1
$\begingroup$

Let $X$ be a positive random variable (let's say finitely supported). Let $\mu_r = \mathbb{E} X^r$ be the $r$-th moment. Is it true that $$\mu_3 \geq \mu_1 \cdot \mu_2$$

$\endgroup$

1 Answer 1

7
$\begingroup$

Yes! An easy way to see this: If $X\ge0$ then $X$ and $X^2$ are positively correlated (draw a picture). Thus $$\operatorname{Cov}(X,X^2)\ge0$$ which means $$E(XX^2)-E(X)E(X^2)\ge0.$$


EDIT: More rigorously, the general result is: If $h$ is a nondecreasing function, then $\operatorname{Cov}(X,h(X))\ge0$. [Here $X\ge0$ so the function $h(x):=x^2$ qualifies.]

Proof: By definition, $$\operatorname{Cov}(X,h(X)):=E(X-EX)(h(X)-Eh(X)).$$ Write $$(X-EX)(h(X)-Eh(X))=(X-EX)(h(X)-h(EX)) + (X-EX)(h(EX)-Eh(X)).\tag1 $$ The first term on the RHS of (1) is a nonnegative random variable, since $h$ is nondecreasing. The second term has expectation zero since $h(EX)$ and $Eh(X)$ are both constant.

$\endgroup$
2
  • 1
    $\begingroup$ Your argument (draw a picture) that $\operatorname{Cov}(X,X^2)\ge0$, is kind of loosey goosey. Can you firm up that up? $\endgroup$ May 11, 2018 at 0:02
  • 1
    $\begingroup$ @MarkL.Stone I have added an edit to tighten up the argument. $\endgroup$
    – grand_chat
    May 11, 2018 at 0:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.