We are asked to show the following:

$E(|X|) \ge \frac{1}{\sqrt{E(X^4)}}$

Given that $E(X^2) = 1$ and $E(X^4) < \infty$

Using Holder's inequality I can show that $E(X^4) \le 1$ assuming $X,Y = X^2,Y^2$ respectively in Holder's inequality. But I am not sure how to proceed


1 Answer 1


Note that by using Hölder's inequalty $$1=(\mathbb EX^2)^2=\left(\mathbb E\sqrt{|X|}|X|^{3/2}\right)^2\leq \mathbb E|X|\cdot \mathbb E |X|^3 $$holds. Then use again Hölder's inequalty to show $$ \mathbb E |X|^3=\mathbb E|X||X|^2\leq\sqrt{\mathbb E X^2 \cdot\mathbb E X^4}. $$


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