Proving an inequality about the deviation of the mean from the median

Question

Let X be a random variable with a median $m$ and mean $\mu$. I should prove that :

$$0.5||X-\mu||_p \leq ||X-m||_p\leq 3||X-\mu||_p$$

I am trying to prove that:

$$||\mu-m||_p \leq ||X-m||_p$$

By the following claim:

$$\left|\left|m-\mu\right|\right|_{p}=\left|\left|E\left[m-X\right]\right|\right|_{p}\leq E\left[\left|\left|m-X\right|\right|_{p}\right]=\left|\left|m-X\right|\right|_{p}$$

The inequality follows from Jensen's and the fact that the norm is convex.

The question :

1. Is the argument valid?
2. Does anyone know what I can do on the second part? I tried to follow a similar argument but got a scalar of 2 instead of 3.

Thanks

Answer 1

Suppose that $X\in L_p(\mathbb{P})$, $p\geq1$, and let $\mu_X=\mathbb{E}[X]$. For any median $m$ of $X$, showing that
$$|m-\mu_X|\leq\|X-m\|_p$$ is just an application of Hölder's inequality: \begin{align} |m-\mu_X|\leq\mathbb{E}\big[|X-m|\big]\leq\|X-m\|_p\tag{0}\label{zero} \end{align}

Often, this bound is not optimal. Another rather well known and useful relationship between the mean and a median is: \begin{align} |\mu_X-m|\leq \sqrt[p]{2}\|X-\mu_X\|_p\tag{1}\label{one} \end{align}

To see that, notice that if $m=\mu_X$ then the statement holds trivially. On the other hand, if $m\neq\mu_X$, then \begin{align*} \frac{|m-\mu_X|^p}{2}&\leq|m-\mu_X|^p\min\{\mathbb{P}[X\geq m],\mathbb{P}[X\leq m]\}\\ &\leq |m-\mu_X|^p\Pr\big[|X-\mu_X|\geq|m-\mu_X|\big]\leq\|X-\mu_X\|^p_p \end{align*}

Putting things together $$|\mu_X-m|\leq\min\big(\sqrt[p]{2}\|X-\mu_X\|_p,\|X-m\|_p\big)$$