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

Question

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

Answer 1

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

Answer 2

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

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