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 $$\mu_3 \geq \mu_1 \cdot \mu_2$$


1 Answer 1


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.

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    $\begingroup$ 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
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    $\begingroup$ @MarkL.Stone I have added an edit to tighten up the argument. $\endgroup$
    – grand_chat
    May 11, 2018 at 0:47

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