# Is it true that $E[|X-E[X]|^j] \le E[|X|^j]$?

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

• Is there a specific motivation ? Dec 28, 2019 at 11:17

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 $$\mathbb{E}|X|=1/3<4/9=\mathbb{E}|X-1/3|$$.

• 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 $\mathbb E[|X-x|^j]$? Dec 27, 2019 at 9:52
• 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. Dec 27, 2019 at 10:17
• 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. Dec 27, 2019 at 11:08
• 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? Dec 28, 2019 at 7:25
• @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.) 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

$$E[|X-\mu|^3]=7,$$

but we have

$$E[|X-3/2|^3]=6.3125.$$