal A}, \mu)$ and $p<q$. $$\int_\Omega X^p d\mu = \int_\Omega (X^q)^{p/q} d\mu \le \int_\Omega X^q d\mu$$ by Jensen's inequality and the concavity of the function $y\mapsto y^\alpha$ for $\alpha=p/q<1$. Raise both sides to the power $1/p$ and you get $$\left(\int_\Omega {X^p dx}\right)^{1/p} < \left(\int_\Omega{X^q dx}\right)^{1/q}.$$ Note that equality holds if and only if $X$ is a constant or the concave function in the equality is linear. Since both are false, it's a strict inequality.

In particular, when $p=\alpha$ and $q=\alpha+1$, we have $[E(X^{\alpha+1})]^{\alpha}-[E(X^{\alpha})]^{\alpha+1} > 0$.

• Is this the Lyapunov's inequality? By the way thank you so much for the answer. Jan 13 '18 at 13:14
• Yes, it is. Jan 13 '18 at 13:42