Let $X$ be a random variable and let $j\in\mathbf{N}$ whith $j >2$, is it true that $$E[|X-E[X]|^j] \leq E[|X|^j]\quad?$$

  • $\begingroup$ Is there a specific motivation ? $\endgroup$
    – Olivier
    Dec 28, 2019 at 11:17

2 Answers 2


The inequality always holds for $j=2$ (assuming the mean of $X$ exists) but does not hold in general for other $j$.

If $X$ is a random variable with mean $\mu$ and finite variance, then $\mu$ is the value of $x$ which minimises $\mathbb{E}[(X-x)^2]$. This implies the inequality for $j=2$.

However, $x=\mu$ does not in general minimise $\mathbb{E}[|X-x|^j]$ for $j$ other than $2$.

For example, taking $j=1$, the quantity that minimises $\mathbb{E}|X-x|$ is not the mean, but the median. So for example your property will fail for any random variable whose median is $0$ and which has a non-zero (and finite) mean. Consider for example a random variable that takes value $0$ with probability $2/3$, and value $1$ with probability $1/3$. The median is $0$ while the mean is $1/3$. We have $\mathbb{E}|X|=1/3<4/9=\mathbb{E}|X-1/3|$.

  • 1
    $\begingroup$ Your counterexample is right; but I don't see how the minimization bit is relevant. Why couldn't the inequality hold without the left-hand side minimizing $\mathbb E[|X-x|^j]$? $\endgroup$
    – joriki
    Dec 27, 2019 at 9:52
  • $\begingroup$ Indeed it could, and sometimes it does. I interpreted the question as asking whether the inequality always holds (but perhaps that's not what the OP meant). I'll edit. Anyway, if you have any $Y$ with finite mean where $y=y^*$ minimises $E[|Y-y|^j]$ and $y=E[Y]$ does not, then $X=Y-y^*$ is a counterexample. $\endgroup$ Dec 27, 2019 at 10:17
  • $\begingroup$ I also interpreted the question as asking whether the inequality always holds. It wasn't obvious to me that the minimization property implies that $X=Y-y^*$ is a counterexample. I think you should edit that into the answer; that's really the gist of your answer. $\endgroup$
    – joriki
    Dec 27, 2019 at 11:08
  • $\begingroup$ Thank you for your counter-example! However, my original question, which was edited without my approval, asked for j>1. Is the claim still false? $\endgroup$
    – Raphaël
    Dec 28, 2019 at 7:25
  • $\begingroup$ @Raphaël: The counterexample remains a counterexample if you choose $j=1+\epsilon$. (If the question is edited without your approval and the meaning changes, you should change it back.) $\endgroup$
    – joriki
    Dec 28, 2019 at 8:25

Another counter-example is $j=3$. For a Geometric random variable, $X$, with $\mu=E[X]=1$ we have


but we have



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