Prove an inequality between second and fourth moments of a random variable

Question

Let $X$ be a random variable such that the second moment of $X$ equals 1 and the fourth moment exists. It cannot be taken as given that $\mathbb{E}(X) \neq 0$ (although I feel that this will be irrelevant).

Prove that $$\mathbb{E}|X| \geq \frac{1}{\sqrt{\mathbb{E}(X^4)}}.$$

A weaker statement would be to remove the absolute value from inside the expectation on the left, however, I think that may be a challenge to do since this seems to be a question for $L_p$-norms (and $\mathbb{E}|X|$ is the 1-norm. That and the fact that such a statement would likely have numerous counter examples).

This seems to be something where a well known inequality like Jensen's inequality or Hölder's inequality can be applied. Any thoughts as to how to go about this?

Answer 1

Hint: Holder's inequality essentially gives: $$E[|X|]^2\cdot E[X^4]\geqslant (E[X^2])^3 = 1$$

Answer 2

Let $X$ be a random variable. Assume that $E(X^2) = 1$ and that $E(X^4) < \infty$.

Note that $X^2 = |X|^{4/3}|X|^{2/3}$. Thusly, let $Y = |X|^{4/3}$ and $Z = |X|^{2/3}$. Then $X^2 = YZ$. Let $p = 3$ and $q = \frac{3}{2}$. Note that $1/p + 1/q = 1$. We apply Holder's inequality as follows: $$\lVert X^2\rVert_{1} = \lVert YZ\rVert_{1} \leq \lVert Y\rVert_{p}\lVert Z\rVert_{q}$$ Using the fact that $\lVert X^2\rVert_{1} = E|X^2| = E(X^2) =1$, we have the following: \begin{eqnarray*} 1 & \leq & \lVert Y\rVert_{p}\lVert Z\rVert_{q}\\ &=& \left(E(|X|^{4/3})^3\right)^{1/3}\left(E(|X|^{2/3})^{\frac{3}{2}} \right)^{2/3}\\ &=& \left(E(|X|^{4})\right)^{1/3}\left(E(|X|) \right)^{2/3}\\ \end{eqnarray*} Note that $\left(E(|X|^{4})\right)^{1/3}\left(E(|X|) \right)^{2/3}$ is by necessity positive. If we place both sides of the inequality that we developed above to the power of $3/2$, we get the following: $$ 1^{3/2} = 1 \leq \left(E(|X|^{4})\right)^{1/2}E |X|$$

If we divide the $\sqrt{E(X^4)}$ part over from the right to the left, we get $$\frac{1}{\sqrt{E(X^4)}} \leq E|X|.$$