2
$\begingroup$

Though it might be a trivial problem, I am stuck on proving $$ \mathbb{E}[X]\leq\frac{\mathbb{E}[X^2]}{\sqrt[3]{\mathbb{E}[X^3]}} $$ where $X$ is a nonnegative random variable with finite third moment.

Can anybody help (or give a counterexample)?

  • 1
    you have restricted to finite third moments, but by truncation it would seem to imply that any r.v. with finite first and second moments and infinite third has expected value 0.2012-10-31
  • 0
    Thanks for the help. It suffices for me to assume that all moments of $X$ are finite. Is the inequality valid for such distributions?2012-10-31

2 Answers 2

1

Let $X$ be a Pareto distributed random variable, $X \sim \operatorname{Pareto}(1,\alpha)$, with $\alpha > 3$. Then, for $0 \leqslant r < \alpha$: $$ \mathsf{E}(X^r) = \frac{\alpha}{\alpha-r} $$ Hence: $$ \frac{\mathsf{E}(X^2)}{\sqrt[3]{\mathsf{E}(X^3)}} = \frac{\alpha^{2/3} \sqrt[3]{\alpha-3}}{\alpha-2}, \qquad \mathsf{E}(X) = \frac{\alpha}{\alpha-1} $$ For $\alpha>3$ we clearly have: $$ \frac{\mathsf{E}(X) \sqrt[3]{\mathsf{E}(X^3)}}{\mathsf{E}(X^2)} = \sqrt[3]{\frac{\alpha}{\alpha-3}} \frac{\alpha-2}{\alpha-1} \stackrel{x=\alpha-3}{=} \sqrt[3]{1+\frac{3}{x}} \frac{x+1}{x+2} > 1 $$ Indeed the function is decreasing function of $x$, and $$\lim_{x \to \infty} \sqrt[3]{1+\frac{3}{x}} \frac{x+1}{x+2} = 1$$

1

Following the spirit of @mike's comment, it is sufficient to take a variable $X$ with distribution: $$ f(x)=\frac{3/\pi}{x^6+1}$$ to get a counterexample. You have: $$ E[X]=1/\sqrt{2},\quad E[X^2]=1/2,\quad E[X^3]=1/\sqrt{3}.$$