# Chebyshev’s inequality: difference between median and mean

I have Chebyshev’s inequality $$P\left(\big|X-E[X]\big|\geq\epsilon\right)\leq\frac{V[X]}{\epsilon^2},$$ where $$E[X]$$ is the expected value and $$V[X]$$ the variance.

Given I know that the median is $$M$$, $$E[X]=\mu$$ and $$E[X^2]<\infty$$, I have to show that: $$|M-\mu|\leq\sqrt{2V[X]}$$

I thought about setting $$\epsilon=\frac{1}{2}$$, as I then will obtain the right side: $$\sqrt{P\left(\big|X-\mu\big|\geq\frac{1}{2}\right)}\leq\sqrt{2V[X]},$$ however, this leads to a blind end for me.

Then I thought about wanting to use the information about the median to set: $$\frac{V[X]}{\epsilon^2}=\frac{1}{2}\quad\quad\Rightarrow\quad\quad \epsilon=\sqrt{2V[X]},$$ and I will then get: $$P\left(\big|X-\mu\big|\geq\sqrt{2V[X]}\right)\leq\frac{1}{2},$$ where I thought I could use the median, when I have $$\frac{1}{2}$$ on the right side. However, I'm still quite stuck.

Any help regarding this inequality will be greatly appreciated.

Note that for any real number $$y$$, $$|X-y|\geq |M-y|$$ with probability at least $$\frac{1}{2}$$. Now you can apply the thing you showed to $$y=\mathbb{E}X.$$